Physics for Scientists & Engineers with Modern Physics [4 ed.] 0321666682, 9780321666680, 0131495089, 9780131495081, 0132274000, 9780132274005, 0132273594, 9780132273596, 0132273586, 9780132273589, 0132275597, 9780132275590

Table of contents :
Contents
APPLICATIONS LIST
PREFACE
AVAILABLE SUPPLEMENTS AND MEDIA
NOTES TO STUDENTS (AND INSTRUCTORS) ON THE FORMAT
COLOR USE: VECTORS, FIELDS, AND SYMBOLS
CHAPTER1: INTRODUCTION, MEASUREMENT, ESTIMATING
1-1 The Nature of Science
1-2 Models, Theories, and Laws
1-3 Measurement and Uncertainty; Significant Figures
1-4 Units, Standards, and the SI System
1-5 Converting Units
1-6 Order of Magnitude: Rapid Estimating
*1-7 Dimensions and Dimensional Analysis
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
PART 1: MECHANICS
CHAPTER 2: DESCRIBING MOTION: KINEMATICS IN ONE DIMENSION
2-1 Reference Frames and Displacement
2-2 Average Velocity
2-3 Instantaneous Velocity
2-4 Acceleration
2-5 Motion at Constant Acceleration
2-6 Solving Problems
2-7 Freely Falling Objects
*2-8 Variable Acceleration; Integral Calculus
*2-9 Graphical Analysis and Numerical Integration
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 3: KINEMATICS IN TWO OR THREE DIMENSIONS; VECTORS
3-1 Vectors and Scalars
3-2 Addition of Vectors-Graphical Methods
3-3 Subtraction of Vectors, and Multiplication of a Vector by a Scalar
3-4 Adding Vectors by Components
3-5 Unit Vectors
3-6 Vector Kinematics
3-7 Projectile Motion
3-8 Solving Problems Involving Projectile Motion
3-9 Relative Velocity
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 4: DYNAMICS: NEWTON'S LAWS OF MOTION
4-1 Force
4-2 Newton's First Law of Motion
4-3 Mass
4-4 Newton's Second Law of Motion
4-5 Newton's Third Law of Motion
4-6 Weight-the Force of Gravity; and the Normal Force
4-7 Solving Problems with Newton's Laws: Free-Body Diagrams
4-8 Problem Solving-A General Approach
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 5: USING NEWTON'S LAWS: FRICTION, CIRCULAR MOTION, DRAG FORCES
5-1 Applications of Newton's Laws Involving Friction
5-2 Uniform Circular Motion-Kinematics
5-3 Dynamics of Uniform Circular Motion
5-4 Highway Curves: Banked and Unbanked
*5-5 Nonuniform Circular Motion
*5-6 Velocity-Dependent Forces: Drag and Terminal Velocity
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 6: GRAVITATION AND NEWTON'S SYNTHESIS
6-1 Newton's Law of Universal Gravitation
6-2 Vector Form of Newton's Law of Universal Gravitation
6-3 Gravity Near the Earth's Surface; Geophysical Applications
6-4 Satellites and "Weightlessness"
6-5 Kepler's Laws and Newton's Synthesis
*6-6 Gravitational Field
6-7 Types of Forces in Nature
*6-8 Principle of Equivalence; Curvature of Space; Black Holes
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 7: WORK AND ENERGY
7-1 Work Done by a Constant Force
7-2 Scalar Product of Two Vectors
7-3 Work Done by a Varying Force
7-4 Kinetic Energy and the Work-Energy Principle
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 8: CONSERVATION OF ENERGY
8-1 Conservative and Nonconservative Forces
8-2 Potential Energy
8-3 Mechanical Energy and Its Conservation
8-4 Problem Solving Using Conservation of Mechanical Energy
8-5 The Law of Conservation of Energy
8-6 Energy Conservation with Dissipative Forces: Solving Problems
8-7 Gravitational Potential Energy and Escape Velocity
8-8 Power
*8-9 Potential Energy Diagrams; Stable and Unstable Equilibrium
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 9: LINEAR MOMENTUM
9-1 Momentum and Its Relation to Force
9-2 Conservation of Momentum
9-3 Collisions and Impulse
9-4 Conservation of Energy and Momentum in Collisions
9-5 Elastic Collisions in One Dimension
9-6 Inelastic Collisions
9-7 Collisions in Two or Three Dimensions
9-8 Center of Mass (CM)
9-9 Center of Mass and Translational Motion
*9-10 Systems of Variable Mass; Rocket Propulsion
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 10: ROTATIONAL MOTION
10-1 Angular Quantities
10-2 Vector Nature of Angular Quantities
10-3 Constant Angular Acceleration
10-4 Torque
10-5 Rotational Dynamics; Torque and Rotational Inertia
10-6 Solving Problems in Rotational Dynamics
10-7 Determining Moments of Inertia
10-8 Rotational Kinetic Energy
10-9 Rotational Plus Translational Motion; Rolling
*10-10 Why Does a Rolling Sphere Slow Down?
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 11: ANGULAR MOMENTUM; GENERAL ROTATION
11-1 Angular Momentum-Object Rotating About a Fixed Axis
11-2 Vector Cross Product; Torque as a Vector
11-3 Angular Momentum of a Particle
11-4 Angular Momentum and Torque for a System of Particles; General Motion
11-5 Angular Momentum and Torque for a Rigid Object
11-6 Conservation of Angular Momentum
*11-7 The Spinning Top and Gyroscope
*11-8 Rotating Frames of Reference; Inertial Forces
*11-9 The Coriolis Effect
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 12: STATIC EQUILIBRIUM; ELASTICITY AND FRACTURE
12-1 The Conditions for Equilibrium
12-2 Solving Statics Problems
12-3 Stability and Balance
12-4 Elasticity; Stress and Strain
12-5 Fracture
*12-6 Trusses and Bridges
*12-7 Arches and Domes
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 13: FLUIDS
13-1 Phases of Matter
13-2 Density and Specific Gravity
13-3 Pressure in Fluids
13-4 Atmospheric Pressure and Gauge Pressure
13-5 Pascal's Principle
13-6 Measurement of Pressure; Gauges and the Barometer
13-7 Buoyancy and Archimedes' Principle
13-8 Fluids in Motion; Flow Rate and the Equation of Continuity
13-9 Bernoulli's Equation
13-10 Applications of Bernoulli's Principle: Torricelli, Airplanes, Baseballs, TIA
*13-11 Viscosity
*13-12 Flow in Tubes: Poiseuille's Equation, Blood Flow
*13-13 Surface Tension and Capillarity
*13-14 Pumps, and the Heart
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 14: OSCILLATIONS
14-1 Oscillations of a Spring
14-2 Simple Harmonic Motion
14-3 Energy in the Simple Harmonic Oscillator
14-4 Simple Harmonic Motion Related to Uniform Circular Motion
14-5 The Simple Pendulum
*14-6 The Physical Pendulum and the Torsion Pendulum
14-7 Damped Harmonic Motion
14-8 Forced Oscillations; Resonance
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 15: WAVE MOTION
15-1 Characteristics of Wave Motion
15-2 Types of Waves: Transverse and Longitudinal
15-3 Energy Transported by Waves
15-4 Mathematical Representation of a Traveling Wave
*15-5 The Wave Equation
15-6 The Principle of Superposition
15-7 Reflection and Transmission
15-8 Interference
15-9 Standing Waves; Resonance
*15-10 Refraction
*15-11 Diffraction
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 16: SOUND
16-1 Characteristics of Sound
16-2 Mathematical Representation of Longitudinal Waves
16-3 Intensity of Sound: Decibels
16-4 Sources of Sound: Vibrating Strings and Air Columns
*16-5 Quality of Sound, and Noise; Superposition
16-6 Interference of Sound Waves; Beats
16-7 Doppler Effect
*16-8 Shock Waves and the Sonic Boom
*16-9 Applications: Sonar, Ultrasound, and Medical Imaging
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 17: TEMPERATURE, THERMAL EXPANSION, AND THE IDEAL GAS LAW
17-1 Atomic Theory of Matter
17-2 Temperature and Thermometers
17-3 Thermal Equilibrium and the Zeroth Law of Thermodynamics
17-4 Thermal Expansion
*17-5 Thermal Stresses
17-6 The Gas Laws and Absolute Temperature
17-7 The Ideal Gas Law
17-8 Problem Solving with the Ideal Gas Law
17-9 Ideal Gas Law in Terms of Molecules: Avogadro's Number
*17-10 Ideal Gas Temperature Scale-a Standard
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 18: KINETIC THEORY OF GASES
18-1 The Ideal Gas Law and the Molecular Interpretation of Temperature
18-2 Distribution of Molecular Speeds
18-3 Real Gases and Changes of Phase
18-4 Vapor Pressure and Humidity
*18-5 Van der Waals Equation of State
*18-6 Mean Free Path
*18-7 Diffusion
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 19: HEAT AND THE FIRST LAW OF THERMODYNAMICS
19-1 Heat as Energy Transfer
19-2 Internal Energy
19-3 Specific Heat
19-4 Calorimetry-Solving Problems
19-5 Latent Heat
19-6 The First Law of Thermodynamics
19-7 Applying the First Law of Thermodynamics; Calculating the Work
19-8 Molar Specific Heats for Gases, and the Equipartition of Energy
19-9 Adiabatic Expansion of a Gas
19-10 Heat Transfer: Conduction, Convection, Radiation
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 20: SECOND LAW OF THERMODYNAMICS
20-1 The Second Law of Thermodynamics-Introduction
20-2 Heat Engines
20-3 Reversible and Irreversible Processes; the Carnot Engine
20-4 Refrigerators, Air Conditioners, and Heat Pumps
20-5 Entropy
20-6 Entropy and the Second Law of Thermodynamics
20-7 Order to Disorder
20-8 Unavailability of Energy; Heat Death
*20-9 Statistical Interpretation of Entropy and the Second Law
*20-10 Thermodynamic Temperature Scale; Absolute Zero and the Third Law of Thermodynamics
*20-11 Thermal Pollution, Global Warming, and Energy Resources
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
PART Two: ELECTRICITY AND MAGNETISM
CHAPTER 21: ELECTRIC CHARGE AND ELECTRIC FIELD
21-1 Static Electricity; Electric Charge and Its Conservation
21-2 Electric Charge in the Atom
21-3 Insulators and Conductors
21-4 Induced Charge; the Electroscope
21-5 Coulomb's Law
21-6 The Electric Field
21-7 Electric Field Calculations for Continuous Charge Distributions
21-8 Field Lines
21-9 Electric Fields and Conductors
21-10 Motion of a Charged Particle in an Electric Field
21-11 Electric Dipoles
*21-12 Electric Forces in Molecular Biology; DNA
*21-13 Photocopy Machines and Computer Printers Use Electrostatics
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 22: GAUSS'S LAW
22-1 Electric Flux
22-2 Gauss's Law
22-3 Applications of Gauss's Law
*22-4 Experimental Basis of Gauss's and Coulomb's Law
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 23: ELECTRIC POTENTIAL
23-1 Electric Potential Energy and Potential Difference
23-2 Relation between Electric Potential and Electric Field
23-3 Electric Potential Due to Point Charges
23-4 Potential Due to Any Charge Distribution
23-5 Equipotential Surfaces
23-6 Electric Dipole Potential
23-7 E Determined from V
23-8 Electrostatic Potential Energy; the Electron Volt
23-9 Cathode Ray Tube: TV and Computer Monitors, Oscilloscope
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 24: CAPACITANCE, DIELECTRICS, ELECTRIC ENERGY STORAGE
24-1 Capacitors
24-2 Determination of Capacitance
24-3 Capacitors in Series and Parallel
24-4 Electric Energy Storage
24-5 Dielectrics
*24-6 Molecular Description of Dielectrics
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 25: ELECTRIC CURRENTS AND RESISTANCE
25-1 The Electric Battery
25-2 Electric Current
25-3 Ohm's Law: Resistance and Resistors
25-4 Resistivity
25-5 Electric Power
25-6 Power in Household Circuits
25-7 Alternating Current
25-8 Microscopic View of Electric Current: Current Density and Drift Velocity
*25-9 Superconductivity
*25-10 Electrical Conduction in the Nervous System
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 26: DC CIRCUITS
26-1 EMF and Terminal Voltage
26-2 Resistors in Series and in Parallel
26-3 Kirchoff's Rules
26-4 EMFs in Series and in Parallel; Charging a Battery
26-5 Circuits Containing Resistor and Capacitor (RC Circuits)
26-6 Electric Hazards
*26-7 Ammeters and Voltmeters
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 27: MAGNETISM
27-1 Magnets and Magnetic Fields
27-2 Electric Currents Produce Magnetic Fields
27-3 Force on an Electric Current in a Magnetic Field; Definition of
27-4 Force on an Electric Charge Moving in a Magnetic Field
27-5 Torque on a Current Loop; Magnetic Dipole Moment
*27-6 Applications: Galvanometers, Motors, Loudspeakers
27-7 Discover and Properties of the Electron
*27-8 The Hall Effect
*27-9 Mass Spectrometer
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 28: SOURCES OF MAGNETIC FIELD
28-1 Magnetic Field Due to a Straight Wire
28-2 Force between Two Parallel Wires
28-3 Definitions of the Ampere and the Coulomb
28-4 Ampere's Law
28-5 Magnetic Field of a Solenoid and a Toroid
28-6 Biot-Savart Law
*28-7 Magnetic materials-Ferromagnetism
*28-8 Electromagnets and Solenoids-Applications
*28-9 Magnetic Fields in Magnetic Materials; Hysteresis
*28-10 Paramagnetism and Diamagnetism
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 29: ELECTROMAGNETIC INDUCTION AND FARADAY'S LAW
29-1 Induced EMF
29-2 Faraday's Law of Induction; Lenz's Law
29-3 EMF Induced in a Moving Conductor
29-4 Electric Generators
*29-5 Back EMF and Counter Torque; Eddy Currents
29-6 Transformers and Transmission of Power
29-7 A Changing Magnetic Flux Produces an Electric Field
*29-8 Applications of Induction: Sound Systems, Computer Memory, Seismograph, GFCI
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 30: INDUCTANCE, ELECTROMAGNETIC OSCILLATIONS,
AND AC CIRCUITS
30-1 Mutual Inductance
30-2 Self-Inductance
30-3 Energy Stored in a Magnetic Field
30-4 LR Circuits
30-5 LC Circuits and Electromagnetic Oscillations
30-6 LC Oscillations with Resistance (LRC Circuit)
30-7 AC Circuits with AC Source
30-8 LRC Series AC Circuit
30-9 Resonance in AC Circuits
*30-10 Impedance Matching
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 31: MAXWELL'S EQUATIONS AND ELECTROMAGNETIC WAVES
31-1 Changing Electric Fields Produce Magnetic Fields; Ampere's Law and Displacement Current
31-2 Gauss's Law for Magnetism
31-3 Maxwell's Equations
31-4 Production of Electromagnetic Waves
*31-5 Electromagnetic Waves, and Their Speed, from Maxwell's Equations
31-6 Light as an Electromagnetic Wave and the Electromagnetic Spectrum
31-7 Measuring the Speed of Light
31-8 Energy in EM Waves; the Poynting Vector
*31-9 Radiation Pressure
*31-10 Radio and Television; Wireless Communication
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 32: LIGHT: REFLECTION AND REFRACTION
32-1 The Ray Model of Light
32-2 The Speed of Light and Index of Refraction
32-3 Reflection; Image Formation by a Plane Mirror
32-4 Formation of Images by Spherical Mirrors
32-5 Refraction: Snell's Law
32-6 Visible Spectrum and Dispersion
32-7 Total Internal Reflection; Fiber Optics
*32-8 Refraction at a Spherical Surface
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 33: LENSES AND OPTICAL INSTRUMENTS
33-1 Thin Lenses; Ray Tracing
33-2 The Thin Lens Equation; Magnification
33-3 Combinations of Lenses
33-4 Lensmaker's Equation
33-5 Cameras, Film and Digital
33-6 The Human Eye; Corrective Lenses
33-7 Magnifying Glass
33-8 Telescopes
*33-9 Compound Microscope
*33-10 Aberrations of Lenses and Mirrors
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 34: THE WAVE NATURE OF LIGHT; INTERFERENCE
34-1 Waves Versus Particles; Huygens' Principle and Diffraction
34-2 Huygens' Principle and the Law of Refraction
34-3 Interference-Young's Double-Slit Experiment
34-4 Intensity in the Double-Slit Interference Pattern
34-5 Interference in Thin Films
*34-6 Michelson Interferometer
*34-7 Luminous Intensity
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 35: DIFFRACTION AND POLARIZATION
35-1 Diffraction by a Single Slit or Disk
35-2 Intensity in Single-Slit Diffraction Pattern
35-3 Diffraction in the Double-Slit Experiment
35-4 Limits of Resolution; Circular Apertures
35-5 Resolution of Telescopes and Microscopes; the ? Limit
*35-6 Resolution of the Human Eye and Useful Magnification
35-7 Diffraction Grating
*35-8 The Spectrometer and Spectroscopy
*35-9 Peak Widths and Resolving Power for a Diffraction Grating
*35-10 X-Rays and X-Ray Diffraction
35-11 Polarization
*35-12 Liquid Crystal Displays (LCD)
*35-13 Scattering of Light by the Atmosphere
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
PART Three: MODERN PHYSICS
CHAPTER 36: SPECIAL THEORY OF RELATIVITY
36-1 Galilean-Newtonian Relativity
*36-2 The Michelson-Morley Experiment
36-3 Postulates of the Special Theory of Relativity
36-4 Simultaneity
36-5 Time Dilation and the Twin Paradox
36-6 Length Contraction
36-7 Four-Dimensional Space-Time
36-8 Galilean and Lorentz Transformations
36-9 Relativistic Momentum and Mass
36-10 The Ultimate Speed
36-11 Energy and Mass; E=mc2
36-12 Doppler Shift for Light
36-13 The Impact of Special Relativity
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 37: EARLY QUANTUM THEORY AND MODELS OF THE ATOM
37-1 Planck's Quantum Hypothesis
37-2 Photon Theory of Light and the Photoelectric Effect
37-3 Photons and the Compton Effect
37-4 Photon Interactions; Pair Production
37-5 Wave-Particle Duality; the Principle of Complementarity
37-6 Wave Nature of Matter
*37-7 Electron Microscopes
37-8 Early Models of the Atom
37-9 Atomic Spectra: Key to the Structure of the Atom
37-10 The Bohr Model
37-11 DeBroglie's Hypothesis Applied to Atoms
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 38: QUANTUM MECHANICS
38-1 Quantum Mechanics-A New Theory
38-2 The Wave Function and Its Interpretation; the Double-Slit Experiment
38-3 The Heisenberg Uncertainty Principle
38-4 Philosophic Implications; Probability Versus Determinism
38-5 The Schrodinger Equation in One Dimension-Time-Independent Form
*38-6 Time-Dependent Schrodinger Equation
38-7 Free Particles; Plane Waves and Wave Packets
38-8 Particle in an Infinitely Deep Square Well Potential (a Rigid Box)
*38-9 Finite Potential Well
38-10 Tunneling through a Barrier
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 39: QUANTUM MECHANICS OF ATOMS
39-1 Quantum-Mechanical View of Atoms
39-2 Hydrogen Atom: Schrodinger Equation and Quantum Numbers
39-3 Hydrogen Atom Wave Functions
39-4 Complex Atoms; the Exclusion Principle
39-5 The Periodic Table of Elements
39-6 X-Ray Spectra and Atomic Number
*39-7 Magnetic Dipole Moments; Total Angular Momentum
*39-8 Fluorescence and Phosphorescence
*39-9 Lasers
*39-10 Holography
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 40: MOLECULES AND SOLIDS
40-1 Bonding in Molecules
40-2 Potential-Energy Diagrams for Molecules
40-3 Weak (van der Waals) Bonds
40-4 Molecular Spectra
40-5 Bonding in Solids
40-6 Free-Electron Theory of Metals
40-7 Band Theory of Solids
40-8 Semiconductors and Doping
*40-9 Semiconductor Diodes
*40-10 Transistors and Integrated Circuits
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 41: NUCLEAR PHYSICS AND RADIOACTIVITY
41-1 Structure and Properties of the Nucleus
41-2 Binding Energy and Nuclear Forces
41-3 Radioactivity
41-4 Alpha Decay
41-5 Beta Decay
41-6 Gamma Decay
41-7 Conservation of Nucleon Number and Other Conservation Laws
41-8 Half-Life and Rate of Decay
41-9 Decay Series
41-10 Radioactive Dating
41-11 Detection of Radiation
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 42: NUCLEAR ENERGY: EFFECTS AND USES OF RADIATION
42-1 Nuclear Reactions and the Transmutations of Elements
42-2 Cross Section
42-3 Nuclear Fission; Nuclear Reactors
42-4 Fusion
42-5 Passage of radiation through matter; Radiation Damage
42-6 Measurement of Radiation-Dosimetry
*42-7 Radiation Therapy
*42-8 Tracers
*42-9 Imaging by Tomography: CAT Scans, and Emission Tomography
*42-10 Nuclear Magnetic Resonance (NMR) and Magnetic Resonance Imaging (MRI)
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 43: ELEMENTARY PARTICLES
43-1 High-Energy Particles
43-2 Particle Accelerators and Detectors
43-3 Beginnings of Elementary Particle Physics-Particle Exchange
43-4 Particles and Antiparticles
43-5 Particle Interactions and Conservation Laws
43-6 Particle Classification
43-7 Particle Stability and Resonances
43-8 Strange Particles
43-9 Quarks
43-10 The "Standard Model": Quantum Chromodynamics (QCD) and the Electroweak Theory
43-11 Grand Unified Theories
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
CHAPTER 44: ASTROPHYSICS AND COSMOLOGY
44-1 Stars and Galaxies
44-2 Stellar Evolution; the Birth and Death of Stars
44-3 General Relativity: Gravity and the Curvature of Space
44-4 The Expanding Universe
44-5 The Big Bang and the Cosmic Microwave Background
44-6 The Standard Cosmological Model: Early History of the Universe
44-7 The Future of the Universe?
SUMMARY
QUESTIONS
PROBLEMS
GENERAL PROBLEMS
APPENDICES
A
B
C
D
E
F
HINTS AND BRIEF RESPONSES TO ODD-NUMBERED QUESTIONS
ANSWERS TO ODD-NUMBERED PROBLEMS
INDEX
PHOTO CREDITS

Citation preview

PHYSICS

;CIENTISTS ith Mo |

FNGINFFRS

GIANCOLI

IEIE

Ql.l;mtity

Symbol

Speed of light in vacuum

c

Avogadro’s number

Na

Gravitational constant

Gas constant

Boltzmann’s constant

2.99792458 X 108 m/s

6.02 X 10% mol™!

6.02214179(30) X 10> mol !

R k

8.314J/mol-K = 1.99 cal/mol-K = 0.0821 L-atm/mol-K 1.38 X 10 2 J/K

o

567 X 10

Mo

47 X

Permittivity of free space

€0 = (1/c*uy)

Planck’s constant Electron rest mass

3.00 X 108 m/s

6.67 X 10 "' N-m?/kg?

e

Permeability of free space

Current Best Value'

G

Charge on electron

Stefan-Boltzmann constant

Approximate Value

1.60 x 10717

my,

Neutron rest mass

my

8.314472(15) J/mol-K

1.3806504(24) X 10 2 J/K

C

1.602176487(40) x 10717C

8W/m?-K*

5.670400(40) x 10 W/m?-K*

885 X 10"2C?/N-m’

h me

Proton rest mass

6.6728(67) x 107" N-m?/kg?

107

8.854187817... X 1072 C?/N-m’

T-m/A

1.2566370614 ...

10 °T-m/A

6.62606896(33) X 103*J-s 9.10938215(45) x 103 kg = 5.4857990943(23) X 10 *u 1.672621637(83) X 10 *"kg

1.6749 X 107*" kg = 1.008665 u = 939.57 MeV/c? 1.6605 X 10?7 kg = 931.49 MeV/c?

1.674927211(84) x 10* kg

= 938.27 MeV/c?

Atomic mass unit (1 u)

X

6.63 X 10 *J-s 9.11 x 103" kg = 0.000549 u = 0.511 MeV/c? 1.6726 X 10* kg = 1.00728u

= 1.00727646677(10) u = 1.00866491597(43) u

1.660538782(83) x 10?7 kg = 931.494028(23) MeV/c?

"CODATA (3/07), Peter J. Mohr and Barry N. Taylor, National Institute of Standards and Technology. Numbers in parentheses indicate one-standarddeviation experimental uncertainties in final digits. Values without parentheses are exact (i.e., defined quantities).

~ Other Useful Data

The Greek Alphabet

Joule equivalent (1 cal)

4.1861]

Absolute zero (0 K) Acceleration due to gravity

—=273.15°C

at Earth’s surface (avg.) Speed of sound in air (20°C)

9.80m/s’ (= g) 343 m/s

Alpha

A

a

Nu

N

v

Beta Gamma

B I

B Y

Xi Omicron

=) O

3 o

Delta Epsilon

A E

5 €.

Pi Rho

I p

T p

Density of air (dry)

1.29 kg/m?

Zeta

7

{

Sigma

5

o

Earth:

Mass Radius (mean)

5.98 X 10** kg 6.38 X 103 km

Eta Theta

H o

n 0

Tau Upsilon

T Y

T v

Radius (mean) Mass

1.74 X 10° km 1.99 x 10*kg

Kappa Lambda

K A

& A

Chi Psi

X v

% W

Moon: Mass Sun:

7.35 X 107 kg

Radius (mean)

Earth—Sun distance (mean) Earth—Moon distance (mean)

lota

6.96 X 10° km

149.6 X 384 X

Mu

10°km 10°km

I

¢

M

"

Phi

®

Omega

)

b.¢ ©

‘Values of Some Numbers 7 = 3.1415927 e = 2.7182818

V2 = 1.4142136 V3 = 1.7320508

In2 = 0.6931472 In 10 = 2.3025851

Mathematical Signs and Symbols o< ~

is proportional to is equal to is approximately equal to

pY

#

is not equal to

x

>

is greater than

>>

is much greater than

< = 11b/in.? = 6.895 X 10° 1Pa=1N/m?> = 1450

1 knot = 1.151 mi/h = 0.5144 m/s

1 atm

Angle 1 radian (rad) = 57.30° = 57°18’ 1° = 0.01745 rad

Il

r

1 rev/min (rpm) = 0.1047 rad/s =

=

=

Metric (SI) Multipliers e

Sl Derived Units and Their Abbreviations Quantity

1.01325 X 10’ N/m? 760 torr N/m? X 10 *Ib/in.2

In Terms of

Prefix

Abbreviation

tt

Y

7

102!

E P

10 10

iot' a

Base Units'

o

Value

10%

Unit

Abbreviation

Force

newton

N

kg-m/s

exa peta

Energy and work

joule

J

kg-m?/s’

tt.:ra

g

]0:

Power

watt

W

kg-m?/s?

Pressure

pascal

Pa

kg/(m-s?)

gga mega

G M

N 10°

Frequency

hertz

Hz

s

hecto

h

102

il

"

8 ,

109

Electric charge

coulomb

C

A-s

deka

da

10!

Electric potential

volt

\%

kg-m*/(A-s®

deci

d

10!

Electric resistance

ohm

Q

kg

centi

¢

1072

Capacitance o

farad

F

Magnetic field )

tesla

T

A5t/ (kg-m?) y

milli micro

m I

10 10°°

Magnetic flux

weber

Wb

nano

n

10

Inductance

henry

H

"kg = kilogram (mass), m = meter (length),

mz/(A2

kg/(A ‘S ) ,

s*)

N

kg-m?/(A-s?) kg-m?

(52 A?)

s = second (time), A = ampere (electric current).

e

-3 5

pico

p

1012

femto

f

10715

atto

a

yocto

y

zepto

z

10"

1072

107

HYSICS for

SCIENTISTS & ENGINEERS - with Modern

Physics

'

DOUGLAS

¢

|

C. GIANCOLI

Upper Saddle River, New Jersey 07458

Library of Congress Cataloging-in-Publication Data Giancoli, Douglas C.

Physics for scientists and engineers with modern physics / Douglas C. Giancoli.—4th ed. p. cm. Includes bibliographical references and index. ISBN 0-13-149508-9 1. Physics—Textbooks. 1. Title. QC21.3.G539 2008 530—dc22 2006039431

President, Science: Paul Corey

Sponsoring Editor: Christian Botting Executive Development Editor: Karen Karlin Production Editor: Clare Romeo Senior Managing Editor: Scott Disanno Art Director and Interior & Cover Designer: John Christiana Manager, Art Production: Sean Hogan Copy Editor: Jocelyn Phillips

Proofreaders: Karen Bosch, Gina Cheselka, Traci Douglas, Nancy Stevenson,

and Susan Fisher Senior Operations Specialist: Alan Fischer Art Production Editor: Connie Long Ilustrators: Audrey Simonetti and Mark Landis Photo Researchers: Mary Teresa Giancoli and Truitt & Marshall Senior Administrative Coordinator: Trisha Tarricone Composition: Emilcomp/Prepare Inc.;

Pearson Education/Lissette Quifiones, Clara Bartunek

Photo credits appear on page A-72 which constitutes a continuation of the copyright page.

© 2009, 2000, 1989, 1984 by Douglas C. Giancoli Published by Pearson Education, Inc. MPVNIOIR Pearson Prentice Hall Pearson Education, Inc. ranfice

Pre RS

Upper Saddle River, NJ 07458

All rights reserved. No part of this book may be reproduced, in any form or by any means, without permission in writing from the publisher. Pearson Prentice Hall™ is a trademark of Pearson Education, Inc. Printed in the United States of America

ISBN-13: ISBN-10: Pearson Pearson Pearson Pearson

978-0-13-149508-1 0-13-149508-9

Education Education Education Education

LTD., London Australia PTY, Limited, Sydney Singapore, Pte. Ltd. North Asia Ltd., Hong Kong

Pearson Education Canada, Ltd., Toronto Pearson Educacion de Mexico, S.A. de C.V.

Pearson Education—Japan, Tokyo Pearson Education Malaysia, Pte. Ltd.

21 2021

Contents

KINEMATICS IN TWO OR THREE DIMENSIONS; VECTORS

51

Vectors and Scalars Addition of Vectors—Graphical Methods

52 52

Multiplication of a Vector by a Scalar Adding Vectors by Components Unit Vectors Vector Kinematics Projectile Motion Solving Problems: Projectile Motion Relative Velocity

54 55 59 39 62 64 71

Subtraction of Vectors, and

SUMMARY PROBLEMS

APPLICATIONS LIST PREFACE To STUDENTS USE oF COLOR

xii Xiv xviii Xix

P

1 1-1

1-2 1-3 1-4

1-5 1-6 *1-7

1

The Nature of Science

2

Models, Theories, and Laws Measurement and Uncertainty; Significant Figures

Units, Standards, and the SI System

2

2-1 2-2 2-3

14 14

QUESTIONS

14

GENERAL PROBLEMS

5-2

16

18

Variable Acceleration; Integral Calculus Graphical Analysis and Numerical Integration

39

44

GENERAL PROBLEMS

*5-5

*5-6

SUMMARY PROBLEMS

48

30 34 40

83 84 84 86 86 89

102 104

QuUEsTIONS 103 GENERAL PROBLEMS

109

USING NEWTON's Laws: FRICTION,

CIRCULAR MOTION, DRAG FORCES 112 Applications of Newton’s Laws Involving Friction

113

Dynamics of Uniform Circular Motion Highway Curves: Banked and Unbanked

122 126

Uniform Circular Motion—Kinematics

Nonuniform Circular Motion

Velocity-Dependent Forces: Drag and Terminal Velocity PROBLEMS

Solving Problems Freely Falling Objects

PROBLEMS

5-3 5-4

80

Weight—the Force of Gravity; the Normal Force 92 Solving Problems with Newton’s Laws: Free-Body Diagrams 95 Problem Solving—A General Approach 102

SUMMARY

19 20 22 24 28

QUESTIONS 43

5

3

RTerence Frames and Displacement Average Velocity Instantaneous Velocity Acceleration Motion at Constant Acceleration

43

Force Newton’s First Law of Motion Mass Newton’s Second Law of Motion Newton’s Third Law of Motion

5-1

8 9 12

DESCRIBING MOTION: KINEMATICS IN ONE DIMENSION

SUMMARY

4-1 4-2 4-3 4-4 4-5

2

6

C%lnverting Units Order of Magnitude: Rapid Estimating Dimensions and Dimensional Analysis SUMMARY PROBLEMS

DYNAMICS: NEWTON'S LAWS OF MOTION

4-8

INTRODUCTION, MEASUREMENT, ESTIMATING

QUESTIONS 75 GENERAL PROBLEMS

4

4-6 4-7

Volume 1

74 75

130 132

QUESTIONS

131

GENERAL PROBLEMS

119 128

129 136

(GRAVITATION AND

NEWTON'S SYNTHESIS

139

Newton’s Law of Universal Gravitation

140

Vector Form of Newton’s Law of

Universal Gravitation Gravity Near the Earth’s Surface; Geophysical Applications Satellites and “Weightlessness” Kepler’s Laws and Newton’s Synthesis Gravitational Field Types of Forces in Nature Principle of Equivalence; Curvature of Space; Black Holes SUMMARY

PROBLEMS

157

158

QUESTIONS

157

GENERAL PROBLEMS

143 143 146 149 154 155 155

9

9-1 9-2

9-3

9-4 9-5 9-6 9-7

9-8 9-9

*9-10

160

Displacement

i~

7-1 7-2 7-3 7-4

Work Done by Scalar Product Work Done by Kinetic Energy

163

a Constant Force of Two Vectors a Varying Force and the

164 167 168

Work-Energy Principle SUMMARY PROBLEMS

176 177

QUESTIONS 177 GENERAL PROBLEMS

172 180

CONSERVATION OF ENERGY

183

Conservative and Nonconservative Forces

Potential Energy Mechanical Energy and Its Conservation Problem Solving Using Conservation of Mechanical Energy The Law of Conservation of Energy Energy Conservation with Dissipative Forces: Solving Problems Gravitational Potential Energy and Escape Velocity Power Potential Energy Diagrams; Stable and Unstable Equilibrium SUMMARY PROBLEMS

iv

CONTENTS

205 207

QUESTIONS 205 GENERAL PROBLEMS

211

1

11-1 11-2 11-3 11-4

190 196

204

Center of Mass (cM)

230

Center of Mass and Translational Motion

234

Systems of Variable Mass; Rocket Propulsion 236

SUMMARY PROBLEMS

239 240

QUESTIONS 239 GENERAL PROBLEMS

245

248 249 254

Constant Angular Acceleration Torque Rotational Dynamics; Torque and Rotational Inertia Solving Problems in Rotational Dynamics Determining Moments of Inertia Rotational Kinetic Energy

255 256

258 260 263 265

Rotational Plus Translational Motion; Rolling 267‘\ SUMMARY PROBLEMS

11-6 *11-7 *11-8 *11-9

199 201

222 222 225 227

*10-10 Why Does a Rolling Sphere Slow Down?

11-5

196

Momentum in Collisions Elastic Collisions in One Dimension Inelastic Collisions Collisions in Two or Three Dimensions

Angular Quantities Vector Nature of Angular Quantities

184

186 189

224

Conservation of Energy and

10-1 10-2

10-9

215

217\

Collisions and Impulse

ROTATIONAL MOTION

10-6 10-7 10-8

214

Momentum and Its Relation to Force Conservation of Momentum

10 10-3 10-4 10-5

WORK AND ENERGY

LINEAR MOMENTUM

274 °~ QUESTIONS 275 276 GENERAL PROBLEMS

281

ANGULAR MOMENTUM;

GENERAL ROTATION

Angular Momentum—Objects Rotating About a Fixed Axis Vector Cross Product; Torque as a Vector Angular Momentum of a Particle Angular Momentum and Torque for

a System of Particles; General Motion Angular Momentum and Torque for a Rigid Object Conservation of Angular Momentum The Spinning Top and Gyroscope Rotating Frames of Reference; Inertial Forces The Coriolis Effect

SUMMARY 302 QUESTIONS 303 PROBLEMS 303 GENERAL PROBLEMS 308

27.

284 285 289 291 292 294 297 299 300 301

14

OSCILLATIONS

14-1 14-2 14-3

Oscillations of a Spring Simple Harmonic Motion Energy in the Simple Harmonic Oscillator Simple Harmonic Motion Related to Uniform Circular Motion The Simple Pendulum The Physical Pendulum and the Torsion Pendulum Damped Harmonic Motion Forced Oscillations; Resonance

14-4 14-5 *14-6 14-7 14-8

SUMMARY

PROBLEMS

12

12-1 12-2 12-3 12-4

12-5

*12-6

@127

StATIC EQUILIBRIUM;

ELASTICITY AND FRACTURE

311

The Conditions for Equilibrium Solving Statics Problems Stability and Balance Elasticity; Stress and Strain

312 313 317 318

Trusses and Bridges

324

Fracture

Arches and Domes SUMMARY

PROBLEMS

329

330

QUESTIONS

329

GENERAL PROBLEMS

| 13 FLUIDS 13-1

13-2 13-3 13-4 13-5 13-6 13-7 13-8

322

327

334

339

Phéses of Matter

»

346 348 352

13-9 Bernoulli’s Equation

354

13-10 Applications of Bernoulli’s Principle: Torricelli, Airplanes, Baseballs, TTA

356

*13-11 Viscosity *13-12 Flow in Tubes: Poiseuille’s Equation, Blood Flow *13-13 Surface Tension and Capillarity (™13-14 Pumps, and the Heart SUMMARY

PROBLEMS

361

363

QUESTIONS

362

GENERAL PROBLEMS

358 358 359 361 367

QUESTIONS GENERAL

388

PROBLEMS

370 372 377 379 379 381 382 385 392

395

15-1 15-2

Characteristics of Wave Motion Types of Waves: Transverse and Longitudinal 15-3 Energy Transported by Waves 15-4 Mathematical Representation of a Traveling Wave *15-5 The Wave Equation 15-6 The Principle of Superposition 15-7 Reflection and Transmission 15-8 Interference 15-9 Standing Waves; Resonance *15-10 Refraction *15-11 Diffraction SUMMARY PROBLEMS

340 341 345 346

388

15 WAVE MOTION

340

Dehsity and Specific Gravity Pressure in Fluids Atmospheric Pressure and Gauge Pressure Pascal’s Principle Measurement of Pressure; Gauges and the Barometer Buoyancy and Archimedes’ Principle Fluids in Motion; Flow Rate and the Equation of Continuity

387

369

417 418

QUESTIONS 417 GENERAL PROBLEMS

16 SOUND 16-1 16-2 16-3 16-4 *16-5 16-6

16-7 *16-8 *16-9

396 398 402 404 406 408 409 410 412 415 416 422

424

Characteristics of Sound Mathematical Representation of Longitudinal Waves Intensity of Sound: Decibels Sources of Sound: Vibrating Strings and Air Columns

425

Superposition

436

Doppler Effect Shock Waves and the Sonic Boom

439 443

426 427 431

Quality of Sound, and Noise;

Interference of Sound Waves; Beats

437

Applications: Sonar, Ultrasound,

Medlcal Imagmg

SUMMARY PROBLEMS

446 448

QUESTIONS 447 GENERAL PROBLEMS

444 451

CONTENTS

v

19

HEAT AND THE FIRST LAW

OF THERMODYNAMICS

496

19-1 19-2 19-3 19-4 19-5 19-6 19-7

Heat as Energy Transfer Internal Energy Specific Heat Calorimetry—Solving Problems Latent Heat The First Law of Thermodynamics The First Law of Thermodynamics Applied; Calculating the Work 19-8 Molar Specific Heats for Gases, and the Equipartition of Energy 19-9 Adiabatic Expansion of a Gas 19-10 Heat Transfer: Conduction, Convection, Radiation SUMMARY

PROBLEMS

1

20

TEMPERATURE, THERMAL EXPANSION,

20-1

7 AND THE IDEAL GAS Law

454

17-1 17-2 17-3

Atomic Theory of Matter Temperature and Thermometers Thermal Equilibrium and the Zeroth Law of Thermodynamics 17-4 Thermal Expansion *17-5 Thermal Stresses 17-6 The Gas Laws and Absolute Temperature 17-7 The Ideal Gas Law 17-8 Problem Solving with the Ideal Gas Law 17-9 Ideal Gas Law in Terms of Molecules: Avogadro’s Number *17-10 Ideal Gas Temperature Scale— a Standard SUMMARY

PROBLEMS

18

471

QUESTIONS

471

GENERAL PROBLEMS

KINETIC THEORY OF GASES

18-1

18-2 18-3 18-4 *18-5 *18-6 *18-7

vi

470

455 456 459 459 463 463 465 466 468 469 474

476

The Ideal Gas Law and the Molecular Interpretation of Temperature Distribution of Molecular Speeds Real Gases and Changes of Phase Vapor Pressure and Humidity Van der Waals Equation of State Mean Free Path Diffusion SUMMARY PROBLEMS

CONTENTS

490 492

QUESTIONS 491 GENERAL PROBLEMS

476 480 482 484 486 487 489 494

20-2 20-3 20-4

520 522

QUESTIONS GENERAL

521

PROBLEMS

SECOND LAW OF

THERMODYNAMICS

497 -~ 498 499 500 502 505 507 il 514 515 526

528

The Second Law of Thermodynamics— Introduction Heat Engines Reversible and Irreversible

529 530

Processes; the Carnot Engine Refrigerators, Air Conditioners, and

533

Heat Pumps Entropy Entropy and the Second Law of Thermodynamics 20-7 Order to Disorder 20-8 Unavailability of Energy; Heat Death *20-9 Statistical Interpretation of Entropy and the Second Law *20~10 Thermodynamic Temperature; Third Law of Thermodynamics *20-11 Thermal Pollution, Global Warming, and Energy Resources

536 539

20-5 20-6

SUMMARY PROBLEMS

551 552

QUESTIONS 552 GENERAL PROBLEMS

541 544 545 546 548 549 556

23

Volume 2 f*21 21-1

21-2 21-3 21-4

E{.ECTRIC CHARGE AND

ECTRIC FIELD

559

Static Electricity; Electric

Charge and Its Conservation Electric Charge in the Atom In%ulators and Conductors

560 561 561

Induced Charge; the Electroscope

562

21-5 Coulomb’s Law

563

21-6 21-7

The Electric Field Electric Field Calculations for Continuous Charge Distributions 21-8 Field Lines 21-9 Electric Fields and Conductors 21-10 Motion of a Charged Particle in an Electric Field 21-11 Electric Dipoles *21-12 Electric Forces in Molecular Biology; DNA

568 572 575 577 578 579 581

*21-13 Photocopy Machines and Computer Printers Use Electrostatics

582

ELECTRIC POTENTIAL

23-1

Electric Potential Energy and

23-2

Relation between Electric Potential and Electric Field

23-3

234 23-5 23-6 23-7 23-8 *23-9

Potential Difference

PROBLEMS

24-1 24-2 24-3 24-4 24-5 *24-6

€o

610

612

Potential Due to Any Charge Distribution Equipotential Surfaces Electric Dipole Potential E Determined from V Electrostatic Potential Energy; the Electron Volt Cathode Ray Tube: TV and Computer Monitors, Oscilloscope SUMMARY

24

607

Electric Potential Due to Point Charges

622 623

QUESTIONS

622

GENERAL PROBLEMS

614 616 617 617 619 620

626

CAPACITANCE, DIELECTRICS,

ELECTRIC ENERGY STORAGE

628

Capacitors + Determination of Capacitance Capacitors in Series and Parallel Electric Energy Storage Dielectrics Molecular Description of Dielectrics SUMMARY

PROBLEMS Qencl

607

643

644

QUESTIONS

643

GENERAL PROBLEMS

628 630 633 636 638 640 648

2 5 ELECTRIC CURRENTS AND RESISTANCE

651

25-1

The Electric Battery

652

25-4

Resistivity

658

25-2 25-3 25-5 25-6

25-7 25-8

Electric Current Ohm’s Law: Resistance and Resistors

654 655

Electric Power Power in Household Circuits

660 662

Alternating Current Microscopic View of Electric Current: Current Density and Drift Velocity *25-9 Superconductivity *#25-10 Electrical Conduction in the Nervous System SUMMARY PROBLEMS

671 672

QUESTIONS 671 GENERAL PROBLEMS

22-1 22-2 22-3 (.*22—4

Gauss’s Law

591

Electric Flux Gauss’s Law Applications of Gauss’s Law Experimental Basis of Gauss’s and Coulomb’s Laws SUMMARY

PROBLEMS

601

601

QUESTIONS

601

GENERAL PROBLEMS

592 593 595 600 605

666 668 669

675

DC Circuirs

22

664

677

26-1

EMF and Terminal Voltage

678

264 26-5

Series and Parallel EMFs; Battery Charging Circuits Containing Resistor and Capacitor (RC Circuits)

686

26-2 26-3

26-6 *26-7

Resistors in Series and in Parallel Kirchhoff’s Rules

679 683

687

Electric Hazards Ammeters and Voltmeters SUMMARY PROBLEMS

698 699

QUESTIONS 698 GENERAL PROBLEMS

692 695 704

CONTENTS

vii

29

29-1 29-2 29-3 29-4 *29-5 29-6 29-7 *29-8

ELECTROMAGNETIC INDUCTION

AND FARADAY's Law

758

Induced EMF Faraday’s Law of Induction; Lenz’s Law EMF Induced in a Moving Conductor Electric Generators Back EMF and Counter Torque; Eddy Currents Transformers and Transmission of Power A Changing Magnetic Flux Produces an Electric Field Applications of Induction: Sound Systems, Computer Memory, Seismograph, GFCI

SUMMARY PROBLEMS

27 MAGNETISM 27-1 27-2 27-3 27-4 27-5 *27-6 27-7 27-8 *27-9

Magnets and Magnetic Fields Electric Currents Produce Magnetic Fields Force on an Electric Current in a Magnetic Field; Definition of B Force on an Electric Charge Moving in a Magnetic Field Torque on a Current Loop; Magnetic Dipole Moment Apf)lications: Motors, Loudspeakers, Galvanometers Discovery and Properties of the Electron The Hall Effect Mass Spectrometer SUMMARY

PROBLEMS

28 28-1 28-2 28-3 284 28-5 28-6 28-7 *28-8 *28-9

707

725

727

QUESTIONS

726

GENERAL PROBLEMS

SOURCES OF MAGNETIC FIELD

viii

CONTENTS

751

720 721 723 724

736 737 741 743 746

}

GENERAL PROBLEMS

718

734 735

747 748 749

Hysteresis PROBLEMS

714

733

Magnetic Materials—Ferromagnetism Electromagnets and Solenoids— Applications Magnetic Fields in Magnetic Materials

*28-10 Paramagnetism and Diamagnetism SUMMARY 750 QUESTIONS 751

710

730

Magnetic Field Due to a Straight Wire Force between Two Parallel Wires Definitions of the Ampere and the Coulomb Ampere’s Law Magnetic Field of a Solenoid and aToroid

Biot-Savart Law

707 710

755

30

777 778

QUESTIONS 777 GENERAL PROBLEMS

30-6

LC Oscillations with Resistance

31-1 31-2 31-3 31-4 31-5 31-6 31-7 31-8 31-9

773 775

782

OsCILLATIONS, AND AC CIrRcuUITS 785

Mutual Inductance Self-Inductance

31

768 770

INDUCTANCE, ELECTROMAGNETIC

30-1 30-2 30-3 304 30-5

30-7 30-8 30-9 *30-10 *30-11

759 760“ 765 766

786 788 790 790

Energy Stored in a Magnetic Field LR Circuits

LC Circuits and Electromagnetic Oscillations

793

795 796" 799 802 802 803

(LRC Circuit)

AC Circuits with AC Source LRC Series AC Circuit Resonance in AC Circuits

Impedance Matching Three-Phase AC SUMMARY PROBLEMS

804 805

QUESTIONS 804 GENERAL PROBLEMS

MAXWELL'S EQUATIONS AND

ELECTROMAGNETIC WAVES

809

812

Changing Electric Fields Produce

Magnetic Fields; Ampéere’s Law and

Displacement Current Gauss’s Law for Magnetism Maxwell’s Equations Production of Electromagnetic Waves Electromagnetic Waves, and Their Speed, from Maxwell’s Equations Light as an Electromagnetic Wave and the Electromagnetic Spectrum Measuring the Speed of Light Energy in EM Waves; the Poynting Vector

813 816 817 817

Wireless Communication SUMMARY 832 QUESTIONS

829

Radiation Pressure 31-10 Radio and Television; PROBLEMS

833

832

GENERAL PROBLEMS

819 823 825 826 828

835

32 f‘

LIGHT: REFLECTION

AND REFRACTION

32-1 32-2 32-3 32-4 32-5 32-6 32-7 *32-8

837

e Ray Model of Light Reflection; Image Formation by a Plane Mirror Formation of Images by Spherical Mirrors Index of Refraction Refraction: Snell’s Law Visible Spectrum and Dispersion Total Internal Reflection; Fiber Optics Refraction at a Spherical Surface IMMARY PROBLEMS

858 860

QUESTIONS 859 GENERAL PROBLEMS

838 838 842 850 850 852 854 856 864

3 4 THE WAVE NATURE OF LIGHT;

INTERFERENCE

’“"‘"M T~ "”w'l.,'_.'if,'g‘}n?u?:p‘r'fiafifii‘i 'Jfl':f.f ‘.’ni'iglf Ji.’.‘fd it

i

A part (b). in 3

'-.n

ower,

obiee

whick—" -

1} cation Oor magn / angle subtended by b obj ecgt R en uaick” e

unaided eye,

normal eyeF ‘

o

with the object at the neals=

0

M = P

!

e

length by noting that

§ = h/N

(Fig.

=

. subtended using the

\

33-1

33-2 33-3 *33-4 33-5 33-6 33-7 33-8 *33-9 ™ *33-10

34-2

Principle and Diffraction Huygens’ Principle and the Law of

34-3

Interference— Young’s Double-Slit

%344

Al

34-5

it M in e ot he foca

*34-6

33— /#gles are small 50 6 and ¢’ equal

*34-7

-:;'da,‘,‘z,"fi:‘f,blfw‘{‘ivffieb{‘

S

f

and

8' = h/f .‘,'

-

‘

33

g

' e eve (N = 25em fora

\ /1 is the height of the object and e assy/4 rve focai pom. see Fip- 3314

e

1 Waves Versus Particles; Huygens’

LENSES AND

INSTRUMENTS P

\

Refraction

901

)

=

Thin Lenses; Ray Tracing

866

..

867

The Thin Lens Equation; Magnification 870 Cpmbinations of Lenses 874 L nsmaker"s Equatio'n. 876 Cameras: Film and Digital 878 e Human Eye; Corrective Lenses 882 Magnifying Glass 885 Telescopes 887 Compound Microscope 890 Aberrations of Lenses and Mirrors 891 SUMMARY 892 QUESTIONS 893 PROBLEMS 894 GENERAL PROBLEMS 897

903

Intensity in the Double-Slit Interference Pattern

906

Mlch'elson Interfc?rometer

914

Interference in Thin Films Luminous

SUMMARY

909

IntenSIty

915

PROBLEMS 916

915

QUESTIONS 916

GENERAL PROBLEMS 918

3 5 DIFFRACTION AND POLARIZATION

OPTICAL

902

.

y

Experiment

900

921

35-1 *35-2

Diffraction by a Single Slit or Disk Intensity in Single-Slit Diffraction

*35-3

Diffraction in the Double-Slit Experiment 927

35-4

35-5

4356 35-7 35-8 *35-9 35-10 35-11 *35-12 *35-13

922

Pattern

924

Limits of Resolution; Circular Apertures

929

Microscopes; the A Limit

931

Resolution of Telescopes and

Resolution of the Human Eye

and Useful Magnification 932 Diffraction Grating 933 The Spectrometer and Spectroscopy 935 Peak Widths and Resolving Power for a Diffraction Grating 937 X-Rays and X-Ray Diffraction 938 Polarization 940 Liquid Crystal Displays (LCD) 943 Scattering of Light by the Atmosphere 945 SUMMARY 945 QUESTIONS 946 PROBLEMS 946 GENERAL PROBLEMS 949 CONTENTS

ix

Volume 3 36

SPECIAL THEORY OF RELATIVITY

951

36-1 *36-2 36-3 364 36-5 36-6 36-7 36-8 36-9 36-10

Galilean—-Newtonian Relativity The Michelson-Morley Experiment Postulates of the Special Theory of Relativity Simultaneity Time Dilation and the Twin Paradox Length Contraction Four-Dimensional Space-Time Galilean and Lorentz Transformations Relativistic Momentum and Mass The Ultimate Speed

952 954 957 958 960 964 967 968 971 974

36-11 E = mc* Mass and Energy

974

*36-12 Doppler Shift for Light 36-13 The Impact of Special Relativity SUMMARY

PROBLEMS

981

982

QUESTIONS

981

GENERAL PROBLEMS

978 980 985

EARLY QUANTUM THEORY AND

37 MODELS OF THE ATOM 37-1 37-2

37-3 37-4 37-5 37-6

987

Blackbody Radiation; Planck’s Quantum Hypothesis

987

Photon Theory; Photoelectric Effect

989

Photon Energy, Mass, and Momentum 993 Compton Effect 994 Photon Interactions; Pair Production 996 Wave-Particle Duality; the Principle of Complementarity 997 37-7 Wave Nature of Matter 997 *37-8 Electron Microscopes 1000 37-9 Early Models of the Atom 1000 37-10 Atomic Spectra: Key to Atomic Structure 1001 37-11 The Bohr Model

1003

37-12 deBroglie’s Hypothesis Applied to Atoms 1009 SUMMARY PROBLEMS

38

1010 1012

QuEsTIONS 1011 GENERAL PROBLEMS

QUANTUM MECHANICS

38-1 38-2

X

CONTENTS

1039 1040

oQuEesTIiONS 1039 GENERAL PROBLEMS

39-1 39-2

1017 1018 1018 1020 1024 1025 1027 1028 1030 1035 1036

1042

1044

Quantum-Mechanical View of Atoms

Hydrogen Atom: Schrédinger Equation and Quantum Numbers 39-3 Hydrogen Atom Wave Functions 39-4 Complex Atoms; the Exclusion Principle 39-5 Periodic Table of Elements 39-6 X-Ray Spectra and Atomic Number *39-7 Magnetic Dipole Moment; Total Angular Momentum 39-8 Fluorescence and Phosphorescence 39-9 Lasers *39-10 Holography SUMMARY PROBLEMS

1014

Quantum Mechanics—A New Theory The Wave Function and Its Interpretation; the Double-Slit Experiment 38-3 The Heisenberg Uncertainty Principle 384 Philosophic Implications; Probability Versus Determinism 38-5 The Schrodinger Equation in One Dimension—Time-Independent Form *38-6 Time-Dependent Schrodinger Equation 38-7 Free Particles; Plane Waves and Wave Packets 38-8 Particle in an Infinitely Deep Square Well Potential (a Rigid Box) 38-9 Finite Potential Well 38-10 Tunneling through a Barrier SUMMARY PROBLEMS

39

QUANTUM MECHANICS OF ATOMS

40

1066 1067

QUESTIONS 1066 GENERAL PROBLEMS

1098 1099

QUESTIONS 1099 GENERAL PROBLEMS

-

1057 1060 1061 1064

1071

Bonding in Molecules Potential-Energy Diagrams for Molecules 40-3 Weak (van der Waals) Bonds 40-4 Molecular Spectra 40-5 Bonding in Solids 40-6 Free-Electron Theory of Metals; Fermi Energy 40-7 Band Theory of Solids 40-8 Semiconductors and Doping 40-9 Semiconductor Diodes 40-10 Transistors and Integrated Circuits (Chips) SUMMARY PROBLEMS

1045 1049 1052 1053 1054

1069

MOLECULES AND SOLIDS

40-1 40-2

1045

1071 1074 1077 1080 1085 1086 1090 1093 1094 1097

1102

41 ™

41-1 41-2 41-3 41-4 41-5

41-6 41-7

NUCLEAR PHYSICS AND RADIOACTIVITY Structure and Properties of the Nucleus Binding Energy and Nuclear Forces Radioactivity Alpha Decay Beta Decay

Gamma Decay

Conservation of Nucleon Number and Other Conservation Laws 41-8 Half-Life and Rate of Decay 41-9 Decay Series 41-10 Radioactive Dating 41-11 Detection of Radiation SUMMARY

PROBLEMS

42 42-1

1126 1127

QUESTIONS GENERAL

1126

PROBLEMS

NUCLEAR ENERGY; EFFECTS AN D USES OF RADIATION Nuclear Reactions and the

42-4 42-5 42-6

™42-7

*42-8 *42-9

43

1105 1108 1110 1111 1114

43-1 43-2

1117 1117 1121 1122 1124

43-7 43-8 43-9 43-10

1116

1129

1132

Nuclear Fission; Nuclear Reactors

1136

'Tracers in Research and Medicine Imaging by Tomography: CAT Scans

and Emission Tomography

*42-10 Nuclear Magnetic Resonance (NMR); Magnetic Resonance Imaging (MRI) SUMMARY

PROBLEMS

1159

1160

QUESTIONS GENERAL

1159

PROBLEMS

434

43-5 43-6

1164

ELEMENTARY PARTICLES

High-Energy Particles and Accelerators 1165 Beginnings of Elementary Particle Physics—Particle Exchange 1171 Particles and Antiparticles 1174

Particle Interactions and Conservation Laws 1175 Neutrinos—Recent Results 1177 Particle Classification 1178

Particle Stability and Resonances Strangeness? Charm? Towards a New Model Quarks The Standard Model: QCD and Electroweak Theory 43-11 Grand Unified Theories 43-12 Strings and Supersymmetry SUMMARY PROBLEMS

1189 1190

QuUEsTIONS 1190 GENERAL PROBLEMS

1180 1181 1182 1184 1187 1189

1191

1135

Nuclear Fusion 1141 Passage of Radiation Through Matter; Radiation Damage 1146 Measurement of Radiation—Dosimetry 1147

R%diation Therapy

43-3

1131

Transmutation of Elements

42-2 Cross Section 42-3

1104

1150

1151

1153 1156 1162

44 44-1

44-2

ASTROPHYSICS AND COSMOLOGY 1193 Stars and Galaxies

1194

Stellar Evolution: Nucleosynthesis,

and the Birth and Death of Stars Distance Measurements General Relativity: Gravity and the Curvature of Space 44-5 The Expanding Universe: Redshift and Hubble’s Law 44-6 The Big Bang and the Cosmic Microwave Background 44-7 The Standard Cosmological Model: Early History of the Universe 44-8 Inflation 44-9 Dark Matter and Dark Energy 44-10 Large-Scale Structure of the Universe 44-11 Finally . ..

1197 1203

44-3 44-4

SUMMARY PROBLEMS

APPENDICES

1225 1226

QUESTIONS 1226 GENERAL PROBLEMS

1205 1209 1213 1216 1219 1221 1224 1224 1227

A B C D

MATHEMATICAL FORMULAS DERIVATIVES AND INTEGRALS MORE ON DIMENSIONAL ANALYSIS GRAVITATIONAL FORCE DUE TO A SPHERICAL MASS DISTRIBUTION E DIrrERENTIAL FORM OF MAXWELL'S EQUATIONS F SELECTED ISOTOPES ANSWERS TO ODD-NUMBERED PROBLEMS INDEX PHOTO CREDITS CONTENTS

A-1 A-6 A-8 A-9 A-12 A-14 A-18 A-47 A-72 xi

APPLICATIONS (SELECTED) Chapter 1 The 8000-m peaks Estimating volume of a lake Height by triangulation Radius of the Earth Heartbeats in a lifetime Particulate pollution (Pr30) Global positioning satellites (Pr39) 16 Lung capacity (Pr65) Chapter 2 Airport runway design Automobile air bags Braking distances CD error correction (Pr10) CD playing time (Pr13) Golfing uphill or down (Pr79) Rapid transit (Pr83) Chapter 3 Kicked football 66, 69 Ball sports (Problems) 77,81,82 Extreme sports (Pr41) 77 Chapter 4 Rocket acceleration What force accelerates a car? How we walk Elevator and counterweight Mechanical advantage of pulley Bear sling (Q24) High-speed elevators (Pr19) 105 Mountain climbing (Pr31, 82,83)

106,110

Oil/mineral exploration

144,420

City planning, cars on hills (Pr71) 109 Bicyclists (Pr72,73) 109 “Doomsday” asteroid (Pr84) 110 Chapter § Push or pull a sled? 116 Centrifugation 122 Not skidding on a curve 126-7 Banked highways 127 Simulated gravity (Q18, Pr48) 131,134 “Rotor-ride” (Pr82) 136 Chapter 6 Artificial Earth satellites 146 Geosynchronous satellites 147 Weightlessness 148 Free fall in athletics 149 Planet discovery, extrasolar planets 152 Black holes 156 Asteroids (Pr44, 78) 159,162 Navstar GPS (Pr58) 160 Black hole, galaxy center (Pr61, 64) 160, 161 Tides (Pr75) 162 Chapter 7

Car stopping distance o< v”

Lever (Pr6) Spiderman (Pr54) Bicycling on hills, gears (Pr85) Child safety in car (Pr87) Rock climber’s rope (Pr90) Chapter 8 Downbhill ski runs Roller coaster Pole vault Toy dart gun

xii

APPLICATIONS

174

177 179 181 181 182 183 191,198 192-3 193

Escape velocity from Earth or Moon 201 Stair climbing power 202 Power needs of car 202-3 Cardiac treadmill (Pr104) 213 Chapter 9 Tennis serve 216 Rocket propulsion 219,236-8 Rifle recoil 220 Karate blow 221

Billiards/bowling Nuclear collisions

223,228 225,228

Ballistic pendulum 226 Conveyor belt 237 Gravitational slingshot (Pr105) 246 Crashworthiness (Pr109) 247 Asteroids, planets (Pr110,112,113) 247 Chapter 10 Hard drive and bit speed 253 Wrench/tire iron 256 Flywheel energy 266,281 Yo-yo 271 Car braking forces 272-3 Bicycle odometer calibration (Q1) 275

Tightrope walker (Q11)

Triceps muscle and throwing

(Pr38,39)

CD speed (Pr84) Bicycle gears (Pr89) Chapter 11

275

278 281 281

Rotating skaters, divers

284,286,309

Blood flow

353,357,361

Neutron star collapse 287 Auto wheel balancing 296 Top and gyroscope 299-300 Coriolis effect 301-2 Hurricanes 302 SUV possible rollover (Pr67) 308 Triple axel jump (Pr79) 309 Bat’s “sweet spot” (Pr82) 310 Chapter 12 Tragic collapse 311,323 Lever’s mechanical advantage 313 Cantilever 315 Biceps muscle force 315 Human balance with loads 318 Trusses and bridges 324-6,335 Architecture: arches and domes ~ 327-8 Forces on vertebrae (Pr87) 337 Chapter 13 Lifting water 345,348 Hydraulic lift, brakes 346 Pressure gauges 346-7 Hydrometer 351 Helium balloon lift 352,368

Airplane wings, lift

Sailing against the wind Baseball curve Blood to the brain, TIA Blood flow and heart disease Surface tension, capillarity Walking on water Pumps and the heart Reynolds number (Pr69) Chapter 14 Car shock absorbers Resonance damage

356

357 357 357 359 359-60 360 361 366 383 386

Chapter 15 Echolocation by animals

400‘

Earthquake waves

401, 403,416

Chapter 16 Distance from lightning Autofocus camera Wide range of human hearing Loudspeaker response Stringed instruments Wind instruments Tuning with beats Doppler blood flow meter Sonar: sonic boom Ultrasound medical imaging Motion sensor (Pr5)

425 426 427-8,431 428 432-3 433-6 439 442,453 444 445-6 448

Chapter 17 Hot air balloon 454 Expansion joints, highways 456, 460, 463 Gas tank overflow 462 Life under ice 462 Cold and hot tire pressure 468 Molecules in a breath 469 Thermostat (Q10) 471 Scuba/snorkeling (Pr38,47,82,85) 473,475 Chapter 18 Chemical reactions, temperature dependence 481 Superfluidity 483 Evaporation cools

484,505

Humidity, weather 485-6 Chromatography 490 Pressure cooker (Pr35) 493 Chapter 19 Working off the calories 498 Cold floors 516 Heat loss through windows 516 How clothes insulate 516-7 R-values for thermal insulation 517 Convective house heating 17 Human radiative heat loss 518 Room comfort and metabolism 519 Radiation from Sun 519 Medical thermography 519 Astronomy—size of a star 520 Thermos bottle (Q30) 521 Weather, air parcel, adiabatic lapse rate (Pr56) 525 Chapter 20 Steam engine 530 Internal combustion engine 531, 535-6 Car efficiency 532 Refrigerators, air conditioners 537-8 Heat pump 538 Biological evolution, development 545 Thermal pollution, global warming 549-51 Energy resources 550 Diesel engine (Pr7)

Chapter 21 Static electricity Photocopiers

553

560, 589 (Pr78) 569, 582-3

Electric shielding, safety 577 581-2 DNA structure and replication Biological cells: electric forces and kinetic theory 581-2,617 Laser & inkjet printers 583

"™

Chapter 23 Breakdown voltage Lightning rods, corona CRT, oscilloscopes,

612 612

TV monitors

620-1, 723

f Photocell (Pr75)

626

Geiger counter (Pr83) Van de Graaff (Pr84) Chapter 24 ‘

627 627,607

Very high capacitance Computer key

631 631

Capacitor uses

628, 631

Camera flash

636

Heart defibrillator 638 DRAM (Pr10,57) 644, 647 Electrostatic air cleaner (Pr20) 645 CMOS circuits (Pr53) 647 Chapter 25 Light bulb 651,653, 660 Battery construction 653 Loudspeaker wires 659 Resistance thermometer 660 Heating elements, bulb filament 660 Why bulbs burn out at turn on 661 Lightning bolt 662 Household circuits, shorts 662-3 Fuses, circuit

breakers

662-3,747,776

Extension cord danger 663 Nervous sysfiem, conduction 669-70 Strain gauge (Pr 24) 673 Chapter 26 Car battery charging, jump start 686, 687 RC applications: flashers, wipers 691 Heart pacemaker 692,787 Electric hazards 6924 Proper grounding 693-4 Heart fibrillation 692 Meters, analog and digital 695-7 Potentiometers and bridges (Pr) 704,705 Chapter 27 Compass and declination 709 Aurora borealis 717 Motors, lou peakers, galvanometers 720-1 Mass spectrometer 724-5 Electromagnetic pumping (Q14) 726 Cyclotron (Pr66) 731 Beam steering (Pr67) 731 Chapter 28 Coaxial cable 740, 789 Solenoid switches: car starters, doorbell

Circuit breakers, magnetic Relay (Q16 Atom trap (Pr73) Chapter 29 Induction stove EM blood-flow meter Power plant generators Car alternators Motor overload Airport metal detector Eddy current damping Transformers and uses, power Car ignition, bulb ballast Microphon Read/write on disks and tape Digital coding Credit card swipe

Ground fault circuit interrupter (GFCI) Inductive battery charger (Pr24) Betatron (Pr55) Search coil (Pr68) Chapter 30 Spark plug Pacemaker Surge protector LC oscillators, resonance Capacitors as filters Loudspeaker cross-over Impedance matching Three-phase AC Q-value (Pr86, 87) Chapter 31 Antennas

776 780 782 783 785 787 792 794, 802 799 799 802-3 803 810 824,831

Phone call lag time Solar sail Optical tweezers

825 829 829

Wireless: AM/FM, TV, tuning,

cell phones, remotes Chapter 32 How tall a mirror do you need Close up and wide-view mirrors

829-32 840-1

842, 849, 859

Where you can see yourself in a concave mirror Optical illusions

Apparent depth in water Rainbows Colors underwater Prism binoculars Fiber optics in

848

851,903

852 853 854 855

telecommunications

855-6, 865

Medical endoscopes 856 Highway reflectors (Pr86) 865 Chapter 33 Where you can see a lens image 869 Cameras, digital and film 878 Camera adjustments 879-80 Pixels and resolution 881 Human eye 882-5, 892 Corrective lenses 883-5 Contact lenses 885 Seeing under water 885 Magnifying glass 885-7 Telescopes Microscopes

887-9,931-2 890-1, 931,933

Chapter 34

Bubbles, reflected color

Mirages

Colors in thin soap film, details

Lens coatings Multiple coating (Pr52) Chapter 35

Lens and mirror resolution Hubble Space Telescope Eye resolution, useful magnification Radiotelescopes Telescope resolution, A rule Spectroscopy X-ray diffraction in biology Polarized sunglasses LCDs—liquid crystal displays Sky color

900, 912-3

903

912-3

9134 919 929-30 930 930, 932-3 931 931 935-6 939 942 943-4 945

Chapter 36 Space travel Global positioning system (GPS) Chapter 37 Photocells Photodiodes Photosynthesis Measuring bone density Electron microscopes Chapter 38 Tunneling through a QM barrier Scanning tunneling electron microscope Chapter 39 Fluorescence analysis Fluorescent bulbs Phosphorescence, watch dials Lasers DVD and CD players Barcodes Laser surgery Holography

963 964 992 992 993 995 1000 ~ 1038 1038-9 1060 1060 1061 1061-5 1063 1063 1064 1064-5

Chapter 40

Cell energy—activation energy, ATP 1075-7

Weak bonds in cells, DNA

1077-8

Protein synthesis Transparency

Semiconductor diodes, transistors

1079-80 1092 1094-8

Rectifier circuits 1096 LED displays; photodiodes 1096 Integrated circuits (Chips) 1098 Chapter 41 Smoke detectors 1114 Carbon-14 dating 1122-3 Archeological, geological dating 11234 Oldest Earth rocks and earliest life 1124 Chapter 42 Nuclear reactors and power plants 1138-40 Manbhattan Project 1141 Stellar fusion 1142-3 Fusion energy reactors 1131,1144-6 Biological radiation damage 1146-7 Radiation dosimetry 1147-9 Radon

Human radiation exposure Radiation sickness Radiation therapy Proton therapy Tracers in medicine and biology X-ray imaging CAT scans Emission tomography: PET and SPET NMR and MRI Chapter 43 Antimatter

Chapter 44 Stars and galaxies Star evolution

1148,1150

1148-9 1149 1150-1 1151 1151-2 1153 1153-5 1156 1156-9

1174-5,1188

1194-9 1200-2

Supernovae

1201, 1202, 1203

Black holes

1202, 1208-9

Star distances

1194,1203+4

Curved space 1207-8 Big Bang 1212,1213-6 Evolution of universe 1216-9 Dark matter and dark energy 1221-3

APPLICATIONS

xiii

Preface I was motivated from the beginning to write a textbook different from others that present physics as a sequence of facts, like a Sears catalog: “here are the facts and you better learn them.” Instead of that approach in which topics are begun formally and dogmatically, I have sought to begin each topic with concrete observations and experiences students can relate to: start with specifics and only then go to the great generalizations and the more formal aspects of a topic, showing why we believe what we believe. This approach reflects how science is actually practiced.

Why a Fourth Edition? Two recent trends in physics texbooks are disturbing: (1) their revision cycles have become short—they are being revised every 3 or 4 years; (2) the books are getting larger, some over 1500 pages. I don’t see how either trend can be of benefit to students. My response: (1) It has been 8 years since the previous edition of this book. (2) This book makes use of physics education research, although it avoids the detail a Professor may need to say in class but in a book shuts down the reader. And this book still remains among the shortest. This new edition introduces some important new pedagogic tools. It contains new physics (such as in cosmology) and many new appealing applications (list on previous page). Pages and page breaks have been carefully formatted to make the physics easier to follow: no turning a page in the middle of a derivation or Example. Great efforts were made to make the book attractive so students will want to read it. Some of the new features are listed below.

-

What's New

F, v, B

xiv

PREFACE

Chapter-Opening Questions: Each Chapter begins with a multiple-choice question, whose responses include common misconceptions. Students are asked to answer before starting the Chapter, to get them involved in the material and to get any preconceived notions out on the table. The issues reappear later in the Chapter, usually as Exercises, after the material has been covered. The Chapter-Opening Questions also show students the power and usefulness of Physics. APPROACH paragraph in worked-out numerical Examples: A short introductory paragraph before the Solution, outlining an approach and the steps we can take to get started. Brief NOTES after the Solution may remark on the Solution, may give an alternate approach, or mention an application. Step-by-Step Examples: After many Problem Solving Strategies (more than 20 in the book), the next Example is done step-by-step following precisely the steps just seen. Exercises within the text, after an Example or derivation, give students a chance to see if they have understood enough to answer a simple question or do a simple calculation. Many are multiple choice. Greater clarity: No topic, no paragraph in this book was overlooked in the search to improve the clarity and conciseness of the presentation. Phrases and sentences that may slow down the principal argument have been eliminated: keep to the essentials at first, give the elaborations later. Vector notation, arrows: The symbols for vector quantities in the text and Figures now have a tiny arrow over them, so they are similar to what we write by hand. Cosmological Revolution: With generous help from top experts in the field‘\ readers have the latest results.

Page layout: more than in the previous edition, serious attention has been paid to how each page is formatted. Examples and all important derivations and arguments are on facing pages. Students then don’t have to turn back and forth. Throughout, readers see, on two facing pages, an important slice of physics. ’"‘Vew Appltcattons LCDs, digital cameras and electronic sensors (CCD, CMOS), electric hazards, GFClISs, photocopiers, 1nk]et and laser printers, metal detectors, underwater vision, curve balls, airplane wings, DNA, how we actually see images. (Turn back a page to see a longer list.) Examples modified: more math steps are spelled out, and many new Examples added. About 10% of all Examples are Estimation Examples. This Book is Shorter than other complete full-service books at this level. Shorter explanations are easier to understand and more likely to be read.

Content and Organizational Changes o ¢ e e e o r

e e

Rot*ionfl Motion: Chapters 10 and 11 have been reorganized. All of angular momentum is now in Chapter 11. First law of thermodynamics in Chapter 19, has been rewritten and extended. The ull form is given: AK + AU + AE,,, = Q — W, where internal energy is Ei., and U is potential energy; the form Q — W is kept so that dW = P dV. Kinematics and Dynamics of Circular Motion are now treated together in Cha{t':er 3 Work and Energy, Chapters 7 and 8, have been carefully revised. Work done by friction is discussed now with energy conservation (energy terms due to friction). Cha;;ters on Inductance and AC Circuits have been combined into one: Chapter 30. Graphical Analysis and Numerical Integration is a new optional Section 2-9. Problems requiring a computer or graphing calculator are found at the end of most Chapters. Length of an object is a script £ rather than normal /, which looks like 1 or I

Versions of this Book

(moment of inertia, current), as in F = I£B. Capital L is for angular mo"{::ntum, latent heat, inductance, dimensions of length [L]. Newton’s law of gravitation remains in Chapter 6. Why? Because the 1/r* law is too important to relegate to a late chapter that might not be covered at all late in the semester; furthermore, it is one of the basic forces in nature. In Chapter 8 we can treat real gravitational potential energy and have a fine

Complete version: 44 Chapters including 9 Chapters of modern physics.

e

New Appendices include the differential form of Maxwell’s equations and more on dimensional analysis.

°

Protjlem Solving Strategies are found on pages 30, 58, 64, 96, 102, 125, 166,

3 Volume version: Available separately or packaged together (Vols. 1 & 2 or all 3 Volumes):

o

instance of using U = — [F-dl.

198,229, 261, 314, 504, 551, 571, 600, 685, 716, 740, 763, 849, 871, and 913.

Organization Some instr‘uctors may find that this book contains more material than can be covered in their courses. The text offers great flexibility. Sections marked with a star * are |considered optional. These contain slightly more advanced physics material, or material not usually covered in typical courses and/or interesting applications; they contain no material needed in later Chapters (except perhaps in later optional Sections). For a brief course, all optional material could be dropped as well as Iajor parts of Chapters

1, 13, 16, 26, 30, and 35, and selected parts of

r.Chapters 9,12, 19, 20, 33, and the modern physics Chapters. Topics not covered in lass can be a valuable resource for later study by students. Indeed, this text can serve as a 4seful reference for years because of its wide range of coverage.

Classic version: 37 Chapters including one each on relativity and quantum theory.

Volume 1: Chapters 1-20 on mechanics, including fluids, oscillations, waves, plus heat and thermodynamics. Volume 2: Chapters 21-35 on electricity and magnetism, plus light and optics. Volume 3: Chapters 36—44 on modern physics: relativity, quantum theory, atomic physics, condensed matter, nuclear physics, elementary particles, cosmology and astrophysics. PREFACE

Xxv

Thanks Many physics professors provided input or direct feedback on every aspect of this textbook. They are listed below, and I owe each a debt of gratitude. Mario Affatigato, Coe College Lorraine Allen, United States Coast Guard Academy Zaven Altounian, McGill University

Jim LaBelle, Dartmouth College M.A K. Lodhi, Texas Tech Bruce Mason, University of Oklahoma

Michael Barnett, Lawrence Berkeley Lab Anand Batra, Howard University

Linda McDonald, North Park College Bill McNairy, Duke University Raj Mohanty, Boston University Giuseppe Molesini, Istituto Nazionale di Ottica Florence Lisa K. Morris, Washington State University Blaine Norum, University of Virginia Alexandria Oakes, Eastern Michigan University

Bruce Barnett, Johns Hopkins University

Cornelius Bennhold, George Washington University Bruce Birkett, University of California Berkeley Dr. Robert Boivin,

Auburn University

Subir Bose, University of Central Florida David Branning, Trinity College Meade Brooks, Collin County Community College

Bruce Bunker,

University of Notre Dame

Grant Bunker, Illinois Institute of Technology

Wayne Carr, Stevens Institute of Technology Charles Chiu, University of Texas Austin

Robert Coakley, University of Southern Maine

David Curott, University of North Alabama Biman Das, SUNY Potsdam

Bob Davis, Taylor University Kaushik De,

University of Texas Arlington

Michael Dennin, University of California Irvine Kathy Dimiduk, University of New Mexico

John DiNardo,

Drexel University

Scott Dudley, United States Air Force Academy

John Essick,

Reed College

Cassandra Fesen, Dartmouth College Alex Filippenko, University of California Berkeley

Richard Firestone, Lawrence Berkeley Lab Mike Fortner, Northern Illinois University Tom Furtak, Colorado School of Mines Edward Gibson, California State University Sacramento John Hardy, Texas A&M J. Erik Hendrickson, University of Wisconsin Eau Claire Laurent Hodges, Iowa State University David Hogg, New York University

Mark Hollabaugh, Normandale Community College Andy Hollerman, University of Louisiana at Lafayette William Holzapfel, University of California Berkeley Bob Jacobsen, University of California Berkeley Teruki Kamon,

Texas A&M

Daryao Khatri, University of the District of Columbia Jay Kunze, Idaho State University

Dan Mazilu,

“

Virginia Tech

Michael Ottinger,

Missouri Western State University

Lyman Page, Princeton and WMAP Bruce Partridge,

Haverford College

R. Daryl Pedigo, University of Washington Robert Pelcovitz, Brown University

Vahe Peroomian, UCLA James Rabchuk, Western Illinois University Michele Rallis, Ohio State University

Paul Richards, University of California Berkeley Peter Riley, University of Texas Austin Larry Rowan, University of North Carolina Chapel Hill Cindy Schwarz, Vassar College Peter Sheldon, Randolph-Macon Woman'’s College Natalia A. Sidorovskaia, University of Louisiana at Lafayette

James Siegrist,

UC Berkeley, Director Physics Division LBNL

George Smoot, University of California Berkeley Mark Sprague, East Carolina University

Michael Strauss,

University of Oklahoma

Laszlo Takac, University of Maryland Baltimore Co. Franklin D. Trumpy, Des Moines Area Community College Ray Turner, Clemson University Som Tyagi, Drexel University John Vasut, Baylor University Robert Webb,

S

Texas A&M

Robert Weidman, Michigan Technological University Edward A. Whittaker, Stevens Institute of Technology John Wolbeck, Orange County Community College Stanley George Wojcicki, Stanford University Edward Wright, UCLA Todd Young, Wayne State College William Younger, College of the Albemarle Hsiao-Ling Zhou, Georgia State University

I owe special thanks to Prof. Bob Davis for much valuable input, and especially for working out all the Problems and producing the Solutions Manual for all Problems, as well as for providing the answers to odd-numbered Problems at the end of this book. Many thanks also to J. Erik Hendrickson who collaborated with Bob Davis on the solutions, and to the team they managed (Profs. Anand Batra, Meade Brooks, David Currott, Blaine Norum, Michael

Ottinger, Larry Rowan,

Ray Turner, John Vasut,

William Younger). I am grateful to Profs. John Essick, Bruce Barnett, Robert Coakley, Biman Das, Michael Dennin, Kathy Dimiduk, John DiNardo, Scott Dudley, David Hogg, Cindy Schwarz, Ray Turner, and Som Tyagi, who inspired many of the Examples, Questions, Problems, and significant clarifications. Crucial for rooting out errors, as well as providing excellent suggestions, were Profs. Kathy Dimiduk, Ray Turner, and Lorraine Allen. A huge thank you to them and to Prof. Giuseppe Molesini for his suggestions and his exceptional photographs™ for optics. xvi

PREFACE

For Chapters 43 and 44 on Particle Physics and Cosmology and Astrophysics, I was fortunate to receive generous input from some of the top experts in the field, to whom I owe a debt of gratitude: George Smoot, Paul Richards, Alex Filippenko, James Siegrist, and William Holzapfel (UC Berkeley), Lyman Page (Princeton and MAP), Edward Wright (UCLA and WMAP), and Michael Strauss (University of Oklahoma). I especially wish to thank Profs. Howard Shugart, Chair Frances Hellman, and many others at the University of California, Berkeley, Physics Department for helpful disc*xssions, and for hospitality. Thanks also to Prof. Tito Arecchi and others at the Istituto Nazionale di Ottica, Florence, Italy. Finally, I am grateful to the many people at Prentice Hall with whom I worked on this project, especially Paul Corey, Karen Karlin, Christian Botting, John Christiana, and Sean Hogan. The firLal responsibility for all errors lies with me. I welcome comments, corrections, and suggestions as soon as possible to benefit students for the next reprint. email: [email protected]

D.C.G.

Post: Paul Corey

One Lake Street Upper Saddle River, NJ 07458

About the Author Douglas C. Giancoli obtained his BA in physics (summa cum laude) from the University of California, Berkeley, his MS in physics at the Massachusetts Institute of Technology, and his PhD in elementary particle physics at the University of California, Berkeley. He spent 2 years as a post-doctoral fellow at UC Berkeley’s Virus lab developing skills in molecular biology and biophysics. His mentors include r»lobel winners Emilio Segré and Donald Glaser. He has taught a wide range of undergraduate courses, traditional as well as innovative ?nes, and continues to update his textbooks meticulously, seeking ways to better provide an understanding of physics for students. Doug’s favorite spare-time activity is the outdoors, especially climbing peaks (here on a dolomite summit, Italy). He says climbing peaks is like learning physics: it takes effort and the rewards are great.

Online Supplements (partial list)

Student Study Guide & Selected Solutions Manual (Volume I: 0-13-227324-1, Volumes II & III: 0-13-227325-X) by Frank Wolfs

tutoring by responding to their wrong answers and providing hints for solving multi-step problems when they get stuck. It gives them immediate and up-to-date assessment of their progress, and shows where they need to practice more. MasteringPhysics provides instructors with a fast and effective way to assign triedand-tested online homework assignments that comprise a range of problem typ;»es. The powerful post-assignment diagnostics allow instructors éjo assess the progress of their class as a whole as well as individual students, and quickly identify areas of difficulty.

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WebAssign (www.webassign.com) CAPA and LON-CAPA (www.lon-capa.org)

Mathematics for Physics with Calculus (0-13-191336-0) by Biman Das

MIT, MasteringPhysics provides students with individualized online

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Student Supplements (partial list)

MnsteringP‘;ysicsTM (www.masteringphysics.com) is a sophisticated online tutoring and homework system developed specially for courses using calculus-based physics. Originally developed by David Pritchard and collaborators at

Student Pocket Companion (0-13-227326-8) by Biman Das

Tutorials in Introductory Physics (0-13-097069-7)

Education Group at the University of Washington Physlet® Physics (0-13-101969-4) by Wolfgang Christian and Mario Belloni

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PREFACE

xvii

To Students HOW TO STUDY 1. Read the Chapter. Learn new vocabulary and notation. Try to respond tc™™ questions and exercises as they occur. 2. Attend all class meetings. Listen. Take notes, especially about aspects you do not remember seeing in the book. Ask questions (everyone else wants to, but maybe you will have the courage). You will get more out of class if you read the Chapter first. 3. Read the Chapter again, paying attention to details. Follow derivations and worked-out Examples. Absorb their logic. Answer Exercises and as many of the end of Chapter Questions as you can. 4. Solve 10 to 20 end of Chapter Problems (or more), especially those assigned. In doing Problems you find out what you learned and what you didn’t. Discuss them with other students. Problem solving is one of the great learning tools. Don’t just look for a formula—it won’t cut it. NOTES ON THE FORMAT AND PROBLEM SOLVING 1. Sections marked with a star (*) are considered optional. They can be omitted without interrupting the main flow of topics. No later material depends on them except possibly later starred Sections. They may be fun to read, though. 2. The customary conventions are used: symbols for quantities (such as m for mass) are italicized, whereas units (such as m for meter) are not italicized. Symbols for vectors are shown in boldface with a small arrow above: F. 3. Few equations are valid in all situations. Where practical, the limitations of important equations are stated in square brackets next to the equation. The equations that represent the great laws of physics are displayed with a tan background, as are a few other indispensable equations. 4. At the end of each Chapter is a set of Problems which are ranked as Level I, II, or

II, according to estimated difficulty. Level I Problems are easiest, Level II arc« standard Problems, and Level III are “challenge problems.” These ranked

Problems are arranged by Section, but Problems for a given Section may depend on earlier material too. There follows a group of General Problems, which are not arranged by Section nor ranked as to difficulty. Problems that relate to optional Sections are starred (*). Most Chapters have 1 or 2 Computer/Numerical Problems at the end, requiring a computer or graphing calculator. Answers to odd-numbered Problems are given at the end of the book. 5. Being able to solve Problems is a crucial part of learning physics, and provides a powerful means for understanding the concepts and principles. This book contains many aids to problem solving: (a) worked-out Examples and their solutions in the text, which should be studied as an integral part of the text; (b) some of the worked-out Examples are Estimation Examples, which show how rough or approximate results can be obtained even if the given data are sparse (see Section 1-6); (c) special Problem Solving Strategies placed throughout the text to suggest a step-by-step approach to problem solving for a particular topic—but remember that the basics remain the same; most of these “Strategies” are followed by an Example that is solved by explicitly following the suggested steps; (d) special problem-solving Sections; (e) “Problem Solving” marginal notes which refer to hints within the text for solving Problems; (f) Exercises within the text that you should work out immediately, and then check your response against the answer given at the bottom of the last page of that Chapter; (g) the Problems themselves at the end of each Chapter (point 4 above). 6. Conceptual Examples pose a question which hopefully starts you to think and come up with a response. Give yourself a little time to come up with your own response before reading the Response given. 7. Math review, plus some additional topics, are found in Appendices. Useful data, conversion factors, and math formulas are found inside the front and back covers. xviii

PREFACE

USE OF COLOR Vectors A general vector

Ii

resultant vector (sum) is slightly thicker

LR

components of any vector are dashed

-

Displacement ( 5, r)

e

Velocity (V)

e

Acceleration (8)

iz

Force

(F)

s

Force on second or

e

third object in same figure

masmsm

—

e

Momentum (p or mv)

et

Angular momentum (L)

I

Angular velocity (w)

i

Torque (7)

———

Electric field (E)

mm——————

Magnetic field (B)

e

Electricity and magnetism Electric field lines

m——

Electric circuit symbols

——

Wire, with switch S

—g—

——=====

Resistor

MWW~

Magnetic field lines

——

Capacitor

—4| }—

Electric charge (+)

@

oL @ ¥

Inductor

LETV-

Electric charge ()

(o)

or @-

Battery

——-|}—

Ground

J_

Equipotential lines

=~

Optics Light rays

Other =

i

Object

'

Real image

‘

Virtual image

)

(dashed)

(dashed and paler)

. H

:

Energy level (atom, etc.) Measurement lines .

+—1.0 m—»

Path of a moving object

—-_—————

Direction of motion

=~ ——

or current

PREFACE

xix

Image of the Earth from a NASA satellite. The sky

appears black from out in space because there are so few molecules to reflect light. (Why the sky appears blue to us on Earth has to do with scattering of light by molecules of the atmosphere, as discussed in Chapter 35.) Note the storm off the coast of Mexico.

Q

Introduct M

easurement,

E

v

Tek

®

o

. . stlmatlng

—Guess now! Suppose you wanted to actually measure the radius of the Earth, at least roughly, rather than taking other people’s word for what it is. Which response below describes the best approach? Give up; it is impossible using ordinary means. Use an extremely long measuring tape. It is only possible by flying high enough to see the actual curvature of the Earth. Use a standard measuring tape, a step ladder, and a large smooth lake. Use a laser and a mirror on the Moon or on a satellite. We start each Chapter with a Question, like the one above. Try to answer it right away. Don’t worry bout getting the right answer now—the

>

idea is to get your preconceived notions out on the table. If they

are misconceptions, we expect them to be cleared up as you read the Chapter. You will usually get another

chance at the Question later in the Chapter when the appropriate material has been covered. These Chapter-Opening Questions will also help you to see the power and usefulness of physics.]

CONTENTS 1-1 The Nature of Science 1-2 Models, Theories, and Laws 1-3 Measurement and Uncertainty; Significant Figures 1-4 Units, Standards, and the SI System 1-5 Converting Units =8

grd?é %f l\./[agn'ltude: apid

bstimating

*#1-7 Dimensions and Dimensional Analysis

hysics is the most basic of the sciences. It deals with the behavior and structure of matter. The field of physics is usually divided into classical physics which includes motion, fluids, heat, sound, light, electricity and magnetism; and modern physics which includes the topics of relativity, atomic structure, condensed matter, nuclear physics, elementary particles, and cosmology and astrophysics. We will cover all these topics in this book, beginning with motion (or mechanics, as it is often called) and ending with the most recent results in our study of the cosmos. An understanding of physics is crucial for anyone making a career in science or technology. Engineers, for example, must know how to calculate the forces within a structure to design it so that it remains standing (Fig. 1-1a). Indeed, in Chapter 12 we will see a worked-out Example of how a simple physics calculation—or even intuition based on understanding the physics of forces—would have saved hundreds of lives (Fig. 1-1b). We will see many examples in this book of how physics is useful in many fields, and in everyday life.

1-1

«

The Nature of Science

The principal aim of all sciences, including physics, is generally considered to be

the search for order in our observations of the world around us. Many people

FIGURE 1-1 (a) This Roman aqueduct was built 2000 years ago and still stands. (b) The Hartford Civic Center collapsed in 1978, just two years after it was built.

think that science is a mechanical process of collecting facts and devising theories. But it is not so simple. Science is a creative activity that in many respects resembles other creative activities of the human mind. One important aspect of science is observation of events, which includes the design and carrying out of experiments. But observation and experiment require imagination, for scientists can never include everything in a description of what they observe. Hence, scientists must make judgments about what is relevant in their observations and experiments. Consider, for example, how two great minds, Aristotle (384-322 B.C.) and Galileo (1564-1642), interpreted motion along a horizontal surface. Aristotle noted that objects given an initial push along the ground (or on a tabletop) always slow down and stop. Consequently, Aristotle argued that the natural state of an object is to be at rest. Galileo, in his reexamination of horizontal motion in the 1600s, imagined that if friction could be eliminated, an object given an initial push along a horizontal surface would continue to move indefinitely without stopping. He concluded that for an object to be in motion was just as natural as for it to be at rest. By inventing a new approach, Galileo founded our modern view of motion (Chapters 2, 3, and 4), and he did so with a leap of the imagination. Galileo made this leap conceptually, without actually eliminating friction. Observation, with careful experimentation and measurement, is one side of the scientific process. The other side is the invention or creation of theories to explain and order the observations. Theories are never derived directly from observations. Observations may help inspire a theory, and theories are accepted or rejected based on the results of observation and experiment. The great theories of science may be compared, as creative achievements, with great works of art or literature. But how does science differ from these other creative activities? One important difference is that science requires testing of its ideas or theories to see if their predictions are borne out by experiment. Although the testing of theories distinguishes science from other creative fields, it should not be assumed that a theory is “proved” by testing. First of all, no measuring instrument is perfect, so exact confirmation is not possible. Furthermore, it is not possible to test a theory in every single possible circumstance. Hence a theory cannot be absolutely verified. Indeed, the history of science tells us that long-held theories can be replaced by new ones.

1-2 2

CHAPTER 1

Models, Theories, and Laws

When scientists are trying to understand a particular set of phenomena, they often make use of a model. A model, in the scientist’s sense, is a kind of analogy or mental image of the phenomena in terms of something we are familiar with. One

“

“

example is‘ the wave model of light. We cannot see waves of light as we can water waves. But it is valuable to think of light as made up of waves because experiments indicate that light behaves in many respects as water waves do. The purpose of a model is to give us an approximate mental or visual picture— r something to hold on to—when we cannot see what actually is happening. Models often give us a deeper understanding: the analogy to a known system (for instance, water waves in the above example) can suggest new experiments to perform and can provide ideas about what other related phenomena might occur. You may wonder what the difference is between a theory and a model. Usually a model is relatively simple and provides a structural similarity to the phenomena being studied. A theory is broader, more detailed, and can give quantitatively testable predictions, often with great precision. It is important, however, not to confuse a model or a theory with the real system or the phenomena themselves. Scientists give the title law to certain concise but general statements about how nature behaves (that energy is conserved, for example). Sometimes the statement takes the form of a relationship or equation between quantities (such as Newton’s second law, F = ma). To be called a law, a statement must be found experimentally valid over a wide range of observed phenomena. For less general statements, the term principle is often used (such as Archimedes’ principle). Scientific laws are different from political laws in that the latter are prescriptive: they tell us how we ought to behave. Scientific laws are descriptive: they do not say how nature should behave, but rather are meant to describe how nature does behave. As with theories, laws cannot be tested in the infinite variety of cases poLsible. So we cannot be sure that any law is absolutely true. We use the term “law” when its validity has been tested over a wide range of cases, and when any limitations and the range of validity are clearly understood. Scientists normally do their research as if the accepted laws and theories were rue. But they are obliged to keep an open mind in case new information should alter the validity of any given law or theory.

1-3

Measurement and Uncertainty; Significant Figures

In the quest to understand the world around us, scientists seek to find relationships among physical quantities that can be measured. Uncertairgz Reliable measurements are an important part of physics. But no measurement is FIGURE 1-2 Measuring the width absolutely precise. There is an uncertainty associated with every measurement. Among of a board with a centimeter ruler. the most important sources of uncertainty, other than blunders, are the limited accuracy ~ The uncertainty is about + 1 mm. of every measuring instrument and the inability to read an instrument beyond some fraction of the smallest division shown. For example, if you were to use a centimeter ruler to measure the width of a board (Fig. 1-2), the result could be claimed to be precise to about 0.1 cm (1 mm), the smallest division on the ruler, although half of this value might be a valid claim as well. The reason is that it is difficult for the observer to estimate (or interpolate) between the smallest divisions. Furthermore, the ruler itself may not have been manufactured to an accuracy very much better than this. When giving the result of a measurement, it is important to state the estimated uncertainty in the measurement. For example, the width of a board might be written as 8.8 + 0.1cm. The + 0.1 cm (“plus or minus 0.1 cm”) represents the estimated uncertainty in the measurement, so that the actual width most likely lies between 8.7 and 8.9 cm. The percent uncertainty is the ratio of the uncertainty rto the measured value, multiplied by 100. For example, if the measurement is 8.8 and the uncertainty about 0.1 cm, the percent uncertainty is 0.1 33 X 100% ~ 1%, where ~ means “is approximately equal to.”

SECTION 1-3

3

Often the uncertainty in a measured value is not specified explicitly. In such cases, the uncertainty in a numerical value is assumed to be one or a few units in the last digit specified. For example, if a length is given as 5.6 cm, the uncertainty is assumed to be about 0.1 cm or 0.2cm, or possibly 0.3 cm. It is important in this case that you do not write 5.60 cm, for this implies an uncertainty on the order of 0.01 or 0.02 cm; i assumes that the length is probably between about 5.58 cm and 5.62 cm, when actually you believe it is between about 5.4 and 5.8 cm.

Significant Figures

i 2

CT-T-1- 1

3

oESS (b)

(a) FIGURE

1-3

These two calculators

show the wrong number of significant figures. In (a), 2.0 was divided by 3.0. The correct final result should be 0.67. In (b), 2.5 was multiplied by 3.2. The correct result is 8.0,

The number of reliably known digits in a number is called the number of significant figures. Thus there are four significant figures in the number 23.21 cm and two in the number 0.062 cm (the zeros in the latter are merely place holders that show where the decimal point goes). The number of significant figures may not always be clear. Take, for example, the number 80. Are there one or two significant figures? We need words here: If we say it is roughly 80km between two cities, there is only one significant figure (the 8) since the zero is merely a place holder. If there is no suggestion that the 80 is a rough approximation, then we can often assume (as we will in this book) that it has two significant figures, so it is 80 km within an accuracy of about 1 or 2 km. If it is precisely 80 km, to within * 0.1 km, then we write 80.0 km (three significant figures). When making measurements, or when doing calculations, you should avoid the temptation to keep more digits in the final answer than is justified. For example, to calculate the area of a rectangle 11.3 cm by 6.8 cm, the result of multiplication would

be 76.84 cm® But this answer is clearly not accurate to 0.01 cm?, since (using the outer limits of the assumed uncertainty for each measurement) the result could be

between 11.2cm X 6.7cm = 75.04cm? and 11.4cm X 6.9cm = 78.66cm? At best,

Z

PROBLEM

SOLVING

Significant figure rule: Number of significant figures in final result should be same as the least significant input value

we can quote the answer as 77 cm?, which implies an uncertainty of about 1 or 2 cm? The other two digits (in the number 76.84 cm?) must be dropped because they are not significant. As a rough general rule (i.e., in the absence of a detailed consideration of uncertainties), we can say that the final result of a multiplication or division should, have only as many digits as the number with the least number of significant figures used in the calculation. In our example, 6.8 cm has the least number of significant figures, namely two. Thus the result 76.84 cm® needs to be rounded off to 77 cm?, EXERCISE A The area of a rectangle 4.5 cm by 3.25 cm is correctly given by (a) 14.625 cm?;

(b) 14.63 cm?; (c) 14.6 cm?; (d) 15 cm?.

A

CAUTION

Calculators err with significant figures

“ PROBLEM

SOLVING

Report only the proper number of significant figures in the final result. Keep extra digits during the calculation

FIGURE 1-4 Example 1-1. A protractor used to measure an angle. E

B

.

e

: pgay

i

>B

$: i e13 1

ie

=

4

CHAPTER 1

X3 bkl

CONCEPTUAL

EXAMPLE

1-1 | Significant figures. Usinga protractor (Fig. 1-4),

you measure an angle to be 30°. (¢) How many significant figures should you quote in this measurement? (b) Use a calculator to find the cosine of the angle you measured.

i 5

When adding or subtracting numbers, the final result is no more precise than the least precise number used. For example, the result of subtracting 0.57 from 3.6 is 3.0 (and not 3.03). Keep in mind when you use a calculator that all the digits it produces may not be significant. When you divide 2.0 by 3.0, the proper answer is 0.67, and not some such thing as 0.666666666 (Fig. 1-3a). Digits should not be quoted in a result unless they are truly significant figures. However, to obtain the most accurate result, you should normally keep one or more extra significant figures throughout a calculation, and round off only in the final result. (With a calculator, you can keep all its digits in intermediate results.) Note also that calculators sometimes give too few significant figures. For example, when you multiply 2.5 X 3.2, a calculator may give the answer as simply 8. But the answer is accurate to two significant figures, so the proper answer is 8.0. See Fig. 1-3b.

S50 SN

RESPONSE (a) If you look at a protractor, you will see that the precision with which you can measure an angle is about one degree (certainly not 0.1°). So you can quote two significant figures, namely, 30° (not 30.0°). (b) If you enter cos 30° in your calculator, you will get a number like 0.866025403. However, the angle you entered is known only to two significant figures, so its cosine is correctly given by 0.87; you must round your answer to two significant figures. NOTE Cosine and other trigonometric functions are reviewed in Appendix A.

“

| EXERCISE B Do 0.00324 and 0.00056 have the same number of significant figures? Be carefu]{ not to confuse significant figures with the number of decimal places.

'

EXERCISE C For each of the following numbers, state the number of significant figures and the number of decimal places: (a) 1.23; (b) 0.123; (c) 0.0123.

Be careful too about digital read-outs. If a digital bathroom scale shows 85.6, do not assume the uncertainty is £ 0.1 or £ 0.2; the scale was likely manufactured with an accuracy of perhaps 1% or so: that is, £ 1 or 2. For scientific instruments, the instruction manual should state the accuracy.

Scientific We

Notation

commjonly write numbers

in “powers

of ten,” or “scientific” notation—for

instance 36,900 as 3.69 X 10% or 0.0021 as 2.1 X 1073, One advantage of scientific

notation is that it allows the number of significant figures to be clearly expressed. For example, it is not clear whether 36,900 has three, four, or five significant figures. With powers of ten notation the ambiguity can be avoided: if the number is known to three significant figures, we write 3.69 X 10 but if it is known to four,

we write 3,690 X 10%

EXERCISE D Write each of the following in scientific notation and state the number of significant| figures for each: (a) 0.0258, (b) 42,300, (c) 344.50.

Percent Uncertainty versus Significant Figures The significant figures rule is only approximate, and in some cases may underestimate the accuracy (or uncertainty) of the answer. Suppose for example we divide 97 by 92: 97 n - 1.05 ~ 1.1 Both 97 and 92 have two significant figures, so the rule says to give the answer as 1.1. Yet/the numbers 97 and 92 both imply an uncertainty of * 1 if no other uncertainty is stated. Now 92 £ 1 and 97 = 1 both imply an uncertainty of about 1% (1/92 ~ 0.01 = 1%). But the final result to two significant figures is 1.1, with an implied uncertainty of =+ 0.1, which is an uncertainty of 0.1/1.1 = 0.1 = 10%. In this case it is better to give the answer as 1.05 (which is three significant figures). Why? Because 1.05 implies an uncertainty of + 0.01 which is 0.01/1.05 = 0.01 ~ 1%, just like the uncertainty in the original numbers 92 and 97. SUGGESTION: Use the significant figures rule, but consider the % uncertainty too, and add an extra digit if it gives a more realistic estimate of uncertainty.

Approximations Much of physics involves approximations, often because we do not have the means to solve a problem precisely. For example, we may choose to ignore air resistance or friction in doing a Problem even though they are present in the real world, and| then our calculation is only an approximation. In doing Problems, we should be aware of what approximations we are making, and be aware that the precision of our answer may not be nearly as good as the number of significant figures given in the result.

Accuracy versus Precision There is a technical difference between “precision” and “accuracy.” Precision in a strict sense refers to the repeatability of the measurement using a given instrument. For example, if you measure the width of a board many times, getting results like 8.81 cm, 8.85cm, 8.78 cm, 8.82 cm (interpolating between the 0.1 cm marks as best as possible each time), you could say the measurements give a precision a bit better than 0.1 cm. Accuracy refers to how close a measurement is to the true value. For example, if the ruler shown in Fig. 1-2 was manufactured with a 2% error, the accuracy of its measurement of the board’s width (about 8.8 cm) would be about 2% of 8.8cm or about £ 0.2 cm. Estimated uncertainty is meant to take both accuracy and precision into account.

\ \

SECTION 1-3

Measurement, Uncertainty; Significant Figures

5

1-4 Units, Standards, and the SI System TABLE 1-1 Some Typical Lengths or Distances (order of magnitude) Length

Meters

(or Distance)

(approximate)

Neutron or proton ISmetir) Ahqm :

1077 m 10-10

V(irLasn[lSee:r;ig i

i

)

2

pape;'

The measurement of any quantity is made relative to a particular standard or unit, and this unit must be specified along with the numerical value of the quantity. For‘\ example, we can measure length in British units such as inches, feet, or miles, or in the metric system in centimeters, meters, or kilometers. To specify that the length of a particular object is 18.6 is insufficient. The unit must be given, because 18.6 meters is very different from 18.6 inches or 18.6 millimeters. For any unit we use, such as the meter for distance or the second for time, we need to define a standard which defines exactly how long one meter or one second is. It is important that standards be chosen that are readily reproducible so that anyone needing to make a very acc_urate measurement can refer to the standard in

the laboratory and communicate with other people.

(thickness)

107% m

Height of Mt. Everest [see Fig. 1-5b] Earth diameter

Earth to Sun

Earth to nearest star Earth to nearest galaxy Earth to farthest

galaxy visible

1072 m 10°

8

Finger width : Football field length

Length The first truly international standard was the meter (abbreviated m) established as

the standard of length by the French Academy of Sciences in the 1790s. The stan-

10*

m

dard meter was originally chosen to be one ten-millionth of the distance from the Earth’s equator to either pole,” and a platinum rod to represent this length was

10"

m

of your finger, with arm and hand stretched out to the side.) In 1889, the meter was

10%

m

107

10" 10?2

m

m m

made. (One meter is, very roughly, the distance from the tip of your nose to the tip defined more precisely as the distance between two finely engraved marks on a particular bar of platinum-iridium alloy. In 1960, to provide greater precision and reproducibility, the meter was redefined as 1,650,763.73 wavelengths of a particular

orange light emitted by the gas krypton-86. In 1983 the meter was again redefined,

this time in terms of the speed of light (whose best measured value in terms of the

older definition of the meter was 299,792,458 m/s, with an uncertainty of 1 m/s).

FIGURE 1-5

Some lengths:

(a) viruses (about 10~ m long)

attacking a cell; (b) Mt. Everest’s

height is on the order of 10*m (8850 m, to be precise).

The new definition reads: “The meter is the length of path traveled by light in vacuum during a time interval of 1/299,792,458 of a second.”* British units of length (inch, foot, mile) are now defined in terms of the ™™

meter. The inch (in.) is defined as exactly 2.54 centimeters (cm; 1cm = 0.01 m).

Other conversion factors are given in the Table on the inside of the front cover

of this book. Table 1-1 presents some typical lengths, from very small to very large, rounded off to the nearest power of 10. See also Fig. 1-5. [Note that the abbreviation for inches (in.) is the only one with a period, to distinguish it from the word “in”.

Time The standard unit of time is the second (s). For many years, the second was defined as 1/86,400 of a mean solar day (24 h/day X 60 min/h X 60s/min = 86,400 s/day). The standard second is now defined more precisely in terms of the frequency of radiation emitted by cesium atoms when they pass between two particular states. [Specifically, one second is defined as the time required for 9,192,631,770 periods of this radiation.] There are, by definition, 60 s in one minute (min) and 60 minutes in one hour (h). Table 1-2 presents a range of measured time intervals, rounded off to the nearest power of ten. Mass

The standard unit of mass is the kilogram platinum—iridium cylinder, kept at the Measures near Paris, France, whose mass masses is presented in Table 1-3. [For 2.2 pounds on Earth.]

(b) 6

CHAPTER 1

(kg). The standard mass is a particular International Bureau of Weights and is defined as exactly 1kg. A range of practical purposes, 1kg weighs about

"Modern measurements of the Earth’s circumference reveal that the intended length is off by about one-fiftieth of 1%. Not bad! *The new definition of the meter has the effect of giving the speed of light the exact value of

299,792,458 m/s.

Introduction, Measurement, Estimating

TABLE 1-2 Some Typical Time Intervals Time Interval r

TABLE 1-3 Some Masses Seconds (approximate)

Lifetime of very unstable subatomic particle Lifetime of radioactive elements

Lifetime of muon Time between human heartbeats One day One year Human life span Length of recorded history Humans cj! Earth Life on Earth Age of Universe

10735

1072 5t010% s

1076 10° 10° 310" 2X10° 1 0% 10" 108

s s(=15s) s "5 s g 5

Object

Electron

=

1.6605

X

1073 kg

Proton, neutron

107% kg

DNA molecule Bacterium Mosquito Plum Human Ship Earth Sun Galaxy

When dealing with atoms and molecules, we usually use the unified atomic mass unit (1). In terms of the kilogram, 1u

Kilograms (approximate)

1072 ke.

107" 1075 107 107! 10> 10% 6 X 10* 2 X 10® 104

kg kg kg kg kg kg kg kg kg

TABLE 1-4 Metric (S1) Prefixes Abbreviation

Value

The definitions of other standard units for other quantities will be given as we ~ Yotta encounter them in later Chapters. (Precise values of this and other numbers are Zzetta given inside the front cover.) exa

Y Z E

10% 10 101:

Unit Prefixes

;),

1812

G

10

s

£

102

deka

da

10

In the metric system, the larger and smaller units are defined in multiples of 10 from . ) . . : the standard unit, and this makes calculation particularly easy. Thus 1 kilometer (km)

Prefix

f:rt: z

giga

is 1000m, 1 centimeter is 5y m, 1 millimeter (mm) is [agm or ;5cm, and so on.

o2

not only to units of length but to units of volume, mass, or any other metric unit.

~ Decto

The prefixes “centi-,” “kilo-,” and others are listed in Table 1-4 and can be applied

r For example, a centiliter (cL) is 155 liter (L), and a kilogram (kg) is 1000 grams (g).

An 8.2 megapixel camera has a detector with 8,200,000 pixels (picture elements). In common usage, 1 um is called 1 micron.

Systems of Units

When dealing with the laws and equations of physics it is very important touse a consistent set of units, Several systems of units have been in use over the years. Today the most important is the Systéme International (French for International System), which is abbreviated SI. In SI units, the standard of length is the meter,

deci centi

:uil:o* -~ p fstio

the standard for time is the second, and the standard for mass is the kilogram. This ~ 2tt©

system used to be called the MKS (meter-kilogram-second) system. A second metric system is the egs system, in which the centimeter, gram, and second are the standard units of length, mass, and time, as abbreviated in the title. The British engineering system has as its standards the foot for length, the pound for force, and the second for time. We use ST units almost exclusively in this book.

Base versus Derived Quantities Physical qu antities can be divided into two categories: base quantities and derived quantities. The corresponding units for these quantities are called base units and derived units. A base quantity must be defined in terms of a standard. Scientists, in the interest of simplicity, want the smallest number of base quantities possible consistent with a full description of the physical world. This number turns out to be seven, and those used i the SI are given in Table 1-5. All other quantities can be defined in terms of these se n base quantities,” and hence are referred to as derived quantities. An example of derived quantity is speed, which is defined as distance divided by the time fit takes to avel that distance. A Table inside the front cover lists many derived quantities d their units in terms of base units. To define any quantity, whether base or derived, we n specify a rule or procedure, and this is called an operational definition. "The only exc ptions are for angle (radians—see Chapter 10) and solid angle (steradian). No general agreement has been reached as to whether these are base or derived quantities.

zepto yocto f

M

105

9

-

h

101

d c

107 10'2

Z‘

ig_fi

i

P f

1079 1012 18 10_18

z y

107 107

8

10 A

wis the Greek letter “mu.”

TABLE 1-5

S| Base Quantities and Units Quantity

Unit

Unit Abbreviation

Length

meter

m

Time

second

s

Mass Electric current Temperature Amount of substance Luminous intensity

kilogram

kg

ampere kelvin

A K

mole

mol

candela

cd

SECTION

1-4

7

1-5 Converting Units Any quantity we measure, such as a length, a speed, or an electric current, consists of a number and a unit. Often we are given a quantity in one set of units, but we “N

want it expressed in another set of units. For example, suppose we measure that a table is 21.5 inches wide, and we want to express this in centimeters. We must use a conversion factor, which in this case is (by definition) exactly lin.

=

2.54cm

or, written another way, 1

=

2.54cm/in.

Since multiplying by one does not change anything, the width of our table, in cm, is ! 21.5inches

=

. cm (21.51n.) X = (12in)?(0.0254 m/in.)* =

0.0929m?. So 880 ft> = (880 ft?)(0.0929 m?/ft*) ~ 82 m’.

NOTE As a rule of thumb, an area given in ft? is roughly 10 times the number of square meters (more precisely, about 10.8 X). DGUAMEL S Speeds. Where the posted speed limit is 55 miles per hour (mi/h or, mph), what is this speed (a) in meters per second (m/s) and (b) in kilometers per hour (km/h)? APPROACH

We again use the conversion factor

1in. = 2.54cm,

and we recall

that there are 5280 ft in a mile and 12 inches in a foot; also, one hour contains (60min/h) X (60s/min) = 3600s/h. SOLUTIC N (a) We can write 1 mile as

1mi = (52801()(12E>< ft

=

100 enr

) ~ 1609 m.

We also know that 1 hour contains 3600 s, so

L

2y

ol

BRI

s

m

T (55 k>. Which of these could possibly be

““ This quantity, tp, is called the Planck time and is thought to be the earliest time, after the creation of the Universe, at

which the currently known laws of physics can be applied.

correct according to a dimensional check?

i General Problems 39; Global positioning satellites (GPS) can be used to determine your position with great accuracy. If one of the satellites is 20,000 km from you, and you want to know your position to £ 2 m, what percent uncertainty in the distance is required? How many significant figures are needed in the distance? . Computer chips (Fig. 1-13) are etched on circular silicon wafers of thickness 0.300 mm that are sliced from a solid cylindrical silicon crystal of length 25cm. If each wafer can hold 100 chips, what is the maximum number of chips that can be produced from one entire cylinder?

. Estimate the number of gumballs in the machine of Fig. 1-14.

FIGURE 1-14 Problem 48. Estimate the number of gumballs in the machine. FIGURE 1-13 Problem 40. The wafer held by the hand (above) is shown below, enlarged and illuminated by colored light. Visible are rows of integrated circuits (chips). 41. (a) How many seconds are there in 1.00 year? (b) How many nanoseconds are there in 1.00 year? (c¢) How many years are there in 1.00 second? 42. American football uses a field that is 100.0 yd long, whereas a regulation soccer field is 100.0 m long. Which field is longer, and by how much (give yards, meters, and percent)? 43. A typical adult human lung contains about 300 million tiny cavities called alveoli. Estimate the average diameter of a single alveolus. . One

hectare

is

defined

as

1.000 X 10*m?%

One

4.356 x 10*ft?>. How many acres are in one hectare?

acre

is

4s. Estimate the number of gallons of gasoline consumed by the total of all automobile drivers in the United States, per year. . Use Table 1-3 to estimate the total number of protons or neutrons in (a) a bacterium, (b) a DNA molecule, (c) the human body, (d) our Galaxy. 47. An average family of four uses roughly 1200L (about

300 gallons) of water per day (1L = 1000 cm®). How much

depth would a lake area of 50 km? and of 40,000 people? neglect evaporation

16

CHAPTER 1

lose per year if it uniformly covered an supplied a local town with a population Consider only population uses, and and so on.

Introduction, Measurement, Estimating

49. Estimate how many kilograms of laundry soap are used i. the U.S. in one year (and therefore pumped out of washing machines with the dirty water). Assume each load of laundry takes 0.1 kg of soap. 50. How big is a ton? That is, what is the volume of something that weighs a ton? To be specific, estimate the diameter of a 1-ton rock, but first make

a wild guess: will it be 1 ft across,

3 ft, or the size of a car? [Hint: Rock has mass per volume

about 3 times that of water, which is 1kg per liter (10° cm?)

or 621b per cubic foot.] 51 A certain audio compact disc (CD) contains 783.216 megabytes of digital information. Each byte consists of exactly 8 bits. When played, a CD player reads the CD’s digital information at a constant rate of 1.4 megabits per second. How many minutes does it take the player to read the entire CD? 52, Hold a pencil in front of your eye at a position where its blunt end just blocks out the Moon (Fig. 1-15). Make appropriate measurements to estimate the diameter of the Moon, given

that the Earth-Moon distance is 3.8 X 10° km. FIGURE 1-15 Problem 52. How big is the Moon? 53. A heavy rainstorm dumps 1.0 cm of rain on a city 5 km wide and 8km long in a 2-h period. How many metric tons

(1 metric ton = 10°kg)

water has a mass of of water was this?

of water fell on the city? (1 cm’

1g = 10kg)

How

many

gallo.

54. Noah'’s ark was ordered to be 300 cubits long, 50 cubits wide,

and 30 cubits high. The cubit was a unit of measure equal to

60. The diameter of the Moon is 3480 km. What is the volume of the Moon? How many Moons would be needed to create a volume equal to that of Earth?

meters, and estimate its volume (m®).

61. Determine the percent uncertainty in 6, and in sin 6, when (a) 6 =15.0° £ 0.5°, (b) 6 = 75.0° £ 0.5°

the length of a human forearm, elbow to the tip of the longest finger. Express the dimensions of Noah’s ark in 35. Estimate how many days it would take to walk around the world, assuming 10 h walking per day at 4 km/h. 56.

One liter (1000 cm®) of oil is spilled onto a smooth lake. If the oil spreads out uniformly until it makes an oil slick just one molecule thick, with adjacent molecules just touching, estimate the diameter of the oil slick. Assume the oil mole-

62. If you walked north along one of Earth’s lines of longitude until you had changed latitude by 1 minute of arc (there are 60 minutes per degree), how far would you have walked (in miles)? This distance is called a nautical mile. . Make a rough estimate of the volume of your body (in m®). . Estimate the number of bus drivers (a) in Washington, D.C., and (b) in your town.

cules have a diameter of 2 X 107 m.

57. Jean stands beside a river and wonders how wide it is. She spots a large rock on the bank directly across from her. She then walks upstream 120 strides and judges that the angle between her and the rock, which she can still see, is now at an angle of 30° downstream (Fig. 1-16). Jean measures her stride to be about 0.8m lon%. Estimate the width of the river.

. The following formula estimates an average person’s lung

capacity V (in liters, where V

=

41H

1L = 10°cm’®):

— 00184

— 2.69,

where H and A are the person’s height (in meters), and age (in years), respectively. In this formula, what are the units of the numbers 4.1, 0.018, and 2.69?

66. The density of an object is defined as its mass divided by its volume. Suppose the mass and volume of a rock are

measured to be 8¢ and 2.8325 cm®. To the correct number of significant figures, determine the rock’s density (kg/m?).

FIGURE 1-16 Problem 57.

fi&

A watch manufacturer claims that its watches gain or lose no more |than 8seconds in a year. How accurate is this watch, expressed as a percentage?

59. An

angstrom

(symbol A) is a unit of length, defined as

107 m, which is on the order of the diameter of an atom.

(a) How many nanometers are in 1.0 angstrom? (b) How many femtometers or fermis (the common unit of length in nuclear physics) are in 1.0angstrom? (¢) How many angstromj are in 1.0 m? (d) How many angstroms are in 1.0 light-year (see Problem 21)?

Answers

'

to

67. To the correct number of significant figures, use the information inside the front cover of this book to determine the ratio of (a) the surface area of Earth compared to the surface area of the Moon; (b) the volume of Earth compared to the volume of the Moon. 68.

One mole of atoms consists of 6.02 X 10? individual atoms. If amole of atoms were spread uniformly over the surface of the Earth, how many atoms would there be per square meter?

69. Recent findings in astrophysics suggest that the observable Universe can be modeled as a sphere of radius

R =137 x 10° light-years (= 13.0 X 10®m) with an average total mass density of about 1 X 1072 kg/m®. Only

about 4% of total mass is due to “ordinary” matter (such as protons, neutrons, and electrons). Estimate how much ordinary matter (kg) there is in the observable Universe.

Exercises

A

(d).

B:

No: they l;re 3 and 2, respectively. All three

have three significant figures, although the

number of idecimal places is (a) 2, (b) 3, (¢) 4.

D: (a) 2.58 X 1072, 3; (b) 4.23 X 10% 3 (probably);

(c) 3.4450 x 10%, 5.

E:

Mt. Everest, 29,035 ft; K2, 28,251 ft; Kangchenjunga, 28,169 ft.

F: No:15m/s

~ 34 mi/h.

! General Problems

17

A high-speed car has released a parachute to reduce its speed quickly. The directions of the car’s velocity and

acceleration are shown by the green (V) and gold (&) arrows.

Motion is described using the concepts of velocity and acceleration. In the case shown here, the acceleration @ is in the opposite direction from the velocity ¥, which means the object is slowing down. We examine in detail motion with constant acceleration, including the vertical motion of objects falling under gravity.

Describing Motion:

Kinematics in One Dimension CONTENTS 2-1

Reference Frames and

2-3

Instantaneous Velocity

2-4

Acceleration

2-5

Motion at Constant Acceleration

2-6

Solving Problems

Displacement 2-2 Average Velocity

2-7 Freely Falling Objects *#2-8 Variable Acceleration; Integral Calculus

#2-9 Graphical Analysis and Numerical Integration

CHAPTER-OPENING QUESTION—Guess now! [Don’t worry about getting the right answer now—you will get another chance later in the Chapter. See also p. 1 of Chapter 1 for more explanation.)

Two small heavy balls have the same diameter but one weighs twice as much as the other. The balls are dropped from a second-story balcony at the exact same time. The time to reach the ground below will be: (a) twice as long for the lighter ball as for the heavier one. (b) longer for the lighter ball, but not twice as long. (¢) twice as long for the heavier ball as for the lighter one. (d) longer for the heavier ball, but not twice as long.

(e) nearly the same for both balls.

he motion of objects—baseballs, automobiles, joggers, and even the Sun and Moon—is an obvious part of everyday life. It was not until the sixteenth and seventeenth centuries that our modern understanding of motion was established. Many individuals contributed to this understanding, particularly Galileo Galilei (1564-1642) and Isaac Newton (1642-1727) 4

20 30 40 50 Distance (m)

60

we can say that the runner’s average velocity is 6.50 m/s to the left.

SECTION 2-2

Average Velocity

21

DGV

EPE

Distance a cyclist travels. How far can a cyclist travel in

2.5h along a straight road if her average velocity is 18 km/h? APPROACH SOLUTION

We want to find the distance traveled, so we solve Eq. 2-2 for Ax.

“

We rewrite Eq.2-2 as Ax = v A¢, and find

Ax = 5Ar = (18km/h)(2.5h) = 45km. EXERCISEB A car travels at a constant 50km/h for 100km. It then speeds up to 100 km/h and is driven another 100 km. What is the car’s average speed for the 200 km trip? (a) 67 km/h; (b) 75 km/h; (c) 81 km/h; (d) S0km/h.

2-3

Instantaneous Velocity

If you drive a car along a straight road for 150 km in 2.0 h, the magnitude of your average velocity is 7Skm/h. It is unlikely, though, that you were moving at precisely 75 km/h at every instant. To describe this situation we need the concept of instantaneous velocity, which is the velocity at any instant of time. (Its magnitude is the number, with units, indicated by a speedometer, Fig. 2-8.) More precisely, the instantaneous velocity at any moment is defined as the average velocity over an infinitesimally short time interval. That is, Eq. 2-2 is to be evaluated in the limit of Af becoming extremely small, approaching zero. We can write the definition of instantaneous velocity, v, for one-dimensional motion as Ax

(2-3)

v = lim-= FIGURE 2-8 Car speedometer showing mi/h in white, and km/h in orange.

The notation lim,,_,, means the ratio Ax/At is to be evaluated in the limit of At -~

approaching zero. But we do not simply set Af = 0 in this definition, for then Ax would also be zero, and we would have an undefined number. Rather, we are considering the ratio Ax/At, as a whole. As we let Af approach zero, Ax approaches zero as well. But the ratio Ax/Af approaches some definite value, which is the instantaneous velocity at a given instant. In Eq. 2-3, the limit as At — 0 is written in calculus notation as dx/dt and is called the derivative of x with respect to f:

U= FIGURE 2-9 Velocity of a car as a function of time: (a) at constant velocity; (b) with varying velocity.

< 60+ E

= I

=E 40 +

2

S0ttt 0

01

02

03

Time (h)

04

05

60 i o

Average velocity

[ (=]

Velocity (km/h)

(a)

0

0

(b)

22

!

01

1

|

02 03 04 05 Time (h)

CHAPTER 2

A

Ax

dx

T oar

-4

This equation is the definition of instantaneous velocity for one-dimensional motion. For instantaneous velocity we use the symbol v, whereas for average velocity we use v, with a bar above. In the rest of this book, when we use the term “velocity” it will refer to instantaneous velocity. When we want to speak of the average velocity, we will make this clear by including the word “average.” Note that the instantaneous speed always equals the magnitude of the instantaneous velocity. Why? Because distance traveled and the magnitude of the displacement become the same when they become infinitesimally small. If an object moves at a uniform (that is, constant) velocity during a particular time interval, then its instantaneous velocity at any instant is the same as its average velocity (see Fig. 2-9a). But in many situations this is not the case. For example, a car may start from rest, speed up to 50 km/h, remain at that velocity for a time, then slow down to 20 km/h in a traffic jam, and finally stop at its destination after traveling a total of 15km in 30 min. This trip is plotted on the graph of.‘ Fig. 2-9b. Also shown on the graph is the average velocity (dashed line), which is v = Ax/At = 15km/0.50h = 30 km/h.

Describing Motion: Kinematics in One Dimension

To better understand instantaneous velocity, let us consider a graph of the

position of a particular particle versus time (x vs. ), as shown in Fig. 2-10. (Note that this is different from showing the “path” of a particle on an x vs. y plot.) The particle is at position x; at a time #;, and at position x, at time ¢,. P; and P, reprer sent these two points on the graph. A straight line drawn from point P, (x,, #,) to point P, (xz‘, 1,) forms the hypotenuse of a right triangle whose sides are Ax and Af. The ratio Ax/At is the slope of the straight line P;P,. But Ax/At is also the average veiocity of the particle during the time interval At = f, — t;. Therefore, we conclude that the average velocity of a particle during any time interval At = t, — t; is equal to the slope of the straight line (or chord) connecting the two

points (x;, #;) and (x,,%,) on an x vs. t graph.

Consider now a time f;, intermediate between #; and t,, at which time the particle is at x; (Fig. 2—-11). The slope of the straight line P, P, is less than the slope of P, P, in this case. Thus the average velocity during the time interval t; — ¢, is less than during the time interval ¢, — ¢,. Now let us imagine that we take the point P; in Fig. 2-11 to be closer and closer to p oint P;. That is, we let the interval ¢, — ¢;, which we now call Af, to become smaller and smaller. The slope of the line connecting the two points becomes cl pser and closer to the slope of a line tangent to the curve at point P;. The average velocity (equal to the slope of the chord) thus approaches the slope of the tangent at point P;. The definition of the instantaneous velocity (Eq. 2-3) is the limiting value of the average velocity as Af approaches zero. Thus the instantaneous wvelocity equals the slope of the tangent to the curve at that point (which we can simply call “the slope of the curve” at that point). Because the velocity at any instant equals the slope of the tangent to the x vs. graph at that instant, we can obtain the velocity at any instant from such a graph. For example, in Fig. 2—-12 (which shows the same curve as in Figs. 2-10 and 2-11), as our object moves from x; to x;,, the slope continually increases, so the velocity is increasing. For times after #,, however, the slope begins to decrease and in fact

0

h

FIGURE position straight average

[5)

2-10 Graph of a particle’s x vs. time ¢. The slope of the line P; P, represents the velocity of the particle during

the time interval At =1, — f;.

FIGURE 2-11 Same position vs. time curve as in Fig. 2-10, but note that the average velocity over the time

interval #; — #; (which is the slope of

P, P)) is less than the average velocity over the time interval £, — #;. The slope of the thin line tangent to the curve at point P; equals the instantaneous velocity at time #;.

reaches zero (so v = 0) where x has its maximum value, at point P; in Fig. 2-12.

Beyond this point, the slope is negative, as for point P,. The velocity is therefore negative, which makes sense since x is now decreasing—the particle is moving toward decreasing values of x, to the left on a standard xy plot. If an object moves with constant velocity over a particular time interval, its instantaneous velocity is equal to its average velocity. The graph of x vs. ¢ in this case will be a straight line whose slope equals the velocity. The curve of Fig. 2-10 has no straight sections, so there are no time intervals when the velocity is constant. X

P3

*

FIGURE 2-12 Same x vs.f curve as in Figs.2-10 and 2-11, but here showing the slope at four different points: At Py, the slope is zero, so v = 0. At P, the slope is negative,so v < 0.

X

0

‘ EXERCISE

direction?

141

%)

13

C What is your speed at the instant you turn around to move in the opposite (a) Depends

on

how

negative; (d) none of the above.

quickly

you

turn

around;

(b)

always

zero;

(c) always

The derivatives of various functions are studied in calculus courses, and this

book gives a summary in Appendix (which we dse a lot) are:

~

d

yr (Ct")n == nCt

_

n-1

and

B. The derivatives of polynomial functions dcC

—=

_—

=

=0,

where C is any constant.

SECTION 2-3

Instantaneous Velocity

23

Given x as a function of . A jet engine moves along an

—r

0

10

!

20\ 30

40

x1

?!’x(m)

50

X2

|60

(a) Tangent at P, whose slope is vy =21.0 m/s

x (m)

60 — Slope of l 50 —cirmioht 13 straight line 40|—2=16.8 m/s

’

30 20

lel

|

=

A

Ax= 133.6m

equation x = Af* + B, where

APPROACH

-/,

0

1

| I

2

FIGURE 2-13

3

(b)

4

We substitute values for #; and £, in the given equation for x to obtain

x; and x,. The average velocity can be found from Eq. 2-2. We take the derivative of the given x equation with respect to ¢ to find the instantaneous velocity, using the formulas just given. SOLUTION (@) At t;, = 3.00s, the position (point P; in Fig. 2—-13b) is

x; = A

J/Al =2.00s

5

!

6

t (S)

Example 2-3.

At t, = 5.00s,

+ B = (210m/s?*)(3.00s)> + 2.80m

= 21.7m.

the position (P, in Fig. 2-13b) is

x; = (2.10m/s?)(5.005)*> + 2.80m

= 553 m.

The displacement is thus X, — x;

(a) Engine traveling on a straight track.

(b) Graph of x vs.7: x = At* + B.

A =2.10m/s*> and B = 2.80m, and this equa-

tion is plotted in Fig. 2-13b. () Determine the displacement of the engine during “ the time interval from # = 3.00s to f, = 5.00s. (b) Determine the average velocity during this time interval. (¢) Determine the magnitude of the instantaneous velocity at ¢ = 5.00s.

e

10 0

b ¥

experimental track (which we call the x axis) as shown in Fig. 2-13a. We will treat the engine as if it were a particle. Its position as a function of time is given by the

=

553m

— 21.7m

=

33.6m.

(b) The magnitude of the average velocity can then be calculated as _

Ax At

33.6m 2.00s

X2 — X

L — 4

=

16.8m/s.

This equals the slope of the straight line joining points P; and P, shown in Fig. 2-13b. (c) The instantaneous velocity at t = f, = 5.00s equals the slope of the tangent to the curve at point P, shown in Fig. 2-13b. We could measure this slope off the .2

The acceleration in this case is constant; it does not depend on time. Figure 2-18

shows graphs of (a) x vs. ¢ (the same as Fig. 2-13b), (b) v vs. £, which is linearly increasing as calculated above, and (c) a vs. t, which is a horizontal straight line because

e

a = constant.

Like velocity, acceleration is a rate. The velocity of an object is the rate at

which its displacement changes with time; its acceleration, on the other hand, is the

rate at which its velocity changes with time. In a sense, acceleration is a “rate of a rate.” This can be expressed in equation form as follows: since a = dv/dt and v = dx/dt, then

o vdt _ dfdx) _ dxdr? dt \ dt

6-..

Here d’x/dt* is the second derivative of x with respect to time: we first take the derivative of x with respect to time (dx/dt), and then we again take the derivative with respect‘to time, (d/dt)(dx/dt), to get the acceleration.

&

a =420 m/s?

4T =

2.._

EXERCISE F The position of a particle is given by the following equation:

x = (200m/s*) + (2.50m/s)t.

What

is the acceleration of the particle at

(c) 24.0 m/sz‘; (d) 2.00 m/s2.

¢ = 2.00s?

(a) 13.0m/s%

(b) 22.5m/s%

0

()

f—t—p—f—t—+— 123456

SECTION 2-4

Acceleration

£ (5)

27

EXAMPLE

2-8 | Analyzing with graphs. Figure 2-19 shows

100}-----5=5--—===3

the velocity as a function of time for two cars accelerating from 0 to 100 km/h in a time of 10.0s. Compare (a) the average acceleration; (b) instantaneous acceleration; and (c) total distance traveled for the two cars.

v (km/h)

CONCEPTUAL

RESPONSE (a) Average acceleration is Av/At. Both cars have the same Av‘ (100 km/h) and the same At (10.05s), so the average acceleration is the same for both cars. (b) Instantaneous acceleration is the slope of the tangent to the v vs. ¢

8 FIGURE 2-19 Example 2-8.

10

£(s)

curve. For about the first 4s, the top curve is steeper than the bottom curve, so

car A has a greater acceleration during this interval. The bottom curve is steeper during the last 65, so car B has the larger acceleration for this period. (c¢) Except at t =0 and ¢ = 10.0s, carA is always going faster than car B. Since it is going faster, it will go farther in the same time.

2-5

Motion at Constant Acceleration

We now examine the situation when the magnitude of the acceleration is constant and the motion is in a straight line. In this case, the instantaneous and average accelerations are equal. We use the definitions of average velocity and

acceleration to derive a set of valuable equations that relate x, v, a, and t when a is

constant, allowing us to determine any one of these variables if we know the others. To simplify our notation, let us take the initial time in any discussion to be zero, and we call it #,: t; = t, = 0. (This is effectively starting a stopwatch at f,.) We can

then let #, = ¢ be the elapsed time. The initial position (x;) and the initial velocity (v;)

of an object will now be represented by x, and v, since they represent x and v at t = 0. At time ¢ the position and velocity will be called x and » (rather than x, and v,). The average velocity during the time interval ¢ — f, will be (Eq. 2-2) . Ax x-x x-Xx

At

since we chose

t—t,

t

£, = 0. The acceleration, assumed constant in time, is (Eq. 2-5) a

=

vV

“™

—

t

A common problem is to determine the velocity of an object after any elapsed time ¢, when we are given the object’s constant acceleration. We can solve such problems by solving for » in the last equation to obtain: v = 1y, + at. [constant acceleration] (2-7) If an object starts from rest (v, = 0) and accelerates at 4.0 m/s?, after an elapsed

time ¢ = 6.0s its velocity will be v = at = (4.0m/s?)(6.0s) = 24 m/s.

Next, let us see how to calculate the position x of an object after a time ¢ when it undergoes constant acceleration. The definition of average velocity (Eq. 2-2) is

v = (x — xo)/t, which we can rewrite as X

A

CAUTION

Average velocity, but only if a = constant

=

xp + vt

Because the velocity increases at a uniform rate, the average velocity, v, will be midway between the initial and final velocities:

v+ v

v

=

xp

(2-9)

+ vt

=x0+


v

=

Equations 2—12 are valid only when

the acceleration is constant, which we

600 " /82

assume in this Example

= 24.5m/s.

This runway length is not sufficient. (b) Now \t'e want to find the minimum length of runway, x — x,, given v =27.8m/s and a = 2.00m/s>. So we again use Eq. 2-12c, but rewritten as v’ — v} (278m/s)* — 0

.~ ~ xp) %) =

2a

=

2(2.00 m/s?)

= 193 m

A 200-m runway is more appropriate for this plane. ANOTE We did this Example as if the plane were a particle, so we round off our answer to 200

m.

EXERCISE G A car starts from rest and accelerates at a constant 10 m/s?> during a } mile (402 m) race, How

fast is the car going at the finish line? (a) 8090m/s;

(c) 81 m/s; (d) 809 m/s.

(b) 90m/s;

SECTION 2-5

29

2—06 Solving Problems Before doing more worked-out Examples, let us look at how to approach problem solving. First, it is important to note that physics is not a collection of equations tq be memorized. Simply searching for an equation that might work can lead you to fl wrong result and will surely not help you understand physics. A better approach is to use the following (rough) procedure, which we put in a special “Problem Solving Strategy.” (Other such Problem Solving Strategies, as an aid, will be found throughout the book.)

> Q) ~

LV,

@

1. Read and reread the whole problem carefully before trying to solve it.

Q

o)

2. Decide what object (or objects) you are going to

A

study, and for what time interval. choose the initial time to be t = 0.

v

You

can

often

3. Draw a diagram or picture of the situation, with coordinate axes wherever applicable. [ You can place the origin of coordinates and the axes wherever you like to make your calculations easier.] 4. Write down what quantities are “known” or “given,”

and then what you want to know. Consider quantities both at the beginning and at the end of the chosen time interval.

5. Think about which principles of physics apply in this problem. Use common sense and your own experiences. Then plan an approach. 6. Consider which equations (and/or definitions) relate the quantities involved. Before using them, be sure their range of validity includes your problem (for example, Egs. 2-12 are valid only when the acceleration is constant). If you find an applicable equation that involves only known quantities and one desired unknown, solve the equation algebraically for the

GV

R[N

unknown. Sometimes several sequential calculations, or a combination of equations, may be needed. It is often preferable to solve algebraically for the desired unknown before putting in numerical values. . Carry out the calculation if it is a numerical problem. Keep one or two extra digits during the calculations, but round off the final answer(s) to the correct number of significant figures (Section 1-3). . Think carefully about the result you obtain: Is it reasonable? Does it make sense according to your own intuition and experience? A good check is to do a rough estimate using only powers of ten, as discussed in Section 1-6. Often it is preferable to do a rough estimate at the start of a numerical problem because it can help you focus your attention on finding a path toward a solution. . A very important aspect of doing problems is keeping track of umits. An equals sign implies the units on each side must be the same, just as the numbers must. If the units do not balance, a mistake has no doubt been made. This can serve as a check on your solution (but it only tells you if you’re wrong, not if you’re right). Always use a consistent set of units.

o

Acceleration of a car. How long does it take a car to cross

a 30.0-m-wide intersection after the light turns green, if the car accelerates from

rest at a constant 2.00 m/s*?

FIGURE 2-20

APPROACH We follow the Problem Solving Strategy above, step by step. SOLUTION 1. Reread the problem. Be sure you understand what it asks for (here, a time interval). 2. The object under study is the car. We choose the time interval: ¢ = 0, the

Example 2-10.

a = 2.00 m/s2

ez

;i__._.:_fi_ XO =0 Vo =

30

0

CHAPTER 2

initial time, is the moment the car starts to accelerate from rest (v = 0); the time ¢ is the instant the car has traveled the full 30.0-m width of the

a=12.00 rfi“

X

=

30.0 m

3

intersection.

Draw a diagram: the situation is shown in Fig. 2-20, where the car is shown\ moving along the positive x axis. We choose x, = 0 at the front bumper of th. car before it starts to move.

Describing Motion: Kinematics in One Dimension

4. The “knowns” and the “wanted” are shown in the Table in the margin, and we choose x, = 0. Note that “starting from rest” means v = 0 at t = 0; that is,

v = 0.

5. The physics: the motion takes place at constant acceleration, so we can use the kinematic equations, Eqgs. 2-12. 6. Equations: we want to find the time, given the distance and acceleration; Eq. 2-12b is perfect since the only unknown quantity is ¢. Setting v, = 0 and xo = 0|in Eq.2-12b (x = xy + vpt + %atz), we can solve for £: x

Xg

=

Wanted

0

t

x=300m

a = 2.00m/s? v =0

%atz,

=

|

SO

=

Known

2L, a

2x

t =

=a

7. The cafbulation: |

2x

2(30.0m)

t = /=a = \Jo—% = e548s. DM

This is our answer. Note that the units come out correctly. 8. We can check the reasonableness of the answer by calculating the final velocity

v = at = (2.00m/s?)(5.485) = 10.96 m/s,

and then finding x = x, + vt =

0 + 5(10.96 m/s + 0)(5.48s) = 30.0m, which is our given distance. 9. We checked the units, and they came out perfectly (seconds). NOTE

In ‘steps

6

and

7, when

we

took

the

square

root,

we

should

have

written = *\/2x/a = £ 5.48s. Mathematically there are two solutions. But the second solution, ¢ = —5.48s, is a time before our chosen time interval and makes no sense physically. We say it is “unphysical” and ignore it. f‘

We explicitly followed the steps of the Problem Solving Strategy for Example 2-10. In upcomin# Examples, we will use our usual “Approach” and “Solution” to avoid being wordT

STV

EITN

ESTIMATE | Air bags. Suppose you want to design an air-

bag system that can protect the driver at a speed of 100 km/h (60 mph) if the car hits a bril?: wall. Estimate how fast the air bag must inflate (Fig. 2-21) to effectively protect the driver. How does the use of a seat belt help the driver?

APPROACJ;3 We assume the acceleration is roughly constant, so we can use Eqgs. 2-12. Both Egs. 2-12a and 2-12b contain ¢, our desired unknown. They both contain a, so we must first find a, which we can do using Eq. 2-12c¢ if we know the distance x over which the car crumples. A rough estimate might be about 1 meter. We choose th% time interval to start at the instant of impact with the car moving at vy = 100km/h, and to end when the car comes to rest (v = 0) after traveling 1 m. SOLUTION We convert the given initial speed to SI units: 100km/h = 100 x 10°m/3600s = 28 m/s. We then find the acceleration from Eq. 2—12c:

|

‘a

=

v3

o T

(28i m/s)? L

RN

om

(K)PHYSIcs

Car safety—air bags

FIGURE 2-21

APPLIED

Example 2-11.

An air bag deploying on impact.

SO 390 m/s".2

This enormous acceleration takes place in a time given by (Eq. 2-12a):

‘

t =

v—wv

a

To be effective, the air bag What does the air bag (to avoid puncture of the | person in a stable position

=

0 - 28m/s

~390 m/s’

=

0.07s.

;

would need to inflate faster than this. do? It spreads the force over a large area of the chest chest by the steering wheel). The seat belt keeps the against the expanding air bag.

SECTION 2-6

Solving Problems

31

Q~c

FIGURE 2-22 Example 2-12: stopping distance for a braking car.

@PHYSics

APPLIED

Braking distances

Known

Wanted

t =050s

X

v

=

14 m/s

v =14m/s a=0 Xp

=

Known

Wanted

xo=70m v = 14m/s

x

v=20

FIGURE 2-23 Example 2-12. Graphs of (a) v vs.f and (b) x vs. L.

2l

|

v6_

\

81

5 4]

3¢ 0

a

|

t

05

0 (b)

speed

is constant,

32

so

a = 0.

(2) The

second

SOLUTION is reacting Fig. 2-22 t = 0.50s multiplied

time

interval

is the

actual

Part (1). We take x, = 0 for the first time interval, when the driver (0.50s): the car travels at a constant speed of 14m/s so a = 0. See and the Table in the margin. To find x, the position of the car at (when the brakes are applied), we cannot use Eq. 2—-12c because x is by a, which is zero. But Eq. 2-12b works: =

yt+0

=

(14m/s)(0.50s)

=

7.0m.

Thus the car travels 7.0m during the driver’s reaction time, until the instant the brakes are applied. We will use this result as input to part (2). Part (2). During the second time interval, the brakes are applied and the car is brought to rest. The initial position is x, = 7.0m (result of part (1)), and other variables are shown in the second Table in the margin. Equation 2-12a doesn’t contain x; Eq. 2-12b contains x but also the unknown f. Equation 2-12c, the final position of the car (when it stops):

1

} +—— 15,20, 25

10

t(s)

X

=

v — v}

X

2

N

5.7 Ul

2(-6.0m/s?)

;

t=0.50s

|

10

CHAPTER 2

15 1(s)

20

2

P

Ol .. i

70m

+ 16m

'

=

05

minimum

braking period when the vehicle slows down (a # 0) and comes to a stop. The stopping distance depends on the reaction time of the driver, the initial speed of the car (the final speed is zero), and the acceleration of the car. For a dry road and good tires, good brakes can decelerate a car at a rate of about 5m/s?® to 8 m/s%. Calculate the total stopping distance for an initial velocity of 50km/h (= 14m/s ~ 31 mi/h) and assume the acceleration of the car is —6.0m/s? (the minus sign appears because the velocity is taken to be in the positive x direction and its magnitude is decreasing). Reaction time for normal drivers varies from perhaps 0.3 s to about 1.0s; take it to be 0.50s. APPROACH During the “reaction time,” part (1), the car moves at constant speed of 14 m/s,s0 a = 0. Once the brakes are applied, part (2), the acceleration is @ =—6.0m/s> and is constant over this time interval. For both parts a 1‘\

'

20 ~15 10 5

the

v* — v} = 2a(x — x,), is what we want; after setting x, = 7.0m, we solve for x,

:

|

distances. Estimate

stopping distance for a car, which is important for traffic safety and traffic design. The problem is best dealt with in two parts, two separate time intervals. (1) The first time interval begins when the driver decides to hit the brakes, and ends when the foot touches the brake pedal. This is the “reaction time” during which

x

a = —6.0m/s

~10 1

a = —6.0 m/s?

constant, so we can use Eqgs. 2-12.

0]

Part 2: Braking

14 4

v decreases from 14 m/s to zero

Braking

the

Part 1: Reaction time

= constant = 14 m/s =0.50s =0

-12m/s? =

23m.

The car traveled 7.0 m while the driver was reacting and another 16 m during the braking period before coming to a stop, for a total distance traveled of 23 m. Figure 2-23 shows graphs of (a) v vs. t and (b) x vs. 1. NOTE From the equation above for x, we see that the stopping distance after £ driver hit the brakes (= x — x,) increases with the square of the initial speed, not just linearly with speed. If you are traveling twice as fast, it takes four times the distance to stop.

SIS

EN

ESTIMATE | Two

Moving

Objects:

Police and

Speeder.

A car speeding at 150 km/h passes a still police car which immediately takes off

in hot pursuit. Using simple assumptions, such as that the speeder continues at

constant speed, estimate how long it takes the police car to overtake the speeder. f Then estimate the police car’s speed at that moment and decide if the assumptions were reasonable. APPROACH When the police car takes off, assumption is that its acceleration is constant. let’s see what happens. We can estimate the automobile ads, which claim cars can accelerate the average acceleration of the police car could

%=

_

100km/h

TT0r

] 2

it accelerates, and the simplest This may not be reasonable, but acceleration if we have noticed from rest to 100 km/h in 5.0s. So be approximately

km/h [ 1000 m

&

1h

o

5

( 1km )(3600s> = Simyer.

SOLUTION We need to set up the kinematic equations to determine the unknown

quantities, and equations. We position by xp. vehicles arTve

since there are two moving denote the speeding car’s Because we are interested at the same position on the

xs = vt | xp

=

where we lJ'lave set

vopl

objects, we need two separate sets of position by xs and the police car’s in solving for the time when the two road, we use Eq. 2-12b for each car:

+ sast>

=

(150km/h)t

+

=

%(5.61’1’1/52)[2,

%aptz

vpp = 0 and

as = 0

=

(42m/s)t

(speeder assumed to move

at constant

speed). We want the time when the cars meet, so we set xg = xp and solve for #:

(42m/s)t

= (2.8m/s%)i%

The solutions are 42 t=0andt=L/Sz=15s.

A

The first solution corresponds to the instant the speeder passed the police car. The second solution tells us when the police car catches up to the speeder, 15s later. This iT our answer, but is it reasonable? The police car’s speed at £ = 155 is

vp = vp + apt

|

= 0 + (5.6m/s?)(15s)

= 84m/s

or 300 km/h (= 190 mi/h). Not reasonable, and highly dangerous. NOTE More reasonable is to give up the assumption of constant acceleration. The

A CAUTION Initial assumptions need to be

police car shrely cannot maintain constant acceleration at those speeds. Also, the ~ checked out for reasonableness speeder, if *a reasonable person, would slow down upon hearing the police siren. Figure 2-24 shows (a) x vs. f and (b) v vs. f graphs, based on the original assumption of ap = constant, whereas (c) shows v vs. t for more reasonable assumptions.

]

v

v Speeder

Speeder

Police

t

(b) \

Example 2-13.

Police

t

0

FIGURE 2-24

0

!

(c) SECTION 2-6

Solving Problems

33

2-7

FIGURE 2-25 (1564-1642). A

One of the most common examples of uniformly accelerated motion is that of an object allowed to fall freely near the Earth’s surface. That a falling object is accel-=my erating may not be obvious at first. And beware of thinking, as was widely believea before the time of Galileo (Fig. 2-25), that heavier objects fall faster than lighter objects and that the speed of fall is proportional to how heavy the object is. Galileo made use of his new technique of imagining what would happen in idealized (simplified) cases. For free fall, he postulated that all objects would fall with the same constant acceleration in the absence of air or other resistance. He showed that this postulate predicts that for an object falling from rest, the distance traveled will be proportional to the square of the time (Fig. 2-26); that is, d o 1. We can see

Galileo Galilei

CAUTION

this from Eq. 2-12b; but Galileo was the first to derive this mathematical relation.

A freely falling object increases in speed, but not in proportion

to its mass or weight ~

FIGURE 2-26

Freely Falling Objects

Multiflash

photograph of a falling apple, at equal time intervals. The apple falls farther during each successive interval, which means it is accelerating.

To support his claim that falling objects increase in speed as they fall, Galileo

made

use of a clever argument: a heavy stone dropped

from a height of 2 m will

drive a stake into the ground much further than will the same stone dropped from a height of only 0.2 m. Clearly, the stone must be moving faster in the former case.

Galileo claimed that all objects, light or heavy, fall with the same acceleration, at

Jeast in the absence of air. If you hold a piece of paper horizontally in one hand and

a heavier object—say, a baseball—in the other, and release them at the same time as in Fig. 2-27a, the heavier object will reach the ground first. But if you repeat the experiment, this time crumpling the paper into a small wad (see Fig. 2-27b), you will find that the two objects reach the floor at nearly the same time. Galileo was sure that air acts as a resistance to very light objects that have a large surface area. But in many ordinary circumstances this air resistance is negligible. In a chamber from which the air has been removed, even light objects like a feather or a horizontally held piece of paper will fall with the same acceleration as any other object (see Fig. 2-28). Such a demonstration in vacuum was not possible in Galileo’s time, which makes Galileo’s achievement all the greater. Galileo is often called the “father of modern science,” not only for the content of his science (astronomical discoveries, inertia, free fall) but also for his approach to science (idealization and simplification, mathematization of theory, theories that have testable consequences, experiments to test theoretical predictions). Galileo’s specific contribution to our understanding of the motion of falling objects can be summarized as follows: at a given location on the Earth and in the absence of air resistance, all objects fall with the same constant acceleration.

We call this acceleration the acceleration due to gravity on the surface of the Earth, and we give it the symbol g. Its magnitude is approximately Acceleration due to gravity

g

=

[at surface of Earth]

9.80m/s%

In British units g is about 32 ft/s% Actually, g varies slightly according to latitude and elevation, but these variations are so small that we will ignore them for most

B

FIGURE 2-27 (a)A balland a light piece of paper are dropped at the same time. (b) Repeated, with

the paper wadded up.

FIGURE 2-28 A rock and a feather are dropped simultaneously (a) in air, (b) in a vacuum.

(a) 34

CHAPTER 2

(b) Describing Motion: Kinematics in One Dimension

Air-filled tube (a)

Evacuated tube

(b)

-

purposes. The effects of air resistance are often small, and we will neglect them for the most part. However, air resistance will be noticeable even on a reasonably heavy object if the velocity becomes large.” Acceleration due to gravity is a vector as is any acceleration, and its direction is downward, toward the center of the Earth. r When dealing with freely falling objects we can make use of Eqs. 2-12, where for a we use the value of g given above. Also, since the motion is vertical we will substitute y in place of x, and y; in place of x,. We take y, = 0 unless otherwise specified. It is arbitrary whether we choose y to be positive in the upward direction or in the d‘ownward direction; but we must be consistent about it throughout a problem’s solution.

“PROBLEM

SOLVING

You can choose y to be positive either up or down

EXERCISE : Return to the Chapter-Opening Question, page 18, and answer it again now. Try to expl ‘ in why you may have answered differently the first time.

Falling

from

a tower. Suppose

that a ball is dropped

(vp = 0) from a tower 70.0m high. How far will it have fallen after a time t, = 1.00 s\ t, = 2.00s, and f; = 3.00s? Ignore air resistance. APPROAC}I Let us take y as positive downward, so the acceleration is a=g=+980m/s>. Weset vy =0 and y, = 0. We want to find the position y of the ball after three different time intervals. Equation 2-12b, with x replaced by y, relajes the given quantities (¢, a, and vp) to the unknown y.

SOLUTION

We set ¢ = f; = 1.00s in Eq.2-12b:

n

o= wh t tar? = 0+ 1ar? = 1(9.80m/s?)(1.005)?

The ball Pas_ f_allen a distance of 4.90m during t}}e' tirpe interval t; = 1.00s, Similarly, after 2.00s (= 1,), the ball’s position is

= 4.90m. =0

Acceleration due to gravity &

'

to

y, = 3at} = 3(9.80m/s?)(2.005)> = 19.6m. Finally, after 3.00s (= ;), the ball’s position is (see Fig. 2-29)

y; = a3

FIGURE 2-29 Example 2-14. (a) An object dropped from a tower falls with progressively greater speed and covers greater distance with each successive second. (See also Fig. 2-26.) (b) Graph of y vs.?.

rrrT-——3=0

y;=4.90m

(After 1.00 s)

l +y

y,=19.6m (After 2.00 s)

= $(9.80m/s?)(3.00s)> = 44.1m.

®y;=44.1m

Thrown down from a tower. Suppose the ball in Example 2-14

(After 3.00 s)

is thrown downward with an initial velocity of 3.00 m/s, instead of being dropped. (@) What then would be its position after 1.00s and 2.00s? (b) What would its speed be ai‘fter 1.00s and 2.00s? Compare with the speeds of a dropped ball. APPROACI"I Again we use Eq. 2-12b, but now v, is not zero, it is vy = 3.00 m/s. SOLUTION (a) At ¢ = 1.00s, the position of the ball as given by Eq. 2-12b is

y =‘ vt + ar* = (3.00m/s)(1.00s) + 1(9.80m/s?)(1.00s)*> = 7.90m.

At t =2.00s,

"

+y

(a)

(time interval f = 0 to ¢ = 2.00s), the position is

y = vt + 3ar* = (3.00m/s)(2.00s) + 3(9.80 m/s?)(2.00s)> = 25.6m.

As expected, the ball falls farther each second than if it were dropped with v, = 0. (b) The velocity is obtained from Eq. 2-12a: =

vy t+ oat

= 3.00m/s + (9.80m/s?)(1.00s)

= 3.00m/s + (9.80m/s?)(2.00s)

In Example 2-14, when the ball was dropped these equations was zero, so v

=0+

at

= (9.80m/s?)(1.00s) = (9.80m/s?)(2.00s)

= 9.80m/s = 19.6m/s.

= 12.8m/s

[att; = 1.005s]

= 22.6m/s. [atf, = 2.005]

I

v

(v, = 0), the first term (vp) in [at #; = 1.005s] [at £, = 2.005]

NOTE For both Examples 2-14 and 2-15, the speed increases linearly in time by 9.80 m/s during each second. But the speed of the downwardly thrown ball at any instant is always 3.00m/s (its initial speed) higher than that of a dropped ball.

—

|

"The speed of an object falling in air (or other fluid) does not increase indefinitely. If the object falls far enough, it will reach a maximum velocity called the terminal velocity due to air resistance.

SECTION 2-7

Freely Falling Objects

35

m

Ball thrown upward, I. A person throws a ball upward into the

air with an initial velocity of 15.0m/s. Calculate (a) how high it goes, and (b) how long the ball is in the air before it comes back to the hand. Ignore air resistance.

APPROACH We are not concerned here with the throwing action, but only wit the motion of the ball after it leaves the thrower’s hand (Fig. 2-30) and until 1. comes back to the hand again. Let us choose y to be positive in the upward direction and negative in the downward direction. (This is a different convention from that used in Examples 2-14 and 2-15, and so illustrates our options.) The acceleration due to gravity is downward and so will have a negative sign, a = —g = —9.80m/s As the ball rises, its speed decreases until it reaches the highest point (B in Fig. 2-30), where its speed is zero for an instant; then it descends, with increasing speed. SOLUTION (a) We consider the time interval from when the ball leaves the thrower’s hand until the ball reaches the highest point. To determine the maximum height, we calculate the position of the ball when its velocity equals zero (v = 0 at the highest point). At f =0 (point A in Fig. 2-30) we have ¥ =0, v=150m/s, and a = —9.80m/s>. At time ¢ (maximum height), v =0, a=—980m/s’, and we wish to find y. We use Eq. 2-12c, replacing x

with y: v* = v} + 2ay. We solve this equation for y:

y

FIGURE 2-30 An object thrown into the air leaves the thrower’s hand at A, reaches its maximum height at B, and returns to the original position at C. Examples 2-16,2-17,2-18, and 2-19.

=

v? — v}

2a

=

0 — (15.0m/s)?

2(~9.80m/s?)

= 11.5m. m

The ball reaches a height of 11.5 m above the hand. (b) Now we need to choose a different time interval to calculate how long the ball is in the air before it returns to the hand. We could do this calculation in two parts by first determining the time required for the ball to reach its highest point, and then determining the time it takes to fall back down. However, it is simpler to consider the time interval for the entire motion from A to B to C (Fig. 2-30) in one step and use Eq. 2-12b. We can do this because y represents position or displacement, and not the total distance traveled. Thus, at both points A and C

y = 0. We use Eq. 2-12b with a = —9.80m/s? and find Yy

=

Y

+

vt

+

%at?'

0 = 0+ (15.0m/s)t + 3(—9.80 m/s?)r% This equation is readily factored (we factor out one #):

(15.0m/s — 4.90m/s*t)t

= 0.

There are two solutions:

t =0

and

¢ =

=

3.06s.

The first solution (# = 0) corresponds to the initial point (A) in Fig. the ball was first thrown from y = 0. The second solution, ¢ = 3.06 s, to point C, when the ball has returned to y = 0. Thus the ball is in the NOTE We have ignored air resistance, which could be significant, so only an approximation to a real, practical situation.

A CAUTION

Quadpratic equations have two solutions. Sometimes only one corresponds to reality, sometimes both

36

CHAPTER 2

2-30, when corresponds air 3.06s. our result is

We did not consider the throwing action in this Example. Why? Because during the throw, the thrower’s hand is touching the ball and accelerating the ball at a rate unknown to us—the acceleration is not g. We consider only the time when the ball is in the air and the acceleration is equal to g. Every quadratic equation (where the variable is squared) mathematically produces two solutions. In physics, sometimes only one solution corresponds to the real situation, as in Example 2-10, in which case we ignore the “unphysical’ solution. But in Example 2-16, both solutions to our equation in ¢* are physicall, meaningful: = 0 and ¢ = 3.06s.

Describing Motion: Kinematics in One Dimension

CONCEPTUAL EXAMPLE

A

2-17 | Two possible misconceptions. Give examples

to show the error in these two common misconceptions: (1) that acceleration and velocity are always in the same direction, and (2) that an object thrown upward has zero acceleration at the highest point (B in Fig. 2-30). RESPONSE Both are wrong. (1) Velocity and acceleration are not necessarily in the same direction. When the ball in Example 2-16 is moving upward, its velocity

A\ CAUTION (1) Velocity and acceleration are

acceleration also zero at this point? No. The velocity near the top of the arc points upward, then becomes zero (for zero time) at the highest point, and then points downward. Gravity does not stop acting, so a = —g = —9.80m/s> even there. Thinking that @ = 0 at point B would lead to the conclusion that upon reaching goint B, the ball would stay there: if the acceleration (= rate of change of velocity) were zero, the velocity would stay zero at the highest point, and the ball would stay up there without falling. In sum, the acceleration of gravity always points down toward the Earth, even when the object is moving up.

1(7;)";1; (())?ggen at the highest point of a trajectory

is positive (upward), whereas the acceleration is negative (downward). (2) At the highest point (B in Fig. 2-30), the ball has zero velocity for an instant. Is the

1ot always in the same direction; ’he.accsle’ ation (of gravity) always

Ball thrown upward, Il. Let us consider again the ball thrown

upward of Example 2-16, and make more calculations. Calculate (¢) how much time it takes for the ball to reach the maximum height (point B in Fig. 2-30), and (b) the veiocity of the ball when it returns to the thrower’s hand (point C). APPROACH Again we assume the acceleration is constant, so we can use Egs. 2—14 We have the height of 11.5 m from Example 2-16. Again we take y as positive upward.

r

SOLUTION (¢) We consider the time vy = 15.0m/s) and the top of the path find ¢t. The acceleration is constant at @ and 2-12b contain the time ¢ with other

interval between the throw (¢ =0, (y = +11.5m, » = 0), and we want to = —g = —9.80m/s>. Both Egs. 2-12a quantities known. Let us use Eq. 2-12a

with @ = —9.80m/s? v, = 15.0m/s, and v = O: 'D='U0+at;

setting v }= 0 and solving for ¢ gives

This is just half the time it takes the ball to go up and fall back to its original position [3.06s, calculated in part (b) of Example 2-16]. Thus it takes the same time to reach the maximum height as to fall back to the starting point. (b) Now we consider the time interval from the throw (t =0, v, = 15.0m/s) until the ball’s return to the hand, which occurs at ¢ = 3.06s (as calculated in Example 2-16), and we want to find v when ¢ = 3.06s: |

v =1

+a

= 150m/s — (9.80m/s?)(3.065)

= —15.0m/s.

NOTE The ball has the same speed (magnitude of velocity) when it returns to the starting point as it did initially, but in the opposite direction (this is the meaning of the negative sign). And, as we saw in part (@), the time is the same up as down. Thus the motion is symmetrical about the maximum height.

The acceleration of objects such as rockets and fast airplanes is often given as a multiple of g = 9.80m/s%. For example, a plane pulling out of a dive and under-

(™ going 3.00¢’s would have an acceleration of (3.00)(9.80 m/s?) = 29.4 m/s>

| EXERCISE I If a car is said to accelerate at 0.50 g, what is its acceleration in m/s2?

SECTION 2-7

Freely Falling Objects

37

FM

Ball thrown upward, llI; the quadratic formula. For the

ball in Example 2-18, calculate at what time ¢ the ball passes a point 8.00m above the person’s hand. (See repeated Fig. 2-30 here).

APPROACH We choose the time interval from the throw ( = 0, v, = 15.0m/s

until the time ¢ (to be determined) when the ball is at position y = 8.00 m, using

Eq. 2-12b. SOLUTION

We

want to find ¢, given

a = —9.80m/s’>. We use Eq.2-12b: y

=

800m

Y

+ vt

y = 8.00m, y, =0,

v, = 15.0m/s,

and

+ jat?

= 0 + (15.0m/s)t + 3(—9.80m/s?)s%

To solve any quadratic equation of the form ar®> + bt + ¢ = 0, where a, b, and ¢ are constants (a is not acceleration here), we use the quadratic formula: t

—b £

=

\/b* — dac

2a We rewrite our y equation just above in standard form, ar®> + bt + ¢ = 0

(4.90m/s*)r> — (15.0m/s)t + (8.00m)

= 0.

So the coefficient a is 4.90 m/s%, b is —15.0 m/s, and c is 8.00 m. Putting these into the quadratic formula, we obtain

15.0m/s + \/(15.0m/s)> — 4(4.90 m/s2)(8.00 m)

FIGURE 2-30

(Repeated for Example 2-19)

2(4.90 m/s)

which gives us ¢ = the ball passes y comes down (¢ = NOTE Figure 2-31

’

0.69s and ¢ = 2.37s. Are both solutions valid? Yes, because = 8.00m when it goes up (f = 0.69s) and again when it 2.37s). shows graphs of (@) y vs. f and (b) v vs. t for the ball thrown

upward in Fig. 2-30, incorporating the results of Examples 2-16, 2-18, and 2—19’

FIGURE 2-31

Graphs of (a) y vs.#, (b) v vs. f for a ball thrown upward,

Examples 2-16,2-18, and 2-19.

L

L\

10

o z

t=1.53s\ |

8

4 2

2.37s\

0695

/

(a)

’U?

E

\

1 15:

1(s)

2 25

Ball thrown

{

10 s

.

1=153s

-5

AN

+

-15

\

;

o 0 05

15

£

t=

t=

20

y=115m

3 35‘ upward

-20 0 05 (b)

1 15

t(s)

u

2 25

\

3 35

at edge of cliff. Suppose that the

person of Examples 2-16, 2-18, and 2-19 is standing on the edge of a cliff, so that the ball can fall to the base of the cliff 50.0 m below as in Fig. 2-32. (a) How long does it take the ball to reach the base of the cliff? (b) What is the total distance traveled by the ball? Ignore air resistance (likely to be significant, so our result is an approximation). APPROACH

We again use Eq. 2-12b, but this time we set

y = —50.0m,

bottom of the cliff, which is 50.0 m below the initial position (y, = 0).

38

CHAPTER 2

Describing Motion: Kinematics in One Dimension

the

SOLUTION and

(a) We use Eq. 2-12b with a = —9.80 m/s?,

Vo

y = —50.0m:

Yy

rk

=

Yot

vt

=

15.0 m/s,

Yo

=

0,

+ %al‘2

~-50.0m = 0 + (15.0m/s)t — 3(9.80m/s?)¢? Rewriting in the standard form we have

(4.90m/s?)t* — (15.0m/s)t — (50.0m)

0.

Using the quadratic formula, we find as solutions ¢ = 5.07s and ¢ = —2.01s. The first solution, ¢ = 5.07s, is the answer we are seeking: the time it takes the ball to rise to its highest point and then fall to the base of the cliff. To rise aqd fall back to the top of the cliff took 3.06s (Exarnple 2-16); so it took an additional 2.01 s to fall to the base. But what is the meaning of the other solution, ¢ = —2.01s? This is a time before the throw, when our calculation | begins, so it isn’t relevant here.” | (b) From Example 2-16, the ball moves up 11.5m, falls 11.5 m back down to the top of the cliff, and then down another 50.0m to the base of the cliff, for a total distance traveled of 73.0m. Note that the displacement, however, was —50.0 m. Figure 2-33 shows the y vs. f graph for this situation. EXERCISE J Two balls are thrown from a down, each with the same initial speed, ball hits the ground at the greater speed: downward, or (c) both the same? Ignore

cliff. One is thrown directly up, the other directly and both hit the ground below the cliff. Which (a) the ball thrown upward, (b) the ball thrown air resistance.

*2—8 Variabl eleration; Integral Calcul us Variable Acceleration; In this brief optional Section we use integral calculus to derive the kinematic equations for constant acceleration, Egs. 2—-12a and b. We also show how calculus can be used when the acceleration is not constant. If you have not yet studied simple ntegration in your calculus course, you may want to postpone reading this Section until you have. We discuss integration in more detail in Section 7-3, where we begin to use it in the physics. First we derive Eq. 2-12a, assuming as we did in Section 2-5 that an object has velocity vy at ¢ = 0 and a constant acceleration a. We start with the definition of instantaneous acceleration, |

L dv

=

a = dv/dt,

which we rewrite as

FIGURE 2-32 Example 2-20. The person in Fig. 2-30 stands on the edge of a cliff. The ball falls to the base of the cliff, 50.0 m below. FIGURE 2-33

the y vs.? graph.

Hand

10 0

-0 E 20

= 30 -40 /500

Base

of cliff

adt.

Example 2-20,

=\ \ 5.07s 1

2

3

4

5

6

t(s)

We take the definite integral of both sides of this equation, using the same notation we did in Section 2-5: v

J dv

t

=

f adt

=1,

1=0

which gives, since a = constant, v — vy

=

at.

This is Eq. 2-12a, v = v, + at. Next we derive Eq. 2-12b starting with the velocity, Eq. 2-4, v = dx/dt. We rewrite this as dx = vdt

definition

of instantaneous

or

dx

(/UO

+

at)dt

where we substituted in Eq. 2-12a. "The solution |t =-201s could be meaningful in a different physical situation. Suppose that a person standing on top of a 50.0-m-high cliff sees a rock pass by him at ¢ = 0 moving upward at 15.0m/s; at what time did the rock leave the base of the cliff, and when did it arrive back at the base r f the cliff? The equations will be precisely the same as for our original Example, and the answers = —201s and ¢ = 5.07s will be the correct answers. Note that we cannot put all the information for a problem into the mathematics, so we have to use common sense in interpreting results. |

*SECTION 2-8

Variable Acceleration; Integral Calculus

39

Now we integrate: X

t

de

= j(vo + at)dt

x=x,

X

—

t

X9

=

=0

t

t

j’vodt‘l’

jatdt

t=0

x — Xy

=

«

=0

vt + Lat?

since v, and a are constants. This result is just Eq. 2-12b, x = xo + vyt + 3at’. Finally let us use calculus to find velocity and displacement, given an acceleration that is not constant but varies in time.

Integrating a time-varying acceleration. An experimental

vehicle starts from rest

(v, =0)

at r = 0 and accelerates at a rate given by

a = (7.00m/s*)t. What is (a) its velocity and (b) its displacement 2.00's later?

APPROACH

We cannot use Eqgs. 2-12 because a is not constant. We integrate the

acceleration

a = dv/dt

over

time

to find

v as a function

of time; and

integrate v = dx/dt to get the displacement. SOLUTION From the definition of acceleration, a = dv/dt, we have dv = adt. We take the integral of both sides from trary time #: v

then

» = 0 at ¢t = 0 to velocity v at an arbi-

4

Jdv

=

Jadt

Il


. Find the position ¥ where the object comes to rest (momentarily).

Kinematics in Two or Three Dimensions; Vectors

27. (IT) A particle’s position as a function of time ¢ is given

by ¥ = (50t + 6.0t2)mi + (7.0 — 3.0£*) mj. At t =505,

find the magnitude and direction of the particle’s displacement vector AF relative to the point ¥ = (0.0 + 7.0j)m.

r'\-7 and 3-8 Projectile Motion [neglect air resistance]

28. (I) A tiger leaps horizontally from a 7.5-m-high rock with a speed of 3.2 m/s. How far from the base of the rock will she land? 29. (I) A diver running 2.3 m/s dives out horizontally from the edge of a vertical cliff and 3.0s later reaches the water below. How high was the cliff and how far from its base did the diver hit the water? 30. (II) Estimate how much farther a person can jump on the Moon as compared to the Earth if the takeoff speed and angle are the same. The acceleration due to gravity on the Moon is Bne-sixth what it is on Earth.

31

(II) A fire hose held near the ground shoots water at a speed of‘fi .Sm/s. At what angle(s) should the nozzle point in order that the water land 2.5m away (Fig. 3-40)? Why are‘ there two different angles? Sketch the two trajectories.

\

FIGURE 3-40 Pro%lem 31.

39. (II) In Example 3-11 we chose the x axis to the right and y axis up. Redo this problem by defining the x axis to the left and y axis down, and show that the conclusion remains

the same—the football lands on the ground 40.5m to the right of where it departed the punter’s foot. 40. (IT) A grasshopper hops down a level road. On each hop, the grasshopper launches itself at angle 6, = 45° and achieves a range R = 1.0m. What is the average horizontal speed of the grasshopper as it progresses down the road? Assume that the time spent on the ground between hops is negligible. 41. (IT) Extreme-sports enthusiasts have been known to jump off the top of El Capitan, a sheer granite cliff of height 910m in Yosemite National Park. Assume a jumper runs horizontally off the top of El Capitan with speed 5.0m/s and enjoys a freefall until she is 150m above the valley floor, at which time she opens her parachute (Fig. 3-41). (a) How long is the jumper in freefall? Ignore air resistance. (b) It is important to be as far away from the cliff as possible before opening the parachute. How far from the cliff is this jumper when she opens her chute?

i e—25m——+

32. (II) A ball is thrown horizontally from the roof of a building 9.0m tall and lands 9.5 m from the base. What was the ball’s initial speed? 33. (II) A football is kicked at ground level with a speed of

18.0m/s !at an angle of 38.0° to the horizontal. How much

later does it hit the ground? . (II) A ball thrown horizontally at 23.7 m/s from the roof of a building lands 31.0m from the base of the building. How high is the building? 3s. (IT) A shot-putter throws the shot (mass = 7.3kg) with an initial speed of 14.4m/s at a 34.0° angle to the horizontal. Calculate the horizontal distance traveled by the shot if it leaves the athlete’s hand at a height of 2.10m above the ground. 36. (II) Show that the time required for a projectile to reach its highest ppint is equal to the time for it to return to its original height if air resistance is neglible.

FIGURE 3-41 Problem 41.

42. (IT) Here is something to try at a sporting event. Show that the maximum height /4 attained by an object projected into the air, such as a baseball, football, or soccer ball, is approx-

imately given by

h~128m where ¢ is the total time of flight for the object in seconds. Assume that the object returns to the same level as that from which it was launched, as in Fig. 3-42. For example, if you

count to find that a baseball was in the air for ¢ = 5.0s,

maximum

height attained was

the

& = 1.2 X (5.0)> = 30m.

The beauty of this relation is that 4 can be determined without knowledge of the launch speed vy or launch angle 6.

37. (IT) You buy a plastic dart gun, and being a clever physics student you decide to do a quick calculation to find its maximum horizontal range. You shoot the gun straight up, and it takes 4.0s for the dart to land back at the barrel. What is the maximum horizontal range of your gun? 38. (I) A baseball is hit with a speed of 27.0m/s at an angle of 45.0°. Tt lands on the flat roof of a 13.0-m-tall nearby building. If the ball was hit when it was 1.0m above the ground,

what

horizontal

lands on the building?

distance

does

it travel

before

it FIGURE 3-42

Problem 42.

Problems

77

43. (II) The pilot of an airplane traveling 170 km/h wants to drop supplies to flood victims isolated on a patch of land 150m below. The supplies should be dropped how many seconds before the plane is directly overhead? 44. (II) (a) A long jumper leaves the ground at 45° above the horizontal and lands 8.0m away. What is her “takeoff” speed vg? (b) Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0m away horizontally and 2.5 m, vertically below. If she long jumps from the edge of the left bank at 45° with the speed calculated in (a), how long, or short, of the opposite bank will she land (Fig. 3-43)?

47. (II) Suppose the kick in Example 3-7 is attempted 36.0 m from the goalposts, whose crossbar is 3.00m above the ground. If the football is directed perfectly between the goalposts, will it pass over the bar and be a field goal? Show why or why not. If not, from what horizontal distance must« this kick be made if it is to score? 48. (II) Exactly 3.0s after a projectile is fired into the air from the ground, it is observed to have a velocity v = (8.6i +4.8§)m/s, where the x axis is horizontal and the y axis is positive upward. Determine (a) the horizontal range of the projectile, (b) its maximum height above the ground, and (c) its speed and angle of motion just before it strikes the ground. 49. (II) Revisit Example 3-9, and assume that the boy with the slingshot is below the boy in the tree (Fig. 3-45) and so aims upward, directly at the boy in the tree. Show that again the boy in the tree makes the wrong move by letting go at the moment the water balloon is shot.

- 10.0m FIGURE 3-43

Problem 44.

45. (II) A high diver leaves the end of a 5.0-m-high diving board and strikes the water 1.3 s later, 3.0 m beyond the end of the board. Considering the diver as a particle, determine

(a) her initial velocity, ¥p; (b) the maximum height reached; and (c) the velocity ¥; with which she enters the water. 46. (II) A projectile is shot from the edge of a cliff 115 m above ground level with an initial speed of 65.0m/s at an angle of 35.0° with the horizontal, as shown in Fig. 3-44. (a) Deter-

Problem 49.

50. (II) A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp (Fig. 3-46). (a) With what minimum speed must he drive off the horifl zontal ramp? The vertical height of the ramp is 1.5m abovc the cars and the horizontal distance he must clear is 22 m. (b) If the ramp is now tilted upward, so that “takeoff angle” is 7.0° above the horizontal, what is the new minimum speed? f

22m Must clear this point! FIGURE 3-46

51.

.

mine the time taken by the projectile to hit point P at ground level. (b) Determine the distance X of point P from the base of the vertical cliff. At the instant just before the projectile hits point P, find (c) the horizontal and the vertical components of its velocity, (d) the magnitude of the velocity, and (e) the angle made by the velocity vector with the horizontal. (f) Find the maximum height above the cliff top reached by the projectile.

FIGURE 3-45

Problem 50.

(II) A ball is thrown horizontally from the top of a cliff with initial speed v (at t = 0). At any moment, its direction of motion makes an angle 6 to the horizontal (Fig. 3-47). Derive a formula for 6 as a function of time, ¢, as the ball

follows a projectile’s path.

FIGURE 3-44

78

CHAPTER 3

Problem 46.

Kinematics in Two or Three Dimensions; Vectors

FIGURE 3-47

Problem 51.

52. (II) At what projection angle will the range of a projectile equal its maximum height? 53 (IT) A projectile is fired with an initial speed of 46.6 m/s at an angle of 42.2° above the horizontal on a long flat firing range. Determine (a) the maximum height reached by the projectile, (b) the total time in the air, (c) the total horizontal distance covered (that is, the range), and (d) the velocity of the projectile 1.50s after firing, 54. (II) An athlete executing a long jump leaves the ground at a 27.0° an‘gle and lands 7.80 m away. () What was the takeoff speed? (b) If this speed were increased by just 5.0%, how much longer would the jump be? 55. (III) A person stands at the base of a hill that is a straight incline making an angle ¢ with the horizontal (Fig. 3-48). For a given initial speed vy, at what angle 6 (to the horizontal) should objects be thrown so that the distance d they land up the hill is as large as possible?

61. (II) A child, who is 45 m from the bank of a river, is being carried helplessly downstream by the river’s swift current of 1.0m/s. As the child passes a lifeguard on the river’s bank, the lifeguard starts swimming in a straight line until she reaches the child at a point downstream (Fig. 3-50). If the lifeguard can swim at a speed of 2.0 m/s relative to the water, how long does it take her to reach the child? How far

downstream does the lifeguard intercept the child?

FIGURE 3-48 Problem 55. Given ¢ and vy, determine 6 to make d maximum,. 56. (III) Derive a formula for the horizontal range R, of a projectile when it lands at a height 4 above its initial point. (For h < 0, it lands a distance —h below the starting point.) Assume it is projected at an angle 6y with initial speed vy.

'3-9

~

Relative Velocity

57. (I) A person going for a morning jog on the deck of a cruise ship is running toward the bow (front) of the ship at 2.0m/s while the ship is moving ahead at 8.5 m/s. What is the velocity of the jogger relative to the water? Later, the jogger is moving toward the stern (rear) of the ship. What is the jogger’s velocity relative to the water now? 58. (I) Huck Finn walks at a speed of 0.70m/s across his raft (that is, he walks perpendicular to the raft’s motion relative to the shore). The raft is traveling down the Mississippi River at a speed of 1.50m/s relative to the river bank (Fig. 3-49). What is Huck’s velocity (speed and direction) relative to the river bank?

FIGURE 3-50

Problem 61.

62. (IT) A passenger on a boat moving at 1.70 m/s on a still lake walks up a flight of stairs at a speed of 0.60 m/s, Fig. 3-51. The stairs are angled at 45° pointing in the direction of motion as shown. Write the vector velocity of the passenger relative to the water.

FIGURE 3-51

Problem 62.

63. (II) A person in the passenger basket of a hot-air balloon throws a ball horizontally outward from the basket with speed 10.0m/s (Fig. 3-52). What initial velocity (magnitude and direction) does the ball have relative to a person standing on the ground (a) if the hot-air balloon is

rising at 5.0m/s rela-

FIGURE 3-49

Problem 58,

R

M’Ww«

59. (IT) Determine the speed of the boat shore in Example 3-14. 60. (II) Two planes approach each other speed of 780 km/h, and they spot each initially 12.0 km apart. How much time take evasive action?

FERTORNES

S

with respect to the head-on. Each has a other when they are do the pilots have to

tive to the ground during this throw, (b) if the hot-air balloon is descending at 5.0m/s relative to the ground. FIGURE 3-52

Problem 63.

Problems

79

. (II) An airplane is heading due south at a speed of 580 km/h. If a wind begins blowing from the southwest at a speed of 90.0km/h (average), calculate (a) the velocity (magnitude and

(b)

direction)

how

of the

far from

plane,

relative

its intended

to

position

the

ground,

and

it will be after

11.0min if the pilot takes no corrective action. [Hint: First

draw a diagram.]

65. (II) In what direction should the pilot aim the plane in Problem 64 so that it will fly due south? . (II) Two cars approach a street corner at right angles to each other (see Fig. 3-35). Car 1 travels at 35km/h and car 2 at 45km/h. What is the relative velocity of car 1 as seen by car 2? What is the velocity of car 2 relative to car 1?7 67. (II) A swimmer is capable of swimming 0.60m/s in still water. (a) If she aims her body directly across a 55-m-wide

69. (II) A motorboat whose speed in still water is 3.40 m/s must aim upstream at an angle of 19.5° (with respect to a line perpendicular to the shore) in order to travel directly across the stream. (a) What is the speed of the current? (b) What is the resultant speed of the boat with respect to the shore? (See Fig. 3-31.) 70. (IT) A boat, whose speed in still water is 2.70 m/s, must cros. a 280-m-wide river > and arrive at a point 120m upstream from where it starts (Fig. 3-53). To do so, the pilot must head the boat at a 45.0° upstream angle. What is the speed of the river’s current?

FIGURE 3-53

river whose current is 0.50 m/s, how far downstream (from a

point opposite her starting point) will she land? (b) How long will it take her to reach the other side?

68. (II) (a) At what upstream angle must the swimmer in Problem 67 aim, if she is to arrive at a point directly across the stream? (b) How long will it take her?

Problem 70.

;

- Start

_

71. (III) An airplane, whose air speed is 580 km/h, is supposed to fly in a straight path 38.0° N of E. But a steady 72 km/h wind is blowing from the north. In what direction should the plane head?

| General Problems 72.

Two vectors, V; and V,, add to a resultant V = V; + V.

Describe Vy and V, if (a) V = V; + Vo, (b) V2 = Vi + V3, N

+V,=V, - V.

73 A plumber steps out of his truck, walks 66 m east and 35m

south, and then takes an elevator 12 m into the subbasement

of a building where a bad leak is occurring. What is the displacement of the plumber relative to his truck? Give your answer in components; also give the magnitude and angles, with respect to the x axis, in the vertical and horizontal plane. Assume x is east, y is north, and z is up. 74. On mountainous downhill roads, escape routes are sometimes placed to the side of the road for trucks whose brakes might fail. Assuming a constant upward slope of 26°, calculate the horizontal and vertical components of the acceleration of a

TI. Romeo is chucking pebbles gently up to Juliet’s window, and he wants the pebbles to hit the window with only a horizontal component of velocity. He is standing at the edge of a rose garden 8.0m below her window and 9.0m from the base of the wall (Fig. 3-55). How fast are the pebbles going when they hit her window?

truck that slowed from 110 km/h to rest in 7.0s. See Fig. 3-54.

FIGURE 3-55

Problem 77.

FIGURE 3-54

75 A light plane is headed due south with a speed relative to of 185km/h.

After

1.00h,

the

pilot

notices

CHAPTER 3

{

FIGURE 3-56 Problem 78.

that

they have covered only 135km and their direction is not south but southeast (45.0°). What is the wind velocity? 76. An Olympic long jumper is capable of jumping 8.0m. Assuming his horizontal speed is 9.1 m/s as he leaves the ground, how long is he in the air and how high does he go? Assume that he lands standing upright—that is, the same way he left the ground.

80

9.0m

78. Raindrops make an angle § with the vertical when viewed through a moving train window (Fig. 3-56). If the speed of the train is vy, what is the speed of the raindrops in the reference frame of the Earth in which they are assumed to fall vertically?

Problem 74. still air

;

79. Apollo astronauts took a “nine iron” to the Moon and hit a golf ball about 180m. Assuming that the swing, launch angle, and so on, were the same as on Earth where the same

astronaut could hit it only 32 m, estimate the acceleratio due to gravity on the surface of the Moon. (We neglect axn resistance in both cases, but on the Moon there is none.)

Kinematics in Two or Three Dimensions; Vectors

80. A hunter aims directly at a target (on the same level) 68.0m

rl.

away. (a) If the bullet leaves the gun at a speed of 175m/s, by how much will it miss the target? (b) At what angle should the gun be aimed so the target will be hit? The cliff hivers of Acapulco push off horizontally from rock platformsl about 35m above the water, but they must

clear rocky outcrops at water level

that extend out into the water 5.0m from the base of the cliff directly under their launch point. See Fig. 3-57. What minimum pushoff speed is necessary to clear the

in the air? ?

rocks? How long are they

|

|

I

/

,4" ik

pod

|' | I I !

86. A basketball leaves a player’s hands at a height of 2.10m above the floor. The basket is 3.05m above the floor. The player likes to shoot the ball at a 38.0° angle. If the shot is made from a horizontal distance of 11.00m and must be accurate to + 0.22m (horizontally), what is the range of initial speeds allowed to make the basket?

87. A particle has a velocity of v = (=2.0i + 3.5¢j)m/s. The

particle starts at ¥ = (1.5i — 3.1j)m at ¢ = 0. Give the position and acceleration as a function of time. What is the shape of the resulting path? 88. A projectile is launched from ground level to the top of a cliff which is 195 m away and 135 m high (see Fig. 3-60). If the projectile lands on top of the cliff 6.6 s after it is fired, find the initial velocity of the projectile (magnitude and direction). Neglect air resistance.

FIGURE 3-57

Landing point

Problem 81.

82. When

IiEbe Ruth hit a homer

over the 8.0-m-high right-

field fen e 98m from home plate, roughly what was the speed of the ball when it left the bat? Assume the minimu ball was hit 1.0m above the ground and its path initially

made a 36° angle with the ground. 83. The speed of a boat in still water is v. The boat is to make a round trip in a river whose current travels at speed u. Derive a formula for the time needed to make a round trip of total distance D if the boat makes the round trip by moving (a) upstream and back downstream, and (b) directly across the river and back. We must assume u < v; why? 84. At serve, a tennis player aims to hit the ball horizontally. F What minimum speed is required for the ball to clear the 0.90-m-h igh net about 15.0m from the server if the ball is “launched” from a height of 2.50m? Where will the ball land if 1 just clears the net (and will it be “good” in the sense that it lands within 7.0 m of the net)? How long will it be in the air? See Fig. 3-58.

FIGURE 3-60

Problem 88.

89. In hot pursuit, Agent Logan of the FBI must get directly across a 1200-m-wide river in minimum time. The river’s current is 0.80 m/s, he can row a boat at 1.60m/s, and he can run 3.00 m/s. Describe the path he should take (rowing plus running along the shore) for the minimum crossing time, and determine the minimum time. 90. A boat can travel 2.20m/s in still water. (a) If the boat points its prow directly across a stream whose current is 1.30m/s, what is the velocity (magnitude and direction) of

the boat relative to the shore? (b) What will be the position

2.50 mQ)

LN

150 m-— FIGURE 3-58

e

70m—u

Problem 84.

85. Spymaster Chris, flying a constant 208 km/h horizontally in a low-fl ng helicopter, wants to drop secret documents into her con ct’s open car which is traveling 156km/h on a level hig way 78.0m below. At what angle (with the horizontal) ould the car be in her sights when the packet is released (Fig. 3-59)?

of the boat, relative to its point of origin, after 3.00 s? 91. A boat is traveling where there is a current of 0.20 m/s east (Fig. 3-61). To avoid some offshore rocks, the boat must clear a buoy that is NNE (22.5°) and 3.0km away. The boat’s speed through still water is 2.1 m/s. If the boat wants to pass the right side of the buoy by 0.15 km, at what angle should the boat head?

FIGURE 3-61

Problem 91.

fIGURE 3-59 Problem 85.

92. A child runs down a 12° hill and then suddenly jumps upward at a 15° angle above horizontal and lands 1.4m down the hill as measured along the hill. What was the child’s initial speed?

General Problems

81

93. A basketball is shot from an initial height of 2.4m (Fig. 3-62) with an initial speed vy = 12m/s directed at an angle 6, = 35° above the horizontal. (¢) How far from the basket was the player if he made a basket? (b) At what angle to the horizontal did the ball enter the basket?

98. Att = 0 a batter hits a baseball with an initial speed of 28 m/s at a 55° angle to the horizontal. An outfielder is 85 m from

the batter at £ = 0 and, as seen from home plate, the line

of sight to the outfielder with the plane in which What speed and direction ball at the same height from with respect to the outfielder’s line of sight to home plate.

makes a horizontal angle of 22° the ball moves (see Fig. 3-64), must the fielder take to catch the which it was struck? Give the angle

FIGURE 3-64

FIGURE 3-62 Problem 93.

Problem 98.

. You are driving south on a highway at 25m/s (approximately 55 mi/h) in a snowstorm. When you last stopped, you noticed that the snow was coming down vertically, but it is passing the windows of the moving car at an angle of 37° to the horizontal. Estimate the speed of the snowflakes relative to the car and relative to the ground. 95. A rock is kicked horizontally at 15m/s from a hill with a 45° slope (Fig. 3-63). How long does it take for the rock to hit the ground?

FIGURE 3-63 Problem 95.

96. A batter hits a fly ball which leaves the bat 0.90m above the ground at an angle of 61° with an initial speed of 28 m/s heading toward centerfield. Ignore air resistance. (¢) How far from home plate would the ball land if not caught? (b) The ball is caught by the centerfielder who, starting at a distance of 105 m from home plate, runs straight toward home plate at a constant speed and makes the catch at ground level. Find his speed. 97. A ball is shot from the top of a building with an initial velocity of 18 m/s at an angle 6 = 42° above the horizontal. (a) What are the horizontal and vertical components of the initial velocity? (b) If a nearby building is the same height and 55 m away, how far below the top of the building will the ball strike the nearby building?

*Numerical/Computer *99.

(II) Students shoot a plastic ball horizontally from a projectile launcher. They measure the distance x the ball travels horizontally, the distance y the ball falls vertically, and the total time ¢ the ball is in the air for six different heights of the projectile launcher. Here is their data. Time,

Horizontal distance,

Vertical distance,

0.217 0.376 0.398 0.431 0.478 0.491

0.642 1.115 1.140 1.300 1.420 1.480

0.260 0.685 0.800 0.915 1.150 1.200

t(s)

x (m)

y (m)

(a) Determine the best-fit straight line that represents x as a function of ¢. What is the initial speed of the ball obtained from the best-fit straight line? (b) Determine the best-fit quadratic equation that represents y as a function of t. What is the acceleration of the ball in the vertical direction? *100. (IIT) A shot-putter throws from a height 4 = 2.1 m above the ground as shown in Fig. 3-65, with an initial speed of vo = 13.5m/s. (a) Derive a relation that describes how the distance traveled d depends on the release angle 6. (b) Using the given values for vy and A, use a graphing calculator or computer to plot d vs. 6y. According to your plot, what value for 6, maximizes d?

FIGURE 3-65

Problem 100.

Answers to Exercises A: When the two vectors D; and D, point in the same direction.

B: 3\/2 = 4.24. C: (a).

82

CHAPTER 3

D: (d). E:

Both balls reach the same height, so are in the air for the

same length of time. F: (¢).

Kinematics in Two or Three Dimensions; Vectors

The

space

shuttle

Discovery

is

carried out into space by powerful rockets.

They

are

accelerating,

increasing in speed rapidly. To do so, a force must be exerted on them according to Newton’s second law, SF = ma. What exerts this force? The rocket engines exert a force on the gases they push out (expel) from the rear of the rockets (labeled Fgg). According to Newton’s third law, these ejected gases exert an equal and opposite force on the rockets in the forward direction. It is this “reaction” force exerted on the rockets by the gases, labeled Frg, that accelerates the rockets forward.

-~

Q

Dynamics:

N ewton s Laws of Motion CHAPTER-OPENING

QUESTIONS—Guess now!

A 150-kg football player collides head-on with a 75-kg running back. During the collision, the heavier player exerts a force of magnitude F, on the smaller player. If the smaller player exerts a force Fy back on the heavier player, which response is most accurate? (a) Fg = F,. (b) Fg Fjy.

Second

CONTENTS

(e) No force—they are being poetic.

Motion

4-6 Weight—the Force of Gravity;

and the Normal Force 4-7 Solving Problems with Newton’s Laws: Free-Body Diagrams

4-8 Problem Solving—A

General Approach

e have discussed how motion is described in terms of velocity and

acceleration. Now we deal with the question of why objects move as they do: What makes an object at rest begin to move? What causes an object to accelerate or decelerate? What is involved when an object moves in a curved path? We can answer in each case that a force is required. In thi Chapter’, we will investigate the connection between force and motion, which is the subject called dynamics.

4—-1

FIGURE 4-1

Aforce exerted on a

grocery cart—in this case exerted by RpETSOL

Force

Intuitively, we experience force as any kind of a push or a pull on an object. When you push a stalled car or a grocery cart (Fig. 4-1), you are exerting a force on it. When a motor lifts an elevator, or a hammer hits a nail, or the wind blows the leaves of a tree, a force is being exerted. We often call these contact forces because the force is exerted when one object comes in contact with another object. On the other hand, we say that an object falls because of the force of gravity. If an object is at rest, to start it moving requires force—that is, a force is needed to accelerate an object from zero velocity to a nonzero velocity. For an object already moving, if you want to change its velocity—either in direction or in magnitude—a force is required. In other words, to accelerate an object, a force is always required. In Section 4-4 we discuss the precise relation between acceleration and net force, which is Newton’s second law. One way to measure the magnitude (or strength) of a force is to use a spring

scale (Fig. 4-2). Normally, such a spring scale is used to find the weight of an

gbject; by weight we mean the force of gravity acting on the object (Section 4-6).

The spring scale, once calibrated, can be used to measure other kinds of forces as well, such as the pulling force shown in Fig. 4-2. A force exerted in a different direction has a different effect. Force has direction as well as magnitude, and is indeed a vector that follows the rules of vector addition discussed in Chapter 3. We can represent any force on a diagran‘\ by an arrow, just as we did with velocity. The direction of the arrow is the direction of the push or pull, and its length is drawn proportional to the magnitude of the force.

/ 012345678910\ FIGURE 4-2

A spring scale used to

s e

measure a force.

4-7

R

T

P e o~

Newton’s First Law of Motion

What is the relationship between force and motion? Aristotle (384-322 B.C.) believed that a force was required to keep an object moving along a horizontal plane. To Aristotle, the natural state of an object was at rest, and a force was believed necessary to keep an object in motion. Furthermore, Aristotle argued, the greater the force on the object, the greater its speed. Some 2000 years later, Galileo disagreed: he maintained that it is just as natural for an object to be in motion with a constant velocity as it is for it to be at rest. To understand Galileo’s idea, consider the following observations involving motion along a horizontal plane. To push an object with a rough surface along a "We treat everyday objects in motion here; the treatment of the submicroscopic world of atomr and molecules, and when velocities are extremely high, close to the speed of light (3.0 X 108m/s), are treated using quantum theory (Chapter 37 ff), and the theory of relativity (Chapter 36).

84

CHAPTER 4

Dynamics: Newton's Laws of Motion

\

but these two forces are in opposite directions, so the net force on the object (the

vector sum of the two forces) is zero. This is consistent with Galileo’s viewpoint, for the object moves with constant speed when no net force is exerted on it. Upon this foundation laid by Galileo, Isaac Newton (Fig. 4-4) built his great theory of motion. Newton’s analysis of motion is summarized in his famous “three laws of motion.” In his great work, the Principia (published in 1687), Newton readily acknowledged his debt to Galileo. In fact, Newton’s first law of motion is close to Galileo’s conclusions. It states that Every object continues in its state of rest, or of uniform velocity in a straight line, as long as no net force acts on it. The tendency of an object to maintain its state of rest or of uniform velocity in a straight line is called inertia. As a result, Newton’s first law is often called the law of inertia. .

CONCEPTUAL

EXAMPLE

4-1|

Newton's first law. A school bus comes to a

PHYSICS

=

R

tabletop at constant speed requires a certain amount of force. To push an equally heavy object with a very smooth surface across the table at the same speed will require less force. If a layer of oil or other lubricant is placed between the surface of the object and the table, then almost no force is required to keep the object moving. f Notice that in each successive step, less force is required. As the next step, we imagine that the object does not rub against the table at all —or there is a perfect lubricant between the object and the table —and theorize that once started, the object would move across the table at constant speed with no force applied. A steel ball bearing rolling on a hard horizontal surface approaches this situation. So does a puck on an air table, in which a thin layer of air reduces friction almost to zero. It was Galileo’s genius to imagine such an idealized world—in this case, one where there is no friction —and to see that it could lead to a more accurate and richer understanding of the real world. This idealization led him to his remarkable conclusion that if no force is applied to a moving object, it will continue to move with constant speed in a straight line. An object slows down only if a force is exerted on it. Galileo thus interpreted friction as a force akin to ordinary pushes and pulls. To push an object across a table at constant speed requires a force from your hand that can balance out the force of friction (Fig. 4-3). When the object moves at constant speed, your pushing force is equal in magnitude to the friction force,

FIGURE 4-3 F represents the force applied by the person and Fy

represents the force of friction.

NEWTON’S FIRST LAW OF MOTION FIGURE 4-4 Isaac Newton (1642-1727).

sudden stop, and all of the backpacks on the floor start to slide forward. What force causes them to do that? RESPONSE It isn’t “force” that does it. By Newton’s first law, the backpacks

continue their state of motion, maintaining their velocity. The backpacks slow down if a force is applied, such as friction with the floor.

Inertial Reference Frames Newton’s first law does not hold in every reference frame. For example, if your reference frame is fixed in an accelerating car, an object such as a cup resting on the dashboard(may begin to move toward you (it stayed at rest as long as the car’s velocity remained constant). The cup accelerated toward you, but neither you nor anything else exerted a force on it in that direction. Similarly, in the reference frame of the decelerating bus in Example 4-1, there was no force pushing the backpacks forward. In accelerating reference frames, Newton’s first law does not hold. Reference frames in which Newton’s first law does hold are called inertial reference frames (the law of inertia is valid in them). For most purposes, we usually make the approximation that a reference frame fixed on the Earth is an inertial frame. This is not precisely true, due to the Earth’s rotation, but usually it is close enough. Any reference frame that moves with constant velocity (say, a car or an airplane) relative to an inertial frame is also an inertial reference frame. Reference frames where the law of inertia does not hold, such as the accelerating reference

frames discussed above, are called noninertial reference frames. How can we be sure a reference frame is inertial or not? By checking to see if Newton’s first law holds. Thus Newton’s first law serves as the definition of inertial reference frames. SECTION 4-2

Newton'’s First Law of Motion

85

4-3 Newton’s

Mass second law, which

we come

to in the next Section, makes

use of the

concept of mass. Newton used the term mass as a synonym for quantity of matter sy This intuitive notion of the mass of an object is not very precise because the concept “quantity of matter” is not very well defined. More precisely, we can say that mass is a measure of the inertia of an object. The more mass an object has, the greater the force needed to give it a particular acceleration. It is harder to start it moving from rest, or to stop it when it is moving, or to change its velocity sideways out of a straight-line path. A truck has much more inertia than a baseball moving at the same speed, and a much greater force is needed to change the truck’s velocity at the same rate as the ball’s. The truck therefore has much more mass.

To quantify the concept of mass, we must define a standard. In SI units, the

A CAUTION

Distinguish mass from weight

unit of mass is the kilogram (kg) as we discussed in Chapter 1, Section 1-4. The terms mass and weight are often confused with one another, but it is important to distinguish between them. Mass is a property of an object itself (a measure of an object’s inertia, or its “quantity of matter”). Weight, on the other hand, is a force, the pull of gravity acting on an object. To see the difference, suppose we take an object to the Moon. The object will weigh only about one-sixth as much as it did on Earth, since the force of gravity is weaker. But its mass will be the same. It will have the same amount of matter as on Earth, and will have just as much inertia—for in the absence of friction, it will be just as hard to start it moving on the Moon as on Earth, or to stop it once it is moving. (More on weight in Section 4-6.)

4-4

FIGURE 4-5

The bobsled

accelerates because the team exerts a force.

Newton's Second Law of Motion

Newton’s first law states that if no net force is acting on an object at rest, the object remains at rest; or if the object is moving, it continues moving with constant speed in a straight line. But what happens if a net force is exerted on an object? Newton perceived that the object’s velocity will change (Fig. 4-5). A net force exerted on an object may make its velocity increase. Or, if the net force is in a direction opposite to the motion, the force will reduce the object’s velocity. If the net force acts sideways on a moving object, the direction of the object’s velocity changes (and the magnitude may as well). Since a change in velocity is an acceleration (Section 2-4), we can say that a net force causes acceleration. What precisely is the relationship between acceleration and force? Everyday experience can suggest an answer. Consider the force required to push a cart when friction is small enough to ignore. (If there is friction, consider the net force, which is the force you exert minus the force of friction.) If you push the cart with a gentle but constant force for a certain period of time, you will make the cart accelerate from rest up to some speed, say 3 km/h. If you push with twice the force, the cart will reach 3 km/h in half the time. The acceleration will be twice as great. If you triple the force, the acceleration is tripled, and so on. Thus, the acceleration of an object is directly proportional to the net applied force. But the acceleration depends on the mass of the object as well. If you push an empty grocery cart with the same force as you push one that is filled with groceries, you will find that the full cart accelerates more slowly. The greater the mass, the less the acceleration for the same net force. The mathematical relation, as Newton argued, is that the acceleration of an object is inversely proportional to its mass. These relationships are found to hold in general and can be summarized as follows:

NEWTON’S SECOND LAW OF MOTION

The acceleration of an object is directly proportional to the net force acting on it, and is inversely proportional to the object’s mass. The direction of the acceleration is in the direction of the net force acting on the object. This is Newton’s second law of motion.

86

CHAPTER4

Dynamics: Newton's Laws of Motion

-

Newton’s second law can be written as an equation:

I

a

=

—»

m flhere a stands for acceleration, m for the mass, and EF for the net force The symbol = (Greek “sigma”) stands for “sum of”; F stands for force, the vector sum of all forces acting on the object, which we define as the We rearrange this equation to obtain the familiar statement second law:

3F = ma

on so net of

the object. =F means force. Newton’s

(4-1a)

Newton’s second law relates the description of motion (acceleration) to the cause of motion ( orce). It is one of the most fundamental relationships in physics. From Newton’s second law we can make a more precise definition of force as an action capable of accelerating an object. Every force F is a vector, with magnitude and direction. Equation 4-1a is a vector equation valid in any inertial reference frame. It can be written in component form in rectangular coordinates as - 3F, = may, 2Fy (4-1b)

where

\

A F = Fi+

X R Fj + Fk

The component of acceleration in each direction is affected only by the component of the net force in that direction. In SI units, with the mass in kilograms, the unit of force is called the newton (N).

NEWTON’S SECOND OF MOTION

LAW

TABLE 4-1

Units for Mass and Force System

Mass

Force

One newton, then, is the force required to impart an acceleration of 1 m/s” to a mass of 1kg. Thus 1N = 1kg-m/s%

SI

kilogram

newton (N) :

In cgs units, the unit of mass is the gram (g) as mentioned earlier.” The unit of force is the dyne, which is defined as the net force needed to impart an acceleration of 1 cm/s?

cgs

gram (g)

dyne

to a mass of 1 g Thus 1dyne = 1g-cm/s?. Itis easy to show that 1dyne = 10° N,

British

In the British system, the unit of force is the pound (abbreviated Ib), where ‘b = 4.44T222 N ~ 4.45N. The unit of mass is the slug, which is defined as that mass which will undergo an acceleration of 1 ft/s* when a force of 11b is applied to it. Thus 11b = 1slug-ft/s>. Table 4-1 summarizes the units in the different systems. It is very important that only one set of units be used in a given calculation or problem, with the SI being preferred. If the force is given in, say, newtons, and the mass in grams, then before attempting to solve for the acceleration in SI units, we must change the mass to kilograms. For example, if the force is given as 2.0 N along the x axis and the mass is 500 g, we change the latter to 0.50 kg, and the acceleration will then automatically come out in m/s* when Newton’s second law is used:

Conversion factors:

SF,

20N

m

050kg

_ 2.0kg'm/s’ 0.50kg

(kg)

slug

Z| PROBLEM

(=kg'm/s) (=g cm/s)

pound (Ib) 1dyne

= 1075 N;

11b ~ 445N.

SOLVING

Use a consistent set of units

= 4.0m/s%

DU JNEPR ESTIMATE | Force to accelerate a fast car. Estimate the net force needed to accelerate (a) a 1000-kg car at 3 g; (b) a 200-g apple at the same rate. APPROACH We use Newton’s second law to find the net force needed for each object. This is an estimate (the 3 is not said to be precise) so we round off to one significant figure. SOLUTth‘I ‘ (a) The car’s acceleration is a = 5 = 3(9.8m/s?) ~ Sm/s%. We use | Newton’s second law to get the net force needed to achieve this acceleration:

SF

= ma

~

(1000kg)(5m/s*)

= 5000 N.

(If you are used to British units, to get an idea of what a 5000-N force is, you can divide by 4.45 N/Ib and get a force of about 1000 1b.) (b) For the apple, m = 200g = 0.2kg, so

L

- IF

= ma

~

(02kg)(5m/s?)

=

1IN.

"Be careful not to confuse g for gram with g for the acceleration due to gravity. The latter is always italicized (or boldface when a vector).

SECTION 4-4

Newton's Second Law of Motion

87

m Force to stop a car. What average net force is required to bring a 1500-kg car to rest from a speed of 100 km/h within a distance of 55 m? APPROACH

We use Newton’s second law,

2F = ma,

to determine the force,

but first we need to calculate the acceleration a. We assume the acceleration is ™ constant, so we can use the kinematic equations, Egs. 2-12, to calculate it.

vy =100 km/

FIGURE 4-6

v=0

-

Example 4-3.

SOLUTION We assume the motion is along the +x axis (Fig. 4-6). We are given the initial velocity v, = 100km/h = 27.8 m/s (Section 1-5), the final velocity v = 0, and the distance traveled x — xy = 55m. From Eq. 2-12c, we have

v

= v} + 2a(x — xp),

SO

= T2

v — v}

%) =

0 — (27.8m/s)?

2(55m)

= —7.0m/s m/s

The net force required is then SF

=

ma

=

(1500kg)(-7.0m/s?*)

=

—1.1 X 10*N.

The force must be exerted in the direction opposite to the initial velocity, which is what the negative sign means. NOTE If the acceleration is not precisely constant, then we are determining an “average” acceleration and we obtain an “average” net force. Newton’s second law, like the first law, is valid only in inertial reference frames (Section 4-2). In the noninertial reference frame of an accelerating car, for

example, a cup on the dashboard starts sliding—it accelerates—even though the net force on it is zero; thus SF = mi doesn’t work in such an accelerating refer-

ence frame

(EF = 0, but d # 0 in this noninertial frame).

EXERCISE A Suppose you watch a cup slide on the (smooth) dashboard of an accelerating car as we just discussed, but this time from an inertial reference frame outside the car, on the

street. From your inertial frame, Newton’s laws are valid. What force pushes the cup off the dashboard?

Precise Definition of Mass As mentioned in Section 4-3, we can quantify the concept of mass using its definition as a measure of inertia. How to do this is evident from Eq. 4-1a, where we see that the acceleration of an object is inversely proportional to its mass. If the same net force TF acts to accelerate each of two masses, m, and m,, then the ratio of their masses can be defined as the inverse ratio of their accelerations: ny

a;

my

a,

If one of the masses is known (it could be the standard kilogram) and the two accelerations are precisely measured, then the unknown mass is obtained from this definition. For example, if m; = 1.00kg, and for a particular force a; = 3.00m/s’

and a, = 2.00m/s%, then m, = 1.50kg.

88

CHAPTER4

Dynamics: Newton's Laws of Motion

.

4-5

Newton’s Third Law of Motion

Newton’s second law of motion describes quantitatively how forces affect motion. But where, we may ask, do forces come from? Observations suggest that a force exerted on rany object is always exerted by another object. A horse pulls a wagon, a person pushes a grocery cart, a hammer pushes on a nail, a magnet attracts a paper clip. In each of these examples, a force is exerted on one object, and that force is exerted by another object. For example, the force exerted on the nail is exerted by the hammer. But N¢wt0n realized that things are not so one-sided. True, the hammer exerts

a force on the nail (Fig. 4-7). But the nail evidently exerts a force back on the hammer as well, for the hammer’s speed is rapidly reduced to zero upon contact. Only a str%ng force could cause such a rapid deceleration of the hammer. Thus,

said Newt n, the two objects must be treated on an equal basis. The hammer exerts a force on the nail, and the nail exerts a force back on the hammer. This is the essencé of Newton’s third law of motion:

FIGURE 4-7 A hammer striking a nail. The hammer exerts a force on the nail and the nail exerts a force back on the hammer. The latter force decelerates

the hammer and brings it to rest.

Whenever one object exerts a force on a second object, the second exerts an equal force in the opposite direction on the first. This law is sometimes paraphrased as “to every action there is an equal and opposite reaction.” This is perfectly valid. But to avoid confusion, it is very important to remember that the “action” force and the “reaction” force are acting on different objects. As evidence for the validity of Newton’s third law, look at your hand when you push a&ainst the edge of a desk, Fig. 4-8. Your hand’s shape is distorted, clear evidence that a force is being exerted on it. You can see the edge of the desk pressing into your hand. You can even feel the desk exerting a force on your hand; it hurts! The harder you push against the desk, the harder the desk pushes back on your hand. (You only feel forces exerted on you; when you exert a force on another object, what you feel is that object pushing back on you.)

Force exerted | on hand by desk

iy

NEWTON’S THIRD LAW OF MOTION

A

CAUTION

Action and reaction forces act on different objects

FIGURE 4-8 If your hand pushes against the edge of a desk (the force vector is shown in red), the desk pushes back against your hand (this force vector is shown in a different color, violet,

Force exerted

to remind us that this force acts on a different object).

on desk by hand

The force the desk exerts on your hand has the same magnitude as the force your hand exerts on the desk. This is true not only if the desk is at rest but is true even if the desk is accelerating due to the force your hand exerts. As another demonstration of Newton’s third law, consider the ice skater in Fig. 4-9. There is very little friction between her skates and the ice, so she will

FIGURE 4-9 An example of Newton’s third law: when an ice skater pushes against the wall, the wall pushes back and this force causes her to accelerate away.

move freely if a force is exerted on her. She pushes against the wall; and then she

starts moving backward. The force she moving, for that force acts on the wall. start her moving, and that force could force with which the wall pushes on her site to the force she exerts on the wall.

r

When 5 person starts moving in the package exerts an propels the person

exerts on the wall cannot make her start Something had to exert a force on her to only have been exerted by the wall. The is, by Newton’s third law, equal and oppo-

o/

Force

on skater

Force on Wall

throws a package out of a small boat (initially at rest), the boat opposite direction. The person exerts a force on the package. The equal and opposite force back on the person, and this force (and the boat) backward slightly. SECTION 4-5

Newton's Third Law of Motion

89

Rocket propulsion also is explained using Newton’s third law (Fig. 4-10). A common misconception is that rockets accelerate because the gases rushing out the back of the engine push against the ground or the atmosphere. Not true. What happens, instead, is that a rocket exerts a strong force on the gases, expelling

them; and the gases exert an equal and opposite force on the rocket. It is thi=™

L]

FIGURE 4-10 Another example of Newton’s third law: the launch of a rocket. The rocket engine pushes the gases downward, and the gases exert an equal and opposite force upward on the rocket, accelerating it upward.

latter force that propels the rocket forward—the force exerted on the rocket by the gases (see Chapter-Opening photo, page 83). Thus, a space vehicle is maneuvered in empty space by firing its rockets in the direction opposite to that in which it needs to accelerate. When the rocket pushes on the gases in one direction, the gases push back on the rocket in the opposite direction. Jet aircraft too accelerate because the gases they thrust out backwards exert a forward force on the engines (Newton’s third law). Consider how we walk. A person begins walking by pushing with the foot backward against the ground. The ground then exerts an equal and opposite force forward on the person (Fig. 4-11), and it is this force, on the person, that moves the person forward. (If you doubt this, try walking normally where there is no friction, such as on very smooth slippery ice.) In a similar way, a bird flies forward by exerting a backward force on the air, but it is the air pushing forward (Newton’s third law) on the bird’s wings that propels the bird forward.

(A rocket does not accelerate as a result of its propelling gases pushing against the ground.)

makes a car go forward?

FIGURE 4-11 We can walk forward because, when one foot pushes backward against the ground, the ground pushes forward on that foot (Newton’s third law). The two forces shown act on different objects.

RESPONSE A common But it is not so simple. are on slick ice or deep the tires push backward law, the ground pushes forward.

Horizontal force exerted on the ground

by person’s foot

Horizontal force exerted

o1 e person’s foot by the ground

Fgp

NEWTON’S THIRD LAW

OF MOTION

90

CHAPTER 4

CONCEPTUAL

EXAMPLE

4-4 | What exerts the force to move a car? What

answer is The engine mud, they against the on the tires

that the engine makes the car move forward. makes the wheels go around. But if the tires just spin. Friction is needed. On firm ground, ground because of friction. By Newton’s third in the opposite direction, accelerating the car

We tend to associate forces with active objects such as humans, animals, engines, or a moving object like a hammer. It is often difficult to see how an inanimate objec*™ at rest, such as a wall or a desk, or the wall of an ice rink (Fig. 4-9), can exert & force. The explanation is that every material, no matter how hard, is elastic (springy) at least to some degree. A stretched rubber band can exert a force on a wad of paper and accelerate it to fly across the room. Other materials may not stretch as readily as rubber, but they do stretch or compress when a force is applied to them. And just as a stretched rubber band exerts a force, so does a stretched (or compressed) wall, desk, or car fender. From the examples discussed above, we can see how important it is to remember on what object a given force is exerted and by what object that force is exerted. A force influences the motion of an object only when it is applied on that object. A force exerted by an object does not influence that same object; it only influences the other object on which it is exerted. Thus, to avoid confusion, the two prepositions on and by must always be used—and used with care. One way to keep clear which force acts on which object is to use double subscripts. For example, the force exerted on the Person by the Ground as the person walks in Fig. 4-11 can be labeled Fpi. And the force exerted on the ground by the person is Fgp. By Newton’s third law Fop

=

_FPG-

4-2 4-2)

Fgp and Fpg have the same magnitude (Newton’s third law), and the minus sign reminds us that these two forces are in opposite directions. Note carefully that the two forces shown in Fig. 4-11 act on different objects—hence we used slightly different colors for the vector arrows representing these forces. These two forces would never appear together in a sum of forces in Newton’s second law, 2F = mi. Why not? Because they act on different objects: i is the acceleration of one particular object, and SF must include only the forces on that one object.

Dynamics: Newton's Laws of Motion

Force on sled exerted by assistant

: VFr-ictovfi

force on

sled exerted

by ground

Force on assistant exerted

by sled

Force on

Force on

Force on

exerted

exerted

exerted

ground

by sled

ground

assistant

by assistant

by ground

=-Fyg)

CONCEPTUAL

EXAMPLE

FIGURE 4-12 Example 4-5, showing only horizontal forces. Michelangelo has selected a fine block of marble for his next sculpture. Shown here is his assistant pulling it on a sled away from the quarry. Forces on the assistant are shown as red (magenta) arrows. Forces on the sled are purple arrows. Forces acting on the ground are orange arrows. Action-reaction forces that are equal and opposite are labeled by the same subscripts but reversed (such as Fga and Fog) and are of different colors because they act on different objects.

4-5 | Third law clarification. Michelangelo’s assistant

has been assigned the task of moving a block of marble using a sled (Fig. 4-12). He says to his boss, “When I exert a forward force on the sled, the sled exerts an equal and opposite force backward. So how can I ever start it moving? No matter how hard I pull, the backward reaction force always equals my forward force, so the net force must be zero. I’ll never be able to move this load.”

Is he correct?

RESPONSE No. Although it is true that the action and reaction forces are equal in magnitude, the assistant has forgotten that they are exerted on different objects. The forward (“action”) force is exerted by the assistant on the sled (Fig. 4-1ZD whereas the backward “reaction” force is exerted by the sled on the assistant. To determine if the assistant moves or not, we must consider only the r‘ forces on the assistant and then apply =F = ma, where XF is the net force 1 on the assistant, 4 is the acceleration of the assistant, and m is the assistant’s mass.

There are two forces on the assistant that affect his forward motion; they are shown as bright red (magenta) arrows in Figs. 4-12 and 4-13: they are (1) the horizontal force Fng exerted on the assistant by the ground (the harder he pushes backward against the ground, the harder the ground pushes forward on hlm—Ne]Eton s third law), and (2) the force Fns exerted on the assistant by the sled, pulling backward on him; see Fig. 4-13. If he pushes hard enough on the ground, the force on him exerted by the ground, F,g, will be larger than the sled pulhng back, Fus, and the assistant accelerates forward (Newton’s second law). The sled, on the other hand, accelerates forward when the force on it exerted by the assistant is greater than the frictional force exerted backward on it by lkhe ground (that is, when Fs, has greater magnitude than Fgg in Fig. 4-12).

Using double subscripts to clarify Newton’s third law can become cumbersome, and we won’t usually use them in this way. We will usually use a single subscript referring to what exerts the force on the object being discussed. Nevertheless, if there is any confusion in your mind about a given force, go ahead and use two 'subscripts to identify on what object and by what object the force is exerted.

)

PROBLEM

SOLVING

A study of Newton’s second and third laws Force on assistant exerted by sled

Force on assistant exerted

by ground

FIGURE 4-13 Example 4-5.The horizontal forces on the assistant.

EXERCISE B Return to the first Chapter-Opening Question, page 83, and answer it again now. Try to explain why you may have answered differently the first time. EXERCISE C A massive truck collides head-on with a small sports car. (a) Which vehicle experlenceé the greater force of impact? (b) Which experiences the greater acceleration during the 1‘mpact? (c) Which of Newton’s laws are useful to obtain the correct answers? r!

EXERCISE D If you push on a heavy desk, does it always push back on you? (a) Not unless ‘ someone else also pushes on it. (b) Yes, if it is out in space. (c) A desk never pushes to start with. (d) No. (e) Yes.

SECTION 4-5

Newton's Third Law of Motion

91

4—6

Weight—the Force of Gravity; and the Normal

Force

As we saw in Chapter 2, Galileo claimed that all objects dropped near the surface*™ of the Earth would fall with the same acceleration, g, if air resistance was negligible. The force that causes this acceleration is called the force of gravity or gravitational force. What exerts the gravitational force on an object? It is the Earth, as we will discuss in Chapter 6, and the force acts vertically’ downward, toward the center of the Earth. Let us apply Newton’s second law to an object of mass m falling freely due to gravity. For the acceleration, &, we use the downward acceleration due to gravity, g. Thus, the gravitational force on an object, F;, can be written as

F; = mg

4-3)

The direction of this force is down toward the center of the Earth. The magnitude of the force of gravity on an object, mg, is commonly called the object’s weight.

In SI units, g = 9.80m/s? = 9.80 N/kg, * so the weight of a 1.00-kg mass on

(b) FIGURE 4-14 (a) The net force on an object at rest is zero according to Newton’s second law. Therefore the

downward force of gravity (Fg) on

an object at rest must be balanced by an upward force (the normal force if'N)_.exerted by the table in this case. (b) FY is the force exerted on the table by the statue and is the reaction force to Fy by Newton’s third law. (FY is shown in a different color to remind us it acts on a different object.) The reaction force to Fg is not shown.

A

CAUTION

Weight and normal force are not action—reaction pairs

Earth is 1.00kg X 9.80m/s> = 9.80N. We will mainly be concerned with the weight of objects on Earth, but we note that on the Moon, on other planets, or in space, the weight of a given mass will be different than it is on Earth. For example, on the Moon the acceleration due to gravity is about one-sixth what it is on Earth, and a 1.0-kg mass weighs only 1.6 N. Although we will not use British units, we note that for practical purposes on the Earth, a mass of 1 kg weighs about 2.2 Ib. (On the Moon, 1 kg weighs only about 0.4 1b.) The force of gravity acts on an object when it is falling. When an object is at rest on the Earth, the gravitational force on it does not disappear, as we know if we weigh it on a spring scale. The same force, given by Eq. 4-3, continues to act. Why, then, doesn’t the object move? From Newton’s second law, the net force on an object that remains at rest is zero. There must be another force on the object to balance the gravitational force. For an object resting on a table, the table exerts this upward force; see Fig. 4-14a. The table is compressed slightly beneath the object, and due to its elasticity, it pushes up on the object as shown. The force exerted by the table is often called a contact force, since it occurs when two objects are in contact. (The force of your hand pushing on a cart is also a contact force.) When a contact force acts perpendicular to the common surface of contact, it is referred to as the normal force (“normal” means perpendicular); hence it is labeled Fy in Fig. 4-14a. The two forces shown in Fig. 4-14a are both acting on the statue, which remains at rest, so the vector sum of these two forces must be zero (Newton’s second law). Hence Fg and Fy must be of equal magnitude and in opposite directions. But they are not the equal and opposite forces spoken of in Newton’s third law. The action and reaction forces of Newton’s third law act on different objects, whereas the two forces shown in Fig. 4-14a act on the same object. For each of the forces shown in Fig. 4-14a, we can ask, “What is the reaction force?” The upward force, Fy, on the statue is exerted by the table. The reaction to this force is a force exerted by the statue downward on the table. It is shown in Fig. 4-14b, where it is labeled Fy. This force, Fy, exerted on the table by the statue, is the reaction force to Fy in accord with Newton’s third law. What about the other force on the statue,

the force of gravity F; exerted by the Earth? Can you guess what the reaction is to this force? We will see in Chapter 6 that the reaction force is also a gravitational force, exerted on the Earth by the statue. EXERCISE E Return to the second Chapter-Opening Question, page 83, and answer it again now. Try to explain why you may have answered differently the first time. “The concept of “vertical” is tied to gravity. The best definition of vertical is that it is the direction in which objects fall. A surface that is “horizontal,” on the other hand, is a surface on which a round object #™ won’t start rolling: gravity has no effect. Horizontal is perpendicular to vertical. *Since 1N = 1kg-m/s® (Section 4-4), then 1m/s> = 1N/kg.

92

CHAPTER4

Dynamics: Newton's Laws of Motion

F

Weight, normal force, and a box. A friend has given you a

special gift, a box of mass 10.0kg with a mystery surprise inside. The box is resting on the smooth (frictionless) horizontal surface of a table (Fig. 4-15a). (a) Determine the weight of the box and the normal force exerted on it by the table. (b) Now your friend pushes down on the box with a force of 40.0 N, as in Fig. 4— 15b Again determine the normal force exerted on the box by the table. (c) It you{ friend pulls upward on the box with a force of 40.0 N (Fig. 4-15¢), what now is the normal force exerted on the box by the table?

&

‘

mg

APPROACH The box is at rest on the table, so the net force on the box in each case is zero (Newton’s second law). The weight of the box has magnitude mg in all three cases. SOLUTION (a) The weight of the box is mg = (10.0kg)(9.80m/s?) = 98.0N, and this force acts downward. The only other force on the box is the normal force exerted upward on it by the table, as shown in Fig. 4-15a. We choose the upward direction as the positive y direction; then the net force XF, on the box is 2Fy, = Fy — mg; the minus sign means mg acts in the negative y direction (m and g are magnitudes). The box is at rest, so the net force on it must be zero (Newton’s second law, 2F, = ma,, and a, = 0). Thus |

FN

i

EFy

=

may

mg

=

0,

(@) 2Fy=Fy—mg=0

so we hav Fx

=

mg.

The normal force on the box, exerted by the table, is 98.0N upward, and has magnitude{equal to the box’s weight. (b) Your friend is pushing down on the box with a force of 40.0N. So instead of only two forces acting on the box, now there are three forces acting on the box, as shown in Fig. 4-15b. The weight of the box is still mg = 98.0 N. The net force A is 2F),=V‘N — mg — 40.0N, and is equal to zero because the box remains at rest ( P). Newton’s second law gives

2F,

=

iy — mg — 400N

= 0.

We solve this equation for the normal force:

" Fy = mg + 400N

= 980N + 400N

= 138.0N,

which is greater than in (a). The table pushes back with more force when a person pushes down on the box. The normal force is not always equal to the weight! (c) The bofi s weight is still 98.0N and acts downward. The force exerted by your friend and the normal force both act upward (positive direction), as shown in Fig. 4-15c. The box doesn’t move since your friend’s upward force is less than the weight. The net force, again set to zero in Newton’s second law because a = 0, is

' SF, = Fx — mg + 400N SO

Fy = mg — 400N

= 0,

= 980N — 400N

mg (c) SF, = Fy — mg + 400N = 0 FIGURE 4-15 Example 4-6. (a) A 10-kg gift box is at rest on a table. (b) A person pushes down on the box with a force of 40.0N. (c) A person pulls upward on the box with a force of 40.0 N. The forces are all assumed to act along a line;

= S80N.

The table does not push against the full weight of the box because of the upward pull exerted by your friend. NOTE The weight of the box (= mg) does not change as a result of your friend’s push or pull. Only the normal force is affected.

they are shown slightly displaced in order to be distinguishable. Only forces acting on the box are shown.

A

CAUTION

A

CAUTION

The normal force is not always equal to the weight

b

? ~ Recall that the normal force is elastic in origin (the table in Fig. 4-15 sags shghtly under the weight of the box). The normal force in Example 4-6 is vertical, fierpendlculqr to the horizontal table. The normal force is not always vertical, owever. When you push against a wall, for example, the normal force with flhlch the wall pushes back on you is horizontal (Fig. 4-9). For an object on a lane mchned at an angle to the horizontal, such as a skier or car on a hill, the normal force acts perpendicular to the plane and so is not vertical.

SECTION 4-6

The normal

force, Fy , is

not necessarily vertical

Weight—the Force of Gravity; and the Normal Force

93

'

"

DOV

Y

Accelerating the box. What happens when a person pulls

upward on the box in Example 4-6¢ with a force equal to, or greater than, the box’s weight? For example, let F, = 100.0N (Fig. 4-16) rather than the 40.0 N shown in Fig. 4-15c.

i s )

APPROACH We can start just as in Example 4-6, but be ready for a surprise. SOLUTION The net force on the box is ZFy

=

FN_mg+Fp

Fy

mg FIGURE 4-16

(98.0N)

Example 4-7.

>

2F,

=

Fpb — mg

A

e

+

100.0N,

and if we set this equal to zero (thinking the acceleration might be zero), we would get Fy = —2.0N. This is nonsense, since the negative sign implies Fy points downward, and the table surely cannot pull down on the box (unless there’s glue on the table). The least Fy can be is zero, which it will be in this case. What really happens here is that the box accelerates upward because the net force is not zero. The net force (setting the normal force Fy = 0) is

The box accelerates upward because F

— 980N

=

1000N

— 98.0N

= 20N upward. See Fig. 4-16. We apply Newton’s second law and see that the box moves upward with an acceleration ;

2F,

Y

_

m

20N

10.0 kg

= 0.20m/s?.

Apparent

weight

loss. A 65-kg woman

descends in e

elevator that briefly accelerates at 0.20g downward. She stands on a scale that reads in kg. (¢) During this acceleration, what is her weight and what does the scale read? (b) What does the scale read when the elevator descends at a constant speed of 2.0m/s? FIGURE 4-17

Example 4-8.The

acceleration vector is shown in gold to distinguish it from the red force vectors.

P

——

APPROACH

Figure 4-17 shows all the forces that act on the woman (and only

those that act on her). The direction of the acceleration is downward, so we choose the positive direction as down (this is the opposite choice from Examples 4-6 and 4-7). SOLUTION (@) From Newton’s second law, mg

2F

=

ma

— Fy

=

m(0.20g).

=

— 020mg

We solve for Fy: Fy

mg

=

0.80mg,

and it acts upward. The normal force Fy is the force the scale exerts on the person, and is equal and opposite to the force she exerts on the scale: Fy = 0.80mg downward. Her weight (force of gravity on her) is still

mg = (65kg)(9.8m/s?) = 640 N. But the scale, needing to exert a force of only

0.80mg, will give a reading of 0.80m = 52kg. (b) Now there is no acceleration, a =0, so by Newton’s second law, mg — Fy = 0 and Fy = mg. The scale reads her true mass of 65 kg. NOTE The scale in (¢) may give a reading of 52 kg (as an “apparent mass”), bu™ her mass doesn’t change as a result of the acceleration: it stays at 65 kg.

94

CHAPTER 4

Dynamics: Newton's Laws of Motion

|

4—7

Solving Problems with Newton's Laws:

Free-Body Diagrams

r‘Newton’s iecond law tells us that the acceleration of an object is proportional to the net force acting on the object. The net force, as mentioned earlier, is the vector sumT of all forces acting on the object. Indeed, extensive experiments have shown that forces do add together as vectors precisely according to the rules we developed in Chapter 3. For example, in Fig. 4-18, two forces of equal magnitude (100 N each) are shown acting on an object at right angles to each other. Intuitively, we can see that the object will start moving at a 45° angle and thus the net force acts at a 45° angle. This is just what the rules of vector addition ! give. From the theorem of Pythagoras, the magnitude of the resultant force is

Fr = \/(100N)?* + (100N)? = 141N, DOV

ELR

exerted on

Adding force vectors. Calculate the sum of the two forces

the boat by workers A and B in Fig. 4-19a.

(a) FIGURE 4-18

(a) Two forces, F

FIGURE 4-19

Example 4-9: Two

and Fy, exerted by workers A and B, act on a crate. (b) The sum, or resultant, of F5 and Fy is Fg.

force vectors act on a boat.

APPROA#H We add force vectors like any other vectors as described in Chapter 3. The first eFtep is to choose an xy coordinate system (see Fig. 4-19a), and then resolve vectors into their components. SOLUTION The two force vectors are shown resolved into components in Fig. 4-19b. We add the forces using the method of components. The components of F, are Fay

=

Fapco0s45.0°

=

(40.0N)(0.707)

=

283N,

Fpy

=

Fasind5.0°

=

(40.0N)(0.707)

=

28.3N.

The com#onents of Fy are

Fzy

=

+Fgcos37.0°

=

+(30.0N)(0.799)

=

+24.0N,

Fgy

=

—Fgsin37.0°

=

—(30.0N)(0.602)

=

—18.1N.

Fgy is nethive because it points along the negative y axis. The components of the resultant force are (see Fig. 4-19c) ‘

Fry

=

Fax + Fz

=

283N

+ 240N

52.3N,

Fry

=

Fay + Fgy

=

283N

— 181N

10.2N.

To find the magnitude of the resultant force, we use the Pythagorean theorem

- Fe = VFh + Fiy = V(5237 + (102N = 533N, The only remaining question is the angle 6 that the net force Fr makes with the x axis. We use:

\ tan

0

=

Fry P

em——

D

102N E

e—m——

2

0.195, .

\ and tan"'(0.195) = 11.0°. The net force on the boat has magnitude 53.3N and acts at an 11.0° angle to the x axis.

| S—

When s!olving problems involving Newton’s laws and force, it is very important to draw a diagram showing all the forces acting on each object involved. Such a diagram is called a free-body diagram, or force diagram: choose one object, and draw an arrow to represent each force acting on it. Include every force acting on that object.Do not show forces that the chosen object exerts on other objects. To help you identify each and every force that is exerted on your chosen object, ask ‘yourself what other objects could exert a force on it. If your problem involves more than one object, a separate free-body diagram is needed for each object. For r]}ow, the likely forces that could be acting are grawity and contact forces (one Jbject pushing or pulling another, normal force, friction). Later we will consider air resistance, drag, buoyancy, pressure, as well as electric and magnetic forces.

SECTION 4-7

ZJPROBLEM

Free-body diagram

SOLVING

Solving Problems with Newton's Laws: Free-Body Diagrams

95

Motion

Motion

FIGURE 4-20 Example 4-10. Which is the correct free-body diagram for a hockey puck sliding across frictionless ice?

(a)

CONCEPTUAL

VFg

(c) ' Fg

EXAMPLE

4-10 | The hockey puck. A hockey puck is sliding at

constant velocity across a flat horizontal ice surface that is assumed to be frictionless. Which of the sketches in Fig. 4-20 is the correct free-body diagram for this puck? What would your answer be if the puck slowed down? RESPONSE Did you choose (@)? If so, can you answer the question: what exerts the horizontal force labeled F on the puck? If you say that it is the force needed to maintain the motion, ask yourself: what exerts this force? Remember that another object must exert any force—and there simply isn’t any poss1b111ty here. Therefore, (a) is wrong. Besides, the force F in Fig. 4-20a would give rise to an acceleration by Newton’s second law. It is (b) that is correct. No net force acts on the puck, and the puck slides at constant velocity across the ice. In the real world, where even smooth ice exerts at least a tiny friction force, then (c) is the correct answer. The tiny friction force is in the direction opposite to the motion, and the puck’s velocity decreases, even if very slowly. Here now is a brief summary of how to approach solving problems involving

{O ~

xQ

©

2 ¢

>

Newton’s laws.

SOLV’,‘,

=

Newton's Laws; Free-Body Diagrams 1. Draw a sketch of the situation.

2. Consider only one object (at a time), and draw a free-body diagram for that object, showing all the forces acting on that object. Include any unknown forces that you have to solve for. Do not show any forces that the chosen object exerts on other objects. Draw the arrow for each force vector reasonably

accurately for direction and magnitude. Label each force acting on the object, including forces you must solve for, as to its source (gravity, person, friction, and so on). If several objects are involved, draw a free-body diagram for each object separately, showing all the forces acting on that object (and only forces acting on that

A CAUTION

Treating an object as a particle

CHAPTER 4

object). For each (and every) force, you must be clear about: on what object that force acts, and by what object that force is exerted. Only forces acting on a given object

can be included in =F = ma for that object.

. Newton’s second law involves vectors, and it is usually important to resolve vectors into components. Choose x and y axes in a way that simplifies the calculation. For example, it often saves work if you choose one coordinate axis to be in the direction of the acceleration. . For each object, apply Newton’s second law to the x and y components separately. That is, the x component of the net force on that object is related to the x component of that object’s acceleration: 2 Fy = ma,, and similarly for the y direction. 5. Solve the equation or equations for the unknown(s).

This Problem Solving Strategy should not be considered a prescription. Rather it is a summary of things to do that will start you thinking and getting involved in the problem at hand. When we are concerned only about translational motion, all the forces on a given object can be drawn as acting at the center of the object, thus treating the object as a point particle. However, for problems involving rotation or statics, the place where each force acts is also important, as we shall see in Chapters 10, 11, and 12. In the Examples that follow, we assume that all surfaces are very smooth so the friction can be ignored. (Friction, and Examples using it, are discussed in Chapter 5).

Dynamics: Newton's Laws of Motion

Pulling the mystery box. Suppose a friend asks to examine | theEXAMPLE 411 10.0-kg box you were given (Example 4-6, Fig. 4-15), hoping to guess what is

inside; and you respond, “Sure, pull the box over to you.” She then pulls the box flby the att ached cord, as shown in Fig, 4-21a, along the smooth surface of the ~ table. The magnitude of the force exerted by the personis Fp = 40.0N, and it is exerted at a 30.0° angle as shown. Calculate (@) the acceleration of the box, and (b) the m agnitude of the upward force Fy exerted by the table on the box. Assume that friction can be neglected. APPROAC H We follow the Problem Solving Strategy on the previous page. SOLUTION | 1. Draw a sketch: The situation is shown in Fig. 4-21a; it shows the box and the force ap plied by the person, Fp. 2. Free-body diagram: Figure 4-21b shows the free-body diagram of the box. To draw it correctly, we show all the forces acting on the box and only the forces acting on the box. They are: the force of gravity mg; the normal force exerted by the table F, «; and the force exerted by the person Fp. We are interested only in translat onal motion, so we can show the three forces acting at a point, Fig. 4-21c. 3. Choose axes and resolve vectors: We expect the motion to be horizontal, so we choose the x axis horizontal and the y axis vertical. The pull of 40.0N has components

Fpy Fpy

(40.0N)(cos 30.0°) (40.0N)(sin 30.0°)

= (40.0N)(0.866) = (40.0N)(0.500)

= 34.6N, = 20.0N.

In the horizontal (x) direction, Fy and mg have zero components. Thus horizon tal component of the net force is Fp,.

e

5. (a) Solv

y Newton’s second law to determine the x component of the acceleration: Foy = ma,.

®

4. (a) Ap

the

T

_

Fy

(346N)

T

T (100ke)

= 3.46m/s%.

The acc eleration of the box is 3.46 m/s* to the right. (b) Next e want to find Fy. 4. (b) Apply Newton’s second law to the vertical (y) direction, with upward as positive: FN

-

mg

§. (b) Soh ve: We have above, Fpy = 20.0N.

2Fy,

=

ma,

pr

=

may.

+

mg = (10.0kg)(9.80m/s?) = 98.0N

and, from point 3

Furthermore, since Fp, < mg, the box does not move vertically, so ay, = 0. Thus Fy — 980N + 200N = 0,

FIGURE 4-21

(a) Pulling the box,

Example 4-11; (b) is the free-body diagram for the box, and (c) is the free-body diagram considering all the forces to act at a point (transla-

tional motion only, which is what we

have here).

SO

Fy

=

78.0N.

NOTE F s less than mg: the table does not push against the full weight of the box becau se part of the pull exerted by the person is in the upward direction.

b

EXERCISE F A 10.0-kg box is dragged on a horizontal frictionless surface by a horizontal

force of 10.0N. If the applied force is doubled, the normal force on the box will (a) increase ; (b) remain the same; (c) decrease.

Tension ir n a Flexible Cord When a flexible cord pulls on an object, the cord is said to be under tension, and the force it exerts on the object is the tension Fr. If the cord has negligible mass, the force ex erted at one end is transmitted undiminished to each adjacent piece of cord along the entire length to the other end. Why? Because SF = ma = 0 for *he cord if t he cord’s mass m is zero (or negligible) no matter what @ is. Hence the orces pullin g on the cord at its two ends must add up to zero (F; and — Fy). Note that flexible cords and strings can only pull. They can’t push because they bend.

SECTION 4-7

PROBLEM

SOLVING

Cords can pull but can’t push; tension exists throughout a cord

Solving Problems with Newton's Laws: Free-Body Diagrams

97

FIGURE 4-22 Example 4-12. (a) Two boxes, A and B, are connected by a cord. A person pulls horizontally on box A with force Fp = 40.0N. (b) Free-body diagram for box A. (c) Free-body diagram for box B. Box B

Box A

Our next Example involves two boxes connected by a cord. We can refer to this group of objects as a system. A system is any group of one or more objects we choose to consider and study.

Two boxes connected by a cord. Two boxes, A and B, are

connected by a lightweight cord and are resting on a smooth (frictionless) table. The boxes have masses of 12.0 kg and 10.0kg. A horizontal force Fp of 40.0N is applied to the 10.0-kg box, as shown in Fig. 4-22a. Find (a) the acceleration of each box, and (b) the tension in the cord connecting the boxes. (b)

APPROACH We streamline our approach by not listing each step. We have two boxes so we draw a free-body diagram for each. To draw them correctly, we must consider the forces on each box by itself, so that Newton’s second law can be applied to each. The person exerts a force Fp on box A. Box A exerts a force Fr on the connecting cord, and the cord exerts an opposite but equal magnitude force Fr back on box A (Newton’s third law). These two horizontal forces on box A are shown in Fig. 4-22b, along with the force of gravity m, g downward and the normal force F,y exerted upward by the table. The cord is light, so we neglect its mass. The tension at each end of the cord is thus the same. Hence the cord exerts a force F; on the second box. Figure 4-22c shows the forces on box B, which are Fp,mgg, and the normal force Fgy . There will be only horizontal motion. We take the positive x axis to the right. SOLUTION (a) We apply XF, = ma, to box A: EFx

=

Fp

=

FT

=

MpQp.

[bOX

A]

[bOX

B]

For box B, the only horizontal force is Fr, so EFx

==

FT

=

mMmpgdg.

The boxes are connected, and if the cord remains taut and doesn’t stretch, then the two boxes will have the same acceleration a. Thus a, = ag = a. We are given m, = 10.0kg and mg = 12.0kg. We can add the two equations above to

eliminate an unknown (Fy) and obtain (mA+mB)a or

S

=

e

R

=

FP_FT+FT

40.0N _ 220kg -

—

—3

=

.

Fp

BT/ 2

2.

This is what we sought.

Alternate Solution We would have obtained the same result had we considered

a single system, of mass m, + mg, acted on by a net horizontal force equal to Fp. (The tension forces Fp would then be considered internal to the system as a whole, and summed together would make zero contribution to the net force on the whole system.) (b) From the equation above for box B (Fr = mp ag), the tension in the cord is

Fr = mga A

CAUTION

For any object, use only

e FORCEY Gk INGT 0L ect. iy OERE

98

CHAPTER 4

°=

= (12.0kg)(1.82m/s*)

= 21.8N.

Thus, Fyis less than Fp (= 40.0 N), as we expect, since Fr acts to accelerate only mg.

NOTE It might be tempting to say that the force the person exerts, Fp, acts not

only on box A but also on box B. It doesn’t. Fp acts only on box A. It affects

box B via the tension in the cord, Fr, which acts on box B and accelerates it.

Dynamics: Newton's Laws of Motion

“

SOV

-

EI RN

Elevator and counterweight (Atwood's machine). A system

of two objects suspended over a pulley by a flexible cable, as shown in Fig. 4-23a, is sometimes referred to as an Arwood’s machine. Consider the real-life application of

PHYSICS APPLIED Elevator (as Atwood’s machine)

an elevator (mg) and its counterweight (mc). To minimize the work done by the motor to raise and lower the elevator safely, mg and m are made similar in mass. We leave the motor out of the system for this calculation, and assume that the cable’s mass is negligible and that the mass of the pulley, as well as any friction, is small and ignorlable. These assumptions ensure that the tension Fry in the cable has the same magnitude on both sides of the pulley. Let the mass of the counterweight be m¢ = 1000 kg. Assume the mass of the empty elevator is 850 kg, and its mass when carrying four passengers is my = 1150kg. For the latter case (my = 1150 kg), calculate (@) the acceleration of the elevator and (b) the tension in the cable. |

APPROACH Again we have two objects, and we will need to apply Newton’s second lawiw to each of them separately. Each mass has two forces acting on it: gravity denward and the cable tension pulling upward, F;. Figures 4-23b and ¢ show the free-body diagrams for the elevator (mg) and for the counterweight (mc). The elevator, being the heavier, will accelerate downward, whereas the counterweight will accelerate upward. The magnitudes of their accelerations will be equal (we assume the cable doesn’t stretch). For the

counterweight, mcg = (1000 kg)(9.80m/s*) = 9800N, than

mgg

9800N

(in order

that mc

will accelerate

= (1150 kg)(9.80 m/sz) = 11,300N,

which

so F; must be greater

upward).

must

have

For

Elevator car

—

e

R

1150 kg

Counterweight

' 1 me = 1000kg

the elevator,

greater

magnitude

than Fy so that mg accelerates downward. Thus our calculation must give Fy

between 9800 N and 11,300 N.

SOLUTION (a) To find Fy as well as the acceleration a, we apply Newton’s second law, =F = ma, to each object. We take upward as the positive y direction for both objects. With this choice of axes, ac = a because m accelerates upward, and ag = —a because myg accelerates downward. Thus n

Fr — mgg

=

mgag

=

—mga

FT

=

Mclc

=

+mca.

-

Mcg

(a)

S

y

Fr

F1

We can subtract the first equation from the second to get (mE

—

mc)g

=

(mE

+

mc)a,

where a is now the only unknown. We solve this for a: Mg — Mc mE+mC

_

1150kg

— 1000kg

~ 1150kg + 1000kg &

0.070g

= 0.68 m/s%

\ The elevator (mg) accelerates downward (and the counterweight mc upward) at

a = 0.070g = 0.68 m/s>.

(b) The tension in the cable Fy can be obtained from either of the two

equations, setting @ = 0.070g = 0.68 m/s% ET

=

mgg

—

mga

=

mg(g

—

=

1150kg (9.80m/s* — 0.68 m/s?)

=

me(g

SF = ma

a)

=

| mEg (b)

(c)

mcg

FIGURE 4-23 Example 4-13. (a) Atwood’s machine in the form of an elevator—counterweight system. (b) and (c) Free-body diagrams for the two objects.

10,500N,

or

Fr

‘

=

mcg

+ mca

+ a)

= 1000kg (9.80m/s* + 0.68m/s?>) = 10,500 N,

which are consistent. As predicted, our result lies between 9800 N and 11,300 N, NOTE We can check our equation for the acceleration 4 in this Example by

noting th‘f\t if the masses were equal (mg = mc), then our equation above for a ™

would give

a = 0, as we should expect. Also, if one of the masses is zero (say,

mc = 0), then the other mass (mg # 0) would be predicted by our equation to

liccelerat

at a = g, again as expected.

SECTION 4-7

PROBLEM

SOLVING

Check your result by seeing if it

works in situations where the

answer is easily guessed

Solving Problems with Newton’s Laws: Free-Body Diagrams

99

CONCEPTUAL

EXAMPLE

4-14 | The advantage of a pulley. A mover is trying

to lift a piano (slowly) up to a second-story apartment (Fig. 4-24). He is using a rope looped over two pulleys as shown. What force must he exert on the rope to slowly lift the piano’s 2000-N weight?

RESPONSE The magnitude of the tension force Fy within the rope is the same at any point along the rope if we assume we can ignore its mass. First notice the forces acting on the lower pulley at the piano. The weight of the piano pulls down on the pulley via a short cable. The tension in the rope, looped through this pulley, pulls up twice, once on each side of the pulley. Let us apply Newton’s second law to the pulley—piano combination (of mass ), choosing the upward direction as positive: 2Fr — mg

FIGURE 4-24

Example 4-14.

=

ma.

To move the piano with constant speed (set @ = 0 in this equation) thus requires a tension in the rope, and hence a pull on the rope, of Fy = mg/2. The

mover can exert a force equal to half the piano’s weight.

NOTE We say the pulley has a mechanical advantage of 2, since without the pulley the mover would have to exert twice the force.

@

PHYSICS

FIGURE 4-25

APPLIED Accelerometer

Example 4-15.

Accelerometer. A small mass m hangs from a thin string and can swing like a pendulum. You attach it above the window of your car as shown in Fig. 4-25a. When the car is at rest, the string hangs vertically. What angle 6

does the string make

(@) when the car accelerates at a constant

and (b) when the car moves at constant velocity, » = 90 km/h?

a = 1.20m/s?,

APPROACH The free-body diagram of Fig. 4-25b shows the pendulum at some angle 6 relative to the vertical, and the forces on it: mg downward, and the tension Fy in the cord (including its components). These forces do not add up to zero if 6 # 0, and since we have an acceleration a, we expect § # 0. . This difference is within the expected precision (number of significant figures, Section 1-3).

117

(case i)

(a) FIGURE 5-9 Example 5-7.Note choice of x and y axes.

(case ii)

(b)

(©)

A ramp, a pulley, and two boxes. A box of mass m, = 10.0kg

rests on a surface inclined at 6 = 37° to the horizontal. It is connected by a lightweight cord, which passes over a massless and frictionless pulley, to a second box of mass my, which hangs freely as shown in Fig. 5-9a. (a) If the coefficient of static friction is us = 0.40, determine what range of values for mass my will keep the system at rest. (b) If the coefficient of kinetic friction is u, = 0.30, and my = 10.0kg, determine the acceleration of the system. APPROACH Figure 5-9b shows force of friction can be either up the box slides: (i) if mg = 0 or is incline, so Fy; would be directed tend to be pulled up the plane,

two free-body diagrams for box m, because the or down the slope, depending on which direction sufficiently small, m , would tend to slide down the up the incline; (ii) if my is large enough, m, will so Fy; would point down the plane. The tension

force exerted by the cord is labeled Fr.

SOLUTION (@) For both cases (i) and (ii), Newton’s second law for the y directionfl (perpendicular to the plane) is the same: Fy

— mpagcosf

=

mpa,

=

0

since there is no y motion. So Fy

=

mpgcosé.

Now for the x motion. We consider case (i) first for which magsing

— Fr — Fy

=

£F = ma

gives

mpay.

We want a, = 0 and we solve for F; since Fr is related to my (whose value we are seeking) by Fp = mgg (see Fig. 5-9c). Thus maygsing

— Fy, =

Fr

=

mgpg.

We solve this for my and set Fj, at its maximum

value

ugFy

= psma

find the minimum value that my can have to prevent motion (a, = 0): mg

=

gcosf

to

mpsSin® — pgmpy cos o

(10.0kg)(sin37° — 0.40cos37°) = 2.8kg. Thus if mp < 2.8 kg, then box A will slide down the incline. Now for case (ii) in Fig. 5-9b, box A being pulled up the incline. Newton’s second law is magsing

118

CHAPTER5

+ Fyp — Fr

=

mpya,

Using Newton's Laws: Friction, Circular Motion, Drag Forces

=

0.

Then the maximum value mjy can have without causing acceleration is given by Fr

=

mgg

=

magsin®

+ ugmpgcosb

flor

masinf

my

+ pgomp cos o

(10.0kg)(sin 37° + 0.40cos37°) = 9.2kg. Thus, to prevent motion, we have the condition 2.8kg < myg < 9.2kg. (b) If mg = 10.0kg and py = 0.30, then my will fall and m, will rise up the plane (case ii). To find their acceleration a, we use XF = ma for box A: maa

=

FT

-

mAgSinO

-

I“'kFN'

Since my accelerates downward, Newton’s second law for box B (Fig. 5-9c¢) tells us mpa = mpgg — Fr, or Fr = mgg — mga, and we substitute this into the equation above: maya

=

mgg

— mga

— mygsiné

— u Fy.

We solve for the acceleration a and substitute A = mg = 10.0kg, to find Mmgg — MAgSin® ma

‘ =

and then

— uxmpy gcosb mg

(10.0kg)(9.80 m/s?)(1 — sin37° — 0.30 cos 37°) 20.0 kg Il

&

+

Fy = m,gcos6,

0.079g = 0.78m/s%

NOTE It is worth comparing this equation for acceleration a with that obtained in Example 5-5: if here we let 6 = 0, the plane is horizontal as in Example 5-5, and we obtain a = (mgg — pxmag)/(ma + mg) just as in Example 5-5.

| S—

5-2 Uniform Circular Motion—Kinematics ~ FISURes-10 A smlobiec

moving in a circle, showing how the

An object moves in a straight line if the net force on it acts in the direction of motion, or the net force is zero. If the net force acts at an angle to the direction of motion at any moment, then the object moves in a curved path. An example of the latter is projectile motion, which we discussed in Chapter 3. Another important case is that of an object moving in a circle, such as a ball at the end of a string revolving around one’s head, or the nearly circular motion of the Moon about the Earth. An oh_]ect that moves in a circle at constant speed v is said to experxence uniform circular motion. The magnitude of the velocity remains constant in this case, but the direction of the velocity continuously changes as the object moves around the circle (Fig. 5-10). Because acceleration is defined as the rate of change of velocity, a change in direction of velocity constitutes an acceleration, just as a change in magnitude of velocity does. Thus, an object revolving in a circle is continuously accelerating, even when the speed remains constant (v, = v, = v). We now investigate this acceleration quantitatively. SECTION 5-2

velocity changes. At each point, the instantaneous velocity is in a direction tangent to the circular path.

Uniform Circular Motion—Kinematics

119

Acceleration is defined as I Av dv a = Atl—n;lo

At

dt

’

where AV is the change in velocity during the short time interval Az, We will eventually consider the situation in which Af approaches zero and thus obtain the instantaneous acceleration. But for purposes of making a clear drawing (Fig. 5-11), we consider a nonzero time interval. During the time interval Af, the particle in Fig. 5-11a moves from point A to point B, covering a distance Al along the arc which subtends an angle A#. The change in the velocity vector is ¥, — ¥; = AV, and is shown in Fig. 5-11b. Now we let At be very small, approaching zero. Then Af and A6 are also very small, and v, will be almost parallel to v, (Fig. 5-11c); Av will be essentially perpendicular to them. Thus AV points toward the center of the circle. Since &, by definition, is in the same direction as AV, it too must point toward the center of the circle. Therefore, this acceleration is called centripetal acceleration (“centerpointing” acceleration) or radial acceleration (since it is directed along the radius, toward the center of the circle), and we denote it by ag. We next determine the magnitude of the radial (centripetal) acceleration, ag. Because CA in Fig. 5-11a is perpendicular to ¥, and CB is perpendicular to v,, it follows that the angle A6, defined as the angle between CA and CB, is also the angle between v, and v,. Hence the vectors v, V,, and AV in Fig. 5-11b form a triangle that is geometrically similar’ to triangle CAB in Fig. 5-11a. If we take A6 to be very small (letting At be very small) and setting » = v, = v, because the magnitude of the velocity is assumed not to change, we can write Av Al v

r

or

Av ~ ZAL r

This is an exact equality when Af approaches zero, for then the arc length Af equals the chord length AB. We want to find the instantaneous acceleration, ag, so we use the expression above to write ' ar

e

()

FIGURE 5-11 Determining the change in velocity, AV, for a particle moving in a circle. The length Al is the distance along the arc, from A to B.

A

CAUTION

In uniform circular motion, the speed is

constant, but the acceleration is not zero

FIGURE 5-12 For uniform circular motion, a is always perpendicular to V.

=

.

Av

Alzlin»oE

=

v Al

MS0FAP

Then, because LY

e

is just the linear speed, v, of the object, we have for the centripetal (radial) acceleration [centripetal (radial) acceleration]

(5-1)

Equation 5-1 is valid even when v is not constant. To summarize, an object moving in a circle of radius r at constant speed v has an acceleration whose direction is toward the center of the circle and whose magnitude is ag = v*/r. It is not surprising that this acceleration depends on v» and r. The greater the speed v, the faster the velocity changes direction; and the larger the radius, the less rapidly the velocity changes direction. The acceleration vector points toward the center of the circle. But the velocity vector always points in the direction of motion, which is tangential to the circle. Thus the velocity and acceleration vectors are perpendicular to each other at every point in the path for uniform circular motion (Fig. 5-12). This is another example that illustrates the error in thinking that acceleration and velocity are always in the same direction. For an object falling vertically, @ and v are indeed parallel. But in circular motion, @ and v are perpendicular, not parallel (nor were they parallel in projectile motion, Section 3-7). EXERCISE C Can Equations 2-12, the kinematic equations for constant acceleration, be used for uniform circular motion? For example, could Eq. 2-12b be used to calculate the time for the revolving ball in Fig. 5-12 to make one revolution?

120

CHAPTER 5

"Appendix A contains a review of geometry.

P

Circular motion is often described in terms of the frequency f, the number of revolutions per second. The period 7 of an object revolving in a circle is the time required for one complete revolution. Period and frequency are related by 1 |

(5-2)

7

T

For example, if an object revolves at a frequency of 3 rev/s, then each revolution takes 1s. For an object revolving in a circle (of circumference 27rr) at constant speed v, we can write 27y T since in one revolution the object travels one circumference. \

ST

Acceleration of a revolving ball. A 150-g ball at the end of a

string is revolving uniformly in a horizontal circle of radius 0.600 m, as in Fig. 5-10 or 5-12.The ball makes 2.00 revolutions in a second. What is its centripetal acceleration? \ APPROACH The centripetal acceleration is ag = v*/r. We are given r, and we can find the speed of the ball, v, from the given radius and frequency. SOLUTION If the ball makes two complete revolutions per second, then the ball travels in a complete circle in a time interval equal to 0.500s, which is its period T. The distance traveled in this time is the circumference of the circle, 27rr, where r is the radius of the circle. Therefore, the ball has speed

‘

|

2mr 2m(0.600m) o == 2o, Al T = (0.500s)

i

aR

The centi‘ripetal acceleration’ is 2 7.54m/s)?

r

=

U_

r

=

———'—‘( /S)

(0.600 m)

=

=

7.54

Skt

94.7

)

m/S2.

\ EXERCISE D If the radius is doubled to 1.20m but the period stays the same, by what factor will the centripetal acceleration change? (a) 2, (b) 4, (c) 1, (d) 1, (e) none of these.

Moon’s centripetal acceleration. The Moon’s nearly circular

orbit about the Earth has a radius of about 384,000 km and a period

days. Determine the acceleration of the Moon toward the Earth.

T of 27.3

APPROACH Again we need to find the velocity v in order to find ag. We will need to convert to SI units to get v in m/s. SOLUTION In one orbit around the Earth, the Moon travels a distance 27r, where r = 3.84 X 10°m is the radius of its circular path. The time required for one complete orbit is the Moon’s period of 27.3 d. The speed of the Moon in its orbit al?out the Earth is o =2mr/T. The period T in seconds is T = (27.? d)(24.0h/d)(3600s/h) = 2.36 X 10°s. Therefore,

\ |

=

2.

=

Q2w

%

4nkr

4n%(3.84 X 10°m)

T?

(236 X 10°s)?

I

|

v

0.00272m/s*>

We can write this acceleration in terms of g = 9.80m/s?> gravity at the Earth’s surface) as

| ||

a

= 2.72 X 10 m/s% (the acceleration of

e 2.72 X 10~ 3/t —8 m/s

&

o

Gravitation and

Q

Newton's Synthesis CHAPTER-OPENING QUESTION—Guess now! A space stafiion revolves around the Earth as a satellite, 100 km above Earth’s surface. What is the net force on an astronaut at rest inside the space station?

CONTENTS 6-1

Newton’s Law of Universal Gravitation

(b) A littlf# less than her weight on Earth.

6-2

Vector Form of Newton’s Law of Universal Gravitation

(d) Zero (she is weightless).

6-3 Gravity Near the Earth’s Surface; Geophysical

(a) Equal to her weight on Earth. (¢)

Less than half her weight on Earth.

(e)

Somewhat larger than her weight on Earth.

Applications

ir Isaac Newton not only put forth the three great laws of motion that serve as the foundation for the study of dynamics. He also conceived of another great law to describe one of the basic forces in nature, gravitation, and he apphed it to understand the motion of the planets. This new law, published

in 1687 in

his book

Philosophiae Naturalis Principia Mathematica

(the Principia,

for short), is called Newton’s law of universal gravitation. It was the capstone of filewton’s analysis of the physical world. Indeed, Newtonian mechanics, with its aree laws of motion and the law of universal gravitation, was accepted for centuries as a mechanical basis for the way the universe works.

6-4

Satellites and “Weightlessness”

6-5

Kepler’s Laws and Newton’s Synthesis

#6-6 Gravitational Field 6-7

Types of Forces in Nature

#6-8

Principle of Equivalence; Curvature of Space; Black Holes

139

6—1

FIGURE 6-1

Anywhere on Earth,

whether in Alaska, Australia, or Peru, the force of gravity acts

downward toward the center of the

Earth

Newton’s Law of Universal Gravitation

Among his many great accomplishments, Sir Isaac Newton examined the motion of the heavenly bodies—the planets and the Moon. In particular, he wondereqfl about the nature of the force that must act to keep the Moon in its nearly circula. orbit around the Earth. Newton was also thinking about the problem of gravity. Since falling objects accelerate, Newton had concluded that they must have a force exerted on them, a force we call the force of gravity. Whenever an object has a force exerted on it, that force is exerted by some other object. But what exerts the force of gravity? Every object on the surface of the Earth feels the force of gravity, and no matter where the object is, the force is directed toward the center of the Earth (Fig. 6-1). Newton concluded that it must be the Earth itself that exerts the gravitational force on objects at its surface. According to legend, Newton noticed an apple drop from a tree. He is said to

have been struck with a sudden inspiration: If gravity acts at the tops of trees, and

even at the tops of mountains, then perhaps it acts all the way to the Moon! With this idea that it is Earth’s gravity that holds the Moon in its orbit, Newton devel-

oped his great theory of gravitation. But there was controversy at the time. Many thinkers had trouble accepting the idea of a force “acting at a distance.” Typical forces act through contact—your hand pushes a cart and pulls a wagon, a bat hits a ball, and so on. But gravity acts without contact, said Newton: the Earth exerts a force on a falling apple and on the Moon, even though there is no contact, and the two objects may even be very far apart. Newton set about determining the magnitude of the gravitational force that the Earth exerts on the Moon as compared to the gravitational force on objects at the Earth’s surface. At the surface of the Earth, the force of gravity accelerates objects at 9.80 m/s% The centripetal acceleration of the Moon is calculated from ag = v*/r (see Example 5-9) and gives ag = 0.00272m/s’>. In terms of the acceleration of gravity at the Earth’s surface, g, this is equivalent to

dg

R

0.00272 m/s?

=

980m/s

&~

~

1

36008

:

That is, the acceleration of the Moon toward the Earth is about g as great as the acceleration of objects at the Earth’s surface. The Moon is 384,000 km from the Earth, which is about 60 times the Earth’s radius of 6380 km. That is, the Moon is 60 times farther from the Earth’s center than are objects at the Earth’s surface. But 60 X 60 = 60° = 3600. Again that number 3600! Newton concluded that the gravitational force F exerted by the Earth on any object decreases with the square of its distance, r, from the Earth’s center:

FIGURE 6-2

1

The gravitational

force one object exerts on a second

F & 7

object is directed toward the first

The Moon is 60 Earth radii away, so it feels a gravitational force only & = 360

Moon

as we have seen. According to Newton’s third law, when the Earth exerts its gravitational force on any object, such as the Moon, that object exerts an equal and

object, and is equal and opposite to the force exerted by the second object on the first. o

grr“’:‘:;a;‘e‘;?:; o Moon by Earth

times as strong as it would if it were a point at the Earth’s surface. Newton realized that the force of gravity on an object depends not only on distance but also on the object’s mass. In fact, it is directly proportional to its mass, opposite

force on the Earth

Gravitational force

exerted on Earth

by the Moon

140

CHAPTER 6

of this symmetry, Newton

masses. Thus

Figg Earth

(Fig. 6-2). Because

reasoned, the magnitude of the force of gravity must be proportional to both the

where

my

)

mg Mg

=i r

is the mass of the Earth, my

distance from the Earth’s center to the center of the other object.

Gravitation and Newton's Synthesis

)

the mass of the other object, and r the

«

Newton went a step further in his analysis of gravity. In his examination of the orbits of the planets, he concluded that the force required to hold the different planets in their orbits around the Sun seems to diminish as the inverse square of their distance from the Sun. This led him to believe that it is also the gravitational @™ orce that acts between the Sun and each of the planets to keep them in their orbits. And if gravity acts between these objects, why not between all objects? Thus he proposec‘i his law of universal gravitation, which we can state as follows: Every particle in the universe attracts every other particle with a force that is proport‘onal to the product of their masses and inversely proportional to the square lflf the distance between them. This force acts along the line joining the two particles. The magnltlLde of the gravitational force can be written as

F =

mym,

r

643,

(6-1)

where m; and m, are the masses of the two particles, r is the distance between them, and G is a universal constant which must be measured experimentally and has the same numerical value for all objects. Equation 6-1 is called an inverse square law because the force is inversely proportional to r2. The value of G must be very small, since we are not aware of any force of attraction between ordinary-sized objects, such as between two baseballs. The force between two ordinary objects was first measured by Henry Cavendish in 1798, over 100 years after Newton published his law. To detect and measure the incredibly small force between ordinary objects, he used an apparatus like that shown in Fig. 6-3. Cavendish confirmed Newton’s hypothesis that two objects attract one another and that Eq. 6-1 accurately describes this force. In addition, because Cavendish could measure F, m;, m,, and r accurately, he was able to determine the value of the constant G as well. The accepted value today is

G

= 6.67 X 107" N-m?/kg?.

See Table ii side front cover for values of all constants to highest known precision.) Strictly peakmg, Eq. 6-1 gives the magnitude of the gravitational force that one particle exerts on a second particle that is a distance » away. For an extended object (that is, not a point), we must consider how to measure the distance r. You might think that » would be the distance between the centers of the objects. This is true for two spheres, and is often a good approximation for other objects. A correct calculation treats each extended body as a collection of particles, and the total force is the sum of the forces due to all the particles. The sum over all these particles is often best done

using integral calculus, which Newton

himself invented. When

NEWTON'’S LAW OF UNIVERSAL GRAVITATION FIGURE 6-3 Schematic diagram of Cavendish’s apparatus. Two spheres are attached to a light horizontal

rod, which is suspended at its center

by a thin fiber. When a third sphere (labeled A) is brought close to one of the suspended spheres, the

gravitational force causes the latter to move, and this twists the fiber

slightly. The tiny movement is magnified by the use of a narrow light beam directed at a mirror mounted on the fiber. The beam reflects onto a scale. Previous determination of how large a force will twist the fiber a given amount then allows the experimenter to determine the magnitude of the gravitational force between two objects.

extended

bodies are small compared to the distance between them (as for the Earth—Sun system), little inaccuracy results from considering them as point particles. Newton was able to show (see derivation in Appendix D) that the gravitational force exerted on a particle outside a sphere, with a spherically symmetric mass distribution, is the same as if the entire mass of the sphere was concentrated at its center. Thus Eq. 6-1 gives the correct force between two uniform spheres where r is the distance between their centers. EXAMPLE 6-1 Can you attract another person gravitationally? A 50-kg person and a 70-kg person are sitting on a bench close to each other. Estimate the magnitude of the gravitational force each exerts on the other. APPROACH Thls is an estimate: we let the distance between the centers of the two people be 4m (about as close as you can get). SOLUTION We use Eq. 61, which gives

A

kg) (70k _— 6.67 X 107" N-m?/kg?)(50 /kg’)(0kg)(T0ke) . (0.5m)?

Light \ source

(narrow beam)

1o

rounded off to an order of magnitude. Such a force is unnoticeably small unless extremely sensitive instruments are used.

\lOTE As a fraction of their weight, this force is (107° N)/(70kg)(9.8 m/s?) ~ 107°.

SECTION 6-1

141

o (Motion

Spacecraft at 2r;. What is the force of gravity acting on a

2000-kg spacecraft when it orbits two Earth radii from the Earth’s center (that is, a distance rg = 6380km above the Earth’s surface, Fig. 6-4)? The mass of the

Earth is my = 5.98 x 10*kg.

APPROACH We could plug all the numbers into Eq. 6-1, but there is a simple,« approach. The spacecraft is twice as far from the Earth’s center as when it is at the surface of the Earth. Therefore, since the force of gravity decreases as the square of the distance (and 5‘- = 1), the force of gravity on the satellite will be only one-fourth its weight at the Earth’s surface. SOLUTION At the surface of the Earth, Fg = mg. At a distance from the Earth’s center of 2r, Fg is | as great: FIGURE 6-4

Example 6-2.

FIGURE 6-5 Example 6-3. Orientation of Sun (S), Earth (E), and Moon (M) at right angles to

each other (not to scale). 7 Moon

ME

@ Earth

Fs

L

= ymg Force

= $(2000kg)(9.80m/s?) on

the

Moon. Find

= 4900N. the net force

APPROACH The forces on our object, the Moon, are the gravitational force exerted on the Moon by the Earth Fy: and the force exerted by the Sun Fyg, as shown in the free-body diagram of Fig. 6-5. We use the law of universal gravitation to find the magnitude of each force, and then add the two forces as vectors. (the gravitational force on the Moon due to the Earth) is

M Sun

the Moon

right angles to each other as in Fig. 6-5.

SOLUTION The Earth is 3.84 X 10°km = 3.84 X 10®m

Fus

on

my = 7.35 X 10”kg) due to the gravitational attraction of both the Earth mg = 5.98 X 10*kg) and the Sun (ms = 1.99 x 10*kg), assuming they are at

from the Moon, so Fy

~ (6.67 x 10" N-m?/kg?)(7.35 X 10 kg)(5.98 X 10* kg)

(3.84 X 10°m)?

=1.99 x 10°N.

The Sun is 1.50 X 10®km from the Earth and the Moon, so Fyg (the gravitational

force on the Moon due to the Sun) is 6.67 X 10" N-m?/kg?)(7.35 x 10%kg)(1.99 x 10¥k

% il

/ke’)

(1.50 x 10" m)’

o)

D) _ s

100

-

The two forces act at right angles in the case we are considering (Fig. 6-5), so we can apply the Pythagorean theorem to find the magnitude of the total force:

F = V(199 x 10°N) + (434 x 10°N}? = 477 X 10°N. The force acts at an angle 6 (Fig. 6-5) given by 6 = tan(1.99/4.34) = 24.6°. NOTE The two forces, Fy and Fys, have the same order of magnitude (10 N). This may be surprising. Is it reasonable? The Sun is much farther from Earth than

the Moon (a factor of 10" m/10°m ~ 10°), but the Sun is also much more massive (a factor of 10¥kg/10® kg ~ 107). Mass divided by distance squared (107/10°) comes out within an order of magnitude, and we have ignored factors of

3 or more. Yes, it is reasonable. CAUTION

:

e

g

.

. Distinguish between Note carefully that the law of universal gravitation describes a particular force Newton’s second law and ~ (gravity), whereas Newton’s second law of motion (F = ma) tells how an object the law of universal gravitation — accelerates due to any type of force.

A

* Spherical Shells Newton was able to show, using the calculus he invented for the purpose, that a thin uniform spherical shell exerts a force on a particle outside it as if all the shell’s mass were at its center; and that such a thin uniform shell exerts zero force on a particle

inside the shell. (The derivation is given in Appendix D.) The Earth can be modelled as a series of concentric shells starting at its center, each shell uniform bu perhaps having a different density to take into account Earth’s varying density in various layers. As a simple example, suppose the Earth were uniform throughout; what is

142

CHAPTER 6

Gravitation and Newton's Synthesis

the gravitational its surface? Only particle. The mass inside r =3rg .

.

=5

l

force on a particle placed exactly halfway from Earth’s center to the mass inside this radius » = § r, would exert a net force on this of a saphere is proportional to its volume V = § 773 so the mass m is (3} =§ the mass of the entire Earth. The gravitational force .

l

—

l

.

.

.

™ on the particle at r = ! re, which is proportional to m/r? (Eq. 6-1), is reduced to

(3)/() =1 the gravitational force it would experience at Earth’s surface.

6G-2

Vector Form of Newton’s Law of Universal

Gravitation

We can write Newton’s law of universal gravitation in vector form as

Fi, = -G——tu, 3

(6-2)

. where F,, is the vector force on particle 1 (of mass m,) exerted by particle 2 (of mass nfz), which is a distance r,; away; I, is a unit vector that points from particle 2 ‘{oward particle 1 along the line joining them so that t,; = F,;/r, where T,; is the displacement vector as shown in Fig. 6-6. The minus sign in Eq. 6-2 is necessary because the force on particle 1 due to particle 2 points toward m,, in the direction opposite to t,,. The displacement vector ¥,, is a vector of the same magnitude as ¥,;, but it points in the opposite direction so that I

=

—Ih.

By Newton’s third law, the force F,; acting on m, exerted by m; must have the same magnitude as F;, but acts in the opposite direction (Fig. 6-7), so that 3 Ky

=

3 mym; . -F,=G—"1y rz1 mymy .

= -G

——

ri2

FIGURE 6-6 The displacement vector ¥, points from particle of mass m; to particle of mass my. The unit vector shown, t, is in the same direction as ¥, but is defined as

having length one. FIGURE 6-7

By Newton’s third

law, the gravitational force on

particle 1 exerted by particle 2, F;,, is equal and opposite to that on_ particle 2 exerted by particle 1, F;;

Iy,

The force of gravity exerted on one particle by a second particle is always lirected toward the second particle, as in Fig. 6-6. When many particles interact, the total g{“avitational force on a given particle is the vector sum of the forces exerted by each of the others. For example, the total force on particle number 1 is FF=F,+F;+F,++F,

mp

thatis F,q = —F,,.

EEi (6-3) = where F;; means the force on particle 1 exerted by particle i, and n is the total number of particles. This vector notation can be very helpful, especially when sums over many particles are needed. However, in many cases we do not need to be so formal and we can deal with directions by making careful diagrams.

6-3 When object of the which weight

=

Gravity Near the Earth’s Surface; Geophysical Applications Eq. ‘6—1 is applied to the gravitational force between at it§ surface, m, becomes the mass of the Earth mg, m, object m, and r becomes the distance of the object from is th‘c radius of the Earth r. This force of gravity due of the object, which we have been writing as mg. Thus, mg

We can solye

=

the Earth and an becomes the mass the Earth’s center, to the Earth is the

mmg

G=a=

rg this for g, the acceleration of gravity at the Earth’s surface: mg

§ = G

r'e Thus, the acceleration of gravity

at the surface

(-4

of the Earth, g, is determined

by mg and r. (Don’t confuse G with g; they are very different quantities, but are related by Eq. 6-4.)

SECTION 6-3

A CAUTION Distinguish

G from g

Gravity Near the Earth’s Surface; Geophysical Applications

143

Until G was measured, the mass of the Earth was not known. But once G was measured, Eq. 6-4 could be used to calculate the Earth’s mass, and Cavendish was the first to do so. Since g = 9.80m/s’ and the radius of the Earth is

re = 6.38 X 10°m, then, from Eq. 6-4, we obtain

grt

"G

(9.80m/s?)(6.38 X 10°m)’

T Il

ME T

-

\

667 x 10 N-m¥/kg? 5.98 X 10* kg

for the mass of the Earth.

Equation 6—4 can be applied to other planets, where g, m, and r would refer to that planet.

EXAMPLE 6-4

Gravity

on

Everest. Estimate

the

effective

value of g on the top of Mt. Everest, 8850 m (29,035 ft) above sea level (Fig. 6-8). That is, what is the acceleration due to gravity of objects allowed to fall freely at this altitude? APPROACH

The force of gravity (and the acceleration due to gravity g) depends

on the distance from the center of the Earth, so there will be an effective value g’

on top of Mt. Everest which will be smaller than g at sea level. We assume the Earth is a uniform sphere (a reasonable “estimate”). SOLUTION

We

use

Eq.

6389 km = 6.389 X 10°m:

¢

oM

r’

6-4,

with

rg replaced

by

r = 6380km

+ 8.9km =

(6.67 x 107" N-m?/kg?)(5.98 x 10*kg)

(6.389 x 105m )’

9.77 m/s?, FIGURE 6-8

Example 6-4. Mount

Everest, 8850 m (29,035 ft) above sea

level; in the foreground, the author with sherpas at 5500 m (18,000 ft).

TABLE 6-1

Acceleration Due to Gravity

at Various Locations on Earth Location New York

Elevation

(m)

0 0

9.800

1650 4300

9.796 9.789

D

9798

0

9.832

San Francisco Qenver Pikes Peak Sydney,

g

(m/s?) 9.803

sl

Equator

North Pole

(calculated)

0

9.780

PHYSICS APPLIED Geology—mineral and oil exploration

144

CHAPTER 6

which is a reduction of about 3 parts in a thousand (0.3%).

-~

NOTE This is an estimate because, among other things, we ignored the mass

accumulated under the mountaintop.

—_——

Note that Eq. 6-4 does not give precise values for g at different locations

because the Earth is not a perfect sphere. The Earth not only has mountains and valleys, and bulges at the equator, but also its mass is not distributed uniformly (see Table 6-1). The Earth’s rotation also affects the value Example 6-5). However, for most practical purposes when an object is fsa’r’;h s surface, we will simply use g = 9.80 m/s* and write the weight of )

.

.

-

2

.

.

precisely of g (see near the an object .

& EXERCISE A Suppose you could double the mass of a planet but kept its volume the same. How would the acceleration of gravity, g, at the surface change?

The value of g can vary locally on the Earth’s surface because of the presence

of irregularities and rocks of different densities. Such variations in g, known as “gravity anomalies,” are very small—on the order of 1 part per 10° or 107 in the

value of g. But they can be measured (“gravimeters” today can detect variations in g to 1 part in 10°). Geophysicists use such measurements as part of their investigations into the structure of the Earth’s crust, and in mineral and oil exploration. Mineral deposits, for example, often have a greater density than surrounding material; because of the greater mass in a given volume, g can have a slightly greater value on top of such a deposit than at its flanks. “Salt domes,” under which petroleum is often found, have a lower than average density an searches for a slight reduction in the value of g in certain locales have led to th. discovery of oil.

Gravitation and Newton’s Synthesis

Effect of Earth’s rotation on g. Assuming the Earth is a

perfect sphere, determine how the Earth’s rotation affects the value of g at the equator cqmpared to its value at the poles.

APPROACH Figure 6-9 shows a person of mass m standing on a doctor’s scale at two places on the Earth. At the North Pole there are two forces acting on the mass m: the force of gravity, F5 = mg, and the force with which the scale pushes up on the mass, w. We call this latter force w because it is what the scale reads as the weight of the object, and by Newton’s third law it equals the force with which the mass pushes down on the scale. Since the mass is not accelerating, Newton’s second law tells us mg

—w = 0,

so w = mg. Thus the weight w that the spring registers equals mg, which is no surprise. Next, at the equator, there is an acceleration because the Earth is rotating. The same magnitude of the force of gravity F; = mg acts downward (we are letting g represent the acceleration of gravity in the absence of rotation and we ignore

the slight bulging of the equator). The scale pushes upward with a force w'; w’ is also the force with which the person pushes down on the scale (Newton’s third law) and hence is the weight registered on the scale. From Newton’s second law we now have (see Fig. 6-9) mg 8 —w

=

FIGURE 6-9

Example 6-5.

m—, e

because the person of mass m now has a centripetal acceleration due to Earth’s rotation; r; = 6.38 X 10°m is the Earth’s radius and v is the speed of m due to the Earth’s dai‘ly rotation. SOLUTION First we determine the speed v of an object at rest on the Earth’s equator, rembering that Earth makes one rotation (distance = circumference of

’*Earth =

27rg) in 1 day = (24 h)(60 min/h)(60 s/min) = 8.64 X 10%s: 2mre 1 day

(6.283)(6.38 X 10°m) (8.64 X 10%s) = 4.640 X 10°m/s.

The effect}ve weight is w’ = mg' where g’ is the effective value of g, and so g' = w'/m. Solving the equation above for w’, we have w' SO

Hence

’

=

g oW =y .

T

m

= == | 1)2

rg

T eY ,r

T

T

(4640 X 10°m/s)’

T8 x 10°m)

= 0.0337 m/s’, which is about Ag ~ 0.003g, a difference of 0.3%. NOTE In Table 6-1 we see that the difference in g at the pole and equator is actually greater than this: (9.832 — 9.780) m/s* = 0.052 m/s”. This discrepancy is due mainly to the Earth being slightly fatter at the equator (by 21 km) than at the poles. NOTE The calculation of the effective value of g at latitudes other than at the Apoles or equator is a two-dimensional problem because Fg acts radially toward the Earth’s center whereas the centripetal acceleration is directed perpendicular to the axis of rotation, parallel to the equator and that means that a plumb line (the I effective direction of g) is not precisely vertical except at the equator and the poles.

SECTION 6-3

Gravity Near the Earth’s Surface; Geophysical Applications

145

Earth as Inertial Reference Frame We often make the assumption that reference frames fixed on the Earth are inertial reference frames. Our calculation in Example 6-5 above shows that this assumption can result in errors no larger than about 0.3% in the use of Newton’s second law, for example. We discuss the effects of Earth’s rotation and reference frames in more detail in Chapter 11, including the Coriolis effect.

6—4 Satellites and “Weightlessness” Satellite Motion @

PHYSICS APPLIED Artificial Earth satellites

Artificial satellites circling the Earth are now commonplace (Fig. 6-10). A satellite is put into orbit by accelerating it to a sufficiently high tangential speed with the use of rockets, as shown in Fig. 6-11. If the speed is too high, the spacecraft will not be confined by the Earth’s gravity and will escape, never to return. If the speed is too low, it will return to Earth. Satellites are usually put into circular (or nearly circular) orbits, because such orbits require the least takeoff speed. 27,000 km/h circular

30,000 km/h elliptical

i 2T

27 ‘/

%

6

"I\\

2

FIGURE 6-10 A satellite, the International Space Station, circling the Earth.

FIGURE 6-12

A moving satellite

“falls” out of a straight-line path toward the Earth. Without

gravity %—\:_ it With™

5, gravity \\

'

P

\

N\

\

\

A

3

VN\ \

= 7

|

40,000 km/h escape

/

/

“

!

N -7

FIGURE 6-11 Artificial satellites launched at different speeds.

It is sometimes asked: “What keeps a satellite up?” The answer is: its high speed. If a satellite stopped moving, it would fall directly to Earth. But at the very high speed a satellite has, it would quickly fly out into space (Fig. 6-12) if it

weren’t for the gravitational force of the Earth pulling it into orbit. In fact, a

satellite is falling (accelerating toward Earth), but its high tangential speed keeps it from hitting Earth. For

satellites

that

move

in

a circle

(at

least

approximately),

the

needed

acceleration is centripetal and equals v*/r. The force that gives a satellite this acceleration is the force of gravity exerted by the Earth, and since a satellite may be at a considerable distance from the Earth, we must use Newton’s law of

universal gravitation (Eq. 6-1) for the force acting on it. When we apply Newton’s

second law,

X F G

= mayg in the radial direction, we find 2

m;’;E = m"’T,

(6-5)

where m is the mass of the satellite. This equation relates the distance of the satellite from the Earth’s center, r, to its speed, v, in a circular orbit. Note that only one~™ force—gravity—is acting on the satellite, and that r is the sum of the Earth’s radius rg plus the satellite’s height 4 above the Earth: r = rg + h.

146

CHAPTER 6

Gravitation and Newton's Synthesis

DCUILINELR

A

Geosynchronous satellite. A geosynchronous satellite is one

that stays above the same point on the Earth, which is possible only if it is above a point on the equator. Such satellites are used for TV and radio transmission, for weather forecasting, and as communication relays. Determine (a) the height above the Earth s surface such a satellite must orbit, and (b) such a satellite’s speed. (c) Compare to the speed of a satellite orbiting 200 km above Earth’s surface.

@PHYSlcs

APPLIED

Geosynchronous satellites

APPROACH To remain above the same point on Earth as the Earth rotates, the satellite must have a period of 24 hours. We can apply Newton’s second law, F = ma,

where a = v*/r if we assume the orbit is circular.

SOLUTION (a) The only force on the satellite is the force of universal gravitation due to the Earth. (We can ignore the gravitational force exerted by the Sun. Why?) We apply Eq. 6-5, assuming the satellite moves in a circle: 3 Mgat Mg () G:‘—Z

=

mSatT'

This equation has two unknowns, r and v». But the satellite revolves around the

Earth with the same period that the Earth rotates on its axis, namely once in

24 hours. Thus the speed of the satellite must be 2ar

beo swemat |

T

where T = 1day = (24h)(3600s/h) = 86,400s. We substitute this into “satellite equation” above and obtain (after canceling mg,, on both sides) G _ (27r)? r

the

rT?

After cancelling an r, we can solve for r*; GmgT? (6.67 X 107" N-m*/kg?)(5.98 x 10** kg)(86,400s)? 4mr r

7.54 X 102 m®

47

We take t he cube root and find r

=

4.23 X 10" m,

or 42,300 km from the Earth’s center. We subtract the Earth’s radius of 6380 km to find th at a geosynchronous satellite must orbit about 36,000 km (about 6 rg)

above the Earth’s surface. (b) We solve for v in the satellite equation, Eq. 6-5: v =

/

(6.67 X 107"

N-m?/kg?)(5.98 X 10*k /ke’) )

(423 x 10’ m)

We get the same result if we use v = 27r/T. (c) The equation in part (b) for v shows v o« V1/r. 6380 km + 200 km = 6580 km, we get v=

r (42,300 km) ”\/: = (3070m/8) |~ 6550 kcm)

So for =

=

3070 m/s.

r=rp+

h =

7780m/s.

NOTE Th e center of a satellite orbit is always at the center of the Earth; so it is not possible to have a satellite orbiting above a fixed point on the Earth at any latitude other than 0°. CONCEPT AL _EXAMPLE 6-7 | Catching a satellite. You are an astronaut in the space shut tle pursuing a satellite in need of repair. You find yourself in a circular orbit of the samy e radius as the satellite, but 30 km behind it. How will you catch up with it? RESPONSE We saw in Example 6-6 (or see Eq. 6-5) that the veloc1ty is proportional to 1‘/ Vr. Thus you need to aim for a smaller orbit in order to increase your f speed. Noke that you cannot just increase your speed without changing your orbit. tifter passing the satellite, you will need to slow down and rise upward again.

SECTION 6-4

Satellites and “Weightlessness”

147

EXERCISE B Two satellites orbit the Earth in circular orbits of the same radius. One satellite is twice as massive as the other. Which of the following statements is true about the speeds of these satellites? (a) The heavier satellite moves twice as fast as the lighter one. (b) The two satellites have the same speed. (c) The lighter satellite moves twice as fast as the heavier one. (d) The heavier satellite moves four times as fast as the lighter one. fl FIGURE 6-13

eleyator B

(a) An object in an

PoR ) (g

spring scale equal to its weight.

b) In an elevator accelerating

o

upward

gt)% g, the object’s apparent %vefght ;s 11 times larger than its true weight. (c) In a freely falling elevator, the object experiences “weightlessness”: the scale reads zero.

Weightlessness

People and other objects in a satellite circling the Earth are said to experience p ; g : apparent weightlessness. Let us first look at a simpler case, that of a falling . : \ . .

clevator. In Fig. 6-13a, an elevator is at rest with a bag hanging from a spring scale.

The scale reading indicates the downward force exerted on it by the bag. This force, exerted on the scale, is equal and opposite to the force exerted by the scale upward on the bag, and we call its magnitude w. Two forces act on the bag: the downward gravitational force and the upward force exerted by the scale equal to w. Because the bag is not accelerating (a = 0) when we apply £F = ma to the bag in Fig. 6-13a we obtain w— mg = 0,

where mg is the weight of the bag. Thus, w = mg, and since the scale indicates the force w exerted on it by the bag, it registers a force equal to the weight of the bag, as we expect. Now let the elevator have an acceleration, a. Applying Newton’s second law, 2F = ma, to the bag as seen from an inertial reference frame (the elevator itself is not an inertial frame) we have w — mg

=

ma.

Solving for w, we have w

=

mg + ma.

[ais + upward]

We have chosen the positive direction up. Thus, if the acceleration a is up, a is posi:fl tive; and the scale, which measures w, will read more than mg. We call w the apparent weight of the bag, which in this case would be greater than its actual weight (mg). If the elevator accelerates downward, a will be negative and w, the apparent weight, will be less than mg. The direction of the velocity v doesn’t matter. Only the direction of the acceleration & (and its magnitude) influences the scale reading. Suppose, for example, the elevator’s acceleration is 3 g upward; then we find

w = mg + mzg) = img.

© 148

} a=g(down)

CHAPTER 6

That is, the scale reads 13 times the actual weight of the bag (Fig. 6-13b). The apparent weight of the bag is 13 times its real weight. The same is true of the person: her apparent weight (equal to the normal force exerted on her by the elevator floor) is 11 times her real weight. We can say that she is experiencing 11 g’s, just as astronauts experience so many g’s at a rocket’s launch. If, instead, the elevator’s acceleration is a = —ig (downward), then w = mg — $mg = 1mg. That is, the scale reads half the actual weight. If the elevator is in free fall (for example, if the cables break), then a = —g and w = mg — mg = 0. The scale reads zero. See Fig. 6-13c. The bag appears weightless. If the person in the elevator accelerating at —g let go of a pencil, say, it would not fall to the floor. True, the pencil would be falling with acceleration g. But so would the floor of the elevator and the person. The pencil would hover right in front of the person. This phenomenon is called apparent weightlessness because in the reference frame of the person, objects don’t fall or seem to have weight—yet gravity does not disappear. Gravity is still acting on each object, whose weight is still mg. The person and other objects seem weightless onl

because the elevator is accelerating in free fall, and there is no contact force on the person to make her feel the weight.

Gravitation and Newton's Synthesis

The “weightlessness” experienced by people in a satellite orbit close to the Earth (Fig. 6-14) is the same apparent weightlessness experienced in a freely falling elevator. It may seem strange, at first, to think of a satellite as freely falling. But a satellite is indeed falling toward the Earth, as was shown in Fig. 6-12. The force of r gravity causes it to “fall” out of its natural straight-line path. The acceleration of the satellite must be the acceleration due to gravity at that point, since the only force acting on it is gravity. (We used this to obtain Eq. 6-5.) Thus, although the force of gravity acts on objects within the satellite, the objects experience an apparent weightlessness because they, and the satellite, are accelerating together as in free fall. EXERCISE C Return to the Chapter-Opening Question, page 139, and answer it again now. Try to explain why you may have answered differently the first time.

Figure 6-15 shows some examples of “free fall,” or apparent weightlessness, experienced by people on Earth for brief moments. A completely different situation occurs if a spacecraft is out in space far from the Earth, the Moon, and other attracting bodies. The force of gravity due to the Earth and other heavenly bodies will then be quite small because of the distani:es involved, and persons in such a spacecraft would experience real weightlessness. EXERCISE D Could astronauts in a spacecraft far out in space easily play catch with a bowling ball (m = 7kg)?

FIGURE 6-15

FIGURE 6-14

This astronaut is

moving outside the International Space Station. He must feel very free because he is experiencing apparent weightlessness.

Experiencing “weightlessness” on Earth.

(b)

6-5

Kepler’s Laws and Newton'’s Synthesis

More than a half century before Newton proposed his three laws of motion and his law of universal gravitation, the German astronomer Johannes Kepler (1571-1630) had worked out a detailed description of the motion of the planets about the Sun. Kepler’s work resulted in part from the many years he spent examining data collected by Tycho Brahe (1546-1601) on the positions of the planets in their motion through the heavens.

SECTION 6-5

Kepler's Laws and Newton's Synthesis

149

FIGURE 6-16 Kepler’s first law. An ellipse is a closed curve such that the sum of the distances from any point P on the curve to two fixed points (called the foci, F; and F,) remains constant. That is, the sum of the distances, FiP + F,P, is the same for all points on the curve. A circle is a special case of

Planet

an ellipse in which the two foci coincide, at the center of the circle. The

Q

semimajor axis is s (that is, the long axis is 2s) and the semiminor axis is b, as shown. The eccentricity, e, is defined as the ratio of the distance from either focus to the center divided by the semimajor axis s. Thus es is the distance from the center to either focus, as shown. For a circle, e = 0. The Earth and most of the other planets have nearly circular orbits. For Earth e = 0.017.

Sun ;

b {

|

€5 —re—es—+| b

F,

N

R

Among Kepler’s writings were three empirical findings that we now refer to as Kepler’s laws of planetary motion. These are summarized as follows, with additional explanation in Figs. 6-16 and 6-17. FIGURE 6-17 Kepler’s second law. The two shaded regions have equal

areas. The planet moves from point 1 to point 2 in the same time as it takes to move from point 3 to point 4.

Planets move fastest in that part of their orbit where they are closest to the Sun. Exaggerated scale.

Kepler’s first law: The path of each planet about the Sun is an ellipse with the Sun at one focus (Fig. 6-16). Kepler’s second law: Each planet moves so that an imaginary line drawn from the Sun to the planet sweeps out equal areas in equal periods of time (Fig. 6-17). Kepler's third law: The ratio of the squares of the periods of any two planets revolving about the Sun is equal to the ratio of the cubes of their semimajor axes. [The semimajor axis is half the long (major) axis of the orbit, as shown in Fig. 6-16, and represents the planet’s mean distance from the Sun.'] That is, if 7} and 7, represent the periods (the time needed for one revolution about the Sun) for any two planets, and s, and s, represent their semimajor axes, then

3

We can rewrite this as

si_

-

s

T?

T}

meaning that s*/7? should be the same for each planet. Present-day data are given in Table 6-2; see the last column.

$*/T*

gl

Mercury Venus Earth

fio’&'fi“"" Period, T (10“5_1!_13)

oo

57.9

ok b

108.2 149.6

Mars Jupiter Saturn

227.9 7783 1427

Uranus

2870

Pluto

5900

Neptune 4497

yr?

Kepler arrived at his laws through careful analysis of experimental data. Fifty years later, Newton was able to show that Kepler’s laws could be derived mathematically from the law of universal gravitation and the laws of motion. He also showed that for any reasonable form for the gravitational force law, only one that depends on the inverse square of the distance is fully consistent with Kepler’s laws. He thus used Kepler’s laws as evidence in favor of his law of universal

gravitation, Eq. 6-1. We

will derive Kepler’s second

law later, in Chapter

11. Here

we

derive

Kepler’s third law, and we do it for the special case of a circular orbit, in which

0.241

3.34

case the semimajor axis is the radius r of the circle. (Most planetary orbits are

0.615 1.0

335 335

close to a circle.) First, we write Newton’s second law of motion, 2F = ma. For F we use the law of universal gravitation (Eq. 6-1) for the force between the Sun

1.88 11.86 295 84.0

165

248

335 335 334 335

334

3.34

and a planet of mass m;, and for a the centripetal acceleration, v*/r. We assume the mass of the Sun Mg is much greater than the mass of its planets, so we ignore the effects of the planets on each other. Then 2F

my Mg

GT

=

ma

=y

v}

"The semimajor axis is equal to the planet’s mean distance from the Sun in the sense that it equals half the sum of the planet’s nearest and farthest distances from the Sun (points Q and R in Fig. 6-16). Most planetary orbits are close to circles, and for a circle the semimajor axis is the radius of the circle.

150

CHAPTER 6

Gravitation and Newton’s Synthesis

)

Here m; is the average speed tion of the Sun time required r:ircurnference

mass of a particular planet, r; its distance from the Sun, and v, its in orbit; Mg is the mass of the Sun, since it is the gravitational attracthat keeps each planet in its orbit. The period 7; of the planet is the for one complete orbit, which is a distance equal to 2mr, the of a circle. Thus 1

v

2mr T,

=

—

We substitute this formula for v, into the equation above: G

m; 1448 M

4mr

1

We rearrange this to get fi

41 GMS

r%

.

(

6-6

)

We derived this for planet 1 (say, Mars). The same derivation would apply for a second planet (say, Saturn) orbiting the Sun, T3 r%

4x’ B

GMS’

where 7, 1nd r, are the period and orbit radius, respectively, for the second planet. Since the right sides of the two previous equations are equal, we have

T3/ = T%J\/r% or, rearranging,

7 - () T1

rwhich

2

r

@

3

is Kepler’s third law. Equations 6-6 and 6-7 are valid also for elliptical

orbits if we replace r with the semimajor axis s. The derivations of Eqs. 6-6 and 6-7 (Kepler’s third law) compared two planets revolving around the Sun. But they are general enough to be applied to other systems. For example, we could apply Eq. 6-6 to our Moon revolving around Earth (then Mg would be replaced by Mg, the mass of the Earth). Or we could apply Eq. 6-7 to compare two moons revolving around Jupiter. But Kepler’s third law, Eq. 6-7, applies only to objects orbiting the same attracting center. Do not use Eq. 6-7 to compare, say, the Moon’s orbit around the Earth to the orbit of Mars around the F,un because they depend on different attracting centers. In the following Examples, we assume the orbits are circles, although it is not

A CAUTION Compare orbits of objects only around the same center

quite true in general.

Where is Mars? Mars’ period (its “year”) was first noted by

Kepler to be about 687 days (Earth-days), which is (687 d/365d) = 1.88 yr (Earth years). Determine the mean distance of Mars from the Sun using the

Earth as a reference.

APPROACH We are given the ratio of the periods of Mars and Earth. We can find the distance from Mars to the Sun using Kepler’s third law, given the

Earth-Sun distance as 1.50 X 10" m (Table 6-2; also Table inside front cover). SOLUTION

Let the distance of Mars from the Sun be ryg, and the Earth—Sun

distance be rgs = 1.50 X 10" m. From Kepler’s third law (Eq. 6-7): r

:

L TEs T:

1lyr

So Mars is 1.52 times the Earth’s distance from the Sun, or 2.28 X 10" m.

SECTION 6-5

Kepler's Laws and Newton's Synthesis

151

@PHYSlcs

APPLIED

Determining the Sun’s mass

DGUVIREEN The Sun’s mass determined. Determine the mass of the Sun given the Earth’s distance from the Sun as rgg = 1.5 X 10" m. APPROACH Equation 6-6 relates the mass of the Sun Mg to the period and distance of any planet. We use the Earth.

SOLUTION The

Earth’s

period

is

3.16 X 107s. We solve Eq. 6-6 for M: Mg

s

=

4rris GTt

=

7 = 1yr = (3653d)(24 h/d)(3600s/h) =

47%(1.5 x 10" m)’ (6.67 x 107" N-m?/kg?)(3.16 x 107s)’

=

2,0 X 10¥kg,

&

EXERCISE E Suppose there were a planet in circular orbit exactly halfway between the orbits of Mars and Jupiter. What would its period be in Earth-years? Use Table 6-2.

@PHYSlcs

APPLIED

Perturbations and discovery of planets

@PHYSlcs

APPLIED Planets around other stars

Accurate measurements on the orbits of the planets indicated that they did not precisely follow Kepler’s laws. For example, slight deviations from perfectly elliptical orbits were observed. Newton was aware that this was to be expected because any planet would be attracted gravitationally not only by the Sun but also (to a much lesser extent) by the other planets. Such deviations, or perturbations, in the orbit of Saturn were a hint that helped Newton formulate the law of universal gravitation, that all objects attract gravitationally. Observation of other perturbations later led to the discovery of Neptune and Pluto. Deviations in the orbit of Uranus, for example, could not all be accounted for by perturbations due to the other known planets. Careful calculation in the nineteenth century indicated that these deviations could be accounted for if another planet existed farther out in the solar system. The position of this planet was predicted from the deviations in the orbit of Uranus, and telescopes focused on that region of the sky quickly found it; the new planet was called Neptune. Similar but much smaller perturbations of Neptune’s orbit led to the discovery of Pluto in 1930. Starting in the mid 1990s, planets revolving about distant stars (Fig. 6-18) were inferred from the regular “wobble” of each star due to the gravitational attraction of the revolving planet(s). Many such “extrasolar” planets are now known. The development by Newton of the law of universal gravitation and the three laws of motion was a major intellectual achievement: with these laws, he was able to describe the motion of objects on Earth and in the heavens. The motions of heavenly bodies and objects on Earth were seen to follow the same laws (not recognized previously). For this reason, and also because Newton integrated the results of earlier scientists into his system, we sometimes speak of Newton’s synthesis. The laws formulated by Newton are referred to as causal laws. By causality we mean the idea that one occurrence can cause another. When a rock strikes a window, we infer that the rock causes the window to break. This idea of “cause and effect” relates to Newton’s laws: the acceleration of an object was seen to be caused by the net force acting on it. As a result of Newton’s theories the universe came to be viewed by many scientists and philosophers as a big machine whose parts move in a deterministic way. This deterministic view of the universe, however, had to be modified by scientists in the twentieth century (Chapter 38).

(a) FIGURE 6-18 Our solar system (a) is compared to recently discovered planets orbiting (b) the star 47 Ursae Majoris and (c) the star Upsilon Andromedae with at

least three planets. m; is the mass of

(b)

© CHAPTER 6

@é

-

3

e

@

o

Jupiter

(&) ny

Jupiter. (Sizes not to scale.)

152

Sun ‘

c?d\\? o @é & Q)é‘

47 Ursae ‘

Planet ©

Majoris Upsilon 1

Andromedae

Gravitation and Newton's Synthesis

3m,

&

e

-

@

’.

Lagrange Point. The mathematician Joseph-Louis Lagrange discovered five special points in the vicinity of the Earth’s orbit about the Sun where a small satellite (mass m) can orbit the Sun with the same period 7T as Earth’s (= 1year). One of these “Lagrange Points,” called L1, lies between the Earth (mass Mg) and Sun (mass Ms), on the line connecting them (Fig. 6-19). That is, the Earth and the satellite are always separated by a distance d. If the Earth’s orbital radius is Rgg, then the satellite’s orbital radius is (Rgs — d). Determine d. APPROACH We use Newton’s law of universal gravitation and set it equal to the mass times the centripetal acceleration. But how could an object with a smaller orbit than Earth’s have the same period as Earth? Kepler’s third law clearly tells us a smaller orbit around the Sun results in a smaller period. But that law depends on only the Sun’s gravitational attraction. Our mass m is pulled by both the Sun and the Earth. SOLUTION Because the satellite is assumed to have negligible mass in comparison to the masses of the Earth and Sun, to an excellent approximation the Earth’s orbit will be determined solely by the Sun. Applying Newton’s second law to the Earth gives

GMgMs _ Ry

or

o

T

o _ Mg (2nRes)

""Res

GMS

47T2RES

Ris

T

R

T?

S,

—

/

!

]

/

/

/

#

/ / i

l\ \

-

2

|

\

/

\

\

\

~

-

=TTy

S W

7

,

N

5

\

\

\

\

\

\

\

\

\

\

mu—da;

:

|

\

\

-

~

\

\

Earth

RES/ /

N

Sel

-

-7

’

/

’

~

/!

/

// 7

\

U

SN _———

FIGURE 6-19 Finding the position of the Lagrange Point L1 for a satellite that can remain along the revolving line between the Sun and Earth, at distance d from the Earth.

Thus a mass m at L1 has the same period around the Sun as the Earth has. (Not to scale.)

(i) .

Next we apply Newton’s second law to the satellite 7 (which has the same period T as Earth), including the pull of both Sun and Earth (see simplified form, Eq. (i))

GM, (RES

N

GMg

41*(Rgs — d)

d2

T2

d)Z

which we rewrite as

We now use the binomial expansion (1 + x)" ~ 1 + nx, if x Ffar

of the Earth, each of Earth to the Sun, the the Sun. Show that forces on these two

= 1.000171.

S

(c) Show that the ratio of the Sun’s average gravitational force on the Earth compared to that of the Moon’s is

Note that the Moon’s smaller force varies much more across the Earth’s diameter than the Sun’s larger force. (d) Estimate the resulting “force difference” (the cause of the tides) ¥ AF=F

near

—F..=F

far

far(

near

Frar

1)

~

F.

avg

(

near

Fiar

1>

. for the Moon and for the Sun. Show that the ratio of the

tide-causing force differences due to the Moon compared to the Sun is AF A—;I s

~

2.3.

Thus the Moon’s influence on tide production is over two times as great as the Sun’s.

*76. A particle is released at a height rg (radius of Earth) above the Earth’s surface. Determine its velocity when it hits the Earth. Ignore air resistance. [Hint: Use Newton’s second law, the law of universal gravitation, the chain rule, and integrate. | 7. Estimate the value of the gravitational constant G in Newton’s law of universal gravitation using the following data: the acceleration due to gravity at the Earth’s surface is about

10m/s?;

the

Earth

has

a circumference

of about

40 x 10%m; rocks found on the Earth’s surface typically have densities of about 3000 kg/m? and assume this density is constant throughout (even though you suspect it is not true). 78. Between the orbits of Mars and Jupiter, several thousand small objects called asteroids move in nearly circular orbits around the Sun. Consider an asteroid that is spherically shaped with radius » and density 2700 kg/m®. (a) You find yourself on the surface of this asteroid and throw a baseball at a speed of 22 m/s (about 50 mi/h). If the baseball is to travel around the asteroid in a circular orbit, what is the largest radius asteroid on which you are capable of accomplishing this feat? (b) After you throw the baseball, you turn around and face the opposite direction and catch the baseball. How much time 7 elapses between your throw and your catch?

*Numerical/Computer *79. (IT) The accompanying table shows the data for the mean distances of planets (except Pluto) from the Sun in our solar system, and their periods of revolution about the Sun. Planet ane Mercury Venus Earth

Mean A

Dist AU Datance (AL) 0.387 0.723

Period

(Years ( ) 0.241 0.615

Mars

1.000

1.000

1.524

Jupiter Saturn

5.203 9.539

1.881

Uranus Neptune

19.18 30.06

‘

11.88 29.46

84.01 164.8

(a) Graph the square of the periods as a function of the cube of the average distances, and find the best-fit straight line. (b) If the period of Pluto is 247.7 years, estimate the mean distance of Pluto from the Sun from the best-fit line.

Answers to Exercises : No; even though they are experiencing weightlessness, the massive ball would require a large force to throw and to decelerate when caught (inertial mass, Newton’s second law). -

A: g would double. B: (b). C: (b).

162

E: 6.17 yr.

CHAPTER 6

Gravitation and Newton's Synthesis

Y Displacement

This baseball pitcher is about to accelerate the baseball to a high velocity by exerting a force on it. He will be doing work on the ball as he exerts the force over a displacement of several meters, from behind his head until he releases the ball with arm outstretched in front of him. The total work done on the ball will be equal to the kinetic

energy (3 mv?) acquired by the ball, a result known as the work-energy principle.

” b4 i

T

E

&

L

BY

Q

CHAPTEB-OPENING QUESTION—Guess now! You push very hard on a heavy desk, trying to move it. You do work on the desk: (a) Whether or not it moves, as long as you are exerting a force. (b) Only if it starts moving,. (¢) Only if it doesn’t move. (d) Never—it does work on you. (e) None of the above. ntil now we have been studying the translational motion of an object in terms of Newton’s three laws of motion. In that analysis, force has

played a central role as the quantity determining the motion. In this

‘Chapter and the two that follow, we discuss an alternative analysis of

the translational \

v

motion e

of objects

in terms

of the3

quantities

energy

and

momentum. The 51gn1f1cance of energy and.momentum is that they are conser?)e'zd. In quite general circumstances they remain constant. That conserved quantities

exist gives us not only a deeper insight into the nature of the world but also gives us another way to approach solving practical problems. The

conservation

laws of energy

and momentum

are especially valuable

in

CONTENTS

)

l‘;v"rk Done by a Constant

vog

Salarknalntnl-tee Vectors

Sor]ce

P

7-3 Work Done by a Varying Force

7-4 Kinetic Energy and the Work-Energy Principle

dealing with systems of many objects, in which a detailed consideration of the forces involved would be difficult or impossible. These laws are applicable to a wide range of phenomena, including the atomic and subatomic worlds, where Newton’s laws cannot be applied. This Chapter is devoted to the very important concept of energy and the losely related concept of work. These two quantities are scalars and so have no direction associated with them, which often makes them easier to work with than vector quantities such as acceleration and force.

163

FIGURE 7-1 A person pulling a crate along the floor. The work done by the force Fis W = Fdcos#, where d is the displacement. Fcos 6=F,

7—1 Work Done by a Constant Force The word work has a variety of meanings in everyday language. But in physics, work is given a very specific meaning to describe what is accomplished when a force acts on an object, and the object moves through a distance. We consider only translational motion for now and, unless otherwise explained, objects are assumed to be rigid with no complicating internal motion, and can be treated like particles. Then the work done on an object by a constant force (constant in both magnitude and direction) is defined to be the product of the magnitude of the displacement times the component of the force parallel to the displacement. In equation form, we can write

W= R

where F| is the component of the constant force F parallel to the displacement d. We can also write W

=

Fdcos®,

7-1'%

where F is the magnitude of the constant force, d is the magnitude of the displacement of the object, and 6 is the angle between the directions of the force and the dlsplacement (Fig. 7-1). The cos 6 factor appears in Eq. 7-1 because F cos (= F) is the component of F that is parallel to d. Work is a scalar quantity—it has only magnitude, which can be positive or negative. Let us consider the case in which the motion and the force are in the same direction, so

FIGURE 7-2

The person does no,

work on the bag of groceries since Fp

is perpendicular to the displacementd.

6 = 0

and

cosf = 1;

in this case,

W

= Fd.

For example, if you

push a loaded grocery cart a distance of 50m by exerting a horizontal force of

30N on the cart, you do 30N X 50m = 1500 N-m of work on the cart.

As this example shows, in SI units work is measured in newton-meters (N-m).

5 special name is given to this unit, the joule (J): [In the cgs system, the unit of work lerg = 1dyne-cm. In British units, work is show that 1J = 10" erg = 0.7376 ft-1b.] A force can be exerted on an object heavy bag of groceries in your hands at rest, force on the bag, but the displacement

1J = 1N-m.

is called the erg and is defined as measured in foot-pounds. It is easy to and yet do no work. If you hold a you do no work on it. You do exert a

of the bag is zero, so the work done

by

you on the bagis W = 0. You need both a force and a displacement to do work. You also do no work on the bag of groceries if you carry it as you walk horizontally across the floor at constant velocity, as shown in Fig. 7-2. No horizontal force is required to move the bag at a constant velocity. The person shown in Fig. 7-2 does exert an upward force Fp on the bag equal to its weight. But this upward force is perpendicular to the horizontal displacement of the bag and thus is doing no work. This conclusion comes from our definition of work, Eq. 7-1: W = 0, because 6 = 90° and cos 90° = Thus, when a particular force is perpendicular to the displacement, no work is don by that force. When you start or stop walking, there is a horizontal acceleration ana you do briefly exert a horizontal force, and thus do work on the bag.

164

CHAPTER7

Work and Energy

When we deal with work, as with force, it is necessary to specify whether you are talking about work done by a specific object or done on a specific object. It is also important to specify whether the work done is due to one particular force (and which one), or the total (net) work done by the net force on the object.

A CAUTION

State that work is done on or by an object

Work done on a crate. A person pulls a 50-kg crate 40 m along a

horizontal floor by a constant force Fp» = 100 N, which acts at a 37° angle as shown in Fig. 7-3.The floor is rough and exerts a friction force, Fi; = 50 N. Determine (@) the work done by each force acting on the crate, and (b) the net work done on the crate.

e

X (40 m)

Sy

FIGURE 7-3 Example 7-1. A 50-kg crate is pulled along a floor.

APPROACH We choose our coordinate system so that % can be the vector that represents the 40-m displacement (that is, along the x axis). Four forces act on the crate, as shown in Fig. 7-3: the force exerted by the person Fp; the friction force F,; the gravitational force exerted by the Earth, mg; and the normal force Fy exerted upward by the flc;)or. The net force on the crate is the vector sum of these four forces. SOLUTIQN (a) The work done by the gravitational and normal forces is zero, since they are perpendicular to the displacement X (6 = 90° in Eq. 7-1): Ws

=

mgxcos90°

=

0

and

Wy

=

Fyxcos90°

=

0.

The work done by Fp is ~

W

=

Fpxcosé

=

(100N)(40m)cos37°

=

32001J.

The work done by the friction force is Wi

=

Frxcos180°

=

The angle between the displacement % site dire;,ctions. Since the force of cos 180° = —1), it does negative work (b) The net work can be calculated in (1) The net work done on an object each force, since work is a scalar:

(S0N)(40m)(—1)

=

—200017.

and Fy, is 180° because they point in oppofriction is opposing the motion (and on the crate. two equivalent ways: is the algebraic sum of the work done by

Wiet = Wg + Wy + Wp + Wy, =

0+ 0 +3200J

— 2000

=

120017.

(2) The net work can also be calculated by first determining the net force on

the object and then taking its component along the displacement: Fpcos @ — Fy. Then the net work is Whet

=

(F,,et)xx

=

(Fp cosf

=

(100N cos37° — 50N)(40m)

(Fy), =

— Ff,)x

=

12001J.

In the vertical (y) direction, there is no displacement and no work done. In Example 7-1 we saw that friction did negative work. In general, the work done

by a force is negative whenever the force (or the component of the force, F)) acts

in the direction opposite to the direction of motion.

A

CAUTION

Negative work

EXERCISE A A box is dragged a distance d across a floor by a force Fp which makes an angle 6 with the horizontal as in Fig. 7-1 or 7-3. If the magnitude of Fp is held constant but the angle 6 is increased, the work done by Fp (a) remains the same; (b) increases; (c) decreases; (d) first iliucreases, then decreases.

EXERCISE B Return to the Chapter-Opening Question, page 163, and answer it again now. Try to explain why you may have answered differently the first time.

SECTION 7-1

Work Done by a Constant Force

165

Q)

>

~

x

C

-2 [~

s LV, Work

@

) B Draw a free-body diagram showing all the forces acting on the object you choose to study. 2. Choose an xy coordinate system. If the object is in motion, it may be convenient to choose one of the coordinate directions as the direction of one of the forces, or as the direction of motion. [Thus, for an object on an incline, you might choose one coordinate axis to be parallel to the incline.| . Apply Newton’s laws to determine any unknown forces.

4. Find the work done by a specific force on the object by using W = Fdcos6 for a constant force. Note that the work done is negative when a force tends to oppose the displacement. . To find the net work done on the object, either (a) find the work done by each force and add the results algebraically; or (b) find the net force on the object, F,.;, and then use it to find the net work done, which for constant net force is: Wiet = Fperd cos 6.

Work on a backpack. (¢) Determine the work a hiker must

FIGURE 7-4

Example 7-2.

do on a 15.0-kg backpack to carry it up a hill of height # = 10.0m, as shown in Fig. 7-4a. Determine also (b) the work done by gravity on the backpack, and (c) the net work done on the backpack. For simplicity, assume the motion is smooth and at constant velocity (i.e., acceleration is zero). APPROACH We explicitly follow the steps of the Problem Solving Strategy above. SOLUTION 1. Draw a free-body diagram. The forces on the backpack are shown in Fig. 7-4b: the force of gravity, mg, acting downward; and Fy, the force the hiker must exert upward to support the backpack. The acceleration is zero, so horizontal forces on the backpack are negligible. 2. Choose a coordinate system. We are interested in the vertical motion of the backpack, so we choose the y coordinate as positive vertically upward. 3. Apply Newton’s laws, Newton’s second law applied in the vertical direction to the backpack gives EFy

Fy—mg

a, = 0. Hence,

Fy

= mg

=0

=

may

(15.0kg)(9.80m/s?)

“Z PROBLEM

SOLVING

Work done by gravity depends on the height ofthe hill and not on the angle of incline

166

CHAPTER 7

=

147N.

Work done by a specific force. (¢) To calculate the work done by the hiker on the backpack, we write Eq. 7-1 as Wy = Fy(dcosb), and we note from Fig. 7-4a that d cos @ = h. So the work done by the hiker is Wy = Fy(dcosb) = Fyh = mgh = (147N)(10.0m) = 14701J. Note that the work done depends only on the change in elevation and not on the angle of the hill, 8. The hiker would do the same work to lift the pack vertically the same height . (b) The work done by gravity on the backpack is (from Eq. 7-1 and Fig. 7-4c) Ws = Fgdcos(180° — 6). Since cos(180° — 6) = —cos 6, we have Wg = Fgd(—cos@) = mg(—dcosb) = —mgh Il

>

since

=

—(15.0kg)(9.80m/s?)(10.0m)

=

—1470J.

NOTE The work done by gravity (which is negative here) doesn’t depend on the angle of the incline, only on the vertical height / of the hill. This is because gravity acts vertically, so only the vertical component of displacement contributes to work done. 5. Net work done. (c) The net work done on the backpack is Wy, = 0, since the net force on the backpack is zero (it is assumed not to accelerate significantly). We can also determine the net work done by adding the work done by each force: Woet = Wg + Wy = —1470J + 1470 = 0. NOTE Even though the net work done by all the forces on the backpack is zero, the hiker does do work on the backpack equal to 1470 J.

>

CONCEPTUAL

EXAMPLE

7-3 | Does the Earth do work on the Moon? The

v

Moon revolves around the Earth in a nearly circular orbit, with approximately constant tangential speed, kept there by the gravitational force exerted by the Earth. Does gravity do (@) positive work, (b) negative work, or (¢) no work at all on the Moon? RESPONSE

The gravitational force Fg on the Moon (Fig. 7-5) acts toward the Earth

and provides its centripetal force, inward along the radius of the Moon’s orbit. The

Moon’s displacement at any moment is tangent to the circle, in the direction of its velocity, perpendicular to the radius and perpendicular to the force of gravity. Hence the angle 6 between the force Fy and the instantaneous displacement of the Moon is 90°, and the work done by gravity is therefore zero (cos90° = 0). This is why the Moon, as well as artificial satellites, can stay in orbit without expenditure of fuel: no work needs to be done against the force of gravity.

7—-2 Scalar Product of Two Vectors

///’

//

|

/

/

Moon

/-FG

\\\\

\

/

\

\

\

7

I

\\

1

g

\

\

Earth \

&

-

~o

~

.

_—

’/’

// s

/

|

|

/

FIGURE 7-5

Example 7-3.

FIGURE 7-6

The scalar product, or

Although work is a scalar, it involves the product of two quantities, force and displacement, both of which are vectors. Therefore, we now investigate the multiplication of vectors, which will be useful throughout the book, and apply it to work. Because vectors have direction as well as magnitude, they cannot be multiplied in the same way that scalars are. Instead we must define what the operation of vector multiplication means. Among the possible ways to define how to multiply vectors, there are three ways that we find useful in physics: (1) multiplication of a vector by a scalar, which was discussed in Section 3-3; (2) multiplication of one vector by a second vector to produce a scalar; (3) multiplication of one vector by a second vector to produce another vector. The third type, called the vector product, will be discussed later, in Section 11-2. f‘ We now discuss the second type, called the scalar product, or dot product (because a dot is used to indicate the multiplication). If we have two vectors, A and B, therjl their scalar (or dot) product is defined to be

A-B

= ABcoso,

(7-2)

where A and B are the magnitudes of the vectors and 6 is the angle (< 180°) between them when their tails touch, Fig. 7-6. Since A, B, and cos 6 are scalars, then so is the scalar product A - B (read “A dot B”). This d?finition, Eq. 7-2, fits perfectly with our definition of the work done by a constant force, Eq. 7-1. That is, we can write the work done by a constant force as the scaler product of force and displacement:

- W = F-d = Fdcosé.

(7-3)

dot product, of two vectors Aand B

is A+ B = ABcos 6. The scalar product can be interpreted as the magnitude of one vector (B in this case) times the projection of the other vector, A cos 6, onto B.

Indeed, the definition of scalar product, Eq. 7-2, is so chosen because many physically important quantities, such as work (and others we will meet later), can be described as the scalar product of two vectors. An equivalent definition of the scalar product is that it is the product of the magnitude of one vector (say B) and the component (or projection) of the other vector along the direction of the first (A cos 8). See Fig. 7-6. Since A, B, and cos@ are scalars, it doesn’t matter in what order they are multiplied. Hence the scalar product is commutative: A‘B

=

B-A

[commutative property]

f‘It is also easy to show that it is distributive (see Problem 33 for the proof):

A-B+C)

=A-B+A-C

[distributive property] SECTION 7-2

Scalar Product of Two Vectors

167

Let us write our vectors A and B in terms of their rectangular components using unit vectors (Section 3-5, Eq. 3-5) as

A = A+ —

A)j+ Ak

-

B = B, + B)j + Bk We will take the scalar product, A - B, of these two vectors, remembering that the unit vectors, i, j, and k, are perpendicular to each other

P=jj=kk=1

F=Tb=jE

=0

Thus the scalar product equals

A-B

= (A,d + A)j + Ak)- (B + B)j + B;k) =

AyBy

+ AyB,

+ A;B,.

(7-4)

Equation 7-4 is very useful. _

If A is perpendicular to B, then Eq. 7-2 tells us A B = ABcos90° = 0. But the converse, given that A - B = 0, can come about in three different ways: A=0,B=0, or ALB.

FIGURE 7-7 Example 7-4. Work done by a force Fp acting at an angle 6

to the ground is W = Fp - d.

«

Using the dot product. The force shown in Fig. 7-7 has

magnitude Fp = 20N and makes an angle of 30° to the ground. Calculate the work done by this force using Eq. 7-4 when the wagon is dragged 100 m along the ground. APPROACH We choose the x axis horizontal to the right and the y axis vertically upward, and write Fp and d in terms of unit vectors.

SOLUTION

Fp = Fi+ whereas

Fj

= (Fcos30°)i + (Fpsin30°)j

= (17N)i + (10N)j,

d = (100m)i. Then, using Eq. 7-4,

W = Fp-d = (17N)(100m) + (10N)(0) + (0)(0) = 17001. Note that by choosing the x axis along d we simplified the calculation because d then has only one component.

7-3 Work Done by a Varying Force If the force acting on an object is constant, the work done by that force can be calculated using Eq. 7-1. In many cases, however, the force varies in magnitude or direction during a process. For example, as a rocket moves away from Earth, work is done to overcome the force of gravity, which varies as the inverse square of the distance from the Earth’s center. Other examples are the force exerted by a spring, which increases with the amount of “® stretch, or the work done by a varying force exerted to pull a box or cart up an uneven hill.

168

CHAPTER7

Work and Energy

FIGURE 7-8 A particle acted on by a variable force, F, moves along the path shown from point a to point b. X

Figur . 7-8 shows the path of an object in the xy plane as it moves from point a to point b. The path has been divided into short intervals each of length M{,, AL,,...,Al;. A force F acts at each point on the path, and is indicated at two points as F, and F5. During each small interval Af, the force is approxime:ltely constant. For the first interval, the force does work AW of approximz}tely (see Eq. 7-1) AW

=

Fl COSO1AB].

In the second interval the work done is approximately F; cos 6, Af,, and so on. The total work done in moving the particle the total distance £ = Al + A, + ... + A,

is the sum of all these terms: W

~

3

> Fcosb;AL;.

(7-5)

i=1

We can examine this graphically by plotting F cos 6 versus the distance £ along the path as shown in Fig. 7-9a. The distance £ has been subdivided into the same seven intervals (see the vertical dashed lines). The value of

= FIGURE 7-9 Work done by a force Fis (a) approximately equal to

lines. Each of the shaded rectangles has an area (Fjcos8)(Af), which is a good

rectangles, (b) exactly equal to the

~ Fcosf

at the center of each interval is indicated by the horizontal dashed

the sum of the areas of the

estimate of the work done during the interval. The estimate of the work done along ~ area under the curve of F cos 6 vs. L.

W

=

lim SFcos6;Al;

AL—0

b | Fcosfdl.

=

(7-6)

.

300

0

|

rThis

is a! general

JF-dI.

(a)

.

D i

Fscos b5

e : 5

|

£ :

l

app ) AL, Akyly AL Al Als 5 Al Qtg AL,DAl Distance, £

S

FcosB(N)

300 +

0

(b)

Distance, £

(7-7)

a

definition

il

.

S

In this limit as Af approaches zero, the total area of the rectangles (Fig. 7-9a)

In the limit as A€ approaches zero, the infinitesimal distance df equals the magnitude of the infinitesimal displacement vector df. The direction of the vector df is along the tangent to the path at that point, so 6 is the angle between F and df at any point. Thus we can rewrite Eq. 7-6, using dot-product notation: b

fadidl L

e

symbol for the integral, [, is an elongated S to indicate an infinite sum; and A£ has been repl \ced by df, meaning an infinitesimal distance. [We also discussed this in the optional Section 2-9.]

between those two points.

e

1001

This limitas Af;. — 0 is the" integral of (F cos;g 6 df) from Ay point a to point b. The

approachés the area between the (F cos®) curve and the £ axis from a to b as shown shzded in Fig. 7-9b. That is, the work done by a variable force in moving an object between two points is equal to the area under the (F cos 0) versus (£) curve

Fycos@

g 200 £

DR

~ is constant over each interval is more accurate). Letting each A; approach zero (sowe approach an infinite number of intervals), we obtain an exact result for the work done:

o

o

smaller, the estimate of the work done becomes more accurate (the assumption that F

] .

rthe entire ipath given by Eq. 7-5, equals the sum of the areas of all the rectangles. If we subdivide the distance into a greater number of intervals, so that each Af; is

of work.

In

this

equation,

a and

b represent

two points in space, (X,, ya, z,) and (xy, yp, 2p). The integral in Eq. 7-7 is called a line integral since it is the integral of F cos 6 along the line that represents the path of the object. (Equation 7-1 for a constant force is a special case of Eq. 7-7.)

SECTION 7-3

169

In rectangular coordinates, any force can be written

F =Fi+Fj+Fk and the displacement dZ is

di = dxi + dyj + dzk.

“

Then the work done can be written

W

=

Xb

Jdex

+

Xa

Yo

JFydy a

-

4]

Jdez. Za

To actually use Eq. 7-6 or 7-7 to calculate the work, there are several options: (1) If Fcos#@ is known as a function of position, a graph like that of Fig. 7-9b can be made and the area determined graphically. (2) Another possibility is to use numerical integration (numerical summing), perhaps with the aid of a computer or calculator. (3) A third possibility is to use the analytical methods of integral calculus, when

it is doable. To do so, we must be able to write F as a function of

position, F(x, y, z), and we must know the path. Let’s look at some specific examples.

Work Done by a Spring Force Let us determine the work needed to stretch or compress a coiled spring, such as that shown in Fig. 7-10. For a person to hold a spring either stretched or compressed an amount x from its normal (relaxed) length requires a force F; that is directly proportional to x. That is,

(a) Unstretched

Fp

=

kx,

where k is a constant, called the spring constant (or spring stiffness constant), and is a measure of the stiffness of the particular spring. The spring itself exerts a force in the opposite direction (Fig. 7-10b or c):

(b) Stretched

FS

(c) Compressed FIGURE 7-10 (a) Spring in normal (unstretched) position. (b) Spring is stretched by a person exerting a force Fp to the right (positive direction). The spring pulls back with a force Fs where Fg = —kx. (c) Person compresses the spring (x < 0), and the spring pushes back with a force Fg = —kx where Fs > 0 because x < 0.

=

_kx.

(7_8)

This force is sometimes called a “restoring force” because the spring exerts its force in the direction opposite the displacement (hence the minus sign), and thus acts to return the spring to its normal length. Equation 7-8 is known as the spring equation or Hooke’s law, and is accurate for springs as long as x is not too great (see Section 12—-4) and no permanent deformation occurs. Let us calculate the work a person does to stretch (or compress) a spring from its normal (unstretched) length, x, = 0, to an extra length, x, = x. We assume the stretching is done slowly, so that the acceleration is essentially zero. The force Fj is exerted parallel to the axis of the spring, along the x axis, so Fp and dl are parallel. Hence, since df = dxi in this case, the work done by

the person is’

Wy = J [Fo(x)i] - [d] = ijP(x)dx - j vk = b 0

FIGURE 7-11 Work done to stretch a spring a distance x equals the triangular area under the curve F = kx. The areaof a

triangle is 5 X base X altitude, so W =1(x)(kx)

= Jkx?,

F

0

0

(As is frequently done, we have used x to represent both the variable of integration, and the particular value of x at the end of the interval x, =0 to x, = x.) Thus we see that the work needed is proportional to the square of the distance stretched (or compressed), x. This same result can be obtained by computing the area under the graph of Fvs. x (with cos® = 1 in this case) as shown in Fig. 7-11. Since the area is a triangle of altitude kx and base x, the work a person does to stretch or compress a spring an amount x is

W = 3(x)(kx) = 3k,

which is the same result as before. Because W o x°,2 it takes the same amount of work to stretch a spring or compress it the same amount x. See the Table of Integrals, Appendix B.

170

CHAPTER 7

Work and Energy

“

SO

AR

Work done on a spring. (a) A person pulls on the spring in

Fig. 7-10, stretching it 3.0 cm, which requires a maximum

force of 75 N. How

much work does the person do? (b) If, instead, the person compresses the

spring 3.0 cm, how much work does the person do? ’

\

APPROACH

The force Fp = kx holds at each point, including x,.,. Hence Fy,x

occurs at ‘x = Xmax-

SOLUTION

(a) First we need to calculate the spring constant k: k

=

Finax "~

SN 0.030m

=

=

2.52.5X 10°N/m 10°N/m.

‘| Then the rvork done by the person on the spring is

l W = lkaty = 4(25 X 10°N/m)(0.030m)? = 1.1J.

|

Wp

x=-0.030m

Xx=-0.030m

= kadx

F JFp(x)dx x=0

0

=

D=

(b) The force that the person exerts is still F, = kx, though now both x and Fp are negative (x is positive to the right). The work done is Lkx?

—0.030 m

0

= 1(25 x 1°N/m)(-0.030m)?

|

= 1.1J,

\ which is the same as for stretching it. NOTE We}: cannot use W = Fd (Eq. 7-1) for a spring because the force is not constant..

f‘

A More ¢ompla Force Law—Robot Arm Force

as function

of x. A robot

arm that controls

the

position of a video camera (Fig. 7-12) in an automated surveillance system is manipulated by a motor that exerts a force on the arm. The force is given by | |

1

where

F(x)

Fé, = 20N,

=

F0

54

X,

B

FIGURE 7-12

Robot arm positions

FIGURE 7-13

Example 7-6.

a video camera.

20.0 T F(N)

(%,

= F| OJ dxdx + =% 6x2Jx

dx

17.5 + 15.0 + 10.0 T

fg

S

=

(=118}

e

=

A

|

|

2)w|x,

12.5 ¢

75 T

We put in the values given and obtain

e

-

2.0N[(0.050m ~ 0.010m) +

(0.050m)* — (0.010m)? ] = 0.361. (3)(6)(0.0070 m)? SECTION 7-3

5.0+ L

0

0.01

|

I

0.02

0.03

L

:x(m)

0.04 0.05

Work Done by a Varying Force

171

7—-4 Kinetic Energy and the Work-Energy Principle

Energy is one of the most important concepts in science. Yet we cannot give a“™h simple general definition of energy in only a few words. Nonetheless, each specific type of energy can be defined fairly simply. In this Chapter we define translational kinetic energy; in the next Chapter, we take up potential energy. In later Chapters we will examine other types of energy, such as that related to heat (Chapters 19 and 20). The crucial aspect of all the types of energy is that the sum of all types, the total energy, is the same after any process as it was before: that is, energy is a conserved quantity. For the purposes of this Chapter, we can define energy in the traditional way as “the ability to do work.” This simple definition is not very precise, nor is it really valid for all types of energy.” It works, however, for mechanical energy which we discuss in this Chapter and the next. We now define and discuss one of the basic types of energy, kinetic energy. A moving object can do work on another object it strikes. A flying cannonball does work on a brick wall it knocks down; a moving hammer does work on a nail

it drives into wood. In either case, a moving object exerts a force on a second object which undergoes a displacement. An object in motion has the ability to do work and thus can be said to have energy. The energy of motion is called kinetic energy, from the Greek word kinetikos, meaning “motion.” To obtain a quantitative definition for kinetic energy, let us consider a simple rigid object of mass m (treated as a particle) that is moving in a straight line with an initial speed v;. To accelerate it uniformly to a speed v,, a constant net force F,. is exerted on it parallel to its motion over a displacement d, Fig. 7-14.

FIGURE 7-14

A constant net force

-

Fret accelerates a car from speed vy to speed v, over a displacement d. The net work done is Wyer = Fperd.

Then

the net work

done

on the object is

second law, F,, = ma, and use Eq.2-12c

W,

= F,td.

We

apply Newton’s

(v3 = v} + 2ad), which we rewrite as

v3 — v} 2d where v, is the initial speed and v, the final speed. Substituting this into F, = ma, we determine the work done: 2 _ .2 2 _ .2 a

=

ey,

¥

W, = Foud = mad = md = m

» Ay

T

E

®

Q

Conservation of Energy —Guess now! A skier starts at the top of a hill. On which run does her gravitational potential energy change the most: (a), (b), (¢), or (d); or are they (¢) all the same? On which run would her speed at the bottom be the fastest if the runs are icy and we assume no friction? Recognizing that

CONTENTS

answer the above tions again. List answers now.

8-4

Problem Solving Using Conservation of

8-5

The Law of Conservation of Energy

8-6

Energy Conservation with

there

is

always

some

friction,

two quesyour four

8-1

Conservative and Nonconservative Forces

8-2

Potential Energy

8-3 Mechanical Energy and Its Conservation Mechanical Energy

Dissipative Forces: Solving Problems 8-7 Gravitational Potential

Energy and Escape Velocity

8-8 B ¢

44

Intermediate Difficult

Very difficult

#8-9

Power Potential Energy Diagrams; Stable and Unstable Equilibrium

183

his chapter continues the discussion of the concepts of work and energy begun in Chapter 7 and introduces additional types of energy, in particular potential energy. Now we will see why the concept of energy is so important. The reason, ultimately, is that energy is conserved—the total energy always remains constant in any process. That a quantity can be defined ™

which remains constant, as far as our best experiments can tell, is a remarkable

statement about nature. The law of conservation of energy is, in fact, one of the great unifying principles of science. The law of conservation of energy also gives us another tool, another approach, to solving problems. There are many situations for which an analysis based on Newton’s laws would be difficult or impossible—the forces may not be known or accessible to measurement. But often these situations can be dealt with using the law of conservation of energy. In this Chapter we will mainly treat objects as if they were particles or rigid objects that undergo only translational motion, with no internal or rotational motion.

8—-1

Conservative and Nonconservative

Forces

We will find it important to categorize forces into two types: conservative and nonconservative. By definition, we call any force a conservative force if the work done by the force on an object moving from one point to another depends only on the initial and final positions of the object, and is independent of the particular path taken. A conservative force can be a function only of position, and cannot depend on other variables like time or velocity. We can readily show that the force of gravity is a conservative force. The gravitational force on an object of mass m near the Earth’s surface is F = mg, where g is a constant. The work done by this gravitational force on an object that falls a vertical distance h is Wg; = Fd = mgh (see Fig. 8-1a). Now suppose that instead of moving vertically downward or upward, an object follows some arbitrary path in the xy plane, as shown in Fig. 8—1b. The object starts at a vertical height y, and reaches a height y,, where y, — y; = h. To calculate the work done by gravity, W, we use Eq. 7-7:

FIGURE 8-1 Object of mass m: (a) falls a height 4 vertically; (b) is raised along an arbitrary two-dimensional path.

[ Q

2

|

WG

!

(a)

==

J

FG

-

J mg cos 6 df.

1

N di

2

1

We now let ¢ = 180° — 6 be the angle between df and its vertical component dy, as shown in Fig. 8-1b. Then, since cosf = —cos¢ and dy = dlcos ¢, we have

\Y

"

Wg

=

—J

¥z

mg dy

= —mg(y, = »)-

(8-1)

Since (y, — y,) is the vertical height 4, we see that the work done depends only on

0

184

(b) CHAPTER 8

x

the vertical height and does not depend on the particular path taken! Hence, by definition, gravity is a conservative force. Note that in the case shown in Fig. 8-1b, y, > y, and therefore the work done #%y by gravity is negative. If on the other hand y, < y, so that the object is falling,

then W is positive.

Conservation of Energy

We can give the definition of a conservative force in another, completely equivalent way: r

a force is conservative if the net work done by the force on an object moving around any closed path is zero. To see w}:fl this is equivalent to our earlier definition, consider a small object that moves from point 1 to point 2 via either of two paths labeled A and B in Fig. 8-2a. If we assume a conservative force acts on the object, the work done by this force is the same whether the object takes path A or path B, by our first definition. This work to get from point 1 to point 2 we will call W. Now consider the round trip shown in Fig. 8-2b. The object :moves from 1 to 2 via path A and our force does work W. Our object then returns to point 1 via path B. How much work is done during the return? In going from 1 to 2 via path B the work done is W, which by definition equals flz F- c{f. In doing the reverse, going from 2 to 1, the force F at each point is the same, but df is directed in precisely the opposite direction. Consequently F + df has the opposite sign at each point so tII:Je total work done in making the return trip from 2 to 1 must be —W. Hence the ‘Fotal work done in going from 1 to 2 and back to 1is W + (—=W) =0, which proves the equivalence of the two above definitions for a conservative force. The se;éond definition of a conservative force illuminates an important aspect of such a force: the work done by a conservative force is recoverable in the sense that if positive work is done by an object (on something else) on one part of a closed path, an equivalént amount of negative work will be done by the object on its return. As we‘saw above, the force of gravity is conservative, and it is easy to show that the elastic force (F = —kx) is also conservative.

(a)

1

1

(b)

FIGURE 8-2 (a) A tiny object moves between points 1 and 2 via two different paths, A and B. (b) The

object makes a round trip, via path A from point 1 to point 2 and via path B back to point 1.

FIGURE 8-3 A crate is pushed slowly at constant speed across a rough floor from position 1 to position 2 via two paths: one straight and one curved. The pushing force Fp is in the direction of motion at each point. (The friction force opposes the motion.) Hence for a constant magnitude pushing force, the work it does is W = Fpd, soif d is greater (as for the curved path), then W is greater. The work done does not depend only on points 1 and 2; it also depends on the path taken.

r

Many forces, such as friction and a push or pull exerted by a person, are nonconservative forces since any work they do depends on the path. For example, if you push a crate across a floor from one point to another, the work you do depends on whether the path taken is straight, or is curved. As shown in Fig. 8-3, if a crate is bushed slowly from point 1 to point 2 along the longer semicircular path, you do more work against friction than if you push it along the straight path. This is because the distance is greater and, unlike the gravitational force, the pushing force Fp is in the direction of motion at each point. Thus the work done by the person in Fig. 8-3 dqes not depend only on points 1 and 2; it depends also on the path taken. The force of kinetic friction, also shown in Fig. 8-3, always opposes the motion; it too is a nonconservative

force,

and

we

discuss

how

to treat

it later in this Chapter

(Section 8-6). Table 8-1 lists a few conservative and nonconservative forces. SECTION 8-1

TABLE 8-1 Conservative and Nonconservative Forces Conservative Forces

Nonconservative Forces

Gravitational

Friction

Elastic

Alir resistance

Electric

Tension in cord Motor or rocket

propulsion Push or pull by a person

Conservative and Nonconservative Forces

185

8—2 Potential Energy In Chapter 7 we discussed the energy associated with a moving object, which is its kinetic energy K = }mwv?. Now we introduce potential energy, which is the energy associated with forces that depend on the position or configuration 05 an object (or objects) relative to the surroundings. Various types of potential energy can be defined, and each type is associated with a particular conservative force. The wound-up spring of a toy is an example of potential energy. The spring acquired its potential energy because work was done on it by the person winding the toy. As the spring unwinds, it exerts a force and does work to make the toy move.

Gravitational Potential Energy

d

F...

(exerted

*! by hand)

Perhaps the most common example of potential energy is gravitational potential energy. A heavy brick held high above the ground has potential energy because of its position relative to the Earth. The raised brick has the ability to do work, for if it is released, it will fall to the ground due to the gravitational force, and can do

work on a stake, for example, driving it into the ground. Let us seek the form for the gravitational potential energy of an object near the surface of the Earth. For an object of mass m to be lifted vertically, an upward force at least equal to its weight, mg, must be exerted on it, say by a person’s hand. To lift the object without acceleration, ammuning gravity is the only other force acting on it, the person exerts an “external force”,

Py FIGURE 8-4 A person exerts an upward force Foyy = mg tolifta brick from y; to y,.

F

=

mg.

If it is raised a

vertical height A, from position y; to y, in Fig. 8-4 (upward direction chosen positive), a person does work equal to the product of the “external” force she exerts, Fext = mg upward, multiplied by the vertical displacement 4. That is,

West =

Fexi+d = mghcos0° = mgh = mg(y, — »)

where both F,, and d point upward. Gravity is also acting on the object as it moves from y, to y,, and does work on the object equal to

Wg

= Fg+d

= mghcos180°

=

—mgh

=

—-mg(y, — y),

-

where 6 = 180° because Fy and d point in opposite directions. Since Fg i downward and d is upward, W;; is negative. If the object follows an arbitrary path, as in Fig. 8-1b, the work done by gravity still depends only on the change in

vertical height (Eq. 8-1): Wi

=

—-mg(y, — y;) =

—mgh.

Next, if we allow the object to start from rest and fall freely under the action of gravity, it acquires a velocity given by v* = 2gh (Eq.2-12c) after falling a height 4. It then has kinetic energy 3mv? = 1m(2gh) = mgh, and if it strikes a stake it can do work on the stake equal to mgh. Thus, to raise an object of mass m to a height 4 requires an amount of work equal to mgh. And once at height A, the object has the ability to do an amount of work equal to mgh. We can say that the work done in lifting the object has been stored as gravitational potential energy. Indeed, we can define the change in gravitational potential energy U, when an object is moved from a height y, to a height y,, as equal to the work done by the net external force that accomplishes this without acceleration: AU

=

U, - Uy

=

W

AU

= Uy - Uy

=

W5

Ugay

= mgy.

=

"18()’2 - Y1)-

A more direct way to define the change in gravitational potential energy, AU, is that it is equal to the negative of the work done by gravity itself in the process (note Eq. 8-1): Equation 8-2 defines object of mass m moves The gravitational potential reference point (the origin

= mg(y, — »n).

(8-2)

the change in gravitational potential energy when an between two points near the surface of the Earth.' energy U at any point a vertical height y above some of the coordinate system) can be defined as

[gravity only]

(8-3)

Note that the potential energy is associated with the force of gravity between th Earth and the mass m. Hence Uy, represents the gravitational potential energy, not simply of the mass m alone, but of the mass—Earth system. 186

CHAPTER 8

*Section 8-7 deals with the 1/r dependence of Newton’s law of universal gravitation.

Gravitational potential energy depends on the wvertical height of the object above some reference level, U = mgy. Sometimes you may wonder from what point to measure y. The gravitational potential energy of a book held high above a table, for example, depends on whether we measure y from the top of the table, from the floor,

4\ CAUTION Change in potential energy is

» from some other reference point. What is physically important in any situation is ~ what is physically meaningful

the change ‘in potential energy, AU, because that is what is related to the work done, and it is AU that can be measured. We can thus choose to measure y from any referenke point that is convenient, but we must choose the reference point at the starj and be consistent throughout any given calculation. The change in potential energy between any two points does not depend on this choice. Potenti |1 energy belongs to a system, and not to a single object alone. Potential energy is associated with a force, and a force on one object is always exerted by some other object. Thus potential energy is a property of the system as a whole. For an object raised to a height y above the Earth’s surface, the change in gravitational potential energy is mgy. The system here is the object plus the Earth, and properties of both arcj;nvolved: object (m) and Earth (g). In general, a system is one or more objects that we choose to study. The choice of what makes up a system is always ours, and often try to choose a simple system. Below, when we deal with the potential energy of an object in contact with a spring, our system will be the object and the spring.

4\ CAUTION Potential energy belongs to a system, not to a single object

Potential energy changes for a roller coaster. A 1000-kg

roller-coaster car moves from point 1, Fig. 85, to point 2 and then to point 3. (@) What is the gravitational potential energy at points 2 and 3 relative to point 1? That is, take y = 0 at point 1. (b) What is the change in potential energy when the car goes from point 2 to point 3? (c¢) Repeat parts () and (b), but take the reference point (y = 0) to be at point 3. APPROACH We are interested in the potential energy We take upward as the positive y direction, and use the potential energy to calculate the potential energy. SOLUTION (a) We measure heights from point 1 (y; = that the gravitational potential energy is zero. At point \

At point

U, = mgy, 3, y; = —15m,

Uy = mgy;

of the car—Earth system. definition of gravitational

FIGURE 8-5

Example 8-1.

0), which means initially 2, where y, = 10m,

= (1000kg)(9.8m/s?)(10m)

= 9.8 x 10*J.

since point 3 is below point 1. Therefore,

=

(1000kg)(9.8m/s?)(—15m)

=

—1.5 X 10°J.

(b) In going from point 2 to point 3, the potential energy change (Ufina] - Uinma,) is

Uy — Uy, = (-1.5 x10°]) — (9.8 X 10*]) = —2.5 X 10°J. The gravitational potential energy decreases by 2.5 X 10°J. (c) Now we set y; = 0. Then y; = +15m at point 1, so the potential energy initially (at point 1) is

Uy = (1000kg)(9.8 m/s?)(15m) At point 2, y, = 25m, U,

=

= 1.5 x 10°J.

so the potential energy is

25 %X 10°1.

At point 3, y; = 0, so the potential energy is zero. The change in potential energy going from point 2 to point 3 is |

Uy = U,

=

0-25%x10°]

=

-25x10°],

which is the same as in part (b). NOTE Wd‘rk done by gravity depends only on the vertical height, so changes \i_ngravita‘cional potential energy do not depend on the path taken. r

EXERCISE I‘\ By how much does the potential energy change when a 1200-kg car climbs to

| the top of a 300-m-tall hill? (a) 3.6 X 10°J; (b) 3.5 X 10°J; (c) 4J: (d) 40J; (e) 39.2J.

SECTION 8-2

Potential Energy

187

Potential Energy in General We have defined the change in gravitational potential energy (Eq. 8-2) to be equal to the negative of the work done by gravity when an object moves from height y, to y,, which we now write as

AU = -W;

2

fl

= —JFG-dl. 1

There are other types of potential energy besides gravitational. Each type of potential energy is associated with a particular conservative force, and can be defined just as for gravity. In general, we define the change in potential energy associated with a particular conservative force ¥ as the negative of the work done by that force: :

A CAUTION S

;

—

only for conservative forces

FIGURE 8-6

(a) Spring in natural

(uncompressed) position (x = 0). force F,y, to the left. (¢) The spring

pushes back with a force Fy = —kx, where Fs > 0 because x < 0. (d) The spring can do work on a ball when released.

TaYAYaVaYaYk f;

(@)

x=0

-w,

(8-4) .

whclan. an object goes from position 1 to position 2. However, we cannot use this

definition to define a potential energy for every force. It makes sense only for

conservative forces such as gravity, for which the integral depends only on the end points and not on the path taken. It does not apply to nonconservative forces like friction, because the integral in Eq. 8-4 would not have a unique value depending on the end points 1 and 2. Thus the concept of potential energy cannot be defined and is meaningless for a nonconservative force. We can also define the change of potential energy in an equivalent way as the work done by a net external force to move the object without acceleration from one position to another.

X

We now discuss potential energy associated with elastic materials, which includes a great variety of practical applications. Consider the simple coil spring shown in Fig. 8-6, whose mass is so small that we can ignore it. The spring has potential energy when it is compressed becauselfl when it is released, it can do work on a ball. Like other elastic materials, a spring i< described by Hooke’s law (see Section 7-3) as long as the displacement x is not too great. Let us take our coordinate system so the end of the uncompressed spring is at x = 0 (Fig. 8-6a) and x is positive to the right. To hold the spring compressed, or stretched, a distance x from its natural (unstretched) length requires the person’s hand to exert an external force F,, = kx on the spring (Fig. 8-6b), where k is the

spring constant (Section 7-3). The spring itself exerts a force (Newton’s third law) in the opposite direction on the hand (Fig. 8-6c):

Fs =

The negative displacement compressed (where x can

—kx.

[spring force]

sign appears because the force Fg is in the direction opposite to the x. From Eq. 8-4, the change in potential energy when the spring is or stretched from x; = 0 (its uncompressed position) to x, = x be + or —) is .

AU

(c)

= U, - U, = —JF-di= » 1 -

Elastic Potential Energy

(b) Spring is compressed (x < 0) by a person exerting an external

‘ R

AU

= U(x) - U(0)

2

= —J Fs-dl 1

=

—J (—kx)dx 0

= 1kx2

Here, U(x) means the potential energy at x, and U(0) means U at x = 0. It is usually convenient to choose the potential energy at x = 0 to be zero: U(0) = 0, so the potential energy of a spring compressed or stretched an amount x from equilibrium is

Ug(x)

= 3kx*

[elastic spring]

(8-5)

Potential Energy Related to Force (1-D) In the one-dimensional case, where a conservative force can be written as a function

of x, say, the potential energy can be written as an indefinite integral

U(x) = —J’F(x) dx + C, where the constant C represents the value of U at

188

CHAPTER 8

Conservation of Energy

(8-6)" x = 0;

we often choose our

T

coordinates so that C = 0. Equation 8-6 tells us how to obtain U(x) when given F(x). If, instead, we are given U(x), we can obtain F(x) by inverting the above equation: that is, we take the derivative of both sides, remembering that integration and differentiation are inverse operations:

AP

ir J Thus

(x)dx

=

F(x).

F(x) = SOITE

SN

:

-7

Determine F from U. Suppose U(x) = —ax/(b* + x?), where

a and b are constants. What is F as a function of x?

APPROACH Since U (x) depends only on x, this is a one-dimensional problem. SOLUTION Equation 8-7 gives

dx

dx|

TEAR

b*+ x?

B a(b2 —xz)-

ax

a

__ax

_i

dU

Gy

()

* Potential Energy in Three Dimensions In three dimensions, we can write the relation between F(x, y, z) and U as: F

or

*

=

oU

-

ax

)

Fy

=

oU

Kz

—

oUu

F

ay

;0U

£

=

-

oU 0z

)

~ U

= =lg==j—=k—=

rHere, d/dx, /dy and 9/dz are called partial derivatives; /dx, for example, means nat although U may be a function of x, y, and z, written U(x, y, z), we take the derivative only with respect to x with the other variables held constant.

8—3 Mechanical Energy and Its Conservation Let us consi‘der a conservative system (meaning only conservative forces do work) in which eriergy is transformed from kinetic to potential or vice versa. Again, we must consider a system because potential energy does not exist for an isolated object Ourlsystem might be a mass m oscillating on the end of a spring or moving in the Earth s gravitational field. Accordmg to the work-energy principle (Eq. 7-11), the net work W,,.; done on an object is equal to its change in kinetic energy: Wnct

(If more than represent the write the work change in total

=

AK.

one object of our system has work done on it, then W,,.; and AK can sum for all of them.) If we assume a conservative system we can done on an object or objects by the conservative forces in terms of the potential energy (see Eq. 8-4) between points 1 and 2: 2

AUga

=

-L

net cdl

=

—Wiet -

(8-8)

We combine the previous two equations, letting U be the total potential energy: r.

‘

AK

+ AU

=

0

[conservative forces only]

(8-9a)

=0,

[conservative forces only]

(8-9b)

s

(K-

Ky) + (U,

- Uy)

SECTION 8-3

Mechanical Energy and Its Conservation

We now define a quantity E, called the total mechanical energy of our system, as the sum of the kinetic energy plus the potential energy of the system at any moment E

=

K+

U.

We can rewrite Eq. 8-9b as CONSERVATION OF MECHANICAL

ENERGY

o

K, +U

=

Ky + U

[conservative forces only]

(8-10a)

)

E, = E, = constant. [conservative forces only] (8-10b) Equations 8-10 express a useful and profound principle regarding the total mechanical energy of a system—namely that it is a conserved quantity. The total mechanical energy E remains constant as long as no nonconservative forces do work. That is, the quantity E = K + U at some initial time 1 is equal to K + U at any later time 2. To say it another way, consider Eq. 8-9a which tells us AU

= —AK;

that is, if the

kinetic energy K of a system increases, then the potential energy U must decrease by an equivalent amount to compensate. Thus the total, K + U, remains constant: CONSERVATION OF MECHANICAL ENERGY

If only conservative forces do work, the total mechanical energy of a system neither increases nor decreases in any process. It stays constant—it is conserved. This is the principle of conservation of mechanical energy for conservative forces. We now see the reason for the term “conservative force”—because for such forces, mechanical energy is conserved. If only one object of a system" has significant kinetic energy, then Egs. 8—10 become E

=

imv® + U

=

constant.

[conservative forces only]

(8-11a)

If we let v; and U, represent the velocity and potential energy at one instant, and v, and U, represent them at a second instant, then we can rewrite this as

Imd

+ U

= Imvd + U,.

[conservative system]

(8-11b)

From this equation we can see again that it doesn’t make any difference where we choose the potential energy to be zero: adding a constant to U merely adds a constant to both sides of Eq. 8-11b, and these cancel. A constant also doesn’ affect the force obtained using Eq. 8-7, F = —dU/dx, since the derivative of a constant is zero. Only changes in the potential energy matter.

8—4 Problem Solving Using FIGURE 8-7 The rock’s potential energy changes to kinetic energy as it falls. Note bar graphs representing

potential energy U and kinetic

energy K for the three different

positions.

A simple example of the conservation of mechanical energy (neglecting air resistance) is a rock allowed to fall due to Earth’s gravity from a height 4 above the ground, as shown in Fig. 8-7. If the rock starts from rest, all of the initial energy is potential energy. As the rock falls, the potential energy mgy decreases (because y decreases),

U K

all potential energy

Conservation o echamcal Energy

but the rock’s kinetic energy increases to compensate, so that the sum of the two remains constant. At any point along the path, the total mechanical energy is given by E

=

K+U

=

§e i smv° + mgy

where y is the rock’s height above the ground at a given instant and v is its speed at that point. If we let the subscript 1 represent the rock at one point along its path (for example, the initial point), and the subscript 2 represent it at some other point, then we can write

total mechanical energy at point 1 = total mechanical energy at point 2 or (see also Eq. 8-11b)

Imt :

y,=0

all kinetic energy 190

CHAPTER

2

8

Just before

+ mgy,

= tmv} + mgy,.

the rock hits the ground, where

[gravity only] we chose

y = 0,

potential energy will have been transformed into kinetic energy.

(8-12)

all of the initial

“For an object moving under the influence of Earth’s gravity, the kinetic energy of the Earth can usually be ignored. For a mass oscillating at the end of a spring, the mass of the spring, and hence its kinetic energy, can often be ignored.

“

GV

P

R

ERN

Falling rock. If the original height of the rock in Fig. 8-7 is

v = h = 3.0m, find the rock’s speed when it has fallen to 1.0 m above the ground. APPROACH We apply the principle of conservation of mechanical energy, Eq. 8-12, with only gravity acting on the rock. We choose the ground as our reference level

(y =0).

SOLUTION At the moment of release (point 1) the rock’s position is y; = 3.0m and it is at rest: v; = 0. We want to find v, when the rock is at position ¥ =1.0 rix Equation 8-12 gives ‘

%mv%

+ mgy,

=

%mv%

+ mgy;.

The m’s cbncel out; setting v1 = 0 and solving for v, we find

= V2

— y») = V2(9.8m/s)[(3.0m) — (1.0m)]

= 63m/s.

The rock 5 speed 1.0 m above the ground is 6.3 m/s downward. NOTE The velocity of the rock is independent of the rock’s mass. EXERCISE B In Example 8-3, what is the rock’s speed just before it hits the ground? (a) 6.5m/s; (b) 7.0m/s; (¢) 7.7m/s; (d) 8.3 m/s; (e) 9.8 m/s.

Equation 8-12 can be applied to any object moving without friction under the action of gravity. For example, Fig. 8-8 shows a roller-coaster car starting from rest at the top of a hill, and coasting without friction to the bottom and up the hill on the other side. True, there is another force besides gravity acting on the car, the normal force exerted by the tracks. But the normal force acts perpendicular to the direction of motion at each point and so does zero work. We ignore rotational motion of the car’s wheels and treat the car as a particle undergoing simple translation. Initially, the car has only potential energy. As it coasts down the hill, it loses potential energy and

gains in kinetic energy, but the sum of the two remains constant. At the bottom of

the hill it has its maximum kinetic energy, and as it climbs up the other side the kinetic energy changes back to potential energy. When the car comes to rest again at rthe same héight from which it started, all of its energy will be potential energy. Given that the gravitational potential energy is proportional to the vertical height, energy conservation tells us that (in the absence of friction) the car comes to rest at a height equal to its original height. If the two hills are the same height, the car will ]ust barely reach the top of the second hill when it stops. If the second hill is lower than the first, not all of the car’s kinetic energy will be transformed to potential energy so the car will have speed to continue over the top and down the other side. If the second hill is higher, the car will reach a maximum height on it equal to its original height on the first hill. This is true (in the absence of friction) no matter how steep the hill is, since potential energy depends only on the vertical height.

DGV

ELN

FIGURE 8-8

A roller-coaster car

moving without f“fcnon 11111us_treites t}:le ;:onservatlon Ot mechamca RIS

Roller-coaster car speed using energy conservation. Assuming

the height of the hill in Fig. 8-8 is 40m, and the roller-coaster car starts from rest at the top, calculate (@) the speed of the roller-coaster car at the bottom of the hill, and (b) at what height it will have half this speed. Take y = 0 at the bottom of the hill.

APPROACH

We choose point 1 to be where the car starts from rest (v, = 0) at

the top ofi the hill (y; = 40 m). Point 2 is the bottom of the hill, which we choose as our reference level, so y, = 0. We use conservation of mechanical energy. SOLUTIO#I (a) We use Eq. 8-12 with »; = 0 and y, = 0, which gives or

(b) Now but now

~

-

mgy,

=

=

i:

mv;

i v, = V2gy

2

= \/2(9.8m/s?)(40m)

= 28m/s.

y, is the unknown. We again use conservation of energy,

imo + mgy, = 3mvi + mgy,,

v, = $(28m/s)

3

= 14m/s

v;

and

v, = 0. Solving for the unknown y, gives

(14m/s)>

yz—yl—z—40m—m—30m That is, the car has a speed of 14 m/s when it is 30 vertical meters above the lowest igint, both when descending the left-hand hill and when ascending the right-hand hill.

SECTION 8-4

191

The mathematics of the roller-coaster Example 8-4 is almost the same as for the rock in Example 8-3. But there is an important difference between them. In Example 8-3 the motion is all vertical and could have been solved using force, acceleration, and the kinematic equations (Eqs. 2-12). But for the roller coaster of Example 84, where the motion is not vertical, we could not have used Egs. 2-12 becaus“™ a is not constant on the curved track; energy conservation gives us the answer readily.

Paul

2 9’5’7&.

I

.

Kathleen

\(-(Li}

FIGURE 8-9

Example 8-5.

CONCEPTUAL

EXAMPLE

8-5 | Speeds on two water slides. Two water slides

at a pool are shaped differently, but start at the same height 4 (Fig. 8-9). Two riders, Paul and Kathleen, start from rest at the same time on different slides. (a) Which rider, Paul or Kathleen, is traveling faster at the bottom? (b) Which rider makes it to the bottom first? Ignore friction and assume both slides have the same path length. RESPONSE (@) Each rider’s initial potential energy mgh gets transformed to kinetic energy, so the speed v at the bottom is obtained from 1mw? = mgh. The mass cancels and so the speed will be the same, regardless of the mass of the rider. Since they descend the same vertical height, they will finish with the same speed. (b) Note that Kathleen is consistently at a lower elevation than Paul at any instant, until the end. This means she has converted her potential energy to kinetic energy earlier. Consequently, she is traveling faster than Paul for the whole trip, and because the distance is the same, Kathleen gets to the bottom first. EXERCISE C Two balls are released from the same height above the floor. Ball A falls freely through the air, whereas ball B slides on a curved frictionless track to the floor. How do the speeds of the balls compare when they reach the floor?

@ PROBLEM SOLVING " Use energy, or Newton’s laws? @

PHYSICS

APPLIED Sports

FIGURE 8-10 Transformation of energy during a pole vault.

You may wonder sometimes whether to approach a problem using work and energy, or instead to use Newton’s laws. As a rough guideline, if the force(s) involved are constant, either approach may succeed. If the forces are not constant, and/or the path is not simple, energy is probably the better approach. There are many interesting examples of the conservation of energy in sport “ such as the pole vault illustrated in Fig. 8—10. We often have to make approximations, but the sequence of events in broad outline for the pole vault is as follows. The

initial kinetic energy of the running athlete is transformed into elastic potential energy of the bending pole and, as the athlete leaves the ground, into gravitational

potential energy. When the vaulter reaches the top and the pole has straightened out again, the energy has all been transformed into gravitational potential energy (if we ignore the vaulter’s low horizontal speed over the bar). The pole does not supply any energy, but it acts as a device to store energy and thus aid in the transformation of kinetic energy into gravitational potential energy, which is the net result. The energy required to pass over the bar depends on how high the center of mass (cM) of the vaulter must be raised. By bending their bodies, pole vaulters keep their cM so low that it can actually pass slightly beneath the bar (Fig. 8-11), thus enabling them to cross over a higher bar than would otherwise be possible. (The cM is where we can treat the object as a “point” for translational motion, as discussed in Chapter 9.)

FIGURE 8-11 By bending their bodies, pole vaulters can keep their center of mass so low that it may even pass below the bar.

192

CHAPTER 8

Conservation of Energy

EXAMPLE 8-6

Pole vault. Estimate the kinetic energy and the

speed required for a 70-kg pole vaulter to just pass over a bar 5.0 m high. Assume the vaulter’s center of mass is initially 0.90m off the ground and reaches its Amaximum height at the level of the bar itself. APPROACH We equate the total energy just before the vaulter places the end of the pole qnto the ground (and the pole beglns to bend and store potential energy) with the vaulter’s total energy when passing over the bar (we ignore the small amount of kinetic energy at this point). We choose the initial position of the vaulter’s center of mass to be y; = 0. The vaulter’s body must then be raised to a height y, = 50m — 09m = 4.1m.

SOLUTION We use Eq. 8-12, ‘%mv%+0 = 0 + mgy, SO

‘

Ky

= jmv?

The speed is

= mgy, = (70kg)(9.8m/s*)(41m)

/2K1

/ 2(28007) 70k

=

~ ~

89m/s

= 2.8 X 10°J.

9m/s.

NOTE This is an approx1mat10n because we have ignored such things as the vaulter’s speed while crossing over the bar, and mechanical energy transformed when the pole is planted in the ground. All would increase the needed initial kinetic energy. As another example of the conservation of mechanical energy, let us consider an object of mass m connected to a compressed horizontal spring (Fig. 8—6) whose own mass can be neglected and whose spring constant is k. After the spring is released, the mass m has speed v at any moment. The potential energy of the system (object plus spring) is 3 kx?, where x is the displacement of the spring from its unstretched length. If neither friction nor any other force is acting, conservation ‘f mechanical energy tells us that

tmvl + 3kx? = imod + Lkad,

[elastic PE only]

(8-13)

where the slbscrlpts 1 and 2 refer to the velocity and displacement at two different moments.

Toy dart gun. A dart of mass 0.100kg is pressed against the

spring of a toy dart gun as shown in Fig. 8—12a. The spring (with spring stiffness constant k¥ = 250 N/m and ignorable mass) is compressed 6.0 cm and released. If the dart detaches from the spring when the spring reaches its natural length (x = 0), what speed does the dart acquire? APPROACH The dart is initially at rest (point 1), so K; = 0. We ignore friction and use co}nservation of mechanical energy; the only potential energy is elastic. SOLUTION We use Eq. 8-13 with point 1 being at the maximum compression of the springl so v; = 0 (dart not yet released) and x; = —0.060m. Point 2 we choose to be the instant the dart flies off the end of the spring (Fig. 8-12b), so x, = 0 and we want to find v,. Thus Eq. 8-13 can be written

\

Then

0+ 1kx?

and

v, A

2

=

(a) E=1 kx?

= imid + 0. U

kx? ——-ml

=

Example 8-7. FIGURE 8-12 (a) A dart is pushed against a spring, compressing it 6.0 cm. The dart is then released, and in (b) it leaves the spring at velocity v,.

\/(250N/m)(—0.060rn)2

(0.100kg)

=

20 3

s

:

K

(b) E= mv~

NOTE In the horizontal direction, the only force on the dart (neglecting friction) was the force exerted by the spring. Vertically, gravity was counterbalanced by ‘the normal force exerted on the dart by the gun barrel. After it leaves the barrel, w dart will follow a projectile’s path under gravity.

SECTION 8-4

Problem Solving Using Conservation of Mechanical Energy

193

f

L

‘ h

@

] h

(&)

FIGURE 8-13

.

Example 8-8.

Two kinds of potential energy. A ball of mass m = 2.60 kg,

starting from rest, falls a vertical distance & = 55.0cm before striking a vertical

coiled spring, which it compresses an amount Y = 15.0cm (Fig. 8-13). Determine the spring constant k of the spring. Assume the spring has negligible mass, and ignore air resistance. Measure all distances from the point where the ball first touches the uncompressed spring (let y = 0 at this point). APPROACH The forces acting on the ball are the gravitational pull of the Earth and the elastic force exerted by the spring. Both forces are conservative, so we can use conservation of mechanical energy, including both types of potential energy. We must be careful, however: gravity acts throughout the fall (Fig. 8-13), whereas the elastic force does not act until the ball touches the spring (Fig. 8—13b). We choose y positive upward, and y = 0 at the end of the spring in its natural (uncompressed) state. SOLUTION We divide this solution into two parts. (An alternate solution follows.) Part I: Let us first consider the energy changes as the ball falls from a height y = h = 055m, Fig. 8-13a,to y, = 0, just as it touches the spring, Fig. 8-13b. Our system is the ball acted on by gravity plus the spring (which up to this point doesn’t do anything). Thus

jmvt + mgy, = 3mvi + mgy, 0

We

+ mgh

solve for

Imvi +

0.

v, = V2gh =1/2(9.80m/s?*)(0.550 m) = 3.283m/s ~ 3.28 m/s.

This is the speed of the ball just as it touches the top of the spring, Fig. 8-13b. Part 2: As the ball compresses the spring, Figs. 8-13b to c, there are two conservative forces on the ball—gravity and the spring force. So our conservation of energy equation is E, (ball touches spring)

=

E; (spring compressed)

Imvd + mgy, + tkyl = jmv} + mgys + 1ky3. Substituting y, = 0, v, = 3.283 m/s, v; = 0 (the ball comes to rest for an instant), and y; = =Y = —0.150m, we have

Im3 +0+0

=0-

mgY + 1k(-Y)>

We know m, v,, and Y, so we can solve for k: 2

m

k = W[%mv% + mgY| (2.60kg)

- m[(s.m :

“JPROBLEM

SOLVING Alternate Solution

= W[v% + 2gY] m/s)? + 2(9.80 m/s?)(0.150m)|

= 1590N/m.

Alternate Solution Instead of dividing the solution into two parts, we can do it all at once. After all, we get to choose what two points are used on the left and right of the energy equation. Let us write the energy equation for points 1 and 3 in Fig. 8-13. Point 1 is the initial point just before the ball starts to fall (Fig. 8-13a), so v, = 0, and y, = h = 0.550m. Point 3 is when the spring is fully compressed (Fig. 8-13c),s0 v; =0, y; = =Y = —0.150m. The forces on the ball in this process are gravity and (at least part of the time) the spring. So conservation of energy tells us

bmot + mgy, + 3k(0)* = 3mv} + mgys + 1ky3 0

+mgh+

0

=

0

- mgY + 3kY?

where we have set y = 0 for the spring at point 1 because it is not acting and is not compressed or stretched. We solve for k:

o

2l +Y) | 2(2.60kg)(9.80m/s%)(0.550 2260kg)(080m/) 053 m Y?

just as in our first method of solution.

194

CHAPTER 8

Conservation of Energy

(0.150 m)

+ 0.150m

S

-

DGV

r

BN

A

swinging

pendulum. The

simple pendulum

shown

in .

Fig. 8-14 consists of a small bob of mass m suspended by a massless cord of length £. The bob is released (without a push) at f = 0, where the cord makes an angle 0 = 6, to the vertical. (a) Describe the motion of the bob in terms of kinetic energy and potential energy. Then determine the speed of the bob (b) as a function of position 6 as it swings back and forth, and (c) at the lowest point of the swingH (d) Find the tension in the cord, F;. Ignore friction and air resistance. APPROACH We use the law of conservation of mechanical energy (only the conservative force of gravity does work), except in () where we use Newton’s second law. SOLUTION (@) At the moment of release, the bob is at rest, so its kinetic energy K =0 the bob moves down, it loses potential energy and gains kinetic energy. At the IOTNeSt point its kinetic energy is a maximum and the potential energy

is a minimum. The bob continues its swing until it reaches an equal height and angle (6,) on the opposite side, at which point the potential energy is a maximum and K = 0. It continues the swinging motion as U — K — U and so on, can never go higher than 6 = + 6, (conservation of mechanical energy). (b) The cord is assumed to be massless, so we need to consider only the kinetic energy, and the gravitational potential energy. The bob has two acting on it at any moment: gravity, mg, and the force the cord exerts on

but it

piGuURE 8-14

Example 8-9: a

gimple pendulum; y is measured positive upward.

bob’s forces it, Fy.

The latter (a constraint force) always acts perpendicular to the motion, so it does

no work. We need be concerned only with gravity, for which we can write the potential energy. The mechanical energy of the system is

E

= 1mv* + mgy,

where y is the vertical height of the bob at any moment. We take lowest point of the bob’s swing. Hence at ¢ = 0,

y =y

= € — Lcosfy

y = 0 at the

= {1 — cosby)

as can be seen from the diagram. At the moment of release

-

E

= mgy,,

E

= imv* + mgy

since v = vy = 0. At any other point along the swing We solve this for v:

v = V2w

= mgy,.

- ).

In terms of the angle 6 of the cord, we can write

SO

(=

y) = (£ — LcosBy) — (£ — Lcosh)

v =

\/ZgE(cos 6—cos 6).

= £(cos® — cosby)

(c) At the lowest point, y = 0, so or

v = /28y v = \V/2gl(1 — coséy).

(d) The tension in the cord is the force F; that the cord exerts on the bob. As

we've seen, there is no work done by this force, but we can calculate the force simply byrusing Newton’s second law F = ma and by noting that at any point the acceleration of the bob in the inward radial direction is v*/£, since the bob is constrained to move in an arc of a circle of radius £. In the radial direction, F; acts inward, and a component of gravity equal to mg cos 6 acts outward. Hence ,UZ

my We solve

r_

=

Fr — mgcos#.

for Fr and use the result of part (b) for v* ?)2

L

Fr = m(—e— + gcose)

= 2mg(cos® — cosby) + mg cos b

(3cos® — 2cosby)mg. SECTION 8-4

Problem Solving Using Conservation of Mechanical Energy

195

8—5 The Law of Conservation of Energy

FIGURE 8-15 The burning of fuel (a chemical reaction) releases energy to boil water in this steam

engine. The steam produced expands against a piston to do work in

turning the wheels.

We now take into account nonconservative forces such as friction, since they are important in real situations. For example, consider again the roller-coaster car inem, Fig. 8-8, but this time let us include friction. The car will not in this case reach the same height on the second hill as it had on the first hill because of friction. In this, and in other natural processes, the mechanical energy (sum of the kinetic and potential energies) does not remain constant but decreases. Because frictional forces reduce the mechanical energy (but not the total energy), they are called dissipative forces. Historically, the presence of dissipative forces hindered the formulation of a comprehensive conservation of energy law until well into the nineteenth century. It was not until then that heat, which is always produced when there is friction (try rubbing your hands together), was interpreted in terms of energy. Quantitative studies in the nineteenth-century (Chapter 19) demonstrated that if heat is considered as a transfer of energy (sometimes called thermal energy), then the total energy is conserved in any process. A .bl'o.ck sl.idin'g

freely agross a table, for exa.mple, comes to rest because of friction. Its initial quetlc

energy is all transformed into thermal energy. The block and table are a little

warmer as a result of this process. According to the atomic theory, thermal energy represents kinetic energy of rapidly moving molecules. We shall see in Chapter 18 that a rise in temperature corresponds to an increase in the average kinetic energy of the molecules. Because thermal energy represents the energy of atoms and molecules that make up an object, it is often called internal energy. Internal energy, from the atomic point of view, includes kinetic energy of molecules as well as potential energy (usually electrical in nature) due to the relative positions of atoms within molecules. For example the energy stored in food, or in a fuel such as gasoline, is regarded as electric potential energy (chemical bonds). The energy is released through chemical reactions (Fig. 8-15). This is analogous to a compressed spring which, when released, can do work. -~ To establish the more general law of conservation of energy, nineteenth-century physicists had to recognize electrical, chemical, and other forms of energy in addition to heat, and to explore if in fact they could fit into a conservation law. For each type of force, conservative or nonconservative, it has always been possible to define a type of energy that corresponds to the work done by such a force. And it has been found experimentally that the total energy E always remains constant. That is, the change in total energy, kinetic plus potential plus all other forms of energy, equals zero:

LAWOF CONg?}fiErlég};l,

AK

+ AU + [change in all other forms of energy]

=

0.

(8-14)

This is one of the most important principles in physics. It is called the law of conservation of energy and can be stated as follows: LAweF

CONgfi%‘}/\’/“E%g})\l,

The total energy is neither increased nor decreased in any process. Energy can

be transformed from one form to another, and transferred from one object to another, but the total amount remains constant.

For conservative mechanical systems, this law can be derived from Newton’s laws (Section 8-3) and thus is equivalent to them. But in its full generality, the validity of the law of conservation of energy rests on experimental observation. Even though Newton’s laws have been found to fail in the submicroscopic world of the atom, the law of conservation of energy has been found to hold there and in every experimental situation so far tested.

8—6

196

CHAPTER 8

Energy Conservation with Dissipative Forces: Solving Problems

In Section 8—4 we discussed several Examples of the law of conservation of energy-m, for conservative systems. Now let us consider in detail some examples that involve nonconservative forces. Suppose, for example, that the roller-coaster car rolling on the hills of Fig. 8-8 is subject to frictional forces. In going from some point 1 to a second point 2, the

energy dissipated by the friction force F}; acting on the car (treating it as a particle) is If F,, is constant in magnitude, Wy, = — F;,£, where £ is the actual distance f12 Ffr » di along the path traveled by the car from point 1 to point 2. (The minus sign appears because K, and df are in opposite directions.) From the work-energy principle (Eq. 7-11), rthe net work W,; done on an object is equal to the change in its kinetic energy: AK

=

u/nct'

We can separate W, into two parts:

Whet = We + Wyc, where W is the work done by conservative forces (gravity for our car) and Wyc is the work done by nonconservative forces (friction for our car). We saw in Eq. 84 that the work done by a conservative force can be written in terms of potential energy:

We Thus

AK

s

=

Wnet

2

JF‘-df

=

1

means

AK + AU

AK

=

=

—AU. WC

=

WNC

=

—AU

+

WNC'

Hence

Wye.

(8-15a)

This equa ion represents the general form of the work-energy principle. It also represents the conservation of energy. For our car, Wyc is the work done by the friction force and represents production of thermal energy. Equation 8-15a is valid in general. Based on our derivation from the work-energy principle, Wyc on the right side of Eq. 8-15a must be the total work done by all forces that are not included "J the potential energy term, AU, on the left side. Let u rewrite Eq. 8-15a for our roller-coaster car example of Fig. 8-8, setting W+C = —F; £ as we discussed above: W, ne or

r

=

—Fgxl

=

AK

+ AU

=

(%mv%

b‘ + gy, = dmid 4 mgy, + Rt

- %mv%)

+

(mgy2

= mgyl)

| ER ity Gand | g g

On the le t we have the mechanical energy of the system initially. It equals the mechanica energy at any subsequent point along the path plus the amount of

thermal (or internal) energy produced in the process. Total energy is conserved. Other ‘nonconservative forces can be treated similarly. If you are not sure about the sign of the extra term ([ F - dl), use your intuition: is the mechanical energy increased or decreased in the process.

EXERCISE D Return to the Chapter-Opening Question, page 183, and answer it again now. Try to exp lain why you may have answered differently the first time.

Work-Energy versus Energy Conservation The law of conservation of energy is more general and more powerful than the work-

energy principle. Indeed, the work-energy principle should not be viewed as a statement of conservation of energy. It is nonetheless useful for some mechanical problems; and whether yo u use it, or use the more powerful conservation of energy, can depend on your choice of the system under study. If you choose as your system a particle or rigid object on which external forces do work, then you can use the work-energy principle: the work done by the external forces on your object equals the change in its kinetic energy. On the other hand, if you choose a system on which no external forces do work, then you need to apply conservation of energy to that system directly. Consider, for example, a spring connected to a block on a frictionless table (Fig. 8-16) If you choose the block as your system, then the work done on the block by the spring equals the change in kinetic energy of the block: the work-energy principle. Energy conservation does not apply to this system—the block’s energy changes.) [f instead you choose the block plus the spring as your system, no external forces do work (since the spring is part of the chosen system). To this system you need to apply conservation of energy: if you compress the spring and then release it, the spring still exerts a force on the block, but the subsequent motion

can be discussed in terms of kinetic energy (3mv*) plus potential energy (3kx?), whose total remains constant.

FIGURE 8-16 A spring connected to a block on a frictionless table. If you choose your system to be the block plus spring, then E

=

1mv® + Lka?

is conserved.

AV k

SECTION 8-6

197

Problem solving is not a process that can be done by simply following a set of rules. The following Problem Solving Strategy, like all others, is thus not a prescription, but is a summary to help you get started solving problems involving energy.

&o ~

@ ) o« ¢

sOLVs,,

&

Conservation

of Energy

L Draw a picture of the physical situation. 2. Determine the system for which you will apply energy conservation: the object or objects and the forces acting. . Ask yourself what quantity you are looking for, and choose initial (point 1) and final (point 2) positions. . If the object under investigation changes its height during the problem, then choose a reference frame with a convenient y = 0 level for gravitational potential energy; the lowest point in the problem is often a good choice.

If springs are involved, choose the unstretched

spring position to be x (or y) = 0.

FIGURE 8-17 Example 8-10. Because of friction, a roller-coaster car does not reach the original height on the second hill. (Not to scale)

DN

A

BRI

5. Is mechanical energy conserved? If no friction or other nonconservative forces act, then conservation of mechanical energy holds: K1+U1=K2+U2.

. Apply conservation of energy. If friction (or other nonconservative forces) are present, then an additional term of the form [ F + d€ will be needed. For a constant friction force acting over a distance £ K1+Ul

=

K2+U2+Ff,f.

For other nonconservative forces use your intuition for the sign of [ F - di: is the total mechanical energy increased or decreased in the process?

. Use the equation(s) you develop to solve for the unknown quantity.

ESTIMATE

| Friction on the

roller-coaster car. The roller-

coaster car in Example 8-4 reaches a vertical height of only 25 m on the second hill before coming to a momentary stop (Fig. 8-17). It traveled a total distance of ™y 400 m. Determine the thermal energy produced and estimate the average friction force (assume it is roughly constant) on the car, whose mass is 1000 kg. APPROACH We explicitly SOLUTION 1. Draw a picture. See Fig. 2. The system. The system the gravitational force).

follow the Problem Solving Strategy above. 8-17. is the roller-coaster car and the Earth (which exerts The forces acting on the car are gravity and friction.

(The normal force also acts on the car, but does no work, so it does not affect

the energy.)

3. Choose initial and final positions. We take point 1 to be the instant when the car started coasting (at the top of the first hill), and point 2 stopped 25 m up the second hill. 4. Choose a reference frame. We choose the lowest point in y = 0 for the gravitational potential energy. 5. Is mechanical energy conserved? No. Friction is present. 6. Apply conservation of energy. There is friction acting on conservation of energy in the form of Eq. 8-15, with v, =0, y, =25m, and £ = 400 m. Thus

0 + (1000kg)(9.8m/s?)(40m)

to be the instant it the motion to be

the car, so we use »; =0, y; = 40m,

= 0 + (1000kg)(9.8 m/s?)(25m)

+ FyL.

7. Solve. We solve the above equation for Fi,£, the energy dissipated to thermal

energy:

Fyf = (1000kg)(9.8m/s?)(40m — 25m) = 147,000].

The average

force of friction was Fj; = (1.47 X 10°J)/400m = 370N. [This result is only a rough average: the friction force at various points depends on the normal force, which varies with slope. |

198

CHAPTER 8

>

Conservation of Energy

Friction with a spring. A block of mass m sliding along a rough horizontal surface is traveling at a speed v, when it strikes a massless spring head-on (see Fig. 8-18) and compresses the spring a maximum distance X.

rl If the spring has stiffness constant , determine the coefficient of kinetic friction between tTlock and surface.

g o n ok = j\{’%’wwfiyfiwp\w&y}w i J’ \ (a)

APPROACH At the moment of collision, the block has K = 3mv} and the spring is presumably uncompressed, so U = 0. Initially the mechanical energy of the syste% is 1mv}. By the time the spring reaches maximum compression,

K=0

ad‘d U = 1kX2 |

i

In the meantime, the friction force (= uyFy = pxmg)

has transformed energy

Fi X = uxmgX

b

;

:

into thermal energy.

3

¥

.

MMMV

SOLUTION Conservation of energy allows us to write |

energy (initial)

=

energy (final)

Imv

=

3kX*

:

_

%

g

VVMUA MUV

WMMWWWW'MMfl (b)

+ wemgX.

FIGURE 8-18

We solve for uy and find Hx

ko

E xamplele 8-11.

kX

2¢X

2mg

8-7 Gravitational Potential Energy and hiscape Velocity We have been dealing with gravitational potential energy so far in this Chapter assuming the force of gravity is constant, F = mg. This is an accurate assumption for ordinary objects near the surface of the Earth. But to deal with gravity more generally, for points not close to the Earth’s surface, we must consider that the avitationll force exerted by the Earth on a particle of mass m decreases inversely as the square of the distance r from the Earth’s center. The precise relationship is given by Newton’s law of universal gravitation (Sections 6-1 and 6-2): F

=

—GmAZ/IEf' r

[r>rE]

where Mg {s the mass of the Earth and r is a unit vector (at the position of m) directed raflially away from the Earth’s center. The minus sign indicates that the force on m is directed toward the Earth’s center, in the direction opposite to . This ~ FIGURE 8-19 Arbitrary path of equation can also be used to describe the gravitational force on a mass m in the particle of mass m moving from vicinity of ofiher heavenly bodies, such as the Moon, a planet, or the Sun, in which case ~ point 1 to point 2. Mg must be replaced by that body’s mass. A Supposb an object of mass m moves from one position to another along an y S arbitrary plath (Fig. 8-19) so that its distance from the Earth’s center changes d,{ from r to H%. The work done by the gravitational force is 2

2.4

T}l

W=JF-d?=-GmMEJrr2dl’ 1

i

1

where dl (epresents an infinitesimal displacement. Since component of df along

or

(see Fig. 8-19), then

=

"GmME

W

=

GmME

GmME

r

n

1 !

£

W

r

5

=

GmME

1

-

r

-

-df = dr

7

7

is the

1

—

r

|

Because the value of the integral depends only on the position of the end points (r, and r,) and not on the path taken, the gravitational force is a conservative force.

SECTION 8-7

Gravitational Potential Energy and Escape Velocity

199

We can therefore use the concept of potential energy for the gravitational force. Since the change in potential energy is always defined (Section 8-2) as the negative of the work done by the force, we have AU U

01

: :

=

y

U, - U,

=

_

GmME

u(r)

FIGURE 8-20 Gravitational

Earth’s center. Valid only for points r > rg, the radius of the Earth.

GmME

(8-16)

n

=

-

GmM m

£+,

where C is a constant. It is usual to choose

potential energy plotted as a function of r, the distance from

+

rn

From Eq. 8-16 the potential energy at any distance » from the Earth’s center can be written:

| |

i

=

ur)y

C = 0, so that

GmMy, ————-

=

gravity

r

8-17)

(r > re)

With this choice for C, U = 0 at r = co. As an object approaches the Earth, its potential energy decreases and is always negative (Fig. 8-20).

Equation 8-16 reduces to Eq. 8-2, AU = mg(y, — y), for objects near the

surface of the Earth (see Problem 48). For a particle of mass m, which experiences only the force of the Earth’s gravity, the total energy is conserved because gravity is a conservative force. Therefore we can write -

n

Package

M

BYE _ constant,

St | Q

mME

- G

I\.)Sy—-

im?

dropped

n

from

high-speed

.

[gra"“y] (8-18) only rocket. A box of

empty film canisters is allowed to fall from a rocket traveling outward from Earth at a speed of 1800 m/s when 1600 km above the Earth’s surface. The package eventually falls to the Earth. Estimate its speed just before impact. Ignore air resistance.

APPROACH We use conservation of energy. The package initially has a speed relative to Earth equal to the speed of the rocket from which it falls. SOLUTION Conservation of energy in this case is expressed by Eq. 8-18: Imv

mME

- G

r

=

1

smv;

2

mME

— G

r

where v; = 1.80 X 10°m/s, r, = (1.60 X 10°m) + (6.38 X 10°m) = 7.98 X 10°m, and

r, = 6.38 X 10°m Dy

\/v] 2

(the radius of the Earth). We solve for v,:

1 ZGME(r1e

1 erz)

(1.80 x 103m/s)* — 2(6.67 x 107" N-m?/kg?)(5.98 X 10*kg)

"

1

798 X 10°m

B

1

6.38 X 10°m

)

5320 m/s.

NOTE In reality, the speed will be considerably less than this because of air resistance. Note, incidentally, that the direction of the velocity never entered into this calculation, and this is one of the advantages of the energy method. The rocket could have been heading away from the Earth, or toward it, or at some other angle, and the result would be the same. 200

CHAPTER 8

Conservation of Energy

v

Escape Velocity When an o bject is projected into the air from the Earth, it will return to Earth unless

its speed is very high. But if the speed is high enough, it will continue out into space never to return to Earth (barring other forces or collisions). The minimum initial r velocity needed to prevent an object from returning to the Earth is called the escape velocity from Earth, v... To determine v from the Earth’s surface (ignoring air resistance) we use Eq. 8-18 with v; = v and r; = ry = 6.38 X 10°m, the radius of the Earth. Since we want the minimum speed for escape, we need the object to reach r, = oo with merely zero speed, v, = 0. Applying Eq. 8-18 we have

M

Imiye = GTEe = 0+ 0

or

Vese = V/2GMg/re = 112 X 10*m/s

(8-19)

or 11.2km/s. It is important to note that although a mass can escape from the Earth (or s olar system) never to return, the force on it due to the Earth’s gravitational field is never actually zero for a finite value of r.

DGV

ESEN

Escaping the Earth or the Moon. (¢) Compare the escape

velocities

of a rocket from the Earth and from the Moon.

rM

X 10m,

energies required to launch the rockets. For the Moon, =

1.7

(b) Compare

the

My, = 7.35 X 102 kg and and for Earth, Mg = 5.98 X 10*kg and r; = 6.38 X 10°m.

APPROACH We use Eq. 8-19, replacing My and rg with My, and ry for finding Vege IO the Moon. SOLUTIO (a) Using Eq. 8-19, the ratio of the escape velocities is Vesc(Earth) vee(Moon)

/flr_M —

=

a7

N My re

To escape Earth requires a speed 4.7 times that required to escape the Moon. (‘

(b) The fuel that must be burned provides energy proportional to v? (K = mv?);

so to launch a rocket to escape Earth requires as to escape from the Moon.

8-8

(4.7) = 22 times as much energy

Power

Power is defined as the rate at which work is done. The average power, P, equals the work W done divided by the time ¢ it takes to do it: =P = = w (8-20a)

Since the work

done in a process involves the transformation of energy from one type (or object) to another, power can also be defined as the rate at which energy is transformed: B W energy transformed i

The in stantaneous power, P, is P

~

=

dw %

—_——

time (8-20b) —

The work done in a process is equal to the energy transferred from one object to another. For example, as the potential energy stored in the spring of Fig. 8-6c¢ is transformed to kinetic energy of the ball, the spring is doing work on the ball. Similarly, when you throw a ball or push a grocery cart, whenever work is done, energy is being transferred from one body to another. Hence we can also say that power is the rate at which energy is transformed:

p = 9E T

(8-20c) 8-2

The power of a horse refers to how much work it can do per unit of time. SECTION 8-8

Power

201

The power rating of an engine refers to how much chemical or electrical energy can be transformed into mechanical energy per unit of time. In SI units, power is . measured in joules per second, and this unit is given a special name, the watt (W): 1W = 1J/s. We are most familiar with the watt for electrical devices: the rate at which an electric lightbulb or heater changes electric energy into light or therma®™y energy. But the watt is used for other types of energy transformation as well. In the British system, the unit of power is the foot-pound per second (ft-1b/s). For practical purposes a larger unit is often used, the horsepower. One horsepower’ (hp) is defined as 550ft-1b/s, which equals 746 watts. An engine’s power is usually

specified in hp or in kW (1kW ~ 1}hp).

To see the distinction between energy and power, consider the following example. A person is limited in the work he or she can do, not only by the total energy required, but also by how fast this energy is transformed: that is, by power. For example, a person may be able to walk a long distance or climb many flights of stairs before having to stop because so much energy has been expended. On the other hand, a person who runs very quickly up stairs may feel exhausted after only a flight or two. He or she is limited in this case by power, the rate at which his or her body can transform chemical energy into mechanical energy.

DOV

EAEN

Stair-climbing power. A 60-kg jogger runs up a long flight

of stairs in 4.0 s (Fig. 8-21). The vertical height of the stairs is 4.5 m. (¢) Estimate the jogger’s power output in watts and horsepower. (b) How much energy did this require? APPROACH The work done by the jogger is against gravity, and W = mgy. To find her average output, we divide W by the time it took. SOLUTION (a) The average power output was

oW

_ mey t

FIGURE 8-21

@

Example 8-14.

PHYSICS APPLIED Power needs of a car

2 _ (60 kg)(9.8 m/s?)(4.5m)

t

equals

.

4.0s

Since there are 746 W in 1 hp, the jogger is doing work at a rate of just under 1 hpfl A human cannot do work at this rate for very long. (b) The energy required is E = Pt = (660J/s)(4.0s) = 2600J. This result equals W = mgy. NOTE The person had to transform more energy than this 2600 J. The total energy transformed by a person or an engine always includes some thermal energy (recall how hot you get running up stairs). Automobiles do work to overcome the force of friction (and air resistance), to climb hills, and to accelerate. A car is limited by the rate it can do work, which is why automobile engines are rated in horsepower. A car needs power most when it is climbing hills and when accelerating. In the next Example, we will calculate how much power is needed in these situations for a car of reasonable size. Even when a car travels on a level road at constant speed, it needs some power just to do work to overcome the retarding forces of internal friction and air resistance. These forces depend on the conditions and speed of the car, but are typically in the range 400 N to 1000 N.

.

It is often convenient to write the power in terms of the net force F applied to an object and its velocity v. Since P = dW/dt and dW =F - dl (Eq. 7-7), then

aw _g.d _ g, dt

P=a

(8-21)

*The unit was chosen by James Watt (1736-1819), who needed a way to specify the power of his newly developed steam engines. He found by experiment that a good horse can work all day at an averag rate of about 360 ft-1b/s. So as not to be accused of exaggeration in the sale of his steam engines, he

multiplied this by roughly 13 when he defined the hp.

202

CHAPTER 8

Conservation of Energy

Power needs of a car. Calculate the power required of a

1400-kg car under the following circumstances: (a) the car climbs a 10° hill (a fairly steep hill) at a steady 80 km/h; and (b) the car accelerates along a level road from 90 to 110km/h in 6.0s to pass another car. Assume that the average r‘ retarding force on the car is Fr = 700N throughout. See Fig. 8-22. APPROACH First we must be careful not to confuse F, which is due to air resistance and‘ friction that retards the motion, with the force F needed to accelerate the car, whlch is the frictional force exerted by the road on the tires—the reaction to the motor-driven tires pushing against the road. We must determine the latter force F before calculating the power. SOLUTIO! (a) To move at a steady speed up the hill, the car must, by Newton’s second law, exert a force F equal to the sum of the retarding force, 700 N, and the component of gravity parallel to the hill, mg sin 10°. Thus F

=

700N

+ mgsin10°

=

700N

+ (1400kg)(9.80m/s?)(0.174)

Since v = 80km/h = 22m/s

P =Fp

=

=

Fy

mg sin 10°

FIGURE 8-22 Example 8-15: Calculation of power needed for a

car to climb a hill.

3100N.

and is parallel to F, then (Eq. 8-21) the power is

(3100N)(22m/s)

= 6.80 X 10'W

= 68.0kW

= 91hp.

(b) The cér accelerates from 25.0 m/s to 30.6 m/s (90 to 110 km/h). Thus the car must exert a force that overcomes the 700-N retarding force plus that required to give it the acceleration |

=

(30.6 m/sfi; S25.0 m/s)

=

B

We apply Newton’s second law with x being the direction of motion: ma,

=

ZF,

=

F — Fy.

We solve for the force required, F: f/

F

=

|-

ma,

+ Fy

= (1400 kg)(0.93m/s?) + 700N

= 1300N + 700N

= 2000 N.

Since P T F - ¥, the required power increases with speed and the motor must be able to provide a maximum power output of

P = (2000N)(30.6m/s)

= 6.12 X 10°'W = 612kW

= 82hp.

NOTE Even taking into account the fact that only 60 to 80% of the engine’s power output reaches the wheels, it is clear from these calculations that an engine of 75 to 100 kW (100 to 130 hp) is adequate from a practical point of view. We mentioned in the Example above that only part of the energy output of a car englne reaches the wheels. Not only is some energy wasted in getting from the engine to the wheels, in the engine itself much of the input energy (from the gasoline) does not end up doing useful work. An important characteristic of all engines is their overall efficiency e, defined as the ratio of the useful power output of the engine, Fou, to the power input, Py: } \

e

= —

F out ]

Py,

The efficiefcy is always less than 1.0 because no engine can create energy, and in fact, cannot even transform energy from one form to another without some going to frictionithermal energy, and other nonuseful forms of energy. For example, an automobile engine converts chemical energy released in the burning of gasoline into mechanical energy that moves the pistons and eventually the wheels. But fnearly 85% of the input energy is “wasted” as thermal energy that goes into the cooling system or out the exhaust pipe, plus friction in the moving parts. Thus car engines are roughly onl‘y about 15% efficient. We discuss efficiency in detail in Chapter 20. SECTION 8-8

Power

203

“8-9 Ux) E, E

\ \ !

AN

A

E, : Ep 0—

XS

/

T NKZS menail L1

X4 .X.'3

FIGURE 8-23 diagram.

xO

x1x2

: : ; : Ly

x6

A potential energy

Potential Energy Diagrams;

Stable and Unstable Equilibrium

If only conservative forces do work on an object, we can learn a great deal™™ about its motion simply by examining a potential energy diagram—the graph ofU(x) versus x. An example of a potential energy diagram is shown in Fig. 8-23. The rather complex curve represents some complicated potential energy U(x). The total energy E = K + U is constant and can be represented as a horizontal line on this graph. Four different possible values for E are shown, labeled Ey, E;, E,, and E;. What the actual value of E will be for a given system depends on the initial conditions. (For example, the total energy E of a mass oscillating on the end of aspring depends on the amount the spring is initially compressed or stretched.) Kinetic energy K = mv? 2 cannot be less than zero (v would be imaginary), and because E = U + K = constant, then U(x) must be less than or equal to E for all situations: U(x) = E. Thus the minimum value which the total energy can take for the potential energy shown in Fig. 8-23 is that labeled E,. For this value of E, the mass can only be at rest at x = x,. The system has potential energy but no kinetic energy at this position. If the system’s total energy E is greater than Ey, say it is £; on our plot, the system can have both kinetic and potential energy. Because energy is conserved,

:

K

:

= E - U(x).

Since the curve represents U(x) at each x, the kinetic energy at any value of x is represented by the distance between the E line and the curve U(x) at that value of x. In the diagram, the kinetic energy for an object at x;, when its total energy is Ey, is indicated by the notation K. An object with energy E, can oscillate only between the points x, and x3. This is because if x > x, or x < x3, the potential energy would be greater than E,

meaning

K = }mv’ < 0 and v would be imaginary, and so impossible. At x, andfl

x5 the velocity is zero, since E = U at these points. Hence x; and x; are called the. turning points of the motion. If the object is at x,, say, moving to the right, its kinetic energy (and speed) decreases until it reaches zero at x = x,. The object then reverses direction, proceeding to the left and increasing in speed until it passes x, again. It continues to move, decreasing in speed until it reaches x = x3, where again v = 0, and the object again reverses direction. If the object has energy E = E, in Fig. 8-23, there are four turning points. The object can move in only one of the two potential energy “valleys,” depending on where it is initially. It cannot get from one valley to the other because of the barrier between them—for example at a point such as x,,U > E,, which means v would be imaginary.” For energy E;, there is only one turning point since U(x) < E; for all x > x5. Thus our object, if moving initially to the left, varies in speed as it passes the potential valleys but eventually stops and turns around at x = xs. It then proceeds to the right indefinitely, never to return. How do we know the object reverses direction at the turning points? Because of the force exerted on it. The force F is related to the potential energy U by Eq. 8-7, F = —dU/dx. The force F is equal to the negative of the slope of the U-versus-x curve at any point x. At x = x,, for example, the slope is positive so the force is negative, which means it acts to the left (toward decreasing values of x). At x = x, the slope is zero,so F = 0. At such a point the particle is said to be in equilibrium. This term means simply that the net force on the object is zero. Hence, its acceleration is zero, and so if it is initially at rest, it remains at rest. If the object at rest at x = x, were moved slightly to the left or right, a nonzero force would act on it in the direction to move it back toward x,. An object that returns

tAlthough this is true according to Newtonian physics, modern quantum mechanics predicts thatfl objects can “tunnel” through such a barrier, and such processes have been observed at the atomic and subatomic level.

204

CHAPTER 8

Conservation of Energy

toward its equilibrium point when displaced slightly is said to be at a point of stable equilibrium. Any minimum in the potential energy curve represents a point of stable equilibrium. An object at x = x;, would also be in equilibrium, since F = —dU/dx = 0. the object were displaced a bit to either side of x4, a force would act to pull the ubject away from the equilibrium point. Points like x4, where the potential energy curve has a maximum, are points of unstable equilibrium. The object will not return to equilibrium if displaced slightly, but instead will move farther away. When an object is in a region over which U is constant, such as near x = x4 in Fig. 8-23, the force is zero over some distance. The object is in equilibrium and if displaced slightly to one side the force is still zero. The object is said to be in neutral equilibrium in this region.

| Summary A conservative force is one for which the work done by the force in moving an object from one position to another depends only on the two positions and not on the path taken. The work

done by a conservative force is recoverable, which is not true for

nonconservative forces, such as friction. Potential energy, U, is energy associated with conservative forces that depend on the position or configuration of objects. Gravitational potential energy is Ugrav

=

mgy,

8-3)

where the mass m is near the Earth’s surface, a height y above

some reference point. Elastic potential energy is given by

‘ Uy

©

= L1kx?

Jemical, ele&trical, and nuclear energy.

Potential energy is always associated with a conservative the

change

in potential

energy,

AU,

between

-U

2

= —JF-di. 1

(8-4)

Inversely, we can write, for the one-dimensional case,

‘

F

=

—m

dU(x) dx

conservative

forces

act, the

total mechanical

=

K

+ U

=

constant.

(8-10)

If nonconservative forces also act, additional types of energy are involved, such as thermal energy. It has been found experimentally that, when all forms of energy are included, the total energy is conserved. This is the law of conservation of energy: AK

+ AU

+ A(other energy types)

=

0.

(8-14)

The gravitational force as described by Newton’s law of universal gravitation is a conservative force. The potential energy of an object of mass m due to the gravitational force exerted on it by the Earth is given by

u(r)

=

GmME r

P or

dW

8-17)

9

where M is the mass of the Earth object from the Earth’s center (r = Power is defined as the rate at rate at which energy is transformed

8-7)

Only changes in potential energy are physically meaningful, so the position where U = 0 can be chosen for convenience.

only

E

two

points under the action of a conservative force F is defined as the negative of the work done by the force:

AU =,

When

energy, E, defined as the sum of kinetic and potential energies, is conserved:

(8-5)

for a spring with stiffness constant k stretched or compressed a ‘splacement x from equilibrium. Other potential energies include force, and

Potential energy is not a property of an object but is associated with the interaction of two or more objects.

and r is the distance of the radius of Earth). which work is done or the from one form to another:

dE o

(8-20)

.

P =F-v

(8-21)

| Questions 1. List some

everyday forces that are not conservative, and

explain wby they aren’t.

2. You lift a heavy book from a table to a high shelf. List the forces on the book

during this process, and state whether

each is conservative or nonconservative. 3. The net t’orce acting on a particle is conservative and increases ‘he kinetic energy by 300 J. What is the change in (a) the potential energy, and (b) the total energy, of the r particle? ‘ » When a “superball” is dropped, can it rebound to a greater height than its original height?

5. A hill has a height 4. A child on slides down starting from rest at the at the bottom depend on the angle icy and there is no friction, and (b) snow)? 6. Why is it tiring to push though no work is done?

hard

a sled (total mass m) top. Does the velocity of the hill if (a) it is there is friction (deep

against a solid wall even

7. Analyze the motion of a simple swinging pendulum in terms of energy, (a) ignoring friction, and (b) taking friction into account. Explain why a grandfather clock has to be wound up. Questions

205

8. Describe precisely what is “wrong” physically in the famous Escher drawing shown in Fig. 8-24.

17. Recall from Chapter 4, Example 4-14, that you can use a pulley and ropes to decrease the force needed to raise a heavy load (see Fig. 8-26). But for every meter

the load is raised, how

much rope must be pulled up? Account for this, using energy concepts.

FIGURE 8-26 Question 17.

18

Two identical arrows, one with twice the speed of the other,

are fired into a bale of hay. Assuming

the hay exerts a

constant “frictional” force on the arrows, the faster arrow

FIGURE 8-24 Question 8.

19

9. In Fig. 8-25, water balloons are tossed from the roof of a building, all with the same speed but with different launch angles. Which one has the highest speed when it hits the ground? Ignore air resistance.

will penetrate how much farther than the slower arrow? Explain. A bowling ball is hung from the ceiling by a steel wire (Fig. 8-27). The instructor pulls the ball back and stands against the wall with the ball against his nose. To avoid injury the instructor is supposed to release the ball without pushing it. Why? FIGURE 8-27

Question 19,

20. A pendulum is launched from a point that is a height H above its lowest point in two different ways (Fig. 8-2& During both launches, the pendulum is given an initial speed’ of 3.0m/s.

FIGURE 8-25

On

the

first launch,

Question 9.

10

11.

13.

14.

. 16.

206

A coil spring of mass m rests upright on a table. If you compress the spring by pressing down with your hand and then release it, can the spring leave the table? Explain using the law of conservation of energy. What happens to the gravitational potential energy when water at the top of a waterfall falls to the pool below? Experienced hikers prefer to step over a fallen log in their path rather than stepping on top and jumping down on the other side. Explain. (a) Where does the kinetic energy come from when a car accelerates uniformly starting from rest? (b) How is the increase in kinetic energy related to the friction force the road exerts on the tires? The Earth is closest to the Sun in winter (Northern Hemisphere). When is the gravitational potential energy the greatest? Can the total mechanical energy E = K + U ever be negative? Explain. Suppose that you wish to launch a rocket from the surface of the Earth so that it escapes the Earth’s gravitational field. You wish to use minimum fuel in doing this. From what point on the surface of the Earth should you make the launch and in what direction? Do the launch location and direction matter? Explain.

CHAPTER8

Conservation of Energy

;

]

h

O“i

/

¢

)

N3

Explain. p

&

AR

(Second launch)

21

of the

lowest point Y (First launch) ~ of itlS swing?

;

|

the initial velocity

pendulum is directed upward along the trajectory, and on the second launch it is directed downward along the trajectory. Which launch will cause the highest speed when the pendulum bob passes the

\

FIGURE 8-28 20 Gliientian

Describe the energy transformations when a child hops around on a pogo stick. 22, Describe the energy transformations that take place when a skier starts skiing down a hill, but after a time is brought to rest by striking a snowdrift. 23. Suppose you lift a suitcase from the floor to a table. The work you do on the suitcase depends on which of the following: (a) whether you lift it straight up or along a more complicated path, (b) the time the lifting takes, (c) the height of the table, and (d) the weight of the suitcase? Repeat Question 23 for the power needed instead of tL work.

25. Why is it easier to climb a mountain via a zigzag trail rather

than to climb straight up? *26. Figure 8-29 shows a potential energy curve, U(x). (a) At which point does the force have greatest magnitude? (b) For each labeled point, state whether the force acts to the left or r to the right, or is zero. (c) Where is U there

ehuilibrium

and of what type is

it?

B

vl

\

G

C

F /\

FIGURE 8-29

N\t

| Problems | 8-1 and 8-2 Conservative

Forces and Potential Energy

Qu‘;stion 26.

Explain why or the form of the . (ID) If = 3x? . (II) A particle along the x axis |

| P& (x)

(a)

Is

this

force

conservative?

why not. (b) If it is conservative, determine potential energy function. + 2xy + 4y%z, what is the force, F? is constrained to move in one dimension and is acted upon by a force given by

FIGURE 8-30

Question 28.

Asin(kx)i

8-3 and 8-4

Conservation of Mechanical Energy

11. (I) A novice skier, starting from rest, slides down a frictionless 13.0° incline whose vertical height is 125 m. How fast is she going when she reaches the bottom? 12, (I) Jane, looking for Tarzan, is running at top speed (5.0m/s) and grabs a vine hanging vertically from a tall tree in the jungle. How high can she swing upward? Does the length of the vine affect your answer? 13. (IT) In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole. With what minimum speed must the athlete leave the ground in order to lift his center of mass 2.10m and cross the bar with a speed of 0.70 m/s? 14. (IT) A sled is initially given a shove up a frictionless 23.0° incline. It reaches a maximum vertical height 1.12m higher than where it started. What was its initial speed? 15. (IT) A 55-kg bungee jumper leaps from a bridge. She is tied to a bungee cord that is 12m long when unstretched, and falls a total of 31 m. (a) Calculate the spring constant k of the bungee cord assuming Hooke’s law applies. (b) Calculate the maximum acceleration she experiences. 16. (II) A 72-kg trampoline artist jumps vertically upward from the top of a platform with a speed of 45m/s. (a) How fast is he going as he lands on the

trampoline,

2.0m

below

(Fig. 8-31)? (b) If the trampoline behaves like a spring of

spring constant 5.8 X 10 N/m, how far does he depress it?

FIGURE 8-31

= - Al =k

F(x) =

in Fig. 8-30.

D

Problem 16.

where k is a constant with units appropriate to the SI system. Find the potential energy function U(x), if U is arbitrarily defined to be zero at x = 2.0m, so that U(2.0m) = 0. 10. IIm A p‘article constrained to move in one dimension is subject to a force F(x) that varies with position x as r

energy F; in Fig. 8-23 as it moves from x¢ to x5 and back to xg. (b) Where is its kinetic energy the greatest and the least? *28. Name the type of equilibrium for each position of the balls

.

1. (1) A spring has a spring constant k of 82.0 N/m. How much must this spring be compressed to store 35.0J of potential energy? . (I) A 6.0-kg monkey swings from one branch to another 1.3 m higher. What is the change in gravitational potential energy? . (II) A spring with k = 63 N/m hangs vertically next to a ruler. The end of the spring is next to the 15-cm mark on the ruler. If a 2.5-kg mass is now attached to the end of the spring, where wil the end of the spring line up with the ruler marks? . (I) A 56.5-kg hiker starts at an elevation of 1270m and climbs to the top of a 2660-m peak. (a) What is the hiker’s change 131 potential energy? (b) What is the minimum work required of the hiker? (¢) Can the actual work done be greater than this? Explain. . (II) A 1.60-m tall person lifts a 1.95-kg book off the ground so it isJZ.ZOm above the ground. What is the potential energy of the book relative to (a) the ground, and (b) the top of the person’s head? (c¢) How is the work done by the person related to the answers in parts (a) and (b)? . (IT) A 1200-kg car moving on a horizontal surface has speed v = 75km/h when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.2 m. What is the spring stiffness constant of the spring? . (II) A particular a spring obeys the force law F =

(—kx + ;ax3 +bx*)i.

*27. (a) Describe in detail the velocity changes of a particle that has

~

where A and k are constants. What is the potential energy function U(x), if we take U = 0 at the point x = 0?

17. (II) The total energy E of an object of mass m that moves in one dimension under the influence of only conservative forces can be written as

E = %mvz + U. Use conservation of energy, dE/dt second law. 18. (IT) A 0.40-kg ball is thrown with upward angle of 36°. (a) What is its and (b) how high does it go? (Use

= 0, to predict Newton’s a speed of 8.5m/s at an speed at its highest point, conservation of energy.)

Problems

207

19. (II) A vertical spring (ignore its mass), whose spring constant is 875N/m, is attached to a table and is compressed down by 0.160 m. (a) What upward speed can it give to a 0.380-kg ball when released? (b) How high above its original position (spring compressed) will the ball fly? 20. (IT) A roller-coaster car shown in Fig. 8-32 is pulled up to point 1 where it is released from rest. Assuming no friction, calculate the speed at points 2, 3, and 4.

FIGURE 8-32 Problems 20 and 34.

21. (II) When a mass m sits at rest on a spring, the spring is compressed by a distance d from its undeformed length (Fig. 8-33a). Suppose instead that the mass is released from rest when it barely touches the undeformed

spring (Fig. 8-33b).

;At

Find the distance D that the spring is compressed before it is able to stop the

)Ungompressed

At rest

mass. Does D = d?

Spring

If not, why not?

FIGURE 8-33

Problem 21.

@)

'

rest

'

(b)

22, (II) Two masses are connected by a string as shown in Fig. 8-34. Mass m = 4.0kg rests on a frictionless inclined plane, while mp = 5.0kg is initially held at a height of h = 0.75m above the floor. (a) If my is allowed to fall, what will be the resulting acceleration of the masses? (b) If the masses were initially at rest, use the kinematic equations (Egs. 2-12) to find their velocity just before my hits the floor. (¢) Use conservation of energy to find the velocity of the masses just before mp hits the floor. You should get the same answer as in part (b).

. (II) A cyclist intends to cycle up a 9.50° hill whose vertical height is 125m. The pedals turn in a circle of diameter 36.0 cm. Assuming the mass of bicycle plus person is 75.0 kg, (a) calculate how much work must be done against gravity. (b) If each complete revolution of the pedals moves thg bike 5.10m along its path, calculate the average force th must be exerted on the pedals tangent to their circular path. Neglect work done by friction and other losses. 25, (II) A pendulum 2.00m long is released (from rest) at an angle 6y = 30.0° (Fig. 8-14). Determine the speed of the 70.0-g bob: (a) at the lowest point (6 = 0); (b) at 6 = 15.0°, (c) at & = —15.0° (i.e., on the opposite side). (d) Determine the tension in the cord at each of these three points. (e) If the bob is given an initial speed vy = 1.20m/s when released at 6 = 30.0°, recalculate the speeds for parts (a), (b), and (c). . (IT) What should be the spring constant k of a spring designed to bring a 1200-kg car to rest from a speed of 95km/h so that the occupants undergo a maximum acceleration of 5.0 g? 27. (III) An engineer is designing a spring to be placed at the bottom of an elevator shaft. If the elevator cable breaks when the elevator is at a height 4 above the top of the spring, calculate the value that the spring constant k should have so that passengers undergo an acceleration of no more than 5.0 g when brought to rest. Let M be the total mass of the elevator and passengers. 28. (II1) A skier of mass m starts from rest at the top of a solid sphere of radius r and slides down its frictionless surface. (a) At what angle 6 (Fig. 8-36) will the skier leave the sphere? (b) If friction were present, o would the skier fly off at a greater or lesser angle?

FIGURE 8-36 Problem 28.

8-5 and 8-6

Law of Conservation of Energy

29. (I) Two railroad cars, each of mass 56,000kg, are traveling 95 km/h toward each other. They collide head-on and come to rest. How much thermal energy is produced in this collision? 30. (I) A 16.0-kg child descends a slide 2.20 m high and reaches the bottom with a speed of 1.25m/s. How much thermal energy due to friction was generated in this process? 31 (II) A ski starts from rest and slides down a 28° incline 85 m

long. (a) If the coefficient of friction is 0.090, what is the ski’s

FIGURE 8-34 Problem 22.

23. (IT) A block of mass m is attached to the end of a spring (spring stiffness constant k), Fig. 8-35. The mass is given an initial displacement x, from equilibrium, and an initial speed vy. Ignoring friction and the mass of the spring, use energy methods to find (a) its maximum speed, and (b) its maximum stretch from equilibrium, in terms of the given quantities.

speed at the base of the incline? (b) If the snow is level at the foot of the incline and has the same coefficient of friction, how far will the ski travel along the level? Use energy methods. 32. (IT) A 145-g baseball is dropped from a tree 14.0m above the ground. (a) With what speed would it hit the ground if air resistance could be ignored? (b) If it actually hits the ground with a speed of 8.00 m/s, what is the average force of air resistance exerted on it? 33. (IT) A 96-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 350 N. For the first 15 m the floor is frictionless, and for the next 15 m the coefficient

FIGURE 8-35

208

CHAPTER 8

Problems 23,37, and 38.

Conservation of Energy

of friction is 0.25. What is the final speed . (I) Suppose the roller-coaster car in Fig. with a speed of 1.70 m/s. If the average equal to 0.23 of its weight, with what point 2? The distance traveled is 45.0 m.

of the crate? 8-32 passes point 1 force of friction speed will it rea.

3s. (IT) A skier traveling 9.0m/s reaches the foot of a steady upward 19° incline and glides 12 m up along this slope before coming to rest. What was the average coefficient of friction? . (II) Consider the track shown in Fig. 8-37. The section AB is one quadrant of a circle of radius 2.0 m and is frictionless. Bto Cis ‘a horizontal span 3.0m long with a coefficient of kinetic fn'c‘tion ux = 0.25. The section CD under the spring is frictionless. A block of mass 1.0 kg is released from rest at A. After sliding on the track, it compresses the spring by 0.20 m. D‘ termine: (a) the velocity of the block at point B; (b) the thermal energy produced as the block slides from B to C; (c) t}{e velocity of the block at point C; (d) the stiffness constant k for the spring. m=10kg

A

:

r=20m

B

‘

C

§°

FIGURE 8-37

Problem 36.

37. (I A 0.6%20-kg wood block is firmly attached to a very light horizontal spring (k = 180N/m) as shown in Fig. 8-35. This block-spring system, when compressed 5.0cm and released, stretches out 2.3 cm beyond the equilibrium position befo}‘e stopping and turning back. What is the coefficient of ki‘netic friction between the block and the table?

38. Im A 18?-g wood block is firmly attached to a very light horizontal spring, Fig. 8-35. The block can slide along a table where the coefficient of friction is 0.30. A force of 25 N compressés the spring 18 cm. If the spring is released from this positibn, how far beyond its equilibrium position will it stretch on its first cycle? . (IT) You drop a ball from a height of 2.0 m, and it bounces back toa heigh{ of 1.5m. (a) What fraction of its initial energy is lost during the bounce? (b) What is the ball’s speed just before and just after the bounce? (c) Where did the energy go? 40. (IT) A 56-kg skier starts from rest at the top of a 1200-mlong trail ‘which drops a total of 230m from top to bottom. At

the bottom,

the skier is moving

11.0m/s.

How

much

energy was dissipated by friction? 41. (IT) How much does your gravitational energy change when you jurnp‘as high as you can (say, 1.0m)? 42. (IIT) A spring (k = 75N/m) has an equilibrium length of 1.00 m. The spring is compressed to a length of 0.50 m and a mass of 2.0 kg is placed at its free end on a frictionless slope which makes an angle of 41° with respect to the horizontal (Fig. 8-38). The spring is then released. (a) If the mass is not attached to the spring, how far up the slope will the mass move before coming to rest? (b) If the mass is attached to the spring, how far up the slope will the mass move before coming to rest? (c¢) Now the incline has a coefficient of kinetic friction wy. If the block, attached to the spring, is observed to stop just as it reaches the spring’s equilibrium position, That is the coefficient of friction wy?

43. (IIT) A 2.0-kg block slides along a horizontal surface with a coefficient of kinetic friction ui = 0.30. The block has a speed v = 1.3m/s when it strikes a massless spring head-on (as in Fig. 8-18). (a) If the spring has force constant k = 120N/m, how far is the spring compressed? (b) What minimum value of the coefficient of static friction, ug, will assure that the spring remains compressed at the maximum compressed position? (c) If ug is less than this, what is the speed of the block when it detaches from the decompressing spring? [Hint: Detachment occurs when the spring reaches its natural length (x = 0); explain why.] . (IIT) Early test flights for the space shuttle used a “glider” (mass of 980 kg including pilot). After a horizontal launch at 480 km/h at a height of 3500 m, the glider eventually landed at a speed of 210 km/h. (a) What would its landing speed have been in the absence of air resistance? (b) What was the average force of air resistance exerted on it if it came in at a constant glide angle of 12° to the Earth’s surface?

8-7

Gravitational Potential Energy

45. (I) For a satellite of mass mg in a circular orbit of radius rg around the Earth, determine (a) its kinetic energy K, (b) its potential energy U (U = 0 at infinity), and (¢) the ratio K/U. 46. (I) Jill and her friends have built a small rocket that soon after lift-off reaches a speed of 850 m/s. How high above the Earth can it rise? Ignore air friction. 47. (I) The escape velocity from planet A is double that for planet B. The two planets have the same mass. What is the ratio of their radii, r5/rg? . (IT) Show that Eq. 8-16 for gravitational potential energy

reduces to Eq. 8-2, AU = mg(y, — y;), for objects near the surface of the Earth.

49. (II) Determine

object

(a)

at

the escape

the

Sun’s

velocity from the Sun for an

surface

(r = 7.0 X 10°km,

M =20 x 10*kg), and (b) at the average distance of the

Earth (1.50 X 108km). Compare to the speed of the Earth

in its orbit.

50. (IT) Two Earth satellites,A and B, each of mass m = 950 kg, are launched into circular orbits around the Earth’s center. Satellite A orbits at an altitude of 4200 km, and satellite B orbits at an altitude of 12,600 km. (a) What are the potential energies of the two satellites? (b) What are the kinetic energies of the two satellites? (¢) How much work would it require to change the orbit of satellite A to match that of satellite B? 51. (IT) Show that the escape velocity for any satellite in a circular orbit is V2 times its velocity. 52. (IT) (a) Show that the total mechanical energy of a satellite (mass m) orbiting at a distance r from the center of the Earth (mass Mg) is E

=

=

l GmME

2

r

)

if U =20 at r = co. (b) Show that although friction causes the value of E to decrease slowly, kinetic energy must actually increase if the orbit remains a circle. rk

6:=41° FIGURE 8-38 Problem 42.

. (IT) Take into account the Earth’s rotational speed (1 rev/day) and determine the necessary speed, with respect to Earth, for a rocket to escape if fired from the Earth at the equator in a direction (a) eastward; (b) westward; (c) vertically upward.

Problems

209

. (II) (a) Determine a formula for the maximum height 4 that a rocket will reach if launched vertically from the Earth’s

surface with speed vy (< vesc). Express in terms of vy, rg, Mg,

and G. (b) How high does a rocket go if vy = 8.35km/s? Ignore air resistance and the Earth’s rotation. 55. (II) (a) Determine the rate at which the escape velocity from

61. (IIT) To escape must overcome Earth and Sun. solar system. (a)

v =

the Earth changes with distance from the center of the Earth,

dvesc/dr. (b) Use the approximation Av =~ (dv/dr) Ar to determine the escape velocity for a spacecraft orbiting the Earth at a height of 320 km. 56. (IT) A meteorite has a speed of 90.0 m/s when 850 km above the Earth. It is falling vertically (ignore air resistance) and strikes a bed of sand in which it is brought to rest in 3.25m. (a) What is its speed just before striking the sand? () How much work does the sand do to stop the meteorite (mass = 575 kg)? (c) What is the average force exerted by the sand on the meteorite? (d) How much thermal energy is produced? 57. (IT) How much work would be required to move a satellite of mass m from a circular orbit of radius r; = 2rg about the Earth to another circular orbit of radius r, = 3rg? (rg is the radius of the Earth.) 58. (II) (a) Suppose we have three masses, my, my, and ms, that initially are infinitely far apart from each other. Show that the work needed to bring them to the positions shown in Fig. 8-39 is

W = -G

(222022 mymy

+

myms

V)

+

-

3 23 (b) Can we say that this formula also gives the potential energy of the system, or the potential energy of one or two of the objects? (c) Is W equal to the binding energy of 5 m the system—that is, equal m, 12 to the energy required to separate the components by an infinite distance? 3 23 Explain. FIGURE 8-39

Problem 58.

mj

59. (IT) A NASA satellite has just observed an asteroid that is on a collision course with the Earth. The asteroid has an

estimated

mass, based

on its size, of 5 x 10°kg.

It is

approaching the Earth on a head-on course with a velocity of 660 m/s

relative

to the Earth

and is now

5.0 X 106 km

away. With what speed will it hit the Earth’s surface, neglecting friction with the atmosphere? . (II) A sphere of radius r; has a concentric spherical cavity of radius r, (Fig. 8-40). Assume this spherical shell of thickness ry — ry is uniform and has a total mass M. Show that the gravitational potential energy of a mass m at a distance r from the center of the shell

(r > r)) is given by

U= -

GmM r

the solar system, an interstellar spacecraft the gravitational attraction of both the Ignore the effects of other bodies in the Show that the escape velocity is

/vt + (vs — 9

= 16.7km/s,

fl

where: vg is the escape velocity from the Earth (Eq. 8-19); vg = V2GMs/rsg is the escape velocity from the gravitational field of the Sun at the orbit of the Earth but far from the Earth’s influence (rgg is the Sun—Earth distance); and v is the Earth’s orbital velocity about the Sun. (b) Show that the energy required is 1.40 X 10%J per kilogram of spacecraft mass. [Hint: Write the energy equation for escape from Earth with v’ as the velocity, relative to Earth, but far from Earth;

then let »' + vy be the escape velocity from the Sun.]

8-8

Power

62. (I) How long will it take a 1750-W motor to lift a 335-kg piano to a sixth-story window 16.0 m above? 63. (I) If a car generates

18 hp when

traveling at a steady

95 km/h, what must be the average force exerted on the car

due to friction and air resistance?

. (I) An 85-kg football player traveling 5.0 m/s is stopped in 1.0s by a tackler. (a) What is the original kinetic energy of the player? (b) What average power is required to stop him? 65. (IT) A driver notices that her 1080-kg car slows down from 95 km/h to 65 km/h in about 7.0 s on the level when it is in neutral. Approximately what power (watts and hp) is needed to keep the car traveling at a constant 80 km/h? . (II) How much work can a 3.0-hp motor do in 1.0 h? 67. (IT) An outboard motor for a boat is rated at 55 hp. If it ca. move a particular boat at a steady speed of 35 km/h, what is the total force resisting the motion of the boat? . (II) A 1400-kg sports car accelerates from rest to 95 km/h 7.4s. What is the average power delivered by the engine?

in

69. (II) During a workout, football players ran up the stadium stairs in 75 s. The stairs are 78 m long and inclined at an angle of 33°. If a player has a mass of 92kg, estimate his average power output on the way up. Ignore friction and air resistance. 70. (II) A pump lifts 21.0 kg of water per minute through a height of 3.50 m. What minimum output rating (watts) must the pump motor have? 71. (IT) A ski area claims that its lifts can move 47,000 people per hour. If the average lift carries people about 200m (vertically) higher, estimate the maximum total power needed. 72. (II) A 75-kg skier grips a moving rope that is powered by an engine and is pulled at constant speed to the top of a 23° hill. The skier is pulled a distance x = 220m along the incline and it takes 2.0 min to reach the top of the hill. If the coefficient of kinetic friction between the snow and skis is wi = 0.10, what horsepower engine is required if 30 such skiers (max) are on the rope at one time? 73. (IIT) The position of a 280-g object is given (in meters) by x = 501> — 801> — 44¢, where ¢ is in seconds. Determine the net rate of work done on this object (a) at ¢ = 2.0s

FIGURE 8-40 Problem 60.

210

CHAPTER 8

and

(b) at t = 4.0s. (c) What is the average net power inpv during the interval from ¢ =0s to ¢ =20s, and in the interval from ¢ = 2.0s to 4.0s?

Conservation of Energy

74. (III) A bicyclist coasts down a 6.0° hill at a steady speed of 4.0m/s. Assuming a total mass of 75 kg (bicycle plus rider), what must be the cyclist’s power output to climb the same hill at the same speed?

(A,-Q

Potential Energy Diagrams

*75. (II) Draw a potential energy diagram, U vs. x, and analyze the motion of a mass m resting on a frictionless horizontal

table and connected to a horizontal spring with stiffness

constant k. The mass is pulled a distance to the right so that the spring is stretched a distance x initially, and then the mass is released from rest. *f76. (Im) The‘ spring of Problem 75 has a stiffness constant k = 160N/m. The mass m = 5.0kg is released from rest

when the spring is stretched xp = 1.0m

from equilibrium.

Determine (a) the total energy of the system; (b) the kinetic energy when x =3x; (c) the maximum Kkinetic energy; (d) the maximum speed and at what positions it occurs; (e) the maximum acceleration and where it occurs.

¥

(III) The potential energy of the two atoms in a diatomic (two-atom) molecule can be written

Ur) = —% + r—ll’z, where r is the distance between the two atoms and a and b are positive constants. (a) At what values of r is U(r) a minimum? A maximum? (b) At what values of r is U(r) = 0? (c) Plot U(r) as a function of r from r =0 to r at a value large enough for all the features in (a) and (b) to show. (d) Describe the motion of one atom with respect to the second atom when E < 0, and when E > 0. (e) Let F be the force one atom exerts on the other. For what values of ris

F>0,F

0. Find a point of equilibrium for the particle and demonstrate that it is stable.

*Numerical/Computer

FIGURE 6-48

*110. (IIT) The two atoms in a diatomic molecule exert an attractive force on each other at large distances and a repulsive force at short distances. The magnitude of the force between two atoms in a diatomic molecule can be approximated by the Lennard-Jones force, or

F(r) —= Fy[2(a/r)

! Problem 104.

13 _

7 (o/r)7],

: where r is; the separation

between the two atoms, and o and F; are constant. For an

oxygen molecule (which is diatomic) F, = 9.60 X 107'' N and o =350 X 10" m. (a) Integrate the equation for

105. (a) If a volcano spews a 450-kg rock vertically upward a distance of 320m, what was its velocity when it left the volcano? (b) If the volcano spews 1000 rocks of this size every minute, estimate its power output.

F(r) to determine the potential energy U(r) of the oxygen molecule. (b) Find the equilibrium distance r, between the two atoms. (¢) Graph F(r) and U(r) between 0.9 ryand 2.5 r.

Answers t¢) Exercises P‘\:

(b).

B: (¢).

‘

C:

Equal speeds.

D: (e), (e); (), (o). General Problems

213

Conservation of linear momentum is another great conservation law of physics. Collisions, such as between billiard or pool balls, illustrate this vector law very nicely: the total vector momentum just before the collision equals the total vector momentum just after the collision. In this photo, the moving cue ball strikes the 11 ball at rest. Both balls move

after the collision, at angles, but the sum of their vector momenta equals

the initial vector momentum of the incoming cue ball, We will consider both elastic collisions (where kinetic energy is also conserved) and inelastic collisions. We also examine the concept of center of mass, and how it helps us in the study of complex motion.

#

X4

> S}

Linear Momentum CHAPTER-OPENING

CONTENTS 9-1

Momentum and Its Relation to Force Conservation of Momentum Collisions and Impulse

Conservation of Energy

and Momentum in Collisions

Elastic Collisions in One Dimension 9-6

Inelastic Collisions

9-7

Collisions in Two or Three Dimensions

9-8

Center of Mass (CM)

9-9

Center of Mass and Translational Motion

*9-10

214

Systems of Variable Mass;

Rocket Propulsion

QUESTIONS—Guess

now!

1. A railroad car loaded with rocks coasts on a level track without friction. A worker on board starts throwing the rocks horizontally backward from the car. Then what happens? (a) The car slows down. (b) The car speeds up. (¢) First the car speeds up and then it slows down. (d) The car’s speed remains constant. (e) None of these. 2. Which answer would you choose if the rocks fall out through a hole in the floor of the car, one at a time? he law of conservation of energy, which we discussed in the previous Chapter, is one of several great conservation laws in physics. Among the other quantities found to be conserved are linear momentum, angular momentum, and electric charge. We will eventually discuss all of these because the conservation laws are among the most important ideas in science. In this Chapter, we discuss linear momentum, and its conservation. The law o conservation of momentum is essentially a reworking of Newton’s laws that give. us tremendous physical insight and problem-solving power.

We make use of the laws of conservation of linear momentum and of energy

to analyze collisions. Indeed, the law of conservation of momentum is particularly

useful when dealing with a system of two or more objects that interact with each ther, such as in collisions of ordinary objects or nuclear particles. Our focus up to now has been mainly on the motion of a single object, often thought of as a “particle” in the sense that we have ignored any rotation or internal motion. In this Chapter we will deal with systems of two or more objects, and toward Fhe end of the Chapter, the concept of center of mass.

9-1 Momentum and Its Relation to Force The linear r‘nomentum (or “momentum” for short) of an object is defined as the product of fis mass and its velocity. Momentum (plural is momenta) is represented by the symbol p. If we let m represent the mass of an object and V represent its velocity, then its momentum p is defined as

RN

©-1)

Velocity is a vector, so momentum too is a vector. The direction of the momentum is the dlrectlon of the velocity, and the magnitude of the momentum is p = mw. Because veFomty depends on the reference frame, so does momentum; thus the reference frame must be spe01f1ed The unit of momentum is that of mass X velocity, which in SI units is kg m/s. There is no special name for this unit. Everyday usage of the term momentum is in accord with the definition above. According to Eq. 9-1, a fast-moving car has more momentum than a slow-moving car of the same mass; a heavy truck has more momentum than a small car moving with the same speed. The more momentum an object has, the harder it is to stop it, and the greater effect it will have on another object if it is brought to rest by striking that object. A football player is more likely to be stunned if tackled by a eavy opponent running at top speed than by a lighter or slower-movmg tackler. A .reavy, fast- rnovmg truck can do more damage than a slow-moving motorcycle. A forc? is required to change the momentum of an object, whether it is to increase the momentum, to decrease it, or to change its direction. Newton originally stated his second law in terms of momentum (although he called the product mo the quantlty of motion”). Newton’s statement of the second law of motion, translated into modern language, is as follows: The rate‘ of change of momentum of an object is equal to the net force applied to it. o

.

.

We can write this as an equation,

SF = —

(9-2)

where =F 14 the net force applied to the object (the vector sum of all forces acting on it). We can readily derive the familiar form of the second law, F = ma, from

Eq. 9-2 for the case of constant mass. If ¥ is the velocity of an object at any

moment, then Eq. 9-2 gives

=

dt

I

dt

2 dv Il

™

=

dt

d(mv) |

iiP_ |

~

| NEWTON'S SECOND LAW A4\ CAUTION The change in the momentum vector

s i the direction of the net force

[constant mass]

because, by definition, & = dv/df and we assume m = constant. Newton’s statement, Eq. 9-2, is actually more general than the more familiar one because it includes the situation in which the mass may change. This is important in certain circumstances, such as for rockets which lose mass as they burn fuel (Section 9-10).

rEXERCISE A Light carries momentum, so if a light beam strikes a surface, it will exert a force on that surface. If the light is reflected rather than absorbed, the force will be (a) the | same, (b) less, (c) greater, (d) impossible to tell, (¢) none of these. SECTION 9-1

Momentum

and Its Relation to Force

215

SCUIIREAM

ESTIMATE | Force of a tennis

serve. For a top player, a

tennis ball may leave the racket on the serve with a speed of 55m/s (about 120 mi/h), Fig. 9-1. If the ball has a mass of 0.060 kg and is in contact with the

racket for about 4ms (4 X 10%s), estimate the average force on the ball. Would this force be large enough to lift a 60-kg person? APPROACH

S

We write Newton’s second law, Eq. 9-2, for the average force as o=

VBT

Ap

OAf

mv, — my,

At

where mwv, and mu, are the initial and final momenta. The tennis ball is hit when its initial velocity v, is very nearly zero at the top of the throw, so we set v, = 0, whereas v, = 55 m/s is in the horizontal direction. We ignore all other forces on the ball, such as gravity, in comparison to the force exerted by the tennis racket. SOLUTION The force exerted on the ball by the racket is P

Ap

mvy

— my

At

B 2

™At

FEURES=1

Examples-f,

(0.060kg)(55m/s)

— 0

0.004 s

800 N.

This is a large force, larger than the weight of a 60-kg person, which would

require a force mg = (60kg)(9.8 m/s?) ~ 600N to lift. NOTE The force of gravity acting on the tennis ball is mg = (0.060 kg)(9.8 m/s?) =

0.59N, which justifies our ignoring it compared to the enormous force the racket exerts. NOTE High-speed photography and radar can give us an estimate of the contact time and the velocity of the ball leaving the racket. But a direct measurement of the force is not practical. Our calculation shows a handy technique for determining an unknown force in the real world.

Washing a car: momentum change and force. Water leaves

FIGURE 9-2

Example 9-2.

-

a hose at a rate of 1.5kg/s with a speed of 20m/s and is aimed at the side of a car, which stops it, Fig. 9-2. (That is, we ignore any splashing back.) What is the force exerted by the water on the car? APPROACH The water leaving the hose has mass and velocity, so it has a momentum pjniia in the horizontal (x) direction, and we assume gravity doesn’t pull the water down significantly. When the water hits the car, the water loses this

momentum

(pgnq = 0). We use Newton’s second law in the momentum form to

find the force that the car exerts on the water to stop it. By Newton’s third law, the force exerted by the water on the car is equal and opposite. We have a continuing process: 1.5 kg of water leaves the hose in each 1.0-s time interval. So let us write F = Ap/At where At = 1.0s, and mvya = (1.5kg)(20m/s). SOLUTION The force (assumed constant) that the car must exert to change the momentum of the water is F

—

Ap el At

Pinal = Pinitia _= 0 — 30kg-m/s At

1.0s

=

—30N.

The minus sign indicates that the force exerted by the car on the water is opposite to the water’s original velocity. The car exerts a force of 30N to the left to stop the water, so by Newton’s third law, the water exerts a force of 30N to the right on the car. NOTE Keep track of signs, although common sense helps too. The water is moving to the right, so common sense tells us the force on the car must be to the right. EXERCISE B If the water splashes back from the car in Example 9-2, would the force on the car be larger or smaller? 216

CHAPTER9

Linear Momentum

-

9-2

Conservation of Momentum

The concept of momentum is particularly important because, if no net external r{orce acts on a system, the total momentum of the system is a conserved quantity. Consider, for example,

the

head-on

collision

of two

billiard balls, as shown

in

of ball A and mg¥Vy the momentum

of

Fig. 9-3. Wfi, assume the net external force on this system of two balls is zero—that is, the only rgnificant forces during the collision are the forces that each ball exerts on the other. Although the momentum of each of the two balls changes as a result of the collision, the sum of their momenta is found to be the same before as after the collision. If m, ¥, is the momentum

ball B, both measured just before the collision, then the total momentum of the two balls before the collision is the vector sum maV, + mgVy. Immediately after

‘the c0111s10fi, the balls each have

a different velocity and momentum,

which we

‘designate by a “prime” on the velocity: m, V), and mg¥Vy. The total momentum after the collision is the vector sum m ¥}, + myVj. No matter what the velocities and masses are, experiments show that the total momentum before the collision is the same as afterward, whether the collision is head-on or not, as long as no net external force acts: momentum before

=

momentum after

MmaVa + mpVg = maVa + mp¥i. That

is, the

total vector

conserved: it stays constant.

FIGURE 9-3 Momentum is conserved in a collision of two balls, labeled A and B.

momentum

of the system

[ZFex = 0] (9-3) of two

colliding balls

CONSERVATION OF MOMENTUM (rwo objects colliding)

is

Althou‘gh the law of conservation of momentum was discovered experimen-

tally, it can be derived from Newton’s laws of motion, which we now show.

Let us consider two objects of mass m, and my that have momenta p, and pg before they collide and p)y and pg after they collide, as in Fig. 9-4. During the collision, suppose that the force exerted by object A on object B at any instant is F. rThen by Newton’s third law, the force exerted by object B on object A is —F. Juring the brief collision time, we assume no other (external) forces are acting (or that F 1i much greater than any other external forces acting). Thus we have

.

B =

and

dpy

FIGURE 9-4 Collision of two objects. Their momenta before collision are p and pg, and after collision are p/y and pg. At any moment during the collision each exerts a force on the other of equal magnitude but opposite direction.

dt

Before collision

§ = P

;

A

dt

We add thEJSG two equations together and find At

d(pa + Ps)

collision

constant.

After collision

=

_/

PB

=—

»

A

=3

Er I

dt

which tells

us that Pa

+ Pg

=

/e

The total momentum thus is conserved. We have put this derivation in the context of a collision. As long as no external fjrces act, it is valid over any time interval, and conservation of momentum is always valid as long as no external forces act. In the real world, external forces do act: friction on billiard balls, gravity acting on a tennis ball, and so on. So 1t‘ may seem that conservation of momentum cannot be applied. Or can it? In a collision, the force each object exerts on the other acts only over a very brief time 1fiterval and is very strong relative to the other forces. If we measure the omenta i mediately before and after the collision, momentum will be very riearly conserved. We cannot wait for the external forces to produce their effect before measuring p)y and pg. SECTION 9-2

Conservation of Momentum

217

For example, when a racket hits a tennis ball or a bat hits a baseball, both before and after the “collision” the ball moves as a projectile under the action of gravity and air resistance. However, when the bat or racket hits the ball, during this brief time of the collision those external forces are insignificant compared to the collision force the bat or racket exerts on the ball. Momentum is conserved (o very nearly so) as long as we measure p, and pg just before the collision and p, and py immediately after the collision (Eq. 9-3). Our derivation of the conservation of momentum can be extended to include any number of interacting objects. Let P represent the total momentum of a system of »n interacting objects which we number from 1 to n: =

P

=

mfi'l

+

mzi.’z

gF

2es

o

m,,?’,,

=

Ep,

We differentiate with respect to time:

dP

——

=

T

= 3F,

9-4)

dt dt where F; represents the net force on the i object. The forces can be of two types: (1) external forces on objects of the system, exerted by objects outside the system, and (2) internal forces that objects within the system exert on other objects in the system. By Newton’s third law, the internal forces occur in pairs: if one object exerts a force on a second object, the second exerts an equal and opposite force on the first object. Thus, in the sum

over all the forces in Eq. 9-4, all the internal

forces cancel each other in pairs. Thus we have — dP

NEWTON'S SECOND LAW ’

(for a system of objects)

dt

S

9-5)

=

where SF,y, is the sum of all external forces acting on our system. If the net external force is zero, then dP/dt = 0, so AP = 0 or P = constant. Thus we see that LAW OF CONSERVATION OF LINEAR MOMENTUM

when the net external force on a system of objects is zero, the total momentum of the system remains constant. This is the law of conservation of momentum. It can also be stated as the total momentum of an isolated system of objects remains constant. By an isolated system, we mean one on which no external forces act—the only forces acting are those between objects of the system. If a net

external

force

acts on

a system,

then

the

law

of conservation

of

momentum will not apply. However, if the “system” can be redefined so as to include the other objects exerting these forces, then the conservation of momentum principle can apply. For example, if we take as our system a falling rock, it does not conserve momentum since an external force, the force of gravity exerted by the Earth, is acting on it and its momentum changes. However, if we include the Earth in the system, the total momentum of rock plus Earth is conserved. (This means that the Earth comes up to meet the rock. Since the Earth’s mass is so great, its upward velocity is very tiny.) Although the law of conservation of momentum follows from Newton’s second law, as we have seen, it is in fact more general than Newton’s laws. In the tiny world of the atom, Newton’s laws fail, but the great conservation laws—those of energy, momentum, angular momentum, and electric charge—have been found to hold in every experimental situation tested. It is for this reason that the conservation laws are considered more basic than Newton’s laws. Railroad cars collide: momentum conserved. A 10,000-kg railroad car, A, traveling at a speed of 24.0 m/s strikes an identical car, B, at rest. If the cars lock together as a result of the collision, what is their common

immediately after the collision? See Fig. 9-5.

speed

-

APPROACH We choose our system to be the two railroad cars. We consider a ver. brief time interval, from just before the collision until just after, so that external 218

CHAPTER9

Linear Momentum

(b) After collision

FIGURE 9-5

Example 9-3.

forces such as friction can be ignored. Then we apply conservation of momentum:

Piital = SOLUTION

Panal:

The initial total momentum

Ppitial = MAVA T Mpvg

is

= MpAUA

because car B is at rest initially (vg = 0). The direction is to the right in the +x

direction. After the collision, the two cars become attached, so they will have the same speed, call it v'. Then the total momentum after the collision is Pivat

=

(ma

+ mp)’

We have assumed there are no external forces, so momentum Pritit

=

Prinal

mava

=

(ma

is conserved:

+ my)o’

Solving for v’, we obtain

A

VS

L

my

e + mg A

RS-

S

=

10,000 kg

>

(10,000 kg + 10,000kg ) 2 rom/e) = 120w/ 4.

=

.

3

“to the right. Their mutual speed after collision is half the initial speed of car A because their masses are equal. NOTE We kept symbols until the very end, so we have an equation we can use in other (rel:]\ed) situations. NOTE We haven’t mentioned friction here. Why? Because we are examining speeds just before and just after the very brief time interval of the collision, and during that brief time friction can’t do much—it is ignorable (but not for long: the cars will slow down because of friction). EXERCISE C A 50-kg child runs off a dock at 2.0 m/s (horizontally) and lands in a waiting rowboat of mass 150 kg. At what speed does the rowboat move away from the dock? EXERCISE

larger than

D In Example 9-3, what result would you get if (¢) mg = 3mp,

mp (mg >> my), and (c) mg

(b) Pgas PHYSICS

Procket

APPLIED

Rocket propulsion

A CAUTION

A rocket pushes on the gases released by the fuel, not on the Earth or other objects

Consetvation of Momentum

219

—

DGV

X

(a) Before shooting (at rest) )

—

vR

VB

(b) After shooting

FIGURE 9-7

Example 9-4.

ELR

Rifle recoil. Calculate the recoil velocity of a 5.0-kg rifle that

shoots a 0.020-kg bullet at a speed of 620 m/s, Fig. 9-7.

APPROACH Our system is the rifle and the bullet, both at rest initially, just before the trigger is pulled. The trigger is pulled, an explosion occurs, and we loolgN at the rifle and bullet just as the bullet leaves the barrel. The bullet moves to th right (+x), and the gun recoils to the left. During the very short time interval of the explosion, we can assume the external forces are small compared to the forces exerted by the exploding gunpowder. Thus we can apply conservation of momentum, at least approximately. SOLUTION Let subscript B represent the bullet and R the rifle; the final velocities are indicated by primes. Then momentum conservation in the x direction gives momentum before mgvg

+

MgoR

0

+

0

SO

R

o

=

momentum after mgvg + MR

=

mgvg + MRUR

_mgvh __

T T

T

(0020kg)(620m/s)

(50kg)

_

i

)

Since the rifle has a much larger mass, its (recoil) velocity is much less than that of the bullet. The minus sign indicates that the velocity (and momentum) of the rifle is in the negative x direction, opposite to that of the bullet.

CONCEPTUAL

EXAMPLE

9-5 | Falling on or off a sled. (a) An empty sled is

sliding on frictionless ice when Susan drops vertically from a tree above onto the sled. When she lands, does the sled speed up, slow down, or keep the same speed? (b) Later: Susan falls sideways off the sled. When she drops off, does the sled speed up, slow down, or keep the same speed? RESPONSE (a) Because Susan falls vertically onto the sled, she has no initial horizontal momentum. Thus the total horizontal momentum afterward equals the\ momentum of the sled initially. Since the mass of the system (sled + person) heincreased, the speed must decrease. (b) At the instant Susan falls off, she is moving with the same horizontal speed as she was while on the sled. At the moment she leaves the sled, she has the same momentum she had an instant before. Because momentum

—

is conserved, the sled keeps the same speed.

EXERCISE E Return to the Chapter-Opening Questions, page 214, and answer them again now. Try to explain why you may have answered differently the first time.

FIGURE 9-8 Tennis racket striking a ball. Both the ball and the racket strings are deformed due to the large force each exerts on the other. FIGURE 9-9 Force as a function of time during a typical collision: F can become very large; At is typically milliseconds for macroscopic collisions.

z

0, the object is moving to the right (increasing x), whereas for v < 0, the object is moving to the left (toward decreasing values of x). From conservation of momentum, we have MaAvsA

mA—.

(9-7)

Primed quantities (') mean after the collision, and unprimed mean before the collision, just as in Eq. 9-3 for conservation of momentum. At the atomic level the collisions of atoms and molecules are often elastic. But in the “macroscopic” world of ordinary objects, an elastic collision is an ideal that is never quite reached, since at least a little thermal energy (and perhaps sound and other forms of energy) is always produced during a collision. The collision of two hard elastic balls, such as billiard balls, however, is very close to being perfect]=™ elastic, and we often treat it as such. We do need to remember that even when the kinetic energy is not conserved, the total energy is always conserved. Collisions in which kinetic energy is not conserved are said to be inelastic collisions. The kinetic energy that is lost is changed into other forms of energy, often thermal energy, so that the total energy (as always) is conserved. In this case,

9—-5 m'Aq—>m'§E

[elastic collision]

+ Mg

=

MV,

+ MpvR.

Because the collision is assumed to be elastic, kinetic energy is also conserved:

m.B?'»

Impvh + imgrh

VB X

= Imavf + mpog.

We have two equations, so we can solve for two unknowns. If we know the masse. and velocities before the collision, then we can solve these two equations for the

Linear Momentum

velocities after the collision, vy and vj. We derive a helpful result by rewriting the momentum equation as mA(vA

—

vg)

=

mB(vi;

— vB),

(i)

rand we rewrite the kinetic energy equation as

ma(vh — v3) = my(vh — v}). Noting that algebraically (a — b)(a + b) = a* — b*, we write this last equation as mA(UA

-

v}‘)(vA

+

va‘)

=

mB(’U’B

=

vB)('U’B

+

’UB).

(ii)

We divide Eq. (ii) by Eq. (i), and (assuming v, # v} and vz # vg)" obtain vy t V) = v t+ vg. We can rewrite this equation as Vp

or

—

Vg

=

va — vg =

v

—

Vi

—(v) — vj).

|head-on (1-D) elastic collision]

(9-8)

This is an interesting result: it tells us that for any elastic head-on collision, the relative speed of the two objects after the collision has the same magnitude (but opposite direction) as before the collision, no matter what the masses are. Equation 9-8 was derived from conservation of kinetic energy for elastic collisions, and can be used in place of it. Because the v’s are not squared in Eq. 9-8, it is simpler to use in calculations than the conservation of kinetic energy equation (Eq. 9-7) directly.

r

A

CAUTION

Relative speeds (1-D only)

Equal masses. Billiard ball A of mass m moving with speed v, collides head-on with ball B of equal mass. What are the speeds of the two balls after the collision, assuming it is elastic? Assume (@) both balls are moving initially (v, and vg), (b) ball B is initially at rest (vg = 0). APPROACH There are two unknowns, v}, and vg, so we need two independent equations. We focus on the time interval from just before the collision until just after. No net external force acts on our system of two balls (mg and the normal force cancel), so momentum is conserved. Conservation of kinetic energy applies as well because we are told the collision is elastic.

SOLUTION

(a) The

masses

momentum gives wp

tvp

=

are

U+

equal

(m, = my = m)

so conservation

of

Uh.

We need a second equation, because there are two unknowns. We could use the conservation of kinetic energy equation, or the simpler Eq. 9-8 derived from it: Vo — Vg = v — V4.

We add these two equations and obtain Vg

=

VA

and then subtract the two equations to obtain VA = vg. That is, the balls exchange velocities as a result of the collision: ball B acquires the velocity that ball A had before the collision, and vice versa.

(b) If ball B is at rest initially, so that vz = 0, we have vg

=

vy

and

v,

=

0.

That is, ball A is brought to rest by the collision, whereas ball B acquires the original velocity of ball A. This result is often observed by billiard and pool players, and is valid only if the two balls have equal masses (and no spin is given to the balls). See Fig. 9-14.

"Note that Egs. (i) and (ii), which are the conservation laws for momentum and kinetic energy, are both satisfied by the solution »j = v, and vj = vg. This is a valid solution, but not very interesting. It corresponds to no collision at all—when the two objects miss each other.

FIGURE 9-14 In this multi-flash photo of a head-on collision

between two balls of equal mass, the

white cue ball is accelerated from rest by the cue stick and then strikes the red ball, initially at rest. The white ball stops in its tracks and the (equal mass) red ball moves off with the same speed as the white ball had before the collision. See Example 9-7.

SECTION 9-5

223

Unequal masses, target at rest. A very common practical

situation is for a moving object (m,) to strike a second object (my, the “target”)

at rest (vg = 0). Assume the objects have unequal masses, and that the collision

is elastic and occurs along a line (head-on). (a) Derive equations in terms of the initial velocity v, of mass m, and the masses (b) Determine the final velocities if the moving object is much than the target (m, >> my). (c) Determine the final velocities object is much less massive than the target (m, M. Hence, the last term—which

represents the thrust—is positive and acts to increase the velocity. (d) The time required to reach burnout is the time needed to use up all the fuel (15,000 kg) at a rate of 190 kg/s; so at burnout,

,_ 1s0x10'kg _ o T T 190kgs ¥ If we take v, = 0, then using the result of part (c):

v = =

238

CHAPTER9

Linear Momentum

6000 kge 2 —(9.80m/s?)(79s) + ( _ 2800m/s)(1n e 21,000kg>

2700 m/s. ™

| Summary The linear momentum, p, of an object is defined as the product

of its mass times its velocity, p In terms written as

of

9-1)

mv.

=

momentum,

Newton’s

second

law

can

be

If two colliding objects stick together as the result of a collision, the collision is said to be completely inelastic. For a system of particles, or for an extended object that can

be considered as having a continuous distribution of matter, the

center of mass (CM) is defined as _

9-2) That is, the rate of change of momentum of an object equals the

net force exerted on it.

Em,-x,-

Xem = T’

Yem

_

M

Iem = '_M—'

9-11)

or

When the net external force on a system of objects is zero, the total momentum remains constant. This is the law of conser-

vation of momentum.

Stated another way, the total momentum

of an isolated The law dealing with sion, two (or

system of objects remains constant. of conservation of momentum is very useful in the class of events known as collisions. In a collimore) objects interact with each other for a very

short time,

and the force each exerts on the other during this

time interval is very large compared to any other forces acting. The impulse of such a force on an object is defined as

1.5 Jf“dt and is equal to the change in momentum of the object as long as F is the net Torce on the object:

AP

= BB

12

= J'th = 7. t i

9-6)

Micy

F

Pa + B8 = Pa + Pb. The total energy is also conserved; but this may not be useful unless kinetic energy is conserved, in which case the collision is called an elt{‘sfic collision:

-7

If kinetic energy is not conserved, the collision is called inelastic.

= ZFex,

9-17)

where M is the total mass of the system, d,, is the acceleration of the cM of the system, and SF.y is the total (net) external force acting on all parts of the system. For a system of particles of total linear momentum P = Sm;V; = Mv,,, Newton’s second law is

dp

Total mk)mentum is conserved in any collision:

2 1 2 _ 1 21 2 Imavi + smpvh = Imavd + tmgoff.

where M is the total mass of the system. The center of mass of a system is important because this point moves like a single particle of mass M acted on by the same net external force, 2F.y. In equation form, this is just Newton’s second law for a system of particles (or extended objects):

=

ZFeXt'

(9-5)

[*If the mass M of an object is not constant, then

dv

M:l,‘t‘

=

=

ZFy

_

dM

+ vrcldT

(9-19b)

where V is the velocity of the object at any instant and v, is the relative velocity at which mass enters (or leaves) the object.]

] Questions that momentum is conserved. Yet most moving objects eventually slow down and stop. Explain. 2. Two blocks of mass my and m;, rest on a frictionless table and are connected by a spring. The blocks are pulled apart, stretching the spring, and then released. Describe the subsequent motion of the two blocks. 3. A light object and a heavy object have the same kinetic energy. Which has the greater momentum? Explain. 4. When a person jumps from a tree to the ground, what happens to the momentum of the person upon striking the ground? 5. Explain, on the basis of conservation of momentum,

how a

fish propels itself forward by swishing its tail back and forth. 6. Two children float motionlessly in a space station. The 20-kg girl pusk‘Ies on the 40-kg boy and he sails away at 1.0 m/s. The girl (a) remains motionless; (b) moves in the same r direction at 1.0m/s; (c) moves in the opposite direction at ~ 1.0m/s; (d) moves in the opposite direction at 2.0 m/s; (e) none of these.

7. A truck going 15 km/h has a head-on collision with a small car going 30 km/h. Which statement best describes the situation? (a) The truck has the greater change of momentum because it has the greater mass. (b) The car has the greater change of momentum because it has the greater speed. (c) Neither the car nor the truck changes its momentum in the collision because momentum is conserved. () They both have the same change in magnitude of momentum because momentum is conserved. (¢) None of the above is necessarily true. 8. If a falling ball were to make a perfectly elastic collision with the floor, would it rebound to its original height? Explain. 9. A boy stands on the back of a rowboat and dives into the water. What happens to the rowboat as the boy leaves it? Explain. 10. It is said that in ancient times a rich man with a bag of gold coins was stranded on the surface of a frozen lake. Because the ice was frictionless, he could not push himself to shore and froze to death. What could he have done to save himself had he not been so miserly? Questions

239

together,

or if the

two

cars

collide

and

rebound

backward? Explain. 15. A superball is dropped from a height 4 onto a hard steel plate (fixed to the Earth), from which it rebounds at very nearly its original speed. (a) Is the momentum of the ball conserved during any part of this process? (b) If we consider the ball and the Earth as our system, during what parts of the process is momentum conserved? (c) Answer part (b) for a piece of putty that falls and sticks to the steel plate. 16. Cars used to be built as rigid as possible to withstand collisions. Today, though, cars are designed to have “crumple zones” that collapse upon impact. What is the advantage of this new design? 17. At a hydroelectric power plant, water is directed at high speed against turbine blades on an axle that turns an electric generator. For maximum power generation, should the turbine blades be designed so that the water is brought to a dead stop, or so that the water rebounds?

18

A squash ball hits a wall at a 45° angle as shown in Fig. 9-36. What is the direction (a) of the change in momentum of the ball, (b) of the force on the wall?

8

remain

21. Inelastic and elastic collisions are similar in that () momentum and kinetic energy are conserved in both; (b) momentum is conserved in both; (¢) momentum and potential energy are conserved in both; (d) kinetic energy is conserved in both. 22, If a 20-passenger plane is not full, sometimes passengers ar told they must sit in certain seats and may not move t. empty seats. Why might this be? 23. Why do you tend to lean backward when carrying a heavy load in your arms? 24. Why is the cM of a 1-m length of pipe at its midpoint, whereas this is not true for your arm or leg? 25. Show on a diagram how your cM shifts when you move from a lying position to a sitting position. Describe an analytic way of determining the cM of any thin, triangular-shaped, uniform plate. 27. Place yourself facing the edge of an open door. Position your feet astride the door with your nose and abdomen touching the door’s edge. Try to rise on your tiptoes. Why can’t this be done? If only an external force can change the momentum of the

2

11. The speed of a tennis ball on the return of a serve can be just as fast as the serve, even though the racket isn’t swung very fast. How can this be? 12. Is it possible for an object to receive a larger impulse from a small force than from a large force? Explain. 13. How could a force give zero impulse over a nonzero time interval even though the force is not zero for at least a part of that time interval? 14. In a collision between two cars, which would you expect to be more damaging to the occupants: if the cars collide and

center of mass

29. 30. 31.

32.

FIGURE 9-36

Question 18.

33.

19. Why can a batter hit a pitched baseball farther than a ball he himself has tossed up in the air? 20. Describe a collision in which all kinetic energy is lost.

of an object, how can the internal force of

the engine accelerate a car? A rocket following a parabolic path through the air suddenly explodes into many pieces. What can you say about the motion of this system of pieces? How can a rocket change direction when it is far out in space and essentially in a vacuum? In observations of nuclear B-decay, the electron and recoil nucleus often do not separate along the same line. Use conservation of momentum in two dimensions to explain why this implies the emission of at least one other particle in the disintegration. fl Bob and Jim decide to play tug-of-war on a frictionless (icy surface. Jim is considerably stronger than Bob, but Bob weighs 160 1bs while Jim weighs 145 Ibs. Who loses by crossing over the midline first? At a carnival game you try to knock over a heavy cylinder by throwing a small ball at it. You have a choice of throwing either a ball that will stick to the cylinder, or a second ball of

equal mass and speed that will bounce backward off the cylinder. Which ball is more likely to make the cylinder move?

| Problems 9-1

Momentum

1. (I) Calculate the force exerted on a rocket when the propelling gases are being expelled at a rate of 1300 kg/s

with a speed of 4.5 X 10*m/s.

2. (I) A constant friction force of 25 N acts on a 65-kg skier for 15 s. What is the skier’s change in velocity? 3. (IT) The momentum of a particle, in SI units, is given by p =

4.8%i — 8.0j — 8.9 k. What is the force as a function of time?

4. (II) The force on a particle of mass m is given by F = 261 — 122§

where Fis in N and ¢ in s. What will be the change in the particle’s momentum between ¢ = 1.0s and ¢ = 2.0s?

5. (II) A 145-g baseball, moving along the x axis with speed 30.0m/s, strikes a fence at a 45° angle and rebounds along the y axis with unchanged speed. Give its change in momentum using unit vector notation. 240

CHAPTER9

Linear Momentum

6. (I) A 0.145-kg baseball pitched horizontally at 32.0 m/s strikes a bat and is popped straight up to a height of 36.5 m. If the contact time between bat and ball is 2.5 ms, calculate the average force between the ball and bat during contact. 7. (I) A rocket of total mass 3180kg is traveling in outer space with a velocity of 115 m/s. To alter its course by 35.0°, its rockets can be fired briefly in a direction perpendicular to its original motion. If the rocket gases are expelled at a speed of 1750 m/s, how much mass must be expelled? 8. (III) Air in a 120-km/h wind strikes head-on the face of a building 45 m wide by 65 m high and is brought to rest. If air has a mass of 1.3 kg per cubic meter, determine the average force of the wind on the building.

9-2

Conservation of Momentum

9. (I) A 7700-kg boxcar traveling 18 m/s strikes a second ca The two stick together and move off with a speed ¢ 5.0 m/s. What is the mass of the second car?

10. (I) A 9150-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0m/s. A 4350-kg load,

initially at rest, is dropped onto the car. What will be the

car’s new speed? . (I) An atomic nucleus at rest decays radioactively into an alpha particle and a smaller nucleus. What will be the speed of this recoiling nucleus if the speed of the alpha particle is 2.8 X 10°m/s? Assume the recoiling nucleus has a mass 57 times Ereater than that of the alpha particle. 12. MHA 13d-kg tackler moving at 2.5 m/s meets head-on (and tackles) an 82-kg halfback moving at 5.0 m/s. What will be their mutual speed immediately after the collision? 13. (IT) A child in a boat throws a 5.70-kg package out horizontally with a sp+sed of 10.0 m/s, Fig. 9-37. Calculate the velocity of the boat immediately after, assuming it was initially at rest. The mass of the child is 24.0kg that of boat is 35.0kg.

FIGURE ?—37 Problem

13.

14. (IT) An atomic nucleus initially moving at 420 m/s emits an alpha particle in the direction of its velocity, and the remaining nucleus slows to 350 m/s. If the alpha particle has a mass of 4.0u and the original nucleus has a mass of 222 u,

21. (IIT) A 224-kg projectile, fired with a speed of 116 m/s at a 60.0° angle, breaks into three pieces of equal mass at the highest point of its arc (where its velocity is horizontal). Two of the fragments move with the same speed right after the explosion as the entire projectile had just before the explosion; one of these moves vertically downward and the other horizontally. Determine (a) the velocity of the third fragment immediately after the explosion and (b) the energy released in the explosion.

9-3

Collisions and Impulse

22. (I) A 0.145-kg baseball pitched at 35.0 m/s is hit on a horizontal line drive straight back at the pitcher at 56.0m/s. If the contact time between bat and ball is 5.00 X 1073 s, calcu-

late the force (assumed to be constant) between the ball and bat. 23. (IT) A golf ball of mass 0.045 kg is hit off the tee at a speed of 45m/s. The golf club was in contact with the ball for 3.5 X 10?5, Find (a) the impulse imparted to the golf ball, and (b) the average force exerted on the ball by the golf club. 24. (IT) A 12-kg hammer strikes a nail at a velocity of 8.5m/s and comes to rest in a time interval of 8.0 ms. (a) What is the impulse given to the nail? (b) What is the average force acting on the nail? 25. (IT) A tennis ball of mass m = 0.060kg and speed v = 25m/s strikes a wall at a 45° angle and rebounds with the same speed at 45° (Fig. 9-38). What is the impulse (magnitude and direction) given to the ball?

Y

what speéd does the alpha particle have when it is emitted?

. (IT) An object at rest is suddenly broken apart into two fragments by an explosion. One fragment acquires twice the kinetic energy of the other. What is the ratio of their masses? 16. (II) A 22-g bullet traveling 210m/s penetrates a 2.0-kg block of wood and emerges going 150m/s. If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges? 17. (IT) A rocket of mass m traveling with speed v, along the x axis suddenly shoots out fuel equal to one-third its mass, perpendicular to the x axis (along the y axis) with speed 2v. Express the final velocity of the rocket in i, j, k notation.

18. (IT) The decay of a neutron into a proton, an electron, and a neutrino is an example of a three-particle decay process. Use the vector nature of momentum to show that if the neutron is initially at rest, the velocity vectors of the three must be coplanar (that is, all in the same plane). The result is not true for numbers greater than three. (19, (IT) A mass m, = 2.0kg, moving with velocity v, = (4.0i + 5.0§ — 2.0k) m/s, collides with mass mp = 3.0kg, which is initially at rest. Inmediately after the collision, mass 71, is observed traveling at velocity ¥/ = (—2.0i + 3.0k) m/s. Find the velocity of mass mpg after the collision. Assume no outside force acts on the two masses during the collision. 20. (IT) A 925-kg two-stage rocket is traveling at a speed of 6.60 X 10° m/s away from Earth when a predesigned explosion separates the rocket into two sections of equal mass

that then move with a speed of 2.80 X 10*m/s relative to

each other along the original line of motion. (¢) What is the speed and direction of each section (relative to Earth) after the expl&sion? (b) How much energy was supplied by the explosion? [Hint: What is the change in kinetic energy as a result of the explosion?]

/

i

FIGURE 9-38 Problem 25. 26. (IT) A 130-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with his legs from a 1700-kg space capsule. (@) What is the change in speed of the space capsule? (b) If the push lasts 0.500s, what is the average force exerted by each on the other? As the reference frame, use the position of the capsule before the push. (c¢) What is the kinetic energy of each after the push? 27. (IT) Rain is falling at the rate of 5.0 cm/h and accumulates in a pan. If the raindrops hit at 8.0 m/s, estimate the force on the bottom of a 1.0 m? pan due to the impacting rain which does not rebound. Water has a mass of 1.00 X 10° kg per m". 28. (IT) Suppose the force acting on a tennis ball (mass 0.060kg) points in the +x direction and is given by the graph of Fig. 9-39 as a function of time. Use graphical methods to estimate (a) the total impulse given the ball, and (b) the velocity of the ; ball after being struck,

assuming the ball is being served so it is nearly at rest initially.

FIGURE 9-39

Problem 28.

300

- 5 £ Ry

100 0O

{

a

]

\

0.05

t(s)

Problems

[

0.10

241

29. (IT) With what impulse does a 0.50-kg newspaper have to be thrown to give it a velocity of 3.0 m/s? . (II)

The

force

on

a

bullet

F =[740 — (23 x 10°s7!)t]N

is

given

by

the

formula

over the time interval

t=0 to t =30 X% 1073s. (a) Plot a graph of F versus ¢ for

t=0 to t=30ms. (b) Use the graph to estimate the impulse given the bullet. (¢) Determine the impulse by integration. (d) If the bullet achieves a speed of 260m/s as a result of this impulse, given to it in the barrel of a gun, what must the bullet’s mass be? (e) What is the recoil speed of the 4.5-kg gun? 31 (IT) (a) A molecule of mass m and speed v strikes a wall at right angles and rebounds back with the same speed. If the collision time is A?, what is the average force on the wall during the collision? (b) If molecules, all of this type, strike the wall at intervals a time ¢ apart (on the average) what is the average force on the wall averaged over a long time? 32. (IIT) (a) Calculate the impulse experienced when a 65-kg person lands on firm ground after jumping from a height of 3.0m. (b) Estimate the average force exerted on the person’s feet by the ground if the landing is stiff-legged, and again (c) with bent legs. With stiff legs, assume the body moves 1.0cm during impact, and when the legs are bent, about 50 cm. [Hinr: The average net force on her which is related to impulse, is the vector sum of gravity and the force exerted by the ground.] 33. (III) A scale is adjusted so that when a large, shallow pan is placed on it, it reads zero. A water faucet at height h = 2.5m above is turned on and water falls into the pan at arate R = 0.14kg/s. Determine (a) a formula for the scale reading as a function of time ¢ and (b) the reading for t = 9.0s. (c) Repeat (a) and (b), but replace the shallow pan with a tall, narrow cylindrical container of area A = 20 cm? (the level rises in this case).

9-4 and 9-5

Elastic Collisions

34. (IT) A 0.060-kg tennis ball, moving with a speed of 4.50 m/s, has a head-on collision with a 0.090-kg ball initially moving in the same direction at a speed of 3.00m/s. Assuming a perfectly elastic collision, determine the speed and direction of each ball after the collision. 35. (IT) A 0.450-kg hockey puck, moving east with a speed of 480m/s,

has

a

head-on

collision

with

a

0.900-kg

puck

initially at rest. Assuming a perfectly elastic collision, what will be the speed and direction of each object after the collision? . (II) A 0.280-kg croquet ball makes an elastic head-on collision with a second ball initially at rest. The second ball moves off with half the original speed of the first ball. (a) What is the mass of the second ball? (b) What fraction of the original kinetic energy (AK/K) gets transferred to the second ball? 37. (IT) A ball of mass 0.220kg that is moving with a speed of 7.5 m/s collides head-on and elastically with another ball initially at rest. Immediately after the collision, the incoming ball bounces

backward

with a speed of 3.8 m/s.

Calculate

(a) the velocity of the target ball after the collision, and (b) the mass of the target ball. . (IT) A ball of mass m makes a head-on elastic collision with a second ball (at rest) and rebounds with a speed equal to 0.350 its original speed. What is the mass of the second ball? 242

CHAPTER9

Linear Momentum

39. (IT) Determine the fraction of kinetic energy lost by a neutron (my = 1.01u) when it collides head-on and elastically with a target particle at rest which is (a) {H (m = 1.01u);

(b) 1H (heavy hydrogen, m = 2.01 u); (c) '23C (m = 12.00u); (d) *BPb (lead, m = 208 u).

40. (II) Show that, in general, for any head-on one-dimensional elastic collision, the speeds after collision are 2mA

vp

VAl

VA

Al ———

and

— ma +

mg

| +

ma

—

mp

ma

+

mg

vg|

mpg

—

Ma

———— ma + mg 2mB

| + vg| —— |, ma

+

mg

where v, and vg are the initial speeds of the two objects of mass mu and mg.

41. (IIT) A 3.0-kg block slides along a frictionless tabletop at 8.0m/s toward a second block (at rest) of mass 4.5kg. A coil spring, which obeys Hooke’s law and has spring constant

k = 850N/m,

is attached to the second block in

such a way that it will be compressed when struck by the moving block, Fig. 9-40. () What will be the maximum compression of the spring? (b) What will be the final velocities of the blocks after the collision? (c) Is the collision elastic?

Ignore the mass of the spring.

FIGURE 9-40

9-6

Problem 41.

Inelastic Collisions

42, (I) In a ballistic pendulum experiment, projectile 1 results infl a maximum height # of the pendulum equal to 2.6cm. A second projectile (of the same mass) causes the pendulum to swing twice as high, A, = 5.2cm. The second projectile was how many times faster than the first? 43. (II) (a) Derive a formula for the fraction of kinetic energy lost, AK/K, in terms of m and M for the ballistic pendulum collision of Example 9-11. (b) Evaluate for m = 16.0g and M = 380g. . (IT) A 28-g rifle bullet traveling 210 m/s buries itself in a 3.6-kg pendulum hanging on a 2.8-m-long string, which makes the pendulum swing upward in an arc. Determine the vertical and horizontal components of the pendulum’s maximum displacement. 4s. (IT) An internal explosion breaks an object, initially at rest, into two pieces, one of which has 1.5 times the mass of the other.

If 75007J

is released

in

the

explosion,

how

much

kinetic energy does each piece acquire? . (IT) A 920-kg sports car collides into the rear end of a 2300-kg SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 2.8 m before stopping. The police officer, estimating the coefficient of kinetic friction between tires and road to be 0.80, calculates

the speed of the sports car at impact. What was that speed? 47. (II) You drop a 12-g ball from a height of 1.5m and it only bounces back to a height of 0.75 m. What was the total impulse on the ball when it hit the floor? (Ignore air resistance). . (I) Car A hits car B (initially at rest and of equal mass) from behind while going 35m/s. Immediately after the‘\ collision, car B moves forward at 25 m/s and car A is at rest.

What fraction of the initial kinetic energy is lost in the collision?

49. (IT) A measure of inelasticity in a head-on collision of two objects is the coefficient of restitution, e, defined as

VA ~ VB

9-7

Vg

T U

h

‘ Problem 49.

Measuremenk of coefficient of rest1tut10n1

moving with

deflected off at an angle of 30.0° with a speed vy = 2.10m/s. (a) Taking the x axis to be the original direction of motion of ball A, write down the equations expressing the conservation of momentum for the components in the x and y directions separately. (b) Solve these equations for the speed, vg, and

where v}y — vj is the relative velocity of the two objects after the collision and vg — vy is their relative velocity before it. (a) ShO\‘y that e =1 for a perfectly elastic collision, and e =0 for a completely inelastic collision. (b) A simple method for measuring the coefficient of restitution for an object colliding with a very hard surface like steel is to drop the object onto a heavy steel plate, as shown in Fig. 9-41. Determine a formula for e in terms of the original height & and the maximum height A’ l reached after collision.

FIGURE 9-41

mu = 0.120kg

speed va = 2.80m/s strikes ball B, initially at rest, of mass mp = 0.140kg. As a result of the collision, ball A is

TeEeTEE———

r

Collisions in Two Dimensions

54. (IT) Billiard ball A of mass

¢

|

11

I i

L4

|

‘

50. (IN) A pendulum consists of a mass M hanging at the bottom end of a massless rod of length £, which th a frictionless pivot at its top end. A mass m, moving as shown in Fig. 9-42 with velocity v, impacts M and becomes embedded. What is the smallest value of v sufficient to cause the pendulum (with embedded mass m) to swing clear ovTr the top of its arc?

angle, 63, of ball B. Do not assume the collision is elastic.

. (IT) A

nucleus,

radioactive an

nucleus

electron,

and

a

at rest decays neutrino.

The

into a second electron

and

neutrino are emitted at right angles and have momenta of

9.6 X 102 kg-m/s

and 6.2 X 102 kg m/s, respectively.

Determine the magnitude and the direction momentum of the second (recoiling) nucleus.

of

the

56. (IT) Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is moving along the y axis at +2.0m/s, and ball B is moving to the right along the x axis with speed +3.7 m/s. After the collision (assumed elastic), the second ball is moving along the positive y axis (Fig. 9-43).

What is the final direction of ball A, and

what

the speeds of the two balls?

are

@

@ >

v'B

=

+x

vg=3.7 m/s vy =2.0 m/s FIGURE 9-43 Problem 56. (Ball A after the collision is not shown.)

FIGLRE 9-42 Prorlem 50.

51

(IT) A bullet of mass m = 0.0010kg embeds itself in a wooden block with mass M = 0.999kg, which then compresses a spring (k = 120N/m) by a distance x = 0.050m before coming to rest. The coefficient of kinetic friction between the block and table is u = 0.50. (a) What is the initial speed of the bullet? (b) What fraction of the bullet’s initial kinetic energy is dissipated (in damage to the v'vlFoden block, rising temperature, etc.) in the collision between the bullet and the block? 52. Im A 1J44-g baseball moving 28.0m/s strikes a stationary 5.25-kg brick resting on small rollers so it moves without significant friction. After hitting the brick, the baseball bouncesj straight back, and the brick moves forward at 1.10m/ . (a) What is the baseball’s speed after the collision? (b) Fin the total kinetic energy before and after the collision §3. (IT) A 6.0-kg object moving in the +x direction at 5.5m/s collides head-on with an 8.0-kg object moving in the

—x direction at 4.0 m/s. Find the final velocity of each mass if:

(a) the objects stick together; (b) the collision is elastic; (c) the 6.0-kg object is at rest after the collision; (d) the 8.0-kg object is at rest after the collision; (e) the 6.0-kg object has a velocity of 4.0m/s in the —x direction after the collision. Are the results in (¢), (d), and (e) “reasonable”? Explain.

.

(IT) An atomic nucleus of mass m traveling with speed v collides elastically with a target particle of mass 2m (initially at rest) and is scattered at 90°. (a) At what angle does the target particle move after the collision? (b) What are the final speeds of the two particles? (c¢) What fraction of the initial kinetic energy is transferred to the target particle? 58. (IT) A neutron collides elastically with a helium nucleus (at rest initially) whose mass is four times that of the neutron. The helium nucleus is observed to move off at an angle e

= 45°.

Determine

the

angle

of the

neutron, 6,

and

the speeds of the two particles, v, and vy, after the collision. The neutron’s initial speed is 6.2 X 10° m/s. 59, (III) A neon atom (m = 20.0u) makes a perfectly elastic collision with another atom at rest. After the impact, the neon atom travels away at a 55.6° angle from its original direction and the unknown atom travels away at a —50.0° angle. What is the mass (in u) of the unknown atom? [Hint: You could use the law of sines.] 60. (ITI) For an elastic collision between a projectile particle of mass mp and a target particle (at rest) of mass mp, show that the scattering angle, 0}, of the projectile (a) can take

any value, 0 to 180°, for ma < mp, but (b) has a maximum

angle ¢ given by cos’¢ = 1 — (mg/my)? for my > mg.

61. (ITIT) Prove that in the elastic collision of two objects of identical mass, with one being a target initially at rest, the angle between their final velocity vectors is always 90°. Problems

243

9-8 Center of Mass (cm)

9-9

62. (I) The cM of an empty 1250-kg car is 2.50 m behind the front of the car. How far from the front of the car will the cM be when two people sit in the front seat 2.80 m from the front of the car, and three people sit in the back seat 3.90m from the front? Assume that each person has a mass of 70.0kg. 63. (I) The distance between a carbon atom (m = 12 u) and an oxygen atom (m = 16u) in the CO molecule is

72. (II) Mass M, = 35kg and mass Mp = 25kg. They have velocities (in m/s) Vo = 12i — 16j and vg = —20i + 14}. Determine the velocity of the center of mass of the system. 73. (IT) The masses of the Earth and Moon are 5.98 X 10%k;

1.13 X 107 m, How far from the carbon atom is the center

of mass of the molecule? 64. (II) Three cubes, of side £y, 2{,, and 3£, are placed next to one another (in contact) with their centers along a straight line as shown in Fig. 9-44. What is the position, along this line, of the cM of this

system? Assume the cubes are made of the same uniform material.

=0

FIGURE 9-44

Problem 64.

Llo’l-—flo

_4‘_‘3[0—"

65. (IT) A square uniform raft, 18 m by 18 m, of mass 6200 kg, is used as a ferryboat. If three cars, each of mass 1350 kg, occupy the NE, SE, and SW corners, determine the cM of the loaded ferryboat relative to the center of the raft. 66. (IT) A uniform circular plate of radius 2R has a circular hole of radius R cut out of it. The center C' of the smaller circle is a distance 0.80R from the center C of the larger circle, Fig. 9-45.

What is the position of the center of mass of the plate? [Hint: Try subtraction.] FIGURE 9-45 Problem 66.

67. (IT) A uniform thin wire is bent into a semicircle of radius r. Determine the coordinates of its center of mass with respect to an origin of coordinates at the center of the “full” circle. 68. (IT) Find the center of mass of the ammonia molecule. The chemical formula is NH3. The hydrogens are at the corners of an equilateral triangle (with sides 0.16 nm) that forms the base of a pyramid, with nitrogen at the apex (0.037 nm vertically above the plane of the triangle). 69. (IIT) Determine the cM of a machine part that is a uniform cone of height /4 and radius R, Fig. 9-46. [Hint: Divide the cone into an infinite number of disks of thickness dz, one of which is shown.]

FIGURE 9-46

Problem 69.

70. (III) Determine the cM of a uniform pyramid that has four triangular faces and a square base with equal sides all of length s. [Hint: See Problem 69.] 71.

244

(III) Determine the cM of a thin, uniform, semicircular plate.

CHAPTER9

Linear Momentum

CM and Translational Motion

and 7.35 X 10% kg, respectively, and their centers are sepa-

rated by 3.84 X 10°m. (a) Where is the cM of this system located?

(b) What

Earth-Moon

can you say about the motion of the

system about the Sun, and of the Earth

and

Moon separately about the Sun? 74. (II) A mallet consists of a uniform cylindrical head of mass 2.80kg and a diameter 0.0800m mounted on a uniform cylindrical handle of mass 0.500kg and length 0.240m, as shown in Fig. 9-47. If this mallet is tossed, spinning, into the air, how far above the bottom of the handle is the point that will follow a parabolic trajectory? FIGURE 9-47 —— Problem 74. 8.00 cm 75. (II) A 55-kg woman and a 72-kg man stand 10.0 m apart on frictionless ice. (a) How far from the woman is their cM? (b) If each holds one end of a rope, and the man pulls on the rope so that he moves

2.5m, how far from the woman

will he be now? (c) How far will the man have moved when he collides with the woman? 76. (II) Suppose that in Example 9-18 (Fig. 9-32), my; = 3m;. (a) Where then would my land? (b) What if m; = 3my? 77. (II) Two people, one of mass 85kg and the other of mass 55 kg, sit in a rowboat of mass 78 kg. With the boat initially at rest, the two people, who have been sitting at opposit«a ends of the boat, 3.0m apart from each other, now exchange

seats. How far and in what direction will the boat move? 78. (III) A 280-kg flatcar 25 m long is moving with a speed of 6.0m/s along horizontal frictionless rails. A 95-kg worker starts walking from one end of the car to the other in the direction of motion, with speed 2.0 m/s with respect to the car. In the time it takes for him to reach the other end, how far has the flatcar moved? 79 (IIT) A huge balloon and its gondola, of mass M, are in the air and stationary with respect to the ground. A passenger, of mass m, then climbs out and slides down a rope with speed v, measured with respect to the balloon. With what speed and direction (relative to Earth) does the balloon then move? What happens if the passenger stops?

*9-10

Variable Mass

*80. (II) A 3500-kg rocket is to be accelerated at 3.0 g at take-off from the Earth. If the gases can be ejected at a rate of 27 kg/s, what must be their exhaust speed? *81. (II) Suppose the conveyor belt of Example 9-19 is retarded by a friction force of 150 N. Determine the required output power (hp) of the motor as a function of time from the moment gravel first starts falling (¢ = 0) until 3.0s after the gravel begins to be dumped off the end of the 22-m-long conveyor belt. *82. (IT) The jet engine of an airplane takes in 120 kg of air per second, which is burned with 4.2 kg of fuel per second. The burned gases leave the plane at a speed of 550 m/s (relative to the plane). If

the plane is traveling 270 m/s (600 mi/h), determine: (a) the™™ thrust due to ejected fuel; (b) the thrust due to accelerated air passing through the engine; and (c) the power (hp) delivered.

r

*83. (IT) A rocket traveling 1850 m/s away from altitude of 6400 km fires its rockets, which speed of 1300 m/s (relative to the rocket). the rocket at this moment is 25,000 kg and of

1.5 n?/s2

ejected

is

desired,

at

what

rate

the Earth at an eject gas at a If the mass of an acceleration

must

the

gases

be

*84. (IIT) A sled filled with sand slides without friction down a 32° slope. Sand leaks out a hole in the sled at a rate of 2.0kg/s. If the sled starts from rest with an initial total mass of 40.0 kg, how

along the slope?

long does

it take the sled to travel 120m

| General Problems 85. A novice pool player is faced with the corner pocket shot shown in Fig, 9-48. Relative dimensions are also shown. Should the player worry that this might be

a “scratch shot,” in

4.0

which the cue ball will also fall into a pocket? Give details. Assume equal mass balls and an elastic collision. FIGURE 9-48 Problem 85.

86. During a Chicago storm, speeds of 120 km/h. If the 45 kg/s per square meter the force of the wind on

1.60m high

r

and

0.50m

winds can whip horizontally at air strikes a person at the rate of and is brought to rest, calculate a person. Assume the person is

wide.

Compare

to the typical

maximum force of friction (u ~ 1.0) between the person and the ground, if the person has a mass of 75 kg. 87. A ball 1s dropped from a height of 1.50m and rebounds to a hexght of 1.20m. Approximately how many rebounds will the balleake before losing 90% of its energy? ?8. In order to convert a tough split in bowling, it is necessary to strik% the pin a glancing blow as shown in Fig. 9-49. Assume that the bowling ball, initially traveling at 13.0m/s, has fiv%‘:imes the mass of a pin and that the pin goes off at 75° from the original direction of the ball. Calculate the

speed (a) of the pin and (b) of

the bafi just after collision, and (c) calculate the angle through which the ball was deflected. Assume the collision is elastic and ignore any spin of the ball.

FIGURE 9-49

Problem 88.

embedded in it? !

|

Qv=310m/s I

FIGURE 9-50

Problem 89.

with the tracks, what is the speed of the car after 60.0 min?

(See Section 9-2.) *93. Consider the railroad car of Problem 92, which is slowly filling with snow. (a) Determine the speed of the car as a function of time using Eqgs. 9-19. (b) What is the speed of the car after 60.0min? Does this agree with the simpler calculation (Problem 92)? 94. Two blocks of mass m, and mg, resting on a frictionless table, are connected by a stretched spring and then released (Fig. 9-51). (a) Is there a net external force on the system? (b) Determine the ratio of their speeds, va/vg. (c) What is the ratio of their kinetic energies? (d) Describe the motion of the cm of this system. (¢) How would the presence of friction alter the above results?

FIGURE 9-51

89. A gun fires a bullet vertically into a 1.40-kg block of wood at rest on a thin horizontal sheet, 1.40 kg Fig. 9-50. If the bullet has a mass of 24.0g and a speed of 310m/s,

how high will the block rise into the air after the bullet becomes

90. A hockey puck of mass 4 m has been rigged to explode, as part of a practical joke. Initially the puck is at rest on a frictionless ice rink. Then it bursts into three pieces. One chunk, of mass m, slides across the ice at velocity wi. Another chunk, of mass 2m, slides across the ice at velocity 2vj. Determine the velocity of the third chunk. L) B8 For the completely inelastic collision of two railroad cars that we considered in Example 9-3, calculate how much of the initial kinetic energy is transformed to thermal or other forms of energy. 92. A 4800-kg open railroad car coasts along with a constant speed of 8.60 m/s on a level track. Snow begins to fall vertically and fills the car at a rate of 3.80 kg/min. Ignoring friction

fl

Problem 94.

95. You have been hired as an expert witness in a court case involving an automobile accident. The accident involved car A of mass 1500 kg which crashed into stationary car B of mass 1100kg. The driver of car A applied his brakes 15 m before he skidded and crashed into car B. After the collision, car A slid 18 m while car B slid 30 m. The coeffi-

cient of kinetic friction between the locked wheels and the road was measured to be 0.60. Show that the driver of car A was exceeding the 55-mi/h (90km/h) speed limit before applying the brakes. 96. A meteor whose mass was about 2.0 X 108kg struck the

Earth (mg = 6.0 X 10**kg) with a speed of about 25 km/s

and came to rest in the Earth. (¢) What was the Earth’s recoil speed (relative to Earth at rest before the collision)? (b) What fraction of the meteor’s kinetic energy was transformed to kinetic energy of the Earth? (¢) By how much did the Earth’s kinetic energy change as a result of this collision? General Problems

245

— y

Lx

-

-7

L

Skeet

-

.-~

-

——

T

-~ i

’1

—

!fv=230m/s

.

Pellet

h

o

S~

FIGURE 9-54

Problem 103.

97. Two astronauts, one of mass 65 kg and the other 85 kg, are initially at rest in outer space. They then push each other apart. How far apart are they when the lighter astronaut has moved 12 m? 98. A 22-g bullet strikes and becomes embedded in a 1.35-kg block of wood placed on a horizontal surface just in front of the gun. If the coefficient of kinetic friction between the block and the surface is 0.28, and the impact drives the block a distance of 8.5 m before it comes to rest, what was the muzzle speed of the bullet?

99. Two

balls, of masses

my =45g

and

mg = 65g,

are

FIGURE 9-52

-{A@

Problem 99.

mp mg

100. A block of mass

m = 2.20kg

-

T

suspended as shown in Fig. 9-52. The lighter ball is pulled away to a 66° angle with the vertical and released. (a) What is the velocity of the lighter ball before impact? (b) What is the velocity of each ball after the elastic collision? (c) What will be the maximum height of each ball after the elastic 30cm collision? may,

103. A 0.25-kg skeet (clay target) is fired at an angle of 28° to

the horizontal with a speed of 25 m/s (Fig. 9-54). When it

reaches the maximum height, £, it is hit from below by a 15-g pellet traveling vertically upward at a speed of 230m/s. The pellet is embedded in the skeet. (¢) How much higher, h', did the skeet go up? (b) How much extra distance, Ax,

does the skeet travel because of the collision? 104. A massless spring with spring constant k is placed between a block of mass m and a block of mass 3 m. Initially the blocks are at rest on a frictionless surface and they are held together so that the spring between them is compressed by an amount D from its equilibrium length. The blocks are then released and the spring pushes them off in opposite directions. Find the speeds of the two blocks when they detach from the spring. 105. The gravitational slingshot effect. Figure 9-55 shows the planet Saturn moving in the negative x direction at its

orbital speed (with respect to the Sun) of 9.6 km/s. The

mass of Saturn is 5.69 X 10 kg, A spacecraft with mass

825 kg approaches Saturn. When far from Saturn, it moves in the +x direction at 10.4 km/s. The gravitational attraction of Saturn (a conservative force) acting on the spacecraft causes it to swing around the planet (orbit shown as dashed line) and head off in the opposite direction. Estimate the final speed of the spacecraft after it is far enough away to be considered free of Saturn’s gravitational pull.

\

slides down a 30.0° incline

which is 3.60m high. At the bottom, it strikes a block of mass

M =7.00kg which is at rest on a horizontal surface, Fig. 9-53. (Assume a smooth transition at the bottom of the incline.) If the collision is elastic, and friction can be ignored, determine (a) the speeds of the two blocks after the collision, and (b) how far back up the incline the smaller mass will go. FIGURE 9-55

K 3.60m

A

1300 FIGURE 9-53

Problems 100 and 101.

101. In Problem 100 (Fig. 9-53), what is the upper limit on mass m if it is to rebound from M, slide up the incline, stop, slide down the incline, and collide with M again? 102. After a completely inelastic collision between two objects of equal mass, each having initial speed, v, the two move off together with speed v/3. What was the angle between their initial directions? 246

CHAPTER9

Linear Momentum

Problem 105.

106. Two bumper cars in an amusement park ride collide elastically as one approaches the other directly from the rear (Fig. 9-56). Car A has a mass of 450kg and car B 490 kg, owing to differences in passenger mass. If car A approaches at 4.50 m/s and car B is moving at 3.70m/s, calculate (a) their velocities after the collision,

and (b) the change in momentum of each.

my = 450 kg o 'UA =

FIGURE 9-56

Problem 106: (a) before collision, (b) after collision.

(a)

(b)

'UB =

450m/s

m&j-vz

3.70 m/s

@'—vz

-

107. In a physics lab, a cube slides down a frictionless incline as shown in Fig. 9-57 and elastically strikes another cube at the bottom that is ( only one-half its mass. If the incline is 35cm high

and the table is 95cm off

112. An extrasolar planet can be detected by observing the wobble it produces on the star around which it revolves. Suppose an extrasolar planet of mass mp revolves around its star of mass m, . If no external force acts on this simple two-object system, then its CM is stationary. Assume m, and mpg are in circular orbits with radii r5 and rg about the system’s cM. (a) Show that

mg

the floor, where does each cube land? [Hint: Both

leave the incline horizontally.]

moving

FIGURE 9-57

‘

fa. = ma M (b) Now consider a Sun-like star and a single planet with the

Problem 107.

same characteristics as Jupiter. That is, mg = 1.0 X 107my and the planet has an orbital radius of 8.0 X 10'! m. Deter-

mine the radius 75 of the star’s orbit about the system’s cMm. (c) When viewed from Earth, the distant system appears to wobble over a distance of 2r,. If astronomers are able to detect angular displacements 6 of about 1 milliarcsec (1 arcsec = 5 of a degree), from what distance d (in light-years) can the star’s wobble be detected

(11ly = 9.46 X 10 m)? (d) The star nearest to our Sun is

108. The space shuttle launches an 850-kg satellite by ejecting it from the cargo bay. The ejection mechanism is activated and is in contact with the satellite for 4.0's to give it a velocity of

about 4ly away. Assuming stars are uniformly distributed throughout our region of the Milky Way Galaxy, about how many stars can this technique be applied to in the search for extrasolar planetary systems? 113. Suppose two asteroids strike head on. Asteroid A

of the shuttle is 92,000 kg. (¢) Determine the component of velocity vy of the shuttle in the minus z-direction resulting

collision, and asteroid B (mg = 1.45 X 10"*kg) has velocity 1.4 km/s before the collision in the opposite direction.

from the ejection. (b) Find the average force that the shuttle

If the asteroids

0.30m/s in the z-direction relative to the shuttle. The mass

exerts on the satellite during the ejection.

109. You are the design engineer in charge of the crashworthiness of new automobile models. Cars are tested by smashing

them into fixed, massive barriers at 45 km/h. A new model

of mass 1500 kg takes 0.15 s from the time of impact until it is broukht to rest. (a) Calculate the average force exerted on the car by the barrier. (b) Calculate the average decel-

(ma = 7.5 X 10"?kg) has velocity 3.3km/s before the

. *Numerical/Computer

~ *114. (III) A particle of mass

eration of the car.

(b)

m,

traveling with speed va

Consider

1 + mp/my now a third particle

of mass

mc

at rest

between m, and mp so that my first collides head on with

asteroid has a mass of 3200kg for each cubic meter of volume and moves toward the Earth at 15km/s. How

mc and then mc collides head on with mp. Both collisions are elastic. Show that in this case,

much destructive energy could be released when it embeds itself in the Earth? (b) For comparison, a nuclear bomb could release about 4.0 X 10'%J. How many such bombs would have to explode simultaneously to release the destructive energy of the asteroid collision with the Earth?

oh

mcna ) A(mc + ma)(mg + mc) (c) From the result of part (b), show that for maximum vy, mc = \/mamg. (d) Assume mg = 2.0kg, ma = 18.0kg and vy = 20m/s. Use a spreadsheet to

111. An astronaut of mass 210 kg including his suit and jet pack wants to acquire a velocity of 2.0 m/s to move back toward his spaée shuttle. Assuming the jet pack can eject gas with a

B

= 4

calculate and graph the values of vg from mc = 0.0kg to mc = 50.0kg in steps of 1.0 kg. For what value of m is the value of vg maximum? Does your numerical result agree with your result in part (c)?

of 35 m/s, what mass of gas will need to be ejected?

Exercises

A: (c) because the momentum change is greater.

E: (b); (d).

B: Larger (41) is greater).

F:

rC:

(rnagni-

oy = _2w

could pose a threat to life on Earth. (a) Assume a spherical

Answers to

is the velocity

collides elastically head-on with a stationary particle of smaller mass mg. (a) Show that the speed of mg after the collision is

110. Astronomers estimate that a 2.0-km-wide asteroid collides with the Earth once every million years. The collision

velocity

stick together, what

tude and direction) of the new asteroid after the collision?

0.50m/s.

D: (a) 6.0m/s; (b) almost zero; (c) almost 24.0 m/s.

Xy = —2.0m; yes.

G: (a). H: The boat moves in the opposite direction.

General Problems

247

You too can experience rapid rotation—if your stomach can take the high angular velocity and centripetal acceleration of some of the faster amusement park rides. If not, try the slower merrygo-round or Ferris wheel. Rotating carnival rides have rotational kinetic energy as well as angular momentum, Angular acceleration is produced by a net torque, and rotating objects have rotational kinetic energy.

Rotational Motion CHAPTER-OPENING

Angular Quantities

Vector Nature of Angular Quantities

Constant Angular Acceleration Torque Rotational Dynamics; Torque and Rotational Inertia Solving Problems in

Rotational Dynamics Determining Moments of Inertia Rotational Kinetic Energy Rotational Plus Translational Motion; Rolling Why Does a Rolling Sphere Slow Down?

248

QUESTION—Guess

Now!

A solid ball and a solid cylinder roll down a ramp. They both start from rest at the same time. Which gets to the bottom first? (a) They get there at the same time. (b) They get there at almost exactly the same time except for frictional differences. (¢)

The ball gets there first.

(d) The cylinder gets there first. (e) Can’t tell without knowing the mass and radius of each. ntil now, we have been concerned mainly with translational motion. We discussed the kinematics and dynamics of translational motion (the role of force), and the energy and momentum associated with it. In this Chapter and the next we will deal with rotational motion. We will discuss the kinematics of rotational motion and then its dynamics (involving torque), as well as rotational kinetic energy and angular momentum (the rotational analog of linear momentum).

Our understanding of the world around

us will be

increased significantly—from rotating bicycle wheels and compact discs to “™ amusement park rides, a spinning skater, the rotating Earth, and a centrifuge—and there may be a few surprises.

We will consider mainly the rotation of rigid objects. A rigid object is an object with a definite shape that doesn’t change, so that the particles composing it stay in fixed positions relative to one another. Any real object is capable of vibrating or deforming when a force is exerted on it. But these effects are often very small, so rthe concept of an ideal rigid object is very useful as a good approximation. Our development of rotational motion will parallel our discussion of translational motion: rotational position, angular velocity, angular acceleration, rotational inertia, and the rotational analog of force, “torque.”

Angular Quantities

r

lrev

=

360°

=

Showing the

ST

-~ /

eket

1

axis to any point in the object sweeps out the same angle 6 in the same time interval. To indicate the angular position of the object, or how far it has rotated, we specify the angle 0 of some particular line in the object (red in Fig. 10-1) with respect to some re ference line, such as the x axis in Fig. 10-1. A point in the object, such as P in Fig. 10-1b, moves through an angle 6 when it travels the distance £ measured along the circumference of its circular path. Angles are commonly stated in degrees, but the mathematics of circular motion is much simpler if we use the radian for angular measure. One radian (abbreviated rad) is defined as the angle subtended by an arc whose length is equal to the radius. For example, in Fig. 10-1, point P is a distance R from the axis of rotation, and it has moved a distance £ along the arc of a circle. The arc length £ is said to “subtend” the angle 6. In general, any angle 6 is given by { 9 = —» [0 in radians] (10-1a) R where R is the radius of the circle and £ is the arc length subtended by the angle 6, which is specified in radians. If £ = R, then 6 = 1rad. The radian, being the ratio of two lengths, is dimensionless. We thus do not have to mention it in calculations, although it is usually best to include it to remind us the angle is in radians and not degrees. We can rewrite Eq. 10-1a in terms of arc length £ f = Ro. (10-1b) Radians can be related to degrees in the following way. In a complete citcle there are 360°, which must correspond to an arc length equal to the circumference of the circle, £ = 2wR. Thus 6 = £/R = 2wR/R = 27 rad in a complete circle, so 360° = 2 rad. One radia n is therefore 360°/27 ~ 360°/6.28 ~ 57.3°. An object that makes one complete revolution (rev) has rotated through 360°, or 27 radians:

FIGURE 10-2

distinction between ¥ (the position vector) and R (the distance from the rotation axis) for a point P on the edge of a cylinder rotating about the Z axis.

R

is R, the distance of that point from the axis of rotation. A straight line drawn from the

FIGURE 10-1 Looking at a wheel that is rotating counterclockwise about an axis through the wheel’s center at O (axis perpendicular to the page). Each point, such as point P, moves in a circular path; £ is the distance P travels as the wheel rotates through the angle 6.

S

The motion of a rigid object can be analyzed as the translational motion of its center of mass plus rotational motion about its center of mass (Sections 9-8 and 9-9). We have already discussed translational motion in detail, so now we focus our attention on purely rotational motion. By purely rotational motion of an object about a fixed axis, we mean that all points in the object move in circles, such as the point P on the rotating wheel of Fig. 10-1, and that the centers of these circles all lie on a line called the is of rotation. In Fig. 10-1 the axis of rotation is perpendicular to the sses through point O. We assume the axis is fixed in an inertial reference page and frame, but we will not always insist that the axis pass through the center of mass. For a ree-dimensional rigid object rotating about a fixed axis, we will use the symbol R to represent the perpendicular distance of a point or particle from the axis of rotation We do this to distinguish R from r, which will continue to represent the position of a particle with reference to the origin (point) of some coordinate system. This distinction is illustrated in Fig. 10-2. This distinction may seem like a small point, but not being fully aware of it can cause huge errors when working with rotational otion. For a flat, very thin object, like a wheel, with the origin in the plane of the object (at the center of a wheel, for example), R and r will be nearly the same. Every point in an object rotating about a fixed axis moves in a circle (shown dashed in Fig. 101 for point P) whose center is on the axis of rotation and whose radius

X

10-

ACAUTION Use radians in calculating, not degrees

2 rad.

SECTION 10-1

Angular Quantities

249

Birds of prey—in radians. A particular bird’s eye can just

distinguish objects that subtend an angle no smaller than about 3 X 10~*rad. (a) How many degrees is this? (b) How small an object can the bird just distinguish when flying at a height of 100 m (Fig. 10-3a)?

APPROACH For (a) we use the relation 360° = 27 rad. For (b) we use Eq. 10—1b,fl £ = R0, to find the arc length.

SOLUTION

(a) We convert 3 X 10 rad to degrees:

(3 x 10“‘rad)


x

=

yt + tar®

[constant @, a]

(10-9b)

for constant

v

=

v} + 2ax

[constant a, a]

(10-9¢)

angular acceleration

_ v

=

v+ v ——-5——9

[constant a, a]

(10-9d)

0 _

=

0} + 206

=

w + o ~——2——9

(¥

=

= =0, 8 = 0)

Note that w, represents the angular velocity at ¢ = 0, whereas 6 and w represent the angular position and velocity, respectively, at time . Since the angular acceleratio

is constant,

| EXAMPLE

(UELW

a = a.

Centrifuge acceleration. A centrifuge rotor is accelerated

in 30s. (a) What is its average angular acceleration? (b) Throu, gh how many revolutions has the centrifuge rotor turned during its acceleratic n period, assuming constant angular acceleration?

from

A

rest to 20,000 rpm

APPROACH

To determine

a = Aw/At,

we need

the initial and

@rpHYSICS Centrifuge

APPLIED

final angular

velocities. For (b), we use Egs. 10-9 (recall that one revolution corresponds to 0 = 2 rad). SOLUTION (@) The initial angular velocity is @ = 0. The final angular velocity is w

Then, since

=

27f

@ = Aw/At

— a

_ =

=

(2mrad/rev) ( and

®—w o

At = 30s,

_ =

2100rad/s 305

20,000 rev/min (60 s/mi/n) )

=

2100 rad/s.

we have

— 0

=

9 70rad/s"

That is, ev ery second

the rotor’s angular velocity increases by 70rad/s, or by (70/27r) 11 revolutions per second. (b) To fin d & we could use either Eq. 10-9b or 10-9c, or both to check our answer. Th e former gives

9 = 0 + 3(70rad/s*)(30s)>

= 3.15 x 10°rad,

where we have kept an extra digit because this is an intermediate result. To find

the total number of revolutions, we divide by 27 rad/rev and obtain 3.15 X 10* rad S van fees NOTE

—

5.0 X 107 3 rev.

Let us calculate 6 using Eq. 10-9c:

9 =

’k

0 — W}

2a

=

(2100rad/s)®> — 0

2(70tad/s%)

which checks our answer using Eq. 10-9b perfectly.

|

=

S

3.15

x

10*

et SECTION 10-3

Constant Angular Acceleration

255

10-4 Torque

FIGURE 10-11 Top view of a door. Applying the same force with

different lever arms, R, and Rg. If Ra = 3Rg, then to create the same

effect (angular 5 acceleration), Fg ] needs

to be three times Fp,or Fy = 3F3.

FIGURE 10-12

(a) A tire iron too

can have a long lever arm. (b) A plumber can exert greater torque using a wrench with a long lever arm.

We have so far discussed rotational kinematics—the description of rotational motion in terms of angular position, angular velocity, and angular acceleration\ Now we discuss the dynamics, or causes, of rotational motion. Just as we four. analogies between linear and rotational motion for the description of motion, so rotational equivalents for dynamics exist as well. To make an object start rotating about an axis clearly requires a force. But the direction of this force, and where it is applied, are also important. Take, for example, an ordinary situation such as the overhead view of the door in Fig. 10-11. If you apply a force F, to the door as shown, you will find that the greater the

magnitude, F, the more quickly the door opens. But now if you apply the same magnitude force at a point closer to the hinge—say, Fy in Fig. 10-11—the door

will not open so quickly. The effect of the force is less: where the force acts, as well as its magnitude and direction, affects how quickly the door opens. Indeed, if only this one force acts, the angular acceleration of the door is proportional not only to the magnitude of the force, but is also directly proportional to the perpendicular

distance from the axis of rotation to the line along which the force acts. This distance

is called the lever arm, or moment arm, of the force, and is labeled R, and Ry for the two forces in Fig. 10-11. Thus, if R, in Fig. 10-11 is three times larger than Ry, then the angular acceleration of the door will be three times as great, assuming that the magnitudes of the forces are the same. To say it another way, if Rx = 3Rg, then Fz must be three times as large as F, to give the same angular acceleration. (Figure 10-12 shows two examples of tools whose long lever arms are very effective.) The angular acceleration, then, is proportional to the product of the force times the lever arm. This product is called the moment of the force about the axis, or, more commonly, it is called the torque, and is represented by 7 (Greek lowercase letter tau). Thus, the angular acceleration a of an object is directly proportional to the net applied torque 7 a X T,

Axis of rotation

Axis of rotation

(a)

(b)

FIGURE 10-13

(a) Forces actingat

different angles at the doorknob.

(b) The lever arm is defined as the

perpendicular distance from the axis iiratakom(ihe Ritge) iiieilie et AT e

and we see that it is torque that gives rise to angular acceleration. This is the rotational analog of Newton’s second law for linear motion, a « F. We defined the lever arm as the perpendicular distance from the axis of rotation to the line of action of the force—that is, the distance which is perpendicular both to the axis of rotation and to an imaginary line drawn along the direction of the force. We do this to take into account the effect of forces acting at an angle. It is clear that a force applied at an angle, such as Fc in Fig. 10-13, will be less effective than the

same magnitude force applied perpendicular to the door, such as F, (Fig. 10-13a).

And if you push on the end of the door so that the force is directed at the hinge (the

axis of rotation), as indicated by Fp, the door will not rotate at all. The lever arm for a force such as F is found by drawing a line along the direction of F (this is the “line of action” of F.). Then we draw another line, perpendicular to this line of action, that goes to the axis of rotation and is perpendicular also to it. The length of this second line is the lever arm for Fc and is labeled R in Fig. 10-13b. The lever arm for FA is the full distance from the hinge to the door knob, R, ; thus

L \
? 20kg -m* + 28kg'm?> = 48kg-m?

L]

10.50m

=

SmR?

Il

(b) The 5.0-kg mass is now 0.50m from the axis, and the 7.0-kg mass is 4.50m from the axis. Then I

’

+— |5.0kg | Axis

(5.0kg)(0.50m)* + (7.0kg)(4.5m)? 1.3kg'm? + 142kg-m*> = 143kg-m>

NOTE This Example illustrates two important points. First, the moment of inertia of a given system is different for different axes of rotation. Second, we see in part (b) that mass close to the axis of rotation contributes little to the total moment of inertia; here, the 5.0-kg object contributed less than 1% to the total.

(a)

A

& 7.0kg

(b)

CAUTION

[ depends on axis of rotation and on distribution of mass

That is, the axis is fixed relative to the object and is fixed in an inertial reference frame. This includes an axis moving at uniform velocity in an inertial frame, since the axis can be considered fixed in a second inertial frame that moves with respect to the first,

SECTION 10-5

Rotational Dynamics; Torque and Rotational Inertia

259

Location

Object

of axis

(a)

Thin hoop radius Ry '

Through center

(b)

Thin hoop,

Through

width w

diameter

radius R,

central

(c)

Solid cylinder,

(d)

Hollow cylinder,

radius R,

'

inner radius R, : outer radius R,

(¢)

(f)

Uniform sphere,

radius r

Long uniform rod length ¢ '

Through

center

Through

center

Through center

Through center

Moment of

inertia

MR;

S

i

"

5MRG + sMw?

%MR(Z)

1

T

MR} + RY)

20

s

g Mr 0

1 M

S FIGURE 10-20

Moments of inertia ~ (2)

for various objects of composition. [We use distance from an axis, distance from a point

uniform R for radial and r for (only in e, the

(h)

sphere), as discussed in Fig. 10-2.]

Long l}niform rod, length

end

Through

1 pp2

Rectangular thin plate,

Through center

L2 12

length £, width w

-

.

+ w2

For most ordinary objects, the mass is distributed continuously, and the calculation

of the moment of inertia, 2mR?,

can be difficult. Expressions can, however, be worked

out (using calculus) for the moments of inertia of regularly shaped objects in terms of the dimensions of the objects, as we will discuss in Section 10-7. Figure 10-20 gives these expressions for a number of solids rotated about the axes specified. The only one for which the result is obvious is that for the thin hoop or ring rotated about an axis passing through its center perpendicular to the plane of the hoop (Fig. 10-20a). For this hoop, all the mass is concentrated at the same distance from the axis, R.

Thus SmR? = (Sm)R} = MR}, where M is the total mass of the hoop.

When calculation is difficult, / can be determined experimentally by measuring the angular acceleration « about a fixed axis due to a known net torque, 27, and applying Newton’s second law, / = Z7/a, Eq.10-14.

10-6

Solving Problems in Rotational Dynamics

When working with torque and angular acceleration (Eq. 10-14), it is important t«® use a consistent set of units, which in SI is: @ in rad/s%* 7 in m-N; and the momen.

of inertia, 7, in kg-m?, 260

CHAPTER 10

Rotational Motion

Q)

sOL

>

?

14

o

Rotational Motion

w

.

o

.

1. As always, draw a clear and complete diagram. 2. Choose the object or objects that will be the system

¢

to be

studied.

3. Draw a free-body diagram for the object under consideration (or for each object, if more than one), showing all (an&] only) the forces acting on that object and exactly where they act, so you can determine the torque due to each. Gravity acts at the ¢G of the object (Section 9-8). 4. Identi fy the axis of rotation and determine the torques about it. Choose positive and negative directions of E

Use

consistent

units, which

in SI are: a in rad/s%

7in m-N; and / in kg-m? . Also apply Newton’s second law for translation, SF = ma, and other laws or principles as needed. 7. Solve the resulting equation(s) for the unknown(s). 8. Do a rough estimate to determine if your answer is reasonable.

A heavy pulley. A 15.0-N force (represented by Fy) is applied

to a cord #apped around a pulley of mass M = 4.00 kg and radius R, = 33.0cm, Fig. 10-21. The pulley accelerates uniformly from rest to an angular speed of 30.0rad/s in 3.00s. If there is a frictional torque 7 = 1.10m-N at the axle, determine the moment of inertia of the pulley. The pulley rotates about its center. APPROACH We follow the steps of the Problem Solving Strategy above. SOLUTIO 1. Draw a diagram. The pulley and the attached cord are shown in Fig. 10-21. 2. Choose the system: the pulley. 3. Draw a free-body diagram. The cord exerts a force Fr on the pulley as shown in Fig, 10-21. The friction force retards the motion and acts all around the axle in a clockwise direction, as suggested by the arrow Fy in Fig. 10-21; we are given its torque, which is all we need. Two other forces should be included in the diagram: the force of gravity mg down, and whatever force holds the axle in place. They do not contribute to the torque (their lever arms are zero) and so we omit them for convenience (or tidiness). 4. Deterl}ine the torques. The torque exerted by the cord equals Ry Fr and is counte clockwise, which we choose to be positive. The frictional torque is given s 74 = 1.10m-N; it opposes the motion and is negative. 5. Apply Newton’s second law for rotation. The net torque is

FIGURE 10-21

Example 10-9.

St = RyFy — 7 = (0.330m)(15.0N) — 1.10m-N = 3.85m-N.

The angular acceleration « is found from the given data that it takes 3.0s to accelerate the pulley from rest to w = 30.0rad/s: Aw 300rad/s — 0 5 a = i 3.00s = 10.0rad/s" We can now solve for 7 in Newton’s second law (see step 7). . Other calculations: None needed. 7. Solve for unknowns. We solve for 7 in Newton’s second law 271 = la, and insert our values for 27 and a;

[=))

’i

LUTIERTEC]

rotation (counterclockwise and clockwise), and assign the correct sign to each torque. . Apply Newton’s second law for rotation, 27 = /a. If the moment of inertia is not given, and it is not the unknown sought, you need to determine it first.

I =

27 o

_

38m'N 10.0 rad/s?

for rotation,

= 0.385kg m°

8. Doar ough estimate. We can do a rough estimate of the moment of inertia by assumi ng the pulley is a uniform cylinder and using Fig. 10-20c:

I ~ MR}

= }(4.00kg)(0.330m)?

= 0.218kg-m>*

This is the same order of magnitude as our result, but numerically somewhat less. Tl nis makes sense, though, because a pulley is not usually a uniform

—

cylinde r but instead has more of its mass concentrated toward the outside edge. S uch a pulley would be expected to have a greater moment of inertia than a solid cylinder of equal mass. A thin hoop, Fig. 10-20a, ought to have a greater I than our pulley, and indeed it does: I = MR} = 0.436 kg-m>.

wwwww

@

PROBLEM

SOLVING

Usefulness and power of rough estimates

SECTION 10-6

261

Pulley and bucket. Consider again the pulley in Fig. 10-21 and Example 10-9 with the same friction. But this time, instead of a constant

15.0-N force being exerted on the cord, we now have a bucket of weight w = 150N (mass m = w/g = 1.53kg) hanging from the cord. See Fig. 10-22a We assume the cord has negligible mass and does not stretch or slip on the pulle'«, (a) Calculate the angular acceleration « of the pulley and the linear acceleration a of the bucket. (b) Determine the angular velocity w of the pulley and the linear velocity » of the bucket at t = 3.00s if the pulley (and bucket) start from rest at 1 = 0. APPROACH This situation looks a lot like Example 10-9, Fig. 10-21. But there is a big difference: the tension in the cord is now an unknown, and it is no longer equal to the weight of the bucket if the bucket accelerates. Our system has two parts: the bucket, which can undergo translational motion (Fig. 10-22b is its free-body diagram); and the pulley. The pulley does not translate, but it can rotate. We apply the rotational version of Newton’s second law to the pulley, 27 = /a, and the linear version to the bucket, 2F = ma. SOLUTION (a) Let F; be the tension in the cord. Then a force Fy acts at the edge of the pulley, and we apply Newton’s second law, Eq. 10-14, for the rotation of

the pulley:

la

=

27

=

RyFp — 7.

[pulley]

Next we look at the (linear) motion of the bucket of mass m. Figure 10-22b, the free-body diagram for the bucket, shows that two forces act on the bucket: the force of gravity mg acts downward, and the tension of the cord Fr pulls upward. Applying Newton’s second law, =F = ma, for the bucket, we have (taking downward as positive): FIGURE 10-22 Example 10-10. (a) Pulley and falling bucket of mass m. (b) Free-body diagram for the bucket.

mg — Fr

=

ma.

[bucket]

Note that the tension Fr, which is the force exerted on the edge of the pulley, i not equal to the weight of the bucket (= mg = 15.0N). There must be a ne. force on the bucket if it is accelerating, so F; < mg. Indeed from the last equation above, Fr = mg — ma. To obtain @, we note that the tangential acceleration of a point on the edge of the pulley is the same as the acceleration of the bucket if the cord doesn’t stretch or slip. Hence we can use Eq. 10-5, ay,, = a = Rya. Substituting Fr = mg — ma = mg — mRya into the first equation above (Newton’s second law for rotation of the pulley), we obtain

Ia = 21 = RyF; — 7 = Ry(mg — mRya) — 7, = mgRy — mRja — . The variable «a appears on the left and in the second term on the right, so we bring that term to the left side and solve for a:

* T

mgRy

—

T R

T )

The numerator (mgR, — ;) is the net torque, and the denominator (I + mRp) is the

total

rotational

m = 1.53kg,

and

a =

inertia

7

of

the

system.

Then,

= 1.10m-N (from Example 10-9),

(

15.0N)(0.330

)(2

0.385kg-m*

)

— 1.10m"-N

=

+ (1.53kg)(0.330m)

since

I = 0.385kg-m’

5 = 6.98rad/s%

The angular acceleration is somewhat less in this case than the 10.0 rad/s® of Example 10-9. Why? Because Fr (= mg — ma) is less than the 15.0-N weight of the bucket, mg. The linear acceleration of the bucket is

4 = Rea = (0330m)(698rad/s?) = 2.30m/s NOTE 262

CHAPTER10

Rotational Motion

The tension in the cord F is less than mg because the bucket accelerates.

>

(b) Since the angular acceleration is constant, after 3.00 s

’ o = w, + at = 0+

(6.98rad/s?)(3.00s)

= 20.9rad/s.

The velocity of the bucket is the same as that of a point on the wheel’s edge:

f‘

v =

Ryw

=

(0.330m)(20.9rad/s)

= 6.91m/s.

DGO EANN Rotating rod. A uniform rod of mass M and length £ can pivot freely (i.e., we ignore friction) about a hinge or pin attached to the case of a large machine, as in Fig. 10-23. The rod is held horizontally and then released. At the moment of release (when you are no longer exerting a force holding it up), determine (@) the angular acceleration of the rod and (b) the linear acceleration of the tip of the rod. Assume the force of gravity acts at the center of mass of the rod, as shown. APPROACH (@) The only torque on the rod about the hinge is that due to gravity, which acts with a force F = Mg downward with a lever arm £/2 at the moment of release (the cM is at the center of a uniform rod). There is also a force on the rod at the hinge, but with the hinge as axis of rotation, the lever arm of this force is zero. The moment of inertia of a uniform rod pivoted about its end is

(Fig. 10-20g) I = M.

' FIGURE 10-23

Mg Example 10-11.

In part (b) we use a;, = Ra.

SOLUTION We use Eq. 10-14, solving for « to obtain the initial angular acceleration of the rod:

4

Lo

I

_ 3 _ 3 iMmP

24

As the rod descends, the force of gravity on it is constant but the torque due to this force is not constant since the lever arm changes. Hence the rod’s angular acceleration is not constant. (b) The linear acceleration of the tip of the rod is found from the relation an = Ra (Eq. 10-5) with R = £ gy

=

la

=

%g .

NOTE The tip of the rod falls with an acceleration greater than g! A small object balanced on the tip of the rod would be left behind when the rod is released. In contrast, the cM of the rod, at a distance £/2 from the pivot, has acceleration

aun = (/2) = 38.

10-7 Determining Moments of Inertia By Experiment The moment of inertia of any object about any axis can be determined experimentally, such as by measuring the net torque 27 required to give the object an angular acceleration a. Then, from Eq. 10-14, I = 27/a. See Example 10-9.

Using Calculus For simple systems of masses or particles, the moment of inertia can be calculated directly, as in Example 10-8. Many objects can be considered as a continuous distribution of mass. In this case, Eq. 10-13 defining moment of inertia becomes

I= J R dm,

(10-16)

'here dm represents the mass of any infinitesimal particle of the object and R is the perpendicular distance of this particle from the axis of rotation. The integral is taken over the whole object. This is easily done only for objects of simple geometric shape.

SECTION 10-7

Determining Moments of Inertia

263

Cylinder, solid or hollow. () Show that the moment of inertia

of a uniform hollow cylinder of inner radius R,, outer radius R,, and mass M,

is 1 =3M(R?+ R}), as stated in Fig. 10-20d, if the rotation axis is through

the center along the axis of symmetry. (b) Obtain the moment of inertia for e solid cylinder. APPROACH We know that the moment of inertia of a thin ring of radius R is mR?. So we divide the cylinder into thin concentric cylindrical rings or hoops of thickness dR, one of which is indicated in Fig. 10-24. If the density (mass per unit volume) is p, then dm FIGURE 10-24 Determining the moment of inertia of a hollow cylinder (Example 10-12).

=

pdV,

where dV is the volume of the thin ring of radius R, thickness dR, and height A. Since dV = (2wR)(dR)(h), we have dm

=

2mwphR dR.

SOLUTION (a) The moment of inertia is obtained by integrating (summing) over all these rings: I =

Jdem

R,

=

J27rth3dR Ry

=

27rph[

R

:

—

1

R

h

‘] = T2> (rs - RY),

where we are given that the cylinder has uniform density, p = constant. (If this were not so, we would have to know p as a function of R before the integration

could be carried out.) The volume V of this hollow cylinder is V = (7R3 — wR3)h, S0 its mass M is M

=

pV

=

pm(R3 — R})h.

Since (R} — R}) = (R3 — R})(R} + R?), we have mph

I = —— (R - R})(R: + Ri) = $M(R} + R3), as stated in Fig, 10-20d. (b) For a solid cylinder, R, = 0 and if we set R, = R,

-

then

I = MR, which is that given in Fig. 10-20c for a solid cylinder of mass M and radius R,.

The Parallel-Axis Theorem There are two simple theorems that are helpful in obtaining moments of inertia. The first is called the parallel-axis theorem. It relates the moment of inertia / of an object of total mass M about any axis, and its moment of inertia /., about an axis passing through the center of mass and parallel to the first axis. If the two axes are a distance & apart, then I = FIGURE 10-25

Example 10-13. Axis

Iy + MK

[parallel axis]

(10-17)

Thus, for example, if the moment of inertia about an axis through the cMm is known, the moment of inertia about any axis parallel to this axis is easily obtained.

DOV

OEAER

Parallel axis. Determine the moment of inertia of a solid

cylinder of radius R, and mass M about an axis tangent to its edge and parallel to its symmetry axis, Fig. 10-25. APPROACH SOLUTION

We use the parallel-axis theorem with Since

& = R,,

[ =

Iy + MW

Iy = 3 MR}

(Fig. 10-20c).

Eq.10-17 gives

= 3MR}.

EXERCISE C In Figs. 10-20f and g, the moments of inertia for a thin rod about two different axes are given. Are they related by the parallel-axis theorem? Please show how.

264

CHAPTER10

Rotational Motion

*Proof of the Parallel-Axis Theorem

y

The proof of the parallel-axis theorem is as follows. We choose our coordinate

system so the origin is at the cm, and Iy is the moment of inertia about the z axis. igure 10-26 shows a cross section of an object of arbitrary shape in the xy plane. Ve let I represent the moment of inertia of the object about an axis parallel to the z axis that passes through the point A in Fig. 10-26 where the point A has coordinates xp and ys. Let x;, ¥, and m; represent the coordinates and mass of an arbitrary particle of the object. The square of the distance from this point

to A is [(x i — xa) + (3 — ya)*]. So the moment of inertia, /, about the axis

/\

§

7

£ \ }

M

s

i

‘\

D(x,, ) ¥; (a0 Va)

0

( \

o

’ /’\‘J

7}

through A

(

P

//%\«/

Em;[(x; — xa)f + (3 = ya)] Emy(x} + i) — 2xaZmx;

— 2ya2m;y;

+ (Zmy)(xx + yA).

The first term on the right is just Iy = Sm;(x} + y}) since the cM is at the origin. The second and third terms are zero since, by definition of the cm,

FIGURE

10-26

Derivation of the

parallel-axis theorem.

Imix; = Zm;y; = 0 because xcy = you = 0. The last term is Mh* since and (x4 + yi) = h* where h is the distance of A from the cM. Thus Em,-=M we have proved I = Iy + Mh?, which is Eq. 10-17. *The Perpendicular-Axis Theorem

r\

The parallel-axis theorem can be applied to any object. The second theorem, the perpendicu ar-axis theorem, can be applied only to plane (flat) objects—that is, to two-dimensional objects, or objects of uniform thickness whose thickness can be neglected c ompared to the other dimensions. This theorem states that the sum of the momen ts of inertia of a plane object about any two perpendicular axes in the plane of th e object, is equal to the moment of inertia about an axis through their point of int ersection perpendicular to the plane of the object. That is, if the object is in the xy plane (Fig. 10-27),

I, =

I + 1I,.

[object in xy plane]

FIGURE 10-27

axis theorem.

The perpendicular-

Z

(10-18)

“Here I, I, I, are moments of inertia about the z, x, and y axes. The proof is simple: sin ce I, = Zm;y?, I, = Sm;x?, and I, = Sm;(x} + y?), Eq. 10-18 follows directly.

10-8 Rotational Kinetic Energy The quanti y 1m1? is the kinetic energy of an object undergoing translational motion. An object rotating about an axis is said to have rotational kinetic energy. By analogy with translational kinetic energy, we would expect this to be given by the expression § /w® where I is the moment of inertia of the object and w is its angular velocity. We can indeed show that this is true. Consider any rigid rotating object as made up of many tiny particles, each of mass m;. If we let R; represent the distance of any one particle from the axis of rotation, then its linear velocity is v; = R;w. The total kinetic energy of the whole object will be the sum of the kinetic energies of all its particles:

K

=

ZGmp})

=

S(GmRie?)

==

%E(mlR,z)wz

We have factored out the § and the w? since they are the same for every particle of a rigid object. Since =m; R? = I, the moment of inertia, we see that the kinetic energy, K, of an object rotating about a fixed axis is, as expected,

K rf

= 1l

[rotation about a fixed axis]

(10-19)

the axis is not fixed in space, the rotational kinetic energy can take on a more comphcated form.

SECTION 10-8

Rotational Kinetic Energy

265

The work done on an object rotating about a fixed axis can be written in terms of angular quantities. Suppose a force F is exerted at a point whose distance from the axis of rotation is R, as in Fig. 10-28. The work done by this force is

W

=

= JF-dl

=

>

fFleB,

where dl is an infinitesimal distance perpendicular to R with magnitude df = R db, and F, is the component of F perpendicular to R and parallel to df (Fig. 10-28). But F|, R is the torque about the axis, so FIGURE 10-28 Calculating the work done by a torque acting on a rigid object rotating about a fixed axis.

W

=

6,

deo

(10-20)

6,

is the work done by a torque 7 to rotate an object through the angle 6, — 6. The rate of work done, or power P, at any instant is P

=

—_—=r—

=

T,

(10‘21)

The work-energy principle holds for rotation of a rigid object about a fixed axis. From Eq. 10-14 we have

do _ dt

r=1Ila =

dodd _ —do do dat ~ '“ae

where we used the chain rule and w = df/dt. Then

W

=

)

deo

w,

=

6,

ledw

= o}

7df = Iw dw

— 11e?.

and

(10-22)

oy

This is the work-energy principle for a rigid object rotating about a fixed axis. It stat« that the work done in rotating an object through an angle 6, — 6, is equal to the change in rotational kinetic energy of the object.

@pnvsncs

APPLIED

Energy from a flywheel

EXAMPLE 10-14

Flywheel. Flywheels, which are simply large

rotating disks, have been suggested as a means of storing energy for solarpowered generating systems. Estimate the kinetic energy that can be stored in an 80,000-kg (80-ton) flywheel with a diameter of 10m (a three-story building). Assume it could hold together (without flying apart due to internal stresses) at 100 rpm. APPROACH We use Eq. 10-19, in rad/s. SOLUTION We are given

K = Iw?,

but only after changing 100 rpm to o

w = 100rpm = (100 min re_v)(lmm>

=

The initial kinetic energy is zero. Hence, from the work-energy principle, 17 2 = Lie? 21w Since I = 1M£ for w:

s — foe—olr

Rotating rod. A rod of mass M is pivoted on a frictionless

hinge at one end, as shown in Fig. 10-29. The rod is held at rest horizontally and then released. Determine the angular velocity of the rod when it reaches the vertical position, and the speed of the rod’s tip at this moment.

FIGURE 10-29

Example 10-15.

{ Mg=. g2

for a rod pivoted about its end (Fig. 10-20g), we can solve

_ /%8 “ TN The tip of the rod will have a linear speed (see Eq. 10-4) v =

o

NOTE By comparison, v = V2gl

=

V3gl. an object that falls vertically a height £ has a speed

EXERCISE D Estimate the energy stored in the rotational motion of a hurricane. Model the hurricane as a uniform cylinder 300 km in diameter and 5 km high, made of air whose mass Ais 1.3 kg per m®. Estimate the outer edge of the hurricane to move at a speed of 200 km/h. {

Rotational Plus Translational

Motion; Rolling Rolling W ithout Slipping

The rolling motion of a ball or wheel is familiar in everyday life: a ball rolling across the flc yor, or the wheels and tires of a car or bicycle rolling along the pavement. Rolling with out slipping depends on static friction between the rolling object and the ground. The friction is static because the rolling object’s point of contact with the ground is at rest at each moment. Rolling without slipping involves both rotation and translation. There is a simple relation between the linear speed v of the axle and the angular velocity w of the rotating wheel or sphere: namely,

v = Rw,

where R is the radius, as we now show.

FIGURE 10-30 (a) A wheel rolling to the right. Its center C moves with velocity ¥. Point P is at rest at this instant. (b) The same wheel as seen from a reference frame in which the axle of the wheel Cis at rest—that is, we are moving to the right with velocity ¥ relative to the ground. Point P, which was at rest in (a),

here in (b) is moving to the left with velocity —V as shown. (See also Section 3-9 on relative velocity.)

Figure 10-30a shows a wheel rolling to the right without slipping. At the instant shown, point P on the wheel is in contact with the ground and is momentarily at rest. The velocity of the axle at the wheel’s center C is v. In Fig. 10-30b we have put ourselves in the reference frame of the wheel—that is, we are moving to the right with velocity ¥ relative to the ground. In this reference frame the axle C is at rest, whereas the ground and point P are moving to the left with velocity —v as shown. Here we are seeing pure rotation. We can then use Eq. 10-4 to obtain v = Rw, where R is the radius of the wheel. This is the same » as in Fig. 10-30a, so we see that the linear speed v of the axle relative to the ground is related to the angular velocity o by r

v = Row.

[rolling without slipping]

(b)

This is valid only if there is no slipping.

SECTION 10-9

Rotational Plus Translational Motion; Rolling

267

FIGURE 10-31 (a) A rolling wheel rotates about the instantaneous axis (perpendicular to the page) passing through the point of contact with the ground, P. The arrows represent the

instantaneous velocity of each point. (b) Photograph of a rolling wheel. The spokes are more blurred where the speed is greater.

(b) Instantaneous Axis When a wheel rolls without slipping, the point of contact of the wheel with the ground is instantaneously at rest. It is sometimes useful to think of the motion of the wheel as pure rotation about this “instantaneous axis” passing through that point P (Fig. 10-31a). Points close to the ground have a small linear speed, as they are close to this instantaneous axis, whereas points farther away have a greater linear speed. This can be seen in a photograph of a real rolling wheel (Fig. 10-31b): spokes near the top of the wheel appear more blurry because they are moving faster than those near the bottom of the wheel.

FIGURE 10-32

A wheel rolling

without slipping can be considered

as translation of the wheel as a whole bonthewith Gvelocity V¢ plus rotation abo

€

.

Translation

Rotation

vV

/

Total Kinetic Energy = K¢y + Ko

An object that rotates while its center of mass (cM) undergoes translational

motion will have both translational and rotational kinetic energy. Equation 10-19, = 1% gives the rotational kinetic energy if the rotation axis is fixed. If the object is moving, such as a wheel rolling along the ground, Fig. 10-32, this equation is still valid as long as the rotation axis is fixed in direction. To obtain the total kinetic energy, we note that the rolling wheel undergoes pure rotation about its instantaneous

point

of contact

discussion of Fig. 10-30, speed of a point on the speeds are related to the point P is the same w for Kiot

=

P, Fig. 10-31. As

we

saw

the speed v of the cM relative edge of the wheel relative to radius R by » = wR. Thus the the wheel about its center, and

above

relative

to tl'f«,

to the ground equals the its center. Both of these angular velocity w about the total kinetic energy is

%IP w’,

where I; is the rolling object’s moment of inertia about the instantaneous axis at P. We can write K, in terms of the center of mass using the parallel-axis theorem: I, = Iy + MR? where we have substituted & = R in Eq. 10-17. Thus

Rolling

KtOt

=

%ICMCUZ

+

%MRZC()Z.

But Rw = vy, the speed of the center of mass. So the total kinetic energy of a rolling object is Kt

=

% ey o

+

%MU%M s

(10-23)

where vcy is the linear velocity of the cM, Iy is the moment of inertia about an axis through the cM, w is the angular velocity about this axis, and M is the total mass of the object.

Sphere rolling down an incline. What will be the speed of

a solid sphere of mass M and radius r, when it reaches the bottom of an incline if it starts from rest at a vertical height H and rolls without slipping? See Fig. 10-33. (Assume no slipping occurs because of static friction, which does no work.) Compare your result to that for an object sliding down a frictionless incline. APPROACH We use the law of conservation of energy with gravitational potentiz energy, now including rotational as well as translational kinetic energy. 268

CHAPTER10

Rotational Motion

\

SOLUTION The total energy at any point a vertical distance y above the base of the incline is

IMY + Jlyo® + Mgy,

"where v is the speed of the center of mass, and Mgy is the gravitational potential energy. Applying conservation of energy, we equate the total energy at the top (y=H, v=0, ®=0) to the total energy at the bottom (y = 0):

0+ 0+ MgH

= Mv* + $Iyw® + 0.

FIGURE 10-33 A sphere rolling down a hill has both translational and rotational kinetic energy. Example 10-16.

The moment of inertia of a solid sphere about an axis through its center of mass is Iy = $Mr}, Fig. 10-20e. Since the sphere rolls without slipping, we have o = v/ry (recall Fig. 10-30). Hence 2

MgH

=

%Mv2

+ %(%Mr%)(%)

Canceling the M’s and r,’s we obtain or

0

@+ 50 = gH

We can compare this result for the speed of a rolling sphere to that for an object sliding down a plane without rotating and without friction, $mv* = mgH (see our energy equation above, removing the rotational term). For the sliding object, v = V2gH, which is greater than for a rolling sphere. An object sliding without friction or rotation transforms its initial potential energy entirely into translational kinetic energy (none into rotational kinetic energy), so the speed of its center of mass is greater. NOTE Our result for the rolling sphere shows (perhaps surprisingly) that v is independent of both the mass M and the radius r, of the sphere.

A CAUTION

Rolling objects go slower than sliding objects because of rotational kinetic energy

e

| CONCEPTUAL

EXAMPLE

10-17 | Which is fastest? Several objects roll without

fiipping down an incline of vertical height H, all starting from rest at the same © ' moment. The objects are a thin hoop (or a plain wedding band), a spherical marble, a solid cylinder (a D-cell battery), and an empty soup can. In what order do they reach the bottom of the incline? Compare also to a greased box that slides down an incline at the same angle, ignoring sliding friction. RESPONSE We use conservation of energy with gravitational potential energy plus rotational and translational kinetic energy. The sliding box would be fastest because the potential energy loss (MgH) is transformed completely into translational kinetic energy of the box, whereas for rolling objects the initial potential energy is shared between translational and rotational kinetic energies, and so the speed of the cM is less. For each of the rolling objects we can state that the loss in potential energy equals the increase in kinetic energy:

= iMv* + LIo

For all our rolling objects, the moment of inertia Iy is a numerical factor times the mass M and the radius R? (Fig. 10-20). The mass M is in each term, so the translational speed vcy doesn’t depend on M; nor does it depend on the radius R since w = v/R, so R* cancels out for all the rolling objects. Thus the speed v at the bottom depends only on that numerical factor in I, which expresses how the mass is distributed. The hoop, with all its mass concentrated at radius R (Icy = MR?), has the largest moment of inertia; hence it will have the lowest

vey and will arrive at the bottom behind the D-cell (I = §MR?), which in

be behind the marble (Iy = 3 MR?). The empty can, which is mainly a

hoop plus a small disk, has most of its mass concentrated at R; so it will be a bit faster than the pure hoop but slower than the D-cell. See Fig. 10-34. NOTE The objects do not have to have the same radius: the speed at the bottom does not depend on the object’s mass M or radius R, but only on the shape (and Lhe height of the hill H).

SECTION 10-9

Example 10-17.

Solid cylinder (D-cell) @) _ Sphere (marble) Box (sliding)

T

turn will

FIGURE 10-34

——

MgH

Rotational Plus Translational Motion; Rolling

e

269

If there had been little or no static friction between the rolling objects and the plane in these Examples, the round objects would have slid rather than rolled, or a combination of both. Static friction must be present to make a round object roll. We did not need to take friction into account in the energy equation for the rolling

objects because it is static friction and does no work—the point of contact of ™

FIGURE 10-35 A sphere rolling to the right on a plane surface. The point in contact with the ground at any moment, point P, is momentarily at rest. Point A to the left of P is moving nearly vertically upward at

the instant shown, and point B to the

right is moving nearly vertically

downward. An instant later, point B will touch the plane and be at rest momentarily. Thus no work is done by the force of static friction.

4

A CAUTION Whenis =t = la valid?

sphere at each instant does not slide, but moves perpendicular to the plane (first down and then up as shown in Fig. 10-35) as it rolls. Thus, no work is done by the static friction force because the force and the motion (displacement) are perpendicular. The reason the rolling objects in Examples 10-16 and 10-17 move down the slope more slowly than if they were sliding is not because friction slows them down. Rather, of the gravitional potential s i it is because some ;i ] Penergy is converted tO rotational kinetic energy, leaving less for the translational kinetic energy.

EXERCISE E Return to the Chapter-Opening Question, p. 248, and answer it again now.

Try to explain why you may have answeted differently the first time. Using

STeom

= Tom@en .

.

.

’

.

We can examine objects rolling down a plane not only from the point of view of kinetic energy, as we did in Examples 10-16 and 10-17, but also in terms of forces and torques. If we calculate torques about an axis fixed in direction (even if the axis is accelerating) which passes through the center of mass of the rolling sphere, then Z1em

=

Iemacw

is valid, as we discussed in Section 10-5. See Eq. 10-15, whose validity we will show in Chapter 11. Be careful, however: Do not assume =7 = [ is always valid. You cannot just calculate 7, 7, and @ about any axis unless the axis is (1) fixed in an inertial reference frame or (2) fixed in direction but passes through the cM of the object.

Analysis of a sphere on an incline using forces. Analyze

FIGURE 10-36

Example 10-18.

the rolling sphere of Example 10-16, Fig. 10-33, in terms of forces particular, find the velocity » and the magnitude of the friction force, APPROACH We analyze the motion as translation of the cM plus the cM. Fy is due to static friction and we cannot assume Fy Fiy = psFy. SOLUTION For translation in the x direction we have from

ZF

and torques. In F, , Fig. 10-36. rotation abou! = ugFy, only

= ma,

Mgsin® — F, = Ma, and in the y direction Fy — Mgcos6 = 0 since there is no acceleration perpendicular to the plane. This last equation merely tells us the magnitude of the normal force, Fy = Mgcosé. For the rotational motion about the cMm, we use Newton’s second law for rotation Stem = Iewacy (Eq. 10-15), calculating about an axis passing through the cm but fixed in direction:

Fyro

= (Mrd)a.

The other forces, Fy and Mg, point through the axis of rotation (cM), so have lever arms equal to zero and do not appear here. As we saw in Example 10-16 and Fig. 10-30, o = v/ry where v is the speed of the cM. Taking derivatives of w = v/rg with respect to time we have « = a/ry; substituting into the last equation we find Ff,-

=

%Ma

When we substitute this into the top equation, we get Mgsind — $Ma = Ma, or a

=

5

.

3gsinb.

-

We thus see that the acceleration of the cM of a rolling sphere is less than tha for an object sliding without friction (a = gsin#). The sphere started from rest 270

CHAPTER10

Rotational Motion

at the top

Eq. 2-12c

of the incline (height H). To find the speed v at the bottom we use

where the total distance traveled along the plane is x = H/sin6

(see Fig. T0—36). Thus v

=

V2ax

S H \/2

=

_ =

10 7 gH.

This is the same result obtained in Example 10-16 although less effort was needed there. To get the magnitude of the force of friction, we use the equations obtained above:

Fy

= $Ma

= iM(3gsinf)

= 3Mgsiné.

NOTE If the coefficient of static friction is sufficiently small, or 6 sufficiently large so that Fy, > ug Fy (that is, if " tan 6 > ), the sphere will not simply roll but will slip as it moves down the plane.

*More Advanced Examples Here we do three more Examples, all of them fun and interesting. When they use 27 = [a, we must remember that this equation is valid only if 7, @, and I are calculated about an axis that either (1) is fixed in an inertial reference frame, or (2) passes through the cM of the object and remains fixed in direction.

DOVIERMOESER

A falling yo-yo. String is wrapped around a uniform solid

cylinder (something like a yo-yo) of mass M and radius R, and the cylinder starts falling from rest, Fig. 10-37a. As the cylinder falls, find (a) its acceleration and (b) the tension in the string.

Fr

APPROACH As always we begin with a free-body diagram, Fig. 10-37b, which shows the weight of the cylinder acting at the cM and the tension of the string Fr acting at the edge of the cylinder. We write Newton’s second law for the linear motion (down is positive) Ma = ZF =

Mg

-

FT‘

Since we do not know the tension in the string, we cannot immediately solve for a. So we try Newton’s center of mass:

second law for the rotational motion, calculated about the

21em

=

Iemacu

FrR

=

%MRza.

Because the cylinder “rolls without slipping” down the string, we have the additional relation that a = aR (Eq. 10-5). SOLUTION The torque equation becomes

=

FrR

=

;MR

=

%Ma.

yr(L)

(R)

=t =

2MRa

(b) FIGURE 10-37

Mg

Example 10-19.

SO FT

Substituting this into the force equation, we obtain Ma

=

Mg

— F;

=

Mg

- iMa.

Solving for a, we find that a = %g. That is, the linear acceleration is less than what it would be if the cylinder were simply dropped. This makes sense since gravity is not the only vertical force acting; the tension in the string is acting as well. (b) Since a = %g, Fr = 1Ma = 1 Mg. EXERCISE F Find the acceleration a of a yo-yo whose spindle has radius

moment of inertia is still MR? (ignore the mass of the spindle). "Fir > g Fy is equivalent to tan8 > Zug because

Fyy = 3Mgsin6

R. Assume the

and us Fy = s Mg cos#.

SECTION 10-9

Rotational Plus Translational Motion; Rolling

271

What if a rolling ball slips? A bowling ball of mass M and

radius r, is thrown along a level surface so that initially (¢ = 0) it slides with a linear speed v, but does not rotate. As it slides, it begins to spin, and eventually rolls without slipping. How long does it take to begin rolling without slipping?

FIGURE 10-38

Example 10-20.

APPROACH The free-body diagram is shown in Fig. 10-38, with the ball moving t. the right. The friction force does two things: it acts to slow down the translational motion of the cM; and it immediately acts to start the ball rotating clockwise. SOLUTION Newton’s second law for translation gives

May, = 2F, = —Fr = —mhy

= —mcMg,

where py is the coefficient of kinetic friction because the ball is sliding. Thus a, = —ug. The velocity of the cM is Vem

=

Y

t ayt

=

vy — uxgt.

Next we apply Newton’s second law for rotation about the cM,

The angular acceleration is thus acy = 5 ug g/2r,, angular velocity of the ball is (Eq. 10-9a) Wy

=

Wy

T oacyt

=

0 +

Icyacy = Z7cey:

which is constant. Then the

S8t o

0

The ball starts rolling immediately after it touches the ground, but it rolls and slips at the same time to begin with. It eventually stops slipping, and then rolls without slipping. The condition for rolling without slipping is that Uem

=

Wcmlos

which is Eq. 10-4, and is not valid if there is slipping. This rolling without slipping begins at a time f =t given by vy = wenty and we apply the equations for vey and wgy above:

S pmgh

Vo — Mk8h

= 2—r0r0

SO 2'00

t, =

@PHYSICs

APPLIED

Braking distribution of a car

FIGURE 10-39

Forces on a

braking car (Example 10-21).

F

[*~—1.5m—I=—1.5m—

CHAPTER10

.

FUIITNTEF Il ESTIVATE | Braking a car. When the brakes of a car are

applied, the front of the car dips down a bit; and the force on the front tires is greater than on the rear tires. To see why, estimate the magnitude of the normal forces, Fy; and Fy,, on the front and rear tires of the car shown in Fig. 10-39 when

the car brakes and decelerates at a rate a = 0.50 g. The car has mass M = 1200 kg,

the distance between the front and rear axles is 3.0 m, and its cM (where the force of gravity acts) is midway between the axles 75 cm above the ground.

APPROACH Figure 10-39 is the free-body diagram showing all the forces on the car. F; and F, are the frictional forces that decelerate the car. We let F; be the sum of the forces on both front tires, and F, likewise for the two rear tires. Fy, and Fy, are the normal forces the road exerts on the tires and, for our estimate, we assume the static friction force acts the same for all the tires, so that F, and F, are proportional respectively to Fy; and Fy,: F

272

7px8

Rotational Motion

=

pFy

and

F,

=

uhy.

SOLUTION law gives

P

The friction forces F; and F, decelerate the car, so Newton’s second Fl

+

F2

=

Ma

(1200 kg)(0.50)(9.8 m/s2) = S900N.

)

While the car is braking its motion is only translational, so the net torque on the car is zero. If we calculate the torques about the cMm as axis, the forces Fy, F,, and Fyxs all act to rotate the car clockwise, and only Fy; acts to rotate it counterclockwise; so Fy; must balance the other three. Hence, Fy; must be significantly greater than Fy,. Mathematically, we have for the torques calculated about the cMm:

(1.5m)Fy — (1.5m)Fy, — (0.75m)F, — (0.75m)F, = O. Since F, and F, are proportional” to Fy, and Fy, (F, = pFy,

F, = pFy),

can write this as

(L5m)(Fy

— Fyo) — (0.75m)(u)(Fy

+ Fy)

= 0.

we (ii)

Also, since the car does not accelerate vertically, we have Mg =

Fy

+ by, =

F, + F,

Comparing (iii) to (i), we see that and use p = 0.50 to obtain by Aflus

=

2+

FN2

(iii)

a

u = a/g = 0.50.

=

Now we solve (ii) for Fy;

)

3 Fy.

Fy s 1% times greater than Fy,. Actual magnitudes are

determined from

(iii) and (i): Fy; + Fyo = (5900N)/(0.50) = 11,800 N which equals Fy,(1 + 3);

so Fy, = 4400N and Fy; = 7400 N. NOTE Because the force on the front tires is generally greater than on the rear tires, cars are often designed with larger brake pads on the front wheels than on the rear. Or, to say it another way, if the brake pads are equal, the front ones wear out a lot faster.

*10-10 Why Does a Rolling Sphere Slow Down? A sphere of mass M and radius r, rolling on a horizontal flat surface eventually comes to rest. What force causes it to come to rest? You might think it is friction, but when you examine the problem from a simple straightforward point of view, a paradox seems to arise. Suppose a sphere is rolling to the right as shown in Fig. 10-40, and is slowing =~ FIGURE 10-40 down. By Newton’s second law, SF = Ma, there must be a force F (presumably the right. frictional) acting to the left as shown, so that the acceleration & will also point to the left and » will decrease. Curiously enough, though, if we now look at the torque equation (calculated about the center of mass), Zr1cy = Ioy@, We see that the force F acts to increase the angular acceleration a, and thus to increase the velocity of the sphere. Thus the paradox. The force F acts to decelerate the sphere if we look at the translational motion, but speeds it up if we look at the rotational motion.

Sphere rolling to

Our proportionality constant u is not equal to us, the static coefficient of friction (Fy, =< p Fx), unless the car is just about to skid.

*SECTION 10-10

Why Does a Rolling Sphere Slow Down?

273

The resolution of this apparent paradox is that some other force must be acting. The only other forces acting are gravity, Mg, and the normal force Fy (= —Mg). These act vertically and hence do not affect the horizontal translational motion. If we assume the sphere and plane are rigid, so the sphere is in contact at only one point, these forces give rise to no torques about the cMm either,‘ since they act through the cm. The only recourse we have to resolve the paradox is to give up our idealization that the objects are rigid. In fact, all objects are deformable to some extent. Our sphere flattens slightly and the level surface also acquires a slight depression where the two are in contact. There is an area of contact, not a point. Hence there can be a torque at this area of contact which acts in the opposite direction to the torque associated with F, and thus acts to slow down the rotation of the sphere. This torque is associated with the normal force Fy that the table exerts on the sphere over the whole area of contact. The net effect is that we can consider Fy acting vertically a distance £ in front of the cm as shown in Fig. 10-41 (where the deformation is greatly exaggerated). Is it reasonable that the normal force Fy should effectively act in front of the cM as shown in Fig. 10-41? Yes. The sphere is rolling, and the leading edge strikes the surface with a slight impulse. The table therefore pushes upward a bit more strongly on the front part of the sphere than it would if the sphere were at rest. At the back part of the area of contact, the sphere is starting to move upward and so the table pushes upward on it less strongly than when the sphere is at rest. The table pushing up more strongly on the front part of the area of contact gives rise to the necessary torque and justifies the effective acting point of Fy being in front of the cm. When other forces are present, the tiny torque ry due to Fy can usually be ignored. For example, when a sphere or cylinder rolls down an incline, the force of gravity has far more influence than 7y, so the latter can be ignored. For many purposes (but not all), we can assume a hard sphere is in contact with a hard surface at essentially one point. “-

Mg FIGURE 10-41 The normal force, Fy, exerts a torque that slows down the sphere. The deformation of the sphere and the surface it moves on has been exaggerated for detail.

|Summary When a rigid object rotates about a fixed axis, each point of the

object moves in a circular path. Lines drawn perpendicularly from the rotation axis to different points in the object all sweep out the same angle 6 in any given time interval. Angles are conveniently measured in radians. One radian is the angle subtended by an arc whose length is equal to the radius, or 27 rad

=

360°

SO

lrad

=

57.3°.

All parts of a rigid object rotating about a fixed axis have the same angular velocity w and the same angular acceleration o at any instant, where w

=

de d_l

(10-2b)

a

=

dw i

(10-3b)

and

components of the linear acceleration. The period T are related to w (rad/s) by

frequency

f and

w

=

2uf

10-7)

T

=

1/f.

(10-8)

Angular velocity and angular acceleration are vectors, For a rigid object rotating about a fixed axis, both @ and & point along the rotation axis. The direction of @ is given by the right-hand rule. If a rigid object undergoes uniformly accelerated rotational motion (@ = constant), equations analogous to those for linear motion are valid: w

=

wy

t

at;

P = ol 20

’]

B

wol + %atz; @ & o

(10-9)

2

The units of w and « are rad/s and rad/s.

The torque due to a force F exerted on a rigid object is equal to T = R F = RF, = RFsing, (10-10)

v = Ro an = Ra

where R, called the lever arm, is the perpendicular distance from the axis of rotation to the line along which the force acts, and 6 is the angle between F and R. The rotational equivalent of Newton’s second law is

The linear velocity and acceleration of any point in an object rotating about a fixed axis are related to the angular quantities by

ag = o’R

(10-4) (10-5) (10-6)

where R is the perpendicular distance of the point from the rotation axis, and a,, and ag are the tangential and radial 274

CHAPTER10

Rotational Motion

21

=

la,

(10—14)“

where I = Em; R} is the moment of inertia of the object about the axis of rotation. This relation is valid for a rigid object

rotating about

an axis fixed in an inertial reference

frame, or

when 7, I, and « are calculated about the center of mass of an

object even if the cM is moving.

The rotational kinetic energy of an object rotating about a rf,ixed axis with angular velocity w is

= 11o? For

(10-19)

an

object undergoing both translational and rotational motion, the total kinetic energy is the sum of the translational kinetic energy of the object’s cM plus the rotational kinetic energy of the object about its cMm: Kiot

=

%M D%M

+ %ICsz

(10-23)

as long as the rotation axis is fixed in direction.

The following Table summarizes angular (or rotational) quantities, comparing them to their translational analogs. Translation

Rotation

Connection

X

7}

x = R6

v

0]

v = Rw

a

o

a = Ra

F

T

7 = RFsinf

m

I

K =imv? W =Fd 2F = ma

I = EmR?

11? W =16 27 = la

| Questions 1 A bicycle odometer (which counts revolutions and is calibrated to report distance traveled) is attached near the wheel hub and is calibrated for 27-inch wheels. What happens if you use it on a bicycle with 24-inch wheels? . Suppose a disk rotates at constant angular velocity. Does a point on the rim have radial and/or tangential acceleration? If the disk’s angular velocity increases uniformly, does the point ha1ve radial and/or tangential acceleration? For which cases would the magnitude of either component of linear acceleration change? . Could a nonrigid object be described by a single value of the angular velocity w? Explain, Can a small force ever exert a greater torque than a larger force? Explain. Why is it more difficult to do a sit-up with your hands behind your head than when your arms are stretched out in front of you? A diagram may help you to answer this. . Mammals that depend on being able to run fast have slender lower legs with flesh and muscle concentrated high, close to the body (Fig. 10-42). On the basis of rotational dynamics, explain why this distribution of mass is advantageous.

10. Two solid spheres simultaneously start rolling (from rest) down an incline. One sphere has twice the radius and twice the mass of the other. Which reaches the bottom of the incline first? Which has the greater speed there? Which has the greater total kinetic energy at the bottom? 11. Why do tightrope walkers (Fig. 10-43) carry a long, narrow beam?

FIGURE

FIGURE 10-42 Question 6.

A gazelle. .

7 If the net force on a system is zero, is the net torque also

zero? If the net torque on a system is zero, is the net force zero? . Two inclines have the same height but make different angles with the horizontal. The same steel ball is rolled down each incline. On which incline will the speed of the ball at the bottom be greater? Explain. f9. Two spheres look identical and have the same mass.

However, one is hollow and the other is solid. Describe an

experiment to determine which is which.

10-43

Question 11.

12. A sphere and a cylinder have the same radius and the same mass. They start from rest at the top of an incline. Which reaches the bottom first? Which has the greater speed at the bottom? Which has the greater total kinetic energy at the bottom? Which has the greater rotational kinetic energy? 13. The moment of inertia of this textbook would be the least about which symmetry axis through its center? 14. The moment of inertia of a rotating solid disk about an axis

through its cM is 3 MR? (Fig. 10-20c). Suppose instead that a parallel axis of rotation passes through a point on the edge of the disk. Will the moment

or smaller?

of inertia be the same, larger,

15. The angular velocity of a wheel rotating on a horizontal axle points west. In what direction is the linear velocity of a point on the top of the wheel? If the angular acceleration points east, describe the tangential linear acceleration of this point at the top of the wheel. Is the angular speed increasing or decreasing?

Questions

275

| Problems 10-1

Angular Quantities

1. (I) Express the following angles in radians: (a) 45.0°, (b) 60.0°, (c) 90.0°, (d) 360.0°, and (e) 445°. Give as numerical values and as fractions of . 2. (I) The Sun subtends an angle of about 0.5° to us on Earth, 150 million km away. Estimate the radius of the Sun. 3. (I) A laser beam is directed at the Moon, 380,000 km from Earth. The beam diverges at an angle 6 (Fig. 10-44) of

1.4 X 10 rad. What diameter spot will it make on the

Moon?

12. (II) A 64-cm-diameter wheel accelerates uniformly about its center from 130rpm to 280 rpm in 4.0s. Determine (a) its angular acceleration, and (b) the radial and tangentia«. components of the linear acceleration of a point on the edge of the wheel 2.0s after it has started accelerating, 13. (II) In traveling to the Moon, astronauts aboard the Apollo spacecraft put themselves into a slow rotation to distribute the Sun’s energy evenly. At the start of their trip, they accelerated from no rotation to 1.0 revolution every minute during a 12-min time interval. The spacecraft can be thought of as a cylinder with a diameter of 8.5 m. Determine (a) the angular acceleration, and (b) the radial and tangential components of the linear acceleration of a point on the skin of the ship 7.0 min after it started this acceleration. 14. (IT) A turntable of radius R; is turned by a circular rubber roller of radius R, in contact with it at their outer edges. What is the ratio of their angular velocities, w;/w,?

10-2 Vector Nature of @ and & 15. (II) The axle of a wheel is mounted on supports that rest on a rotating turntable as shown in Fig. 10-46. The wheel has

FIGURE 10-44 Problem 3. 4. (I) The blades in a blender rotate at a rate of 6500 rpm. When the motor is turned off during operation, the blades slow to rest in 4.0s. What is the angular acceleration as the blades slow down? 5. (II) (a) A grinding wheel 0.35m in diameter rotates at 2500 rpm. Calculate its angular velocity in rad/s. (b) What are the linear speed and acceleration of a point on the edge of the grinding wheel? 6. (IT) A bicycle with tires 68 cm in diameter travels 7.2 km. How many revolutions do the wheels make? 7. (II) Calculate the angular velocity of (a) the second hand, (b) the minute hand, and (c) the hour hand, of a clock. State in rad/s. (d) What is the angular acceleration in each case? 8. (II) A rotating merry-go-round makes one complete revolution in 4.0s (Fig. 10-45). (a) What is the linear speed of a child seated 1.2m from the center? (b) What is her acceleration (give components)?

FIGURE 10-45 Problem 8.

9. (II) What is the linear speed of a point (a) on the equator, (b) on the Arctic Circle (latitude 66.5° N), and (c¢) at a latitude of 45.0° N, due to the Earth’s rotation?

10. (IT) Calculate the angular velocity of the Earth (a) in its orbit around the Sun, and (b) about its axis.

11. (II) How fast (in rpm) must a centrifuge rotate if a particle 7.0 cm from the axis of rotation is to experience an acceleration of 100,000 g’s? 276

CHAPTER10

Rotational Motion

angular

velocity

w; = 44.0rad/s

about

its axle,

and

the

turntable has angular velocity w, = 35.0rad/s about a vertical axis. (Note arrows showing these motions in the figure.) (a) What are the directions of @; and @, at the instant shown? (b) What is the resultant angular velocity of the wheel, as seen by an outside observer, at the instant shown? Give the magnitude and direction. (¢) What is the magnitude and direction of the angular acceleration of the wheel at the instant shown? Take the z axis vertically« upward and the direction of the axle at the moment shown to be the X axis pointing to the right. FIGURE 10-46 Problem 15.

10-3 Constant Angular Acceleration 16. (I) An automobile engine slows down from 3500 rpm to 1200rpm in 2.5s. Calculate (@) its angular acceleration, assumed constant, and (b) the total number of revolutions the engine makes in this time. 17. (I) A centrifuge accelerates uniformly from rest to 15,000 rpm in 220s. Through how many revolutions did it turn in this time? 18. (I) Pilots can be tested for the stresses of flying high-speed jets in a whirling “human centrifuge,” which takes 1.0 min to turn through 20 complete revolutions before reaching its final speed. (a) What was its angular acceleration (assumed constant), and (b) what was its final angular speed in rpm? 19. (II) A cooling fan is turned off when it is running at 850 rev/min.

It turns 1350 revolutions before it comes to a

stop. (@) What was the fan’s angular acceleration, assumed constant? (b) How long did it take the fan to come to a complete stop?

20. (II) Using calculus, derive the angular kinematic equation-™, 10-9a and 10-9b for constant angular acceleration. Star. with a = dw/dt.

21. Im A Jmall rubber wheel is used to drive a large pottery wheel. The two wheels are mounted so that their circular edges touch. The small wheel has a radius of 2.0cm and acceler{tes at the rate of 7.2 rad/s?, and it is in contact with the pottery wheel (radius 21.0 cm) without slipping. Calcu’. late (a) the angular acceleration of the pottery wheel, and (b) the time it takes the pottery wheel to reach its required speed of 65 rpm. 22. (IT) The angle through which a rotating wheel has turned in

29. (IT) The bolts on the cylinder head of an engine require tightening to a torque of 75m-N. If a wrench is 28 cm long, what force perpendicular to the wrench must the mechanic exert at its end? If the six-sided bolt head is 15 mm across (Fig. 10-49), estimate the force applied near each of the six points by a socket wrench. 28 cm

time 7 is given by 6 = 8.5¢ — 15.07% + 1.6¢%, where 6 is in

radians and ¢ in seconds. Determine the instantaneous angular velocity w taneous angular acceleration a. (¢) t =3.0s. (d) What is the average (e) the average angular acceleration t=

30

23. (IT) The

an expression (a) for and (b) for the instanEvaluate w and « at angular velocity, and between ¢ = 2.0s and

a?

angular acceleration of a wheel, as a function of

time, is a = 5.0t — 8.5¢,

where « is in rad/s* and ¢ in

seconds, If the wheel starts from rest (6 =0, w =0, at t = 0), determine a formula for (a) the angular velocity @ and (b) the angular position 6, both as a function of time. (c) Evaluate w and 6 at ¢ = 2.0s.

F, on F, on

wrench

bolt

FIGURE 10-49

Problem 29.

30. (IT) Determine the net torque on the 2.0-m-long uniform beam shown in Fig. 10-50. Calculate

about

the cM, and one end.

(a)

point

(b) point

C,

i

P at

130° :

[

65N

10-4 Tor3ue 24. (I) A 62-kg person riding a bike puts all her weight on each pedal when climbing a hill. The pedals rotate in a circle of radius 17 cm. (a) What is the maximum torque she exerts? (b) How could she exert more torque? 25. (I) Calculate the net torque about the axle of the wheel shown in Fig. 10-47. Assume that a friction torque of 0.40m*N opposes the motion.

56N

I

45° C

P FIGURE 10-50 Problem 30.

60°

52N

10-5 and 10-6 Rotational Dynamics FIGURE 10-47 Problem 25.

I8N

26. (IT) A person exerts a horizontal force of 32N on the end of a door 96 cm wide. What is the magnitude of the torque if the force is exerted (a) perpendicular to the door and (b) at a 60.0° angle to the face of the door? 27. (IT) Two blocks, each of mass m, are attached to the ends of a massless rod which pivots as shown in Fig. 10-48. Initially the rod is held in the horizontal position and then released. Calculate the magnitude and direction of the net torque on this system when it is first released. :

¢

%

I


by means of the triceps muscle, as shown. Calculate (a) the torque needed, and (b) the force that must be exerted by the triceps muscle. Ignore the mass of the arm.

25cm™~

\(at elbow)

=

FIGURE 10-52

Triceps

Problems 38 and 39.

%

™

muscle

39. (IT) Assume that a 1.00-kg ball is thrown solely by the action of the forearm, which rotates about the elbow joint under the action of the triceps muscle, Fig. 10-52. The ball is accelerated

uniformly

from

rest

to

8.5m/s

in

0.35s,

at

which point it is released. Calculate (@) the angular acceleration of the arm, and (b) the force required of the triceps muscle. Assume that the forearm has a mass of 3.7kg and rotates like a uniform rod about an axis at its end. . (II) Calculate the moment of inertia of the array of point objects shown in Fig. 10-53 about (a) the vertical axis, and (b) the horizontal axis. Assume m = 2.2kg, M = 3.1kg, and the objects are wired together by very light, rigid pieces of wire. The array is rectangular and is split through the middle by the horizontal axis. (c) About which axis would it be harder to accelerate this array?

-«

~—0.50m-—

|

2

by

1.50 m

:

> m

B

050m —-4-----=- 5 ILA-"lS ---------------- x

l

|

| I

:

278

CHAPTER

10

10-53

radius

7.0m

and

mass

31,000 kg, calculate

Problem 40.

Rotational Motion

the net torque

required to accelerate it. 42. (IT) A 0.72-m-diameter solid sphere can be rotated about a axis through its center by a torque of 10.8m-N which accelerates it uniformly from rest through a total of 180 revolutions in 15.0s. What is the mass of the sphere? 43. (IT) Suppose the force Fr in the cord hanging from the pulley of Example 10-9, Fig. 10-21, is given by the relation Fr =3.00¢ — 0.20 > (newtons) where / is in seconds. If the pulley starts from rest, what is the linear speed of a point on its rim 8.0s later? Ignore friction. . (I) A dad pushes tangentially on a small hand-driven merry-go-round and is able to accelerate it from rest to a frequency of 15 rpm in 10.0 s. Assume the merry-go-round is a uniform disk of radius 2.5 m and has a mass of 760 kg, and

two children (each with a mass of 25 kg) sit opposite each other on the edge. Calculate the torque required to produce the acceleration, neglecting frictional torque. What force is required at the edge? 45. (II) Four equal masses M are spaced at equal intervals, £, along a horizontal straight rod whose mass can be ignored. The system is to be rotated about a vertical axis passing through the mass at the left end of the rod and perpendicular to it. (a) What is the moment of inertia of the system about this axis? (b) What minimum force, applied to the farthest mass, will impart an angular acceleration a? (c¢) What is the direction of this force? . (II) Two blocks are connected by a light string passing over a pulley of radius 0.15m and moment of inertia /. The blocks move (towards the right) with an acceleration o‘fl 1.00m/s? along their frictionless inclines (see Fig. 10-54). (a) Draw free-body diagrams for each of the two blocks and the pulley. (b) Determine Fra and Frg, the tensions in the two parts of the string. (c) Find the net torque acting on the pulley, and determine its moment of inertia, /.

FIGURE 10-54

Problem 46.

47. (IT) A helicopter rotor blade can be considered a long thin rod, as shown in Fig. 10-55. (a) If each of the three rotor helicopter blades is 3.75m long and has a mass of 135 kg, calculate the moment of inertia of the three rotor blades about the axis of rotation. (b) How much torque must the motor apply to bring the blades from rest up to a speed of 5.0rev/s in 8.0s?

M

| FIGURE

41. (IT) A merry-go-round accelerates from rest to 0.68 rad/s in 24s. Assuming the merry-go-round is a uniform disk of

FIGURE 10-55 Problem 47.

/

27

56‘

m=135 kgfl

[

48. (IT) A centrifuge rotor rotating at 10,300 rpm is shut off and is eventually brought uniformly to rest by a frictional torque of 1.20m-N. If the mass of the rotor is 3.80 kg and it can be approxiqaated as a solid cylinder of radius 0.0710 m, through how many revolutions will the rotor turn before coming to rest, and how long will it take? 49. (II) When discussing moments of inertia, especially for unusual or irregularly shaped objects, it is sometimes convenient to work with the radius of gyration, k. This radius is defined so that if all the mass of the object were concentrated

54. (IIT) A thin rod of length £ stands vertically on a table. The rod begins to fall, but its lower end does not slide. (a) Determine the angular velocity of the rod as a function of the angle ¢ it makes with the tabletop. (b) What is the speed of the tip of the rod just before it strikes the table?

the same as that of the original object. Thus, the moment of inertia of any object can be written in terms of its mass M and the radius of gyration as [ = Mk? Determine the radius of gyration for each of the objects (hoop, cylinder, sphere, etc.) shown in Fig. 10-20. 50. (IT) To get a flat, uniform cylindrical satellite spinning at the correct rate, engineers fire four tangential rockets as shown in Fig. 10-56. If the satellite has a mass of 3600 kg, a radius of

56. (IT) Determine the moment of inertia of a 19-kg door that is 2.5m high and 1.0m wide and is hinged along one side. Ignore the thickness of the door. . (II) Two uniform solid spheres of mass M and radius r; are connected by a thin (massless) rod of length ry so that the centers are 3ry apart. (a) Determine the moment of inertia of this system about an axis perpendicular to the rod at its center. (b) What would be the percentage error if the masses of each sphere were assumed to be concentrated at their centers and a very simple calculation made? . (IT) A ball of mass M and radius r; on the end of a thin massless rod is rotated in a horizontal circle of radius R about an axis of rotation AB, as shown in Fig. 10-58.

at this distance from the axis, the moment of inertia would be

4.0m, and the rockets each add

a mass of 250 kg, what required steady of each rocket End view of satellite is to cylindrical satellite

32rpm

in

is the force if the reach

5.0 min,

starting from rest?

acceleration of the masses m and mg, and compare to the situation in which the moment of inertia of the pulley is ignored. [Hint: The tensions Fp, and Frg are not equal. We discussed the Atwood machine in Example 4-13, assuming / = 0 for the pulley.] FIGURE 10-57

Problem 51.

Atwood’s machine.

R vt

through the center, / = {5 M¢?> (Fig. 10-20f and g).

(a) Considering the mass of the ball to be concentrated at its center of mass, calculate its moment of inertia about AB.

the parallel-axis

theorem

and

considering

finite radius of the ball, calculate the moment

(ITI) An Atwood’s machine consists of two masses, m, and mp, which are connected by a massless inelastic cord that passes over a pulley, Fig. 10-57. If the pulley has radius R and moment of O R inertia / about its axle, determine the

55. (I) Use the parallel-axis theorem to show that the moment of inertia of a thin rod about an axis perpendicular to the rod at one end is I = }M{?, given that if the axis passes

(b) Using

FIGURE 10-56

Problem 50.

51

10-7 Moment of Inertia

the ball about AB. (c) Calculate the percentage error introduced by the point mass approximation for r; = 9.0cm and RO

=1.0m.

JLTFTB mg

‘v

52. (IIT) A string passing over a pulley has a 3.80-kg mass hanging from one end and a 3.15-kg mass hanging from the other end. The pulley is a uniform solid cylinder of radius 4.0cm and mass 0.80kg. (a) If the bearings of the pulley were frictionless, what would be the acceleration of the two masses? (b) In fact, it is found that if the heavier mass is given a downward speed of 0.20 m/s, it comes to rest in 6.2 s. What is the average frictional torque acting on the pulley? . (II0) hammer thrower accelerates the hammer (mass = 7.30kg) from rest within four full turns (revolutions) and releases it at a speed of 26.5 m/s. Assuming a uniform rate of increase in angular velocity and a horizontal circular path of radius 1.20m, calculate (a) the angular acceleration, (b) the (linear) tangential acceleration, (c) the centripetal acceleration just before release, (d) the net force being exerted on the hammer by the athlete just before release, and (e) the angle of this force with respect to the radius of the circular motion. Ignore gravity.

B

4>

g

Fra

the

of inertia of

FIGURE 10-58 Problem 38.

A

59. (IT) A thin 7.0-kg wheel of radius 32 cm is weighted to one side by a 1.50-kg weight, small in size, placed 22 cm from the center of the wheel. Calculate (a) the position of the center of mass of the weighted wheel and (b) the moment of inertia about an axis through its cM, perpendicular to its face. 60. (III) Derive the formula for the moment of inertia of a uniform thin rod of length £ about an axis through its center, perpendicular to the rod (see Fig. 10-20f). 61. (IIT) (a) Derive the formula given in Fig. 10-20h for the moment

of inertia of a uniform,

flat, rectangular

plate of

dimensions £ X w about an axis through its center, perpendicular to the plate. (b) What is the moment of inertia about each of the axes through the center that are parallel to the edges of the plate?

10-8 Rotational Kinetic Energy 62. (I) An automobile engine develops a torque of 255m-N at 3750 rpm. What is the horsepower of the engine? 63. (I) A centrifuge rotor has a moment of inertia of 4.25 x 10"2kg-m? How much energy is required to bring it from rest to 9750 rpm?

Problems

279

. (IT) A rotating uniform cylindrical platform of mass 220 kg and radius 5.5 m slows down from 3.8 rev/s to rest in 16 s when the driving motor is disconnected. Estimate the power output of the motor (hp) required to maintain a steady speed of 3.8 rev/s. 65. (IT) A merry-go-round has a mass of 1640 kg and a radius of 7.50 m. How much net work is required to accelerate it from rest to a rotation rate of 1.00 revolution per 8.00 s? Assume it is a solid cylinder. . (II) A uniform thin rod of length £ and mass M is suspended freely from one end. It is pulled to the side an angle 6 and released. If friction can be ignored, what is its angular velocity, and the speed of its free end, at the lowest point? 67. (I) Two masses, mp = 35.0kg and mp = 38.0kg, are connected by a rope that hangs over a pulley (as in Fig. 10-59). The pulley is a uniform cylinder of radius 0381m and mass 3.1kg. Initially m, is on the ground and mpg rests 2.5m above the ground. If the system is released, use conservation of energy to determine the speed of my just before it strikes the ground. Assume the pulley bearing is frictionless.

73. (IT) A sphere of radius 7, = 24.5cm and mass m = 1.20kg starts from rest and rolls without slipping down a 30.0° incline that is 10.0m long. (a) Calculate its translational and rotational speeds when it reaches the bottom. (b) What is the ratio of translational to rotational kinetic energy at the #®y bottom? Avoid putting in numbers until the end so you can answer: (c) do your answers in (a) and (b) depend on the radius of the sphere or its mass? 74. (IT) A narrow but solid spool of thread has radius R and mass M. If you pull up on the thread so that the cm of the spool remains suspended in the air at the same place as it unwinds, (a) what force must you exert on the thread? (b) How much work have you done by the time the spool turns with angular velocity w? 75. (II) A ball of radius rq rolls on the inside of a track of radius R, (see Fig. 10-61). If the ball starts from rest at the vertical edge of the track, what will be its speed when it reaches the lowest point of the track, rolling without slipping?

FIGURE 10-59 Problem 67.

68. (IIT) A 4.00-kg mass and a 3.00-kg mass are attached to opposite ends of a thin 42.0-cm-long horizontal rod (Fig. 10-60). The system is rotating at angular speed w = 5.60rad/s about a vertical axle at the center of the rod. Determine (a) the kinetic energy K of the system, and (b) the net force on each mass. (c) Repeat parts (a) and (b) assuming that the axle passes through the cMm of the system.

@ =]

400 kg

>

~%‘?kg FIGURE 10-60

Problem 68.

69. (III) A 2.30-m-long pole is balanced vertically on its tip. It

starts to fall and its lower end does not slip. What will be the

speed of the upper end of the pole just before it hits the ground? [Hint: Use conservation of energy.]

10-9 Rotational Plus Translational Motion

70. (I) Calculate the translational speed of a cylinder when it reaches the foot of an incline 7.20m high. Assume it starts from rest and rolls without slipping. 71. (I) A bowling ball of mass 7.3 kg and radius 9.0 cm rolls without slipping down a lane at 3.7m/s. Calculate its total kinetic energy. 72. (I) Estimate the kinetic energy of the Earth with respect to the Sun as the sum of two terms, (@) that due to its daily rotation about its axis, and (b) that due to its yearly revolution about the Sun. [Assume the Earth is a uniform sphere

with

mass = 6.0 X 10** kg, radius = 6.4 X 10°m,

1.5 X 10%km from the Sun.]

280

CHAPTER 10

Rotational Motion

and is

FIGURE

10-61

Problems 75 and 81.

76. (II) A solid rubber ball rests on the floor of a railroad car when the car begins moving with acceleration a. Assuming the ball rolls without slipping, what is its acceleratlon relative to (a) the car and (b) the ground? T (IT) A thin, hollow 0.545-kg section of pipe of radius 10.0 cm starts rolling (from rest) down a 17.5° incline 5.60m long. (a) If the pipe rolls without slipping, what will be its speed at the base of the incline? (b) What will be its total kinetic energy at the base of the incline? (¢) What minimum value must the coefficient of static friction have if the pipe is not to slip? *78. (II) In Example 10-20, (a) how far has the ball moved down the lane when it starts rolling without slipping? (b) What are its final linear and rotational speeds? 79, (IIT) The 1100-kg mass of a car includes four tires, each of mass (including wheels) 35 kg and diameter 0.80 m. Assume each tire and wheel combination acts as a solid cylinder. Determine (a) the total kinetic energy of the car when traveling 95 km/h and (b) the fraction of the kinetic energy in the tires and wheels. (¢) If the car is initially at rest and is then pulled by a tow truck with a force of 1500 N, what is the acceleration of the car? Ignore frictional losses. (d) What percent error would you make in part (c) if you ignored the rotational inertia of the tires and wheels? *80. (I1T) A wheel with rotational inertia I = 3 MR? about its horizontal central axle is set spinning with initial angular speed wy. It is then lowered, and at the instant its edge touches the ground the speed of the axle (and cM) is zero. Initially the wheel slips when it touches the ground, but then begins to move forward and eventually rolls without slipping. (a) In what direction does friction act on the slipping wheel? (b) How long does the wheel slip before it begins to roll

without slipping? (c) What is the wheel’s final translational “® speed, vey? [Hint: Use SF = mi, 7oy = Ieyacy, and recall that only when there is rolling without slipping is vcy = @R.]

-

81. (ITIT) A small sphere of radius ry = 1.5cm rolls without slipping on the track shown in Fig. 10-61 whose radius is Ry = 26.0 cm. The sphere starts rolling at a height Ry above the bottom of the track. When it leaves the track after passing through an angle of 135° as shown, (a) what will be its speed, and (b) at what distance D from the base of the

track will the sphere hit the ground?

*10~-10 Rolling Sphere Slows Down

*82. (I) A rolling ball slows down because the normal force does

not pass exactly through the cm of the ball, but passes in front of the cM. Using Fig. 10-41, show that the torque resulting from

the normal force (ry = £Fy in Fig. 10-41) is { of that due to the frictional force, 7

= rgF

that is, show that 7y = Z7g.

where rg is the ball’s radius;

| General Problems 83. A large spool of rope rolls on the ground with the end of the rope lying on the top edge of the spool. A person grabs

the end of the rope and walks a distance £, holding onto it,

Fig. 10-62. The spool rolls behind the person slipping. What length of rope unwinds from the spool? How far does the spool’s center of mass move?

without

FIGURE 10-62 Problem 83.

88. A 1.4-kg grindstone in the shape of a uniform cylinder of radius 0.20m acquires a rotational rate of 18rev/s from rest over a 6.0-s interval at constant angular acceleration. Calculate the torque delivered by the motor. 89. Bicycle gears: (a) How is the angular velocity wg of the rear wheel of a bicycle related to the angular velocity wg of the front sprocket and pedals? Let Ng and Ny be the number of teeth on the front and rear sprockets, respectively, Fig. 10-64. The teeth are spaced the same on both sprockets and the rear sprocket is firmly attached to the rear wheel. (b) Evaluate the ratio wr/wr when the front and rear sprockets have 52 and 13 teeth, respectively, and

On a 12.0-cm-diameter audio compact disc (CD), digital bits of information are encoded sequentially along an outward spiraling path. The spiral starts at radius R; = 2.5cm and winds its way out to radius R, = 5.8 cm. To read the digital information, a CD player rotates the CD so that the player’s readout laser scans along the spiral’s sequence of bits at a constant linear speed of 1.25 m/s. Thus the player must accurately adjust the rotational frequency f of the CD as the laser moves outward. Determine the values for f (in units of rpm) when the laser is located at Ry and when it is at R, . 85. (a) A yo-yo is made of two solid cylindrical disks, each of mass 0.050 kg and diameter 0.075 m, joined by a (concentric) thin solid cylindrical hub of mass 0.0050 kg and diameter 0.010m. Use conservation of energy to calculate the linear speed of the yo-yo just before it reaches the end of its 1.0-m-long string, if it is released from rest. (b) What fraction of its kinetic energy is rotational? 86. A cyclist accelerates from rest at a rate of 1.00m/s?. How fast will a point at the top of the rim of the tire (diameter = 68 cm) be moving after 2.5s? [Hint: At any moment, the lowest point on the tire is in contact with the ground and is at rest—see Fig. 10-63.]

(c) when they have 42 and 28 teeth.

FIGURE 10-64

Problem 89.

90. Figure 10-65 illustrates an H,O molecule. The O —H bond length is 0.096 nm and the H— O — H bonds make an angle of 104°. Calculate the moment of inertia for the H,O molecule about an axis passing through the center of the oxygen atom (a) perpendicular to the plane of the molecule, and (b) in the plane of the molecule, bisecting the

H— O — H bonds.

FIGURE 10-65 Problem 90.

This point on tire at rest momentarily

FIGURE 10-63

Problem 86.

87. Suppose David puts a 0.50-kg rock into a sling 1.5m and begins whirling the rock in a nearly circle, ?celerating it from rest to a rate of 85 rpm What is the torque required to achieve this feat, does the torque come from?

of length horizontal after 5.0s. and where

91. One possibility for a low-pollution automobile is for it to use energy stored in a heavy rotating flywheel. Suppose such a car has a total mass of 1100kg, uses a uniform cylindrical flywheel of diameter 1.50 m and mass 240 kg, and should be able to travel 350 km without needing a flywheel “spinup.” (a) Make reasonable assumptions (average frictional retarding force = 450 N, twenty acceleration periods from rest to 95 km/h, equal uphill and downhill, and that energy can be put back into the flywheel as the car goes downhill), and estimate what total energy needs to be stored in the flywheel. (b) What is the angular velocity of the flywheel when it has a full “energy charge”? (¢) About how long would it take a 150-hp motor to give the flywheel a full energy charge before a trip?

General Problems

281

92. A hollow cylinder (hoop) is rolling on a horizontal surface

at speed

v =33m/s

when

it reaches

a 15°

incline.

(a) How far up the incline will it go? (b) How long will it be on the incline before it arrives back at the bottom? 93. A wheel of mass M has radius R. It is standing vertically on

the floor, and we want to exert a horizontal force F at its

axle so that it will climb a step against which it rests (Fig. 10-66). The step has height 4, where h < R. What minimum force F is needed?

97. If the coefficient of static friction between tires and pavement

is 0.65, calculate the minimum torque that must be applied to the 66-cm-diameter tire of a 950-kg automobile in order to “lay rubber” (make the wheels spin, slipping as the car accelerates). Assume each wheel supports an equaL\ share of the weight. 98. A cord connected at one end to a block which can slide on an inclined plane has its other end wrapped around a cylinder resting in a depression at the top of the plane as

shown in Fig. 10-69. Determine the speed of the block after it has traveled 1.80m along the plane, starting from rest. Assume (a) there is no friction, (b) the coefficient of friction

between all surfaces is u = 0.035. [Hint: In part (b) first determine the normal force on the cylinder, and make any reasonable assumptions needed.]

M=33kg R=020m FIGURE 10-66

Problem 93.

94. A marble of mass m and radius r rolls along the looped rough track of Fig. 10-67. What is the minimum value of the vertical height /4 that the marble must drop if it is to reach the highest point of the loop without leaving the track? (a) Assume r kg-m?,

FIGURE 10-67

Problem 9%4.

A force of 2.5 N is exerted on the end of the roll for 1.3, but the paper does not tear so it begins to unroll. A constant friction torque of 0.11 m-N is exerted on the rol which gradually brings it to a stop. Assuming that the paper’s thickness is negligible, calculate (a) the length of paper that unrolls during the time that the force is applied (1.3s) and (b) the length of paper that unrolls from the time the force ends to the time when the roll has stopped moving.

95. The density (mass per unit length) of a thin rod of length £ increases uniformly from A, at one end to 3Aq at the other end. Determine the moment of inertia about an axis perpendicular to the rod through its geometric center. 96. If a billiard ball is hit in just the right way by a cue stick, the ball will roll without slipping immediately after losing contact with the stick. Consider a billiard ball (radius r, mass M) at rest on a horizontal pool table. A cue stick exerts a constant horizontal force F on the ball for a time ¢ at a point that is a height 4 above the table’s surface (see Fig. 10-68). Assume that the coefficient of kinetic friction between the ball and table is uy . Determine the value for A so that the ball will roll without slipping immediately after losing contact with the stick.

100. A solid uniform disk of mass 21.0 kg and radius 85.0 cm rest flat on a frictionless surface. Figure 10-71 shows a from above. A string is wrapped around the rim of the and a constant force of 35.0N is applied to the string. string does not slip on the rim. (a) In what direction

Rotational Motion

F

is at view disk The does

the cM move? When the disk has moved a distance of 5.5 m,

determine (b) how fast it is moving, (c) how fast it is spinning (in radians per second), and (d) how much string has unwrapped from around the rim.

Problem 100, looking down on the disk.

Problem 96.

CHAPTER10

Problem 99.

FIGURE 10-71

FIGURE 10-68

282

FIGURE 10-70

¢

101. When bicycle and motorcycle riders “pop a wheelie,” a large acceleration causes the bike’s front wheel to leave the ground. Let M be the total mass of the bike-plus-rider system; let x and y be the horizontal and vertical distance of this system’s cM from the rear wheel’s point of contact with the ground (Fig. 10-72). (2) Determine the horizontal acceleration a required to barely lift the bike’s front wheel off of the ground. (b) To minimize the acceleration necessary to pop a wheelie, should x be made as small or as large as possible? How about y? How should a rider position his or her body on the bike in order to achieve these optimal values for x and y? (c) If x = 35cm and y = 95 cm, find a.

103. A thin uniform stick of mass M and length £ is positioned vertically, with its tip on a frictionless table. It is released and allowed to fall. m Determine the speed of its l| CcM just before it hits the || | table (Fig. 10-74). :! I

i

|

!

FIGURE 10-74 Problem 103.

b Vem

*104. (a) For the yo-yo-like cylinder of Example 10-19, we saw that the downward acceleration of its cM was a = % g. If it starts from rest, what will be the cM velocity after it has fallen a distance h? (b) Now use conservation of energy to determine the cylinder’s cM velocity after it has fallen a distance A, starting from rest.

*Numerical/Computer *105. (II) Determine the torque produced about the support A of the rigid structure, shown in Fig. 1075, as a function of the leg angle 6 if a force F = 500N is applied at the point P perpendicular to the leg end. Graph the values of the torque 7 as a function of 6 from

FIGURE 10-72

FIGURE

10-75

Problem 105.

*106. (IT) Use the expression that was derived in Problem 51 for the acceleration of masses on an Atwood’s machine to investigate at what point the moment of inertia of the pulley becomes negligible. Assume m,a = 0.150kg, mp = 0.350kg, and R = 0.040m. (a) Graph the acceleration as a function of the moment of inertia. (b) Find the acceleration of the masses when the moment of inertia goes to zero. (c¢) Using your graph to guide you,

at what

minimum

value

of / does

the

calculated

acceleration deviate by 2.0% from the acceleration found in part (b)? (d) If the pulley could be thought of as a

uniform disk, find the mass of the pulley using the / found

FIGURE 10-73

in part (c).

Problem 102.

Exercises

A: f =0076Hz;

T = 13s.

:FA.

C: Yes; s

to 90°, in 1° increments.

Problem 101.

102. A crucial part of a piece of machinery starts as a flat uniform cylindrical disk of radius Ry and mass M. It then has a circular hole of radius Ry drilled into it (Fig. 10-73). The hole’s center is a distance /4 from the center of the disk. Find the moment of inertia of this disk (with off-center hole) when rotated about its center, C. [Hint: Consider a solid disk and “subtract” the hole; use the parallel-axis theorem.]

Answers t

6 = 0°

|

M + M(3 17 = M.

D: 4 X 10'77. E: (¢). F:

a=

_%g.

General Problems

283

This skater is doing a spin. When her arms are spread outward horizontally, she spins less fast than when her arms are held close to the axis of rotation. This is an example of the conservation of angular momentum. Angular momentum, which we study

in this Chapter,

is conserved

only if no net torque acts on the object or system. Otherwise, the rate of change of angular momentum is proportional to the net applied torque—which,

if zero,

means

the

angular momentum is conserved. In this Chapter we also examine more complicated aspects of rotational motion.

Angular Momentum; General Rotation CONTENTS 11-1

11-2 11-3

11-4 11-5

11-6 #11-7 #11-8 #11-9

284

Angular Momentum—

Objects Rotating About a

Fixed Axis Vector Cross Product; Torque as a Vector Angular Momentum of a Particle

Angular Momentum and

Torque for a System of Particles; General Motion Angular Momentum and

Torque for a Rigid Object Conservation of Angular

Momentum The Spinning Top and Gyroscope Rotating Frames of Reference; Inertial Forces The Coriolis Effect

CHAPTER-OPENING

QUESTION—Guess

now!

You are standing on a platform at rest, but that is free to rotate. You hold a spinning bicycle wheel by its axle as shown here. You then flip the wheel over so its axle points down. What happens then?

(a) The platform starts rotating in the direction the bicycle wheel was originally rotating. (b) The platform starts rotating in the direction

opposite to the original rotation of the bicycle wheel. (¢) The platform stays at rest. (d) The platform turns only while you are flipping the wheel. (e)

None of these is correct.

n Chapter 10 we dealt with the kinematics and dynamics of the rotation of a

rigid object about an axis whose direction is fixed in an inertial reference

frame. We analyzed the motion in terms of the rotational equivalent of Newton’s laws (torque plays the role that force does for translational otion), as well as rotational kinetic energy. To keep the axis of a rotating object fixed, the object must usually be constrained by external supports (such as bearings at the end of an axle). The motion of objects that are not constrained to move about a fixed axis is more difficult to describe and analyze. Indeed, the complete analysis of the general otion of an object (or system of objects) is very complicated, rotational and we will only look at some aspects of general rotational motion in this Chapter. We start this Chapter by introducing the concept of angular momentum, which is the rotational analog of linear momentum. We first treat angular momentum and its conservation for an object rotating about a fixed axis. After that, we examine the vector nature of torque and angular momentum. We will derive some general theorems and apply them to some interesting types of motion.

-1

Angular Momentum — Objects Rotating About a Fixed Axis

In Chapter 10 we saw that if we use the appropriate angular variables, the kinematic and dynami c equations for rotational motion are analogous to those for ordinary linear motion. In like manner, the linear momentum, p = mwv, has a rotational analog. It is called angular momentum, L, and for an object rotating about a fixed axis with an gular velocity w, it is defined as L = lw, 11-1) where [ is the moment of inertia. The SI units for L are kg-m?/s; there is no special nam e for this unit. We saw in Chapter 9 (Section 9-1) that Newton’s second law can be written ot only as 2F = ma, but also more generally in terms of momentum (Eq. 9-2), 3F = dp/dt. In a similar way, the rotational equivalent of Newton’s second law, which we saw in Egs. 10-14 and 10-15 can be written as

27 = [a,

written in terms of angular momentum: since the angular acceleration (Eq. 10-3), then la = I(dw/dt) = d(lw)/dt = dL/dt, so

can also be

a = dw/dt

dL (11-2) dt This derivation assumes that the moment of inertia, /, remains constant. However, Eq. 11-2 is valid even if the moment of inertia changes, and applies also to a system of objects rotating about a fixed axis where Z7 is the net external torque (discussed in Section 11-4). Equation 11-2 is Newton’s second law for rotational motion about a fixed axis, and is also valid for a moving object if its rotation is about an axis passing through its center of mass (as for Eq. 10-15).

2r =

NEWTON’S SECOND

FOR ROTATION

LAW

Conservation of Angular Momentum Angular momentum is an important concept in physics because, under certain conditions, it is a conserved quantity. What are the conditions for which it is conserved? From Eq. 11-2 we see immediately that if the net external torque =7 on an object (or system of objects) is zero, then

dL dt

=

0

and

L

=

[w

=

constant.

[Z7 = 0]

This, then, is the law of conservation of angular momentum for a rotating object: The total angular momentum of a rotating object remains constant if the net external torque acting on it is zero.

CONSERVATION OF ANGULAR MOMENTUM

~

Then, since

=

~— N

w 2

Clutch. A simple clutch consists of two cylindrical plates that

can be pressed together to connect two sections of an axle, as needed, in a piece of machiflery. The two plates have masses M, = 6.0kg and My = 9.0kg, with equal radii R, = 0.60m. They are initially separated (Fig. 11-4). Plate M, is accelerated from rest to an angular velocity o, = 7.2rad/s in time At = 2.0s. Calculate (a) the angular momentum of M, , and (b) the torque required to have accelerated M, from rest to w;. (c) Next, plate My, initially at rest but free to rotate without friction, is placed in firm contact with freely rotating plate M, , and the two plates both rotate at a constant angular velocity w, , which is considerably less than w,. Why does this happen, and what is w, ? APPROACH We use the definition of angular momentum plus Newton’s second law for rotation, Eq. 11-2. SOLUTION (@) The angular momentum of M, will be

Ly = Iao; = $MsR}w;

L = [o

= $(6.0kg)(0.60m)*(7.2rad/s)

FIGURE 11-4

Example 11-2.

(Eq. 11-1)

= 7.8kg-m?/s.

(b) The plate started from rest so the torque, assumed constant, was

AL

T= (c) Initially, M, contact, why is each exerts on tion of angular

3y

7.8kg-m?/s — 0 30s

39m-N.

is rotating at constant w, (we ignore friction). When plate B comes in their joint rotation speed less? You might think in terms of the torque the other upon contact. But quantitatively, it’s easier to use conservamomentum, since no external torques are assumed to act. Thus

angular momentum before

=

angular momentum after

IA(!)1

=

(IA

Solving for w, we find wy

=

IB)wz.

B (6.0kg>

M,

3

_[_1a

+

\15.0kg

n (IA + IB>“’1 = w1 =

~

(7.2rad/s) = 2.9rad/s.

@)pHYSICS

APPLIED

Neutron star

ar is the size of our Sun (r ~ 7 X 10° km) with mass 2.0 times as great as the

s rotating at a frequency of 1.0 revolution every 100 days. If it were to avitational collapse to a neutron star of radius 10 km, what would its rota-

tion frequency be? Assume the star is a uniform sphere at all times, and loses no mass.

APPROACH We assume the star is isolated (no external forces), so we can use conservation of angular momentum for this process. We use r for the radius of a sphere, as compared to R used for distance from an axis of rotation or cylindrical symmetry: see Fig. 10-2. SOLUTION From conservation of angular momentum, Lo,

=

Lo,,

where the subscripts 1 and 2 refer to initial (normal star) and final (neutron star), respectively. Then, assuming no mass is lost in the process, = .

The frequency

e

f2

-

- %

2t

f = w/2m,

_n

2

2

2

2

(5m1r1>w s5myr;

1

=

2

n,

Wi

rz

so

2

r%fl

B (7 X 10° km)z 1.0rev 10 km 100d(24 h/d)(3600s/h) ) ~ 6 X 10%rev/s. SECTION 11-1

Angular Momentum—Objects Rotating About a Fixed Axis

287

Directional Nature of Angular Momentum Angular momentum is a vector, as we shall discuss later in this Chapter. For now we consider the simple case of an object rotating about a fixed axis, and the direction of L is specified by a plus or minus sign, just as we did for one-dimensiona}\ linear motion in Chapter 2. For a symmetrical object rotating about a symmetry axis (such as a cylinder or wheel), the direction of the angular momentum’ can be taken as the direction of the angular velocity @. That is,

L = Ia As a simple example, consider a person standing at rest on a circular platform capable of rotating friction-free about an axis through its center (that is, a simplified merry-go-round). If the person now starts to walk along the edge of the platform, Fig. 11-5a, the platform starts rotating in the opposite direction. Why? One explanation is that the person’s foot exerts a force on the platform. Another explanation (and this is the most useful analysis here) is as an example of the conservation of angular momentum. If the person starts walking counterclockwise, the person’s angular momentum will be pointed upward along the axis of rotation (remember how we defined the direction of @ using the right-hand rule in Section 10-2). The magnitude FIGURE 11-5 (a) A person on a circular platform, both initially at rest, begins walking along the edge at speed ». The platform, assumed to be mounted on frictionfree bearings, begins rotating in the opposite direction, so that the total angular momentum remains zero, as shown in (b).

of the person’s angular momentum will be L = o = (mR?)(v/R), where v is the

person’s speed (relative to the Earth, not the platform), R is his distance from the rotation axis, m is his mass, and mR? is his moment of inertia if we consider him a particle (mass concentrated at one point). The platform rotates in the opposite direction, so its angular momentum points downward. If the initial total angular momentum was zero (person and platform at rest), it will remain zero after the person starts walking. That is, the upward angular momentum of the person just balances the oppositely directed downward angular momentum of the platform (Fig. 11-5b), so the total vector angular momentum remains zero. Even though the person exerts a force (and torque) on the platform, the platform exerts an equal and opposite torque on the person. So the net torque on the system of person pl platform is zero (ignoring friction) and the total angular momentum remains constan.

Running on a circular platform. Suppose a 60-kg person

stands at the edge of a 6.0-m-diameter circular platform, which is mounted on frictionless bearings and has a moment of inertia of 1800 kg-m? The platform is at rest initially, but when the person begins running at a speed of 4.2 m/s (with respect to the Earth) around its edge, the platform begins to rotate in the opposite direction as in Fig. 11-5. Calculate the angular velocity of the platform. APPROACH momentum

We

use

conservation

of angular

momentum.

The

total

angular

is zero initially. Since there is no net torque, L is conserved and will

remain zero, as in Fig. 11-5. The person’s angular momentum is Lye, = (mR?)(v/R),

YT

mRv

_

g

[

~

3

_

Lp]at N—————

0

+

2,

Lper

=

=

SA

So

L

Il

and we take this as positive. The angular momentum of the platform is L,y = —lw. SOLUTION Conservation of angular momentum gives

(60kg)(3.0m)(4.2 m/s)

1800 kg - m?

=

0.42rad/s.

NOTE The frequency of rotation is f = w/27 = 0.067 rev/s and the period 1/f = 15s per revolution.

T =

*For more complicated situations of objects rotating about a fixed axis, there will be a component of L along the direction of @ and its magnitude will be equal to /@, but there could be other components well. If the total angular momentum is conserved, then the component /@ will also be conserved. & our results here can be applied to any rotation about a fixed axis.

288

CHAPTER 11

Angular Momentum; General Rotation

-

CONCEPTUAL

EXAMPLE

11-5 | Spinning bicycle wheel. Your physics teacher

is holding a spinning bicycle wheel while he stands on a stationary frictionless turntable (Fig. 11-6). What will happen if the teacher suddenly flips the bicycle wheel Pover so that it is spinning in the opposite direction? RESPONSE We consider the system of turntable, teacher, and bicycle wheel. The total angular momentum initially is L vertically upward. That is also what the system’s angular momentum must be afterward, since L is conserved when there is no net torque. Thus, if the wheel’s angular momentum after being flipped over is —L downward, then the angular momentum of teacher plus turntable will have to be +2L upward. We can safely predict that the teacher will begin spinning E)und in the same direction the wheel was spinning originally. EXERCISE

In Example 11-5, what if he moves the axis only 90° so it is horizontal? (a) The same direction and speed as above; (b) the same as above, but slower; (c) the opposite res ult.

FIGURE 11-6

Example 11-5.

EXERCISE B Return to the Chapter-Opening Question, page 284, and answer it again now. Try to expla in why you may have answered differently the first time.

EXERCISE C Suppose you are standing on the edge of a large freely rotating turntable. If you walk toward the center, (a) the turntable slows down; (b) the turntable speeds up; (c) its rotati on speed is unchanged; (d) you need to know the walking speed to answer.

-2 f

Vector Cross Product;

Torque as a Vector

/ector Cr¢ nss Product To deal wit h the vector nature of angular momentum and torque in general, we will need th e concept of the vector cross product (often called simply the vector product or ¢ ross product). In general, the vector or cross product of two vectors A and B is defined as another vector C = A X B whose magnitude is (11-3a) where 8 is the angle (< 180°) between A and B, and whose direction is perpendicular

to both A and B in the sense of the right-hand rule, Fig. 11-7. The angle 6 is measured between A and B when their tails are together at the same point. According t o the right-hand rule, as shown in Fig. 11-7, you orient your right hand so your fing ers pomt along A, and when you bend your fingers they point along B. When your hand is correctly oriented in this way, your thumb will point along the direction of C=AXB. _The cr oss product of two vectors, A=A i+ A J + A, k, and B = B,i + By ] + Bk, can be written in component form (see Problern 26) as

AXB

=

i

A, B,

j

A, B,

FIGURE 11-7 The vector C=AXBis perpendicular to the plane containing A and B; its direction is given by the right-hand rule.

Kk

A, B,

(11-3b)

AyB, — A;B))i + (A;B, — A,B))j + (A,B, — A,B))k. (11-3¢) Zquation 1 1-3b is meant to be evaluated using the rules of determinants (and obtain Eq. 11-3c). SECTION 11-2

Vector Cross Product; Torque as a Vector

289

AXA

AxB = -BxA Ax(B+C) = (AxB)+ (AxC) d s dA dB dt(A B) o XBtAX—

YExA FIGURE

11-8

=0

T

Thevector B X A

equals —A X B; compare to Fig. 11-7.

x

=

—_——

—_—

(11-4a)

(11-4b) “ [distributive law] (11-4c) (11-4d) —

Equation 11-4a follows from Egs. 11-3 (since 6 = 0). So does Eq. 11-4b, since the magnitude of B X A is the same as that for A X B, but by the right-hand rule the direction is opposite (see Fig. 11-8). Thus the order of the two vectors is crucial. If you change the order, you change the result. That is, the commutative

law does not hold for the cross product (A X B # B X A), although it does hold

FIGURE

11-9

for the dot product of two vectors and for the product of scalars. Note in Eq. 11-4d that the order of quantities in the two products on the right must not be changed (because of Eq. 11-4b).

Exercise D.

-

A

-

B

EXERCISE D For the vectors A and @ in the plane of the page as shown in Fig. 11-9, in what direction is (i)A - B, (ii) A X B, (iii) B X A? (a) Into the page; (b) out of the page; (c) between A and B; (d) it is a scalar and has no direction; (e) it is zero and has no direction.

The Torque Vector

FIGURE 11-10 The torque due to the force F (in the plane of the wheel) starts the wheel rotating counterclockwise so @ and @ point out of the page.

Torque is an example of a quantity that can be expressed as a cross product. To see this, let us take a simple example: the thin wheel shown in Fig. 11-10 which is free to rotate about an axis through its center at point O. A force F acts at the edge of the wheel, at a point whose position relative to the center O is given by the position vector F as shown. The force F tends to rotate the wheel (assumed initially at rest) counterclockwise, so the angular velocity @ will point out of the page toward the viewer (remember the right-hand rule from Section 10-2). The torque due to F will tend to increase @ so @ also points outward along the rotation axis. The relation between angular acceleration and torque that we developed in Chapter 10 for an object rotating about a fixed axis is 27t

=

la,

(Eq. 10-14) where [ is the moment of inertia. This scalar equation is the rotational

equivalent of 2F = ma,

and we would like to make it a vector equation just

# = ixF

(11-5)

as 2F = ma is a vector equation. To do so in the case of Fig. 11-10 we must have the direction of # point outward along the rotation axis, since @ (= da/dt) has that direction; and the magnitude of the torque must be (see Egs. 10-10 and Fig. 11-10) 7 = rF, = rFsin6. We can achieve this by defining the torque vector to be the cross product of ¥ and F:

FIGURE 11-11

7 =¥ X F, where

T is the position vector. y

From the definition of the cross product above (Eq. 11-3a) the magnitude of 7 will be rF sin 6 and the direction will be along the axis, as required for this special case. We will see in Sections 11-3 through 11-5 that if we take Eq. 11-5 as the general definition of torque, then the vector relation 2% = Ia will hold in general. Thus we state now that Eq. 11-5 is the general definition of torque. It contains both magnitude and direction information. Note that this definition involves the position vector F and thus the torque is being calculated about a point. We can choose that point O as we wish. For a particle of mass m on which a force F is applied, we define the torque about a point O as

F=¥FxF

7 X

e R

(2]

\ 290

CHAPTER 11

where ¥ is the position vector of the particle relative to O (Fig. 11-11). If we have a system of particles (which could be the particles making up a rigid object) the total torque 7 on the system will be the sum of the torques on the individual particles: T

F

=

Z(i",

X

F,'),

where F; is the position vector of the i" particle and F; is the net force on the i™ particle.

Angular Momentum; General Rotation

>

Torque Vector. Suppose the vector ¥ is in the xz plane, as in Fig. 11-11, and is given by ¥ = (12m)i + (12m)k. Calculate the torque vector # if F = (150N)i. APPROACH We use the determinant form, Eq. 11-3b.

SOLUTIO So 7 has EXERCISE

7 =fixF

=

i

j

150N

0

|12m

0

k

12m|

= 0i + (180m-N)j + Ok.

O

magnitude 180 m*N and points along the positive y axis. If F=50Ni

(c) 10m-NKk, (d)

and

r—ZOmJ,

-10m- N], () ~10m-Nk.

what is #? (a) 10m-N,

(b) —10m-N

11-3 Angular Momentum of a Particle The most g eneral way of writing Newton’s second law for the translational motion of a particle (or system of particles) is in terms of the linear momentum p = mv as given by Eq. 9-2 (or 9-5)

Tt

The rotatio nal analog of linear momentum

is angular momentum. Just as the rate of change of p is related to the net force =F, so we might expect the rate of change of angular momentum to be related to the net torque. Indeed, we saw this was true in Section 11-1 for the special case of a rigid object rotating about a fixed axis. Now we wi 1 see it is true in general. We first treat a single particle. Suppos e a particle of mass m has momentum p and position vector F with respect to the origin O in some chosen inertial reference frame. Then the general L, of the particle about point O is the vector

L =FfXp

[particle]

(11-6)

Angular momentum is a vector.” Its direction is perpendicular to both ¥ and p as given by the right-hand rule (Fig. 11-12). Its magnitude is given by L

or

=

rpsin®

L =rp

=nrp

(2]

definition of the angular momentum,

'cross product of ¥ and p:

FIGURE 11-12

The angular

momentum of a particle of mass m is givenby L=F X p =T X mV.

where 6 is the angle between ¥ and p and p, (= psin8) and r, (= rsin6) are the components of p and ¥ perpendicular to ¥ and p, respectively. Now let us find the relation between angular momentum and torque for a particle. If we take the derivative of L with respect to time we have di:i("x*)—flx

But

TR

TACKR

v _ :i_t_xp

=

DRl

o VXmv

=

+

T m(VXVv)

Xdij

S

=0,

since sin § = 0 for this case. Thus

dL

a - "

—_—

=

F

X

dp

—-

If we let ZF represent the resultant force on the particle, then in an inertial reference

frame, 3F = dp/dt and

Lt

FX3F

But ¥ X ZF = 27

e

-

=

dp _ dL

FX—

at

= —

ot

is the net torque on our particle. Hence

21“'=dt

[particle, inertial frame]

(11-7)

The time rate of change of angular momentum of a particle is equal to the net torque applied to it. Equation 11-7 is the rotational equivalent of Newton’s "Actually a pseudovector; see footnote in Section 10-2.

SECTION 11-3

291

second law for a particle, written in its most general form. Equation 11-7 is valid only in an inertial frame since only then is it true that 2F = dp/dt, which was used in the proof.

/. ll \

Pl o

\\

L

FIGURE 11-13

5

=myv

CONCEPTUAL EXAMPLE 11-7 | A particle’s angular momentum. What is th ™ angular momentum of a particle of mass m moving with speed v in a circle of radius r in a counterclockwise direction?

AN ¥ _ o

/

\, /

T The angular

momentum ofaparticlc of massm rotating in a circleof radius ¥ with velocity vis L = F X mv (Example 11-7).

RESPONSE The value of the angular momentum depends on the choice of the point O. Let us calculate L with respect to the center of the circle, Fig. 11-13. Then ¥ is perpendlcular to p so L = [f X p| = rmv. By the right-hand rule, the direction of L is perpendicular to the plane of the circle, outward toward the viewer. Since v = wr and I = mr* for a single particle rotating about an axis a distance r away, we can write S

I

=

mor

=

ity

=

la

'

11-4 Angular Momentum and Torque for a System of Particles; General Motion Relation Between Angular Momentum and Torque Consider a system of n particles which have angular momenta L, L,, ..., L,. The system could be anything from a rigid object to a loose assembly of particles whose positions are not fixed relative to each other. The total angular momentum L of the system is defined as the vector sum of the angular momenta of all the particles in the system:

oA P

(11-8)

i=1

The resultant torque acting on the system is the sum of the net torques acting o™ all the particles: net

==

E Ti.

This sum includes (1) internal torques due to internal forces that particles of the system exert on other particles of the system, and (2) external torques due to forces exerted by objects outside our system. By Newton’s third law, the force each particle exerts on another is equal and opposite (and acts along the same line as) the force that the second particle exerts on the first. Hence the sum of all internal torques adds to zero, and ?nct

=

Z?i

=

Efcxv

Now we take the time derivative of Eq. 11-8 and use Eq. 11-7 for each particle to obtain

dL

d_t

NEWTON’S SECOND LAW

or

dL

s

{retation; sysiem of purtcles)

-

i

2 i

>t

E?CXI

. [inertial reference frame]

(11-9a)

This fundamental result states that the time rate of change of the total angular momentum of a system of particles (or a rigid object) equals the resultant external torque

on the system. It is the rotat10nal equivalent of Eq. 9-5,

dP/dt = 3F.y

for translational motion. Note that L and =7 must be calculated about the same origin O. -~ Equatlon 11-9a s valid when L and 7.y, are calculated with reference to a pomL fixed in an inertial reference frame. (In the derivation, we used Eq. 11-7 which is

292

CHAPTER11

Angular Momentum; General Rotation

valid only in this case.) It is also valid when 7.y, and L are calculated about a point

which is moving uniformly in an inertial reference frame since such a point can be

considered the origin of a second inertial reference frame. It is not valid in general when #.,,

and L are calculated about a point that is accelerating, except for one special

(and very iafilportant) case—when that point is the center of mass (CM) of the system:

dLCM =

D

[even if accelerating]

(11-9b)

NEWTON’S SECOND

ey

LAW Sormalsowrinit

Equation 11-9b is valid no matter how the cM moves, and 27y is the net external torque calculated about the center of mass. The derivation is in the optional subsection below. It is because of the validity of Eq. 11-9b that we are justified in describing the general motion of a system of particles, as we did in Chapter 10, as translational motion of the center of mass plus rotation about the center of mass. Equations 11-9b plus 9-5 (dPgy/dt = SF,y) provide the more general statement of this principle. (See also Section 9-8.)

*Derivation of dLey/dt = ZFcy The proof of Eq. 11-9b is as follows. Let F; be the position vector of the i™ particle in an inertial reference frame, and ¥, be the position vector of the center of mass of the system in this reference frame. The position of the i*" particle with respect to the cM is ¥, where (see Fig. 11-14)

F, = Foy + T

X

If we multiply each term by m; and take the derivative of this equation, we can write J)

P=

o

s

dii

_

= m d (FFw. o

Byt 3m; R? is the moment of inertia 7 of the object about the axis of rotation. There-

fore the component of the total angular momentum along the rotation axis is given by

(11-10)

L, = la,

Note that we would obtain Eq. 11-10 no matter where we choose the point O (for measuring ¥;) as long as it is on the axis of rotation. Equation 11-10 is the same as Eq. 11-1, which we have now proved from the general definition of angular momentum. If the object rotates about a symmetry axis through the center of mass, then L, is the only component of L, as we now show. For each point on one side of the axis there will be a corresponding point on the opposite side. We can see from

Fig. 11-15 that each L; has a component parallel to the axis (L;,) and a component

perpendicular to the axis. The components parallel to the axis add together for each pair of opposite points, but the components perpendicular to the axis for opposite points will have the same magnitude but opposite direction and so will cancel. Hence, for an object rotating about a symmetry axis, the angular momentum vector is parallel to the axis and we can write L

=

I,

[rotation axis = symmetry axis, through cm]

(11-11)

where L is measured relative to the center of mass. The general relation between angular momentum and torque is Eq. 11-9: -

27 =

dL

where =% and L are calculated either about (1) the origin of an inertial reference frame, or (2) the center of mass of the system. This is a vector relation, and must

294

CHAPTER11

Angular Momentum; General Rotation

therefore be valid for each component. Hence, for a rigid object, the component along the rotation axis is r

SToxis

=

dL

dTw.

d

=

E(lw)

dw

=

===

lu,

which is valid for a rigid object rotating about an axis fixed relative to the object; also this axis must be either (1) fixed in an inertial system or (2) passing through the cMm of the object. This is equivalent to Eqs. 10-14 and 10-15, which we now see are spe01al cases of Eq. 11-9, =% = dL/dt.

Atwood’s

machine. An Atrwood machine consists of two

mMasses, mp and my, which are connected by an inelastic cord of negligible mass that passes over a pulley, Fig. 11-16. If the pulley has radius R, and moment of inertia / about its axle, determine the acceleration of the masses m, and mg, and compare to the situ{ation where the moment of inertia of the pulley is ignored. APPROACH We first determine the angular momentum of the system, and then apply Newton’s second law, 7 = dL/dt. SOLUTION The angular momentum is calculated about an axis along the axle through the center O of the pulley. The pulley has angular momentum /w, where o = v/R, and v is the velocity of m, and my at any instant. The angular momentum of my is Ryma v and that of my is Rymgv. The total angular momentum is

The external torque on the system, calculated about the axis O (taking clockwise as positive), is T

=

mBgRO

-

mAgRO.

(The force on the pulley exerted by the support on its axle gives rise to no torque because the lever arm is zero.) We apply Eq. 11-9a:

fl

T

‘

(mB - mA)gRo Solving for a = dv/dt,

FIGURE 11-16 Atwood’s machine, Example 11-8. We also discussed this in Example 4-13.

_dL =

(mA

dv + mB)ROd_t

I dv Ro a

we get

_ dv

T odt

(mp — my)g

(ma + mg) + I/RE

If we were to ignore I, a = (mg — my)g/(mg + m,) and we see that the effect

of the m(%ment of inertia of the pulley is to slow down the system. This is just what we would expect.

11-9 | Bicycle wheel. Suppose you are holding a

bicycle wheel by a handle connected to its axle as in Fig. 11-17a. The wheel is spinning rapidly so its angular momentum L points horizontally as shown. Now you suddenly try to tilt the axle upward as shown by the dashed line in Fig. 11-17a (so the cM moves vertically). You expect the wheel to go up (and it would if it weren’t rotating), but it unexpectedly swerves to the right! Explain. RESPONSE To explain this seemingly odd behavior—you may need to do it to believe it—we only need to use the relation

7,

= dL/dt.

In the short time At,

you exert a net torque (about an axis through your wrist) that points along the x axis perpendicular to L. Thus the change in L is AL =~ #, At so AL must also point (approximately) along the x axis, since #,,¢; does (Fig. 11-17b). Thus the new angular momentum, L + AL, points to the right, looking along the axis of the wheel, as shown in Fig. 11-17b. Since the angular momentum is directed along the axle of the wheel, we see that the axle, which now is along L+ AL, must move sideways to the right, which is what we observe.

|

SECTION 11-5

FIGURE 11-17 When you try to tilt a rotating bicycle wheel vertically upward, it swerves to the side instead.

Rotation ‘ axis for —— spinning wheel Rotation axis for lifting wheel

S

EXAMPLE

L=

CONCEPTUAL

(b)

X

Angular Momentum and Torque for a Rigid Object

295

A CAUTION

L = I® isnot always valid

F

— Bearing

FIGURE 11-18

In this system L

and @ are not parallel. This is an example of rotational imbalance.

FrpHysics

APPLIED

Automobile wheel balancing

FIGURE

11-19

automobile wheel.

Unbalanced

Axis

Although Eq. 11-11, L = @, is often very useful, it is not valid in general unless the rotation axis is along a symmetry axis through the center of mass. Nonetheless, it can be shown that every rigid object, no matter what its shape, has three “principal axes” about which Eq. 11-11 is valid (we will not go into the details here). As an example of a case where Eq. 11-11 is not valid, consider the nonsymmetrical object shown in Fig. 11-18. It consists of two equal masses, m, and mg, attached to the ends of a rigid (massless) rod which makes an angle ¢ with the axis of rotation. We calculate the angular momentum about the cMm at point O. At the moment shown, m, is coming toward the viewer, and mg is moving away,so L, = ¥, X p, and Ly = ¥ X py are as shown. The total angular momentum is L = L, + Ly, which is clearly not along @ if ¢ # 90°. * Rotational Imbalance Let us go one step further with the system shown in Fig. 11-18, since it is a fine illustration of 3# = dL/dt. If the system rotates with constant angular velocity, o, the magnitude of L will not change, but its direction will. As the rod and two masses rotate about the z axis, L also rotates about the axis. At the moment shown in Fig. 11-18, L is in the plane of the paper. A time dt later, when the rod has rotated through an angle d6 = w dt, L will also have rotated through an angle df (it remains perpendicular to the rod). L will then have a component pointing into the page. Thus dL points into the page and so must dL/d!. Because

we see that a net torque, directed into the page at the moment shown, must be applied to the axle on which the rod is mounted. The torque is supplied by« bearings (or other constraint) at the ends of the axle. The forces F exerted by the bearings on the axle are shown in Fig. 11-18. The direction of each force F rotates as the system does, always being in the plane of L and @ for this system. If the torque due to these forces were not present, the system would not rotate about the fixed axis as desired. The axle tends to move in the direction of F and thus tends to wobble as it rotates. This has many practical applications, such as the vibrations felt in a car whose wheels are not balanced. Consider an automobile wheel that is symmetrical except for an extra mass m, on one rim and an equal mass myg opposite it on the other rim, as shown in Fig. 11-19. Because of the nonsymmetry of m, and mg, the wheel bearings would have to exert a force perpendicular to the axle at all times simply to keep the wheel rotating, just as in Fig. 11-18. The bearings would wear excessively and the wobble of the wheel would be felt by occupants of the car. When the wheels are balanced, they rotate smoothly without wobble. This is why “dynamic balancing” of automobile wheels and tires is important. The wheel of Fig. 11-19 would balance statically just fine. If equal masses mc and mp are added symmetrically, below m, and above my, the wheel will be balanced dynamically as well (L will be parallel to @, and 7. = 0).

Torque on imbalanced system. Determine the magnitude

of the net torque 7,; needed to keep the system turning in Fig. 11-18.

APPROACH Figure 11-20 is a view of the angular momentum vector, looking down the rotation axis (z axis) of the object depicted in Fig. 11-18, as it rotates. L cos¢ is the component of L perpendicular to the axle (points to the right in Fig. 11-18). We find dL from Fig. 11-20 and use 7, = dL/dt.

296

CHAPTER11

S

Angular Momentum; General Rotation

SOLUTION

In a time d¢, L changes by an amount (Fig. 11-20 and Eq. 10-2b) dL

where

=

o = df/dt.

¢ L

=

LA

+

LB

=

Lcosd¢ wdt,

Axi

Hence

R

gt Now

(Lcos¢)dd

-

=

FAMA

VA

+

rg

Toet = wLcosd

Angular momentum

vector looking down along the rotation Mgvg

=

rAmA(erSin¢)

+

erB(er

Sin¢)

=

(mari + mgri)wsin ¢. Since I = (myri + myry)sin® ¢ is the moment of inertia about the axis of rotation, then

LD

L cos ¢ (at time ¢)

FIGURE 11-20

—

dt

2l L°°i¢"

L = lw/sin¢.

axis of the system of Fig. 11-18 as it

rotates durmgatimedt

So

= (mari + mpri)o’singcosd = lw?/tan ¢.

The situation of Fig. 11-18 illustrates the usefulness of the vector nature of torque and angular momentum. If we had considered only the components of angular momentum and torque along the rotation axis, we could not have calculated the torque due to the bearings (since the forces F act at the axle and hence produce no torque along that axis). By using the concept of vector angular momentum we have a far more powerful technique for understanding and for attacking problems.

11-6 Conservation of Angular Momentum In Chapter 9 we saw that the most general form of Newton’s second law for the translational motion of a particle or system of particles is EFext

dP

=

_’

where P is the (linear) momentum, defined as mv for a particle, or M¥.y for a ’hsystem of particles of total mass M whose cM moves with velocity vy, and SF.y, is the net external force acting on the particle or system. This relation is valid only in an inertial reference frame. In this Chapter, we have found a similar relation to describe the general rotation of a system of particles (including rigid objects):

st -4

dL

where =7 is the net external torque acting on the system, and L is the total angular momentum. This relation is valid when =7 and L are calculated about a point fixed in an inertial reference frame, or about the cM of the system For translational motion, if the net force on the system is zero, dP/dt = 0, so the total linear momentum of the system remains constant. This is the law of conservation of linear momentum.

For rotational motion, if the net torque on the

system is zero, then

dL

g In words:

0

and

L

=

constant.

[Ei‘ = 0]

(11-12)

|

The total angular momentum

of a system remains constant if the net external

torque acting on the system is zero.

CONSERVATION

OF

ANGULAR MOMENTUM

This is the law of conservation of angular momentum in full vector form. It ranks with the laws of conservation of energy and linear momentum (and others to be discussed later) as one of the great laws of physics. In Section 11-1 we saw some Examples of this important law applied to the special case of a rigid object rotating about a fixed axis. Here we have it in general form. We use it now in interesting Examples.

SECTION 11-6

Conservation of Angular

Momentum

297

Kepler's second law derived. Kepler’s second law states

that each planet moves so that a line from the Sun to the planet sweeps out equal areas in equal times (Section 6-5). Use conservation of angular momentum to show this. APPROACH We determine the angular momentum area swept out with the help of Fig. 11-21. Planet

of a planet in terms of thcfl

SOLUTION The planet moves in an ellipse as shown in Fig. 11-21. In a time dt,

vdt sin 6

FIGURE 11-21 Kepler’s second law of planetary motion (Example 11-11).

the planet moves a distance v dt and sweeps out an area dA equal to the area of a triangle of base r and height v df sin # (shown exaggerated in Fig. 11-21). Hence

dA = }(r)(v dtsing)

and

B

Lrosing

dt

.

‘

The magnitude of the angular momentum L about the Sun is L

SO

But

=

]i" X mi"\

=

mrvsiné,

a4dt _ 12m L = constant,

since the gravitational force F is directed toward the Sun so

the torque it produces is zero (we ignore the pull of the other planets). Hence dA/dt = constant, which is what we set out to prove.

Bullet strikes cylinder edge. A bullet of mass m moving

FIGURE 11-22 Bullet strikes and becomes embedded in cylinder at its edge (Example 11-12).

with velocity v strikes and becomes embedded at the edge of a cylinder of mass M and radius Ry, as shown in Fig. 11-22. The cylinder, initially at rest, begins to rotate about its symmetry axis, which remains fixed in position. Assuming no frictional torque, what is the angular velocity of the cylinder after this colhslon" Is kinetic energy conserved? APPROACH We take as our system the bullet and cylinder, on which there is no net external torque. Thus we can use conservation of angular momentum, and we calculate all angular momenta about the center O of the cylinder. SOLUTION Initially, because the cylinder is at rest, the total angular momentum about O is solely that of the bullet:

L = [fXp|

= Rymo,

since R, is the perpendlcular

cylinder

(I

= $MR3)

distance

of p from

rotates with the bullet

angular velocity w:

L = lIo = (Iy + mR})w

O. After

(I, = mR3)

the collision, the

embedded in it at

= (3M + m)Rjo.

Hence, because angular momentum is conserved, we find that is L mvR, mv w

=

GM + m)R}

=

(M + m)R}

—

(M

+ m)R,

.

Angular momentum is conserved in this collision, but kinetic energy is not: K[

-

Kl

=

zlcy]w

+

(mRz)

-

2vz

= (MRz)w - (mR0 Jo? — smo? =iGM 2(

_

T

+m

)

1M

+

m

mM

T2M

+ 4m "

which is less than zero. Hence K; < K;. This energy is transformed to therma‘fl energy as a result of the inelastic colhslon 298

CHAPTER 11

Angular Momentum; General Rotation

*11-7 The Spinning Top and Gyroscope The motion of a rapidly spinning top, or a gyroscope, is an interesting example of r[otational motion and of the use of the vector equation

Consider a symmetrical top of mass M spinning rapidly about its symmetry axis, as in Fig. 11-23. The top is balanced on its tip at point O in an inertial reference frame. If the axis of the top makes an angle ¢ to the vertical (z axis), when the top is carefully released its axis will move, sweeping out a cone about the vertical as shown by the dashed lines in Fig. 11-23. This type of motion, in which a torque produces a change in the direction of the rotation axis, is called precessnon The rate at which the rotation axis moves about the vertical (z) axis is called the angular velocity of precession, () (capital Greek omega). Let us now try to understand the reasons for this motion, and calculate (). If the top were not spinning, it would immediately fall to the ground when released due to the pull of gravity. The apparent mystery of a top is that when it is spinning, it does not 1mmed1ately fall to the ground but instead precesses—it moves slowly sideways. But this is not really so mysterious if we examine it from the point of view of angular momentum and torque, which we calculate about the point O. When the top is spinning with angular velocity w about its symmetry axis, it has an angular momentum L directed along its axis, as shown in Fig. 11-23. (There is also angular momentum due to the precessional motion, so that the total L is not exactly along the axis of the top; but if Q) (). As w decreases due to friction and air resistance, the gyroscope will begin to fall, just as does a top. FIGURE 11-24

A

toy gyroscope.

*11-8

Rotating Frames of Reference; Inertial Forces

Inertial and Noninertial Reference Frames FIGURE 11-25 Path of a ball released on a rotating merry-go-

Up to now, we have examined the motion of objects, including circular and rotational motion, from the outside, as observers fixed on the Earth. Sometimes it is convenient

the merry-go-round, and (b) in a

Let us examine the motion of objects from the point of view, or frame of reference, of

round (a) in the reference frame of reference frame fixed on the ground. P

People on ground appear to move Hhis ety

e

N\ o =\ " | Path of ball / with respect

L]

to rotating

:

\\_‘_7///

.

/"

0}’;%"“2)0" E

g

Egtg ;’efsbzlgt Do gmun% : (i.e., as seen |

by observers

@ on the ground) /

.

/

Platform rotating

i

sersons seated on a rotating platform such as a merry-go-round. It looks to them as if

the rest of the world is going around them. But let us focus attention on what they observe when they place a tennis ball on the floor of the rotating platform, which we assume is frictionless. frictionless. If th they put the he ball bal d down gently, ly, without with giving iving iti any push, they will observe that it accelerates from rest and moves outward as shown in Fig. 11-25a. According to Newton’s first law, an object initially at rest should stay at rest if no net force acts on it. But, according to the observers on the rotating platform, the ball starts moving even though there is no net force acting on it. To observers on the ground this is all very clear: the ball has an initial velocity when it is released (because the platform is moving), and it simply continues moving in 2™ .

.

o

e

.

A

5.

e

platform (i.e, ~ Straight-line path as shown in Fig. 11-25b, in accordance with Newton’s first law. s seen by But what shall we do about the frame of reference of the observers on the rotating

(a) Rotating reference frame a0}

to place ourselves (in theory, if not physically) into a reference frame that is rotating.

—__~ counterclockwise

(b) Inertial reference frame

platform? Since the ball starts moving without any net force on it, Newton’s first law, the law of inertia, does not hold in this rotating frame of reference. For this reason,

such a frame is called a noninertial reference frame. An inertial reference frame

(as we discussed in Chapter 4) is one in which the law of inertia—Newton’s first

law—does hold, and so do Newton'’s second and third laws. In a noninertial reference

frame, such as our rotating platform, Newton’s second law also does not hold. For instance in the situation described above, there is no net force on the ball; yet, with respect to the rotating platform, the ball accelerates. ol

i

Fictitious (Inertial) Forces

Because Newton’s laws do not hold when observations are made with respect to a rotating frame of reference, calculation of motion can be complicated. However, we can still make use of Newton’s laws in such a reference frame if we make use of a trick. The ball on the rotating platform of Fig. 11-25a flies outward when released (even though no force is actually acting on it). So the trick we use is to write down the equation 2F = ma as if a force equal to mv?/r (or mw’r) were acting radially outward on the object in addition to any other forces that may be acting. This extra force, which might be designated as “centrifugal force” since it seems to act outward, is called a fictitious force or pseudoforce. It is a pseudoforce (“pseudo” means “false”) because there is no object that exerts this force. Furthermore, when viewed

from

an inertial reference frame, the effect doesn’t exist at all. We

have

made up this pseudoforce so that we can make calculations in a noninertial frame using Newton’s second law, £F = ma. Thus the observer in the noninertial frame of Fig. 11-25a uses Newton’s second law for the ball’s outward motion by assuming“™ that a force equal to mv?/r acts on it. Such pseudoforces are also called inertial forces since they arise only because the reference frame is not an inertial one.

300

CHAPTER11

Angular Momentum; General Rotation

The Earth itself is rotating on its axis. Thus, strictly speaking, Newton’s laws are not valid on the Earth. However, the effect of the Earth’s rotation is usually so small that it can be ignored, although it does influence the movement of large air

masses

and ocean currents. Because

of the Earth’s rotation, the material

of the

arth is concentrated slightly more at the equator. The Earth is thus not a perfect sphere but is slightly fatter at the equator than at the poles.

The Coriolis Effect

In a reference frame that rotates at a constant angular speed w (relative to an inertial frame), there exists another pseudoforce known as the Coriolis force. It appears to act on an object in a rotating reference frame only if the object is

.

)

S

.

the man

at B on

the outer

edge

of the platform.

In Fig. 11-26a, we

view

N

N

the

situation from an inertial reference frame. The ball initially has not only the

velocity v radially outward, but also a tangential velocity ¥, due to the rotation of the platform. Now Eq. 10-4 tells us that v, = ryow, where r, is the woman’s radial distance from the axis of rotation at O. If the man at B had this same velocity vy, , the ball would reach him perfectly. But his speed is vy = rgw, which is greater than v, because rg > r,. Thus, when the ball reaches the outer edge of the platform, it passes a point that the man at B has already gone by because his speed in that direction is greater than the ball’s. So the ball passes behind him. Figure 11-26b shows the situation as seen from the rotating platform as frame " reference. Both A and B are at rest, and the ball is thrown with velocity v .oward B, but the ball deflects to the right as shown and passes behind B as previously described. This is not a centrifugal-force effect, for the latter acts radially outward. Instead, this effect acts sideways, perpendicular to ¥, and is called a Coriolis acceleration; it is said to be due to the Coriolis force, which is a fictitious inertial force. Its explanation as seen from an inertial system was given above:itis an effect ofibeing in a rotating system, wherein a point farther from the rotation

axis has a higher linear speed. On the other hand, when viewed from the rotating

system, we can describe the motion using Newton’s second law,

add a “pseudoforce” term corresponding to this Coriolis effect.

SF = ma,

if we

NG

S

:

>

i

‘_"i""f

gx =3

oe,

>

|

\

S 8 8

at distances r, and rp from the axis of rotation (at O). The woman at A throws a

ball with a horizontal velocity ¥ (in her reference frame) radially outward toward

i

bl

moving relafive to that rotating reference frame, and it acts to deflect the object sideways. It, too, is an effect of the rotating reference frame being noninertial and hence is referred to as an inertial force. It too affects the weather. To see how the Coriolis force arises, consider two people, A and B, at rest on a platform rotating with angular speed w, as shown in Fig. 11-26a. They are situated

P

x

*11-9

g

MPFEECUIE] seleeae fEm \\\\

Oe

\

S

By

\

7\ Pathof ball )

(b) Rotating reference frame

~ FIGURE 11-26 The origin of the Coriolis effect. Looking down on a

rotating platform, (a) as seen from a

nonrotating inertial reference frame,

and (b) as seen from the rotating

Let us determine the magnitude of the Coriolis acceleration for the simple ~ Platform as frame of reference.

case described above. (We assume v is large and distances short, so we can ignore gravity.) We do the calculation from the inertial reference frame (Fig. 11-26a). The ball moves radially outward a distance rg — r, at speed v in a short time ¢ given by rg —

rp

=

ot

During this time, the ball moves to the side a distance s, given by SA

=

Va L

The man at B, in this time ¢, moves a distance

| 55 = wvgt. The ball therefore passes behind him a distance s (Fig. 11-26a) given by

s = sgp— sa = (vg — wu)l.

We saw earlier that v, = ryw s

™

We substitute

=

(rB =

and

vz = rgw, so

rA)wt.

rg — r, = ot (see above) and get

s = wut’,

(11-14)

rhis same s equals the sideways displacement as seen from the noninertial rotating system (Fig. 11-26b). *SECTION 11-9

The Coriolis Effect

301

FIGURE 11-27 (a) Winds (moving air masses) would flow directly toward a low-pressure area if the Earth did not rotate. (b) and (c): Because of the Earth’s rotation, the winds are deflected to the right in the Northern Hemisphere (as in Fig. 11-26) as if a fictitious (Coriolis) force were acting.

\ —_—

/ prlgg’sfire

We see acceleration. acceleration in the form

-—

-

immediately that Eq. 11-14 corresponds to motion at constan. For as we saw in Chapter 2 (Eq. 2-12b), y = lar? for a constant (with zero initial velocity in the y direction). Thus, if we write Eq. 11-14 s = Jac,,%, we see that the Coriolis acceleration ac, is Acor

/

\

()

=

200.

This relation is valid for any velocity in the plane of rotation perpendicular to the axis of rotation’ (in Fig. 11-26, the axis through point O perpendicular to the page). Because the Earth rotates, the Coriolis effect has some interesting manifestations on the Earth. It affects the movement of air masses and thus has an influence on weather. In the absence of the Coriolis effect, air would rush directly into a region of low pressure, as shown

W

Low pressure

S———

(b)

(11-15)

in Fig. 11-27a. But because of the Coriolis effect, the

winds are deflected to the right in the Northern Hemisphere (Fig. 11-27b), since the Earth rotates from west to east. So there tends to be a counterclockwise wind pattern around a low-pressure area. The reverse is true in the Southern Hemisphere. Thus cyclones rotate counterclockwise in the Northern Hemisphere and clockwise in the Southern Hemisphere. The same effect explains the easterly trade winds near the equator: any winds heading south toward the equator will be deflected toward the west (that is, as if coming from the east). The Coriolis effect also acts on a falling object. An object released from the top of a high tower will not hit the ground directly below the release point, but will be deflected slightly to the east. Viewed from an inertial frame, this happens because the top of the tower revolves with the Earth at a slightly higher speed than the bottom of the tower. "The Coriolis acceleration can be written in general in terms of the vector cross product a. dcor = —20 X Vv where @ has direction along the rotation axis; its magnitude is acor = 2wv; where v, is the component of velocity perpendicular to the rotation axis.

| Summary The angular momentum L of a rigid object rotating about a fixed axis is given by L

=

lo.

at any instant. The net torque =7 on a particle is related to its angular momentum by

11-1

dL

Newton’s second law, in terms of angular momentum, is Sip

=

dL ar

If the net torque on an object is zero,

2%

11-2) dL/dt

= 0,

so

L =

constant. This is the law of conservation of angular momentum. The vector product or cross product of two vectors A and B is another vector € = A X B whose magnitude is ABsin 6 and whose direction is perpendicular to both A and B in the sense of the right-hand rule. The torque 7 due to a force F is a vector quantity and is always calculated about some point O (the origin of a coordinate system) as follows: =

7 =¥FXF,

(11-5)

where T is the position vector of the point at which the force F acts. Angular momentum is also a vector. For a particle having momentum P = m¥, the angular momentum L about some point O is

L =f%xp

(11-6)

where T is the position vector of the particle relative to the point O

302

CHAPTER 11

Angular Momentum; General Rotation

=

dr

11-7

For a system of particles, the total angular momentum L = L. The total angular momentum of the system is related to the total net torque =7 on the system by 27

=

dL dr

(11-9)

This last relation is the vector rotational equivalent of Newton’s second law. It is valid when L and 27 are calculated about an origin (1) fixed in an inertial reference system or (2) situated at the cM of the system. For a rigid object rotating about a fixed axis, the component of angular momentum about the rotation axis is given by L, = [w. If an object rotates about an axis of symmetry, then the vector relation

not true in general. If the total net total vector

angular

torque

on

momentum

L = I®

a system

L remains

holds, but this is

is zero, then

constant. This

the

is

the important law of conservation of angular momentunfl It applies to the vector L and therefore also to each of ii components.

| Questions

-~

1. If there were a great migration of people toward the Earth’s

equator, would the length of the day (a) get longer because of conservation of angular momentum; (b) get shorter because of conservation of angular momentum; (c) get shorter because of conservation of energy; (d) get longer because of conservation of energy; or (e) remain unaffected? 2. Can the diver of Fig. 11-2 do a somersault without having any initial rotation when she leaves the board? . Suppose you are sitting on a rotating stool holding a 2-kg mass in each outstretched hand. If you suddenly drop the

13. Describe the torque needed if the person in Fig. 11-17 is to 14.

15.

masses, will your angular velocity increase, decrease, or stay

5 6

the same? Explain. When a motorcyclist leaves the ground on a jump and leaves the throttle on (so the rear wheel spins), why does the front of the cycle rise up? Suppose ‘you are standing on the edge of a large freely rotating tumtablel‘ What happens if you walk toward the center? A shortstop may leap into the air to catch a ball and throw

it quickly. As he throws the ball, the upper part of his body

rotates. If you look quickly you will notice that his hips and legs rotate in the opposite direction (Fig. 11-28). Explain.

16.

17.

tilt the axle of the rotating wheel directly upward without it swerving to the side. An astronaut floats freely in a weightless environment. Describe how the astronaut can move her limbs so as to (a) turn her body upside down and (b) turn her body about-face. On the basis of the law of conservation of angular momentum, discuss why a helicopter must have more than one rotor (or propeller). Discuss one or more ways the second propeller can operate in order to keep the helicopter stable. A wheel is rotating freely about a vertical axis with constant angular velocity. Small parts of the wheel come loose and fly off. How does this affect the rotational speed of the wheel? Is angular momentum conserved? Is kinetic energy conserved? Explain. Consider the following vector quantities: displacement, velocity,

acceleration,

see

11-29,

momentum,

angular

Fig.

which

is

drawn

for

the

winter

(December 21). About what axis would you Earth’s rotation axis to precess as a result

of the torque due to the Sun? Does the torque exist three months later?

FIGURE 11-28

Question 6. A shortstop in the air,

throwing the ball.

in direction, how would this alter V; X V,? Name the four different conditions that V]

X VZ

Equator

Explain.

7 If all the components of the vectors V; and V, were reversed could

make

= 0.

. A force F = Fj is applied to an object at a position ¥ = xi+ yj + zk where the origin is at the cM. Does the torque about the cM depend on x? On y? On z? 10. A particle moves with constant speed along a straight line. How does its angular momentum, calculated about any point not on its path, change in time? 11. If the net force on a system is zero, is the net torque also zero? If the net torque on a system is zero, is the net force zero? Give examples. 12. Explain how a child “pumps” on a swing to make it go higher.

momentum,

torque. (a) Which of these are independent of the choice of origin of coordinates? (Consider different points as origin which are at rest with respect to each other.) (b) Which are independent of the velocity of the coordinate system? 18. How does a car make a right turn? Where does the torque come from that is needed to change the angular momentum? *19. The axis of the Earth precesses with a period of about 25,000 years. This is much like the precession of a top. Explain how the Earth’s equatorial bulge gives rise to a torque exerted by the Sun and Moon on the Earth;

FIGURE 11-29 Question 19. (Not to scale.)

’ Sun

Orbit plane

solstice

expect

the

12359/ = :

I I

/

South Pole

*20. Why is it that at most locations on the Earth, a plumb bob does not hang precisely in the direction of the Earth’s center? *21, In a rotating frame of reference, Newton’s first and second laws remain useful if we assume that a pseudoforce equal to ma?r is acting, What effect does this assumption have on the validity of Newton’s third law? *22. In the battle of the Falkland Islands in 1914, the shots of British gunners initially fell wide of their marks because their calculations were based on naval battles fought in the Northern Hemisphere. The Falklands are in the Southern Hemisphere. Explain the origin of their problem.

| Problems 11-1

rxl.

Angular Momentum

(I) What is the angular momentum of a 0.210-kg ball rotating on the end of a thin string in a circle of radius 1.35 m at an angular speed of 10.4 rad/s?

. (I) (a) What is the angular momentum of a 2.8-kg uniform cylindrical grinding wheel of radius 18 cm when rotating at 1300 rpm? (b) How much torque is required to stop it in 6.0s?

Problems

303

(II) A person stands, hands at his side, on a platform that is

rotating at a rate of 0.90 rev/s. If he raises his arms to a horizontal position, Fig. 11-30, the speed of rotation decreases to 0.70 rev/s. (a) Why? (b) By what factor has his moment of inertia changed?

FIGURE 11-30

Problem 3.

—

. (II) A figure skater can increase her spin rotation rate from an initial rate of 1.0 rev every 1.5s to a final rate of 2.5rev/s. If her initial moment of inertia was 4.6 kg m?, what is her final moment of inertia? How does she physically accomplish this change? . (II) A diver (such as the one shown in Fig. 11-2) can reduce her moment of inertia by a factor of about 3.5 when changing from the straight position to the tuck position. If she makes 2.0 rotations in 1.5s when in the tuck position, what is her angular speed (rev/s) when in the straight position? . (II) A uniform horizontal rod of mass M and length { rotates with angular velocity @ about a vertical axis through its center. Attached to each end of the rod is a small mass m. Determine the angular momentum of the system about the axis. . (II) Determine the angular momentum of the Earth (a) about its rotation axis (assume the Earth is a uniform sphere), and (b) in its orbit around the Sun (treat the Earth as a particle orbiting the Sun). The Earth has

mass = 6.0 X 10**kg

and

1.5 x 108km from the Sun.

radius = 6.4 X 10°m,

and is

. (II) (a) What is the angular momentum of a figure skater spinning at 2.8rev/s with arms in close to her body, assuming her to be a uniform cylinder with a height of 1.5 m, a radius of 15 cm, and a mass of 48 kg? (b) How much torque is required to slow her to a stop in 5.0 s, assuming she does not move her arms? . (II) A person stands on a platform, initially at rest, that can rotate freely without friction. The moment of inertia of the person plus the platform is /p. The person holds a spinning bicycle wheel with its axis horizontal. The wheel has moment of inertia /y and angular velocity wy . What will be the angular velocity wp of the platform if the person moves the axis of the wheel so that it points (a) vertically upward, (b) at a 60° angle to the vertical, (c¢) vertically downward? (d) What will wp be if the person reaches up and stops the wheel in part (a)? 10. (II) A uniform disk turns at 3.7 rev/s around a frictionless spindle. A nonrotating rod, of the same mass as the disk and length : equal

to

the

disk’s

diameter,

is

dropped onto the freely spinning

disk, Fig. 11-31. They then turn together around the spindle with their centers superposed. What is the angular frequency in rev/s of the combination? FIGURE 11-31 Problem 10.

304

CHAPTER 11

//

}/

}

: I

Angular Momentum; General Rotation

11. (II) A person of mass 75 kg stands at the center of a rotating merry-go-round platform of radius 3.0 m and moment of inertia 920 kg - m?. The platform rotates without friction with angular velocity 0.95 rad/s. The person walks radially to the edge of the platform. (@) Calculate the angular velocity when t person reaches the edge. (b) Calculate the rotational kinew energy of the system of platform plus person before and after the person’s walk. 12. (II) A potter’s wheel is rotating around a vertical axis through its center at a frequency of 1.5 rev/s. The wheel can be considered a uniform disk of mass 5.0kg and diameter 0.40m. The potter then throws a 2.6-kg chunk of clay, approximately shaped as a flat disk of radius 8.0 cm, onto the center of the rotating wheel. What is the frequency of the wheel after the clay sticks to it? 13. (II) A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.80 rad/s. Its total moment of inertia is 1760 kg - m?. Four people standing on the ground, each of mass 65kg, suddenly step onto the edge of the merry-go-round. What is the angular velocity of the merry-go-round now? What if the people were on it initially and then jumped off in a radial direction (relative to the merry-go-round)? 14. (II) A woman of mass m stands at the edge of a solid cylindrical platform of mass M and radius R. At ¢ = 0, the platform is rotating with negligible friction at angular velocity wy about a vertical axis through its center, and the woman begins walking with speed v (relative to the platform) toward the center of the platform. (a) Determine the angular velocity of the system as a function of time. (b) What will be the angular velocity when the woman reaches the center? 15. (IT) A nonrotating cylindrical disk of moment of inertia / r\ dropped onto an identical disk rotating at angular speed Assuming no external torques, what is the final common angular speed of the two disks? 16. (IT) Suppose our Sun eventually collapses into a white dwarf, losing about half its mass in the process, and winding up with a radius 1.0% of its existing radius. Assuming the lost mass carries away no angular momentum,

what would

the Sun’s new rotation rate be? (Take the Sun’s current period to be about 30 days.) What would be its final kinetic energy in terms of its initial kinetic energy of today? 17. (III) Hurricanes can involve winds in excess of 120 km/h at the outer edge. Make a crude estimate of (a) the energy, and (b) the angular momentum, of such a hurricane, approximating it as a rigidly rotating uniform cylinder of air (density 1.3kg/m®) of radius 85km and height 4.5 km. 18. (IIT) An asteroid of mass 1.0 X 10° kg, traveling at a speed of 35km/s relative to the Earth, hits the Earth at the equator tangentially, in the direction of Earth’s rotation, and is embedded there. Use angular momentum to estimate the percent change in the angular speed of the Earth as a result of the collision. . (IIT) Suppose a 65-kg person stands at the edge of a 6.5-m diameter merry-go-round turntable that is mounted on frictionless bearings and has a moment of inertia of 1850 kg-m?. The turntable is at rest initially, but when the person begins running at a speed of 3.8 m/s (with respect to the turntable) around its edge, the turntable begins to rota in the opposite direction. Calculate the angular velocity ¢ the turntable.

11- 2 Vector Cross Product and Torque 20. (I) If vector A points along the negative x axis and vector B

along the positive z axis, what is the direction of (a) A X B and (b) B X A? (c) What is the magnitude of A X B and

f

B x A?

21.

29. (IT) Use the result of Problem 26 to determine (a) the vector product A x B and (b) the angle between A and B if A =54i — 35j and B = —8.5i + 5.6j + 2.0k. 30. (III) Show that the velocity¥ of any point in an object rotating with angular velocity @ about a fixed axis can be written

()Showthat (a) i xi=jxj=kxk=0 (b)ixj=k

V=aXF

iXk——J,and_]xk=i

22, (I) The directions of vectors A and B are given below for several cases. For each case, state the direction of A X B. (a) A points east, B points south. (b) A points east, B points straight down. (c) A points straight up, B points north. (d) A points straight up, B points straight down. 23. (IT) What

is the angle 6 between

calculated about the origin?

25. (I1) Con51der a particle of a rigid object rotating about a fixed axis. Show that the tangential and radial vector componefits of the linear acceleration are:

26. (IT) = A=

=

@ XTF

and

dg

=

@ X V.

(a) Show that the cross product of two vectors, Ax1+Ay_|+Azk and B = Bxl+ByJ+sz is

AxB = (AyB; - AzBy))i + (ABy - AyBy)j -

-

+ (AxBy — AyBy)k.

AXB =

J

|Ay Ay By B,

tions,

on in

the Fig

the

total

highway 11-32

Ay, B;

that under

the most

adverse

sign will

acting at the cM. What torque does this force exert about the base O?

same _plane

11 32

Show

that

-3 Angular Momentum of a Particle (I) What are the x, y, and z components of the angular momentum of a particle located at T = xi+ yi + zk which has momentum p = pyi + pyj + p k?

33. (I) Show that the kinetic energy K of a particle of mass m, moving

in a circular path, is

K

= 1%/2I,

where

L is its

angular momentum and / is its moment of inertia about the center of the circle. 34. (I) Calculate the angular momentum of a particle of mass m moving with constant velocity v for two cases (see Fig. 11-33): (a) about origin O, and (b) about O'.

have equal but opposite are not traveling along the angular momentum of this choice of origin.

36. (IT) Determine the angular momentum of a 75-g particle about the origin of coordinates when the particle is at x = 4.4m,

y = —6.0m, and it has velocity v = (3.2i — 8.0k) m/s.

37. (II) A particle is at the position (x, y,z) = (1.0,2.0,3.0) m. It is traveling with a vector velocity (—5.0, +2.8, —=3.1) m/s. Its mass is 3.8 kg. What is its vector angular momentum about the origin?

38. (II) An Atwood machine (Fig. 11-16) consists of two masses, ma = 7.0kg and mp = 82kg, connected by a cord that passes over a pulley free to rotate about a fixed axis. The pulley is a solid cylinder of radius Ry = 0.40m and mass 0.80 kg. (¢) Determine the acceleration a of each mass. (b) What

= || =—> 8

Pivot

480 g FIGURE 11-34 Problem 40.

24 cm

41. (IT) Figure 11-35 shows two masses connected by a cord passing over a pulley of radius Ry and moment of inertia /. Mass M slides on a frictionless surface, and Mp hangs freely. Determine a formula for (a) the angular momentum of the system about the pulley axis, as a function of the speed v of mass M, or Mg, and (b) the acceleration of the masses.

FIGURE 11-36 Problems 48 and 83. 49. (II) Suppose a 5.8 X 10°kg at the equator with a

speed

| meteorite

struck

the Earth

v =22 X 10*m/s,

as shown in Fig. 11-37 and remained stuck. By what factor would this affect the rotational frequency of the Earth (1rev/day)? FIGURE 11-37 Problem 49.

50. (IIT) A 230-kg beam 2.7 m in length slides broadside dow

*45. (III) Suppose in Fig. 11-18 that mpg = 0; that is, only one mass, mpa, is actually present. If the bearings are each a distance d from O, determine the forces F, and Fy at the upper and lower bearings respectively. [Hint: Choose an origin—different than O in Fig. 11-18—such that L is parallel to @. Ignore effects of gravity.] * 46. (IIT) Suppose in Fig. 11-18 that mu = mg = 0.60 kg, ra = rg = 0.30m, and the distance between the bearings is 0.23m. Evaluate the force that each bearing must exert on the axle if ¢ = 34.0°, w = 11.0rad/s.

306

CHAPTER 11

Angular Momentum; General Rotation

(Fig. 11-38). A 65-kg man at rest grabs one end as it goes past and hangs on as both he and the beam go spinning down the ice. Assume frictionless motion. (@) How fast does the center of mass of the system move after the collision? (b) With what angular velocity does the system rotate about its cM?

R

CM

FIGURE 11-38 Problem 50.

51. (IIT) A thin rod of mass M and length £ rests on a frictionless table and is struck at a point £/4 from its cM by a clay ball of mass m — moving at speed v (Fig. 11-39). The »

*44, (TIT) What is the magnitude of the force F exerted by each bearing in Fig. 11-18 (Example 11-10)? The bearings are a distance d from point O. Ignore the effects of gravity.

of 18m/s

ball sticks to the rod. Determine the translational and rotational motion of the rod after the collision.

-

43. (IIT) Show that the total angular momentum L = 3§ X pi of a system of particles about the origin of an inertial reference frame can be written as the sum of the angular momentum about the cM, L* (spin angular momentum), plus the angular momentum of the cM about the origin (orbital angular momentum): L = L* + Foy X M¥qy. [Hint: See the derivation of Eq. 11-9b.]

a speed

fe %&

42. (III) A thin rod of length £ and mass M rotates about a vertical axis through its center with angular velocity w. The rod makes an angle ¢ with the rotation axis. Determine the magnitude and direction of L.

ice with

'.

FIGURE 11-35 Problem 41.

the

CcM

FIGURE 11-39

Problems 51 and 84.

™

52. (1I1) On a level billiards table a cue ball, initially at rest at point O on the table, is struck so that it leaves the cue stick with a center-of-mass speed vy and a “reverse” spin of angular speed wy (see Fig. 11-40). A kinetic friction force acts on the ball as it r initially skids across the table. (a) Explain why the ball’s angular momentum is conserved about point O. (b) Using conservation of angular momentum, find the critical angular speed wc such that, if wy = wc,

kinetic friction will bring the

ball to a complete (as opposed to momentary) stop. (¢) If wg is 10% smaller than wc, ie., wy = 0.90wc, determine the ball’s cM velocity vy When it starts to roll without slipping. (d) If wq is 10% larger than wc, i.e., wg = 1.10 wc, determine the ball’s cM velocity vy When it starts to roll without slipping. [Hint: The ball possesses two types of angular momentum, the first due to the linear speed vy, of its cM relative to point O, the second due to the spin at angular velocity w about its own cM. The ball’s total L about O is the sum of these two angular momenta. ]

*11-8

Rotating Reference Frames

*58. (II) If a plant is allowed to grow from seed on a rotating platform, it will grow at an angle, pointing inward. Calculate what this angle will be (put yourself in the rotating frame) in terms of g, r, and w. Why does it grow inward rather than outward? *89. (IIT) Let g’ be the effective acceleration of gravity at a point on the rotating Earth, equal to the vector sum of the “true” value g plus the effect of the rotating reference frame (ma?r term). See Fig. 11-42. Determine the magnitude and direction of g’ relative to a radial line from the center of the Earth (a) at the North Pole, (b) at a latitude of 45.0° north, and (c) at mw?r the equator. Assume that g (if o were zero) is a

constant 9.80 m/s’.

FIGURE 11-40

FIGURE 11-42

Problem 52.

*11-7

Problem 59.

Spinning Top

*53. (IT) A 220-g top spinning at 15 rev/s makes an angle of 25° to the vertical and precesses at a rate of 1.00 rev per 6.5s. If its cM is 3.5 cm from its tip along its symmetry axis, what is the moment of inertia of the top? *54, (IT) A toy gyroscope consists of a 170-g disk with a radius of 5.5cm mounted at the center of a thin axle 21cm long (Fig. 11-41). The gyroscope spins at 45 rev/s. One end of its axle rests on a stand and the other end precesses horizontally about the stand. (¢) How long does it take the gyroscope to precess once around? (b) If all the dimensions of the gyroscope were doubled (radius = 11cm, axle = 42cm), how long would it take to precess once?

~

S Pole

*11-9 Coriolis Effect *60. (II) Suppose the man at B in Fig. 11-26 throws the ball toward the woman at A. (¢) In what direction is the ball deflected as seen in the noninertial system? (b) Determine a formula for the amount of deflection and for the (Coriolis) acceleration in this case. *61. (I1) For what directions of velocity would the Coriolis effect on an object moving at the Earth’s equator be zero? *62. (IIT) We can alter Eqs. 11-14 and 11-15 for use on Earth by considering only the component of v perpendicular to the axis of rotation. From Fig. 11-43, we see that this is v cos A for a vertically falling object, where A is the latitude of the place on the Earth. If a lead ball is dropped vertically from

a

110-m-high

tower

in

Florence,

Italy (latitude = 44°), how far from the base of the tower is it deflected by the Coriolis force?

FIGURE 11-41 A wheel, rotating about a horizontal

axle supported at one end,

precesses. Problems 54, 55,

North Pole ¥ €08 A

and 56.

* 85, (IT) Suppose the solid wheel of Fig. 11-41 has a mass of 300g

and rotates

at

85rad/s;

it has radius 6.0cm

and is

mounted at the center of a horizontal thin axle 25 cm long. At what rate does the axle precess? *56. (IT) If a mass equal to half the mass of the wheel in Problem 55 is placed at the free end of the axle, what will be the precession

rate now? Treat the extra mass as insignificant in size. *§7. (IT) A bicycle wheel of diameter 65 cm and mass m rotates on its axle; two 20-cm-long wooden handles, one on each side of the wheel, act as the axle. You tie a rope to a small hook on the end of one of the handles, and then spin the bicycle wheel with a flick of the hand. When you release the spinning wheel, it precesses about the vertical axis defined by the rope, instead of falling to the ground (as it would if it were not spinning). Estimate the rate and direction of precession if the wheel rotates counterclockwise at 2.0rev/s and its axle remains horizontal.

FIGURE 11-43 Problem 62. Object of mass m falling vertically to Earth at a latitude A.

South Pole

*63. (III) An ant crawls with constant speed outward along a radial spoke of a wheel rotating at constant angular velocity w about a vertical axis. Write a vector equation for all the forces (including inertial forces) acting on the ant. Take the x axis along the spoke, y perpendicular to the spoke pointing to the ant’s left, and the z axis vertically upward. The wheel rotates counterclockwise as seen from above.

Problems

307

| General Problems 64. A thin string is wrapped around a cylindrical hoop of radius R and mass M. One end of the string is fixed, and the hoop

is allowed to fall vertically, starting from rest, as the string

unwinds. (a) about its cM the string as 65. A particle v = (7.0i + L relative

Determine the angular momentum of the hoop as a function of time. () What is the tension in function of time? of mass 1.00kg is moving with velocity 6.0j)m/s. (a) Find the angular momentum to the origin when the particle is at

F = (2.0j + 4.0k) m. (b) At position ¥ a force of F = 4.0Ni

is applied to the particle. Find the torque relative to the origin. 66. A merry-go-round with a moment of inertia equal to

1260 kg-m? and a radius of 2.5m rotates with negligible

friction at 1.70rad/s. A child initially standing still next to the merry-go-round jumps onto the edge of the platform straight toward the axis of rotation causing the platform to slow to 1.25 rad/s. What is her mass? 67. Why might tall narrow SUVs and buses be prone to “rollover”? Consider a vehicle rounding a curve of radius R on a flat road. When just on the verge of rollover, its tires on the inside of the curve are about to leave the ground, so the friction and normal force on these two tires are zero. The total normal force on the outside tires is Fy and the total friction force is Fy;. Assume that the vehicle is not skidding, (a) Analysts define a static stability factor SSF = w/2h, where a vehicle’s “track width” w is the distance between tires on the same axle, and 4 is the height of the cM above the ground. Show that the critical rollover speed is

[Hint: Take

mass

of

the

torques SUV,

about an axis through

parallel

to

its

direction

the center of of

motion.]

(b) Determine the ratio of highway curve radii (minimum possible) for a typical passenger car with SSF = 1.40 and

an SUV with SSF = 1.05 at a speed of 90 km/h.

68. A

spherical

asteroid

with

radius

r = 123m

and

mass

M = 225 x 10'%kg rotates about an axis at four revolutions

per day. A “tug” spaceship attaches itself to the asteroid’s south pole (as defined by the axis of rotation) and fires its engine, applying a force F tangentially to the asteroid’s surface as shown in Fig. 11-44. If F = 265N, how long will it take the tug to rotate the asteroid’s axis of rotation throughan angle of 10.0° by this method?

FIGURE 11-44 Problem 68.

308

CHAPTER11

Angular Momentum; General Rotation

69. The

time-dependent

position

of a point

object

which

moves counterclockwise along the circumference of a circlh (radius R) in the xy plane with constant speed v is given F =

iRcoswr

where the constant

+ szinwt

@ = v/R.

Determine the velocity ¥ and

angular velocity @ of this object and then show that these three vectors obey the relation v = @ X T. 70. The position of a particle with mass m traveling on a helical path (see Fig. 11-45) is given by r =

Rcos(-z—Zé)i

+ Rsin(%—z)j

where R and d are the radius and pitch tively, and z has time dependence z the (constant) component of velocity Determine the time-dependent angular particle about the origin.

Cmm”

“"mme®

FIGURE

~~-—’

11-45

+ 7k

of the helix, respec= v;¢ where v; is in the z direction. momentum L of the

Came®

e

Problem 70.

b B A boy rolls a tire along a straight level street. The tire h.\ mass 8.0kg, radius 0.32m and moment of inertia about its central axis of symmetry of 0.83 kg-m?. The boy pushes the tire forward away from him at a speed of 2.1 m/s and sees that the tire leans 12° to the right (Fig. 11-46). (a) How will the resultant torque affect the subsequent motion of the tire? (b) Compare the change in angular momentum caused by this torque in 0.20s to the original magnitude of angular momentum.

FIGURE 11-46 Problem 71.

72. A 70-kg person stands on a tiny rotating platform with arms outstretched. (a) Estimate the moment of inertia of the person using the following approximations: the body (including head and legs) is a 60-kg cylinder, 12 cm in radius and 1.70m high; and each arm is a 5.0-kg thin rod, 60 cm long, attached to the cylinder. (b) Using the same approximations, estimate the moment of inertia when the arms are at the person’s sides. (c) If one rotation takes 1.5 s when the person’s arms are outstretched, what is the time for each rotation with arms at the sides? Ignore the moment of inertia of the lightweight platform. (d) Determine the change in kinetic energy when the arms are lifted from the sides to the horizontal position. (¢) From your answer 1" part (d), would you expect it to be harder or easier to lift your arms when rotating or when at rest?

’

a2 Water drives a waterwheel (or turbine) of radius R = 3.0m as shown in Fig. 11-47. The water enters at a speed vy = 70m/s and exits from the waterwheel at a speed v =38m/s. (a) If 85kg of water passes through per r second, what is the rate at which the water delivers angular momentum to the waterwheel? (b) What is the torque the water applies to the waterwheel? (c) If the water causes the waterwheel to make one revolution every 5.5s, how much power is delivered to the wheel?

FIGURE 11-47

¥,

Problem 73.

78. A bicyclist traveling with speed » = 9.2m/s on a flat road is making a turn with a radius r = 12m. The forces acting on the cyclist and cycle are the normal force (Fy) and friction

force (Fy;) exerted by the road on the tires and mg, the total weight of the cyclist and cycle. Ignore the small mass of the wheels. (a) Explain carefully why the angle 6 the bicycle makes with the vertical (Fig. 11-48) must be given by tané = Fy/Fy if the cyclist is to maintain balance. (b) Calculate 6 for the values given. [Hint: Consider the “circular” translational motion of the bicycle and rider.] (c¢) If the coefficient of static friction between tires and road is ug = 0.65, what is the minimum turning radius?

s

74. The Moon orbits the Earth such that the same side always faces the Earth. Determine the ratio of the Moon’s spin angular momentum (about its own axis) to its orbital angular momentum. (In the latter case, treat the Moon as a particle orbiting the Earth.) 75. A particle of mass m uniformly accelerates as it moves counterclockwise along the circumference of a circle of radius R: F =

iRcos6

+ jRsin6

i:1fr

with 6 = wyt + 5 ar?, where the constants wy and a are the

initial angular velocity and angular acceleration, respectively. Determine the object’s tangential acceleration &;,, and determine the torque acting on the object using () # = ¥ X F, b) 7 = Ia. . A projectile with mass m is launched from the ground and follows a trajectory given by = r

=

('Ux()t)l

3

t

0 counterclockwise 7 < 0 clockwise

SECTION 12-2

313

> &

~N

4’@ 4. For the torque equation,

Statics

Q @

¢

sO LV,

1. Choose one object at a time for consideration. Make a careful free-body diagram by showing all the forces acting on that object, including gravity, and the points at which these forces act. If you aren’t sure of the direction of a force, choose a direction; if the actual direction is opposite, your eventual calculation will give a result with a minus sign. 2. Choose a convenient coordinate system, and resolve the forces into their components. 3. Using letters to represent unknowns, write down the equilibrium equations for the forces:

=2

3F, = 0 and

ZF, = 0,

3t

APPLIED

Balancing a seesaw

my =30kg

g

Pl A

«~—25m

Ky =m,g

j

4 FN

Phe

X Fg=mgg

rotate the object counterclockwise

are positive, then

tions allow a maximum of three unknowns to be solved

for. They can be forces, distances, or even angles.

Balancing a seesaw. A board of mass

M = 2.0kg

serves

as a seesaw for two children, as shown in Fig. 12-6a. Child A has a mass of 30 kg and sits 2.5m from the pivot point, P (his center of gravity is 2.5m from the pivot). At what distance x from the pivot must child B, of mass 25 kg, place herself to balance the seesaw? Assume the board is uniform and centered over the pivot. APPROACH We follow the steps of the Problem Solving Strategy above. SOLUTION 1. Free-body diagram. We choose the board as our object, and assume it is horizontal. Its free-body diagram is shown in Fig. 12-6b. The forces acting on the board are the forces exerted downward on it by each child, F, and Fg, the upward force exerted by the pivot Fy, and the force of gravity on the board (= Mg) which acts at the center of the uniform board. Coordinate system. We choose y to be vertical, with positive upward, and x horizontal to the right, with origin at the pivot. 3 Force equation. All the forces are in the y (vertical) direction, so

B

(b)

b

2.5 m-*—%x—%

(a) ¢

v

mg=25kg

?)’

0,

those tending to rotate it clockwise are negative. . Solve these equations for the unknowns. Three equa-

assuming all the forces act in a plane.

@rpHysics

=

choose any axis perpendicular to the xy plane that might make the calculation easier. (For example, you can reduce the number of unknowns in the resulting equation by choosing the axis so that one of the unknown forces acts through that axis; then this force will have zero lever arm and produce zero torque, and so won’t appear in the torque equation.) Pay careful attention to determining the lever arm for each force correctly. Give each torque a + or — sign to indicate torque direction. For example, if torques tending to

FIGURE 12-6 (a) Two children ona seesaw, Example 12-3. (b) Free-body diagram of the board.

2F,

Fx — mag — mgg

=0

— Mg = 0,

where Fy, = m,g and Fy = mgg because each child is in equilibrium when the seesaw is balanced. 4. Torque equation. Let us calculate the torque about an axis through the board at the pivot point, P. Then the lever arms for Fy and for the weight of the board are zero, and they will contribute zero torque about point P.

Thus the torque equation will involve only the forces F, and Fg, which are equal to the weights of the children. The torque exerted by each child will be mg times the appropriate lever arm, which here is the distance of each child from the pivot point. Hence the torque equation is 21

or

mag(2.5m)

— myggx + Mg(0m) mpg(25m)

=0

+ Fy(Om) — mggx

0 =

0,

where two terms were dropped because their lever arms were zero.

314

CHAPTER 12

Static Equilibrium; Elasticity and Fracture

5. Solve. We solve the torque equation for x and find 30kg

x = 2Amsm) =

(25m) = 3.0m.

25kg To balance the seesaw, child B must sit so that her cM is 3.0 m from the pivot point. This makes sense: since she is lighter, she must sit farther from the pivot than the heavier child in order to provide equal torque. mg

EXERCISE C We did not need to use the force equation to solve Example 12-3 because of our choice of the axis. Use the force equation to find the force exerted by the pivot.

Figure 12-7 shows a uniform beam that extends beyond its support like a diving board Such a beam is called a cantilever. The forces acting on the beam in Fig. 12-7 are those due to the supports, F, and Fy, and the force of gravity which acts at the cG, 5.0m to the right of the right-hand support. If you follow the procedure of the last Example and calculate F, and Fy, assuming they point upward as shown in Fig. 12-7, you will find that F, comes out negative. If the beam has a mass of 1200 kg and a weight mg = 12,000N, then F; = 15,000N and F, = —3000N (see Problem 9). Whenever an unknown force comes out negative, it merely means that the force actually points in the opposite direction from what you assumed. Thus in Fig. 12-7, F, actually points downward. With a little reflectwn it should become clear that the left-hand support must indeed pull downward qn the beam (by means of bolts, screws, fasteners and/or glue) if the beam is to be in equilibrium; otherwise the sum of the torques about the cG (or about the point where Fy acts) could not be zero. EXERCISE D Return to the Chapter-Opening Question, p. 311, and answer it again now. Try to explain why you may have answered differently the first time.

Force exerted by biceps muscle. How much force must the

biceps muscle exert when a 5.0-kg ball is held in the hand (@) with the arm . horizontal as in Fig. 12-8a, and (b) when the arm is at a 45° angle as in Fig, 12-8b? The biceps muscle is connected to the forearm by a tendon attached 5.0 cm from the elbow joint. Assume that the mass of forearm and hand together is 2.0 kg and

@PHYSICS

APPLIED

Cantilever

Z{PROBLEM

SOLVING

If a force comes out negative

30.0 m —-‘

FIGURE 12-7 A cantilever. The force vectors shown are hypothetical—one may even have a different direction.

F)PHYSICS

APPLIED

Forces in muscles and joints

FIGURE 12-8

Example 12-4.

their CG is as shown.

APPROACH The free-body diagram for the forearm is shown in Fig. 12-8; the forces are the weights of the arm and ball, the upward force Fy; exerted by the muscle, and a force F; exerted at the joint by the bone in the upper arm (all assumed to act vertically). We wish to find the magnitude of Fy, which is done most easily by using the torque equation and by choosing our axis through the joint so that F contributes zero torque. SOLUTION (a) We calculate torques about the point where Fj acts in Fig. 12-8a. The 27 = 0 equation gives (0.050 m)FM = (0.15m)(2.0kg)g — (0.35m)(5.0kg)g = 0. We solve for Fy: (0.15m)(2.0kg)g + (0.35m)(5.0kg)g M 0.050m (41kg)g 400N F

=

=

=

‘

(b) The lever arm, as calculated about the joint, is reduced by the factor cos 45° for all three forces. Our torque equation will look like the one just above, except that each term will have its lever arm reduced by the same factor, which will cancel out. The same result is obtained,

~

Fy = 400 N.

NOTE The force required of the muscle (400N) is quite large compared to the weight of the object lifted (= mg = 49N). Indeed, the muscles and joints of the de are generally subjected to quite large forces.

(b)

EXERCISE E How much mass could the person in Example 124 hold in the hand with a | biceps force of 450N if the tendon was attached 6.0 cm from the elbow instead of 5.0 cm?

SECTION 12-2

Solving Statics Problems

315

Our next Example involves a beam that is attached to a wall by a hinge and is

supported by a cable or cord (Fig. 12-9). It is important to remember that a flexible cable can support a force only along its length. (If there were a component of force perpendicular to the cable, it would bend because it is flexible.) But for a rigid device, such as the hinge in Fig. 12-9, the force can be in any direction and D can know the direction only after solving the problem. The hinge is assumed small and smooth, so it can exert no internal torque (about its center) on the beam.

Hinged beam and cable. A uniform beam, 2.20 m long with

FIGURE 12-9

Example 12-5.

mass m = 25.0kg, is mounted by a small hinge on a wall as shown in Fig. 12-9. The beam is held in a horizontal position by a cable that makes an angle 6 = 30.0°. The beam supports a sign of mass M = 28.0kg suspended from its end. Determine the components of the force Fy that the (smooth) hinge exerts on the beam, and the tension F; in the supporting cable. APPROACH Figure 12-9 is the free-body diagram for the beam, showing all the forces acting on the beam. It also shows the components of F; and a guess for Fy;. We have three unknowns, Fy,, Fyy, and Fr (we are given 6), so we will need all three equations, 2F, = 0, £F, = 0, 27 = 0. SOLUTION The sum of the forces in the vertical (y) direction is 2F, =0 Fyy + Fry — mg — Mg = 0. (i) In the horizontal (x) direction, the sum of the forces is 2F, = 0 Fyy — Fry = 0. (ii) For the torque equation, we choose the axis at the point where Fr and Mg act (so our equation then contains only one unknown, Fyy). We choose torques that tend to rotate the beam counterclockwise as positive. The weight mg of the (uniform) beam acts at its center, so we have 2r =0 S —(FHy)(Z.ZO m) + mg(1.10 m) 0. We solve for Fyy:

_

(110m

_

N

Fyy = (—-——2.20m>mg = (0.500)(25.0kg)(9.80m/s?) Next, since the tension F; in the cable acts along the cable from Fig. 12-9 that tan6 = Fr,/Fry, or Fry = Frytan® = Fry(tan30.0°). Equation (i) above gives

= 123N.

(iii)

(6 = 30.0°), we see

Fry = (m + M)g — Fyy, = (53.0kg)(9.80m/s?) — 123N

(iv)

= 396N;

Equations (iv) and (ii) give

Fry Fux

= Fry/tan30.0° = 686N; = Fry = 686N.

The components of Fy are Fyy = 123N and Fy, = 686N. The tension in the wire is Fr = \/FTx + F%, = \/(686 N)2 + (396 N)2 = 792 N. Alternate Solution 1et us see the effect of choosing a different axis for calculating torques, such as an axis through the hinge. Then the lever arm for Fy is zero, and the torque equation (27 = 0) becomes -mg(1.10m) — Mg(220m) + Fry(2.20m) = 0. We solve this for Fry and find

Fry = % g + Mg = (125kg + 28.0kg)(9.80m/s?) = 397N. We get the same result, within the precision of our significant figures. NOTE It doesn’t matter which axis we choose for £r = 0. Using a second axis

316

CHAPTER 12

|SRR

heck. AA

Ladder. A 5.0-m-long ladder leans against a smooth wall at a

point 4.0m above a cement floor as shown in Fig. 12-10. The ladder is uniform and has mass m = 12.0kg. Assuming the wall is frictionless (but the floor is not), determine the forces exerted on the ladder by the floor and by the wall. APPROACH Figure 12-10 is the free-body diagram for the ladder, showing all the forces acting on the ladder. The wall, since it is frictionless, can exert a force only

perpendicular to the wall, and we label that force Fy, . The cement floor exerts a force F which has both horizontal and vertical force components: Fc., is frictional and Fey

is the normal force. Finally, gravity exerts a force

mg = (12.0kg)(9.80 m/s?) =

118 N on the ladder at its midpoint, since the ladder is uniform. SOLUTION Again we use the equilibrium conditions, XF, =0, ZF, =0, 37 = 0. We will need all three since there are three unknowns: Fy, Fc,, and Fcy. The y component of the force equation is 2F, = Fcy — mg = 0, so immediately we have Fcy = mg = 118N. The x component of the force equation is SF, = Fey — Fy = 0. To determine both Fc, and Fy, we need a torque equation. If we choose to calculate torques about an axis through the point where the ladder touches the cement floor,

FIGURE 12-10 A ladder leaning against a wall. Example 12-6.

then F., which acts at this point, will have a lever arm of zero and so won’t enter the

equation. The ladder touches the floor a distance

x, = \/ (5.0m)? — (4.0m)* =

3.0m from the wall (right triangle, ¢ = a* + b?). The lever arm for mg is half this, or 1.5 m, and the lever arm for Fy, is 4.0 m, Fig. 12-10. We get 3t = (40m)Fy — (1.5m)mg = 0.

Thus

_

(15m)(12.0kg)(9.8 m/s?)

= 44N. A 40m AThen, from the x component of the force equation, Fey = Fy = 44N. Since the components of F. are F., = 44N and Fcy = 118N,

Fe = \/(44N)> + (118N)? = 126N ~ 130N

then

(rounded off to two significant figures), and it acts at an angle to the floor of 6 = tan '(118 N/44N) = 70°. NOTE The force F. does not have to act along the ladder’s direction because the ladder is rigid and not flexible like a cord or cable.

|g \ EXERCISE F Why is it reasonable to ignore friction along the wall, but not reasonable to ignore it along the floor?

12-3

Stability and Balance

An object in static equilibrium, if left undisturbed, will undergo no translational or rotational acceleration since the sum of all the forces and the sum of all the torques acting on it are zero. However, if the object is displaced slightly, three outcomes are possible: (1) the object returns to its original position, in which case it is said to be in stable equilibrium; (2) the object moves even farther from its original position, and it is said to be in unstable equilibrium; or (3) the object remains in its new position, and it is said to be in neutral equilibrium. Consider the following examples. A ball suspended freely from a string is in stable equilibrium, for if it is displaced to one side, it will return to its original position (Fig. 12-11a) due to the net force and torque exerted on it. On the other hand, a pencil standing on its point is in unstable equilibrium. If its center of gravity is directly over its tip (Fig. 12-11b), the net force and net torque on it will be zero. fiut if it is displaced ever so slightly as shown—say, by a slight vibration or tiny air arrent—there will be a torque on it, and this torque acts to make the pencil continue to fall in the direction of the original displacement. Finally, an example of an object in neutral equilibrium is a sphere resting on a horizontal tabletop. If it is placed slightly to one side, it will remain in its new position—no net torque acts on it.

FIGURE 12-11 (a) Stable equilibrium, and (b) unstable equilibrium.

Net force

(a) 2

9

(b) SECTION 12-3

317

—

La [ €

S~

A

A“G'* /., / ' ]f y4 CG/,// £ .7

M% (a) (b)

- % (c)

FIGURE 12-12 Equilibrium of a refrigerator resting on a flat floor.

FIGURE 12-13

Humans adjust

their posture to achieve stability

when carrying loads.

In most situations, such as in the design of structures and in working with the

human body, we are interested in maintaining stable equilibrium, or balance, as we sometimes say. In general, an object whose center of gravity (cG) is below its point of support, such as a ball on a string, will be in stable equilibrium. If the cG is above

the base of support, we have a more complicated situation. Consider a standin;‘fl refrigerator (Fig. 12-12a). If it is tipped slightly, it will return to its original position due to the torque on it as shown in Fig. 12-12b. But if it is tipped too far, Fig. 12—12c, it will fall over. The critical point is reached when the cG shifts from one side of the pivot point to the other. When the G is on one side, the torque pulls the object back onto its original base of support, Fig. 12-12b. If the object is tipped further, the cG goes past the pivot point and the torque causes the object to topple, Fig. 12-12c. In general, an object whose center of gravity is above its base of support will be stable if a vertical line projected downward from the cG falls within the base of support. This is because the normal force upward on the object (which balances out gravity) can be exerted only within the area of contact, so if the force of gravity acts beyond this area, a net torque will act to topple the object.

Stability, then, can be relative. A brick lying on its widest face is more stable

than a brick standing on its end, for it will take more of an effort to tip it over. In

the extreme case of the pencil in Fig. 12-11b, the base is practically a point and the slightest disturbance will topple it. In general, the larger the base and the lower the CG, the more stable the object. In this sense, humans are less stable than four-legged mammals, which have a larger base of support because of their four legs, and most also have a lower center of gravity. When walking and performing other kinds of movement, a person continually shifts the body so that its cG is over the feet, although in the normal adult this requires no conscious thought. Even as simple a movement as bending over requires moving the hips backward so that the ¢G remains over the feet, and you do this repositioning without thinking about it. To see this, position yourself with your heels and back to a wall and try to touch your toes. You won’t be able to do it without falling. People carrying heavy loads automatically adjust thein\ posture so that the cG of the total mass is over their feet, Fig. 12-13.

12—4

Elasticity; Stress and Strain

In the first part of this Chapter we studied how to calculate the forces on objects in equilibrium. In this Section we study the effects of these forces: any object changes shape under the action of applied forces. If the forces are great enough, the object

will break, or fracture, as we will discuss in Section 12-5.

Elasticity and Hooke's Law FIGURE 12-14 Hooke’s law: Af applied force.

If a force is exerted on an object, such as the vertically suspended metal rod shown in Fig. 12-14, the length of the object changes. If the amount of elongation, Af, is small compared to the length of the object, experiment shows that Af is proportional to the force exerted on the object. This proportionality, as we saw in Section 7-3, can be written as an equation:

F = kAL

(12-3)

Here F represents the force pulling on the object, Af is the change in length, and k is a proportionality constant. Equation 12-3, which is sometimes called Hooke’s law" after Robert Hooke (1635-1703), who first noted it, is found to be valid for almost any solid material from iron to bone—but it is valid only up to a point. For if the force is too great, the object stretches excessively and eventually breaks. Figure 12-15 shows a typical graph of applied force versus elongation. Up to a point called the proportional limit, Eq. 12-3 is a good approximation for many "The term “law” applied to this relation is not really appropriate, since first of all, it is only an approxi, mation, and second, it refers only to a limited set of phenomena. Most physicists prefer to reserve tk word “law” for those relations that are deeper and more encompassing and precise, such as Newton’s laws of motion or the law of conservation of energy.

318

CHAPTER 12

Static Equilibrium; Elasticity and Fracture

Young's Modulus The amount of elongation of an object, such as the rod shown in Fig. 12-14, depends not only on the force applied to it, but also on the material of which it is made and on its dimensions. That is, the constant k in Eq. 12-3 can be written in terms of these factors. If we compare rods made of the same material but of different lengths and cross-sectional areas, it is found that for the same applied force, the amount of stretch (again assumed small compared to the total length) is proportional to the original length and inversely proportional to the cross-sectional area. That is, the longer the object, the more it elongates for a given force; and the thicker it is, the less it elongates. These findings can be combined with Eq. 12-3 to yield 1F AL = by, (12-4) E A where £ is the original length of the object, A is the cross-sectional area, and A/ is the change in length due to the applied force F. E is a constant of proportionality’ known as the elastic modulus, or Young’s modulus; its value depends only on the material. The value of Young’s modulus for various materials is given in Table 12-1 (the shear modulus and bulk modulus in this Table are discussed later in this Section). Because E is a property only of the material and is independent of the object’s size or (‘hape, Eq. 12-4 is far more useful for practical calculation than Eq. 12-3.

Ultimate strength Proportional limit

%

plastic €€

Force, F

common materials, and the curve is a straight line. Beyond this point, the graph deviates from a straight line, and no simple relationship exists between F and AZ. Nonetheless, up to a point farther along the curve called the elastic limit, the object will return to its original length if the applied force is removed. The region from the rigin to the elastic limit is called the elastic region. If the object is stretched beyond the elastic limit, it enters the plastic region: it does not return to the original length upon removal of the external force, but remains permanently deformed (such as a bent paper clip). The maximum elongation is reached at the breaking point. The maximum force that can be applied without breaking is called the ultimate strength of the material (actually, force per unit area, as we discuss in Section 12--5).

&

s§

.o

Breaking point

Elastic limit

&

)

Elongation,

A{

FIGURE 12-15 Applied force vs. elongation for a typical metal under tension.

"The fact that E is in the denominator, so 1/E is the actual proportionality constant, is merely a convention. When we rewrite Eq. 12-4 to get Eq. 12-5, E is found in the numerator.

TABLE 12-1 Elastic Moduli Material

1

Young’s Modulus,

Shear Modulus,

100 x 10°

40 x 10°

E (N/m?)

G (N/m?)

Bulk Modulus,

B (N/m?)

Solids

Iron, cast

Steel

Brass

200 x 10°

140 x 10° 70 x 10°

100 x 10°

35 x 10°

Aluminum

70 x 10°

25 x 10°

Concrete

20 X 10°

Brick

90 x 10’

80 x 10°

14 x 10°

80 x 10°

Marble |

50 X 10°

70 X 10°

Granite |

45 x 10°

45 x 10°

Nylon Bone (limb) Liquids

5% 10° 15 x 10°

Wood (pine) (parallel to grain) (perpendicular to grain)

10 x 10° 1% 10°

80 X 10°

Water

2.0 x 10°

Mercury Gases™

2.5 x10°

Alcohol (ethyl)

Air, H,, He, CO,

'At normal atmospheric pressure; no variation in temperature during process.

1.0 x 10°

1.01 X 10° SECTION 12-4

319

DEVIJRRPEYM

Tension in piano wire. A 1.60-m-long steel piano wire has a

diameter of 0.20 cm. How great is the tension in the wire if it stretches 0.25cm when tightened?

APPROACH We assume Hooke’s law holds, and use it in the form of Eq. 12—4\ finding E for steel in Table 12-1. ‘ SOLUTION We solve for F in Eq. 12-4 and note that the area of the wire is

A = mr* = (3.14)(0.0010 m)? = 3.14 X 10°m? Then F=E—A=(20X

10“N/m2)
h)? (a) Assume the force is applied at the top edge as shown. F (in a) (b) Assume the force is applied instead at the wheel’s center.

the building’s

face, and

that the building is not anchored in bedrock.

[Hint: Fg in Fig. 12-77

represents the force that

the Earth would exert on the building in the case where the building would just begin to tip.]

wind (), gravity (mg),

60. A 28-kg round table is supported by three legs equal distances apart on the edge. What minimum mass, placed on the table’s edge, will cause the table to overturn?

CHAPTER 12

acts at the midpoint of

FIGURE 12-77 Forces on a building subjected to

FIGURE 12-75 Problem 59.

334

Suppose a 200-km/h wind exerts a force of 950 N/m? over the 76.0-m-wide face (Fig. 12-77). Calculate the torque about the potential pivot point, the rear edge of the building (where Fg acts in Fig. 12-77), and determine whether the building will topple. Assume the total force of the wind

Static Equilibrium; Elasticity and Fracture

and the force Fg on the building due to the Earth if the building were just about to tip. Problem 62.

i

i

68. A uniform flexible steel cable of weight mg between two points at the same elevation Fig. 12-82, where 6 = 56°. Determine the cable (a) at its lowest point, and (b) at the points (c) What is the direction of the tension force in each

63. The center of gravity of a loaded truck depends on how the truck is packed. If it is 4.0m high and 2.4m wide, and its CG is 2.2 m above the r ground, how steep a slope can the truck be parked on without tipping over (Fig. 12-78)? FIGURE 12-78 Problem 63.

is suspended as shown in tension in the of attachment.

4

case?

i

FIGURE 12-82 Problem 68.

. In Fig. 12-79, consider the right-hand (northernmost) section of the Golden Gate Bridge, which has a length dy = 343 m. Assume the cG of this span is halfway between the tower and anchor. Determine Fry and Fry (which act on the northernmost cable) in terms of mg, the weight of the northernmost span, and calculate the tower height 4 needed for equilibrium. Assume the roadway is supported only by the suspension cables, and neglect the mass of the cables and vertical wires. [Hint: Fr3 does not act on this section.]

69. A 20.0-m-long uniform beam weighing 650 N rests on and B, as shown in Fig. 12-83. (a) Find the maximum of a person who can walk to the extreme end D tipping the beam. Find the forces that the walls A and on the beam when the person is standing: (b) at D; point 2.0m to the right of B; (d) 2.0m to the right of -

20.0m

walls A weight without B exert (c) at a A. |

B

FIGURE 12-83

| |

Problem 69.

{ Suspension cable

70. A cube of side £ rests on a rough floor. It is subjected to a steady

horizontal

floor as shown in will either begin to coefficient of static slide rather than

pull

F, exerted

a distance

4 above

the

Fig. 12-84. As F is increased, the block slide, or begin to tip over. Determine the friction ug so that (@) the block begins to tip; (b) the block

begins to tip. [Hint: Where will the FIGURE

12-79

normal force on the block act if it tips?]

Problems 64 and 65.

fiS.

Assume that a single-span suspension bridge such as the Golden Gate Bridge has the symmetrical configuration indicated in Fig. 12-79. Assume that the roadway is uniform over the length of the bridge and that each segment of the suspension cable provides the sole support for the roadway directly below it. The ends of the cable are anchored to the ground only, not to the roadway. What must the ratio of d, to d; be so that the suspension cable exerts no net horizontal force on the towers? Neglect the mass of the cables and the fact that the roadway isn’t precisely horizontal. 66. When a mass of 25 kg is hung from the middle of a fixed straight aluminum wire, the wire sags to make an angle of 12° with the horizontal as shown in Fig. 12-80. (a) Determine the radius

of the wire. (b) Would the Al wire break under these conditions? If so, what other material (see Table 12-2) might work?

FIGURE 12-80 Problem 66.

e

st

67. The forces acting on a 77,000-kg aircraft flying at constant veloc1ty are shown in Fig. 12-81. The engine thrust,

Fr = 5.0 X 10°N, acts on a line 1.6

r

m below the cM. Deter-

mine the drag force Fp, and the distance above the cum that it acts. Assume Fp and Fr are horizontal. (F_ is the “lift” force on the wing.) FIGURE 12-81 Problem 67.

FIGURE 12-84 Problem 70.

71. A 65.0-kg painter is on a uniform 25-kg scaffold supported from above by ropes (Fig. 12-85). There is a 4.0-kg pail of paint to one side, as shown. Can

the painter ends of the end(s) is close to the safely?

walk safely to both scaffold? If not, which dangerous, and how end can he approach FIGURE 12-85 Problem 71.

72. A man doing Fig. 12-86. His force exerted the floor (a) each hand; (b) each foot.

push-ups pauses in the position shown in mass m = 68kg. Determine the normal by ! on on

FIGURE 12-86 Problem 72. 73

42cm

95 cm

1

A 23-kg sphere rests between two smooth planes-as shown in Fig. 12-87. Determine the magnitude of the force acting on the sphere exerted by each plane. FIGURE 12-87 Problem 73.

General Problems

335

74. A 15.0-kg ball is supported from the ceiling by rope A. Rope B pulls downward and to the side on the ball. If the angle of A to the vertical is 22° and if B makes an angle of 53° to the vertical (Fig. 12-88), find the tensions in ropes A and B. FIGURE 12-88 Problem 74. b

75

Parachutists whose chutes have failed to open have been known to survive if they land in deep snow. Assume that a 75-kg parachutist hits the ground with an area of impact of

0.30m?

at a velocity of 55m/s,

and that the ultimate

strength of body tissue is 5 X 10° N/m? Assume that the person is brought to rest in 1.0m of snow. Show that the person may escape serious injury. 76. A steel wire 2.3 mm in diameter stretches by 0.030% when a mass is suspended from it. How large is the mass? . A 2500-kg trailer is attached to a stationary truck at point B, Fig. 12-89. Determine the normal force exerted by the road on the rear tires at A, and the vertical force exerted on the

trailer by the support B.

81. There is a maximum height of a uniform vertical column made of any material that can support itself without buckling, and it is independent of the cross-sectional area (why?).

Calculate this height for (a) steel (density 7.8 X 10° kg/m?), and (b) granite (density 2.7 X 10° kg/m?).

82. A 95,000-kg train locomotive starts across a 280-m-lox._ bridge at time ¢ = 0. The bridge is a uniform beam of mass 23,000kg and the train travels at a constant 80.0 km/h. What are the magnitudes of the vertical forces, Fa(f) and F3(t), on the two end supports, written as a function of time during the train’s passage? 83. A 23.0-kg backpack is suspended midway between two trees by a light cord as in Fig. 12-50. A bear grabs the backpack and pulls vertically downward with a constant force, so that each section of cord makes an angle of 27° below the horizontal. Initially, without the bear pulling, the angle was 15°; the tension in the cord with the bear pulling is double what it was when he was not. Calculate the force the bear is exerting on the backpack. . A uniform beam of mass M and length £ is mounted on a hinge at a wall as shown in Fig. 12-91. It is held in a horizontal position by a wire making an angle 6 as shown. A mass m is placed on the beam a distance x from the wall, and this distance can be varied. Determine, as a func-

tion of x, (a) the tension in the wire and (b) the components of the force exerted by the beam on the hinge. FIGURE 12-91 Problem 84.

FIGURE 12-89

Problem 77.

78. The roof over a 9.0-m X 10.0-m room in a school has a total mass of 13,600 kg. The roof is to be supported by vertical wooden “2 X 4s” (actually about 4.0cm X 9.0 cm) equally spaced along the 10.0-m sides. How many supports are required on each side, and how far apart must they be? Consider only compression, and assume a safety factor of 12. 79. A 25-kg object is being lifted by pulling on the ends of a 1.15-mm-diameter nylon cord that goes over two 3.00-m-high poles that are 4.0m apart, as shown in Fig. 12-90. How high above the floor will the object be when the cord breaks?

8s. Two identical, uniform beams are symmetrically against each other (Fig. 12-92) on a floor with which they have a coefficient of fricA tion wg = 0.50. What is the minimum angle the beams can make with the floor and still not fall?

set up

/\\

FIGURE 12-92

Problem 85.

A\

A

-

86. If 35 kg is the maximum mass m that a person can hold in a hand when the arm is positioned with a 105° angle at the elbow as shown in Fig. 12-93, what is the maximum force Fnax that the biceps muscle exerts on the forearm? Assume the forearm and hand have a total mass of 2.0kg with a CG that is 15 cm from the elbow, and that \ the biceps muscle attaches 5.0 cm from the elbow. FIGURE 12-90

Problem 79.

80. A uniform 6.0-m-long ladder of mass 16.0 kg leans against a smooth wall (so the force exerted by the wall, Fy, is perpendicular to the wall). The ladder makes an angle of 20.0° with the vertical wall, and the ground is rough. Determine the coefficient of static friction at the base of the ladder if the ladder is not to slip when a 76.0-kg person stands three-fourths of the way up the ladder.

336

CHAPTER 12

Static Equilibrium; Elasticity and Fracture

15¢cm e—35cm—-+

FIGURE 12-93 Problem 86.

>

87. (a) Estimate the magnitude of the force Fy the muscles exert on the back to support the upper body when a person bends forward. Use the model shown in Fig. 12-94b. (b) Estimate the magnitude and

direction

of

the

force

Fy

acting on the fifth lumbar vertebra (exerted by the spine below).

Fifth lumbar vertebra ?

12

a rough floor as in Fig. 12-97. The uniform box weighs 250N and has a coefficient of static friction with the floor of 0.60. What minimum force must be exerted on the box to make it slide? What is the maximum height 4 above the floor that this force can be applied without tipping the box over? Note ~ F that as the box tips, the normal force and the friction force will act at the lowest corner.

Erector spinae muscles

h

FIGURE 12-97

_ L#

|

Problem 91.

(a:rms) wr

(a)

FIGURE 12-94

. w = Total weight of person

= 0.46w (trunk)

Problem 87.

(b)

88. One rod of the square frame shown in Fig. 12-95 contains a

turnbuckle which, when

turned, can put the rod under

tension or compression. If the turnbuckle under a compressive force F, determine the forces produced in the other rods. Ignore the mass of the rods and assume the diagonal rods cross each other freely at the center without friction. [Hint: Use the symmetry of the situation.| FIGURE 12-95 Problem 88.

C

puts rod AB

FIGURE 12-98

D

apart, by what distance did this

interatomic spacing have to change in order to produce the normal force required to support the man? [Note: Neighboring atoms repel each other, and this repulsion accounts for the observed normal force.] 90. A home mechanic wants to raise the 280-kg engine out of a car. The plan is to stretch a rope vertically from the engine to a branch of a tree 6.0m above, and back to the bumper (Fig. 12-96). When the mechanic climbs up a stepladder and pulls‘ horizontally on the rope at its midpoint, the engine rises out of the car. (a) How much force must the mechanic exert to hold the engine 0.50m above its normal position? (b) What is the system’s

advantage?

mechanical

L

Problem 92.

89. A steel rod of radius R = 15cm and length £, stands upright on a firm surface. A 65-kg man climbs atop the rod. (a) Determine the percent decrease in the rod’s length. (b) When a metal is compressed, each atom throughout its bulk moves closer to its neighboring atom by exactly the same fractional amount. If iron atoms in steel are

normally 2.0 X 107m

92. You are on a pirate ship and being forced to walk the plank (Fig. 12-98). You are standing at the point marked C. The plank is nailed onto the deck at point A, and rests on the support 0.75 m away from A. The center of mass of the uniform plank is located at point B, Your mass is A 3.0m 65 kg and the mass of i the plank is 45kg. | What is the minimum | 0.75m downward force the ! nails must exert on the plank to hold it in |L place? T 3

[

Ay

¢

91. A 2.0-m-high box with a 1.0-m-square base is moved across

ianmmssamB

93. A uniform sphere of weight mg and radius ry is tethered to a wall by a rope of length £. The rope is tied to the wall a distance & above the contact point of the sphere, as shown in Fig. 12-99. The rope makes an angle 0 with respect to the wall and is not in line with the ball’s center. The coefficient of static friction between the wall and sphere is . (@) Determine the value of the frictional force on the sphere due to the wall. [Hinr: A wise choice of axis will make this calculation easy.] (b) Suppose the sphere is just on the verge of slipping, Derive an expression for u in terms of 4 and 6. FIGURE 12-99

Problem 93.

*94. Use the method member

shown

of

the

of joints to determine the force in each truss

in Fig. 12-100.

State whether each member is in tension

or compression.

B —{“

6.0m

A

C 4.0 m:t—

r

FIGURE 12-96 Problem 90.

FIGURE 12-100

Problem 9%4.

6.0 m——f

12,000 N

General Problems

337

95. A uniform ladder of mass m and length { leans at an angle 0 against a wall, Fig. 12-101. The coefficients of static friction between ladder—ground and ladder—wall are ug and Mw , respectively. The ladder will be on the verge of slipping when both the static friction forces due to the ground and due to the wall take on their maximum values. (a) Show that the ladder will be stable if 8 = 6,,;,, where the minimum angle 6, is given by tan O,

=

*Numerical/Computer *97. (IIT) A metal cylinder has an original diameter of 1.00 cm and a length of 5.00cm. A tension test was performed on the specimen and the data are listed in the Table. (a) Grapl the stress on the specimen vs. the strain. (b) Considerin only the elastic region, find the slope of the best-fit straight line and determine the elastic modulus of the metal.

'2[.LLG(1 - NGP‘W)-

(b) “Leaning ladder problems” are often analyzed under the

seemingly unrealistic assumption that the wall is frictionless

(see Example 12-6). You wish to investigate the magnitude of error introduced by modeling the wall as frictionless, if in reality it is frictional. Using the relation found in part (a), calculate the true value of 6, for a frictional wall, taking ug = uww = 0.40. Then, determine the approximate value of 6p,;, for the “frictionless wall” model by taking ug = 0.40 and uw = 0. Finally, determine the percent deviation of the approximate value of 6,

from its true

value.

FIGURE 12-101

Problem 95.

5

96. In a mountain-climbing technique called the “Tyrolean traverse,” a rope is anchored on both ends (to rocks or strong trees) across a deep chasm, and then a climber traverses the rope while attached by a sling as in Fig. 12-102. This technique generates tremendous forces in the rope and anchors, so a basic understanding of physics is crucial for safety. A typical climbing rope can undergo a tension force of perhaps 29kN before breaking, and a “safety factor” of 10 is usually recommended. The length of rope used in the Tyrolean traverse must allow for some “sag” to remain in the recommended safety range. Consider a 75-kg climber at the center of a Tyrolean traverse, spanning a 25-m chasm. (a) To be within its recommended safety range, what minimum distance x must the rope sag? (b) If the Tyrolean traverse is set up incorrectly so that the rope sags by only one-fourth the distance found in (a), determine the tension in the rope. Will the rope break?

Load (kN)

Elongation (cm)

0 1.50 4.60 8.00 11.00 11.70 11.80 12.00 16.60 20.00 21.50 19.50 18.50

0 0.0005 0.0015 0.0025 0.0035 0.0050 0.0080 0.0200 0.0400 0.1000 0.2800 0.4000 0.4600

*98. (III) Two springs, attached by a rope, are connected as shown in Fig. 12-103. The length AB is 4.0m and AC = BC. The spring constant of each spring is k£ = 20.0N/m. A force F acts downward at C on the rope. Graph 6 as a function of F from 6 =0 to 75° assuming the springs are unstretched at 6 = 0.

FIGURE 12-103

Problem 98.

FIGURE 12-102 Problem 96.

Answers to Exercises A

7.0kg.

F4 also has a component to balance the sideways force Fy.

B: Yes: cos 8 (angle of bar with ground) appears on both sides

and cancels out.

s

Fy

= mag

+ mpg

D: (a).

338

CHAPTER 12

+

Mg

F:

Static friction at the cement floor (= Fcy) is crucial, or else the

G:

(b).

= 560N.

Static Equilibrium; Elasticity and Fracture

ladder would slip. At the top, the ladder can move and adjust, so we wouldn’t need or expect a strong static friction force there.

Underwater divers and sea creatures experience a buoyant force (FB) that closely balances their weight mg. The buoyant force is equal to the weight of the volume of fluid displaced (Archimedes’ principle) and arises because the pressure increases with depth in the fluid. Sea creatures have a density very close to that of water, so their weight very nearly equals the buoyant force. Humans have a density slightly less than water, so they can float. When fluids flow, interesting effects occur because the pressure in the fluid is lower where the fluid velocity is higher (Bernoulli’s principle).

N

&

. v

T

E

%

Q

ml'g'

Fluids Which container has the largest pressure at the bottom? Assume each container holds the same volume of water.

\e——>/

/

& 1

|

The pressures

are

equal.

Two balloons are tied and hang with their nearest edges about 3 cm apart. If you blow between the balloons (not at the balloons, but at the opening between them), what will happen? Nothing. The balloons will move closer together. The balloons will move farther apart.

CONTENTS 13-1

Phases of Matter

13-2

Density and Specific Gravity

13-3

Pressure in Fluids

13-4

Atmospheric Pressure and Gauge Pressure

13-5

Pascal’s Principle

13-6

Measurement of Pressure;

Gauges and the Barometer

13-7

Buoyancy and Archimedes’ Principle

13-8

Fluids in Motion; Flow Rate and the Equation of Continuity

13-9

Bernoulli’s Equation

13-10

Applications of Bernoulli’s Principle: Torricelli, Airplanes, Baseballs, TIA

*#13-11

Viscosity

#13-12 Flow in Tubes: Poiseuille’s Equation, Blood Flow #13-13

Surface Tension and Capillarity

*#13-14 Pumps, and the Heart

339

n previous Chapters we considered objects that were solid and assumed to maintain their shape except for a small amount of elastic deformation. We sometimes treated objects as point particles. Now we are going to shift our attention to materials that are very deformable and can flow. Such “fluids” include liquids and gases. We will examine fluids both at rest (fluid statics) and ' motion (fluid dynamics). '

13—-1

TABLE 13-1

Densities of Substances’ Density,

Substance

p (kg/m’)

Solids

Aluminum

2.70 x 103

Iron and steel

78

Copper Lead Gold

Concrete Granite

89 113 193

23 2T

% 10°

x10° x10° x 10° X10° 0P

Wood (typical)

0.3-0.9 x 10°

Ice (H,0) Bone

0.917 x 103 1.7-20 x 10°

Glass,common

2.4-2.8 X 10°

Liquids

Water (4°C) Blood, plasma Blood, whole Sea water

Mercury

Alcohol, ethyl Gasoline

1.00 1.03

x 10° x 103

1.05 x 10° 1.025 x 10°

136

0.79 0.68

x10°

x 10° x 10°

Gases

Air Helium Carbon dioxide Steam (water, 100°C)

1.29 0.179 1.98 0.598

Densities are given at 0°C and 1 atm

pressure unless otherwise specified.

340

CHAPTER13

Fluids

Phases of Matter

The three common phases, or states, of matter are solid, liquid, and gas. We can distinguish these three phases as follows. A solid maintains a fixed shape and a fixed size; even if a large force is applied to a solid, it does not readily change in shape or volume. A liquid does not maintain a fixed shape—it takes on the shape of its container—but like a solid it is not readily compressible, and its volume can be changed significantly only by a very large force. A gas has neither a fixed shape nor a fixed volume—it will expand to fill its container. For example, when air is pumped into an automobile tire, the air does not all run to the bottom of the tire as a liquid would; it spreads out to fill the whole volume of the tire. Since liquids and gases do not maintain a fixed shape, they both have ability to flow; they are thus often referred to collectively as fluids. The division of matter into three phases is not always simple. How, for example, should butter be classified? Furthermore, a fourth phase of matter can be distinguished, the plasma phase, which occurs only at very high temperatures and consists of ionized atoms (electrons separated from the nuclei). Some scientists believe that so-called colloids (suspensions of tiny particles in a liquid) should also be considered a separate phase of matter. Liquid crystals, which are used in TV and computer screens, calculators, digital watches, and so on, can be considered a phase of matter intermediate between solids and liquids. However, for our present purposes we will mainly be interested in the three ordinary phases of matter.

13-2

Density and Specific Gravity

It is sometimes said that iron is “heavier” than wood. This cannot really be true since a large log clearly weighs more than an iron nail. What we should say is that iron is more dense than wood. The density, p, of a substance (p is the lowercase Greek letter rho) is defined as its mass per unit volume: -

P=y

m

(13-1)

where m is the mass of a sample of the substance and V is its volume. Density is a characteristic property of any pure substance. Objects made of a particular pure substance, such as pure gold, can have any size or mass, but the density will be the same for each. We will sometimes use the concept of density, Eq. 13-1, to write the mass of an object as m

=

pV,

and the weight of an object as mg

=

pVg.

The SI unit for density is kg/m*. Sometimes densities are given in g/cm®. Note that since 1kg/m?® = 1000g/(100cm)® = 10°g/10°cm® = 10 g/cm? then a density given in g/cm® must be multiplied by 1000 to give the result in kg/m? Thus

the density of aluminum is p = 2.70g/cm?, which is equal to 2700 kg/m>. Th densities of a variety of substances are given in Table 13-1. The Table specific temperature and atmospheric pressure because they affect the density of substances (although the effect is slight for liquids and solids). Note that air is roughly 1000 times less dense than water.

POVIJINEES

r

Mass, given volume and density. What is the mass of a

solid iron wrecking ball of radius 18 cm? APPROACH First we use the standard formula V = $77* (see inside rear cover) to obtain the volume of the sphere. Then Eq. 13-1 and Table 13-1 give us the mass m.

SOLUTION The volume of the sphere is

V = $7r’ = $(3.14)(0.18m)*

= 0.024m’.

From Table 13-1, the density of iron is p = 7800kg/m?,

m = pV

so Eq. 13-1 gives

= (7800kg/m?)(0.024m*) = 190kg.

The specific gravity of a substance is defined as the ratio of the density of that substance to the density of water at 4.0°C. Because specific gravity (abbreviated SG) is a ratio, it is a simple number without dimensions or units. The density of water is

1.00 g/cm® = 1.00 X 10°kg/m?

so the specific gravity of any substance will be

equal numerically to its density specified in g/cm?, or 1073 times its density specified in kg/m?. For example (see Table 13-1), the specific gravity of lead is 11.3, and that of alcohol is 0.79. The concepts of density and specific gravity are especially helpful in the study of fluids because we are not always dealing with a fixed volume or mass.

13—3

Pressure in Fluids

Pressure and force are related, but they are not the same thing. Pressure is defined as force per unit area, where the force F is understood to be the magnitude of the force acting perpendicular to the surface area A: pressure

=

P

=

%

(13-2)

Although force is a vector, pressure is a scalar. Pressure has magnitude only. The SI unit of pressure is N/m? This unit has the official name pascal (Pa), in honor of Blaise Pascal (see Section 13-5); that is, 1Pa = 1 N/m? However, for simplicity, we will often use N/m? Other units sometimes used are dynes/cm?, and Ib/in.? (abbreviated “psi”). Several other units for pressure are discussed, along with conversions between them, in Section 13-6 (see also the Table inside the front cover).

DGR

MEEYH

4\ CAUTION Pressure is a scalar, not a vector

Calculating pressure. The two feet of a 60-kg person cover an

area of 500 cm?, (a) Determine the pressure exerted by the two feet on the ground. (b) If the person stands on one foot, what will the pressure be under that foot?

APPROACH Assume the person is at rest. Then the ground pushes up on her with a force equal to her weight mg, and she exerts a force mg on the ground

where her feet (or foot) contact it. Because 1cm? = (102m)’ = 10#m? 500 cm? = 0.050 m>. SOLUTION

then

(a) The pressure on the ground exerted by the two feet is

pms

A

o= T A

2

(60kg)8 T/ ) (0'050 m )

= 12 X 10°N/m%

(b) If the person stands on one foot, the force is still equal to the person’s

weight, but the area will be half as much, so the pressure will be twice as much: 24 X 10°N/m?

Pressure is particularly useful for dealing with fluids. It is an experimental observation that a fluid exerts pressure in any direction. This is well known to swimmers and divers who feel the water pressure on all parts of their bodies. At any depth in a fluid at rest, the pressure is the same in all directions at a given rdepth. To see why, consider a tiny cube of the fluid (Fig. 13-1) which is so small chat we can consider it a point and can ignore the force of gravity on it. The pressure on one side of it must equal the pressure on the opposite side. If this weren'’t true, there would be a net force on the cube and it would start moving. If the fluid is not flowing, then the pressures must be equal.

FIGURE 13-1

Pressure is the same

in every direction in a nonmoving

fluid at a given depth. If this weren’t

true, the fluid would be in motion.

SECTION 13-3

341

FIGURE 13-2

If there were a

component of force parallel to the . . solid surface of the container, the

liquid would move in response to it. For a liquid at rest, F| = 0,

FIGURE 13-3

For a fluid at rest, the force due to fluid pressure always acts perpendicular to any solid surface it touches. If there were a component of the force parallel to the surface, as shown in Fig. 13-2, then according to Newton’s third law the solid surface would exert a force back on the fluid that also would have a component parallel to the surface. Such a component would cause the fluid to flow, in contradiction to o™ assumption that the fluid is at rest. Thus the force due to the pressure in a fluid at rest is always perpendicular to the surface. Let us now calculate quantitatively how the pressure in a liquid of uniform density varies with depth. Consider a point at a depth 4 below the surface of the liquid, as shown in Fig. 13-3 (that is, the surface is a height 4 above this point). The pressure due to the liquid at this depth 4 is due to the weight of the column of liquid above it. Thus the force due to the weight of liquid acting on the area A is ; i : F = mg = (pV)g T pzfihg, where Ah is the volume of the .column of llqulq, pis the density of the liquid (assumed to be constant), and g is the acceleration of gravity. The pressure P due to the weight of liquid is then

Calculating the

pressure at a depth 4 in a liquid.

P=

i = T l|| i &

pAhg A

P = pgh. el

| JL

F

A

v

%

e

i 4

[liquid]

(13-3)

Note that the area A doesn’t affect the pressure at a given depth. The fluid pressure is directly proportional to the density of the liquid and to the depth within the liquid. In general, the pressure at equal depths within a uniform liquid

is the same.

EXERCISE A Return to Chapter-Opening Question 1, page 339, and answer it again now. Try to explain why you may have answered differently the first time.

Equation 13-3 tells us what the pressure is at a depth 4 in the liquid, due to the

liquid itself. But what if there is additional pressure exerted at the surface of the

FIGURE 13-4 Forces on a flat, slablike volume of fluid for

determining the pressure P as a function of height y in the fluid.

I { 5

(P +dP)A d Fo i

t PA

>

liquid, such as the pressure of the atmosphere or a piston pushing down? And what, if the density of the fluid is not constant? Gases are quite compressible and henc their density can vary significantly with depth. Liquids, too, can be compressed, although we can often ignore the variation in density. (One exception is in the depths of the ocean where the great weight of water above significantly compresses the water and increases its density.) To cover these, and other cases, we now treat the general case of determining how the pressure in a fluid varies with depth. As shown in Fig. 13-4, let us determine the pressure at any height y above

some reference point’ (such as the ocean floor or the bottom of a tank or swimming pool). Within this fluid, at the height y, we consider a tiny, flat, slablike volume of the fluid whose area is A and whose (infinitesimal) thickness is dy, as shown. Let the pressure acting upward on its lower surface (at height y) be P. The pressure acting downward on the top surface of our tiny slab (at height y + dy) is designated P + dP. The fluid pressure acting on our slab thus exerts a force equal to PA upward on our slab and a force equal to (P + dP)A downward on it. The only other force acting vertically on the slab is the (infinitesimal) force of gravity dFg, which on our slab of mass dm is

dFg = (dm)g

68

= pgdV

= pgAdy,

where p is the density of the fluid at the height y. Since the fluid is assumed to be at rest, our slab is in equilibrium so the net force on it must be zero. Therefore we have PA — (P + dP)A — pgAdy = 0,

—

which when simplified becomes

Z—;) = ~p8

(13-4

This relation tells us how the pressure within the fluid varies with height above an reference point. The minus sign indicates that the pressure decreases with a. increase in height; or that the pressure increases with depth (reduced height). 342

CHAPTER 13

.

Fluids

"Now we are measuring y positive upwards, the reverse of what we did to get Eq. 13-3 where we

measured the depth (i.e. downward as positive).

If the pressure at a height y; in the fluid is P;, and at height y, it is P, then we can integrate Eq. 13-4 to obtain j

r

Py Py

dP

=

— f

Y2 N

pg dy

Y2

.

(13-5)

= - J pg dy,

BE—-H

N

where we assume p is a function of height y: p = p(y). This is a general relation, and we apply it now to two special cases: (1) pressure in liquids of uniform density and (2) pressure variations in the Earth’s atmosphere. For liquids in which any variation in density can be ignored, p = constant and Eq. 13-5 is readily integrated:

B~ R = —pgln - 9

be the position of the top surface, then P, represents the atmospheric pressure, Py, at

f

h=y,-y

(13-6a)

For the everyday situation of a liquid in an open container—such as water in a glass, a swimming pool, a lake, or the ocean—there is a free surface at the top exposed to the atmosphere. It is convenient to measure distances from this top surface. That is, we let 4 be the depth in the liquid where & = y, — y, as shown in Fig. 13-5.If we let y,

.

P,=P,

B l FIGURE 13-5

72 Y1 |

ok

Pressure at a depth

& = (y» — y1) inaliquid of density p

the top surface. Then, from Eq. 13—6a, the pressure P (= P,) at adepth 4 in the fluid is ~ is P = Py + pgh, where Py is the P

=

P, + pgh.

[k is depth in liquid]

(13-6b)

Note that Eq. 13-6b is simply the liquid pressure (Eq. 13-3) plus the pressure F, due to the atmosphere above. Pressure at a faucet. The surface of the water in a storage tank is 30 m above a water faucet in the kitchen of a house, Fig. 13-6. Calculate the difference in water pressure between the faucet and the surface of the water in the tank. APPROACH Water is practically incompressible, so p is constant even for h = 30 m when used in Eq. 13-6b. Only /4 matters; we can ignore the “route” of Athe pipe and its bends. SOLUTION We assume the atmospheric pressure at the surface of the water in the storage tank is the same as at the faucet. So, the water pressure difference between the faucet and the surface of the water in the tank is

AP = pgh

= (1.0 X 10°kg/m*)(9.8 m/s*)(30m)

external pressure at the liquid’s top surface.

~ FIGURE 13-6

Example 13-3.

FIGURE 13-7 T 1.0m : 1[

Example 13-4. :

= 2.9 X 10°N/m?

NOTE The height 4 is sometimes called the pressure head. In this Example, the head of water is 30 m at the faucet. The very different diameters of the tank and faucet don’t affect the result—only height does.

Force on aquarium window. Calculate the force due to

water pressure exerted on a 1.0m X 3.0m aquarium viewing window whose top

edge is 1.0m below the water surface, Fig. 13-7. APPROACH At a depth 4, the pressure due to the water is given by Eq. 13-6b. Divide the window up into thin horizontal strips of length £ = 3.0m and thickness dy, as shown in Fig. 13-7. We choose a coordinate system with y = 0 at the surface of the water and y is positive downward. (With this choice, the minus sign in Eq. 13-6a becomes plus, or we use Eq. 13—-6b with y = h.) The force due to water pressure on each strip is dF = PdA = pgyl dy. SOLUTION The total force on the window is given by the integral: ¥,=20m

LOm] J—

y

window

B

% dy

J y=1.0m pgyldy = 3pgl(y; — y?)

= 3(1000 kg/m*)(9.8 m/s*)(3.0m)[(2.0m)? — (1.0m)?] = 44,000 N.

ANOTE

To check our answer, we can do an estimate: multiply the area of the

window (3.0m?) times the pressure at the middle of the window

(h = 1.5m)

using Eq. 13-3, P = pgh = (1000 kg/m*)(9.8 m/s?)(1.5m) ~ 1.5 X 10*N/m?% So

F = PA~ (15 x 10*N/m?)(3.0m)(1.0m) ~ 4.5 X 10*N. Good!

EXERCISE B A dam holds back a lake that is 85 m deep at the dam. If the lake is 20 km

long, how much thicker should the dam be than if the lake were smaller, only 1.0 km long?

SECTION 13-3

343

Now let us apply Eq. 13-4 or 13-5 to gases. The density of gases is normally quite small, so the difference in pressure at different heights can usually be ignored if y, — y; is not large (which is why, in Example 13-3, we could ignore the difference in air pressure between the faucet and the top of the storage tank). Indeed, for most ordinary containers of gas, we can assume that the pressure i\. the same throughout. However, if y, — y; is very large, we cannot make this assumption. An interesting example is the air of Earth’s atmosphere, whose pressure at sea level is about 1.013 X 10° N/m? and decreases slowly with altitude.

Elevation effect on atmospheric pressure. (2) Determine

the variation in pressure in the Earth’s atmosphere as a function of height y above sea level, assuming g is constant and that the density of the air is proportional to the pressure. (This last assumption is not terribly accurate, in part because temperature and other weather effects are important.) (b) At what elevation is the air pressure equal to half the pressure at sea level? APPROACH We start with Eq. 13-4 and integrate it from the surface of the Earth where y = 0and P = Py, up to height y at pressure P. In (b) we choose P =1 P,. SOLUTION (@) We are assuming that p is proportional to P, so we can write

B

B

Po

B

where P, = 1.013 X 10°N/m? is atmospheric pressure at sea level and po = 1.29kg/m? is the density of air at sea level at 0°C (Table 13-1). From the differential change in pressure with height, Eq. 13-4, we have dP

SO

Po

AR

Il

5~

ap _ _m PR

g dy.

We integrate this from y = 0 (Earth’s surface) and where the pressure is P:

[ p P

ln£

P

since InP —InPy P

=

=

Rl

P = F,,

to the height yfl

Rt

~&gy

B

= In(P/PR,). Then Poe_

(POX/PO).V.

So, based on our assumptions, we find that the air pressure in our atmosphere decreases approximately exponentially with height. NOTE The atmosphere does not have a distinct top surface, so there is no natural point from which to measure depth in the atmosphere, as we can do for a liquid.

(b) The constant (pog/P,) has the value

pog _ (1.29kg/m’)(9.80m/s’) = P, (1.013 X 10°N/m?) Then, when we set %

=

=

1.25 X 10~*m™.

P = 1 P, in our expression derived in (a), we obtain ¢ (1.25x10™*m™)y

or, taking natural logarithms of both sides, so (recall In}

Ini = (=125 =

X 10*m™)y

—In2, Appendix A-7, Eq. ii)

y = (In2.00)/(125 X 10*m™)

= 5550 m.

Thus, at an elevation of about 5500 m (about 18,000 ft), atmospheric pressure drops to half what it is at sea level. It is not surprising that mountain climber: often use oxygen tanks at very high altitudes. 344

CHAPTER13

Fluids

13-4 Atmospheric Pressure and Gauge Pressure

™ \tmospheric Pressure

The pressure of the air at a given place on Earth varies slightly according to the weather. At sea level, the pressure of the atmosphere on average is 1.013 X 10°N/m?

(or 14.71b/in.%). This value lets us define a commonly used unit of pressure, the

atmosphere (abbreviated atm): latm

=

1.013 X 10°N/m?

=

101.3 kPa.

Another unit of pressure sometimes used (in meteorology and on weather maps) is the bar, which is defined as

1bar

= 1.000 X 10°N/m?2

Thus standard atmospheric pressure is slightly more than 1 bar. The pressure due to the weight of the atmosphere is exerted on all objects immersed in this great sea of air, including our bodies. How does a human body withstand the enormous pressure on its surface? The answer is that living cells maintain an internal pressure that closely equals the external pressure, just as the pressure inside a balloon closely matches the outside pressure of the atmosphere. An automobile tire, because of its rigidity, can maintain internal pressures much greater than the external pressure. CONCEPTUAL EXAMPLE 13-6 | Finger holds water in a straw. You insert a straw of length £ into a tall glass of water. You place your finger over the top of the straw, capturing some air above the water but preventing any additional ‘air from getting in or out, and then you lift the straw from the water. You find that the straw retains most of the water (see Fig. 13-8a). Does the air in the space between your finger and the top of the water have a pressure P that is greater than, equal to, or less than the atmospheric pressure P, outside the straw? RESPONSE Consider the forces on the column of water (Fig. 13-8b). Atmospheric pressure outside the straw pushes upward on the water at the bottom of the straw, gravity pulls the water downward, and the air pressure inside the top of the straw pushes downward on the water. Since the water is in

equilibrium, the upward force due to atmospheric pressure Py must balance the

two downward forces. The only way this is possible is for the air pressure inside the straw to be less than the atmosphere pressure outside the straw. (When you initially remove the straw from the glass of water, a little water may leave the ‘bottom of the straw, thus increasing the volume of trapped air and reducing its density and pressure.)

@ PHYSICS APPLIED Pressure on living cells

p—9

,

g

‘(

i

,L ‘ ‘ l

-

lPA

{

~

§ R

(@) FIGURE 13-8

’

iz = pgAh

. PuA 0

) Example 13-6.

Gauge Pressure Itis importént to note that tire gauges, and most other pressure gauges, register the pressure above and beyond atmospheric pressure. This is called gauge pressure. Thus, to get the absolute pressure, P, we must add the atmospheric pressure, P, to the gauge pressure, Pg: P

=

PO

+

PG‘

rIf a tire gauge registers 220kPa, the absolute pressure within the -20kPa + 101 kPa = 321 kPa, equivalent to about 3.2atm (22atm pressure).

SECTION 13-4

tire is gauge

Atmospheric Pressure and Gauge Pressure

345

13-5

Pascal’s Principle

The Earth’s atmosphere exerts a pressure on all objects with which it is in contact, including other fluids. External pressure acting on a fluid is transmitted throughout thq\ fluid. For instance, according to Eq. 13-3, the pressure due to the water at a depth «

100m below the surface of a lake is P = pgh = (1000 kg/m?)(9.8 m/s?)(100 m) =

9.8 x 10°N/m?, or 9.7 atm. However, the total pressure at this point is due to the pressure of water plus the pressure of the air above it. Hence the total pressure (if the lake is near sea level) is 9.7 atm + 1.0 atm = 10.7 atm. This is just one example of a general principle attributed to the French philosopher and scientist Blaise Pascal

Master

(1623-1662). Pascal’s principle states that if an external pressure is applied to a

D cylinder

cylinder Disk (b)

! pads

confined fluid, the pressure at every point within the fluid increases by that amount. A number of practical devices make use of Pascal’s principle. One example is the hydraulic lift, illustrated in Fig. 13-9a, in which a small input force is used to exert a large output force by making the area of the output piston larger than the area of the input piston. To see how this works, we assume the input and output pistons are at the same height (at least approximately). Then the external input force F;,, by Pascal’s principle, increases the pressure equally throughout. Therefore, at the same level (see Fig. 13-9a),

attached to wheel

FIGURE 13-9 Applications of Pascal’s principle: (a) hydraulic lift;

Py

@

PHYSICS

appLIED

Fouw Aout

or

_

Fa Am

Fou _ Aou

Hydraulic lift

APPLIED Hydraulic brakes

P

‘vyheri tl}e input quantities are .represented by the §ubscript “in” and the output by out.” Since P = F/A, we write the above equality as

(b) hydraulic brakes in a car.

@erHysics

=

F,

A

The quantity F,/F;, is called the mechanical advantage of the hydraulic lift, and it is equal to the ratio of the areas. For example, if the area of the output piston fl 20 times that of the input cylinder, the force is multiplied by a factor of 20. Thus .. force of 200 Ib could lift a 4000-1b car. Figure 13-9b illustrates the brake system of a car. When the driver presses the ~ brake pedal, the pressure in the master cylinder increases. This pressure increase occurs throughout the brake fluid, thus pushing the brake pads against the disk attached to the car’s wheel.

13—6

Measurement of Pressure;

Gauges and the Barometer

Many devices have been invented to measure pressure, some of which are shown in Fig. 13-10. The simplest is the open-tube manometer (Fig 13-10a) which is a U-shaped tube partially filled with a liquid, usually mercury or water. The pressure P being measured is related to the difference in height A4 of the two levels of the liquid by the relation P = Py + pg Ah, where P, is atmospheric pressure (acting on the top of the liquid in the left-hand tube), and p is the density of the liquid. Note that the quantity pg A is the gauge pressure—the amount by which P exceeds atmospheric pressure Fy. If the liquid in the left-hand column were lower than that in the right-hand column, P would have to be less than atmospheric pressure (and Ak would be negative). Instead of calculating the product pg Ak, sometimes only the change in height Ah is specified. In fact, pressures are sometimes specified as so many “millimeters

of mercury” (mm-Hg) or “mm of water” (mm-H,0). The unit mm-Hg is equivalent to a pressure of 133 N/m?, since pg Ak for 1 mm = 1.0 X 10 m of mercury give™

pg Ah = (13.6 X 10° kg/m*)(9.80 m/s?)(1.00 X 10 m) = 1.33 X 10°N/m*

346

CHAPTER 13

Fluids

The unit mm-Hg is also called the torr in honor of Evangelista Torricelli (1608-1647), a student of Galileo’s who invented the barometer (see next page).

Scale reading,

gauge pressure Atmospheric pressure

T

Air pressure Flexiblel

@

chamgor

5

-

y

\

/

% |

I T

1t

S

(Pressure being measured)

- Spring

¢

(b) Aneroid gauge (used mainly for air pressure and then called an aneroid barometer)

(a) Open-tube manometer

Pressure of /; air in tire &

|

(c) Tire gauge

FIGURE 13-10 Pressure gauges: (a) open-tube manometer, (b) aneroid gauge, and (c) common tire-pressure gauge.

Conversion factors among the various units of pressure (an incredible nuisance!) are given in Table 13-2. It is important that only N/m* = Pa, the proper SI unit, be used in calculations involving other quantities specified in SI units. TABLE 13-{2

PROBLEM SOLVING In calculations, use SI units:

1Pa = 1N/m?

Conversion Factors Between Different Units of Pressure

In Terms of 1Pa = 1 N/m?

1atm = 1.013 X 105 N/m?

1 atm in Different Units

1atm = 1.013 X 10° N/m?

= 1.013 X 10°Pa = 101.3kPa

f

1bar = 1.000 X 10° N/m? 1dyne/cm? = 0.1 N/m? 11b/in2 = 6.90 X 103 N/m? 11b/ft? = 47.9 N/m?

1em-Hg = 1.33 X 10°N/m? 1mm-Hg = 133 N/m? 1 torr = 133 N/m?

1 mm-H,0 (4°C) = 9.80 N/m?

1atm = 1.013 bar

1atm = 1.013 X 10° dyne/cm? 1atm = 14.71b/in.?

latm = 2.12 X 10% 1b/ft? latm = 76.0 cm-Hg 1 atm = 760 mm-Hg 1 atm = 760 torr

1atm = 1.03 X 10*mm-H,0 (4°C)

Another type of pressure gauge is the aneroid gauge (Fig. 13-10b) in which the pointer 1s linked to the flexible ends of an evacuated thin metal chamber. In an electronic gauge the pressure may be applied to a thin metal diaphragm whose resulting dlstortlon is translated into an electrical signal by a transducer. A common tlre gauge is shown in Fig. 13-10c. Atmospheric pressure can be measured by a modified kind of mercury manometer with one end closed, called a mercury barometer (Fig. 13-11). The glass tube 14 completely filled with mercury and then inverted into the bowl of mercury. If the tube is long enough the level of the mercury will drop, leaving a vacuum at the top of the tube, since atmospheric pressure can support a column of mercury only about 76 cm high (exactly 76.0 cm at standard atmospheric pressure). That is, a column of mercury 76 cm high exerts the same pressure as the atmosphere: P = pg Ah r.

FIGURE 13-11

A mercury

barometer, invented by Torricelli, is

shown here when the air pressure is standard atmospheric, 76.0 cm-Hg. P=0\:‘

76.0 cm

P=1atm

= (13.6 x 10°kg/m®)(9.80m/s?)(0.760m) = 1.013 X 10°N/m? = 1.00 atm.

"This calculation confirms the entry in Table 13-2,

1atm = 76.0 cm-Hg.

SECTION 13-6

Measurement of Pressure; Gauges and the Barometer

347

A calculation similar to what we just did will show that atmospheric pressure can maintain a column of water 10.3 m high in a tube whose top is under vacuum (Fig. 13-12). No matter how good a vacuum pump is, water cannot be made to rise more than about 10 m using normal atmospheric pressure. To pump water out of deep mine shafts with a vacuum pump requires multiple stages for depths greater than 10 ™ Galileo studied this problem, and his student Torricelli was the first to explain it. The

point is that a pump does not really suck water up a tube—it merely reduces the pressure at the top of the tube. Atmospheric air pressure pushes the water up the tube if the top end is at low pressure (under a vacuum), just as it is air pressure that pushes (or maintains) the mercury 76 cm high in a barometer. [Force pumps (Section 13-14) that push up from the bottom can exert higher pressure to push water more than 10 m high.] CONCEPTUAL EXAMPLE 13-7 | Suction. A student suggests suction-cup shoes for Space Shuttle astronauts working on the exterior of a spacecraft. Having just studied this Chapter, you gently remind him of the fallacy of this plan. What is it? RESPONSE Suction cups work by pushing out the air underneath the cup. What holds the suction cup in place is the air pressure outside it. (This can be a substantial force when on Earth. For example, a 10-cm-diameter suction cup has an area of

7.9 X 10*m?. The force of the atmosphere on it is (7.9 X 107°m?)(1.0 X 10°N/m?) ~ FIGURE 13-12 A water barometer: a full tube of water is inserted into a tub of water, keeping the tube’s spigot at the top closed. When the bottom end of the tube is unplugged, some water flows out of the tube into the tub, leaving a vacuum between the water’s upper

surface and the spigot. Why? Because air pressure can not support a column of water more than 10 m high.

FIGURE 13-13 Determination of the buoyant force. i

-

800 N, about 1801bs!) But in outer space, there is no air pressure to push the suction cup onto the spacecraft.

We sometimes mistakenly think of suction as something we actively do. For example, we intuitively think that we pull the soda up through a straw. Instead, what we do is lower the pressure at the top of the straw, and the atmosphere pushes the soda up the straw.

13—7

Buoyancy and Archimedes’ Principle

Objects submerged in a fluid appear to weigh less than they do when outside th«™ fluid. For example, a large rock that you would have difficulty lifting off the. ground can often be easily lifted from the bottom of a stream. When the rock breaks through the surface of the water, it suddenly seems to be much heavier. Many objects, such as wood, float on the surface of water. These are two examples of buoyancy. In each example, the force of gravity is acting downward. But in addition, an upward buoyant force is exerted by the liquid. The buoyant force on fish and underwater divers (as in the Chapter-Opening photo) almost exactly balances the force of gravity downward, and allows them to “hover” in equilibrium. The buoyant force occurs because the pressure in a fluid increases with depth. Thus the upward pressure on the bottom surface of a submerged object is greater than the downward pressure on its top surface. To see this effect, consider a cylinder of height Ak whose top and bottom ends have an area A and which is completely submerged in a fluid of density pg, as shown in Fig. 13—13. The fluid exerts a pressure P, = ppgh, at the top surface of the cylinder (Eq. 13-3). The force due to this pressure on top of the cylinderis F; = P, A = ppgh; A, and it is directed downward. Similarly, the fluid exerts an upward force on the bottom of the cylinder equal to F, = P,A = ppgh, A. The net force on the cylinder exerted by the fluid pressure, which is the buoyant force, Fy, acts upward and has the magnitude Fg

=

FKHL-F

=

PFgA(hz

=

pr8A Ah

= =

peVg Mg§,

- h1)

where V = A Ah is the volume of the cylinder, the product pgV is the mass of the fluid displaced, and ppVg = mgpg is the weight of fluid which takes up e volume equal to the volume of the cylinder. Thus the buoyant force on the cylinde is equal to the weight of fluid displaced by the cylinder. 348

CHAPTER13

Fluids

This result is valid no matter what the shape of the object. Its discovery is credited to Archimedes (287?-212 B.C.), and it is called Archimedes’ principle: the buoyant force on an object immersed in a fluid is equal to the weight of the fluid displaced by that object. By “fluid displaced,” we mean a volume of fluid equal to the submerged volume of the object (or that part of the object that is submerged). If the object is placed in a glass or tub initially filled to the brim with water, the water that flows over the top represents the water displaced by the object. We can derive Archimedes’ principle in general by the following simple but clegant argument. The irregularly shaped object D shown in Fig. 13-14a is actedon by the force of gravity (its weight, mg, downward) and the buoyant force, Fy, upward. We wish to determine Fy. To do so, we next consider a body (D’ in Fig. 13-14b), this time made of the fluid itself, with the same shape and size

as the original object, and located at the same depth. You might think of this

s S

K (a)
. If the cube floats so that it is 72% in the water and 28% in the oil, what is the mass of the

cube and what is the buoyant force on the cube? 39. (IT) How many helium-filled balloons would it take to lift a person? Assume the person has a mass of 75kg and that each helium-filled balloon is spherical with a diameter ot 33cm.

40. (IT) A scuba tank, when fully submerged, displaces 15.7 L of seawater. The

tank itself has a mass

of 14.0kg

and, when

“full,” contains 3.00 kg of air. Assuming only a weight and buoyant force act, determine the net force (magnitude and direction) on the fully submerged tank at the beginning of a dive (when it is full of air) and at the end of a dive (when it no longer contains any air). 41. (III) If an object floats in water, its density can be determined by tying a sinker to it so that both the object and the sinker are submerged. Show that the specific gravity is given

by w/(w; — w;,), where w is the weight of the object alone

in air, wy is the apparent weight when a sinker is tied to it and the sinker only is submerged, and wj, is the apparent weight when both the object and the sinker are submerged. 42. (III) A 3.25-kg piece of wood (SG = 0.50) floats on water. What minimum mass of lead, hung from the wood by a string, will cause it to sink?

13-8 to 13-10

Fluid Flow, Bernoulli’s Equation

43. (I) A 15-cm-radius air duct is used to replenish the air of a room 82m X 5.0m X 3.5m every 12 min. How fast does the air flow in the duct? 44. (I) Using the data of Example 13-13, calculate the average speed of blood flow in the major arteries of the body which have a total cross-sectional area of about 2.0 cm?. 45. (I) How fast does water flow from a hole at the bottom of a very wide, 5.3-m-deep storage tank filled with water? Ignore viscosity. 46. (II) A fish tank has dimensions 36 cm wide by 1.0 m long by 0.60m high. If the filter should process all the water in the tank once every 4.0 h, what should the flow speed be in the 3.0-cm-diameter input tube for the filter? . (II) What gauge pressure in the water mains is necessary if a firehose is to spray water to a height of 18 m? (IT) A $-in. (inside) diameter garden hose is used to fill a round swimming pool 6.1 m in diameter. How long will it take to fill the pool to a depth of 1.2m if water flows from

the hose ata speed of 0.40 m/s?

49. (IT) A 180-km/h wind blowing over the flat roof of a house causes the roof to lift off the house. If the house is 62m X 12.4m in size, estimate the weight of the roof. Assume the roof is not nailed down. 50. (II) A 6.0-cm-diameter horizontal pipe gradually narrows to 4.5 cm. When water flows through this pipe at a certain rate, the gaugé pressure in these two sections is 32.0kPa and 24.0 kPa, respectively. What is the volume rate of flow? S1. (II) Estimate the air pressure inside a category 5 hurricane, where the wind speed is 300 km/h (Fig. 13-53).

52. (II) What is the lift (in newtons) due to Bernoulli’s principle on a wing of area 88 m? if the air passes over the top and bottom

surfaces at speeds of 280 m/s and 150 m/s, respectively?

. (II) Show that the power needed to drive a fluid through a pipe with uniform cross-section is equal to the volume rate of flow, Q, times the pressure difference, P

— P,.

54. (IT) Water at a gauge pressure of 3.8 atm at street level flows into an office building at a speed of 0.68 m/s through a pipe 5.0cm in diameter. The pipe tapers down to 2.8 cm in diameter by the top

floor,

18 m

above

(Fig. 13-54), where the faucet has been left open. Calculate the flow velocity and the gauge pressure in the pipe on the top floor. Assume no branch pipes and ignore viscosity.

FIGURE 13-54 Problem 54.

3S. (II) In Fig. 13-55, take into account the speed of the top surface of the tank and show that the speed of fluid leaving the opening at the bottom is

"

B

2gh

(1 - A3/Ad)

where h =y, — y;, and A; and A, are the areas of the opening and of the top surface, respectively. Assume Ay 0 at

because sin(—90°) = —1. Thus our solution for this case is

Acos|\wt

"

— 5

= Asinot,

cos(6 — m/2) = sin6.

The solution in this case is a pure sine

Many other situations are possible, such as that shown in Fig. 14-7. The

constant ¢ is called the phase angle, and it tells us how long after (or before) t = 0 the peak at x = A is reached. Notice that the value of ¢ does not affect the shape of the x(f) curve, but only affects the displacement at some arbitrary time,

t = 0. Simple harmonic motion is thus always sinusoidal. Indeed, simple harmonic

motion is defined as motion that is purely sinusoidal. Since our oscillating mass repeats its motion after a time equal to its period 7, it must be at the. same position and moving in the same direction at t = T as it was at ¢ = 0. Since a sine or cosine function repeats itself after every 27 radians, then from Eq. 14-4, we must have | oT = 2.

r

or

;

:

/\

:

%

eV

FIGURE 14-6

t

Special case of SHM

~ Where the massm starts,at 7 = 0, at

the equilibrium position x = 0 and

hasinitial velocity toward positive

Ralnea it

(o = Bay F=,

FIGURE 14-7

A plot of

~ * = 4 cos{a¥ + &) wheng )

(14-6a)

x

=

Acos(2wft

+ ¢).

(14-6b)

SECTION 14-2

Simple Harmonic Motion

373

Because

w = 27f = Vk/m

(Eq.14-5), then

1 [k m\m

f

(14-7a)

T = 2, /%

(14-7b,“

Note that the frequency and period do not depend on the amplitude. Changing the amplitude of a simple harmonic oscillator does not affect its frequency. Equation 14-7a tells us that the greater the mass, the lower the frequency; and the stiffer the spring, the higher the frequency. This makes sense since a greater mass means more inertia and therefore a slower response (or acceleration); and larger k means greater force and therefore quicker response. The frequency f (Eq. 14-7a) at which a SHO oscillates naturally is called its natural frequency (to distinguish it from a frequency at which it might be forced to oscillate by an outside force, as discussed in Section 14-8). The simple harmonic oscillator is important in physics because whenever we have a net restoring force proportional to the displacement (F = —kx), which is at least a good approximation for a variety of systems, then the motion is simple harmonic—that is, sinusoidal.

@pHYsics

APPLIED Car springs

DGUIZRRMEEYE

Car springs again. Determine the period and frequency of

the car in Example 14-1a after hitting a bump. Assume the shock absorbers are poor, so the car really oscillates up and down.

APPROACH We put m = 1400kg and k = 6.5 X 10*N/m from Example 14-1a into Eqgs. 14-7. SOLUTION From Eq. 14-7b,

T—2\/E—2,/m~——092s "™Vk T V65 x 10°'N/m N

or slightly less than a second. The frequency

f = 1/T = 1.09 Hz.

S

EXERCISE E By how much should the mass on the end of a spring be changed to halve the frequency of its oscillations? (a) No change; (b) doubled; (c) quadrupled; (d) halved; (e) quartered. FIGURE 14-8 Displacement, x, velocity, dx/dt, and acceleration,

d*x/dt?, of a simple harmonic oscillator when

¢ = 0.

0w

~

~

g = d*x

(a) > &

UmaxT

g

G}g

: !

0

S

g

t

*é

_A

™

+

0

2

8

> b/2m.) (b) Show that the fractional energy lost per period is

AE _ =

E

2mb _ 2m

—_———=

mag

]

Q

where wy = Vk/m and Q = mawy/b is called the quality factor or Q value of the system. A larger Q value means the system can undergo oscillations for a longer time.

62. (IIT) A glider either end of same spring (a) Determine

on an air track is connected by springs the track (Fig. 14-39). Both springs have constant, k, and the glider has mass the frequency of the oscillation, assuming

damping, if k =125N/m

and

M =215g.

to the M. no

(b) It is™y

observed that after 55 oscillations, the amplitude of the oscillation has dropped to one-half of its initial value. Estimate the value of vy, using Eq. 14-16. (c¢) How long does it take the amplitude to decrease to one-quarter of its initial value? FIGURE 14-39 Problem 62.

14-8 Forced Oscillations; Resonance 63. (IT) (a) For a forced oscillation at resonance (w = wy), what is the value of the phase angle ¢, in Eq. 14-22? (b) What, then, is the displacement at a time when the driving force Fext is a maximum, and at a time when Foy = 0? (c) What is the phase difference (in degrees) between the driving force and the displacement in this case? . (I) Differentiate Eq. 14-23 to show that the resonant amplitude peaks at 7 R

b2

= 2m? (II) An 1150 kg automobile has springs with £ = 16,000 N/m. One of the tires is not properly balanced; it has a little extra mass on one side compared to the other, causing the car to shake at certain speeds. If the tire radius is 42cm, at what speed will the wheel shake most? *66. (IT) Construct an accurate resonance curve, from o = 0 to o = 2wg, for Q = 6.0. *07. (IT) The amplitude of a driven harmonic oscillator reaches a value of 23.7Fy/k at a resonant frequency of 382 Hz. What is the Q value of this system? 68. (IIT) By direct substitution, show that Eq. 14-22, with Egs. 14-23 and 14-24, is a solution of the equation of motion (Eq. 14-21) for the forced oscillator. [Hint: To find sin ¢ and cos ¢ from tan ¢, draw a right triangle.] *69. (ITI) Consider a simple pendulum (point mass bob) 0.50m long with a Q of 350. (@) How long does it take for the amplitude (assumed small) to decrease by two-thirds? (b) If the amplitude is 2.0 cm and the bob has mass 0.27 kg, what is the initial energy loss rate of the pendulum in watts? (c) If we are to stimulate resonance with a sinusoidal driving force, how close must the driving frequency be to the natural frequency of the pendulum (give Af = f — fy)? 65

-

14-7

56. (IT) A 0.835-kg block oscillates on the end of a spring whose

>

| General Problems 70. A 62-kg person jumps from a window to a fire net 20.0m below, which stretches the net 1.1 m. Assume that the net behaves like a simple spring. (a) Calculate how much it would stretch if the same person were lying in it. (b) How much would it stretch if the person jumped from 38 m? 71. An energy-absorbing car bumper has a spring constant of 430 kN/m. Find the maximum compression of the bumper if the car, with mass 1300 kg, collides with a wall at a speed of 2.0m/s (approximately 5 mi/h).

392

CHAPTER14

Oscillations

72. The length of a simple pendulum is 0.63 m, the pendulum bob has a mass of 295 g, and it is released at an angle of 15° to the vertical. (@) With what frequency does it oscillate? () What is the pendulum bob’s speed when it passes through the lowest point of the swing? Assume SHM. (c¢) What is the total energy stored in this oscillation assuming no losses? 73. A simple pendulum oscillates with frequency f. What is its‘fl frequency if the entire pendulum accelerates at 0.50g (a) upward, and (b) downward?

74. A 0.650-kg mass oscillates according to the equation x = 0.25sin(5.50¢) where x is in meters and £ is in seconds. Determine (a) the amplitude, (b) the frequency, (c) the period, (d) the total energy, and (e) the kinetic energy and potential energy when x is 15 cm. (a) A crane has hoisted a 1350-kg car at the junkyard. The crane’s steel cable is 20.0m long and has a diameter of

(g

6.4 mm. If the car starts bouncing at the end of the cable, what

is the period of the bouncing? [Hint: Refer to Table 12-1.] (b) What amplitude of bouncing will likely cause the cable to snap? (See Table 12-2, and assume Hooke’s law holds all the way up to the breaking point.) 76. An oxygen atom at a particular site within a DNA molecule can be made to execute simple harmonic motion when illuminated by infrared light. The oxygen atom is bound with a spring-like chemical bond to a phosphorus atom, which is rigidly attached to the DNA backbone. The oscillation of the oxygen atom occurs with frequency f = 3.7 X 103 Hz. If the oxygen atom at this site is chemically replaced with a sulfur atom, the spring constant of the bond is unchanged (sulfur is just below oxygen in the Periodic Table). Predict the frequency for a DNA molecule after the sulfur substitution. 7. A “seconds” pendulum has a period of exactly 2.000s. That is, each one-way swing takes 1.000s. What is the length of a seconds pendulum in Austin, Texas, where g = 9.793m/s*?

If the pendulum is moved to Paris, where g = 9.809 m/s?,

by how many millimeters must we lengthen the pendulum? What is the length of a seconds pendulum on the Moon,

where g = 1.62m/s??

78. A 320-kg wooden raft floats on a lake. When a 75-kg man stands

on

the raft, it sinks

3.5cm

deeper

into

the water.

When he steps off, the raft oscillates for a while. (a) What is the frequency of oscillation? (b) What is the total energy of oscillation (ignoring damping)? 79. At what displacement from equilibrium a SHO half the maximum value?

is the speed

of

80. A diving board oscillates with simple harmonic motion of frequency 2.5 cycles per second. What is the maximum amplitude with which the end of the board can oscillate in order that a pebble placed there (Fig. 14-40) does not lose contact with the board during the oscillation?

83. A 1.60-kg table is supported on four springs. A 0.80-kg chunk of modeling clay is held above the table and dropped so that it hits the table with a speed of 1.65 m/s (Fig. 14-42). The clay makes an inelastic collision with the table, and the table and clay oscillate up and down. After a long time the table comes to rest l!“ | 6.0 cm below its original position. (a) What is the effective spring constant of all four springs taken together? (b) With what maximum amplitude does the platform oscillate?

@

FIGURE 14-42 Problem 83. . In some

diatomic molecules, the force each atom exerts on

the other can be approximated by

F = —C/r? + D/r’,

where

small displacements,

where r is the atomic separation and C and D are positive constants. () Graph Fvs.r from r = 0.8D/C to r = 4D/C. (b) Show that equilibrium occurs at » = ry = D/C. (c) Let Ar=r —ry be a small displacement from equilibrium, Ar

A series of compressions and expansions propagate along the spring. The compressions are those areas where the coils are momentarily close together. Expansions (sometimes called rarefactions) are regions where the coils are momentarily far apart. FIGURE 15-5 Production of a Compressions and expansions correspond to the crests and troughs of a transverse sound wave, which is longitudinal, wave. shown at two moments in time about An important example of a longitudinal wave is a sound wave in air. A vibrating a half period (3T apart. drumhead, for instance, alternately compresses and rarefies the air in contact with it, producing a longitudinal wave that travels outward in the air, as shown in Fig. 15-5. As in the case of transverse waves, each section of the medium in which a longitudinal wave passes oscillates over a very small distance, whereas the wave itself can travel large distances. Wavelength, frequency, and wave velocity all have meaning for a longitudinal wave. The wavelength is the distance between Compression Expansion ~ successive compressions (or between successive expansions), and frequency is the membrane number of compressions that pass a given point per second. The wave velocity is the velocity with which each compression appears to move; it is equal to the product of wavelength and frequency, v = Af (Eq.15-1). A longitudinal wave can be represented graphically by plotting the density of air molecules (or coils of a Slinky) versus position at a given instant, as shown in Fig. 15-6. Such a graphical representation makes it easy to illustrate what is happening. Note that the graph looks much like a transverse wave.

FIGURE 15-6 (a) A longitudinal wave with (b) its graphical representation at a particular

%

(b)

398

CHAPTER

15

Wave Motion

J

High +

D C D C

instant in time.

Velocityof Transverse Waves The velocity of a wave depends on the properties of the medium in which it travels. The velocity of a transverse wave on a stretched string or cord, for example, depends on the tension in the cord, Fr, and on the mass per unit length of the cord, u "‘the Greek letter mu, where here u = m/f). For waves of small amplitude, the relationship is Fr

transverse

VEANTT

[wave ona cord]

(15-2)

Before giving a derivation of this formula, it is worth noting that at least qualitatively it makes sense on the basis of Newtonian mechanics. That is, we do expect the tension to be in the numerator and the mass per unit length in the denominator. Why? Because when the tension is greater, we expect the velocity to be greater since each segment of cord is in tighter contact with its neighbor. And, the greater the mass per unit length, the more inertia the cord has and the more slowly the wave would be expected to propagate. EXERCISE C A wave starts at the left end of a long cord (see Fig. 15-1) when someone shakes the cord back and forth at the rate of 2.0 Hz. The wave is observed to move to the right at 4.0 m/s. If the frequency is increased from 2.0 to 3.0 Hz, the new speed of the wave is (a) 1.0m/s, (b) 2.0m/s, (c) 4.0m/s, (d) 8.0m/s, (e) 16.0m/s. };i

Fp i:“ " L

r

(a) y

" h

i '

)

v_;_’

FIGURE 15-7 Diagram of simple wave pulse on a cord for derivation of Eq. 15-2. The vector shown in (b) as the resultant of Fy + Fy has to be directed along the cord because the cord is flexible. (Diagram is not to scale: we assume v’

= 679N.

(b) The frequencies of the second, third, and fourth harmonics are two, three, and four times the fundamental frequency: 262, 393, and 524 Hz, respectively. NOTE The; speed of the wave on the string is not the same as the speed of the sound wave that the piano string produces in the air (as we shall see in Chapter 16). A standing wave does appear to be standing in place (and a traveling wave appears to move). The term “standing” wave is also meaningful from the point of view of energy. Since the string is at rest at the nodes, no energy flows past these points. Hence the energy is not transmitted down the string but “stands” in place in the string. Standing waves are produced not only on strings, but on any object that is struck, such as a drum membrane or an object made of metal or wood. The resonant frequencies depend on the dimensions of the object, just as for a string they depend on its length. Large objects have lower resonant frequencies than small objects. All musical instruments, from stringed instruments to wind instruments (in hich a column of air vibrates as a standing wave) to drums and other percussion .astruments, depend on standing waves to produce their particular musical sounds, as we shall see in Chapter 16. "The term “harmonic” comes from music, because such integral multiples of frequencies “harmonize.”

SECTION 15-9

413

Mathematical Representation of a Standing Wave As we saw, a standing wave can be considered to consist of two traveling waves that move in opposite directions. These can be written (see Egs. 15-10c and 15-13c) Dy(x,t)

=

Asin(kx

— wt)

and

Dy(x,t)

=

Asin(kx

+ wt)

-~

since, assuming no damping, the amplitudes are equal as are the frequencies andwavelengths. The sum of these two traveling waves produces a standing wave which can be written mathematically as D

=D

+ D,

=

From the trigonometric identity

can rewrite this as D

=

A[sin(kx

- wt)

+ sin(kx + ot)].

sin6; + sinf, = 2 sin%(el

+ 02) cos%(é)] = 62), we

2Asin kx cos wt.

(15-18)

If we let x = 0 at the left-hand end of the string, then the x = [ where £ is the length of the string. Since the string is (Fig. 15-26), D(x,t) must be zero at x =0 and at x = already satisfies the first condition (D = 0 at x = 0) and condition if sin kf = 0 which means

right-hand end is at fixed at its two ends £. Equation 15-18 satisfies the second

kt = @, 2w, 3m, -, nmw, where n = aninteger. Since k = 27/A, then A = 2€/n, which is just Eq. 15-17a. Equation 15-18, with the condition A = 2£/n, is the mathematical representation

of a standing wave. We see that a particle at any position x vibrates in simple harmonic motion (because of the factor cos wt). All particles of the string vibrate with the same frequency f = w/2w, but the amplitude depends on x and equals 2Asinkx. (Compare this to a traveling wave for which all particles vibrate with the same amplitude.) The amplitude has a maximum, equal to 24, when kx = m/2,3m/2,5m/2, and so on—that is, at X

=

A

33X

5A

Z,T’T’””

fl

These are, of course, the positions of the antinodes (see Fig. 15-26).

Wave forms. Two waves traveling in opposite directions on

a string fixed at x = 0 are described by the functions D, = (020m)sin(2.0x — 4.0t) and D, = (0.20m)sin(2.0x + 4.0t) (where x is in m, ¢ is in s), and they produce a standing wave pattern. Determine (a) the function for the standing wave, (b) the maximum amplitude at x = 0.45m, (c) where the other end is fixed (x > 0), (d) the maximum amplitude, and where it occurs. APPROACH We use the principle of superposition to add the two waves. The given waves have the form we used to obtain Eq. 15-18, which we thus can use. SOLUTION (@) The two waves are of the form D = Asin(kx + wt), so

FIGURE 15-27 Example 15-9: possible lengths for the string.

p

£=157m

po £=3.14m

k =20m"'

x

*

414

CHAPTER 15

o = 40s".

These combine to form a standing wave of the form of Eq. 15-18:

D = 2Asinkxcoswt

= (0.40m)sin(2.0x) cos(4.0¢),

where x is in meters and ¢ in seconds. (b) At x = 0.45m, D = (0.40m)sin(0.90) cos(4.0¢)

=

(0.31 m) cos(4.0t).

The maximum amplitude at this pointis D = 0.31 m and occurs when cos(4.0f) = 1. (c) These waves make a standing wave pattern, so both ends of the string must be nodes. Nodes occur every half wavelength, which for our string is A

t= (nmr/2.0) m=n(1.57 m)

and

-

127

=

lm

=

1.587m.

fl

If the string includes only one loop, its length is £ = 1.57 m. But without more

information, it could be twice as long, £ = 3.14 m,

or any integral number times

1.57 m, and still provide a standing wave pattern, Fig. 15-27.

(d) The nodes occur at x = 0,x = 1.57m, and, if the string is longer than £=157m, at x=314m, 471m, and so on. The maximum amplitude (antinode) is 0.40 m [from part (@) above] and occurs midway between the nodes. For

£ = 1.57 m,

~15-10

Ray

there is only one antinode, at x = 0.79 m.

Refraction’

When any wave strikes a boundary, some of the energy is reflected and some is transmitted or absorbed. When a two- or three-dimensional wave traveling in one ‘medium crosses a boundary into a medium where its speed is different, the transmitted wave may move in a different direction than the incident wave, as shown in 'Fig. 15-28. This phenomenon is known as refraction. One example is a water wave;

the velocity decreases in shallow water and the waves refract, as shown in Fig. 15-29 below. [When the wave velocity changes gradually, as in Fig. 15-29,

without a sharp boundary, the waves change In Fig. 15-28, the velocity of the wave in this case, the wave front bends so that it boundary. That is, the angle of refraction, 6.,

direction (refract) gradually.] medium 2 is less than in medium 1. In -travels more nearly parallel to the is less than the angle of incidence, 6;.

To see why this is so, and to help us get a quantitative relation between 6, and 6;,

let us think of each wave front as a row of soldiers. The soldiers are marching from firm ground (medium 1) into mud (medium 2) and hence are slowed down after the boundary. The soldiers that reach the mud first are slowed down first, and the ‘row bends as shown in Fig. 15-30a. Let us consider the wave front (or row of soldiers) labeled A in Fig. 15-30b. In the same time ¢ that A; moves a distance £, = v t, we see that A, moves a distance £, = v,¢. The two right triangles in Fig. 15-30b, shaded yellow and green, have the side labeled @ in common. Thus .

sinf;,

=

£

—

a

=

FIGURE 15-28 Refraction of Wwaves passing a boundary. FIGURE

15-29

Water waves

— p. gradually as they approach the shore, as their velocity decreases. There is no distinct boundary, as in

Fig. 15-28, because the wave velocity changes gradually.

vt

——

a

";since a is the hypotenuse, and £ vt

sing,

= 2 a

= 2. a

Dividing these two equations, we obtain the law of refraction: sin 6 2 v —— sin 64

2, vy

(15-19)

Since 6, is the angle of incidence (6;), and 6, is the angle of refraction (6;), Eq. 15-19

gives the quantitative relation between the two. If the wave were going in the opposite direction, the geometry would not change; only 6, and 6, would change roles: 6; would be the angle of refraction and 6, the angle of incidence. Clearly then, if the wave travels into a medium where it can move faster, it will bend the opposite way, 6, > 6;. We see from Eq. 15-19 that if the velocity increases, the angle increases, and vice versa. Earthquake waves refract within the Earth as they travel through rock layers of different densities (and therefore the velocity is different) just as water waves do. Light waves refract as well, and when we discuss light, we shall find Eq. 15-19 very useful. "This Section and the next are covered in more detail in Chapters 32 to 35 on optics.

|

FIGURE 15-30 (a) Soldier analogy to derive (b) law of refraction for waves.

*SECTION 15-10

415

®)PpHYSsics

APPLIED

m

Earthquake wave refraction

Refraction of an earthquake wave. An earthquake P wave

passes across a boundary in rock where its velocity increases from 6.5 km/s to 8.0 km/s. If it strikes this boundary at 30°, what is the angle of refraction? APPROACH We apply the law of refraction, Eq. 15-19, sin 6,/sin 6, = v,/v,. SOLUTION Since sin30° = 0.50, Eq. 15-19 yields S (8.0m/s) sinf, = (0.50) = 0.62. (6.5m/s) So 6, = sin"1(0.62) = 38°. NOTE Be careful with angles of incidence and refraction. As we discussed in Section 15-7 (Fig. 15-21), these angles are between the wave front and the boundary line, or—equivalently—between the ray (direction of wave motion) and the line perpendicular to the boundary. Inspect Fig. 15-30b carefully.

*15-11

FIGURE 15-31 Wave diffraction. The waves are coming from the upper left. Note how the waves, as

they pass the obstacle, bend around it, into the “shadow region” behind

the obstacle.

Diffraction

Waves spread as they travel. When they encounter an obstacle, they bend around it somewhat and pass into the region behind it, as shown in Fig. 15-31 for water waves. This phenomenon is called diffraction. The amount of diffraction depends on the wavelength of the wave and on the size of the obstacle, as shown in Fig. 15-32. If the wavelength is much larger than the object, as with the grass blades of Fig. 15-32a, the wave bends around them almost as if they are not there. For larger objects, parts (b) and (c), there is more of a “shadow” region behind the obstacle where we might not expect the waves to penetrate—but they do, at least a little. Then notice in part (d), where tt}e obstac.le is t_he same as in part (c)‘ but the wavelength is longer, that there is more diffraction into the shadow region. As a rule of thumb, only if the wavelength is smaller than the size of the object will there be _g\ significant shadow region. This rule applies to reflection from an obstacle as we: Very little of a wave is reflected unless the wavelength is smaller than the size ot

the obstacle. A rough guide to the amount of diffraction is f(radians)

=

%,

where 6 is roughly the angular spread of waves after they have passed through an opening of width £ or around an obstacle of width £. FIGURE 15-32 Water waves passing objects of various sizes. Note that the longer the wavelength compared to the size of the object, the more diffraction there is into the “shadow region.”

(a) Water waves passing blades of grass

(b) Stick in water

(c) Short-wavelength waves passing log

(d) Long-wavelength waves passing log

That waves can bend around obstacles, and thus can carry energy to areas behind obstacles, is very different from energy carried by material particles. A clear example is the following: if you are standing around a corner on one side of a building, you cannot be hit by a baseball thrown from the other side, but yo can hear a shout or other sound because the sound waves diffract around th edges of the building. 416

CHAPTER15

Wave Motion

|Summary Vibrating objects act as sources of waves that travel outward om the source. Waves on water and on a string are examples. he wave may be a pulse (a single crest) or it may be continuous (many crests and troughs).

The wavelength of a continuous wave is the distance between two successive crests (or any two identical points on the wave shape). The frequency is the number of full wavelengths (or crests) that pass a given point per unit time. The wave velocity (how fast a crest moves) is equal to the product of wavelength and frequency,

v = Af.

(15-1)

The amplitude of a wave is the maximum height of a crest, or depth of a trough, relative to the normal (or equilibrium) level. In a transverse wave, the oscillations are perpendicular to the direction in which the wave travels. An example is a wave on a string. In a longitudinal wave, the oscillations are along (parallel to) the line of travel; sound is an example.

The velocity of both longitudinal and transverse waves in matter is proportional to the square root of an elastic force factor divided by an inertia factor (or density). Waves carry energy from place to place without matter being carried. The intensity of a wave (energy transported across unit area per unit time) is proportional to the square of the amplitude of the wave. For a wave traveling outward in three dimensions from a point source, the intensity (ignoring damping) decreases with

fl\e

square of the distance from the source,

I x

1 =

(15-8a)

The amplitude decreases linearly with distance from the source. A one-dimensional transverse wave traveling in a medium to the right along the x axis (x increasing) can be represented by a formula for the displacement of the medium from equilibrium at any point x as a function of time as D(x,t)

where

=

Asin[(%’r)(x

=

Asin(kx

— wt)

- vt)]

(15-10a) (15-10¢)

‘ k

=

2 Y

(1s-11)

and

o

=

2uf.

If a wave is traveling toward decreasing values of x, D(x,t)

=

Asin(kx

+ ot).

(15-13¢)

[*Waves can be described by the wave equation, which in

one dimension is 8°D/ax? = (1/v?) 8*D/at?, Eq.15-16.]

When two or more waves pass through the same region of space at the same time, the displacement at any given point will be the vector sum of the displacements of the separate waves. This is the principle of superposition. It is valid for mechanical waves if the amplitudes are small enough that the restoring force of the medium is proportional to displacement. Waves reflect off objects in their path. When the wave front of a two- or three-dimensional wave strikes an object, the angle of reflection equals the angle of incidence, which is the law of reflection. When a wave strikes a boundary between two materials in which it can travel, part of the wave is reflected and part is transmitted. When two waves pass through the same region of space at the same time, they interfere. From the superposition principle, the resultant displacement at any point and time is

the sum of their separate displacements. This can result in

constructive interference, destructive interference, or something in between depending on the amplitudes and relative phases of the waves. Waves traveling on a cord of fixed length interfere with waves that have reflected off the end and are traveling back in the opposite direction. At certain frequencies, standing waves can be produced in which the waves seem to be standing still rather than traveling. The cord (or other medium) is vibrating as a whole. This is a resonance phenomenon, and the frequencies at which standing waves occur are called resonant frequencies. The points of destructive interference (no vibration) are called nodes. Points of constructive interference (maximum amplitude of vibration) are called antinodes. On a cord of length £ fixed at both ends, the wavelengths of standing waves are given by Ap=2/n

(15-17a)

where 7 is an integer. [*Waves change direction, or refract, when traveling from one medium into a second medium where their speed is different. Waves spread, or diffract, as they travel and encounter obstacles. A rough guide to the amount of diffraction is 6 ~ A/f, where A is the wavelength and { the width of an opening or obstacle. There is a significant “shadow region” only if the wavelength A is smaller than the size of the obstacle.]

| Questions 1 Is the frk:quency of a simple periodic wave equal to the

frequency of its source? Why or why not? 2. Explain the difference between the speed of a transverse wave traveling down a cord and the speed of a tiny piece of the cord. 3. You are finding it a challenge to climb from one boat up onto a higher boat in heavy waves. If the climb varies from 2.5m to 4.3 m, what is the amplitude of the wave? Assume the centers of the two boats are a half wavelength apart. ’ 1. What kind of waves do you think will travel down a horizontal metal rod if you strike its end (a) vertically from above and (b) horizontally parallel to its length?

5. Since the density of air decreases with an increase in temperature, but the bulk modulus B is nearly independent of temperature, how would you expect the speed of sound waves in air to vary with temperature? 6. Describe how you could estimate the speed of water waves across the surface of a pond. 7. The speed of sound in most solids is somewhat greater than in

air, yet the density of solids is much greater (10° to 10* times).

Explain. 8. Give two reasons why circular water waves decrease amplitude as they travel away from the source.

Questions

in

417

9 Two linear waves have the same amplitude and speed, and otherwise are identical, except one has half the wavelength of the other. Which transmits more energy? By what factor? 10. Will any function of (x — vt)—see Eq. 15-14—representa wave motion? Why or why not? If not, give an example. . p 11, When a sinusoidal wave crosses the boundary between two 5 . sections of cord as in Fig. 15-19, the frequency does not change (although the wavelength and velocity do change). Explain why, ) i ) ) ;

12. If a sinusoidal wave on a two-section cord (Fig. 15-19) is

inverted upon reflection, does the transmitted wave have a

15. When a standing wave exists on a string, the vibrations of incident and reflected waves cancel at the nodes. Does this mean that energy was destroyed? Explain. 16. Can the amplitude of the standing waves in Fig. 15-25 be greater than the amplitude of the vibrations that cause then (up and down motion of the hand)? o 17. When a cord is vibrated as in Fig. 15-25 by hand or by a . : « 5 g mechanical osc‘lllator: the nqdes are not quite true nodes (at rest). Explain. [Hint: Consider damping and energy flow 18,

longer or shorter wavelength?

13. Is energy Explain.

always

conserved

from hand or oscillator.]

AM radio signals can usually be heard behind a hill, but FM often

when

two

waves

interfere?

14. If a string is vibrating as a standing wave in three segments,

are there any places you could touch it with a knife blade

without disturbing the motion?

cannot.

That

is, AM

signals

bend

more

than

FM.

Explain. (Radio signals, as we shall see, are carried by electromagnetic waves whose wavelength for AM is typically

200 to 600 m and for FM abo‘ft 3m.)

19+ If we knew that energy was being transmitted from one place to another, how might we determine whether the energy was

being carried by particles (material objects) or by waves?

| Problems 15-1

and 15-2

Characteristics of Waves

11. (II) The wave on a string shown in Fig. 15-33 is moving to

1. (I) A fisherman notices that wave crests pass the bow of his

the right with a speed of 1.10m/s. (a) Draw the shape of the

anchored boat every 3.0s. He measures the distance between two crests to be 8.0m. How fast are the waves traveling?

2. (I) A

sound

wave

in air has

a frequency

of 262Hz

string 1.00s later and indicate which parts of the string are

moving up and whlc_h down at that instant. (b)‘ Estimate the

and

yertlcal .speed of point A on the string at the instant shown

travels with a speed of 343 m/s. How far apart are the wave

in the Figure.

crests (compressions)? 3. (I) Calculate the speed of longitudinal waves in (a) water,

(b) granite, and (c) steel. 4. (I) AM radio signals have frequencies between 550 kHz and

P

1600 kHz (kilohertz) and travel with a speed of 3.0 X 10%m/s.

What are the wavelengths of these signals? On FM the frequencies range from 88 MHz to 108 MHz (megahertz) and travel at the same speed. What are their wavelengths? 5. (I) Determine the wavelength of a 5800-Hz sound wave traveling along an iron rod.

6. (IT) A cord of mass 0.65 kg is stretched between two supports 8.0m apart. If the tension in the cord is 140 N, how long will it take a pulse to travel from one support to the other? 7. (II) A 0.40-kg cord is stretched between two supports, 7.8 m apart. When one support is struck by a hammer, a transverse wave travels down the cord and reaches the other support

in 0.85s. What is the tension in the cord?

. (II) A sailor strikes the side of his ship just below the surface of the sea. He hears the echo of the wave reflected from the ocean floor directly below 2.8 s later. How deep is the ocean at this point? . (IT) A ski gondola is connected to the top of a hill by a steel cable of length 660m and diameter 1.5 cm. As the gondola comes to the end of its run, it bumps into the terminal and sends a wave pulse along the cable. It is observed that it took 17 s for the pulse to return. (a) What is the speed of the pulse? (b) What is the tension in the cable? 10. (IT) P and S waves from an earthquake travel at different speeds, and this difference helps locate the earthquake “epicenter” (where the disturbance took place). (a) Assuming typical speeds of 8.5 km/s and 5.5 km/s for P and S waves, respectively, how far away did the earthquake occur if a particular seismic station detects the arrival of these two types of waves 1.7 min apart? (b) Is one seismic station sufficient to determine the position of the epicenter? Explain. 418

CHAPTER

15

Wave Motion

0

N\

N\ Tm

N~—"

A 5 m

FIGURE 15-33 12.

(II)

A

5.0kg

ball

hangs

\

2em 1.

0

\]

-lem

Im

—2cm

Problem 11. from

a

steel

wire

1.00mm

in

diameter and 5.00m long. What would be the speed of a wave in the steel wire?

13. (II) Two children are sending signals along a cord of total mass 0.50 kg tied between tin cans with a tension of 35 N. It takes the vibrations in the string 0.50s to go from one child to the other. How far apart are the children?

*14. (II) Dimensional

analysis. Waves

on the surface of the

ocean do not depend significantly on the properties of water such as density or surface tension. The primary “return force” for water piled up in the wave crests is due to the gravitational attraction of the Earth. Thus the speed v (m/s) of ocean waves depends on the acceleration due to gravity g. It is reasonable to expect that v might also depend on water depth s and the wave’s wavelength A. Assume the wave speed is given by the functional form

v = Cg®hP)",

where a,B,7, and C are numbers without

dimension. (a) In deep water, the water deep below does not affect the motion of waves at the surface. Thus v should be independent of depth 4 (i.e., B = 0). Using only

dimensional analysis (Section 1-7), determine the formula for the speed of surface waves in deep water. (b) In shallow water, the speed of surface waves is found experimentally to be independent of the wavelength (i.e, Y = 0). Using onlyfl dimensional analysis, determine the formula for the speed of waves in shallow water.

15-3

Energy Transported by Waves

15. (I) Two earthquake waves of the same frequency travel through the same portion of the Earth, but one is carrying 3.0 times the energy. What is the ratio of the amplitudes of r the two waves? 16. (IT) What is the ratio of (a) the intensities, and (b) the amplitudes, of an earthquake P wave passing through the Earth and detected at two points 15 km and 45 km from the source? 17, (IT) Show that if damping is ignored, the amplitude A of circular water waves decreases as the square root of the

distance r from the source: A o 1/Vr.

18. (IT) The intensity of an earthquake wave passing through

the Earth is measured to be 3.0 X 10°J/m?s at a distance

of 48 km from the source. (¢) What was its intensity when it passed a point only 1.0km from the source? (b) At what rate did energy pass through an area of 2.0 m? at 1.0 km? 19. (IT) A small steel wire of diameter 1.0 mm is connected to an oscillator and is under a tension of 7.5N. The frequency of the oscillator is 60.0 Hz and it is observed that the amplitude of the wave on the steel wire is 0.50cm. (a) What is the power output of the oscillator, assuming that the wave is not reflectedw‘ back? (b) If the power output stays constant but the frequency is doubled, what is the amplitude of the wave? 20. (Im) Show that the intensity of a wave is equal to the energy density tenergy per unit volume) in the wave times the wave spéed. 21. (I1) (a) iShow that the average rate with which energy is transported along a cord by a mechanical wave of frequency f and amplitude A is

(‘

P

=

2nuvf’A,

where v is the speed of the wave and u is the mass per unit length of the cord. (b) If the cord is under a tension

Fr = 135N

and has mass per unit length 0.10 kg/m, what

power is required to transmit 120-Hz transverse waves of amplitude 2.0 cm?

15-4 Mathematical Representation of Traveling Wave

22. (I) A transverse wave on a wire is given by D(x,f) = 0.015sin(25x — 1200¢) where D and x are in meters and / is in seconds. (a) Write an expression for a wave with the same amplitude, wavelength, and frequency but traveling in the opposite direction. (b) What is the speed of either wave? 23. (I) Suppose at t =0, a wave shape is represented by D = Asin(2mx/A + ¢); that is, it differs from Eq. 15-9 by a constant phase factor ¢. What then will be the equation for a wave traveling to the left along the x axis as a function of x and ¢? . (IT) A transverse traveling wave on a cord is represented by D = 0.22sin(5.6x + 34t) where D and x are in meters and ¢ is in seconds. For this wave determine (@) the wavelength, (b) frequency, (c) velocity (magnitude and direction), (d) amplitude, and (e¢) maximum and minimum speeds of particles of the cord. 25. (II) Consider the point x = 1.00m on the cord of Example 15-5. Determine (a) the maximum velocity of this point, and (b) its maximum acceleration. (c) What is its velocity and acceleration at ¢t = 2.50s? (II) A transverse wave on a cord is given by D(x,t) = 0.12sin(3.0x — 15.0¢), where D and x are in m and ¢ is in s. At t = 0.20s, what are the displacement and velocity of the point on the cord where x = 0.60 m?

27. (IT) A transverse wave pulse travels to the right along a string with a speed v = 2.0m/s. At ¢ = 0 the shape of the pulse is given by the function D = 0.45cos(2.6x + 1.2), where D and x are in meters. (a) Plot D vs. x at ¢ = 0. (b) Determine a formula for the wave pulse at any time ¢ assuming there are no frictional losses. (¢) Plot D(x, ) vs. x at ¢ = 1.0s. (d) Repeat parts (b) and (c) assuming the pulse is traveling to the left. Plot all 3 graphs on the same axes for easy comparison. 28. (IT) A 524-Hz longitudinal wave in air has a speed of 345 m/s. (a) What is the wavelength? (b) How much time is required for the phase to change by 90° at a given point in space? (c¢) At a particular instant, what is the phase difference (in degrees) between two points 4.4 cm apart? 29. (IT) Write the equation for the wave in Problem 28 traveling to the right, if its amplitude is 0.020 cm, and

D = —0.020 cm,

at t=0 and x = 0. 30. (IT) A sinusoidal wave traveling on a string in the negative x direction has amplitude 1.00cm, wavelength 3.00cm, and frequency 245 Hz. At ¢ = 0, the particle of string at x = 0 is displaced a distance D = 0.80cm above the origin and is moving upward. (a) Sketch the shape of the wave at ¢ = 0 and (b) determine the function of x and ¢ that describes the wave.

*15-5

The Wave Equation

*31. (IT) Determine if the function D = Asinkxcosw! is a solution of the wave equation. *32. (IT) Show by direct substitution that the following functions satisfy the wave equation: (a) D(x,t?) = Aln(x + of); (b) D(x,t) = (x — vt)* *33. (IT) Show that the wave forms of Egs. 15-13 and 15-15 satisfy the wave equation, Eq. 15-16. *34. (II) Let two linear waves be represented by Dy = fi(x,1) and D, = fo(x,t). If both these waves satisfy the wave equation (Eq. 15-16), show that any combination D =C{D; + C;D, does as well, where C; and C, are constants.

. (II) Does the function D(x,t) = e k" equation? Why or why not? . (IT) In deriving Eq. 15-2, v = a transverse

wave

on

a string,

V Fy/u, it was

satisfy the wave

for the speed of

assumed

that

the

wave’s amplitude A is much less than its wavelength A. Assuming a sinusoidal wave shape D = Asin(kx — wf), show via the partial derivative o' = dD/ot that the assumption A . Vsnd

The difference between the observed and emitted wavelengths willbe AA = A" — A =

+ Msource/Vsna). The observed frequency of the wave, f' = vgq/A’, will be d

R

(1 +

S,

source)

source moving away from

[stationary observer

] (16-9b)“

Vsnd

440

CHAPTER 16

Sound

If a source emitting at 400 Hz is moving away from a fixed observer at 30 m/s, the observer hears a frequency f' = (400 Hz)/[1 + (30 m/s)/(343 m/s)| = 368 Hz.

S

0,

®

L———vA

—

‘_.

source detects wave crests passing at

=

Vobs

Vsnd

Vsource =0

’

FIGURE 16-20 Observer moving with speed v, toward a stationary

Obse,rver

ource

speed v/ = vgpq + Vops Where vgq is the speed of the sound waves in air.

The Doppler effect also occurs when the source is at rest and the observer is in motion. If the observer is traveling toward the source, the pitch heard is higher than that of the emitted source frequency. If the observer is traveling away from the source, the pitch heard is lower. Quantitatively the change in frequency is different than for the case of a moving source. With a fixed source and a moving observer, the distance between wave crests, the wavelength A, is not changed. But the velocity of the crests with respect to the observer is changed. If the observer is moving toward the source, Fig. 16-20, the speed v’ of the waves relative to the observer is a simple addition of velocities: v’ = vgq + Vops, Where vgg is the velocity of sound in air (we assume the air is still) and v, is the velocity of the observer. Hence, the frequency heard is Vv

Usd

+

A Because

A = vg4/f,

then (Usnd

or

’

if the

Uobs)f

-

Usnd

f

o=

Vo (1 + i) f. is

SO

"=

=

’

: toward] [obsc_erver moving

Vgnd

T Uobs,

DGV

+

f

observer

V' = Vgq

'Uobs.

A

moving

EERES

away

from

the

source,

the

relative

observer moving away

Zobs

_

(1

stationary source

[from stationary source]

Vsnd )f'

(16-10a) velocity

is

_

(16-10b)

A moving siren. The siren of a police car at rest emits at a

predominant frequency of 1600 Hz. What frequency will you hear if you are at rest and the police car moves at 25.0 m/s (a) toward you, and (b) away from you? APPROACH The observer is fixed, and the source moves, so we use Egs. 16-9. The frequency you (the observer) hear is the emitted frequency f divided by

the factor (1 £ vgource/Vsnd) Where vyouree is the speed of the police car. Use the

minus sign when the car moves toward you (giving a higher frequency); use the plus sign when the car moves away from you (lower frequency). SOLUTION (a) The car is moving toward you, so (Eq. 16-9a)

f o=

!

| _ Psouree

=

1

160

Vgnd

H 60HZ __ _ 1726Hz ~ 1730 Hz. 25.0m/s

343m/s

(b) The car is moving away from you, so (Eq. 16-9b)

fo=

1

&

!Usource Vgnd

=

1600 H 1

+

=

25.0 m/s

1491Hz ~ 1490 Hz.

343m/s

EXERCISE G Suppose the police car of Example 16-14 is at rest and emits at 1600 Hz. What frequency would you hear if you were moving at 25.0m/s (a) toward it, and l (b) away from it?

SECTION 16-7

Doppler Effect

441

When a sound wave is reflected from a moving obstacle, the frequency of the reflected wave will, because of the Doppler effect, be different from that of the incident wave. This is illustrated in the following Example.

Two Doppler shifts. A 5000-Hz sound wave is emitted by “™

stationary source. This sound wave reflects from an object moving 3.50 m/s toward the source (Fig. 16-21). What is the frequency of the wave reflected by the moving object as detected by a detector at rest near the source?

Original

o |e

source

velocity,

(a)

Object

Source and

@

Wave

-

-

detector | Vvelocity

Vsou 3.50

m/s

APPROACH There are actually two Doppler shifts in this situation. First, the moving object acts like an observer moving toward the source with speed Vobs = 3.50m/s (Fig. 16-21a) and so “detects” a sound wave of frequency (Eq. 16-10a) f' = f [1 + (vobs/vsnd)]. Second, reflection of the wave from the moving object is equivalent to the object reemitting the wave, acting effectively as a moving source with speed wvuree = 3.50m/s (Fig. 16-21b). The final frequency detected, f", is given by f" = f'/ [1 == vsource/vsnd], Eq. 16-9a. SOLUTION The frequency f’' that is “detected” by the moving object is (Eq. 16-10a):

f = + As sin(i—: x), and D is the displacement of the string at a time ¢ = 0. Imagine plucking this string at its midpoint (Fig. 16-45a) or at a point two-thirds from the left end (Fig. 16-45b). Using a graphing calculator or computer program, show that the above expression can fairly accurately represent the shape in: (a) Fig. 16-45a, if

Obstacle

destructive

A

interference occurs? (Assume A 6.02 X 10%/mol

-

N 138 X 10°BJ/K. /

“For example, the molecular mass of H, gas is 2.0 atomic mass units (u), whereas that of O, gas is 32.0 u. Thus 1 mol of H, has a mass of 0.0020 kg and 1 mol of O, gas, 0.0320 kg. The number of molecules in a mole is equal to the total mass M of a mole divided by the mass m of one molecule; since th*® ratio (M/m) is the same for all gases by definition of the mole, a mole of any gas must contain the same number of molecules.

468

CHAPTER 17

Temperature, Thermal Expansion, and the Ideal Gas Law

DGV AN Hydrogen atom mass. Use Avogadro’s number to determine the mass of a hydrogen atom.

f

APPROACH

The mass of one atom

number of atoms in 1 mol, N,. SOLUTION

equals the mass of 1 mol divided by the

One mole of hydrogen atoms (atomic mass = 1.008 u, Section 17-1

or Appendix F) has a mass of 1.008 X 10 kg and contains 6.02 X 10% atoms. Thus one atom has a mass

1.008 X 103 kg T 6.02 X 107

1.67 x 10 kg, SLYTIVMFESEN

ESTIVIATE | How many molecules in one breath? Estimate

how many molecules you breathe in with a 1.0-L breath of air.

(X)PHYSicS

APPLIED

Molecules in a breath

APPROA} H We determine what fraction of a mole 1.0 L is by using the result of Example 17-10 that 1 mole has a volume of 22.4 L at STP, and then multiply that by N, to get the number of molecules in this number of moles.

SOLUTION

One

mole

(1.0L)/(22.4 L/mol)

corresponds

to

22.4L

at

(0.045 mol)(6.02 X 10 molecules/mol)

#17-10

STP,

= 0.045 mol. Then 1.0 L of air contains

so

1.0L

of

air

is

~ 3 X 10* molecules.

Ideal Gas Temperature Scale—

a Standard

It is important to have a very precisely defined temperature scale so that measurements of temperature made at different laboratories around the world can be accurately compared. We now discuss such a scale that has been accepted by the general scientific community. The standard thermometer for this scale is the constant-volume gas thermometer discussed in Section 17-2. The scale itself is called the ideal gas temperature scale, since

it is based on the property of an ideal gas that the pressure is directly proportional to the absolute temperature (Gay-Lussac’s law). A real gas, which would need to be used in any real constant-volume gas thermometer, approaches this ideal at low density. In other words, the temperature at any point in space is defined as being proportional to the pressure in the (nearly) ideal gas used in the thermometer. To set up a scale we need two fixed points. One fixed point will be P =0 at T = 0K. The second fixed point is chosen to be the triple point of water, which is that point where water in the solid, liquid, and gas states can coexist in equilibrium. This occurs only at a unique temperature and pressure,” and can be reproduced at different laboratories with great precision. The pressure at the triple point of water is 4.58 torr and the temperature is 0.01°C. This temperature corresponds to 273.16 K, since absolute zero is about —273.15°C. In fact, the triple point is now defined to be exactly 273.16 K.

"Liquid water and steam can coexist (the boiling point) at a range of temperatures depending on the pressure. Water boils at a lower temperature when the pressure is less, such as high in the mountains. The triple point represents a more precisely reproducible fixed point than does either the freezing point or boiling point of water at, say, 1 atm. See Section 18-3 for further discussion.

*SECTION 17-10

Ideal Gas Temperature Scale—a Standard

469

1

o

373.00

T

TRt

w 2 o o

T (K)

+ 1Tt

The absolute or Kelvin temperature 7' at any point is then defined, using a constant-volume gas thermometer for an ideal gas, as

T H, TSt 0 200 400 600 800 Plp (torr)

FIGURE 17-17 Temperature readings of a constant-volume gas thermometer for the boiling point of water at 1.00 atm are plotted, for different gases, as a function of the gas pressure in the thermometer at

the triple point (P,p). Note that as

the amount of gas in the thermometer is reduced, so that Py — 0, all gases give the same reading, 373.15 K. For pressure less than 0.10 atm (76 torr), the variation shown is less than 0.07 K.

=

from Eq. 17-5a to be 373.87 K when the gas is O, and

The atomic theory of matter postulates that all matter is made up of tiny entities called atoms, which are typically 107° m in diameter. Atomic and molecular masses are specified on a scale

where ordinary carbon ('2C) is arbitrarily given the value

12.0000 u (atomic mass units). The distinction between solids, liquids, and gases can be attributed to the strength of the attractive forces between the atoms or molecules and to their average speed. Temperature is a measure of how hot or cold something is. Thermometers are used to measure temperature on the Celsius (°C), Fahrenheit (°F), and Kelvin (K) scales. Two standard points on each scale are the freezing point of water (0°C, 32°F, 273.15K) boiling

point

of

water

(100°C,

212°F,

373.15K).

A one-kelvin change in temperature equals a change of one Celsius degree or ¢ Fahrenheit degrees. Kelvins are related to °C by T(K) = T(°C) + 273.15. The change in length, A{, of a solid, when its temperature changes by an amount A7, is directly proportional to the temperature change and to its original length £. That is, Al

=

alyAT,

(17-1a)

where a is the coefficient of linear expansion. The change in volume of most solids, liquids, and gases is proportional to the temperature change and to the original

470

CHAPTER 17

(17-5a)

-

P, = 1000 torr. If the amount

of O, in the bulb is reduced so that at the triple point P, = 500 torr, the boiling point of water from Eq. 17-5a is then found to be 373.51 K. If H, gas is used instead, the corresponding values are 373.07K and 373.11 K (see Fig. 17-17). But now suppose we use a particular real gas and make a series of measurements in which the amount of gas in the thermometer bulb is reduced to smaller and smaller amounts, so that P becomes smaller and smaller. It is found experimentally that an extrapolation of such data to P, = 0 always gives the same value for the temperature of a given system (such as 7 = 373.15K for the boiling point of water at 1.00atm) as shown in Fig. 17-17. Thus the temperature 7 at any point in space, determined using a constantvolume gas thermometer containing a real gas, is defined using this limiting process:

(273.16K) lim at STP) is far beyond the capability of any present

computer. Instead we take a statistical approach and determine averages of certain quantities, and these averages correspond to macroscopic variables. We will, of course, demand that our microscopic description correspond to the macroscopic properties of gases; otherwise our theory would be of little value. Most importantly, we will arrive at an important relation between the average kinetic energy of molecules in a gas and the absolute temperature.

18—1

The Ideal Gas Law and the Molecular Interpretation of Temperature

We make the following assumptions about the molecules in a gas. These assumptions reflect a simple view of a gas, but nonetheless the results they predict correspond well to the essential features of real gases that are at low pressures and far from the liquefaction point. Under these conditions real gases follow the ideal gas law quite closely, and indeed the gas we now describe is referred to as an ideal gas.

476

The assumptlons which represent the basic postulates of the kinetic theory for an ideal gas, are 1. There are a large number of molecules, N, each of mass m, moving in random directions with a variety of speeds. This assumptlon is in accord with our ’A - observation that a gas fills its container and, in the case of air on Earth, is kept from escaping only by the force of gravity. 2. The molecules are, on average, far apart from one another. That is, their average separation is much greater than the diameter of each molecule. 3. The molecules

are assumed

to obey the laws of classical mechanics, and are

assumeci to interact with one another only when they collide. Although molecules exert weak attractive forces on each other between collisions, the potential energy associated with these forces is small compared to the kinetic energy, and we ignore it for now. 4. Collisions with another molecule or the wall of the vessel are assumed to be perfectlL;I elastic, like the collisions of perfectly elastic billiard balls (Chapter 9). We assume the collisions are of very short duration compared to the time between collisions. Then we can ignore the potential energy associated with collisions in comparison to the kinetic energy between collisions. We can see immediately how this kinetic view of a gas can explain Boyle’s law (Section 17—6) The pressure exerted on a wall of a container of gas is due to the constant bombardmefit of molecules. If the volume is reduced by (say) half, the molecules are closer together and twice as many will be striking a given area of the wall per second. Hence we exbect the pressure to be twice as great, in agreement with Boyle’s law. Now let us calculate quantitatively the pressure a gas exerts on its container as based on kinetic theory. We imagine that the molecules are inside a rectangular container (at rest) whose ends have area A and whose length is £, as shown in Fig. 18-1a. The pressure exerted by the gas on the walls of its container is, according to our model, due to the collisions of the molecules with the walls. Let us focus our

y

tention on the wall, of area A, at the left end of the container and examine what

ppens when one molecule strikes this wall, as shown in Fig. 18—1b. This molecule exerts a force on the wall, and according to Newton’s third law the wall exerts an equal and o%posite force back on the molecule. The magnitude of this force on the molecule, according to Newton’s second law, is equal to the molecule’s rate of change of momentum, F = dp/dt (Eq.9-2). Assuming the collision is elastic, only the x component of the molecule’s momentum changes, and it changes from —mwv, (it is moving in the negative x direction) to +muw, . Thus the change in the molecule’s momentum,

A (mw), which is the final momentum

A(mv) for one

=

collision. This

mo,

— (—mvy)

molecule

minus the initial momentum, is

=

2mo,

will make

many

collisions

with

the wall, each

separated by a time At, which is the time it takes the molecule to travel across the container and back again, a distance (x component) equal to 2£. Thus 2¢ = v, At, or

2%

vy

FIGURE 18-1

The time Af between collisions is very small, so the number of collisions per second is very large. Thus the average force—averaged over many collisions—will be equal to the momentum change during one collision divided by the time between collisions (Newton’s second law): F

=

A(mv)

—

2mv,

(b)

(a) Molecules of a

gas moving about in a rectangular container. (b) Arrows indicate the momentum of one molecule as it rebounds from the end wall.

mo}

[due to one molecule] At 2/, 1 During its passage back and forth across the container, the molecule may collide with the tops and sides of the container, but this does not alter its x component of momentum and thus does not alter our result. It may also collide with other lecules, which may change its v,. However, any loss (or gain) of momentum is quired by other molecules, and because we will eventually sum over all the molecules, this effect will be included. So our result above is not altered.

SECTION 18-1

The Ideal Gas Law and the Molecular Interpretation of Temperature

477

The actual force due to one molecule is number of molecules are striking the wall per nearly constant. To calculate the force due to all have to add the contributions of each. Thus the m F

=

T(Uil

+

U2x2

+

eee

4

intermittent, but because a huge second, the force is, on average, the molecules in the container, we net force on the wall is -

’UiN),

-

where v, means v, for molecule number 1 (we arbitrarily assign each molecule a number) and the sum extends over the total number of molecules N in the container. The average value of the square of the x component of velocity is ij

-

Vy1 74x1 + Uy 2%2 + 0+

0

2

xN’

(18—1)

N

where the overbar (7) means “average.” Thus we can write the force as

m

F ®i=p Nvl. We know that the square of any vector is equal to the sum of the squares of its

components (theorem of Pythagoras). Thus v* = v} + v}, + v} for any velocity . Taking averages, we obtain

o=

v+

+ ol

Since the velocities of the molecules in our gas are assumed to be random, there is no preference to one direction or another. Hence

7=

-

Combining this relation with the one just above, we get

v

= 3.

We substitute this into the equation for net force F: F

=

m

TN

v

?

q

The pressure on the wall is then

piEod Nmv? A

Or

3

Al

1 Nmv?

P = T

where

V = [A

(18-2)

is the volume of the container. This is the result we were seeking, the

pressure exerted by a gas on its container expressed in terms of molecular properties. Equation 18-2 can be rewritten in a clearer form by multiplying both sides by V and rearranging the right-hand side:

PV = iN(Em?).

(18-3)

The quantity £ ma” is the average translational kinetic energy K of the molecules in the gas. If we compare Eq. 18-3 with Eq. 174, the ideal gas law PV = NkT, we see that the two agree if

Hmi?) = KT,

TEMPERATURE RELATEDTO| AVERAGE KINETIC ENERGY OF MOLECULES

” K = im? = 3T,

, lideal gas] (18-4)

This equation tells us that the average translational kinetic energy of molecules in random motion in an ideal gas is directly proportional to the absolute temperature of the gas. The higher the temperature, according to kinetic theory, the faster the molecules a. moving on the average. This relation is one of the triumphs of the kinetic theory.

478

CHAPTER 18

Kinetic Theory of Gases

DOV

MEEI N Molecular kinetic energy. What is the average translational

kinetic energy of molecules in an ideal gas at 37°C?

APPROACH We use the absolute temperature in Eq. 18—4. fSOI.UTION We change 37°C to 310K and insert into Eq. 18—4:

K = 3kT NOTE

= 3(1.38 x 102 J/K)(310K)

= 6.42 X 1072'].

A mole of molecules would have a total translational kinetic energy equal

to (6.42 x 1072'J)(6.02 x 10%) = 3860J, a 1-kg stone traveling almost 90 m/s.

which equals the kinetic energy of

EXERCISEl In a mixture of the gases oxygen and helium, which statement is valid: (a) the helium molecules will be moving faster than the oxygen molecules, on average; (b) both kinds of molecules will be moving at the same speed; (c) the oxygen molecules will, on average, be moving more rapidly than the helium molecules; (d) the kinetic energy of the helium will exceed that of the oxygen; (e) none of the above.

Equation 18-4 holds not only for gases, but also applies reasonably accurately to liquids and solids. Thus the result of Example 18-1 would apply to molecules within living cells at body temperature (37°C). We can use Eq. 18-4 to calculate how fast molecules are moving on the average. Notice that the average in Eqs. 18-1 through 18-4 is over the square of the speed. The square root of v’ is called the root-mean-square speed, vy (since we are taking the square root of the mean of the square of the speed):

|

s

DGV

= N

EPE

=

Speeds

%1

(18-5)

of air molecules. What is the rms speed of air

molecules (O, and N,) at room temperature (20°C)?

APPROACH To obtain v, we need the masses of O, and N, molecules and then apply Eq. 18-5 to oxygen and nitrogen separately, since they have different masses. SOLUTION The masses of one molecule of O, (molecular mass = 32 u) and

N, (molecular mass = 28 u) are (where 1u = 1.66 X 10 2 kg) x 10 kg, m(0,;) = (32)(1.66 X 107kg) = 5.3

= (28)(1.66 X 10 kg) = 4.6 x 10 ®kg.

m(N;)

Thus, for oxygen Vrms

=

3kT m

=

(3)(1.38 x 1072 J/K)(293K) — (5.3 x 10 %kg)

=

480m/s,

and for nitrogen the result is v, = 510m/s. These speeds are more than 1700 km/h or 1000 mi/h, and are greater than the speed of sound ~340m/s at 20°C (Chapter 16). NOTE The speed v, is a magnitude only. The velocity of molecules averages to zero: the velocity has direction, and as many molecules move to the right as to the left, as many up as down, as many inward as outward. EXERCISE B Now you can return to the Chapter-Opening Question, page 476, and answer it correctly. Try to explain why you may have answered differently the first time. EXERCISE C If you double the volume of a gas while keeping the pressure and number of moles consl}ant, the average (rms) speed of the molecules (a) doubles, (b) quadruples,

(c) increases by V2, (d) is half, (e) is &+ |

EXERCISE q By what factor must the absolute temperature change to double vyms? (a) V2; | (b) 2; (c) 2VZ; (d) 4; (e) 16.

SECTION 18-1

The Ideal Gas Law and the Molecular Interpretation of Temperature

479

CONCEPTUAL

EXAMPLE

18-3 | Less gas in the tank. A tank of helium is used

to fill balloons. As each balloon is filled, the number of helium atoms remaining in the tank decreases. How does this affect the rms speed of molecules remaining in the tank? RESPONSE The rms speed is given temperature matters, not pressure P a constant (ambient) temperature, though the pressure of helium in the

by Eq. 18-5: v, = V3kT/m. So only tlfl or number of moles ». If the tank remains at then the rms speed remains constant even tank decreases.

In a collection of molecules, the average speed, v, is the average of the magnitudes of the speeds themselves; v is generally not equal to v.y,. To see the difference between the average speed and the rms speed, consider the following Example.

Average speed and rms speed. Eight particles have the

following speeds, given in m/s: 1.0, 6.0, 4.0, 2.0, 6.0, 3.0, 2.0, 5.0. Calculate (a) the average speed and (b) the rms speed.

APPROACH In (a) we sum the speeds and divide by N = 8. In (b) we square each speed, sum the squares, divide by N = 8, and take the square root. SOLUTION (a) The average speed is _

1.0 + 6.0 + 40

v =

+ 2.0 + 6.0 + 3.0 + 2.0 + 5.0

8

(b) The rms speed is (Eq. 18-1):

=

3.6m/s.

\/ (1.0)% + (6.0)2 + (4.0)2 + (2.0)2 + (6.0)2 + (3.0)2 + (2.0)* + (5.0)2

Vrms

8

= 4.0m/s.

m /S

We see in this Example that v and v, are not necessarily equal. In fact, for an ideal gas they differ by about 8%. We will see in the next Section how to calculate v f an ideal gas. We already have the tool to calculate v,,s (Eq. 18-5). ;

* Kinetic Energy Near Absolute Zero Equation 18-4, K = 3kT, implies that as the temperature approaches absolute zero, the kinetic energy of molecules approaches zero. Modern quantum theory, however, tells us this is not quite so. Instead, as absolute zero is approached, the kinetic energy approaches a very small nonzero minimum value. Even though all real gases become liquid or solid near 0 K, molecular motion does not cease, even at absolute zero.

18-2 FIGURE 18-2 Distribution of speeds of molecules in an ideal gas.

Note that 7 and v, are not at the peak of the curve, This is because the curve is skewed to the right: it is not symmetrical. The speed at the peak

of the curve is the “most probable

speed,” vp.

480

The Maxwell

-

Distribution

0

L1 ; : | : : : : : l ! i ] Y \vrms v

CHAPTER 18

Speed v

.

.

.

The molecules in a gas are assumed to be in random motion, which means that =~ many molecules have speeds less than the average speed and others have speeds greater than the average. In 1859, James Clerk Maxwell (1831-1879) worked out a formula for the most probable distribution of speeds in a gas containing N molecules.

We will not give a derivation here but merely quote his result:

f(v) = 4N

=

EZ 5 § % S 2 B "

Distribution of Molecular Speeds

2=

3

e P

2

(18-6) .

T

where f(v) is called the Maxwell distribution of speeds, and is plotted in Fig. 18-2. The quantity f(v) dv represents the number of molecules that have speed between v and v + dv. Notice that f(v) does not give the number of molecules with speed v; f(v) must be multiplied by dv to give the number of molecules (the number of molecules depends on the “width” or “range” of velocities included, dv). In ta formula for f(v), m is the mass of a single molecule, T is the absolute temperatu. and k is the Boltzmann constant. Since N is the total number of molecules in the gas,

Kinetic Theory of Gases

when we sum over all the molecules in the gas we must get N; thus we must have

fo}(v) dv = N.

Just as v, increases with temperature, so the whole distribution curve shifts to the

right at higher temperatures. Figure 18-3 illustrates how kinetic theory can be used to explain why many chemical reactions, including those in biological cells, take place more rapidly as the

temperature increases. Most chemical reactions take place in a liquid solution, and the

molecules in a liquid have a distribution of speeds close to the Maxwell distribution. Two molecules may chemically react only if their kinetic energy is great enough so that when th%ay collide, they partially penetrate into each other. The minimum energy required is called the activation energy, E,, and it has a specific value for each chemical reaction. The molecular speed corresponding to a kinetic energy of E, for a particular reaction is indicated in Fig. 18-3. The relative number of molecules with energy greater than this value is given by the area under the curve to the right

of v(E,), shown in Fig, 18-3 by the two different shadings. We see that the number

Relative number of molecules

0

r’roblem 22 is an exercise to show that this is true.)’ Experiments to determine the distribution of speeds in real gases, starting in the 1920s, confirmed with considerable accuracy the Maxwell distribution (for gases at not too high a pressure) and the direct proportion between average kinetic energy and absolute temperature, Eq. 18-4. The Maxwell distribution for a given gas depends only on the absolute temperature.‘ Figure 18-3 shows the distributions for two different temperatures.

T=273 K (0°C) T=310K (37°C)

Speed FIGURE 18-3

v(Ey)

Distribution of

molecular speeds for two different temperatures.

@ PHYSICS APPLIED How chemical reactions depend on lemperature

of molecules that have kinetic energies in excess of E, increases greatly for only a small increase in temperature. The rate at which a chemical reaction occurs is proportional‘to the number of molecules with energy greater than E,, and thus we see why reacfion rates increase rapidly with increased temperature.

* Calculations Using the Maxwell Distribution 't us see hoiw the Maxwell distribution can be used to obtain some interesting results.

Determining ¢ and v,. Determine formulas for (a) the

average speed, v, and (b) the most probable speed, v, of molecules in an ideal gas at temperature 7. APPROACH (@) The average value of any quantity is found by multiplying each possible value of the quantity (here, speed) by the number of molecules that have that value, and then summing all these numbers and dividing by N (the total number). For (b), we wanr to find where the curve of Fig. 18-2 has zero slope; so we set df/dv = 0. SOLUTION (a) We are given a continuous distribution of speeds (Eq. 18-6), so the sum o‘i‘fer the speeds becomes an integral over the product of v and the number f(v) dv that have speed v:

| orwyan

ov

T N

-

2J kT

4

3N

Ove

dv.

We can integrate by parts or look up the definite integral in a Table, and obtain

_

m

\2(2k*T*\

|8 kT

3

kT

”—4fl%w>(w>—v;7~1“w;

(b) The most probable speed is that speed which occurs more than any others, and thus isthat speed where f(v) has its maximum value. At the maximum of the curve, the slope is zero: df(v)/dv = 0. Taking the derivative of Eq. 18-6 gives

df(v)

|ttt e

—=

47N

( io

kT

)%

2ve

22T

—

D’

——— KT

e

a22kT

=

0.

).Solving forv, we find ‘UP

=

|2k’

_k_T_‘ m

~

141

/

H m

(Another solution is » = 0, but this corresponds to a minimum, not a maximum.)

SECTION 18-2

481

In summary, Most probable

speed, v,

vp

=

21—(mZ

Average speed, v

v

=

%%

=

3;

~

141

~

1.60

%

%

(18-7a)

“

(18-7h)

and from Eq. 18-5 rms speed, Vg

Vrms

kT

~

1.73

[kT o

These are all indicated in Fig. 18-2. From Eq. 18-6 and Fig. 18-2, it the speeds of molecules in a gas vary from zero up to many times speed, but as can be seen from the graph, most molecules have speeds far from the average. Less than 1% of the molecules exceed four times

18-3

is clear that the average that are not vypy;.

Real Gases and Changes of Phase

The ideal gas law PV

4 FIGURE 18-4 PV diagram for a real substance. Cuirves A, B, C,and

D represent the same substance at different temperatures

(Ta > Tg > Te > Tp).

482

CHAPTER 18

=

NkT

is an accurate description of the behavior of a real gas as long as the pressure is not too high and as long as the temperature is far from the liquefaction point. But what happens when these two criteria are not satisfied? First we discuss real gas behavior, and then we examine how kinetic theory can help us understand thi behavior. ; Let us look at a graph of pressure plotted against volume for a given amount of gas. On such a “PV diagram,” Fig. 18—4, each point represents an equilibrium state of the given substance. The various curves (labeled A, B, C, and D) show how the pressure varies, as the volume is changed at constant temperature, for several different values of the temperature. The dashed curve A’ represents the behavior of a gas as predicted by the ideal gas law; that is, PV = constant. The solid

curve A represents the behavior of a real gas at the same temperature. Notice that at high pressure, the volume of a real gas is less than that predicted by the ideal

gas law. The curves B and C in Fig. 18-4 represent the gas at successively lower

temperatures, and we see that the behavior deviates even more from the curves predicted by the ideal gas law (for example, B'), and the deviation is greater the closer the gas is to liquefying. To explain this, we note that at higher pressure we expect the molecules to be closer together. And, particularly at lower temperatures, the potential energy associated with the attractive forces between the molecules (which we ignored before) is no longer negligible compared to the now reduced kinetic energy of the molecules. These attractive forces tend to pull the molecules closer together so that at a given pressure, the volume is less than expected from the ideal gas law, as in Fig. 18-4. At still lower temperatures, these forces cause liquefaction, and the molecules become very close together. Section 18-5 discusses in more detail the effect of these attractive molecular forces, as well as the effect of the volume which the molecules themselves occupy. Curve D represents the situation when liquefaction occurs. At low pressure on curve D (on the right in Fig. 18-4), the substance is a gas and occupies a large volume. As the pressure is increased, the volume decreases until point b is reached. Beyond t the volume decreases with no change in pressure; the substance is gradually changing from the gas to the liquid phase. At point a, all of the substance has changed to

Kinetic Theory of Gases

liquid. Further increase in pressure reduces the volume only slightly—liquids are nearly incompressible—so on the left the curve is very steep as shown. The colored area under the dashed line represents the region where the gas and liquid hases exist together in equilibrium. Curve C in Fig. 18-4 represents the behavior of the substance at its critical temperature; the point c (the one point where curve C is horizontal) is called the critical point. At temperatures less than the critical temperature (and this is the definition of the term), a gas will change to the liquid phase if sufficient pressure is applied. Above the critical temperature, no amount of pressure can cause a gas to change phase and become a liquid. The critical temperatures for various gases are given in Table 18-1. Scientists tried for many years to liquefy oxygen without success. Only after the discovery of the behavior of substances associated with the critical point was it realized that oxygen can be liquefied only if first cooled below its critical témperature of —118°C. Often a distinction is made between the terms “gas” and “vapor”: a substance below its critical temperature in the gaseous state is called a vapor; above the critical temperature, it is called a gas. The behavior of a substance can be diagrammed not only on a PV diagram but also on a PT diagram. A PT diagram, often called a phase diagram, is particularly convenient for comparing the different phases of a substance. Figure 18-5 is the phase diagram for water. The curve labeled £-v represents those points where the liquid and vapor phases are in equilibrium—it is thus a graph of the boiling point versus pressure. Note that the curve correctly shows that at a pressure of 1 atm the boiling point is 100°C and that the boiling point is lowered for a decreased pressure. The curve s-£ represents points where solid and liquid exist in equilibrium and thus is a graph of the freezing point versus pressure. At 1atm, the freezing point of water is 0°C, as shown. Notice also in Fig. 18-5 that at a pressure of 1 atm,

the substance is in the liquid phase if the temperature is between 0°C and 100°C,

but is in the solid or vapor phase if the temperature is below 0°C or above 100°C. e curve labeled s-v is the sublimation point versus pressure curve. Sublimation .cfers to the process whereby at low pressures a solid changes directly into the vapor phase without passing through the liquid phase. For water, sublimation occurs if the pressure of the water vapor is less than 0.0060 atm. Carbon dioxide, which in the solid phase is called dry ice, sublimates even at atmospheric pressure (Fig. 18-6). The intersection of the three curves (in Fig. 18-5) is the triple point. For water this occurs at 7 = 273.16K and P = 6.03 X 10 atm. It is only at the triple point that the three phases can exist together in equilibrium. Because the triple point corresponds to a unique value of temperature and pressure, it is precisely reproduciblé and is often used as a point of reference. For example, the standard of temperature is usually specified as exactly 273.16 K at the triple point of water, rather than 273.15K at the freezing point of water at 1 atm. Notice that the solid liquid (s-£) curve for water slopes upward to the left. This is true only of substances that expand upon freezing: at a higher pressure, a lower temperature is needed to cause the liquid to freeze. More commonly, substances contract upon freezing and the s-£ curve slopes upward to the right, as shown for

TABLE 18-1 Critical

Temperatures and Pressures Critical Temperature

Substance

———————— °C K

Water CO, Oxygen Nitrogen Hydrogen Helium

374 31 —118 —147 —239.9 =267.9

~

g

Critical

Pressure (atm)

647 304 155 126 333 .53

218 72.8 50 33.5 12.8 23

Critical point P

218

i

S

|

10

|

:Gas

0.006

o | poi

0.00 0.01

100

i l

374

FIGURE 18-5 Phase diagram for water (note that the scales are not linear). FIGURE 18-6 Phase diagram for carbon dioxide. Critical point

carbon dioxide (CO,) in Fig. 18-6.

The phase transitions we have been discussing are the common ones. Some substances, however, can exist in several forms in the solid phase. A transition from one phase to another occurs at a particular temperature and pressure, just like ordinary phase changes. For example, ice has been observed in at least eight forms at very high pressure. Ordinary helium has two distinct liquid phases, called helium I and II. They exist only at temperatures within a few degrees of absolute zero. Helium IT exhibits very unusual properties referred to as superfluidity. It has essentially zero viscosity and exhibits strange properties such as climbing up the des of an open container. Also interesting are liquid crystals (used for TV and computer monitors, Section 35-11) which can be considered to be in a phase between liquid and solid.

SECTION 18-3

PHYSICS

APPLIED

Liquid crystals

Real Gases and Changes of Phase

483

18-4 Vapor Pressure and Humidity Evaporation

@rPHYSICS

APPLIED

Evaporation cools

FIGURE 18-7

Vapor appears

If a glass of water is left out overnight, the water level will have dropped t*™ morning. We say the water has evaporated, meaning that some of the water has changed to the vapor or gas phase. This process of evaporation can be explained on the basis of kinetic theory. The molecules in a liquid move past one another with a variety of speeds that follow, approximately, the Maxwell distribution. There are strong attractive forces between these

molecules, which is what keeps them close together in the liquid phase. A molecule

near the surface of the liquid may, because of its speed, leave the liquid momentarily. But just as a rock thrown into the air returns to the Earth, so the attractive forces of the other molecules can pull the vagabond molecule back to the liquid surface—that is, if its velocity is not too large. A molecule with a high enough velocity, however, will escape the liquid entirely (like an object leaving Earth with a high enough speed, Section 8-7), and become part of the gas phase. Only those molecules that have kinetic energy above a particular value can escape to the gas phase. We have already seen that kinetic theory predicts that the relative number of molecules with kinetic energy above a particular value (such as E, in Fig. 18-3) increases with temperature. This is in accord with the well-known observation that the evaporation rate is greater at higher temperatures. Because it is the fastest molecules that escape from the surface, the average speed of those remaining is less. When the average speed is less, the absolute temperature is less. Thus kinetic theory predicts that evaporation is a cooling process. You have no doubt noticed this effect when you stepped out of a warm shower and felt cold as the water on your body began to evaporate; and after working up a sweat on a hot day, even a slight breeze makes you feel cool through evaporation.

above a liquid in a closed container.

Vapor Pressure

TABLE 18-2 Saturated

from evaporation. To look at this process in a little more detail, consider a close

Vapor Pressure of Water Saturated Vapor Pressure Tempe?oté';e e ml-lg) (= 1571.12)

=50 -10

0.030 1.95

0 5 10 15 20 25 30

4.58 6.54 9.21 12.8 175 23.8 318

40

55.3

50 60

2.5;i

149

4.0

2.60 x 10° 6.11 8.72 1.23 1.71 233 317 424

x X x x X % X

10? 107 10° 103 103 10° 10°

737 %:10°

123 g x 00

199 x 10*

container that is partially filled with water (or another liquid) and from which the air has been removed (Fig. 18-7). The fastest moving molecules quickly evaporate into the empty space above the liquid’s surface. As they move about, some of these molecules strike the liquid surface and again become part of the liquid phase: this is called condensation. The number of molecules in the vapor increases until a point is

reached when the number of molecules returning to the liquid equals the number

leaving in the same time interval. Equilibrium then exists, and the space above the

liquid surface is said to be saturated. The pressure of the vapor when it is saturated is called the saturated vapor pressure (or sometimes simply the vapor pressure). The saturated vapor pressure does not depend on the volume of the container. If the volume above the liquid were reduced suddenly, the density of molecules in the vapor phase would be increased temporarily. More molecules would then be striking the liquid surface per second. There would be a net flow of molecules back to the liquid phase until equilibrium was again reached, and this would occur at the same value of the saturated vapor pressure, as long as the temperature had

1Ot changed The saturated vapor pressure of any substance depends on the temperature.

At higher temperatures, more molecules have sufficient kinetic energy to break

700

234

312 X10*

920

526

7.01 x 10*

given in Table 18-2. Notice that even solids—for example, ice—have a measurable

1.99 x 10° 476 %X 10°

In everyday situations, evaporation from a liquid takes place into the air above it rather than into a vacuum. This does not materially alter the discussion above relating to Fig. 18-7. Equilibrium will still be reached when there are sufficie molecules in the gas phase that the number reentering the liquid equals the numbe. leaving. The concentration of particular molecules (such as water) in the gas phase

80

100*

120 150

355

760

1489 3570

473 x 10

1.01 x 10°

"Boiling point on summit of Mt. Everest. “Boiling point at sea level.

484

Air normally contains water vapor (water in the gas phase), and it comes main

CHAPTER 18

from the liquid surface into the vapor phase. Hence equilibrium will be reached at

a higher pressure. The saturated vapor pressure of water at various temperatures is saturated vapor pressure.

Kinetic Theory of Gases

is not affected by the presence of air, although collisions with air molecules may

lengthen the time needed to reach equilibrium. Thus equilibrium occurs at the same

value of the saturated vapor pressure as if air were not there. If the container is large or is not closed, all the liquid may evaporate before

F aturation is reached. And if the container is not sealed—as, for example, a room

in your house—it is not likely that the air will become saturated with water vapor (unless it is raining outside).

Boiling The saturated vapor pressure of a liquid increases with temperature. When the temperature is raised to the point where the saturated vapor pressure at that temperature equals the external pressure, boiling occurs (Fig. 18-8). As the boiling

point is apbroached, tiny bubbles tend to form in the liquid, which indicate a change from the liquid to the gas phase. However, if the vapor pressure inside the bubbles is less than the external pressure, the bubbles immediately are crushed. As the temperature is increased, the saturated vapor pressure inside a bubble eventually becomes equal to or exceeds the external pressure. The bubble will then not collapse but can rise to the surface. Boiling has then begun. A liquid boils when its saturated vapor pressure equals the external pressure. This occurs for water at a pressure of 1 atm (760 torr) at 100°C, as can be seen from Table 18-2. The boiling point of a liquid clearly depends on the external pressure. At high elevations, the boiling point of water is somewhat less than at sea level since the

FIGURE 18-8 Boiling: bubbles of water vapor float upward from the bottom (where the temperature is

5

e

boiling temperatures to be attained.

highest).

,,Partial Pressure and Humidity Vhen we refer to the weather as being dry or humid, we are referring to the water vapor content of the air. In a gas such as air, which is a mixture of several types of gases, the total pressure is the sum of the partial pressures of each gas present.” By partial pressure, we mean the pressure each gas would exert if it alone were present. The partial pressure of water in the air can be as low as zero and can vary up to a maximum equal to the saturated vapor pressure of water at the given temperature. Thus, at 20°C, the partial pressure of water cannot exceed 17.5 torr (see Table‘ 18-2). The relative humidity is defined as the ratio of the partial pressure of water vapor to the saturated vapor pressure at a given temperature. It is usually expressed as a percentage: Relative humidity

=

partial pressure of H,O saturated vapor pressure of H,O

X 100%.

Thus, when the humidity is close to 100%, the air holds nearly all the water vapor it can.

DGR

Relative humidity. On a particular hot day, the temperature

is 30°C and the partial pressure of water vapor in the air is 21.0 torr. What is the relative hr,lmidity? APPROACH

From Table 18-2, we see that the saturated vapor pressure of water

at 30°C is 31.8 torr. SOLUTION

The relative humidity is thus

‘

L

21.0 torr

318 torr X 100%

=

66%.

For example, 78% (by volume) of air molecules are nitrogen and 21% oxygen, with much smaller amounts of water vapor, argon, and other gases. At an air pressure of 1atm, oxygen exerts a partial pressure of 0.21 atm and nitrogen 0.78 atm.

SECTION 18-4

Vapor Pressure and Humidity

485

@

PHYSICS APPLIED Humidity and comfor

@

PHYSICS

FIGURE 18-9

Humans are sensitive to humidity. A relative humidity of 40-50% is generally ~ optimum for both health and comfort. High humidity, particularly on a hot day, reduces the evaporation of moisture from the skin, which is one of the body’s vital mechanisms for regulating body temperature. Very low humidity, on the other hand, can dry the skin and mucous membranes. “Q APPLIED Air is saturated with water vapor when the partial pressure of water in the air is Weather equal to the saturated vapor pressure at that temperature. If the partial pressure of water exceeds the saturated vapor pressure, the air is said to be supersaturated. This situation can occur when a temperature decrease occurs. For example, suppose the temperature is 30°C and the partial pressure of water is 21 torr, which represents a humidity of 66% as we saw in Example 18-6. Suppose now that the temperature falls to, say, 20°C, as might happen at nightfall. From Table 18-2 we see that the saturated vapor pressure of water at 20°C is 17.5 torr. Hence the relative humidity would be greater than 100%, and the supersaturated air cannot hold this much water. The excess water may condense and appear as dew, or as fog or rain (Fig. 18-9). When air containing a given amount of water is cooled, a temperature is reached where the partial pressure of water equals the saturated vapor pressure. This is called the dew point. Measurement of the dew point is the most accurate means of

Fog or mist settling

around a castle where the temperature has dropped below the dew point.

determining the relative humidity. One method uses a polished metal surface in contact with air, which is gradually cooled down. The temperature at which moisture

~ begins to appear on the surface is the dew point, and the partial pressure of water

can then be obtained from saturated vapor pressure Tables. If, for example, on a

given day the temperature is 20°C and the dew point is 5°C, then the partial pressure of water (Table 18-2) in the 20°C air is 6.54 torr, whereas its saturated vapor pressure is 17.5 torr; hence the relative humidity is 6.54/17.5 = 37%.

EXERCISE E As the air warms up in the afternoon, how would the relative humidity change if there were no further evaporation? It would (a) increase, (b) decrease, (c) stay the same.

CONCEPTUAL EXAMPLE 18-7 | Dryness in winter. Why does the air inside heated buildings seem very dry on a cold winter day? RESPONSE Suppose Table 18-2 tells us this air is brought (1.0torr)/(17.5 torr) pressure of 1.95 torr,

the relative humidity outside on a —10°C day is 50%. the partial pressure of water in the air is about 1.0 torr. If indoors and heated to +20°C, the relative humidity is = 5.7%. Even if the outside air were saturated at a partial the inside relative humidity would be at a low 11%.

*18-5 Van der Waals Equation of State

FIGURE 18-10

radius r, colliding.

Molecules, of

In Section 18-3, we discussed how real gases deviate from ideal gas behavior, particularly at high densities or when near condensing to a liquid. We would like to understand these deviations using a microscopic (molecular) point of view. J. D. van der Waals (1837-1923) analyzed this problem and in 1873 arrived at an equation of state which fits real gases more accurately than the ideal gas law. His analysis is based on kinetic theory but takes into account: (1) the finite size of molecules (we previously neglected the actual volume of the molecules themselves, compared to the total volume of the container, and this assumption becomes poorer as the density increases and molecules become closer together); (2) the range of the forces between molecules may be greater than the size of the molecules (we previously assumed that intermolecular forces act only during collisions, when the molecules are “in contact”). Let us now look at this analysis and derive the van der Waals equation of state. Assume the molecules in a gas are spherical with radius r. If we assume these molecules behave like hard spheres, then two molecules collide and bounce off one another if the distance between their centers (Fig. 18-10X gets as small as 2r. Thus the actual volume in which the molecules can move about

486

CHAPTER 18

is somewhat

less than

the

volume

V of the

container

holding

the

gas.

The amount of “unavailable volume” depends on the number of molecules and on their size. Let b represent the “unavailable volume per mole” of gas.

Then in the ideal gas law we replace V by (V' — nb), where n is the number of moles, and we obtain

\ P(V — nb) = nRIT.

f@ we divide through by n, we have |4 P(—— = b) = RT. (18-8) n This relation (sometimes called the Clausius equation of state) predicts that for a given tempe&ature T and volume V, the pressure P will be greater than for an ideal gas. This makes sense since the reduced “available” volume means the number of collisions with the walls is increased. Next we consider the effects of attractive forces between molecules, which are responsible for holding molecules in the liquid and solid states at lower temperatures. These forces are electrical in nature and although they act even when molecules are not touching, we assume their range is small—that is, they act mainly between nearest neighbors. Molecules at the edge of the gas, headed toward a wall of the container, are slowed down by a net force pulling them back into the gas. Thus these molecules will exert less force and less pressure on the wall than if there were no attractive forces. The reduced pressure will be proportional to the density of molecules in the layer of gas at the surface, and also to the density in the next layer, which exerts the inward force.” Therefore we expect the pressure to be reduced by a factor proportional to the density squared (n/V)? here written as moles per volume. If the pressure P is given by Eq. 18-8, then we should reduce this by an amount a(n/V)? where a is a proportionality constant. Thus we have

p =

or

RT

(V/n) = b

__a

(V/n)?

(P + (V;ln)2>

and

b = 4.3 X 1075 m*/mol.

Figure 18-11 shows a typical PV diagram for Eq. 18-9 (a “van der Waals gas”) for four different temperatures, with detailed caption, and it should be compared to Fig. 18-4 for real gases. Neither the van der Waals equation of state nor the many other equations of state that have been proposed are accurate for all gases under all conditions. Yet Eq. 18-9 is a very useful relation. And because it is quite accurate for many situations, its derivation gives us further insight into the nature of gases at the microscopic level. Note that at low densities, a/(V /n)?> X ~

K

RS Calorimetry

¢

1. Be sure you have sufficient information to apply energy conservation. Ask yourself: is the system isolated (or very nearly so, enough to get a good estimate)? Do we know or can we calculate all significant sources of energy transfer? . Apply conservation of energy: heat gained = heat lost. For each substance in the system, a heat (energy) term will appear on either the left or right side of this equation. [Alternatively, use ZQ = 0.] . If no phase changes occur, each term in the energy conservation equation (above) will have the form

o

2

¢

or

Q(gain)

of the

.

.

.

= mc(Ty — T)

Q(lost) = me(T; - T))

where 7; and T are the initial and final temperatures

7.

substance,

and

Determining

a latent

heat. The

mercury is 140J/kg-C°. When 1.0kg —39°C is placed in a 0.50-kg aluminum 20.0°C, the mercury melts and the final to be 16.5°C. What is the heat of fusion APPROACH

SOLUTION

m

and

¢ are

its mass

and

specific heat, respectively. If phase changes do or might occur, there may be terms in the energy conservation equation of the form Q = mL, where L is the latent heat. But before applying energy conservation, determine (or estimate) in which phase the final state will be, as we did in Example 19-5 by calculating the different contributing values for heat Q. Be sure each term appears on the correct side of the energy equation (heat gained or heat lost) and that each AT is positive. Note that when the system reaches thermal equilibrium, the final temperature of each substance will have the same value. There is only one 7;. Solve your energy equation for the unknown.

specific heat of liquid

of solid mercury at its melting point of calorimeter filled with 1.2 kg of water at temperature of the combination is found of mercury in J/kg?

We follow the Problem Solving Strategy above.

1. Is the system isolated? The mercury is placed in a calorimeter, which we assume is well insulated. Our isolated system is the calorimeter, the water, and the mercury. 2. Conservation of energy. The heat gained by the mercury = the heat lost by the water and calorimeter. 3. and 4. Phase changes. There is a phase change (of mercury), plus we use specific heat equations. The heat gained by the mercury (Hg) includes a term representing the melting of the Hg, O(melt solid Hg)

=

myg Ly,

plus a term representing the heating of the liquid Hg from —39°C to +16.5°C:

= mygcpg[16.5°C — (—39°C)] Il

QO(heat liquid Hg)

(1.0kg)(140J/kg-C°)(55.5C°)

All of this heat gained by the mercury calorimeter, which cool down:

is obtained

from

= 77701. the water

and

Ocal + Ow = Mea1Cal(20.0°C — 16.5°C) + mycy(20.0°C — 16.5°C) = (0.50kg)(900J/kg- C°)(3.5C°) + (1.2kg)(4186J/kg-C°)(3.5C°) = 19,2007.

5. Energy equation. The conservation of energy tells us the heat lost by the water and calorimeter cup must equal the heat gained by the mercury: Qcal T Ow = QO(melt solid Hg) + Q(heat liquid Hg) or

19,200) = Mg Ly + 77701 6. Equilibrium temperature. It is given as 16.5°C, and we already used it.

504

CHAPTER 19

Heat and the First Law of Thermodynamics

7. Solve. The only unknown in our latent heat of fusion (or melting) myg = 1.0kg: r

— 77701 Ly, = 19,200)T 0kg

energy equation (point 5) is Ly, the of mercury. We solve for it, putting in

= 11,400J/kg

~ 11kJ/kg,

where we rounded off to 2 significant figures. Evaporation The latent heat to change a liquid to a gas is needed not only at the boiling point. Water can change from the liquid to the gas phase even at room temperature. This process is called evaporation (see also Section 18-4). The value of the heat of vaporization of water increases slightly with a decrease in temperature: at 20°C, for example, it is 2450 kJ/kg (585 kcal/kg) compared to 2260 kJ/kg (= 539 kcal/kg) at 100°C. When water evaporates, the remaining liquid cools, because the energy required (the latent heat of vaporization) comes from the water itself; so its internal energy, and therefore its temperature, must drop.’ Evaporation of water from the skin is one of the most important methods the body uses to control its temperature. When the temperature of the blood rises slightly above normal, the hypothalamus region of the brain detects this temperature mcrease and sends a signal to the sweat glands to increase their production. The energy‘ (latent heat) required to vaporize this water comes from the body, and hence the body cools.

@ PHYSICS Body temperature

APPLIED

Kinetic Theory of Latent Heats We can make use of kinetic theory to see why energy is needed to melt or vaporize a substance. At the melting point, the latent heat of fusion does not act to increase the average kinetic energy (and the temperature) of the molecules in the solid, but r\nstead is used to overcome the potential energy associated with the forces between the molecules. That is, work must be done against these attractive forces to break the molecules loose from their relatively fixed positions in the solid so they can freely roll over one another in the liquid phase. Similarly, energy is required for molecules held close together in the liquid phase to escape into the gaseous phase. This process is a more violent reorganization of the molecules than is melting (the average distance between the molecules is greatly increased), and hence the heat of vaporization is generally much greater than the heat of fusion for a given substance.

19-6 The First Law of Thermodynamics Up to now‘ in this Chapter we have discussed internal energy and heat. But work too is often involved in thermodynamlc processes. In Ch%pter 8 we saw that work is done when energy is transferred from one object to 1nother by mechanical means. In Section 19-1 we saw that heat is a transfer of energy from one object to a second one at a lower temperature. Thus, heat is much like work. To distinguish them, heat is defined as a transfer of energy due to a dtfference in temperature, whereas work is a transfer of energy that is not due to a temperature difference. In Section 19-2, we defined the internal energy of a system as the sum total of all the energy of the molecules within the system. We would expect that the internal energy of a system would be increased if work was done on the system, or if heat were added to it. Similarly the internal energy would be decreased if heat flowed out of the system or if work were done by the system on something in the surroundings.

r

"According to kinetic theory, evaporation is a cooling process because it is the fastest-moving molecules that escape from the surface. Hence the average speed of the remaining molecules is less, so by Eq. 18—4 the temperature is less.

SECTION 19-6

The First Law of Thermodynamics

505

Thus it is reasonable to extend conservation of energy and propose an important law: the change in internal energy of a closed system, AE;;, will be equal to the energy added to the system by heating minus the work done by the system on the surroundings. In equation form we write FIRST LAW OF

K

THERMODYNAMICS A

CAUTION

Heat added is + Heat lost is — Work on system is —

Work by system is =

At where

i

-

-4

Q is the net heat added to the system and W is the net work done by the

system.” We must be careful and consistent in following the sign conventions for Q and W. Because W in Eq. 19-4 is the work done by the system, then if work is done on the system, W will be negative and E;, will increase. Similarly, Q is

sositive for heat added to the system, so if heat leaves the system, Q is negative.

Equation 19-4 is known as the first law of thermodynamics. It is one of the great laws of physics, and its validity rests on experiments (such as Joule’s) to which no exceptions have been seen. Since Q and W represent energy transferred into or out of the system, the internal energy changes accordingly. Thus, the first law of thermodynamics is a great and broad statement of the law of conservation of energy. It is worth noting that the conservation of energy law was not formulated until the nineteenth century, for it depended on the interpretation of heat as a transfer of energy. Equation 19-4 applies to a closed system. It also applies to an open system (Section 19-4) if we take into account the change in internal energy due to the increase or decrease in the amount of matter. For an isolated system (p. 500), no work is done and no heat enters or leaves the system,so W = Q = 0, and hence AEint

=0.

A given system at any moment is in a particular state and can be said to have a certain amount of internal energy, E;,;. But a system does not “have” a certain amount of heat or work. Rather, when work is done on a system (such as compressing a gas), or when heat is added or removed from a system, the state of the system changes. Thus, work and heat are involved in thermodynamic processes that can change the system from one state to another; they are not characteristic O the state itself. Quantities which describe the state of a system, such as interna. energy Ej,., pressure P, volume V, temperature 7, and mass m or number of moles n, are called state variables. Q and W are not state variables. Because Ej, is a state variable, which depends only on the state of the system and not on how the system arrived in that state, we can write AE,

where and Q going It

=

int,2

Eint.l

=Q0-W

E;,; ; and E;, , represent the internal energy of the system in states 1 and 2, and W are the heat added to the system and work done by the system in from state 1 to state 2. is sometimes useful to write the first law of thermodynamics in differential form:

dE, = dQ — dW. Here, dE;, represents an infinitesimal change in internal energy when an infinitesimal amount of heat dQ is added to the system, and the system does an infinitesimal

amount of work dW.*

"This convention relates historically to steam engines: the interest was in the heat input and the work output, both regarded as positive. In other books you may see the first law of thermodynamics written as

AEj,

= Q + W,

in which case W refers to the work done on the system.

*The differential form of the first law is often written dEyn = dQ — aw, where the bars on the differential sign (&) are used to remind us that W and Q are not functions of the state variables (such as P, V, T, n). Internal energy, Ejy, is a function of the state variables, and so dEi,“ represents the differential (called an exact differential) of some function Ej,,. The differentials W an. @Q are not exact differentials (they are not the differential of some mathematical function); they thus only represent infinitesimal amounts. This issue won’t really be of concern in this book.

506

CHAPTER 19

Heat and the First Law of Thermodynamics

Using the first law. 25007 of heat is added to a system, and

1800 J of work is done on the system. What is the change in internal energy of the system? fAPPROACH We apply the first law of thermodynamics, Eq. 19-4, to our system. SOLUTION The heat added to the system is Q = 2500J. The work W done by the system is —1800J. Why the minus sign? Because 1800 ] done on the system (as given) equals —1800J done by the system, and it is the latter we need for the sign conventions we used in Eq. 19-4. Hence AEy,, = 2500J — (—1800J) = 2500J + 1800J = 43001J. You may have intuitively thought that the 2500 J and the 1800 J would need to be added together, since both refer to energy added to the system. You would have been right. EXERCISE D What would be the internal energy change in Example 19-7 if 2500 J of heat is added to the system and 18007 of work is done by the system (i.e., as output)?

*The First Law of Thermodynamics Extended To write the first law of thermodynamics in its complete form, consider a system that has kinetic energy K (there is motion) as well as potential energy U. Then the first law of thermodynamics would have to include these terms and would be written as

AK + AU + AEy, = Q — W.

(19-5)

Kinetic energy transformed to thermal energy. A 3.0-g bullet

traveling at a speed of 400 m/s enters a tree and exits the other side with a speed of 200 m/s. Where did the bullet’s lost kinetic energy go, and what was the energy transferred? APPROACH Take the bullet and tree as our system. No potential energy is involved. ANo work is done on (or by) the system by outside forces, nor is any heat added because no energy was transferred to or from the system due to a temperature difference. Thus the kinetic energy gets transformed into internal energy of the bullet and tree. SOLUTION From the first law of thermodynamics as given in Eq. 19-5, we are given Q = W = AU = 0, so we have AK + AE, = 0 or, using subscripts i and f for initial and final velocities

AE,

=

—-AK

=

—(K; - K))

= im(v} — v})

= 1(3.0 x 10 kg)[(400m/s)?> — (200m/s)?]

= 18017.

NOTE The internal energy of the bullet and tree both increase, as both experience a rise in temperature. If we had chosen the bullet alone as our system, work would be done on it and heat transfer would occur.

19-7

The First Law of Thermodynamics

Applied; Calculating the Work

Let us analyze some simple processes in the light of the first law of thermodynamics.

Isothermal Processes (AT = 0)

First we consider an idealized process that is carried out at constant temperature.

r

Such a process temperature”). fixed amount process follows curve for PV the state of the

is called an isothermal process (from the Greek meaning “same If the system is an ideal gas, then PV = nRT (Eq.17-3),so for a of gas kept at constant temperature, PV = constant. Thus the a curve like AB on the PV diagram shown in Fig. 19-6, which is a = constant. Each point on the curve, such as point A, represents system at a given moment—that is, its pressure P and volume V. At

lower temperature, another isothermal process would be represented by a curve

like A'B’ in Fig. 19-6 (the product PV = nRT = constant is less when T is less). The curves shown in Fig. 19-6 are referred to as isotherms.

SECTION 19-7

FIGURE 19-6

PV diagram for an

ideal gas undergoing isc%trhermal

~ processes at two different temperatures. p A N

0

B Y

Higher T

Lower T

Vv

The First Law of Thermodynamics Applied; Calculating the Work

507

FIGURE 19-7 Anideal gasina cylinder fitted with a movable piston.

Let us assume that the gas is enclosed in a container fitted with a movable piston, Fig. 19-7, and that the gas is in contact with a heat reservoir (a body whose mass is so large that, ideally, its temperature does not change significantly when heat is exchanged with our system). We also assume that the process of compression (volume decreases) or expansion (volume increases) is done quasistatically (“almos statically”), by which we mean extremely slowly, so that all of the gas moves between’ a series of equilibrium states each of which are at the same constant temperature. If an amount of heat Q is added to the system and temperature is to remain constant, the gas will expand and do an amount of work W on the environment (it exerts a force on the piston and moves it through a distance). The temperature and mass are kept constant so, from Eq. 19-1, the internal energy does not change: AE;,, = 3nR AT = 0. Hence, by the first law of thermodynamics, Eq. 19-4, AE, =Q - W =0, so W=0: the work done by the gas in an isothermal ~ process equals the heat added to the gas.

Adiabatic Processes (Q = 0) An adiabatic process is one in which no heat is allowed to flow into or out of the system: Q = 0. This situation can occur if the system is extremely well insulated, or the process happens so quickly that heat—which flows slowly—has no time to flow in or out. The very rapid expansion of gases in an internal combustion engine is one example of a process that is very nearly adiabatic. A slow adiabatic expansion of an

A

P

Isothermal

ideal gas follows a curve like that labeled AC in Fig. 19-8. Since Q = 0, we have

B Adiabatic N ¢

0 4 FIGURE 19-8 PV diagram for adiabatic (AC) and isothermal (AB)

processes on an ideal gas.

FIGURE 19-9

0

A

(a) Isobaric (“same

B

—

P|

v

0

(a) Isobaric

In a diesel engine, the fuel-air mixture is rapidly compressed adiabatically by a factor

of 15 or more; the temperature rise is so great that the mixture ignites spontaneously

pressure”) process. (b) Isovolumetric (“same volume”™) process.

Pl

from Eq. 19-4 that AE;,; = —W. That is, the internal energy decreases if the gas expands; hence the temperature decreases as well (because AE;, = 3nR AT). This is evident in Fig. 19-8 where the product PV (= nRT) is less at point C than at point B (curve AB is for an isothermal process, for which AE;; = 0 and AT = 0). In the reverse operation, an adiabatic compression (going from C to A, for example), work is done on the gas, and hence the internal energy increases and the temperature rises.

l

#A B

Isobaric and Isovolumetric Processes Isothermal and adiabatic processes are just two possible processes that can occur. Two other simple thermodynamic processes are illustrated on the PV diagrams of Fig. 19-9: (a) an isobaric process is one in which the pressure is kept constant, so

the process is represented by a horizontal straight line on the PV diagram,

%

(b) Isovolumetric

Fig. 19-9a; (b) an isovolumetric (or isochoric) process is one in which the volume does not change (Fig. 19-9b). In these, and in all other processes, the first law of

thermodynamics holds.

Work Done in Volume Changes FIGURE 19-10 The work done by a gas when its volume increases by dV =Adl is dW = PdV.

We often want to calculate the work done in a process. Suppose we have a gas confined to a cylindrical container fitted with a movable piston (Fig. 19-10). We must always be careful to define exactly what our system is. In this case we choose our system to be the gas; so the container’s walls and the piston are parts of the environment. Now let us calculate the work done by the gas when it expands quasistatically, so that P and T are defined for the system at all instants.” The gas expands against the piston, whose area is A. The gas exerts a force F = PA on the piston, where P is the pressure in the gas. The work done by the gas to move the piston an infinitesimal displacement dZ is

dW

= F-dl

= PAdl = PdV

(19-6)

since the infinitesimal increase in volume is dV = A dl. If the gas was compressed so that df pointed into the gas, the volume would decrease and dV < 0. The work done by the gas in this case would then be negative, which is equivalent to saying that positive work was done on the gas, not by it. For a finite change in volumc“

508

CHAPTER

19

"If the gas expands or is compressed quickly, there would be turbulence and different parts would be at

different pressure (and temperature).

from Vj to J/B, the work W done by the gas will be W

=

JdW = J

Vs Va

Pav.

19-7)

.quations 19-6 and 197 are valid for the work done in any volume change—by a gas, a liquidL or a solid—as long as it is done quasistatically. In order to integrate Eq. 19-7, we need to know how the pressure varies during the process, and this depends on the type of process. Let us first consider a quasistatic isothermal expansion of an ideal gas. This process is represented by the curve between points A and B on the PV diagram of Fig. 19-11. The work done by the gas in this process, according to Eq. 19-7, is just the area between the PV curve and the V axis, and is shown shaded in Fig. 19-11. We can do the integral in Eq. 19-7 for an ideal gas by using the ideal gas law, P = nRT/V. The work done at constant 7' is

e

dv

JVB q dV v,

= nRT J " av WV

= uRTIn Vs, Va

[isothermal process;] ideal gas

(19-8)

FIGURE 19-11

Work done by an

ideal gasin an isothermal process equals the area under the PV curve.

Let us next consider a different way of taking an ideal gas between the same states A and B. This time, let us lower the pressure in the gas from P, to Py, as

tsfiladed a}rlea etquals tl:ie ;rvork‘(}otne‘l/)y 808 WUsnL expanaat A 10 Ve

be allowed to flow out of the gas so its temperature drops.) expand from V, to Vj at constant pressure (= Pg), which is line DB in Fig. 19-12. (In this isobaric process, heat is added to temperature.) No work is done in the isovolumetric process AD,

FIGURE 19-12 Process ADB consists of an isovolumetric (AD) and an isobaric (DB) process.

indicated by

the line AD

W

=

in Fig. 19-12. (In this isovolumetric process, heat must

0.

Then let indicated the gas to since dV

the gas by the raise its = 0:

[isovolumetric process]

In the isobaric process DB the pressure remains constant, so

W

=

J

Vs Va

Pdv

= Py(Vg — Vu) = PAV.

[isobaric process]

(19-9a)

The work done is again represented on the PV diagram by the area between the rve (ADB) and the V axis, as indicated by the shading in Fig. 19-12. Using the ideal gas law, we can also write

W

=_

PB(VB -V

=_

nRTB(l

- _‘fg VB>'

isobaric process;

[ideal gas

(19-9b)

As can be seen from the shaded areas in Figs. 19-11 and 19-12, or by putting in numbers in Egs. 19-8 and 19-9 (try it for Vz = 2V} ), the work done in these two processes is different. This is a general result. The work done in taking a system from one state to another depends not only on the initial and final states but also on the type of process (or “path”). This result reemphasizes the fact that work cannot be considered a property of a system. The same is true of heat. The heat input required to change the gas from state A to state B depends on the process; for the isothermal process of Fig. 19-11, the heat input turns out to be greater than for the process ADB of Fig. 19-12. In general, the amount of heat added or removed in taking a system from one state to another depends not only on the initial and final states but also on the path or process.

CONCEPTUAL

EXAMPLE

19-9 | Work in isothermal and adiabatic processes.

In Fig. 19-8 we saw the PV diagrams for a gas expanding in two ways, isothermally and adiabatically. The initial volume V, was the same in each case, and the final

volumes were the same (Vi = V(). In which process was more work done by the gas?

RESPONSE Our system is the gas. More work was done by the gas in the isothermal process, which we can see in two simple ways by looking at Fig. 19-8. First, the “average” pressure was higher during the isothermal process AB, so AW = PAV was greater (AV is the same for both processes). Second, we can look at the area under each curve: the area under curve AB, which represents the work done, was greater (since curve AB is higher) than that under AC. J

EXERCISE E Is the work done by the gas in process ADB of Fig, 19-12 greater than, less than, or equal to the work done in the isothermal process AB?

SECTION

19-7

509

First law in isobaric and isovolumetric processes. An ideal

Isovolumetric

1 T

>

|

L1~ >

gas is slowly compressed at a constant pressure of 2.0 atm from 10.0L to 2.0 L. This process is represented in Fig. 19-13 as the path B to D. (In this process, some heat flows out of the gas and the temperature drops.) Heat is then added to the gas, holdin the volume constant, and the pressure and temperature are allowed to rise (line DA until the temperature reaches its original value (7, = Tg). Calculate (a) the total work done by the gas in the process BDA, and (b) the total heat flow into the gas.

Isothermal

|

0

de

T

2

APPROACH (a) Work is done only in the compression process BD. In process DA, the volume is constant so AV = 0 and no work is done. (b) We use the first 1

4

FIGURE 19-13

|

6

I

8

10 V(L)

Example 19-10.

law of thermodynamics, Eq. 19-4.

SOLUTION

(a) During

the

compression

BD,

the

pressure

is

2.0atm =

2(1.01 x 10°N/m?) and the work done is (since 1L = 10°cm® = 1073 m?) W = PAV = (202 X 10°N/m*)(2.0 X 10°m® — 10.0 x 107 m’) =

—-1.6 X 10°J.

The total work done by the gas is —1.6 X 10°J, where the minus sign means that

+1.6 X 10°J of work is done on the gas.

(b) Because the temperature at the beginning and at the end of process BDA is the same, there is no change in internal energy: AE;,; = 0. From the first law of thermodynamics we have 0

=

AE,

=

Q-

W

so Q=W = —1.6 X 10°]. Because Q is negative, 1600J of heat flows out of the gas for the whole process, BDA. EXERCISE F In Example 19-10, if the heat lost from the gas in the process 8.4 x 10°J, what is the change in internal energy of the gas during process BD?

BD

is

Work done in an engine. In an engine, 0.25mol of an

ideal monatomic gas in the cylinder expands rapidly and adiabatically against themy piston. In the process, the temperature of the gas drops from 1150K to 400 K. How much work does the gas do? APPROACH We take the gas as our system (the piston is part of the surroundings). The pressure is not constant, and its varying value is not given. Instead, we can use the first law of thermodynamics because we can determine AE;, given Q = 0 (the process is adiabatic). SOLUTION We determine AE;, from Eq. 19-1 for the internal energy of an ideal monatomic gas, using subscripts f and i for final and initial states:

AEy

= Ept — Emi

= 3nR(T; — T)

= 3(0.25 mol)(8.314 J/mol -K)(400K — 1150K) =

—23001.

Then, from the first law of thermodynamics, Eq. 19-4, W = Q — AE,; = 0 — (-2300J) = 23001J. Table 19-3 gives a brief summary of the processes we have discussed.

Free Expansion One type of adiabatic process is a so-called free expansion in which a gas is allowed to expand in volume adiabatically without doing any work. The apparatus to

TABLE 19-3 Simple Thermodynamic Processes and the First Law Process

510

CHAPTER 19

What is constant:

The first law predicts:

Isothermal

T = constant

AT = 0 makes

Isobaric

P = constant

Q = AEy

Isovolumetric Adiabatic

V = constant Q=0

AV =0 makes AE, = —W

+ W

AEj;y =0,

s0

Q=W

= AEj,; + PAV

W =0,

so Q = AEjy;

accomplish a free expansion is shown in Fig. 19-14. It consists of two well-insulated compartments (to ensure no heat flow in or out) connected by a valve or stopcock. One compartment is filled with gas, the other is empty. When the valve is opened, the gas expands to fill both containers. No heat flows in or out (Q = 0), and no r’vork is dojxe because the gas does not move any other object. Thus Q = W = 0 and by the first law of thermodynamics, AE;,; = 0. The internal energy of a gas does not change in a free expansion. For an ideal gas, AT = 0 also, since E;,; depends only on 7 (Section 19-2). Experimentally, the free expansion has been used to determine if the internal energy of real gases depends only on 7. The experiments are very difficult to do accurately, but it has been found that the temperature of a real gas drobas very slightly in a free expansion. Thus the internal energy of real gases does depend a little, on pressure or volume as well as on temperature. A free expansion can not be plotted on a PV diagram, because the process is rapid, not qua51stat1c The intermediate states are not equ111br1um states, and hence the pressure (and even the volume at some instants) is not clearly defined.

19-8

= FIGURE 19-14

Free expansion.

Molar Specific Heats for Gases,

and the Equipartition of Energy

In Section 19-3 we discussed the concept of specific heat and applied it to solids and liquids. Much more than for solids and liquids, the values of the specific heat for gases depends on how the process is carried out. Two important processes are those in which either the volume or the pressure is kept constant. Although for solids and liquids it matters little, Table 19-4 shows that the specific heats of gases

at constant volume (cy) and at constant pressure (c,) are quite different. Molar Specific Heats for Gases

The difference in specific heats for gases is nicely explained in terms of the first w of thermodynamics and kinetic theory. Our discussion is simplified if we use molar pecific heats, C,, and Cp, which are defined as the heat required to raise 1 mol of the gas by 1 C° at constant volume and at constant pressure, respectively. That is, in analogy to Eq. 19-2, the heat Q needed to raise the temperature of n moles of gas by AT is Q

=

nCyAT

[volume constant]

(19-10a)

QO

=

nCpAT.

[pressure constant]

(19-10b)

It is clear from the definition of molar specific heat (or by comparing Egs. 19-2 and 19-10) that CV

=

MCV

CP

=

MCP,

where M is the molecular mass of the gas (M = m/n in grams/mol). The values for molar specific heats are included in Table 19-4, and we see that the values are nearly

the same for different gases that have the same number of atoms per molecule.

TABLE 19-4 Specific Heats of Gases at 15°C Specific heats

Gas

(keal/kg - K)

ey

cp

Molar specific heats

(cal/mol - K)

Gy

)

Cp= Cy

ol

_Cp

Ky = LGy

Monatomic He

Ne Diatomic N2 r

0.75

1.15

298

497

1.99

1.67

0.148

0.246

298

497

1.99

1.67

0.177

0.248

496

6.95

1.99

1.40

0.155

0.218

5.03

7.03

2.00

1.40

0.153 0.350

0.199 0.482

6.80 6.20

8.83 820

2.03 2.00

1.30 1:32

Trlatomlc

CO, H,0 (100°C)

SECTION 19-8

511

Now we use kinetic theory and imagine that an ideal gas is slowly heated via

two different processes—first at constant volume, and then at constant pressure. In

both of these processes, we let the temperature increase by the same amount, AT. In the process done

at constant volume, no work

is done

since

AV = 0.

Thus,

according to the first law of thermodynamics, the heat added (which we denote t*™ Qy) all goes into increasing the internal energy of the gas: Ov

=

AEy,.

In the process carried out at constant pressure, work is done, and hence the heat added, Qp, must not only increase the internal energy but also is used to do the

work

W

= PAV.

Thus, more

heat

must

be

added

in this process

at constant

pressure than in the first process at constant volume. For the process at constant pressure, we have from the first law of thermodynamics Op = AE,,; + PAV. Since AE;, is the same in the two processes (AT was chosen to be the same), we can combine the two above equations:

Qr — Qy

= PAV.

nCpAT

nCVAT

From the ideal gas law, V = nRT /P, so for a process at constant pressure we have AV = nR AT/P. Putting this into the above equation and using Egs. 19-10, we find =

=

P(

nRAT

P or, after cancellations, Cpr— Cy = R (19-11) Since the gas constant R = 8.314 J/mol-K = 1.99 cal/mol-K, our prediction is that Cp will be larger than Cy, by about 1.99 cal/mol - K. Indeed, this is very close to what is obtained experimentally, as shown in the next to last column of Table 19-4. Now we calculate the molar specific heat of a monatomic gas using kinetic theory. In a process carried out at constant volume, no work is done; so the first law of thermodynamics tells us that if heat Q is added to the gas, the internq\ energy of the gas changes by AEint

=

0.

For an ideal monatomic gas, the internal energy Ej, is the total kinetic energy of all the molecules,

En = N@Emv’) = 3nRT

as we saw in Section 19-2. Then, using Eq. 19-10a, we can write AE;,; = Q in the form

or

AEy,

= 3nRAT

Cy = 3R

= nCyAT

(19-12)

(19-13)

Since R = 8.314J/mol-K = 1.99 cal/mol-K, kinetic theory predicts that C, = 2.98 cal/mol-K for an ideal monatomic gas. This is very close to the experimental values for monatomic gases such as helium and neon (Table 19-4). From Eq. 19-11, Cp is predicted to be about 4.97 cal/mol - K, also in agreement with experiment.

Equipartition of Energy FIGURE 19-15 A diatomic molecule can rotate about two different axes.

Axis

The measured molar specific heats for more complex gases (Table 19-4), such as diatomic (two atoms) and triatomic (three atoms) gases, increase with the increased number of atoms per molecule. We can explain this by assuming that the internal energy includes not only translational kinetic energy but other forms of energy as well. Take, for example, a diatomic gas. As shown in Fig. 19-15 the two atoms can

rotate about two different axes (but rotation about a third axis passing through the two atoms would give rise to very little energy since the moment of inertia is so small). The molecules can have rotational as well as translational kinetic energy. It is useful to introduce the idea of degrees of freedom, by which we mean the number of

512

CHAPTER 19

independent ways molecules can possess energy. For example, a monatomic gas is sai to have three degrees of freedom, since an atom can have velocity along the x axis, th. y axis, and the z axis. These are considered to be three independent motions because a change in any one of the components would not affect any of the others. A diatomic molecule has the same three degrees of freedom associated with translational kinetic

energy plus two more degrees of freedom associated with rotational kinetic energy, for a total of five degrees of freedom. A quick look at Table 19-4 indicates that the Cy for diatomic gases is about 3 times as great as for a monatomic gas — that is, in the same ratio as their degrees of freedom. This result led nineteenth-century physicists ) an important idea, the principle of equipartition of energy. This principle states that energy is shared equally among the active degrees of freedom, and in particular each active degree of freedom of a molecule has on the average an energy equal to LkT. Thus, the average energy for a molecule of a monatomic gas would be 3 kT (which we already knew) and of a diatomic gas 3 k7. Hence the internal energy of a diatomic gas would be Ej, = N(% kT) = 3nRT, where n is the number of moles. Using the same argument we did for monatomic gases, we see that for diatomic gases the molar specific heat at constant volume would be 3R = 4.97 cal/mol K, in accordance with measured values. More complex molecules have even more degrees of freedom and thus greater molar specific heats. The situation was complicated, however, by measurements that showed that for diatomic gases at very low temperatures, Cy, has a value of only 3 R, as if it had only three degrees of freedom. And at very high temperatures, C,, was about 3 R, as if there were seven degrees of freedom. The explanation is that at low temperatures, nearly all molecules have only translational kinetic energy. That is, no energy goes into rotational energy, so o?tly three degrees of freedom are “active.” At very high temperatures, on the other hand all five degrees of freedom are active plus two additional ones. We can interpret the two new degrees of freedom as being associated with the two atoms vibrating as 1‘f they were connected by a spring, as shown in Fig. 19-16. One degree of freedom comes from the kinetic energy of the vibrational motion, and the second comes from the potential energy of vibrational motion (3 kx?). At room temperature, these two degrees of freedom are apparently not active. See Fig, 19-17.

A diatomic FIGURE 19-16 molecule can vibrate, as if the two

atoms were connected by a spring. Of course they are not connected by a spring; rather they exert forces on each other that are electrical in nature, but of a form that resembles a spring force.

4R +Cy -

f

3R +

0 20

R

---------------- S

2R + R T

L

Vibrational energy

Rotational kinetic energy

T T

T

‘

Translational kinetic energy l 1 f f 50

100

200

+

500

g

t

1000

t

2000

+

5000

e =

10,000

T"(K)

FIGURE 19-17 Molar specific heat Cy as a function of temperature for hydrogen molecules (H,). As the temperature is increased, some of the translational kinetic energy can be transferred in collisions into rotational kinetic energy and, at still higher temperature, into vibrational energy. [Note: H; dissociates into two atoms at about 3200 K, so the last part of the curve is shown dashed.]

Just why fewer degrees of freedom are “active” at lower temperatures was eventually explained by Einstein using the quantum theory. [According to quantum theory, energy does not take on continuous values but is quantized —it can have only certain values, and there is a certain minimum energy. The minimum rotational and vibrational energies are higher than for simple translational kinetic energy, so at lower temperatures and lower translational kinetic energy, there is not enough energy to excite the rotational or vibrational kinetic energy.] Calculations based on kinetic theory and the principle of equipartition of energy (as modified by the quantum thTory) give numerical results in accord with experiment.

FIGURE 19-18 Molar specific heats of solids as a function of temperature.

g

% _

*Solids The principle of equipartition of energy can be applied to solids as well. The molar specific heat of any solid, at high temperature, is close to 3R (6.0 cal/mol-K), Fig. 19-18. This is called the Dulong and Petit value after the scientists who first measured it in 1819. (Note that Table 19-1 gives the specific heats per kilogram, not per mole.) At high temperatures, each atom apparently has six degrees of ’ieedom although some are not active at low temperatures. Each atom in a ystalline solid can vibrate about its equilibrium position as if it were connected by springs to each of its neighbors. Thus it can have three degrees of freedom for kinetic energy and three more associated with potential energy of vibration in each of the x, y, and z directions, which is in accord with the measured values. |

. Lead

g

®3

5+

|

3 B

8 h=I

2.

3]

Q&

85 7]

,_5

[=}

=

Copper

[=)

>

Diamond

-

0

i

0

100

! T

300

I T

|

T

Jor

T

500

ot T

700

Temperature (K)

SECTION 19-8

i

T

T

900

513

19-9 Adiabatic Expansion of a Gas The PV curve for the quasistatic (slow) adiabatic expansion (Q = 0) of an ideal gas was shown in Fig. 19-8 (curve AC). It is somewhat steeper than for an isotherma process (AT = 0), which indicates that for the same change in volume the change in: pressure will be greater. Hence the temperature of the gas must drop during an adiabatic expansion. Conversely, the temperature rises during an adiabatic compression. We can derive the relation between the pressure P and the volume V of an ideal gas that is allowed to slowly expand adiabatically. We begin with the first law of thermodynamics, written in differential form: dE, = dQ — dW = —dW = —PdV, since dQ =0 for an adiabatic process. Equation 19-12 gives us a relation between AE;,, and Cy,, which is valid for any ideal gas process since E;, is a function only of 7 for an ideal gas. We write this in differential form: dE,, = nCydT. When we combine these last two equations, we obtain nCy,dT + PdV = 0. We next take the differential of the ideal gas law, PV = nRT, allowing P, V,and T to vary: PdV + VdP = nRdT. We solve for dT in this relation and substitute it into the previous relation and get

, cv< PdV + VadP

)+PdV=0

nR or, multiplying through by R and rearranging,

(Cy + R)PAV + C,VdP

= 0.

We note from Eq. 19-11 that C, + R = Cp, so we have CpPdV + C,VdP = 0, or

Cr Cy

—PdV We define

+ VdP

o

=

-

0.

P

Y = —c,

9-14 (19-14)

so that our last equation becomes dP dv P

+

VT/-

=

(,

This is integrated to become InP + YInV = constant. This simplifies (using the rules for addition and multiplication of logarithms) to

_ quasistatic adiabatic Y (19=155 gas ideal [process; s, = PV This is the relation between P and V for a quasistatic adiabatic expansion or contraction. We will find it very useful when we discuss heat engines in the next Chapter. Table 19-4 (p. 511) gives values of ¥ for some real gases. Figure 19-8 compares an adiabatic expansion (Eq. 19-15) in curve AC to an isothermal expansion (PV = constant) in curve AB. It is important to remember that the ideal gas law, PV = nRT, continues to hold even for an adiabatic expansion (PV” = constant); clearly PV is not constant, meaning 7 is not constant. compressed

P, = 100kPa,

starting

Compressing at point

V, = 1.00m?,

batically to state B

A

and

CHAPTER 19

PV

7, = 300K.

(P = 200kPa).

point B to point C (Ve = 0.50m®)

514

an ideal gas. An ideal monatomic gas is

in the

diagram

of Fig.

19-19,

where

The gas is first compressed adian\

The gas is then further compressed from

in an isothermal process. (@) Determine V3.

(b) Calculate the work done on the gas for the whole process.

APPROACH Volume V3 is obtained using Eq. 19-15. The work done by a gas is given by Eq.19-7, W = JPdV. The work on the gas is the negative of this: W, = —[PdV. A

SOLUTION

PV

In the adiabatic process, Eq. 19-15 tells us PV” = constant. Therefore,

= P,tV,’\ = PV}

where for a monatomic gas ¥ = Cp/Cy = (5/2)/(3/2) =3

(a) Eq. 19-15 gives Vy = Va(Pa/Py)7 = (1.00 m’)(100 kPa/200 kPa)s = 0.66 m’

(b) The pressure P at any instant during the adiabatic process is given by P = P,VAV~". The work done on the gas in going from V} to Vj is B

Wi 4 —f PdV = ~PAV}§J

Vi

V7dy = —PAm(

A A Since ¥ =3, then =Y +1=1-7=

=) 5

-3,s

1

v+

1

)(v%;7 - Vi)

) ]

3

_2

3 2 +5(1OOkPa)(100m)[066 ) 5 - ]

V

P (kPa)

C

Isothermal |

200 F-------->¢-~ 1 100F--=======--| | |

1

|

|

} Adiabatic

i | I I [ I I

'

0 0.50 oo V@ FIGURE 19-19 Example 19-12.

3

_2

Va

+48 kl.

For the isothermal process from B to C, the work is done at constant temperature, so the pressure at any instant during the process is P = nRTy/V and C

Wac = ——j Pdv = —nRTBj B

Ve Vi

B

V

V.

—nRTyIn==

VB

Vi

= —PByVgln—

VB

= +37kl.

The total work done on the gas is 48kJ + 37kJ = 85KkIJ.

19-10

Heat Transfer: Conduction, Convection, Radiation

Heat is transferred from one place or body to another in three different ways: by onduction, convection, and radiation. We now discuss each of these in turn; but in practical situations, any two or all three may be operating at the same time. We start with

Conduction When a metal poker is put in a hot fire, or a silver spoon is placed in a hot bowl of soup, the exposed end of the poker or spoon soon becomes hot as well, even though it is not directly in contact with the source of heat. We say that heat has been condqcted from the hot end to the cold end. Heat conduction in many materials can be visualized as being carried out via molecular collisions. As one end of an object is heated, the molecules there move faster and faster. As they collide with slower-moving neighbors, they transfer some of their kinetic energy to these other molecules, which in turn transfer energy by collision with molecules still farther along the object. In metals, collisions of free electrons are mainly responsible for conduction. Heat Jpnduction from one point to another takes place only if there is a difference in temperature between the two points. Indeed, it is found experimentally that the rate of heat flow through a substance is proportional to the difference in temperature between its ends. The rate of heat flow also depends on the size and shape of the object. To investigate this quantitatively, let us consider the heat flow through a uniform cylinder, as illustrated in Fig. 19-20. It is found experimentally that the heat flow AQ over a time interval At is given by the relation AQ

T

=

T

—= = fA——>2 (19-16a) At { where A is the cross-sectional area of the object, £ is the distance between the two nds, which are at temperatures 7; and 7, and k is a proportionality constant called the thermal conductivity which is characteristic of the material. From Eq. 19-16a, we see that the rate of heat flow (units of J/s) is directly proportional to the cross-sectional area and to the temperature gradient (7, — T5)/L

=~ FIGURE 19-20 Heat conduction between areas at temperatures 7

and 75. If 7} is greater than 7, the

heat flows to the right; the rate is

given by Eq. 19-16a.

SECTION 19-10

515

TABLE 19-5 Thermal Conductivities

In some cases (such as when k or A cannot be considered constant) we need to consider the limit of an infinitesimally thin slab of thickness dx. Then

:

Thermal conductivity, k Substance

s (s'm:C°

Silver Copper Aluminum Steel s Giiiss Brick

10 X 1072 9.2 X 1072 5.0 X 1072 11 X102 5%x10% 20X 10~ 20% 10

A Vehinr

ath i

0x107*

.

10 -4 10_ .

Human tissue 0.5 X 10

Wood :

e X

Fiberglass

! (s:m-C°)

0,042

0.024 0.023 :

Example 19-13.

Tile is a better conductor of heat than the rug. Heat that flows from your foot to the ature of your foot and feels good. But the tile conducts the heat away rapidly and thus

can take more heat from your foot quickly, so your foot’s surface temperature drops.

DEVISURMEESEN

3.0 m2

APPROACH Heat flows by cond.uction through the glass from the higher ins.ide temperature to the lower outside temperature. We use the heat conduction equation, Eq. 19-16a.

516

A = (20m)(1.5m) =3.0m?

Table 19-5 to get k, we have

850

At

.

L-T

and

{ =32 X 107 m.

Using\

(0.84J/s-m-C°)(3.0m?)(15.0°C — 14.0°C)

e

.

=

790J/s.

o

(32 x10°m)

.

3 NOTE This rate of heat flow is equivalent to (790J/s)/(4.19 X 10°J/kcal) = 0.19 kcal/s, or (0.19kcal/s) X (3600s/h) = 680 kcal/h.

You might notice in Example 19-13 that 15°C is not very warm for the living room of a house. The room itself may indeed be much warmer, and the outside might be colder than 14°C. But the temperatures of 15°C and 14°C were specified as those at the window surfaces, and there is usually a considerable drop in temperature of the air in the vicinity of the window both on the inside and the outside. That is, the layer of air on either side of the window acts as an insulator, and normally the major part of the temperature drop between the inside and outside of the house takes place across the air layer. If there is a heavy wind, the air outside a window will constantly be replaced with cold air; the temperature gradient across the glass will be greater PHYSICS APPLIED and there will be a much greater rate of heat loss. Increasing the width of the air Thermal windows — layer, such as using two panes of glass separated by an air gap, will reduce the heat loss more than simply increasing the glass thickness, since the thermal conductivity of air is much less than that for glass. The insulating properties of clothing come from the insulating properties of air. Without clothes, our bodies in still air would heat the air in contact with the skin and would soon become reasonably comfortable because air is a very good insulator. o

@

|-15.0°C

Heat loss through windows. A major source of heat loss

from a house is through the windows. Calculate the rate of heat flow through a glass window 2.0m X 1.5m in area and 3.2 mm thick, if the temperatures at the inner and outer surfaces are 15.0°C and 14.0°C, respectively (Fig. 19-21).

SOLUTION Here

14.0°C—|

o

rug is not conducted away rapidly, so the rug’s surface quickly warms up to the temper-

02

0.040 0.025

FIGURE 19-21

(19-16b

-0l

;

Wool 0.1 x10* Goose down 0.06 X 10~ Polyurethane 0.06 X 10 Air 0055 x 10 .

a

dT pal,

ks 056

420 380 200 40 2 084 084

0.048

01x10™*

do s

where dT/dx is the temperature gradient” and the negative sign is included since ‘ the heat flow is in the direction opposite to the temperature gradient. The thermal conductivities, &, for a variety of substances are given in Table 19-5. Substances for which k is large conduct heat rapidly and are said to be good thermal conductors. Most metals fall in this category, although there is a wide range even among them, as you may observe by holding the ends of a silver spoon and a stainless-steel spoon immersed in the same hot cup of soup. Substances for which k& is small, such as wool, fiberglass, polyurethane, and goose down, are poor conductors of heat and are therefore good thermal insulators. : ; ; The relative magnitudes of k can explain simple phenomena such as why a tile floor is much colder on the feet than a rug-covered floor at the same temperature.

012 X 107"

Cork

Eq. 19-16a becomes

/ —f =32 % 103m

CHAPTER

19

"Equations 19-16 are quite similar to the relations describing diffusion (Section 18-7) and the flow (q fluids through a pipe (Section 13-12). In those cases, the flow of matter was found to be proportional to the concentration gradient dC/dx, or to the pressure gradient (P, — P;)/f. This close similarity is one reason we speak of the “flow” of heat. Yet we must keep in mind that no substance is flowing in

the case of heat—it is energy that is being transferred.

But since air moves—there are breezes and drafts, and people move about—the warm air would be replaced by cold air, thus increasing the temperature difference and the heat loss from the body. Clothes keep us warm by trapping air so it cannot move readily. It is not the cloth that insulates us, but the air that the cloth traps. r Goose down is a very good insulator because even a small amount of it fluffs up and traps a great amount of air. [For practical purposes the thermal properties of building materials, particularly when considered as insulation, are usually specified by R-values (or “thermal resistance”), defined for a given thickness £ of material as: = {/k. The R-value of a given piece of material combines the thickness £ and the thermal conductlvfly k in one number. In the United States, R-values are given in British

@pnvsms an air layer

FPHYSics

TABLE 19-6 R-values Thickness

gives R-values for some common building materials. R-values increase directly with material th‘[ckness for example, 2 inches of fiberglass is R-6, whereas 4 inches is R-12.]

Glass Brick Plywood

Linch 3linches linch

Convection

Fiberglass insulation ~ 4inches

Although quuids and gases are generally not very good conductors of heat, they can transfer he t quite rapidly by convection. Convection is the process whereby heat flows by the mass movement of molecules from one place to another. Whereas conduction involves molecules (and/or electrons) moving only over small distances and colliding, convection involves the movement of large numbers of molecules over large distances. A forced-air furnace, in which air is heated and then blown by a fan into a room, is an example of forced convection. Natural convection occurs as well, and one familiar example is that hot air rises. For instance, the air above a radiator (or other type of heater) expands as it is heated (Chapter 17), and hence its density decreases. Because its density is less than that of the surrounding cooler air, it rises,

just as a ldg submerged in water floats upward because its density is less than that

of water. \L’arm or cold ocean currents, such as the balmy Gulf Stream, represent r"natural convection on a global scale. Wind is another example of convection, and

weather in general is strongly influenced by convective air currents. When a pot of water is heated (Fig. 19-22), convection currents are set up as the heated‘ water at the bottom of the pot rises because of its reduced density. That heated water is replaced by cooler water from above. This principle is used in many heating systems, such as the hot-water radiator system shown in Fig. 19-23. Water is heated in the furnace, and as its temperature increases, it expands and rises as shown. This causes the water to circulate in the heating system. Hot water then

enters the radiators, heat is transferred

by conduction

to the air, and

the

cooled wager returns to the furnace. Thus, the water circulates because of convection; pumps are sometimes used to improve circulation. The air throughout the room also becomes heated as a result of convection. The air heated by the radiators rises and is replaced by cooler air, resulting in convective air currents, as shown by the green arrows in Fig. 19-23. Other types of furnaces also depend on convection. Hot-air furnaces with registers (openings) near the floor often do not have fans but depend on natural convection, which can be appreciable. In other systems, a fan is used. In either case, it is impox;rant that cold air can return to the furnace so that convective currents circulate throughout the room if the room is to be uniformly heated. Convection is not always favorable. Much of the heat from a fireplace, for example, goes up the chimney and not out into the room.

APPLIED

R-values of thermal insulation

Material

units as ftth F°/Btu (for example, R-19 means R = 19 ft>-h-F°/Btu). Table 19-6

APPLIED

Clothes insulate by trapping

R-value

(ft* - h- F°/Btu)

1 0.6-1 0.6 12

Cooler

=\ H

7

FIGURE 19-22 Convection currents in a pot of water being heated on a stove. PHYSICS

APPLIED

Convective home heating FIGURE 19-23 Convection plays a role in heating a house. The circular arrows show the convective air currents in the rooms.

Radiation Convectlon and conduction require the presence of matter as a medium to carry the heat f;om the hotter to the colder region. But a third type of heat transfer roccurs without any medium at all. All life on Earth depends on the transfer of energy from the Sun, and this energy is transferred to the Earth over empty (or nearly empty) space. This form of energy transfer is heat—since the Sun’s surface temperature is much higher (6000 K) than Earth’s—and is referred to as radiation. The warmth we receive from a fire is mainly radiant energy.

SECTION 19-10

517

As we shall see in later Chapters, radiation consists essentially of electromagnetic waves. Suffice it to say for now that radiation from the Sun consists of visible light plus many other wavelengths that the eye is not sensitive to, including infrared (IR) radiation. The rate at which an object radiates energy has been found to be proportionafl

to the fourth power of the Kelvin temperature, T. That is, an object at 2000 K, as

compared to one at 1000 K, radiates energy at a rate 2* = 16 times as much. The rate of radiation is also proportional to the area A of the emitting object, so the rate at which energy leaves the object, AQ/At, is T?

=

eaAT*,

(19-17)

This is called the Stefan-Boltzmann equation, and o is a universal constant called the Stefan-Boltzmann constant which has the value

o = 567X 10°8W/m?-K*

@

PHYSICS APPLIED Dark vs. light clothing ~

The factor e (Greek letter epsilon), called the emissivity, is a number between 0 and 1 that is characteristic of the surface of the radiating material. Very black surfaces, such as charcoal, have emissivity close to 1, whereas shiny metal surfaces have e close to zero and thus emit correspondingly less radiation. The value depends somewhat on the temperature of the material. Not only do shiny surfaces emit less radiation, but they absorb little of the radiation that falls upon them (most is reflected). Black and very dark objects are good emitters (e ~ 1), and they also absorb nearly all the radiation that falls on them—which is why light-colored clothing is usually preferable to dark clothing on a hot day. Thus, a good absorber is also a good emitter. Any object not only emits energy by radiation but also absorbs energy radiated by other objects. If an object of emissivity e and area A is at a temperature 7}, it radiates energy at a rate ea AT{. If the object is surrounded by an environment at temperature 7,, the rate at which the surroundings radiate energy is proportional to 7%, and the rate that energy is absorbed by th«™ object is proportional to T4. The net rate of radiant heat flow from the object is given by

—AA—t— = ecA(T} - T%),

(19-18)

where A is the surface area of the object, 7 its temperature and e its emissivity (at temperature 7}), and 7, is the temperature of the surroundings. This equation is consistent with the experimental fact that equilibrium between the object and its surroundings is reached when they come to the same temperature. That is, AQ/At must equal zero when 7} = 7,, so € must be the same for emission and absorption. This confirms the idea that a good emitter is a good absorber. Because both the object and its surroundings radiate energy, there is a net transfer of energy from one to the other unless everything is at the same temperature.

@rHyYsics

APPLIED

The body’s radiative heat loss

FEITITHTERIE

ESTIMATE | Cooling by radiation. An athlete is sitting

unclothed in a locker room whose dark walls are at a temperature of 15°C. Estimate his rate of heat loss by radiation, assuming a skin temperature of 34°C and e = 0.70. Take the surface area of the body not in contact with the chair to

be 1.5 m? SOLVING

Must use the Kelvin temperature

APPROACH SOLUTION

aAQ At

Il

ZIPROBLEM

We use Eq. 19-18, with Kelvin temperatures. We

eaA(T

have

— T3)

(0.70)(5.67 x 1078 W/m?-K*)(1.5 m?)[ (307 K)* — (288K)*] NOTE The “output” of this resting person is a bit more lightbulb uses.

518

CHAPTER 19

Heat and the First Law of Thermodynamics

= 120 W.

than what a 100-W*™H

A resting person naturally produces heat internally at a rate of about 100 W, which is less than the heat loss by radiation as calculated in Example 19-14. Hence, the person’s temperature would drop, causing considerable discomfort. The body responds to excessive heat loss by increasing its metabolic rate, and shivering f‘ s one method by which the body increases its metabolism. Naturally, clothes help a lot. Example 19-14 illustrates that a person may be uncomfortable even if the temperature of the air is, say, 25°C, which is quite a warm room. If the walls or floor are cold, radiation to them occurs no matter how warm the air is. Indeed, it is estimated that radiation accounts for about 50% of the heat loss from a sedentary person in a normal room. Rooms are most comfortable when the walls and floor are warm an the air is not so warm. Floors and walls can be heated by means of hot-water conduits or electric heating elements. Such first-rate heating systems are becoming more common today, and it is interesting to note that 2000 years ago the Romans, even in houses in the remote province of Great Britain, made use of hot-water a;bd steam conduits in the floor to heat their houses. Heating of an object by radiation from the Sun cannot be calculated using Eq. 19-18 since this equation assumes a uniform temperature, 75, of the environment surrounding the object, whereas the Sun is essentially a point source. Hence the Sun must be treated as a separate source of energy. Heating by the Sun is calculated using the fact that about 1350 J of energy strikes the atmosphere of the Earth from the Sun per second per square meter of area at right angles to the Sun’s rays. This number, 1350 W/m?, is called the solar constant. The atmosphere may absorb as much as 70% of this energy before it reaches the ground, depending on the cloud cover. On a clear day, about 1000 W/m? reaches the Earth’s surface. An object of emissivity e with area A facing the Sun absorbs energy from the Sun at a rate, in watts, of about

AQ At

= (1000 W/m?) e A cos 6,

(19-19)

fvhere 6 is the angle between the Sun’s rays and a line perpendicular to the area A (Fig. 19-24). That is, Acos# is the “effective” area, at right angles to the Sun’s rays. The explanation for the seasons and the polar ice caps (see Fig. 19-25) depends on this cos 0 factor in Eq. 19-19. The seasons are not a result of how close the Earth is to the Sun—in fact, in the Northern Hemisphere, summer occurs when the Earth is farthest from the Sun. It is the angle (i.e., cos#6) that really matters. FuSLthermore, the reason the Sun heats the Earth more at midday than at sunrise or sunset is also related to this cos 6 factor. An int‘Fresting application of thermal radiation to diagnostic medicine is thermography. A special instrument, the thermograph, scans the body, measuring the intensity of radiation from many points and forming a picture that resembles an X-ray (Fig. 19-26). Areas where metabolic activity is high, such as in tumors, can often be detected on a thermogram as a result of their higher temperature and consequent increased radiation.

FIGURE 19-24 Radiant energy striking a body at an angle 6. FIGURE 19-25

(a) Earth’s seasons

arise from the 235° angle Earth’s axis makes with its orbit around (b) June sunlight makes an about 23° with the equator. the southern United States

the Sun. angle of Thus 6 in (A) is

near 0° (direct summer sunlight),

whereas in the Southern Hemisphere (B), 60 is 50° or 60°, and less heat can be absorbed—hence it is winter. Near the poles (C), there is never strong direct sunlight; cos 6 varies from about in summer to 0 in winter; so with little heating, ice can form. Axis 5

5 Equator

Earth

Sun

(June)

Earth

(December)

(a) . ’A;X’S

(A)6=0° (Summer) Sun’s: rays : (June):

> (C)6=90° (Cold)

(B)O =50° (Winter)

(b)

FIGURE 19-26 Thermograms of a healthy person’s arms and hands (a) before and (b) after smoking a

cigarette, showing a temperature decrease due to impaired blood circulation associated with smoking. The thermograms have been color-coded according to temperature; the scale on the right goes from blue (cold) to white (hot).

(b) SECTION

19-10

Heat Transfer: Conduction, Convection, Radiation

519

@pHyYsics

APPLIED

Astronomy—size

of a star

EXAMPLE

19-15

Star radius. The giant star Betelgeuse emits

radiant energy at a rate 10* times greater than our Sun, whereas its surface temperature is only half (2900 K) that of our Sun. Estimate the radius of Betelgeuse, assuming € = 1 for both. The Sun’s radius is ¢ = 7 X 10°m. APPROACH

We assume both Betelgeuse and the Sun are spherical, with surface

area 4mr. SOLUTION We solve Eq. 19-17 for A: 477r2

Then

=

s

2

A

(AQ/At)

=

TT“

(AQ/At)B.T_g

(AQ/Ab) T

= (10*)(2%) = 16 x 10%

Hence rg = \/16 X 10*rg = (400)(7 X 10°m) ~ 3 X 10" m. NOTE If Betelgeuse were our Sun, it would envelop us (Earth is 1.5 X 10! m from the Sun).

EXERCISE G Fanning yourself on a hot day cools you by (a) increasing the radiation rate of the skin; (b) increasing conductivity; (¢) decreasing the mean free path of air; (d) increasing the evaporation of perspiration; (e) none of these.

|Summary Internal energy, E;,:, refers to the total energy molecules in an object. For an ideal monatomic gas,

Eiynt = 3NKT = 3nRT

of all the

19-1)

where N is the number of molecules or n is the number of moles. Heat refers to the transfer of energy from one object to another because of a difference of temperature. Heat is thus measured in energy units, such as joules. Heat and internal energy are also sometimes specified in calories or kilocalories (kcal), where lkcal

=

4.186kJ

is the amount of heat needed to raise the temperature of 1kg of water by 1 C°. The specific heat, c, of a substance is defined as the energy (or heat) required to change the temperature of unit mass of substance by 1 degree; as an equation, QO

=

mcAT,

CHAPTER 19

AE,

=

Q-

W.

19-4)

This important law is a broad restatement of the conservation of energy and is found to hold for all processes. Two simple thermodynamic processes are isothermal, which is a process carried out at constant temperature, and adiabatic, a process in which no heat is exchanged. Two more are isobaric (a process carried out at constant pressure) and isovolumetric (a process at constant volume). The work done by (or on) a gas to change its volume by dV is dW = P dV, where P is the pressure. Work and heat are not functions of the state of a system (as are P, V, T, n, and Ej,;) but depend on the type of process that takes a system from one state to another. The molar specific heat of an ideal gas at constant volume, Cy, and at constant pressure, Cp, are related by

Cp—Cy

(19-2)

where Q is the heat absorbed or given off, AT is the temperature increase or decrease, and m is the mass of the substance. When heat flows between parts of an isolated system, conservation of energy tells us that the heat gained by one part of the system is equal to the heat lost by the other part of the system. This is the basis of calorimetry, which is the quantitative measurement of heat exchange. Exchange of energy occurs, without a change in temperature, whenever a substance changes phase. The heat of fusion is the heat required to melt 1kg of a solid into the liquid phase; it is also equal to the heat given off when the substance changes from liquid to solid. The heat of vaporization is the energy required to change 1kg of a substance from the liquid to the vapor phase; it is also the energy given off when the substance changes from vapor to liquid.

520

The first law of thermodynamics states that the change in internal energy AE;, of a system is equal to the heat added to the system, Q, minus the work, W, done by the system:

= R,

(19-11)

where R is the gas constant. For a monatomic ideal gas, Cy, = 3 R. For ideal gases made up of diatomic or more complex molecules, Cy is equal to 3R times the number of degrees of freedom of the molecule. Unless the temperature is very high, some of the degrees of freedom may not be active and so do not contribute. According to the principle of equipartition of energy, energy is shared equally among the active degrees of freedom in an amount 3 k7 per molecule on average. When an ideal gas expands (or contracts) adiabatically (Q = 0), the relation PV = constant holds, where Cp (19-14) Cy Heat is transferred from one place (or object) to another in Y

=

three different ways: conduction, convection, and radiation.

Heat and the First Law of Thermodynamics

»

In conduction, energy is transferred by collisions between molecules or electrons with higher kinetic energy to slowermoving neighbors. Convection is the transfer of energy by the mass movement f molecules over considerable distances. Radiation, which does not require the presence of matter, is

energy transfer by electromagnetic waves, such as from the Sun.

All objects radiate energy in an amount that is proportional to the

fourth power of their Kelvin temperature (7*) and to their surface

area. The energy radiated (or absorbed) also depends on the nature of the surface, which is characterized by the emissivity, e (dark surfaces absorb and radiate more than do bright shiny ones). Radiation from the Sun arrives at the surface of the Earth

on a clear day at a rate of about 1000 W/m?,

| Questions 1. What happens to the work done on a jar of orange juice when it is vigorously shaken? 2. When a hot object warms a cooler object, does temperature flow between them? Are the temperature changes of the two objects equal? Explain. 3. (a) If two objects of different temperature are placed in contact, will heat naturally flow from the object with higher internal energy to the object with lower internal energy? (b) Is it possible for heat to flow even if the internal energies of the two objects are the same? Explain. In warm regions where tropical plants grow but the temperature may drop below freezing a few times in the winter, the destruction of sensitive plants due to freezing can be reduced by watering them in the evening. Explain. 5 The specific heat of water is quite large. Explain why this fact makes water particularly good for heating systems (that is, hot-water radiators).

. Why does water in a canteen stay cooler if the cloth jacket surrounding the canteen is kept moist? . Explain why burns caused by steam at 100°C on the skin are often more severe than burns caused by water at 100°C. . Explain why water cools (its temperature drops) when it evaporates, using the concepts of latent heat and internal energy. . Will potatoes cook faster if the water is boiling more vigorously? 10. Very high in the Earth’s atmosphere the temperature can be 700°C. Yet an animal there would freeze to death rather than roast. Explain. 11. What happens to the internal energy of water vapor in the air that condenses on the outside of a cold glass of water? Is work done or heat exchanged? Explain. 12. Use the conservation of energy to explain why the temperature -insulated gas increases when it is compressed—say, by pushing down on a piston—whereas the temperature decreases when the gas expands. 13. In an isothermal process, 3700J of work is done by an ideal gas. Is this enough information to tell how much heat has been a%ed

to the system? If so, how much?

14. Explorers on failed Arctic expeditions have survived by covering themselves with snow. Why would they do that? 15. Why is wet sand at the beach cooler to walk on than dry sand? 16. When hot-air furnaces are used to heat a house, why is it important that there be a vent for air to return to the furnace? What happens if this vent is blocked by a bookcase? rl7. Is it pdfsible for the temperature of a system to remain constant even though heat flows into or out of it? If so, give examples.

18. Discuss how the first law of thermodynamics can apply to metabolism in humans. In particular, note that a person does work W, but very little heat Q is added to the body (rather, it tends to flow out). Why then doesn’t the internal energy drop drastically in time? 19. Explain in words why Cp is greater than Cy . 20. Explain why the temperature of a gas increases when it is adiabatically compressed. 21. An ideal monatomic gas is allowed to expand slowly to twice its volume (1) isothermally; (2) adiabatically; (3) isobarically. Plot each on a PV diagram. In which process is AEjy the greatest, and in which is A Ejy; the least? In which is W the greatest and the least? In which is Q the greatest and the least? 22, Ceiling fans are sometimes reversible, so that they drive the air down in one season and pull it up in another season. Which way should you set the fan for summer? For winter? 23. Goose down sleeping bags and parkas are often specified as so many inches or centimeters of /oft, the actual thickness of the garment when it is fluffed up. Explain. 24 Microprocessor chips nowadays have a “heat sink” glued on top that looks like a series of fins. Why is it shaped like that? 25 Sea breezes are often encountered on sunny days at the shore of a large body of water. Explain, assuming the temperature of the land rises more rapidly than that of the nearby water. 26. The Earth cools off at night much more quickly when the weather is clear than when cloudy. Why? 27. Explain why air-temperature readings are always taken with the thermometer in the shade. 28. A premature baby in an incubator can be dangerously cooled even when the air temperature in the incubator is warm. Explain. 29. The floor of a house on a foundation under which the air can flow is often cooler than a floor that rests directly on the ground (such as a concrete slab foundation). Explain. 30. Why is the liner of a thermos bottle silvered (Fig. 19-27), and why does it have a Hitrs Stopper vacuum between its two walls? Outer casing

Air/ Liner

I8 '

Liquid,

hot or cold —— ||| | Vacuum FIGURE

19-27

Question 30.

Insulating

support e

Questions

521

3L A 22°C day is warm, while a swimming pool at 22°C feels cool. Why?

34. Early in the day, after the Sun has reached the slope of a mountain, there tends to be a gentle upward

Heat loss occurs through windows by the following processes: (1) ventilation around edges; (2) through the frame, particularly if it is metal; (3) through the glass panes; and (4) radiation. (a) For the first three, what is (are) the mechanism(s): conduction, convection, or radiation? (b) Heavy curtains reduce which of these heat losses? Explain in detail.

33

37

.

heat a room where the windows face north is much higher than that required where the windows face south. Explain.

w 1]

32. In the Northern Hemisphere the amount of heat required to

movement

of

air. Later, after a slope goes into shadow, there is a gentle downdraft. Explain. A piece of wood lying in the Sun absorbs more heat than piece of shiny metal. Yet the metal feels hotter than thc wood when you pick it up. Explain. An “emergency blanket” is a thin shiny (metal-coated) plastic foil. Explain how it can help to keep an immobile person warm. Explain why cities situated by the ocean tend to have less extreme temperatures than inland cities at the same latitude.

| Problems 19- 1 Heat as Energy Transfer 1. (I) To what temperature will 8700J of heat raise 3.0kg of water that is initially at 10.0°C? . (II) When a diver jumps into the ocean, water leaks into the gap region between the diver’s skin and her wetsuit, forming a water layer about 0.5mm thick. Assuming the total surface area of the wetsuit covering the diver is about

1.0 m?, and that ocean water enters the suit at 10°C and is

warmed by the diver to skin temperature of 35°C, estimate how much energy (in units of candy bars = 300 kcal) is required by this heating process. . (IT) An average active person consumes about 2500 Cal a day. (a) What is this in joules? (b) What is this in kilowatthours? (c) If your power company charges about 10¢ per kilowatt-hour, how much would your energy cost per day if you bought it from the power company? Could you feed yourself on this much money per day? (II) A British thermal unit (Btu) is a unit of heat in the British system of units. One Btu is defined as the heat needed to raise 11b of water by 1 F°. Show that 1Btu

=

0.252kcal

=

10567.

. (II) How many joules and kilocalories are generated when the brakes are used to bring a 1200-kg car to rest from a speed of 95 km/h? (IT) A small immersion heater is rated at 350 W. Estimate how long it will take to heat a cup of soup (assume this is 250 mL of water) from 15°C to 75°C.

19 -3 and 19-4

Specific Heat; Calorimetry

7 (I) An automobile cooling system holds 18 L of water. How much heat does it absorb if its temperature rises from 15°C to 95°C? . (I) What is the specific heat of a metal substance if 135 kJ of heat is needed to raise 5.1 kg of the metal from 18.0°C to 37.2°C? . (II) (@) How much energy is required to bring a 1.0-L pot of water at 20°C to 100°C? (b) For how long could this amount of energy run a 100-W lightbulb? 10. (II) Samples of copper, aluminum, and water experience the same temperature rise when they absorb the same amount of heat. What is the ratio of their masses? 11. (II) How long does it take a 750-W coffeepot to bring to a boil 0.75 L of water initially at 8.0°C? Assume that the part of the pot which is heated with the water is made of 280 g of

12. (IT) A hot iron horseshoe (mass = 0.40kg), just forged (Fig. 19-28), is dropped into 1.05L of water in a 0.30-kg iron pot initially at 20.0°C. If the final equilibrium temperature is 25.0°C, estimate the initial temperature of the hot horseshoe. FIGURE 19-28 Problem 12.

13. (II) A 31.5-g glass thermometer reads 23.6°C before it is placed in 135 mL of water. When the water and thermometer come

CHAPTER 19

reads

39.2°C. What

Q

=

CAT.

(a) Write the heat capacity C in terms of the specific heat, c, of the material. (b) What is the heat capacity of 1.0kg of water? (c) Of 35 kg of water? 1% (II) The 1.20-kg head of a hammer has a speed of 7.5m/s just before it strikes a nail (Fig. 19-29) and is brought to rest. Estimate the temperature rise of a 14-g iron nail generated by 10 such hammer blows done in quick succession. Assume the nail absorbs all the energy.

aluminum, and that no water boils away.

522

to equilibrium, the thermometer

was the original temperature of the water? [Hint: Ignore the mass of fluid inside the glass thermometer.] fl 14. (II) Estimate the Calorie content of 65g of candy from th. following measurements.A 15-g sample of the candy is placed in a small aluminum container of mass 0.325 kg filled with oxygen. This container is placed in 2.00kg of water in an aluminum calorimeter cup of mass 0.624 kg at an initial temperature of 15.0°C. The oxygen-candy mixture in the small container is ignited, and the final temperature of the whole system is 53.5°C. 15. (II) When a 290-g piece of iron at 180°C is placed in a 95-g aluminum calorimeter cup containing 250 g of glycerin at 10°C, the final temperature is observed to be 38°C. Estimate the specific heat of glycerin. 16. (II) The heat capacity, C, of an object is defined as the amount of heat needed to raise its temperature by 1 C°. Thus, to raise the temperature by A7 requires heat Q given by

Heat and the First Law of Thermodynamics

FIGURE 19-29 Problem 17.

19-5

Latent Heat

18. (I) How much heat is needed to melt 26.50 kg of silver that is initially at 25°C? 19. (I) During exercise, a person may give off 180 kcal of heat in 25 min by evaporation of water from the skin. How much water has been lost? . (IT) A 35-g ice cube at its melting point is dropped into an insulated container of liquid nitrogen. How much nitrogen

evaporaq‘es if it is at its boiling point of 77 K and has a latent

heat of vaporization of 200kJ/kg? Assume for simplicity that the Epecific heat of ice is a constant and is equal to its value near its melting point. 21. (II) High-altitude mountain climbers do not eat snow, but always melt it first with a stove. To see why, calculate the energy absorbed from your body if you (a) eat 1.0kg of —10°C snow which your body warms to body temperature of 37°C. (b) You melt 1.0kg of —10°C snow using a stove and drink the resulting 1.0 kg of water at 2°C, which your body has to warm to 37°C. iron boiler of mass 180 kg contains 730 kg of water 22 (II) An at 18°C. A heater supplies energy at the rate of 52,000 kJ/h. How long does it take for the water (a) to reach the boiling point, and (b) to all have changed to steam? 23. (IT) In a hot day’s race, a bicyclist consumes 8.0 L of water over the span of 3.5 hours. Making the approximation that all of the cyclist’s energy goes into evaporating this water as sweat, how much energy in kcal did the rider use during the ride? (Since the efficiency of the rider is only about 20%, most of the energy consumed does go to heat, so our

approximation is not far off.)

. (II) The specific heat of mercury is 138 J/kg- C°. Determine the latent heat of fusion of mercury using the following calorimeter data: 1.00 kg of solid Hg at its melting point of —39.0°C is placed in a 0.620-kg aluminum calorimeter with 0.400kg of water at 12.80°C; the resulting equilibrium temperature is 5.06°C. 25. (II) At a crime scene, the forensic investigator notes that the 7.2-g lead bullet that was stopped in a doorframe apparently melted completely on impact. Assuming the bullet was shot at room temperature (20°C), what does the investigator calculate as the minimum muzzle velocity of the gun? 26. (II) A 58-kg ice-skater moving at 7.5m/s glides to a stop. Assuming the ice is at 0°C and that 50% of the heat generated by friction is absorbed by the ice, how much ice melts?

19-6 and

19-7

First Law of Thermodynamics

27. (I) Sketch a PV diagram of the following process: 2.0 L of ideal gas at atmospheric pressure are cooled at constant pressure to a volume of 1.0 L, and then expanded isothermally back to 2.0L, whereupon the pressure is increased at constant volume until the original pressure is reached. 28. (I) A gas is enclosed in a cylinder fitted with a light frictionless piston maintained at atmospheric pressure. When 1250 kcal of heat is added to the gas, the volume is observed to increase

slowly from 12,0 m? to 18.2 m®. Calculate (a) the work done by

the gas and (b) the change in internal energy of the gas. . (II) The !pressure in an ideal gas is cut in half slowly, while being kelpt in a container with rigid walls. In the process, 365kJ of heat left the gas. (¢) How much work was done during this process? (b) What was the change in internal energy of the gas during this process?

30. (II) A 1.0-L volume of air initially at 3.5 atm of (absolute) pressure is allowed to expand isothermally until the pressure is 1.0 atm. It is then compressed at constant pressure to its initial volume, and lastly is brought back to its original pressure by heating at constant volume. Draw the process on a PV diagram, including numbers and labels for the axes. 3L (II) Consider the following two-step process. Heat is allowed to flow out of an ideal gas at constant volume so that its pressure drops from 2.2 atm to 1.4 atm. Then the gas expands

at constant

pressure,

from

a volume

of 59L

to

9.3L, where the temperature reaches its original value. See Fig. 19-30. Calculate (a) the total work done by the gas

in

(b) the internal the

the

gas

process,

change energy

in of

p

the

2.2 atin F=====x

into

l4atm ----- '2

in

process, and (c) the

totalout heat or of theflowgas.

|

:

I I I I I I

FIGURE 19-30 Problem 31.

| I | | |

|

0

1

59L

93L

Vv

32. (II) The PV diagram in Fig. 19-31 shows two possible states of a system containing 1.55 moles of a monatomic ideal

gas.

(P, = Py =455N/m?,

V;=200m’

V,=800m’)

(a) Draw the process which depicts an isobaric expansion from state 1 to state 2, and label this process A. (b) Find the work done by the gas and the change in internal energy of the gas in process A. (c) Draw the two-step process which depicts an

isothermal

expansion

from

state

1 to the

volume

V,,

followed by an isovolumetric increase in temperature to state 2, and label this process B. (d) Find the change in internal energy of p (N/m2) the gas for the two-step process B. 500 % %

400 300 200

FIGURE 19-31

100

Problem 32. 33, (IT)

Suppose

e

0

2.60mol

2 of

4

an

ideal

6

8 gas

10Vmd of

volume

Vi =350m? at 7, =290K is allowed to expand isothermally to V, = 7.00m> at 7, = 290K. Determine

(a) the work done by the gas, (b) the heat added to the gas, and (c) the change in internal energy of the gas. . (II) In an engine, an almost ideal gas is compressed adiabatically to half its volume. In doing so, 2850J of work is done on the gas. (¢) How much heat flows into or out of the gas?

(b) What

is the change

in internal energy of the

gas? (c) Does its temperature rise or fall? 35. (IT) One and one-half moles of an ideal monatomic gas expand adiabatically, performing 7500J of work in the process. What is the change in temperature of the gas during this expansion? 36. (I) Determine (a) the work done and (b) the change in internal energy of 1.00 kg of water when it is all boiled to steam at 100°C. Assume a constant pressure of 1.00 atm.

Problems

523

37. (II) How much work is done by a pump to slowly compress,

isothermally, 3.50 L of nitrogen at 0°C and 1.00 atm to 1.80 L

at 0°C? 38. (IT) When a gas is taken from a to c along the curved path in Fig. 19-32, the work done by the gas is W = —35] and the heat added to the gas is Q = —63J. Along path abc, the work done is W = —54J. (a) What is Q for path

abc? (b) If P.=3R,,

whatis W for path cda? (c) What

is Q for path cda? (d) What is Ejy , — Einc? Eint.d — Eint,c = 12J, what is Q for path da?

(e)

If

a

Y

A

c

FIGURE 19-32

Problems 38, 39,

and 40.

0

d

39, (IIT) In the process of taking a gas from state a to state c along the curved path shown in Fig. 19-32, 85J of heat leaves the system and 557 of work is done on the system. (a) Determine the change in internal energy, Eint, o — Eint,c(b) When the gas is taken along the path cda, the work done by the gas is W = 38J. How much heat Q is added to the gas in the process cda? (c) If P, = 2.2P4, how much work is done by the gas in the process abc? (d) What is Q for path abc? (e) If Ejpt o — Eint,b = 15J, what is Q for the process bc? Here is a summary of what is given:

Qaye = —85J =

=557

Weda

=

387

Einta = Ein,p

=

157

P,

=

22P;.

40. (IIT) Suppose a gas is taken clockwise around the rectangular cycle shown in Fig. 19-32, starting at b, then to a, to d, to c, and returning to b. Using the values given in Problem 39, (a) describe each leg of the process, and then calculate (b) the net work done during the cycle, (c) the total internal energy change during the cycle, and (d) the net heat flow during the cycle. (¢) What percentage of the

45. (II)

this “rectangular” cycle (give as a percentage)? *41. (III) Determine the work done by 1.00mol of a van der Waals gas (Section 18-5) when it expands from volume V; to V; isothermally.

42. (I) What is the internal energy of 4.50mol of an ideal diatomic gas at 645 K, assuming all degrees of freedom are active? 43. (I) If a heater supplies 1.8 X 10°J/h to a room 35m X 46m X 3.0m containing air at 20°C and 1.0 atm, by how much will the temperature rise in one hour, assuming no losses of heat or air mass to the outside? Assume air is an ideal diatomic gas with molecular mass 29.

524

CHAPTER 19

A

certain

theory

predicts

monatomic

cy = 0.0356 kcal/kg-C°,

temperature range. What What gas is it?

gas

Cy =3inR

has

and

specific

heat

which changes little over a wide*™ is the atomic mass of this gas?

. (II) Show that the work done by n moles of an ideal gas when it expands adiabatically is W = nCy (Ty — T;), where T and T; are the initial and final temperatures, and Cy is the molar specific heat at constant volume.

49. (II) A 2.00 mole sample of N, gas at 0°C is heated to 150°C at constant pressure (1.00 atm). Determine (a) the change in internal energy, (b) the work the gas does, and (c) the heat added to it. 50. (III) A 1.00-mol sample of an ideal diatomic gas at a pressure of 1.00 atm and temperature of 420 K undergoes a process in which its pressure increases linearly with temperature. The final temperature and pressure are 720K and 1.60 atm. Determine (a) the change in internal energy, (b) the work done by the gas, and (c) the heat added to the gas. (Assume five active degrees of freedom.)

19-9 Adiabatic Expansion of a Gas

51. (I) A 1.00-mol sample of an ideal diatomic gas, originally at 1.00 atm

and

20°C, expands

adiabatically

to 1.75 times

its

initial volume. What are the final pressure and temperature for the gas? (Assume no molecular vibration.)

52, (II) Show, using Egs. 19-6 and 19-15, that the work done by a gas that slowly expands adiabatically from pressure P; and volume Vi, to P, and V,, is given by

W = (PV; - BV)/(Y - 1).

53. (I) A 3.65-mol sample of an ideal diatomic gas expands

adiabatically from a volume

of 0.1210m’® to 0.750m?.

Initially the pressure was 1.00 atm. Determine: (a) the initial and final temperatures; (b) the change in internal energy; (c) the heat lost by the gas; (d) the work done on the gas. (Assume no molecular vibration.)

intake heat was turned into usable work: i.e., how efficient is

19-8 Molecular Specific Heat for Gases; Equipartition of Energy

then

Cp=13%(n+2)R

. (II) The specific heat at constant volume of a particular gas is 0.182 kcal/kg-K at room temperature, and its molecular mass is 34. (a) What is its specific heat at constant pressure? (b) What do you think is the molecular structure of this gas? 1%

Wase

freedom,

47. (II) An audience of 1800 fills a concert hall of volume 22,000 m>. If there were no ventilation, by how much would the temperature of the air rise over a period of 2.0 h due to the metabolism of the people (70 W /person)?

P

b

. (I) Show that if the molecules of a gas have n degrees of

54. (II) An ideal monatomic gas, consisting of 2.8 mol of volume

0.086m®,

expands

adiabatically.

The

initial

and

final

temperatures are 25°C and —68°C. What is the final volume of the gas?

S5, (III) A 1.00-mol sample of an ideal monatomic gas, originally at a pressure of 1.00 atm, undergoes a three-step process: (1) it is expanded adiabatically from 7; = 588 K to T, = 389K; (2) it is compressed at constant pressure until its temperature reaches 73; (3) it then returns to its original pressure and temperature by a constant-volume process. (a) Plot these processes on a PV diagram. (b) Determine 73. (c) Calculate the change in internal energy, the work don by the gas, and the heat added to the gas for each process, and (d) for the complete cycle.

Heat and the First Law of Thermodynamics

-

56. (I1I) Conisider a parcel of air moving to a different altitude y in the Earth’s atmosphere (Fig. 19-33). As the parcel changes altitude it acquires the pressure P of the

surrounding air. From Eq. 13-4 we have

r

a _ _ dy where

58. (I) One end of a 45-cm-long copper rod with a diameter of 2.0cm is kept at 460°C, and the other is immersed in water at 22°C. Calculate the heat conduction rate along the rod. . (IT) How long does it take the Sun to melt a block of ice at

0°C with a flat horizontal area 1.0 m? and thickness 1.0 cm?

P8

p is the parcel’s

altitude-dependent

mass

Assume that the Sun’s rays make an angle of 35° with the vertical and that the emissivity of ice is 0.050.

density.

60. (II) Heat conduction to skin. Suppose 150 W of heat flows by conduction from the blood capillaries beneath the skin to the body’s surface area of 1.5m% If the temperature

y

difference

is

0.50 C°,

estimate

capillaries below the skin surface.

the

average

distance

of

61. (IT)A ceramic teapot (e = 0.70) and a shiny one (e = 0.10) each hold 0.55 L of tea at 95°C. (a) Estimate the rate of heat loss from each, and (b) estimate the temperature drop after 30min

“Parcel” of N air molecules

y=0pg FIGURE 19-33 Problem 56.

During this motion, the parcel’s volume will change and, because air is a poor heat conductor, we assume this expansion or contraction will take place adiabatically. (a) Starting with Eq. 19-15, PV? = constant, show that for an ideal

gas undergoing an adiabatic process,

dpP

P dT

P'~"T” = constant.

Cu

(@

=7)(=rs) s

+

P dr;T

YTy e

ar _1-7vmg Y

k

where m is the average mass of an air molecule and k is the Boltzmann constant. (¢) Given that air is a diatomic gas with an average molecular mass of 29, show that dT/dy = —9.8 C°/km. This value is called the adiabatic lapse rate for dry air. (d) In California, the prevailing westerly winds descend from one of the highest elevations (the 4000-m Sierra Nevada mountains) to one of the lowest elevations (Death Valley, —100 m) in the continental United States. If a dry wind has a temperature of —5°C at the top of the Sierra

Nevada, what

is the wind’s

has descended to Death Valley?

19-10

temperature

after it

Conduction, Convection, Radiation

57. (I) (a) How much power is radiated by a tungsten sphere (emissivity € = 0.35) of radius 16 cm at a temperature of 25°C? (b) If the sphere is enclosed in a room whose walls are kept at —5°C, what is the net flow rate of energy out of the sphere?

(

only

T=2

FIGURE 19-34

(b) Use the ideal gas law with the result from part (a) to show that the change in the parcel’s temperature with change in altitude is given by dy

225°C

= 0

and thus "

Consider

radiation,

and

assume

the

62. (1) A copper rod and an aluminum rod of the same length and cross-sectional area are attached end to end (Fig. 19-34). The copper end is placed in a furnace maintained at a constant temperature of 225°C. The aluminum end is placed in an ice bath held at constant temperature of 0.0°C. Calculate the temperature at the point where the two rods are joined.

Then show that the parcel’s pressure and temperature are related by

(A~ 1= W=%)== +v=— + ¥z &

for each.

surroundings are at 20°C.

Al

0.0°C Problem 62.

63. (IT) (a) Using the solar constant, estimate the rate at which the whole Earth receives energy from the Sun. (b) Assume the Earth radiates an equal amount back into space (that is, the Earth is in equilibrium). Then, assuming the Earth is a perfect emitter (e = 1.0), estimate its average surface temperature. [Hint: Use area A = 4mrg, and state why.] 64. (II) A 100-W lightbulb generates 95W of heat, which is dissipated through a glass bulb that has a radius of 3.0cm and is 0.50 mm thick. What is the difference in temperature between the inner and outer surfaces of the glass? 65. (IIT) A house thermostat is normally set to 22°C, but at night it is turned down to 12°C for 9.0 h. Estimate how much more heat would be needed (state as a percentage of daily usage) if the thermostat were not turned down at night. Assume that the outside temperature averages 0°C for the 9.0 h at night and 8°C for the remainder of the day, and that the heat loss from the house is proportional to the difference in temperature inside and out. To obtain an estimate from the data, you will have to make other simplifying assumptions; state what these are. 66.

(III) Approximately how long should it take 9.5kg of ice at 0°C to melt when it is placed in a carefully sealed Styrofoam ice chest of dimensions 25cm X 35cm X 55cm whose walls are 1.5 cm thick? Assume that the conductivity of Styrofoam is double that of air and that the outside temperature is 34°C. Problems

525

67. (ITII) A cylindrical pipe has inner radius R; and outer radius R,. The interior of the pipe carries hot water at

temperature

Tj. The

temperature

outside

is 7, (< Ty).

(a) Show that the rate of heat loss for a length £ of pipe is

68. (I1I) Suppose the insulating qualities of the wall of a house come mainly from a 4.0-in. layer of brick and an R-19 layer of insulation, as shown in Fig. 19-35. What is the total rate of heat loss through such

Brick

(Ry)

a wall, if its total area is 195 f2

Insulation

(Ry)

T2

and the temperature difference

Ty

across it is 12 F°?

where k is the thermal conductivity of the pipe. (b) Suppose the pipe is steel with Ry =33cm, R, =4.0cm, and T, = 18°C. If the pipe holds still water at 77 = 71°C, what will be the initial rate of change of its temperature? (c) Suppose water at 71°C enters the pipe and moves at a speed of 8.0 cm/s. What will be its temperature drop per centimeter of travel?

Heat flow

FIGURE 19-35

Problem 68.

Two layers insulating a wall.

| General Problems 69. A soft-drink can contains about 0.20kg of liquid at 5°C. Drinking this liquid can actually consume some of the fat in the body, since energy is needed to warm the liquid to body temperature (37°C). How many food Calories should the drink have so that it is in perfect balance with the heat needed to warm the liquid (essentially water)? 70. (a) Find the total power radiated into space by the Sun, assuming it to be a perfect emitter at 7 = 5500 K. The Sun’s radius is 7.0 X 108m. (b) From this, determine the

power per unit area arriving at the Earth, 1.5 X 10!'m away.

71. To get an idea of how much thermal energy is contained in the

world’s oceans, estimate the heat liberated when a cube of ocean water, 1 km on each side, is cooled by 1 K. (Approximate

the ocean water as pure water for this estimate.)

72 A mountain climber wears a goose-down jacket 3.5 cm thick with total surface area 0.95 m?. The temperature at the surface of the clothing is —18°C and at the skin is 34°C. Determine the rate of heat flow by conduction through the jacket (a) assuming it is dry and the thermal conductivity k is that of goose down, and (b) assuming the jacket is wet, so k is that of water and the jacket has matted to 0.50 cm thickness. 73. During light activity, a 70-kg person may generate 200 kcal/h. Assuming that 20% of this goes into useful work and

the

other

80%

is

converted

to

heat,

estimate

76. A house has well-insulated walls 19.5 cm thick (assume

conductivity of air) and area 410m? a roof of wood 5.5cm

thick

and

area

280m?

and

uncovered

needs to be heated and that its volume is 750 m>. (c) natural

gas costs

$0.080

per kilogram

velocity of 12 m/s,

that the

surface

area

and that the kinetic energy

the

is 1.5m?2

Compare this to the measured value of about 230W that must be dissipated by a person working lightly. This clearly shows the necessity of convective cooling by the blood. 75. A marathon runner has an average metabolism rate of about 950 kcal/h during a race. If the runner has a mass of 55 kg, estimate how much water she would lose to evaporation

from the skin for a race that lasts 2.2 h.

526

CHAPTER 19

«

lost in the

process heats the ball. What will be the temperature increase of the ball after one bounce? (The specific heat of rubber is about 1200 J/kg- C°.)

thickness of tissue is 4.0 cm, that the skin is at 34°C and the

and

its heat

cost to maintain the house as in part (a) for 24h each day, assuming 90% of the heat produced is used to heat the house? Take the specific heat of air to be 0.24 kcal/kg- C°. 7. In a typical game of squash (Fig. 19-36), two people hit a soft rubber ball at a wall until they are about to drop due to dehydration and exhaustion. Assume that the ball hits the wall at a velocity of 22m/s and bounces back with a

74. Estimate the rate at which heat can be conducted from the interior of the body to the surface. Assume that the

at 37°C,

and

combustion is 5.4 X 107 J/kg, how much is the monthly

temperature rise of the body after 30 min if none of this heat is transferred to the environment.

interior

windows

0.65cm thick and total area 33m?% (a) Assuming that heat is lost only by conduction, calculate the rate at which heat must be supplied to this house to maintain its inside temperature at 23°C if the outside temperature is —15°C. (b) If the house is initially at 12°C, estimate how much heat must be supplied to raise the temperature to 23°C within 30min. Assume that only the air

Heat and the First Law of Thermodynamics

FIGURE 19-36

Problem 77.

78 . A bicycle pump is a cylinder 22cm long and 3.0cm in diameter. The pump contains air at 20.0°C and 1.0 atm. If the outlet at the base of the pump is blocked and the handle is pushed in very quickly, compressing the air to half its original volume, how hot does the air in the pump become? . A micr wave oven is used to heat 250g of water. On its maximum

setting, the oven can raise the temperature of the

liquid water from 20°C to 100°C in 1 min 45 s (= 105s). (a) At

what rate does the oven input energy to the liquid water?

80.

81.

82.

83.

(b) If th‘ power input from the oven to the water remains constant, determine how many grams of water will boil away if the oven is operated for 2 min (rather than just 1 min 45s). The te Perature within the Earth’s crust increases about 1.0C° for each 30m of depth. The thermal conductivity of the crust is 0.80 W/C°m. (a) Determine the heat transferred from the interior to the surface for the entire Earth in 1.0h. (b) Compare this heat to the amount of energy incident on the Earth in 1.0h due to radiation from the Sun. An ice sheet forms on a lake. The air above the sheet is at —18°C, whereas the water is at 0°C. Assume that the heat of fusion of the water freezing on the lower surface is conducted through the sheet to the air above. How much time will it take to form a sheet of ice 15cm thick? An iroq meteorite melts when it enters the Earth’s atmosphere. If its initial temperature was —105°C outside of Earthrs atmosphere, calculate the minimum velocity the meteorite must have had before it entered Earth’s atmosphere. A scuba‘} diver releases a 3.60-cm-diameter (spherical) bubble of air from a depth of 14.0 m. Assume the temperature

8s. The temperature of the glass surface of a 75-W lightbulb is 75°C when the room temperature is 18°C. Estimate the temperature of a 150-W lightbulb with a glass bulb the same size. Consider only radiation, and assume that 90% of the energy is emitted as heat. 86. Suppose 3.0 mol of neon (an ideal monatomic gas) at STP are compressed slowly and isothermally to 0.22 the original volume. The gas is then allowed to expand quickly and adiabatically back to its original volume. Find the highest and lowest temperatures and pressures attained by the gas, and show on a PV diagram where these values occur. 87. At very low temperatures, the molar specific heat of many substances varies as the cube of the absolute temperature:

which is sometimes called Debye’s law. For rock salt, To = 281K and k = 1940J/mol-K. Determine the heat needed to raise 2.75 mol of salt from 22.0K to 48.0 K. 88. A diesel engine accomplishes ignition without a spark plug by an adiabatic compression of air to a temperature above the ignition temperature of the diesel fuel, which is injected into the cylinder at the peak of the compression. Suppose air is taken into the cylinder at 280 K and volume Vj and is compressed adiabatically to 560°C (=~ 1000°F) and volume V;. Assuming that the air behaves as an ideal gas whose ratio of Cp to Cy is 1.4, calculate the compression ratio V;/V; of the engine. 89.

is constant at 298 K, and that the air behaves as an ideal gas.

(a) How large is the bubble when it reaches the surface? (b) Sketch a PV diagram for the process. (c¢) Apply the first law of thermodynamics to the bubble, and find the work done by the air in rising to the surface, the change in its internal energy, and the heat added or removed from the air in the bubble as it rises. Take the density of water to be 1000 kg/m3. . A reciprocating compressor is a device that compresses air by a back-and-forth straight-line motion, like a piston in a cylinder. Consider a reciprocating compressor running at 150 rpm. During a compression stroke, 1.00mol of air is compressed. The initial temperature of the air is 390K, the engine of the compressor is supplying 7.5kW of power to compress the air, and heat is being removed at the rate of 1.5kW. Calculate the temperature change per compreSfiion stroke.

When 6.30 X 10°J of heat is added to a gas enclosed in a cylinder fitted with a light frictionless piston maintained at atmospheric pressure, the volume is observed to increase from 2.2m? to 4.1 m>. Calculate (a) the work done by the gas, and (b) the change in internal energy of the gas. (c) Graph this process on a PV diagram.

90. In a cold environment, a person can lose heat by conduction and radiation at a rate of about 200 W. Estimate how long it would take for the body temperature to drop from 36.6°C to 35.6°C if metabolism were nearly to stop. Assume a mass of 70 kg. (See Table 19-1.)

*Numerical/Computer

*91. (II) Suppose 1.0mol of steam at 100°C of volume 0.50m? is expanded isothermally to volume 1.00m’, Assume steam obeys the van der Waals equation

(P + n?a/V*)(V/n — b) = RT, 0.55N'm*/mol> expression

dW

and

Eq.18-9,

b = 3.0 X 10° m®/mol.

= P dV,

determine

with

numerically

a=

Using the the

total

work done W. Your result should agree within 2% of the result obtained by integrating the expression for dW.

Answers to Exercises A

(b).

B:

(c).

0.21 kg. 2 7001J.

¢ Less:

—6.8 X 10°7. G:

(d).

General Problems

527

There

are

many

uses

for

a

heat

piston

that

engine, such as old steam trains and modern coal-burning power plants. Steam engines produce steam which does work: on turbines to generate electricity,

and

on

a

moves linkage to turn locomotive wheels. The efficiency of any engine—no matter how carefully engineered—is limited by nature as described in the second law of thermodynamics. This great law is best stated in terms of a quantity called entropy, which is unlike any other. Entropy is not conserved, but

instead is constrained always to increase in any real process. Entropy is a measure of disorder. The second law of thermodynamics tells us that as time moves

forward, the disorder

in the universe increases. We discuss many practical matters including heat engines, heat pumps, and refrigeration.

Second Law of Thermodynamics CHAPTER-OPENING

QUESTION—Guess

now!

Fossil-fuel electric generating plants produce “thermal pollution.” Part of the heat produced by the burning fuel is not converted to electric energy. The reason for this waste is (a) The efficiency is higher if some heat is allowed to escape. (b) Engineering technology has not yet reached the point where 100% waste heat recovery is possible. (¢) Some waste heat must be produced: this is a fundamental property of nature when converting heat to useful work.

The Second Law of

Thermodynamics—Introduction Heat Engines Reversible and Irreversible Processes; the Carnot Engine Refrigerators, Air Conditioners, and Heat Pumps Entropy

(d)

Entropy and the Second Law of Thermodynamics

The plants rely on fossil fuels, not nuclear fuel.

(e) None of the above.

Order to Disorder Unavailability of Energy; Heat Death Statistical Interpretation of Entropy and the Second Law Thermodynamic Temperature; Third Law of Thermodynamics Thermal Pollution, Global

Warming, and Energy Resources

528

in

|

n this final Chapter on heat and thermodynamics, we discuss the famous second law of thermodynamics, and the quantity “entropy” that arose from this fundamental law and is its quintessential expression. We also discuss heat engines—the engines that transform heat into work

power

plants,

trains,

that a new law was thermodynamics.

and

needed.

motor

vehicles—because

Finally,

we

briefly

they

discuss

first

the

showed

third

law

uyg

|

20-1 The Second Law of

Thermodynamics

—Introduction

fhe first law of thermodynamics states that energy is conserved. There are, however, many processes we can imagine that conserve energy but are not observed to occur in nature. For example, when a hot object is placed in contact with a cold object, heat flows from the hotter one to the colder one, never spontaneouély the reverse. If heat were to leave the colder object and pass to the hotter one, energy could still be conserved. Yet it does not happen spontaneously.’ As a second example, consider what happens when you drop a rock and it hits the ground. The initial potential energy of the rock changes to kinetic energy as the rock falls. When the rock hits the ground, this energy in turn is transformed into internal energy of the rock and the ground in the vicinity of the impact; the molecules move faster and the temperature rises slightly. But have you seen the reverse happen—a rock at rest on the ground suddenly rise up in the air because the thermal energy of molecules is transformed into kinetic energy of the rock as a whole? Energy could be conserved in this process, yet we never see it happen. There are many other examples of processes that occur in nature but whose reverse does not. Here are two more. (1) If you put a layer of salt in a jar and cover it with a layer of similar-sized grains of pepper, when you shake it you get a thorough m£xture But no matter how long you shake it, the mixture does not separate into two layers again. (2) Coffee cups and glasses break spontaneously if you drop them. But they do not go back together spontaneously (Fig. 20-1).

(a) Initial state.

(b) Later: cup reassembles and rises up.

(c) Later still: cup lands on table.

FIGURE 20-1 Have you ever observed this process, a broken cup spontaneously reassembling and rising up onto a table? This process could conserve energy and other laws of mechanics.

The first law of thermodynamics (conservation of energy) would not be violated if any of these processes occurred in reverse. To explain this lack of reversibility, scientists in the latter half of the nineteenth century formulated a new principle known as the second law of thermodynamics. The second law of thermodynamics is a statement about which processes occur in nature and which do not. It can be stated in a variety of ways, all of which are equivalent. One statement, due to R. J. E. Clausius (1822-1888), is that heat flow spontaneously from a hot object to a cold object; heat will not flow spontaneously from a cold object to a hot object.

SECOND LAW OF THERMODYNAMICS (Clausius statement)

Since this statement applies to one particular process, it is not obvious how it applies to other processes. A more general statement is needed that will include other posmbie processes in a more obvious way. The development of a general statement of the second law of therrnodynarnlcs

was based partly on the study of heat engines. A heat engine is any device that changes thermal energy into mechanical work, such as a steam engine or automobile engine. We now examine heat engines, both from a practical point of view and to show their importance in developing the second law of thermodynamics.

3y spontaneofllsly, we mean by itself without input of work of some sort. (A refrigerator does move heat from a cold environment to a warmer one, but only because its motor does work—Section 20-4.)

SECTION 20-1

The Second Law of Thermodynamics—Introduction

529

20-2

FIGURE 20-2 Schematic diagram of energy transfers for a heat engine.

A

CAUTION New sign convention:

Oy > 0,0, >0,W

>0

Heat Engines

It is easy to produce thermal energy by doing work—for example, by simply rubbing your hands together briskly, or indeed by any frictional process. But to get work from thermal energy is more difficult, and a practical device to do this was invented only about 1700 with the development of the steam engine. The basic idea behind any heat engine is that mechanical energy can be obtained from thermal energy only when heat is allowed to flow from a high temperature to a low temperature. In the process, some of the heat can then be transformed to mechanical work, as diagrammed schematically in Fig. 20-2. We will be interested only in engines that run in a repeating cycle (that is, the system returns repeatedly to its starting point) and thus can run continuously. In each cycle the change in internal energy of the system is AE;,;, = 0 because it returns to the starting state. Thus a heat input Qy at a high temperature Ty is partly transformed into work W and partly exhausted as heat Op at a lower temperature 7; (Fig. 20-2). By conservation of energy, Quy =W + Q,. The high and low temperatures, Ty and 77, are called the ~ operating temperatures of the engine. Note carefully that we are now using a new sign convention: we take Qy, Q; , and W as always positive. The direction of

each energy transfer is shown by the arrow on the applicable diagram, such as Fig. 20-2.

Steam Engine and Internal Combustion Engine @

PHYSICS

APPLIED Engines

The operation of a steam engine is illustrated in Fig. 20-3. Steam engines are of two main types, each making use of steam heated by combustion of coal, oil or gas, or by nuclear energy. In a reciprocating engine, Fig. 20-3a, the heated steam passes through the intake valve and expands against a piston, forcing it to move. As the piston returns to its original position, it forces the gases out the exhaust valve. A steam turbine, Fig. 20-3b, is very similar except that the reciprocating piston i replaced by a rotating turbine that resembles a paddlewheel with many sets q\ blades. Most of our electricity today is generated using steam turbines.” The material that is heated and cooled, steam in this case, is called the working substance. "Even nuclear power plants utilize steam turbines; the nuclear fuel—uranium—merely serves as fuel to heat the steam.

(a) Reciprocating type

(b) Turbine (boiler and

condenser not shown)

High temperature

High-pressure steam, from boiler

Intake valve (open during

FIGURE 20-3

Steam engines.

expansion)

Heat input

Exhaust valve (closed during expansion) Pum

CHAPTER20

Piston

Condenser ~”

530

-

Second Law of Thermodynamics

Low temperature

Low-pressure steam, exhausted to condenser

fl

Intake valve (open) r

Exhaust valve (closed)

/ Gas—air

mixture from carburetor

Both valves closed

Both valves closed

Both valves closed

Exhaust valve (open) % — =

:

T.o exhaust pipe

|

Connect rod

Crankshe

(a) Intake

(b) Compression

(c) Ignition

(d) Expansion

(power stroke)

(e) Exhaust

In an internal combustion engine (used in most automobiles), the high temperature is achieved by burning the gasoline—air mixture in the cylinder itself (ignited by the spark plug), as described in Fig. 20-4.

FIGURE 20-4 Four-stroke-cycle . . ) internal combustion engine: (a) the gasoline—air mixture flows into the

Why a AT Is Needed to Drive a Heat Engine

(b) the piston moves upward and compresses the gas; (c) the brief instant when firing of the spark plug ignites the highly compressed

To see why a temperature difference is required to run an engine, let us examine the steam e ngine. In the reciprocating engine, for example, suppose there were no condenser or pump (Fig. 20-3a), and that the steam was at the same temperature throughout the system. This would mean that the pressure of the gas being exhausted would be the same as that on intake. Thus, although work would be done by the gas on the piston when it expanded, an equal amount of work would have to be done by the piston to force the steam out the exhaust; hence, no net work would be done. In a real engine, the exhausted gas is cooled to a lower mperature and condensed so that the exhaust pressure is less than the intake * ressure. Thus, although the piston must do work on the gas to expel it on the exhaust stroke, it is less than the work done by the gas on the piston during the intake. So a net amount of work can be obtained—but only if there is a difference of temperature. Similarly, in the gas turbine if the gas isn’t cooled, the pressure on each side of the blades would be the same. By cooling the gas on the exhaust side, the pressure on the back side of the blade is less and hence the turbine turns.

cylinder as the piston moves down;

gasoline—air mixture, raising it to a

high temperature; (d) the gases, now at high temperature and pressure, expand against the piston in this, the power stroke; () the burned gases are pushed out to the exhaust pipe; when the piston reaches the top, the exhaust valve closes and the intake valve opens, and the whole cycle repeats. (a), (b), (d), and (e) are the four strokes of the cycle.

Efficiency and the Second Law The efficiency, e, of any heat engine can be defined as the ratio of the work it does, W, to the heat input at the high temperature, Qy (Fig. 20-2): e

_w, On

This is a se sible definition since W is the output (what you get from the engine), whereas Q is what you put in and pay for in burned fuel. Since energy is

conserved,

the heat input Qy must equal the work done plus the heat that flows

out at the low temperature (Q,): Ou

Thus

= W+

0.

W = Oy — O, , and the efficiency of an engine is w

On

e = —

On — O = ==t

O

O =1 - =k

O

: 20-1a) 20-1b -1

f‘o give the efficiency as a percent, we multiply Egs. 20-1 by 100. Note that e could .e 1.0 (or 100%) only if O, were zero—that is, only if no heat were exhausted to the environment.

SECTION 20-2

Heat Engines

531

Car efficiency. An automobile engine has an efficiency of

20% and produces an average of 23,000J of mechanical work per second during operation. (¢) How much heat input is required, and (b) how much heat is discharged as waste heat from this engine, per second?

APPROACH We want to find the heat input Qy as well as the heat output Q; , given W = 23,000] each second and an efficiency e = 0.20. We can use the definition then Q; .

of efficiency,

SOLUTION

Eq.

W

The engine requires

. 0

v

FIGURE 20-5 Exercise A.

| L

Adiabatic process,

=

forms,

to

find

first

Qy

and

we solve for Qy:

23,000]

020

1.15 X 10°T

=

115KkJ.

115kJ/s = 115kW

(b) Now we use Eq.20-1b

A

in its various

(@) From Eq. 20-1a, e = W/Qy,

% =7 =

0L

20-1

(1 —e)Qy

of heat input.

(e = 1 — Q,/Qy)

=

(0.80)115kJ

=

and solve for Q; : 92KkIJ.

The engine discharges heat to the environment at a rate of 92kJ/s = 92kW. NOTE Of the 115kJ that enters the engine per second, only 23kJ does useful work whereas 92 kJ is wasted as heat output. NOTE The problem was stated in terms of energy per unit time. We could just as well have stated it in terms of power, since 1J/s = 1 W. EXERCISE A An adiabatic process is defined as one in which no heat flows in system. If an ideal gas expands as shown in Fig. 20-5 (see also Fig. 19-8), done in this expansion equals the area under the graph, shown shaded. The this process would be e = W/Q, much greater than 100% (= oo since Q = violation of the second law?

or out of the the work W efficiency of 0). Is this a

It is clear from Eq.20-1b, e = 1 — Q;/Qy, that the efficiency of an engine wiq be greater if Q; can be made small. However, from experience with a wide variety of systems, it has not been found possible to reduce Q; to zero. If Q; could be reduced to zero we would have a 100% efficient engine, as diagrammed in Fig. 20-6.

FIGURE 20-6 Schematic diagram of a hypothetical perfect heat engine in which all the heat input would be used to do work.

That such a perfect engine (running continuously in a cycle) is not possible is another way of expressing the second law of thermodynamics: SECOND LAW OF THERMODYNAMICS (Kelvin-Planck statement)

No device is possible whose sole effect is to transform a given amount of heat completely into work.

This is known as the Kelvin-Planck statement of the second law of thermodynamics. Said another way, there can be no perfect (100% efficient) heat engine such as that diagrammed in Fig. 20-6. If the second law were not true, so that a perfect engine could be built, some rather remarkable things could happen. For example, if the engine of a ship did not need a low-temperature reservoir to exhaust heat into, the ship could sail acros the ocean using the vast resources of the internal energy of the ocean wate. Indeed, we would have no fuel problems at all!

532

CHAPTER 20

Second Law of Thermodynamics

20-3

Reversible and Irreversible

Processes; the Carnot Engine

f‘ln the early nineteenth century, the French scientist N. L. Sadi Carnot (1796-1832) studied in detail the process of transforming heat into mechanical energy. His aim had

been

to determine

how

to increase

the efficiency of heat engines, but

his

studies soon led him to investigate the foundations of thermodynamics itself. In 1824, Carnot invented (on paper) an idealized type of engine which we now call the Carnot engine. No Carnot engine actually exists, but as a theoretical idea it played an important role in the establishment and understanding of the second law of thermodynamics.

Reversible and Irreversible Processes The Carnot engine involves reversible processes, so before we discuss it we must discuss wflat is meant by reversible and irreversible processes. A reversible process is one that is carried out infinitely slowly, so that the process can be considered as a series of equilibrium states, and the whole process could be done in reverse with no change in magnitude of the work done or heat exchanged. For example, a gas contained in a cylinder fitted with a tight, movable, but frictionless piston could be compressed isothermally in a reversible way if done infinitely slowly. Not all very slow (quasistatic) processes are reversible, however. If there is friction present, for example (as between the movable piston and cylinder just mentioned), the work done in one direction (going from some state A to state B) will not be the negative of the work done in the reverse direction (state B to state A). Such a process would not be considered reversible. A perfectly reversible process is not possible in reality because it would require an infinite time; reversible processes can be approached arbitrarily closely, however, and they are very important theoretically. All real processes are irreversible: they are not done infinitely slowly. There rcould be turbulence in the gas, friction would be present, and so on. Any process

2ould not be done precisely in reverse since the heat lost to friction would not

reverse itself, the turbulence would be different, and so on. For any given volume there would not be a well-defined pressure P and temperature 7 since the system

would

not

always

be

in an

equilibrium

state. Thus

a real, irreversible, process

cannot be plotted on a PV diagram, except insofar as it may approach an ideal reversible process. But a reversible process (since it is a quasistatic series of equilibrium states) always can be plotted on a PV diagram; and a reversible process, when done in reverse, retraces the same path on a PV diagram. Although all real processes are irreversible, reversible processes are conceptually important, just as the concept of an ideal gas is.

Carnot’s Engine Now let us look at Carnot’s idealized engine. The Carnot engine makes use of a reversible cycle, by which we mean a series of reversible processes that take a given substance (the working substance) from an initial equilibrium state through many other equilibrium states and returns it again to the same initial state, In particular, the Carnot engine utilizes the Carnot cycle, which is illustrated in Fig. 20-7, with the working substance assumed to be an ideal gas. Let us take point a as the initial state. The gas is first expanded isothermally and reversibly, path ab, at temperature Ty, as heat Qy is added to it. Next the gas is expanded

adiabati-

cally and reversibly, path bc; no heat is exchanged and the temperature of the gas is reduced to Ty . The third step is a reversible isothermal compression, path cd, during which heat Q; flows out of the working substance. Finally, the gas is compressed adiabatically, path da, back to its original state. Thus a Carnot cycle consists offiwo isothermal and two adiabatic processes. ’. The net work done in one cycle by a Carnot engine (or any other type of engine asing a reversible cycle) is equal to the area enclosed by the curve representing the cycle on the PV diagram, the curve abed in Fig. 20-7. (See Section 19-7.) SECTION 20-3

0

Vv FIGURE 20-7 The Carnot cycle. Heat engines work in a cycle, and the cycle for the Carnot engine begins at point a on this PV diagram. (1) The gas is first expanded isothermally, with the addition of heat Oy, along the

path ab at temperature 7y. (2) Next the gas expands adiabatically from b to c—no heat is exchanged, but the temperature drops to 7. (3) The gas is then compressed at constant temperature 77 , path cd, and heat QO flows out. (4) Finally, the gas is compressed adiabatically, path da, back to its original state. No Carnot engine actually exists, but as a

theoretical idea it played an important role in the development of thermodynamics.

Reversible and Irreversible Processes; the Carnot Engine

533

Carnot Efficiency and the Second Law of Thermodynamics The efficiency of a Carnot engine, like any heat engine, is given by Eq. 20-1b: e

=

1

—

—=—.

On

For a Carnot engine using an ideal gas, however, we can show that the efficiency depends only on the temperatures of the heat reservoirs, 7y and 77.. In the first isothermal process ab in Fig. 20-7, the work done by the gas is (see Eq. 19-8)

S

Vi

Ws = nRTy 1n7", a

where n is the number of moles of the ideal gas used as working substance. Because the internal energy of an ideal gas does not change when the temperature remains constant, the first law of thermodynamics tells us that the heat added to the gas equals the work done by the gas: Oy H

==

N RTIYh ana

Similarly, the heat lost by the gas in the isothermal process cd is

= nRTiIn eVe

O

The paths bc and da are adiabatic, so we have from Eq. 19-15:

RV{

= BV]

where Y = Cp/Cy ideal gas law,

and

RV}

= BV},

is the ratio of molar specific heats (Eq. 19-14). Also, from the

v, Ty

_EV. T

. 4

EVa_FHAV, I, Ty

-

When we divide these last equations, term by term, into the corresponding set of equations on the line above, we obtain

TuVi'

= BVt

and

TV

= TV,

Next we divide the equation on the left by the one on the right and obtain

W) - ) Vb

or

¥=1

V.

¥=1

“ _ Y%

Vo

Va

Inserting this result in our equations for Qy and Q; above, we obtain

o

-_— on

_ — T,

=_

T,

[ Carnot cycle] ycle]

(20-2)

Hence the efficiency of a reversible Carnot engine can now be written

Q Cideal = 1 — Q_: or

gea

_

Iy

= 1 = Ty

Carnot efficiency;

[Kelvin temperatures | (203

The temperatures 7y, and Ty are the absolute or Kelvin temperatures as measured\ on the ideal gas temperature scale. Thus the efficiency of a Carnot engine depend. only on the temperatures 7} and Ty.

534

CHAPTER 20

Second Law of Thermodynamics

We could imagine other possible reversible cycles that could be used for an ideal reversible engine. According to a theorem stated by Carnot: r

All reversible engines operating between the same two constant temperatures Ty and T have the same efficiency. Any irreversible engine operating between the same two fixed temperatures will have an efficiency less than this.

This is known as Carnot’s theorem.” It tells us that Eq. 20-3, e = 1 — (T./Ty),

applies to any ideal reversible engine with fixed input and exhaust temperatures, Ty and T, and that this equation represents a maximum possible efficiency for a real (i.e., irr versible) engine. In practice, the efficiency of real engines is always less than the Carnot efficiency. \’T’ell-designed engines reach perhaps 60% to 80% of Carnot efficiency.

DOV

following output per APPROAC Eqgs. 20-1. SOLUTION

TP

A

phony

claim? An

engine

manufacturer

makes

the

claims: An engine’s heat input per second is 9.0kJ at 435 K. The heat second is 4.0 kJ at 285 K. Do you believe these claims? H The engine’s efficiency can be calculated from the definition, It must be less than the maximum possible, Eq. 20-3. | The claimed efficiency of the engine is (Eq. 20-1b) _ QL . 40K = 0.56,

¢=1-0.~ " " %0.

or 56%. Hc ywever, the maximum possible efficiency is given by the Carnot efficiency, Eq. 20-3: Cideal

=

i3

Lo,T

285K

sk

_

or 34%. The manufacturer’s claims violate the second law of thermodynamics and

cannot be believed.

r EXERCISE B A motor is running with an intake temperature 7y = 400K and an exhaust temperature engine? (a)

7 = 300 K.

Which

of the following is not a possible efficiency for the

0.10; (b) 0.16; (c) 0.24; (d) 0.30.

It is clear from Eq. 20-3 that a 100% efficient engine is not possible. Only if

the exhaust temperature, 7y, were at absolute zero would 100% efficiency be obtainable. But reaching absolute zero is a practical (as well as theoretical) impossibilit y. Thus we can state, as we already did in Section 20-2, that no device

is possible whose sole effect is to transform a given work. As wi e saw in Section 20-2, this is known as the second aw of thermodynamics. It tells us that efficient) heat engine such as the one diagrammed

amount of heat completely into the Kelvin-Planck statement of there can be no perfect (100% in Fig. 20-6.

EXERCISE C Return to the Chapter-Opening Question, page 528, and answer it again now. Try to explain why you may have answered differently the first time.

*The Otto Cycle The operat on of an automobile internal combustion engine (Fig. 20-4) can be approximat ed by a reversible cycle known as the Otto cycle, whose PV diagram is shown in Fig. 20-8. Unlike the Carnot cycle, the input and exhaust temperatures of the Otto cycle are not constant. Paths ab and cd are adiabatic, and paths bc and da are at constant volume. The gas (gasoline-air mixture) enters the cylinder at point a and is compressed adiabatically (compression stroke) to point b. At b ignition occurs (spark plug) and the burning of the gas adds heat Qy to the system at constant volume (approximately in a real engine). The temperature and pressure rise, .| A . and then in the power stroke, cd, the gas expands adiabatically. In the exhaust stroke, da, heat Q) is ejected to the environment (in a real engine, the gas leaves f‘he engine ?nd is replaced by a new mixture of air and fuel). |

"Carnot’s theorem can be shown to follow directly from either the Clausius or Kelvin-Planck statements of the second lbw of thermodynamics. *This result is known as the third law of thermodynamics, as discussed in Section 20-10.

FIGURE 20-8

The Otto cycle.

c

p

On

b d

a 0

0

g

1% SECTION 20-3

535

The Otto cycle. (¢) Show that for an ideal gas as working

substance, the efficiency of an Otto cycle engine is

o

- () Va

QL 0

v

FIGURE 20-8 (repeated for Example 20-3) The Otto cycle.

1=y

“

where 7 is the ratio of specific heats (¥ = Cp/Cy, Eq.19-14) and V,/V,, is the compression ratio. (b) Calculate the efficiency for a compression ratio V,/V;, = 8.0 assuming a diatomic gas like O, and N,. APPROACH We use the original definition of efficiency and the results from Chapter 19 for constant volume and adiabatic processes (Sections 19-8 and 19-9). SOLUTION The heat exchanges take place at constant volume in the ideal Otto cycle, so from Eq. 19-10a:

Quy = nCy(T,-T,)

and

Qp

= nCy(Ty — T,).

Then from Eq. 20-1b, e

=

1-—

&zl_[u] QH

Tc_Tb'

To get this in terms of the compression ratio, V,/V;, we use the result from Section 19-9, Eq. 19-15, PV” = constant during the adiabatic processes ab and cd. Thus BV!

=

PV}

and

P.V!

We use the ideal gas law, P = nRT/V,

T,Va' = T,vy ' and

=

PiV].

and substitute P into these two equations

T.VI' = Tyvi'.

Then the efficiency (see above) is

T

) P

L

But processes bc and da are at constant volume,so

Ve/Va = Vp/V, and

.

€=

[(vb/va)"l(n - Tb)] 3 T, ~F,

-

e U V., = V;, and V4 = V,. Hence

(vb)y-l g v,)

125

'

(b) For diatomic molecules (Section 19-8), ¥ = Cp/Cy = 1.4 so FIGURE 20-9 Schematic diagram of energy transfers for a refrigerator or air conditioner.

e =1-(80)""

= 1- (80)%

= 0.56.

Real engines do not reach this high efficiency because they do not follow perfectly the Otto cycle, plus there is friction, turbulence, heat loss and incomplete combustion of the gases.

20-4

Refrigerators, Air Conditioners, and Heat Pumps

The operating principle of refrigerators, air conditioners, and heat pumps is just the reverse of a heat engine. Each operates to transfer heat out of a cool environment into a warm environment. As diagrammed in Fig. 20-9, by doing work W, heat is taken from a low-temperature region, 7; (such as inside a refrigerator), and a greater amount of heat is exhausted at a high temperature, 7Ty (the room). You ca often feel this heated air blowing out beneath a refrigerator. The work W is usually done by an electric motor which compresses a fluid, as illustrated in Fig. 20-10.

536

CHAPTER 20

Second Law of Thermodynamics

e Low pressure N

Expansion

ngh

zwvalve

=

FIGURE 20-10

Pl'elssure

=\ Condenser /) coils (outside

refrigerator)

(to outside)

|

||

refrigerator to coils

)

via a valve, to low-pressure tubes on

the inside walls of the refrigerator; the liquid evaporates at this lower pressure and thus absorbs heat (QL) from the inside of the refrigerator. The fluid returns to the compressor where the cycle begins again. (b) Schematic diagram, like Fig. 20-9.

| |

" High pressure

vapor

(a)

Compressor motor

(a) Typical

refrigerator system. The electric compressor motor forces a gas at high pressure through a heat exchanger (condenser) on the rear outside wall of the refrigerator where Qy is given off and the gas cools to become liquid. The liquid passes from a high-pressure region,

» Plug

A perfi ect refrigerator—one in which no work is required to take heat from

the low-ten nperature region to the high-temperature region—is not possible. This

is the Clausius statement of the second law of thermodynamics, already mentioned in Section 2 20-1, which can be stated formally as )

r

No device is possible whose sole effect is to transfer heat from one system at

a temperature 77, into a second system at a higher temperature 7.

To make heat flow from a low-temperature object (or system) to one at a higher temperature, work must be done. Thus, there can be no perfect refrigerator. The coefficient of performance (COP) of a refrigerator is defined as the heat Q. remove d from the low-temperature area (inside a refrigerator) divided by the work W do ne to remove the heat (Fig. 20-9 or 20-10b): COP

=

O

SECOND LAW OF THERMODYNAMICS

(Clausius statement)

@

PHYSICS

APPLIED

@)rpHYSsics

ApPPLIED

Refrigerator

refrigerator and

W

[ air conditioner | (2048

This makes sense because the more heat Q; that can be removed from the inside of the refrigerator for a given amount of work, the better (more efficient) the refrigerator is. Energy is conserved, so from the first law of thermodynamics we can write (see Fig. 20-9 or 20-10b) Q; + W = Oy, or W = Qy — O, . Then Eq. 20-4a becomes

COP

%

=3,

oL

= Or — 0

refrigerator

[air conditioner | (20-4b)

For an ideal refrigerator (not a perfect one, which is impossible), the best we could do would be

_

Iy

El . = Ty — T,

refrigerator

[air conditioner | 20-49)

analagous to an ideal (Carnot) engine (Egs. 20-2 and 20-3). An air conditioner works very much like a refrigerator, although the actual construction details are different: an air conditioner takes heat Q; from inside a r[oorn or building at a low temperature, and deposits heat Qy outside to the envionment at a higher temperature. Equations 20—4 also describe the coefficient of performance for an air conditioner.

SECTION 20-4

Air conditioner

Refrigerators, Air Conditioners, and Heat Pumps

537

GV

IEY S Making ice. A freezer has a COP of 3.8 and uses 200 W of

power. How long would it take this otherwise empty freezer to freeze an ice-cube tray that contains 600 g of water at 0°C?

APPROACH In Eq. 20-4b, Q, is the heat that must be transferred out of thg\ water so it will become ice. To determine Q; , we use the latent heat of fusion L ¢

water and Eq. 19-3, Q = mL. SOLUTION From Table 19-2,

L =333kJ/kg.

Hence

Q =mL

=

200W

=

(0.600kg)(3.33 x 10°J/kg) = 2.0 X 10°J is the total energy that needs to be

removed

from

the

water.

The

freezer

does

work

at

the

rate

200J/s = W/t, which is the work W it can do in t seconds. We t = W/(200J/s). For W, we use Eq.20-4a: W = Q; /COP. Thus

W ©200J/s

of

solve for t:

Q/COP 200]/s

-

(2.0 x 10°7)/(3.8)

2007/s

= 260

or about 43 min. @

PHYSICS

APPLIED Heat pump

Heat naturally flows from high temperature to low temperature. Refrigerators and air conditioners do work to accomplish the opposite: to make heat flow from cold to hot. We might say they “pump” heat from cold areas to hotter areas, against the natural tendency of heat to flow from hot to cold, just as water can be pumped uphill, against the natural tendency to flow downhill. The term heat pump is usually reserved for a device that can heat a house in winter by using an electric motor that does work W to take heat Q; from the outside at low temperature and delivers heat Qy to the warmer inside of the house; see Fig. 20-11. A in a refrigerator, there is an indoor and an outdoor heat exchanger (coils « the refrigerator) and an electric compressor motor. The operating principle is like

FIGURE 20-11 A heat pump uses an electric motor to “pump” heat from the cold outside to the warm inside of a house.

A CAUTION

Heat pumps and air conditioners have different COP definitions

that

for a refrigerator

or air conditioner;

but

the

objective

of a heat

pump is to heat (deliver Qy), rather than to cool (remove Q). Thus, the coefficient of performance of a heat pump is defined differently than for an air conditioner because it is the heat Qy delivered to the inside of the house that is important now:

Oy

COP

= W

[heat pump]

(20-5)

~ 1he COP is necessarily greater than 1. Most heat pumps can be “turned around” ~ and used as air conditioners in the summer. m Heat pump. A heat pump has a coefficient of performance of 3.0 and is rated to do work at 1500 W. () How much heat can it add to a room per second? (b) If the heat pump were turned around to act as an air conditioner in the summer, what would you expect its coefficient of performance to be, assuming all else stays the same? APPROACH We use the definitions of coefficient of performance, which are different for the two devices in (a) and (b). SOLUTION (a) We use Eq. 20-5 for the heat pump, and, since our device does 15007 of work per second, it can pour heat into the room at a rate of Oy

=

COP X W

=

per second, or at a rate of 4500 W.

538

CHAPTER20

Second Law of Thermodynamics

3.0X 1500]

=

4500J

-

(b) If our device is turned around in summer, it can take heat Q; from inside the house, doing 1500 J of work per second to then dump Qy = 4500] per second to the hot outside. Energy is conserved, so Q; + W = Qy (see Fig. 20-11, but reverse the inside and outside of the house). Then

QL

= Oy — W

= 4500J — 1500J

= 3000J.

The coefficient of performance as an air conditioner would thus be (Eq. 20—4a)

_ S

=

_ 30007 _

w1500

o

20-5 Entropy Thus far we have stated the second law of thermodynamics for specific situations. What we really need is a general statement of the second law of thermodynamics

that will cover all situations, including those discussed earlier in this Chapter that

are not observed in nature even though they would not violate the first law of thermodynamics. It was not until the latter half of the nineteenth century that the second law of thermodynamics was finally stated in a general way—namely, in terms of a quantity called entropy, introduced by Clausius in the 1860s. In Section 20-7 we will see that entropy can be interpreted as a measure of the order or disorder of a system. When we deal with entropy—as with potential energy—it is the change in entropy during a process that is important, not the absolute amount. According to Clausius, the change in entropy S of a system, when an amount of heat Q is added to it by a reversible process at constant temperature, is given by AS S

’h

=

Q

=T

20(20-6)

‘here T is the kelvin temperature. If the temperature is not constant, we define entropy S by the relation d as = 2.

(20-7)

Then the change in entropy of a system taken reversibly between two states a and b is

given by’

AS

_= S, — S, _= | (e dS _=

[PdQT

.

[reversible process]

(20-8)

A careful analysis (see next page) shows that the change in entropy when a system moves by a reversible process from any state a to another state b does not depend on the proc ess. That is, AS = S, — S, depends only on the states a and b of the system. Thu s entropy (unlike heat) is a state variable. Any system in a given state has a tempe rature, a volume, a pressure, and also has a particular value of entropy. It is eas y to see why entropy is a state variable for a Carnot cycle. In Eq. 20-2 we saw that Q,/Qy = T;/Ty, which we rewrite as

9 _ 9 T,

Ty

In the PV diagram for a Carnot cycle, Fig. 20-7, the entropy change AS = Q/T in going from state a to state ¢ along path abc (=Qy/Ty + 0) is thus the same as going along the path adc. That is, the change in entropy is path independent—it depends only on the initial and final states of the system. (F.quation 20-8 says nothing about the absolute value of S; it only gives the change in S. This is much * e potential e nergy (Chapter 8). However, one form of the so-called third law of thermodynamics (see also Section 20 ~10) states that as 7 — 0, S — 0.

SECTION 20-5

Entropy

539

* Showing Entropy Is a State Variable In our study of the Carnot cycle we found (Eq. 20-2) that rewrite this as

T

Q;/Qy

= T;/Ty.

We

T

In this relation, both Qy and Q; are positive. But now let us recall our original convention as used in the first law (Section 19-6), that Q is positive when it represents a heat flow into the system (as Q) and negative for a heat flow out of the system (as —Q;). Then this relation becomes

%, &

0 v FIGURE 2012 Any reversible cycle can be approximated as a series of Carnot cycles. (The dashed lines represent isotherms.)

Ly

T

=

0.

[Carnot cycle]

D

(20-9)

Now consider any reversible cycle, as represented by the smooth (oval-shaped) cur;fe 1;_F1g. 23(;1122 Afily rever181b!e cyc}ie can l‘tlae appré)x%nzzitel.d as a series of Ca(ringt cycles. Figure 20-12 shows only six—the isotherms (dashed lines) are connected by adiabatic paths for each—and the approximation becomes better and better if we increase the number of Carnot cycles. Equation 20-9 is valid for each of these cycles, so we can write 2—?—

=0

[Carnot cycles]

(20-10)

for the sum of all these cycles. But note that the heat output Q; of one cycle crosses the boundary below it and is approximately equal to the negative of the heat input, Qy, of the cycle below it (actual equality occurs in the limit of an infinite number of infinitely thin Carnot cycles). Hence the heat flows on the inner paths of all these Carnot cycles cancel out, so the net heat transferred, and the work done, is the same for the series of Carnot cycles as for the original cycle. Hence, in the limit of infinitely many Carnot cycles, Eq. 20-10 applies to any reversible cycle. In this case Eq. 20-10 becomes 3€ %—,Q— =

FIGURE 20-13

The integral, § dS,

of the entropy for a reversible cycle is zero. Hence the difference in entropy between states a and b, Sb — Sa = [, dS, is the same for path I as for path II.

0,

[reversible cycle]

where dQ represents an infinitesimal heat flow." The integral is taken around a closed path; the integral can the path such as at a or b in Fig. 20-12, and proceed divide the cycle of Fig. 20-12 into two parts as indicated J ° @ a T I

+

J ¢ d_Q b T 1

=0

et

b J 49 I

0

%

symbol ¢ means that the be started at any point on in either direction. If we in Fig. 20-13, then

'

The first term is the integral from point a to point b along path I in Fig. 20-13, and the second term is the integral from b back to a along path II. If path II is taken in reverse, dQ at each point becomes

a

(20—11‘

a T

=

J 1l

b

.

a T

—dQ, since the path is reversible. Therefore

[reversible paths]

(20-12)

The integral of dQ/T between any two equilibrium states, a and b, does not depend on the path of the process. By defining entropy as dS = dQ/T (Eq.20-7), we see from Eq. 20-12 that the change in entropy between any two states along a reversible path is independent of the path between two points a and b. Thus entropy is a state variable—its value depends only on the state of the system, and not on the process or the past history of how it got there.* This is in clear distinction to Q and W which are not state variables; their values do depend on the processes undertaken. "dQ is often written d@Q: see footnote at the end of Section 19-6. *Real processes are irreversible. Because entropy is a state variable, the change in entropy AS f an irreversible process can be determined by calculating AS for a reversible process between the sam. two states.

540

CHAPTER 20

Second Law of Thermodynamics

20-6

|

Entropy and the Second Law of

- Thermodynamics

’e have de%ined a new quantity, S, the entropy, which can be used to describe the state of the‘ system, along with P, T, V, E;;, and n. But what does this rather abstract quantity have to do with the second law of thermodynamics? To answer this, let us tthke some examples in which we calculate the entropy changes during particular processes. But note first that Eq. 20-8 can be applied only to reversible processes. How then do we calculate AS = S, — S, for a real process that is irreversible? What we can do is this: we figure out some other reversible process that takes

the system

between

the

same

two

states, and

calculate

AS

for this

reversible process. This will equal AS for the irreversible process since AS depends only on the initial and final states of the system. If the temperature varies during a process, a summation of the heat flow over the changing temperature can often be calculated using calculus or a computer. However, if the temperature change is not too great, a reasonable approximation can be made using the average value of the temperature, as indicated in the next Example.

DGVIANSPHEIN

ESTIMATE | Entropy change when mixing water. A sample

of 50.0 kg of water at 20.00°C is mixed with 50.0 kg of water at 24.00°C. Estimate the change in entropy. APPROACH The final temperature of the mixture will be 22.00°C, since we started with equal amounts of water. We use the specific heat of water and the methods of calorimetry (Sections 19-3 and 19-4) to determine the heat transferred. Then we use the average temperature of each sample of water to estimate the entropy change (AQ/T). SOLUTION

»

A quantity of heat,

Q = mcAT = (50.0kg)(4186J/kg-C°)(2.00C°) = 4.186 x 10°7, flows out of the hot water as it cools down from 24°C to 22°C, and this heat flows

into the cold water as it warms from 20°C to 22°C. The total change in entropy,

AS, will be the sum of the changes in entropy of the hot water, ASy, and that of the cold wqter, ASc:

AS = ASy + ASc. We estimatje entropy changes by writing AS = Q/7, where T is an “average” temperature for each process, which ought to give a reasonable estimate since the temperature change is small. For the hot water we use an average temperature of 23°C (296 K), and for the cold water an average temperature of 21°C (294 K). Thus

ASu ~

4186 X 10°] = —1414J/K 296 K

which is negative because this heat flows out, whereas heat is added to the cold water:

ASe ®

4.186 % 10°]

ok

=

1424J/K.

The entropy of the hot water (S};) decreases since heat flows out of the hot water. But the entropy of the cold water (Sc) increases by a greater amount. The total change in entropy is AS = ASy + ASc = —1414J/K + 1424J/K =~ 10J/K. \ We see thTt although the entropy of one part of the system decreased, the ntropy of the other part increased by a greater amount so that the net change in entropy of }he whole system is positive.

L

SECTION 20-6

Entropy and the Second Law of Thermodynamics

541

We can now show in general that for an isolated system of two objects, the

flow of heat from the higher-temperature (7;;) object to the lower-temperature (Tp) object eventually object (Qy one (Q,=

always come to = —Q, Q), so

results in an increase in the total entropy. The two objects some intermediate temperature Ty - The heat lost by the hotter where Q is posmve) is equal to the heat gained by the colde“ the total change in entropy is

AS = ASy + &S, = —o2 4 L THM

TLM

where Ty is some intermediate temperature between Ty and Ty, for the hot object as it cools from Ty to Ty, and T} is the counterpart for the cold object. Since the temperature of the hot object is, at all times during the process, greater than that of the cold object, then Ty > Ty . Hence

One object decreases in entropy, while the other gains in entropy, but the total change is positive.

DGR

IEYE

Entropy

changes

in

a

free

expansion. Consider

the

adiabatic free expansion of n moles of an ideal gas from volume V; to volume V,, where V, > V] as was discussed in Section 19-7, Fig. 19-14. Calculate the change in entropy (a) of the gas and (b) of the surrounding environment. (c) Evaluate AS for 1.00 mole, with

APPROACH of volume

V, = 2.00 V;.

We saw in Section 19-7 that the gas is initially in a closed container

V;, and, with the opening of a valve, it expands

adiabatically into a

previously empty container. The total volume of the two containers is V,. TH whole apparatus is thermally insulated from the surroundings, so no heat flows into the gas, Q = 0. The gas does no work, W = 0, so there is no change in internal energy, AE;;; = 0, and the temperature of the initial and final states is the same, 7, = T} = T. The process takes place very quickly, and so is irreversible. Thus we cannot apply Eq. 20-8 to this process. Instead we must think of a reversible process that will take the gas from volume V; to V, at the same temperature, and use Eq. 20-8 on this reversible process to get AS. A reversible isothermal process will do the trick; in such a process, the internal energy does not change, so from the first law,

dQ SOLUTION

=

dW

=

PdV.

(a) For the gas, ASges

=

dQ [T

=

1 (% TJVPdV.

The ideal gas law tells us P = nRT/V, Vs ASgas

=

nRTJ

T

Vi

so Vv

_d_Z

%

=

v,:

Since V, > V;, ASg, > 0.

(b) Since no heat is transferred to the surrounding

change

of the state of the environment

environment, there is no

due to this process. Hence

ASc,, = 0.

Note that the total change in entropy, AS,. + AScny, is greater than zero. (c) Since n =1.00 and V, = 2.00V;, then AS,, = RIn2.00 = 5.76 J/K.

542

CHAPTER 20

Second Law of Thermodynamics

DG IEE N Heat transfer. A red-hot 2.00-kg piece of iron at temperature T, = 880K is thrown into a huge lake whose temperature is 7, = 280 K. Assume the lake is so large that its temperature rise is insignificant. Determine the chang'r in entropy (a) of the iron and (b) of the surrounding environment

© (the lake).

APPROACH The process is irreversible, but the same entropy change will occur for a reversible process, and we use the concept of specific heat, Eq. 19-2. SOLUTION (a) We assume the specific heat of the iron is constant at ¢ = 450J/kg-K. Then dQ = mc dT and in a quasistatic reversible process ASion

_[dQ = J T

T, mclnTl

=

T; mclnT2

880K —(2.00kg)(450J/kg-K)In B0K

-

—1030J/K.

=

mcL]

T dT T

=

Putting in numbers, we find ASion

=

(b) The initial and final temperatures of the lake are the same, 7' = 280 K. The

lake receives from the iron an amount of heat

Q = mc(T, - T}) = (2.00kg)(4507/kg-K)(880K — 280K)

= 540kJ.

Strictly sp‘eaking, this is an irreversible process (the lake heats up locally before equilibrium is reached), but is equivalent to a reversible isothermal transfer of heat

Q = 540kJ

at 7 = 280 K.

ASchy

=

540 kJ

280K

Hence

1930 J/K.

Thus, although the entropy of the iron actually decreases, the fotal change in entropy oi iron plus environment is positive: 1930J/K — 1030J/K = 900J/K. AEXERCISE DA 1.00-kg piece of ice at 0°C melts very slowly to water at 0°C. Assume the ice is in contact with a heat reservoir whose temperature is only infinitesimally greater ; than 0°C. [T‘,termine the entropy change of (a) the ice cube and (b) the heat reservoir.

In each of these Examples, the entropy of our system plus that of the environment (or surroundings) either stayed constant or increased. For any reversible process, such as that in Exercise D, the total entropy change is zero. This can be seen in general as follows: any reversible process can be considered as a series of quasistatic isothermal transfers of heat AQ between a system and the environment, which differ in temperature only by an infinitesimal amount. Hence the change in entropy of either the system or environment is AQ/7T and that of the other is —AQ/T, so the total is AS

=

ASyy

+ ASeny

=

0.

[any reversible process]

In Examples 20-6, 20-7, and 20-8, we found that the total entropy of system plus environment increases. Indeed, it has been found that for all real (irreversible) processes, the total entropy increases. No exceptions have been found. We can thus make the general statement of the second law of thermodynamics as follows: The entropy of an isolated system never decreases. It either stays constant (reversible processes) or increases (irreversible processes).

Since all real processes are irreversible, we can equally well state the second law as: The total entropy of any system plus that of its environment increases as a result of any natural process:

AS

= ASyy + ASeq > 0.

(20-13)

SECOND LAW OF THERMODYNAMICS (general statement)

Ithough thF entropy of one part of the universe may decrease in any process (see «e Examples above), the entropy of some other part of the universe always increases by a greater amount, so the total entropy always increases.

SECTION 20-6

Entropy and the Second Law of Thermodynamics

543

Now that we finally have a quantitative general statement of the second law of thermodynamics, we can see that it is an unusual law. It differs considerably from other laws of physics, which are typically equalities (such as F = ma) or conservation laws (such as for energy and momentum). The second law of thermodynamics introduces a new quantity, the entropy, S, but does not tell us it is conserved. Quite the opposit™™ Entropy is not conserved in natural processes. Entropy always increases in time.

“Time’s Arrow” The second law of thermodynamics summarizes which processes are observed in nature, and which are not. Or, said another way, it tells us about the direction processes go. For the reverse of any of the processes in the last few Examples, the entropy would decrease; and we never observe them. For example, we never observe

heat flowing spontaneously from a cold object to a hot object, the reverse of Example 20-8. Nor do we ever observe a gas spontaneously compressing itself into a smaller volume, the reverse of Example 20-7 (gases always expand to fill their containers). Nor do we see thermal energy transform into kinetic energy of a rock so the rock rises spontaneously from the ground. Any of these processes would be consistent with the first law of thermodynamics (conservation of energy). But they are not consistent with the second law of thermodynamics, and this is why we need the second law. If you were to see a movie run backward, you would probably realize it immediately because you would see odd occurrences—such as rocks rising spontaneously from the ground, or air rushing in from the atmosphere to fill an empty balloon (the reverse of free expansion). When watching a movie or video, we are tipped off to a faked reversal of time by observing whether entropy is increasing or decreasing. Hence entropy has been called time’s arrow, for it can tell us in which direction time is going.

20-7

Order to Disorder

The concept of entropy, as we have discussed it so far, may seem rather abstra% But we can relate it to the more ordinary concepts of order and disorder. In fa the entropy of a system can be considered a measure of the disorder of the systent. Then the second law of thermodynamics can be stated simply as: SECOND LAW OF THERMODYNAMICS (general statement)

544

CHAPTER20

Natural processes tend to move toward a state of greater disorder. Exactly what we mean by disorder may not always be clear, so we now consider a few examples. Some of these will show us how this very general statement of the second law applies beyond what we usually consider as thermodynamics. Let us look at the simple processes mentioned in Section 20-1. First, a jar containing separate layers of salt and pepper is more orderly than when the salt and pepper are all mixed up. Shaking a jar containing separate layers results in a mixture, and no amount of shaking brings the layers back again. The natural process is from a state of relative order (layers) to one of relative disorder (a mixture), not the reverse. That is, disorder increases. Next, a solid coffee cup is a more “orderly” and useful object than the pieces of a broken cup. Cups break when they fall, but they do not spontaneously mend themselves (as faked in Fig. 20-1). Again, the normal course of events is an increase of disorder. Let us consider some processes for which we have actually calculated the entropy change, and see that an increase in entropy results in an increase in disorder (or vice versa). When ice melts to water at 0°C, the entropy of the water increases (Exercise D). Intuitively, we can think of solid water, ice, as being more ordered than the less orderly fluid state which can flow all over the place. This change from order to disorder can be seen more clearly from the molecular point of view: the orderly arrangement of water molecules in an ice crystal has changed to the disorderly and somewhat random motion of the molecules in the fluid state. When a hot object is put in contact with a cold object, heat flows from tb\ high temperature to the low until the two objects reach the same intermedic temperature. At the beginning of the process we can distinguish two classes of

Second Law of Thermodynamics

molecules: those with a high average kinetic energy (the hot object), and those with a low average kinetic energy (the cooler object). After the process in which heat flows, all the molecules are in one class with the same average kinetic energy. We no longer have the more orderly arrangement of molecules in two f asses. Order has gone to disorder. Furthermore, the separate hot and cold objects could serve as the hot- and cold-temperature regions of a heat engine, and thus could be used to obtain useful work. But once the two objects are put in contact and reach the same temperature, no work can be obtained. Disorder has increased, since a system that has the ability to perform work must surely be considered to have a higher order than a system no longer able to do work. When a stone falls to the ground, its macroscopic kinetic energy is transformed to thermal energy. Thermal energy is associated with the disorderly random motion of molecules, but the molecules in the falling stone all have the same velocity downward in addition to their own random velocities. Thus, the more orderly kinetic energy of the stone as a whole is changed to disordered thermal energy when the stone strikes the ground. Disorder increases in this process, as it does in all processes that occur in nature.

*Biological Evolution An interesti'ng example of the increase in entropy relates to biological evolution and to grow}h of organisms. Clearly, a human being is a highly ordered organism. The theory of evolution describes the process from the early macromolecules and simple forms of life to Homo sapiens, which is a process of increasing order. So, too, the development of an individual from a single cell to a grown person is a process of increasing order. Do these processes violate the second law of thermodynamics? fio, they do not. In the processes of evolution and growth, and even during the mature life of an individual, waste products are eliminated. These small molecules tl'}at remain as a result of metabolism are simple molecules without much order. Thus they represent relatively higher disorder or entropy. Indeed, the total entropy of the molecules cast aside by organisms during the processes of “olution and growth is greater than the decrease in entropy associated with the * rder of the growing individual or evolving species.

@ PHYSICS APPLIED Biological Evolution and Development

20-8 Unavailability of Energy; Heat Death In the procer of heat conduction from a hot object to a cold one, we have seen that entropy incr ases and that order goes to disorder. The separate hot and cold objects could serve as the high- and low-temperature regions for a heat engine and thus could be used to obtain useful work. But after the two objects are put in contact with each other and reach the same uniform temperature, no work can be obtained from them. With regard to being able to do useful work, order has gone to disorder in this process. The same can be said about a falling rock that comes to rest upon striking the ground. Before hitting the ground, all the kinetic energy of the rock could have been used t do useful work. But once the rock’s mechanical kinetic energy becomes thermal energy, doing useful work is no longer possible. Both these examples illustrate another important aspect of the second law of thermodynamics: in any natural process, some energy becomes unavailable to do useful work. In any process, no energy is ever lost (it is always conserved). Rather, energy becomes less useful—it can do less useful work. As time goes on, energy is degraded, in a sense; it goes from more orderly forms (such as mechanical) eventually to the least orderly form, internal, or thermal, energy. Entropy is a factor here because the amount of energy that becomes unavailable to do work is roportional to the change in entropy during any process.” «t can be shown that the amount of energy that becomes unavailable to do useful work is equal to T AS, where Ty is the lowest available temperature and AS is the total increase in entropy during the process.

SECTION 20-8

Unavailability of Energy; Heat Death

545

A natural outcome of the degradation of energy is the prediction that as time goes on, the universe should approach a state of maximum disorder. Matter would become a uniform mixture, and heat will have flowed from high-temperature regions to low-temperature regions until the whole universe is at one temperature. No work could then be done. All the energy of the universe would have degradetfl to thermal energy. This prediction, called the heat death of the universe, has been much discussed, but would lie very far in the future. It is a complicated subject, and some scientists question whether thermodynamic modeling of the universe is possible or appropriate.

*20-9 Statistical Interpretation of Entropy and the Second Law

The ideas of entropy and disorder are made clearer with the use of a statistical or probabilistic analysis of the molecular state of a system. This statistical approach, which was first applied toward the end of the nineteenth century by Ludwig Boltzmann (1844-1906), makes a clear distinction between the “macrostate” and the “microstate” of a system. The microstate of a system would be specified in giving the position and velocity of every particle (or molecule). The macrostate of a system is specified by giving the macroscopic properties of the system—the temperature, pressure, number of moles, and so on. In reality, we can know only the macrostate of a system. We could not possibily know the velocity and position of every one of the huge number of molecules in a system at a given moment. Nonetheless, we can hypothesize a great many different microstates that can correspond to the same macrostate. Let us take a very simple example. Suppose you repeatedly shake four coins in your hand and drop them on a table. Specifying the number of heads and the number of tails that appear on a given throw is the macrostate of this system. Specifying each coin as being a head or a tail is the microstate of the system. In th*™ following Table we see how many microstates correspond to each macrostate:

Macrostate

e

4 heads HHHH 3heads,1tal HHHT, 2heads2taills. HHTT, 1head 3 tails - - TTTH; 4 tails TTA:L

Possible Microstates

Number of

(H = heads, T = tails) HHTH, HTHH, THHH HTHT. THHT, HTTH, THTH,. TTHT. THTT, -HTTT

Microstates

TTHH

1 4 6 4 1

A basic assumption behind the statistical approach is that each microstate is equally probable. Thus the number of microstates that give the same macrostate corresponds to the relative probability of that macrostate occurring. The macrostate of two heads and two tails is the most probable one in our case of tossing four coins; out of the total of 16 possible microstates, six correspond to two heads and two tails, so the probability of throwing two heads and two tails is 6 out of 16, or 38%. The probability of throwing one head and three tails is 4 out of 16, or 25%. The probability of four heads is only 1 in 16, or 6%. If you threw the coins 16 times, you might not find that two heads and two tails appear exactly 6 times, or four tails exactly once. These are only probabilities or averages. But if you made 1600 throws, very nearly 38% of them would be two heads and two tails. The greater the number of tries, the closer the percentages are to the calculated probabilities. EXERCISE E In the Table above, what is the probability that there will be at least two heads.

(@ 3: (0)153 () §: (d) §5 (o) T 546

CHAPTER 20

Second Law of Thermodynamics

If we toss more coins—say, 100 all at the same time—the relative probability of throwing all heads (or all tails) is greatly reduced. There is only one microstate corresponding

since each

to all heads. For

99 heads

and

1 tail, there

are 100 microstates

of the coins could be the one tail. The relative probabilities for other

1acrostates are given in Table 20-1. About 1.3 X 10* microstates are possible.’ Thus the relative probability of finding all heads is 1 in 10%, an incredibly

unlikely

event!

The

probability

of

obtaining

50

heads

and

50

tails

(see

Table 20-1) is (1.0 X 10%)/1.3 X 10% = 0.08 or 8%. The probability of obtaining anything between 45 and 55 heads is over 70%. TABLE 20-1

Probabilities of Various Macrostates for 100 Coin Tosses

Macrostate

Heads

Number of

Tails

100 99 90

0 10

80

|

20

60 55 50 45 40 20 10 1

Probability

1 1.0 x 10? 17 et

7.9 x 10731 7.9 X 1072 1451079

14 6.1 1.0 6.1 34 5.4 1.7 1.0

0.01 0.05 0.08 0.05 0.01 42 %1071 1.4 x 107V 798 1074 7.9:x:10~

5.4 x 100

40 45 50 55 60 80 9 99 100

r

Microstates

x x x x % % x X 1

42 x 10710

1038 10%8 10% 10%8 10% 10% 101 10%

Thus we see that as the number of coins increases, the probability of obtaining

the most orderly arrangement

(all heads or all tails) becomes extremely unlikely.

The least orderly arrangement (half heads, half tails) is the most probable, and the probability of being within, say, 5% of the most probable arrangement greatly increases as the number of coins increases. These same ideas can be applied to the molecules of a system. For example, the most probable state of a gas (say, the air in a room) is one in which the molecules take up the whole space and move about randomly; this corresponds to the Maxwellian distribution, Fig. 20-14a (and see Section 18-2). On the other hand, the very orderly arrangement of all the molecules located in one corner of the room and all moving with the same velocity (Fig. 20-14b) is extremely unlikely. From these examples, it is clear that probability is directly related to disorder and hence to entropy. That is, the most probable state is the one with greatest entropy or greatest disorder and randomness. Boltzmann showed, consistent with Clausius’s definition (dS = dQ/T), that the entropy of a system in a given (macro) state can be written S

=

klnW,

(k = R/N, = 1.38 X 102 J/K)

L coin

has

two

2t

possibilities,

heads

or

|

tails. Then

the

|

possible

2X2%X2X T =210 = 1,27 x 10% (using a calculator or logarithms).

*SECTION 20-9

number

and W is the

|

of microstates

a gas (Maxwellian, or random);

(b) orderly, but highly unlikely, distribution of speeds in which all =~ molecules have nearly the same speed. 7]

% . = s .§ g “‘0

Speed, v ' @)

-;-'%

number of microstates corresponding to the given macrostate. That is, W' is proportional to the probability of occurrence of that state. W is called the thermodynamic probability, or, sometimes, the disorder parameter. cach

(a) Most probable

QA

(20-14)

where k is Boltzmann’s constant

FIGURE 20-14

distribution of molecular speeds in

|

is

Q

g

s é g

S0

Speed.v (b)

Statistical Interpretation of Entropy and the Second Law

547

Free

expansion—statistical

determination

of entropy.

Use Eq.20-14 to determine the change in entropy for the adiabatic free expansion of a gas, a calculation we did macroscopically in Example 20-7. Assume W, the number of microstates for each macrostate, is the number of possible positions. APPROACH We assume the number of moles is #» = 1, and then the number ¢ moleculesis N = nNy = N,. We let the volume double, just as in Example 20-7. Because the volume doubles, the number of possible positions for each molecule doubles. SOLUTION When the volume doubles, each molecule has two times as many positions (microstates) available. For two molecules, the number of total microstates increases by 2 X 2 = 2% For N, molecules, the total number of microstates increases by a factor of 2 X 2 X 2 X --- = 2M, That is

2Wi = 2%

The change in entropy is, from Eq. 20-14, (.‘W'

AS =S, — S, = k(InW, — InW;) = kanZ = kIn2™M = kN,In2 = RIn2 1

which is the same result we obtained in Example 20-7. In terms of probability, the second law of thermodynamics—which tells us that entropy increases in any process—reduces to the statement that those processes occur which are most probable. The second law thus becomes a trivial statement. However there is an additional element now. The second law in terms of probability does not forbid a decrease in entropy. Rather, it says the probability is extremely low. It is not impossible that salt and pepper could separate spontaneously into layers, or that a broken tea cup could mend itself. It is even possible that a lake could freeze over on a hot summer day (that is, heat flow out of the cold lake into the warmer surroundings). Buey the probability for such events occurring is extremely small. In our col. examples, we saw that increasing the number of coins from 4 to 100 drastically reduced the probability of large deviations from the average, or most probable, arrangement. In ordinary systems we are not dealing with 100 molecules, but with incredibly large numbers of molecules: in 1 mol alone there are 6 X 107 molecules. Hence the probability of deviation far from the average is incredibly tiny. For example, it has been calculated that the probability that a stone resting on the ground should transform 1 cal of thermal energy into mechanical energy and rise up into the air is much less likely than the probability that a group of monkeys typing randomly would by chance produce the complete works of Shakespeare.

*20-10

Thermodynamic Temperature; Third Law of Thermodynamics

In Section 20-3 we saw for a Carnot cycle that the ratio of the heat absorbed Qy from the high-temperature reservoir and the heat exhausted Q; to the low-temperature reservoir is directly related to the ratio of the temperatures of the two reservoirs (Eq. 20-2):

On

Ty

This result is valid for any reversible engine and does not depend on the working substance. It can thus serve as the basis for the Kelvin or thermodynamic temperature scale. We use this relation and the ideal gas temperature scale (Section 17-10) 1 complete the definition of the thermodynamic scale: we assign the value

548

CHAPTER 20

Second Law of Thermodynamics

Ty, = 273.16 K to the triple point of water so that

|

T

=

(273.16 K)(g),

O

vhere Q and Qy, are the magnitudes of the heats exchanged by a Carnot engine with reservoirs at temperatures 7 and Ti,. Then the thermodynamic scale is identical to the ideal gas scale over the latter’s range of validity. Very low temperatures are difficult to obtain experimentally. The closer the temperature is to absolute zero, the more difficult it is to reduce the temperature further, and it is generally accepted that it is not possible to reach absolute zero in any finite number of processes. This last statement is one way to state’ the third law of therrnodynamics. Since the maximum efficiency that any heat engine can have is the Carnot efficiency

and since 7} can never be zero, we see that a 100% efficient heat engine is not possible. FIGURE 20-15 (a) An array of mirrors focuses sunlight on a boiler to produce steam at a solar energy installation. (b) A fossil-fuel steam plant (this one uses forest waste products, biomass). (c) Large cooling towers at an electric generating plant.

*20~11 Thermal Pollution, Global Warming,

and Energy Resources

Much of the energy we utilize in everyday life—from motor vehicles to most of the electricity produced by power plants—makes use of a heat engine. Electricity produced by falling water at dams, by windmills, or by solar cells (Fig. 20-15a) does not involve a heat engine. But over 90% of the electric energy produced in the US. is generated at fossil-fuel steam plants (coal, oil, or gas—see Fig. 20-15b), and they make use of a heat engine (essentially steam engines). In electric power plants, the steam drives the turbines and generators (Fig. 20-16) whose output is electric energy. The various means to turn the turbine are discussed briefly in Table 20-2 (next page), along with some of the advantages and disadvantages of each. Even nuclear power plants use nuclear fuel to run a steam engine. The heat output Q; from every heat engine, from power plants to cars, is referred to as thermal pollution because this heat (Q;) must be absorbed by the environment—such as by water from rivers or lakes, or by the air using large cooling towers (Fig. 20-15c). When water is the coolant, this heat raises the temperature of the water, altering the natural ecology of aquatic life (largely because warmer water holds less oxygen). In the case of air cooling towers, the @output heat\QL raises the temperature of the atmosphere, which affects the weather.

FIGURE 20-16 Mechanical energy is transformed to electric energy with a turbine and generator. Source of energy: water, steam, or wind Electric generator

Electric energy

FPHYSICS

APPLIED

Heat engines and thermal pollution

"See also the statement in the footnote on p. 539.

*SECTION 20-11

Thermal Pollution, Global Warming, and Energy Resources

549

TABLE 20-2 Electric Energy Resources Fors of Electele

% of Production (approx.)

Energy Production

US.

World

Advantages

Fossil-fuel steam plants: burn coal, oil, or natural gas to boil water, producing high-pressure steam that turns a turbine of a generator (Figs. 20-3b, 20-16); uses heat engine.

71

66

We know how to build them; for now relatively inexpensive.

Disadvantages

fi

Air pollution; thermal pollution; limited efficiency; land devastation from extraction of raw materials (mining); global warming; accidents such as oil spills at sea; limited fuel supply (estimates range from a

couple of decades to a few centuries).

Nuclear energy: Fission: nuclei of uranium or plutonium atoms split (“fission”) with release of energy (Chapter 42) that heats steam; uses heat engine. Fusion: energy released when isotopes of hydrogen (or other small nuclei) combine or “fuse” (Chapter 42). Hydroelectric: falling water

turns turbines at the base of a dam.

20

16

0

0

7

16

:

Geothermal: natural steam from inside the Earth comes to the

X 10°N/C= 4.5 X 105N/C, and its dlr‘ectlon is ¢ given by tan¢ = E,y/Exx = 4.4/1.1 = 4.0, so ¢ = 76°. (b) Because B is equidistant from the two equal charges (40 cm by the Pythagorean theorem), ‘the magnitudes of Eg; and Ey, are the same; that is,

4

|

E

==

=

E

e

=

k0 2

—_—=

(9.0 x 10°N-m?/C?)(50 x 1076 C) (0.40 m)?

2.8 X 105N/C.

Also, because of the symmetry, the y components are equal and opposite, and so cancel out. ‘Hence the total field Ej is horizontal and equals Eg; cos 6 + Ey,cosf = 2Eg; cos 0. From the diagram, cos# = 26 cm/40cm = 0.65. Then

)=) PROBLEM

SOLVING

Use symmeltry to save work,

when possible

Ey = 2Eg cosf = 2(2.8 X 10°N/C)(0.65)

‘ = 3.6 X 10°N/C, and the direction of Ej is along the +x direction. NOTE We could have done part (b) in the same way we symmetry‘allowed us to solve the problem with less effort.

did part (a). But

& sOLV, o ;

X

3o

¢

‘ Electrostatics: Electric Forces and T Electnc; Fields Solving electrostatics problems follows, to a large extent, the general problem-solving procedure discussed in Section 4-8. Whether you use electric field or electrostatic forces, the procedure is similar: 1. Draw a careful diagram—namely, a free-body diagram for each object, showing all the forces acting on thaf object, or showing the electric field at a point due to all significant charges present. Determine the direction of each force or electric field physically: like cflarges repel each other, unlike charges attract; fields point away from a + charge, and toward

a — charge. Show and label each vector force or field on your diagram. 2. Apply Coulomb’s law to calculate the magnitude of the force that each contributing charge exerts on a charged object, or the magnitude of the electric field each charge produces at a given point. Deal only with magnitudes of charges (leaving out minus signs), and obtain the magnitude of each force or electric field. 3. Add vectorially all the forces on an object, or the contributing fields at a point, to get the resultant. Use symmetry (say, in the geometry) whenever possible. . Check your answer. Is it reasonable? If a function of distance, does it give reasonable results in limiting cases? SECTION 21-6

The Electric Field

571

21-7

Electric Field Calculations for

Continuous Charge Distributions

In many cases we can treat charge as being distributed continuously.” We can divid up a charge distribution into infinitesimal charges dQ, each of which will act as a tiny point charge. The contribution to the electric field at a distance r from each dQ is d

1

dE =

4e

r_zQ

(21-~6a)

Then the electric field, E, at any point is obtained infinitesimal contributions, which is the integral

E =

by summing

over all the

J dE.

(21-6b)

Note that dE is a vector (Eq.21-6a gives its magnitude). [In situations where Eq. 21-6b is difficult to evaluate, other techniques (discussed in the next two Chapters) can often be used instead to determine E. Numerical integration can also be used in many cases.]

A ring of charge. A thin, ring-shaped object of radius a

holds a total charge +Q distributed uniformly around it. Determine the electric field at a point P on its axis, a distance x from the center. See Fig. 21-28. Let A be the charge per unit length (C/m). dl

y

APPROACH AND SOLUTION We explicitly follow the steps of the Problem Solving Strategy on page 571. 1. Draw a careful diagram. The direction of the electric field due to one infinitesimal length df of the charged ring is shown in Fig. 21-28. 2. Apply Coulomb’s law. The electric field, dE, due to this particular segment of the ring of length df has magnitude

1 dQ

FIGURE 21-28

Example 21-9.

o = dmey 12 N The whole ring has length (circumference) of 27ra, so the charge on a length df is” dQ

where

dl

=

Q(Zfl'a)

A = Q/2ma

-

\dl

is the charge per unit length. Now we write dE as 1

E =

=

dme,

iAdl o0 r?

3 Add vectorially and use symmetry: The vector dE has components dE, along "

“

PROBLEM

SOLVING Use symmetry

when possible

the x axis and dE, perpendicular to the x sum (integrate) around the entire ring. We diametrically opposite the df shown will perpendicular to the x axis will just cancel segments of the ring, so by symmetry E will

need only sum the x components, dE, . The total field is then E

=

E,

b i =

|dE,

Since cos® = x/r, where A

s

=

|dEcosf

SOLVING

Check result by noting that at a great

distance the ring looks like a point charge

4. To

check

reduces to

X

reasonableness,

s g 1

=

A|

dt —cosé.

r = (x* + az)%, we have rg"

1

(dmeo) (2 + a?)i )o PROBLEM

axis (Fig. 21-28). We are going to note that an equal-length segment produce a dE whose component the dE, shown. This is true for all have zero y component, and so we

note

E = Q/(4meyx?).

Ax(27a)

dmeg (x2 + a?)p that

at great

distances,

1

Ox

4meq (x% + d¥): x >> a, this

result

We would expect this result because at great

distances the ring would appear to be a point charge (1/r* dependence). Also note that our result gives £ = 0 at x = 0, as we might expect because al*™

components will cancel at the center of the circle. 572

CHAPTER

21

"Because we believe there is a minimum charge (e), the treatment here is only for convenience; it is nonetheless useful and accurate since e is usually very much smaller than macroscopic charges.

Note in this Example three important problem-solving techniques that can be

used elsewhere: (1) using symmetry to reduce the complexity of the problem; (2) expressing the charge dQ in terms of a charge density (here linear, A = Q/2ma);

and (3) checking the answer at the limit of large r, which serves as an indication

ut not proof) of the correctness of the answer—if the result does not check at large r, your result has to be wrong. CONCEPTUAL EXAMPLE 21-10 | Charge small positive charge placed at the center uniformly distributed negative charge. Is the displaced slightly from the center along the What if the small charge is negative? Neglect electrostatic forces.

&

Z1PROBLEM Use symmetry,

SOLVING

ffa”;b: gfzefsgyb‘;"’l‘i -

at the center of a ring. Imagine a of a nonconducting ring carrying a positive charge in equilibrium if it is axis of the ring, and if so is it stable? gravity, as it is much smaller than the

RESPONSE The positive charge is in equilibrium because there is no net force on it, by symmetry. If the positive charge moves away from the center of the ring along the axis in either direction, the net force will be back towards the center of the ring and so the charge is in stable equilibrium. A negative charge at the center of the ring would feel no net force, but is in unstable equilibrium because if it moved along the ring’s axis, the net force would be away from the ring and the charge would be pushed farther away.

IETTEIEEN

Long line of charge. Determine the magnitude of the

electric field at any point P a distance x from the midpoint 0 of a very long line (a wire, say) of uniformly distributed positive charge, Fig. 21-29. Assume x is much smaller than the length of the wire, and let A be the charge per unit length (C/m).

A

APPROACH We set up a coordinate system so the wire is on the y axis with origin 0 as shown. A segment of wire dy has charge dQ = A dy. The field dE at point P due to this length dy of wire (at y) has magnitude

1

d0

dmey

1

¥

\dy

dme, (x + yz)’

1

=5

where r = (x* + y*)2 as shown in Fig. 21-29. The vector dE has components

dEy and dE\ as shown where dE, = dEcos6 and dE, = dEsin6. SOLUTION Because 0 is at the midpoint of the wire, the y component of E will be zero s'ince there will be equal contributions to E, = [dE, from above and below point 0: E,

=

desinB

=

=

FE,

=

Example 21-11.

0,

Thus we have E

FIGURE 21-29

JdEcos()

-

47re

cos 0 dy

J

x? + y?

The integration here is over y, along the wire, with x treated as constant. We must now write 6 as a function of y, or y as a function of 6. We do the latter: since y = xtanf, then dy = x df/cos?6. Furthermore, because

cos 6 = x/\/x?+ y2, then

1/(x?+ y?)=cos?6/x?

(cos8)(x d8/cos?8)(cos?0/x?) = cos 6 db/x. Hence E

=

A

1

pee

—JcosOdB

dmrey X | o)

=

el

41re (X

and our integrand above is

(sin6)

‘rr/2

-2

1 =

A

27rey X

)

where we have assumed the wire is extremely long in both directions (y — +o0) which corresponds to the limits § = +/2. Thus the field near a long straight wire of uniform charge decreases inversely as the first power of the distance from the wire. NOTE This result, obtained for an infinite wire, is a good approximation for a ire of finite length as long as x is small compared to the distance of P from the flds of the wire.

SECTION 21-7

Electric Field Calculations for Continuous Charge Distributions

573

AE

Uniformly charged disk. Charge is distributed uniformly

over a thin circular disk of radius R. The charge per unit area (C/m?) is o. Calculate the electric field at a point P on the axis of the disk, a distance z above

P

its center, Fig. 21-30.

g

APPROACH We apply the result the rings. SOLUTION For magnitude (see

FIGURE 21-30 Example 21-12; a uniformly charged flat disk of radius R.

can think of the disk as a set of concentric rings. We can ther. of Example 21-9 to each of these rings, and then sum over all the ring of radius r shown in Fig. 21-30, the electric field has result of Example 21-9)

dE =

1 zdQ 4re,, ( 2+ rz)%

where we have written dE (instead of E) for this thin ring of total charge dQ. The ring has area (dr)(2mr) and charge per unit area o = dQ/(2mr dr). We solve this for dQ (= o 2mrr dr) and insert it in the equation above for dE: dE

=

1

zo2mwrdr

dmeg (22 + r2);

_

=

zordr

2eo(2% + 1)

s

Now we sum over all the rings, starting at » = 0 out to the largest with » = R:

This gives the magnitude of E at any point z along the axis of the disk. The direction of each dE due to each ring is along the z axis (as in Example 21-9 and therefore the direction of E is along z. If Q (and o) are positive, E point away from the disk; if Q (and o) are negative, E points toward the disk. If the radius of the disk in Example 21-12 is much greater than the distance of our point P from the disk (i.e., z |-

X 5 ¢

4

L

o

+

£S

+

Lo

4

+

=

e

+

po:

1

CONCEPTUAL

had a component parallel to the

surface, EH , the latter would accelerate electrons into motion. In the static case, EH must be zero, and the electric

field must be perpendicular to the conductor’s surface: E = E 1-

Good conductor

Example 21-14.

FIGURE 21-39 A strong electric field exists in the vicinity of this “Faraday cage,” so strong that stray electrons in the atmosphere are accelerated to the kinetic energy needed to knock electrons out of air atoms, causing an avalanche of charge which flows to (or from) the metal cage. Yet the person inside the cage is not affected.

B

+

4

21-38

FIGURE 21-37 If the electric field E at the surface of a conductor

+

£

H

FIGURE

FIGURE 21-36 A charge inside a neutral spherical metal shell induces charge on its surfaces. The electric field exists even beyond the shell, but not within the conductor itself.

Ll (a)

EXAMPLE

e

LH~———+1

(b)

21-14 | Shielding, and safety in a storm. A neutral

hollow metal box is placed between two parallel charged plates as shown Fig. 21-38a. What is the field like inside the box?

in

RESPONSE If our metal box had been solid, and not hollow, free electrons in the box would have redistributed themselves along the surface until all their individual fields would have canceled each other inside the box. The net field inside the box would have been zero. For a hollow box, the external field is not changed since the electrons in the metal can move just as freely as before to the surface. Hence the field inside the hollow metal box is also zero, and the field lines are shown in Fig. 21-38b. A conducting box used in this way is an effective device for shielding Adelicate instruments and electronic circuits from unwanted external electric fields. We also can see that a relatively safe place to be during a lightning storm is inside a parked car, surrounded by metal. See also Fig. 21-39, where a person inside a porous “cage” is protected from a strong electric discharge. e

SECTION 21-9

@)PHYSICS

Electrical shielding

APPLIED

Electric Fields and Conductors

577

21-10

Motion of a Charged Particle in an Electric Field "

If an object having an electric charge g is at a point in space where the electric fiela. is E, the force on the object is given by F

=

gE

(see Eq. 21-5). In the past few Sections we have seen how to determine E for some particular situations. Now let us suppose we know E and we want to find the force on a charged object and the object’s subsequent motion. (We assume no other forces act.)

Electron accelerated by electric field. An electron (mass

m = 9.1 X 10" kg) is accelerated in the uniform field E (E = 2.0 X 10*N/C)

between two parallel charged plates. The separation of the plates is 1.5 cm. The electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate, Fig. 21-40. (a) With what speed does it leave the hole? (b) Show that the gravitational force can be ignored. Assume the hole is so small that it does not affect the uniform field between the plates. APPROACH We can obtain the electron’s velocity using the kinematic equations of Chapter 2, after first finding its acceleration from Newton’s second law, F = ma. The magnitude of the force on the electronis F = gE and is directed to the right. SOLUTION (a) The magnitude of the electron’s acceleration is FIGURE 21-40

Example 21-15.

F _ aEm

m

Between the plates E is uniform so the electron undergoes uniformly accelerated motion with acceleration a

=

(

1.6 X 107 C)(2.0 x 10*N/C )( m / (9.1 x 1073 kg)

It travels a distance

x = 1.5 X 102m

)

=

35X

10151'1’1/52.

fl

before reaching the hole, and since its

initial speed was zero, we can use the kinematic equation, (Eq.2-12c), with v, = O:

v* = v} + 2ax

v = V2ax = \/2(3.5 X 108 m/s?)(1.5 X 102m) = 1.0 X 10"m/s. There is no electric field outside the plates, so after passing through the hole, the electron moves with this speed, which is now constant. (b) The magnitude of the electric force on the electron is gE

=

(1.6 x 107 C)(2.0 X 10*N/C)

=

3.2 X 107®N.

The gravitational force is

mg

= (9.1 x 103 kg)(9.8 m/s?)

= 8.9 X 107N,

which is 10' times smaller! Note that the electric field due to the electron does not enter the problem (since a particle cannot exert a force on itself).

Electron

FIGURE 21-41

Example 21-16.

moving

perpendicular

to

E. Suppose

an

electron traveling with speed v, enters a uniform electric field E, which is at right angles to ¥, as shown in Fig. 21-41. Describe its motion by giving the equation of its path while in the electric field. Ignore gravity. APPROACH Again we use Newton’s second law, with F = gE, and the kinematic equations from Chapter 2. SOLUTION When the electron enters the electric field (at x = y = 0) it has velocity ¥, = v,i in the x direction. The electric field E, pointing vertically upward, imparts a uniform vertical acceleration to the electron of > F

qE

eE

m

m

m

ay:———:—————Q

578

CHAPTER 21

where we set

¢ = —e

for the electron.

The electron’s vertical position is given by Eq. 2-12b, y r. since the

1

=

antz

=

——

motion is at constant acceleration. The horizontal position is given by T x

=

!t

since a, = 0. We eliminate ¢ between these two equations and obtain ek

=

Y

———x?

2mad

which is the equation of a parabola (just as in projectile motion, Section 3-7).

21-11

Electric Dipoles

3

The combination of two equal charges of opposite sign, +Q and —Q, separated by a distance £, is referred to as an electric dipole. The quantity Qf is called the dipole

moment and is represented” by the symbol p. The dipole moment can be considered to be a vector p, of magnitude QF, that points from the negative to the positive charge as shown in Fig. 21-42. Many molecules, such as the diatomic molecule CO, have a dipole moment (C has a small positive charge and O a small negative charge of equal magnitude), and are referred to as polar molecules. Even

though the molecule as a whole is neutral, there is a separation of charge that

results from

an

uneven

sharing

of electrons

by the two

atoms.”

(Symmetric

:;

-2

~ FIGURE 21-42 A dipole consists of equal but opposite charges, +Q and —Q,

separated by a distance £. The dipole =~ momentis p = Q¢ and points from the negative to the positive charge. FIGURE 21-43

In the water

molecule (H,0), the electrons spend

~ more time around the oxygen atom

diatomic molecules, like O,, have no dipole moment.) The water molecule, with its ~ than around the two hydrogen uneven sharing of electrons (O is negative, the two H are positive), also has a dipole moment—see

Fig, 21-43,

Dipole in an External Field

point from the O toward each H as

First let u# consider a dipole, of dipole moment

p = Qf, that is placed in a

miform electric field E, as shown in Fig. 21—-44. If the field is uniform, the force QF on the flaos1t1ve charge and the force —QE on the negative charge result in no net force on the dipole. There will, however, be a torque on the dipole (Fig. 21-44) which has magnitude (calculated about the center, 0, of the dipole) =

atoms.The net dipole moment p can

be considered as the vector sum of two dipole moments p; and p, that

QF % sind + QF -g sinf

=

pEsiné.

o

=

. ¥ o

(21-9a)

104°

This can be written in vector notation as

= pxX E.

(21-9b)

The effect of the torque is to try to turn the dipole so p is parallel to E. The work done on the dipole by the electric field to change the angle 6 from 6, to 6, is (see Eq. 10-22) W

=

J

0,

6

T do.

We need to write the torque as 7 = —pEsin6 because its direction is opposite to the direction of increasing 6 (right-hand rule). Then

W

=

6,

f 7d0 0,

Positive work done this field. (Recall AU = —W.) If we 6, = 90° so cosf;

U=

=

—pE J

6,

6,

sinfdd

= pEcosf|

by the field decreases the relation between choose U = 0 when p = 0), and setting 6, =

-W

= —pEcosf

0,

6,

= pE(cosf, — cosb,).

FIGURE 21-44

(below) An electric

dipole in a uniform electric field.

the potential energy, U, of the dipole in work and potential energy, Eq. 8-4, is perpendicular to E (that is, choosing 6, then

= —j-E.

(21-10)

If the electric field is not uniform, the force on the +Q of the dipole may not have he same magnitude as on the —Q, so there may be a net force as well as a torque. "Be careful not to confuse this p for dipole moment with p for momentum. *The value of the separated charges may be a fraction of e (say + 0.2¢ or + 0.4e) but note that such charges do not violate what we said about e being the smallest charge. These charges less than e cannot

be isolated and merely represent how much time electrons spend around one atom or the other.

SECTION

21-11

579

Dipole in a field. The dipole moment of a water molecule

is 6.1 X 107 C-m. A water molecule is placed in a uniform electric field with magnitude 2.0 X 10°N/C. (a) What is the magnitude of the maximum torque that the field can exert on the molecule? (b) What is the potential energy wheq\ the torque is at its maximum? (c) In what position will the potential energy tak. on its greatest value? Why is this different than the position where the torque is maximum? APPROACH

Eq. 21-10.

SOLUTION

The

torque

is given

by Eq. 21-9

and

the

potential

energy

by

(@) From Eq. 21-9 we see that 7 is maximized when 6 is 90°. Then

7=pE =(61x10C-m)(2.0 X 10°N/C) = 1.2 X 107N m.

(b) The potential energy for 6 = 90° is zero (Eq. 21-10). Note that the potential energy is negative for smaller values of 6, so U is not a minimum for 6 = 90°. (c) The potential energy U will be a maximum when cosé = —1 in Eq. 21-10, so 6 = 180°, meaning E and p are antiparallel. The potential energy is maximized when the dipole is oriented so that it has to rotate through the largest angle, 180°, to reach the equilibrium position at 6 = 0°. The torque on the other hand is maximized when the electric forces are perpendicular to p.

Electric Field Produced by a Dipole We have just seen how an external electric field affects an electric dipole. Now let us suppose that there is no external field, and we want to determine the electric field produced by the dipole. For brevity, we restrict ourselves to points that are on the perpendicular bisector of the dipole, such as point P in Fig. 21-45 which is a distance r above the midpoint of the dipole. Note that r in Fig. 2145 is not the

distance from either charge to point P; the latter distance is (r* + 22/4)% and this is what must be used in Eq. 21-4. The total field at P is

o -0

E-F

/

FIGURE 21-45

Electric field due

to an electric dipole.

+E.

“>

where E. and E_ are the fields due to the + and magnitudes E. and E_ are equal: E,

=

E_

=

1 dmey

— charges respectively. The

- o — r° + £°/4

Their y components cancel at point P (symmetry again), so the magnitude of the total field E is

B=

or, setting Of = p, E

=

1 0 ¢ = e 2 + /4%

B 1

P

dmey (12 + £ /4)§

.

[on perpendicular bisector] : of dipole

(

21-11

)

Far from the dipole, r >> £, this reduces to _

E =

1

p

dmeg 1P

[

on perpendicular bisector

of dipole; r >> {

@1-12)

So the field decreases more rapidly for a dipole than for a single point charge

(1/r% versus 1/r%), which we expect since at large distances the two opposite

charges appear so close together as to neutralize each other. This dependence also applies for points not on the perpendicular bisector Problem 67).

580

CHAPTER 21

Electric Charge and Electric Field

1/r (sefl

*21-12 Electric Forces in Molecular Biology; DNA

rThe interior of every biological cell is mainly water. We can imagine a cell as a vast sea of molecules continually in motion (kinetic theory, Chapter 18), colliding with

@ PHYSICS Inside a cell:

APPLIED

one another with various amounts of kinetic energy. These molecules interact with ~ kinetic theory plus one another because of electrostatic attraction between molecules.

Indeeci, cellular processes are now considered to be the result of random (“thermal”) molecular motion plus the ordering effect of the electrostatic force. As an example, we look at DNA structure and replication. The picture we present has not been seen “in action.” Rather, it is a model of what happens based on physical theories and experiment. The genetic information that is passed on from generation to generation in all living cells is contained in the chromosomes, which are made up of genes. Each gene contains the information needed to produce a particular type of protein molecule, and that information is built into the principal molecule of a chromosome, DNA (deoxyribonucleic acid), Fig. 21-46. DNA molecules are made up of many small molecules known as nucleotide bases which are each polar due to unequal sharing of electrons. There are four types of nucleotide bases in DNA: adenine (A), cytosine (C), guanine (G), and thymine (T).

electrostatic force

FIGURE 21-46 DNA replicating in a human HeLa cancer cell. This is a

The DNA of a chromosome generally consists of two long DNA strands wrapped ~ false-color image made by a about one another in the shape of a “double helix.” The genetic information is transmission electron microscope contained in the specific order of the four bases (A, C, G, T) along the strand. As shown ~ (TEM: discussed in Chapter 37). in Fig. 21-47, the two strands are attracted by electrostatic forces—that is, by the attraction of positive charges to negative charges that exist on parts of the molecules. We see in Fig. 21-47a that an A (adenine) on one strand is always opposite a T on the other strand; similarly, a G is always opposite a C. This important ordering effect occurs because the shapes of A, T, C, and G are such that a T fits closely only into an A, and a G into a C; and only in the case of this close proximity of the charged portions is the electrostatic force great enough to hold them together even for a short time (Fig. 21-47b), forming what are referred to as “weak bonds.” Thymine (T)

FIGURE 21-47 (a) Section of a DNA double helix. (b) “Close-up” view of the helix, showing how A and T attract each other and how G and C attract each other through electrostatic forces. The + and — signs represent net charges, usually a fraction of e, due to [ e uneven sharing of electrons. The red dots indicate the electrostatic attraction (often called a “weak bond” or r“hydrogen bond”—Section 40-3). Note that there are two weak bonds between A and T, and three between Cand G.

*SECTION 21-12

Electric Forces in Molecular Biology; DNA

581

FIGURE 21-48

Replication of DNA.

When the DNA replicates (duplicates) itself just before cell division, the arrangement of A opposite T and G opposite C is crucial for ensuring that the genetic information is passed on accurately to the next generation, Fig. 21-48. The two strands of DNA separate (with the help of enzymes, which also operate via the electrostatic force), leaving the charged parts of the bases exposed. Once replication starts, let us see how the correct order of bases occurs by looking at the G molecule indicated by the red arrow in Fig. 21-48. Many unattached nucleotide bases of all four kinds are bouncing around in the cellular fluid, and the only type that will experience attraction to our G, if it bounces close to it, will be a C. The charges on the other three bases can not get close enough to those on

the

G

to provide

a significant

attractive

force—remember

that

the

force

decreases rapidly with distance (o 1/r%). Because the G does not attract an A, T,

or G appreciably, an A, T, or G will be knocked away by collisions with other molecules before enzymes can attach it to the growing chain (number 3). But the electrostatic force will often hold a C opposite our G long enough so that an enzyme can attach the C to the growing end of the new chain. Thus we see tha.q electrostatic forces are responsible for selecting the bases in the proper order during replication. This process of DNA replication is often presented as if it occurred in clockwork fashion—as if each molecule knew its role and went to its assigned place. But this is not the case. The forces of attraction are rather weak, and if the molecular shapes are not just right, there is almost no electrostatic attraction, which is why there are few mistakes. Thus, out of the random motion of the molecules, the electrostatic force acts to bring order out of chaos. The random (thermal) velocities of molecules in a cell affect cloning. When a bacterial cell divides, the two new bacteria have nearly identical DNA. Even if the

DNA were perfectly identical, the two new bacteria would not end up behaving in

the same way. Long protein, DNA, and shapes, and even the expression of genes large molecules such as a methyl group collision with another molecule. Hence, their DNA were identical. Indeed, there

*21-13

582

CHAPTER 21

RNA molecules get bumped into different can thus be different. Loosely held parts of (CHj;) can also be knocked off by a strong cloned organisms are not identical, even if can not really be genetic determinism.

Photocopy Machines and Computer Printers Use Electrostatics

Photocopy machines and laser printers use electrostatic attraction to print an image. They each use a different technique to project an image onto a special cylindrical drum. The drum is typically made of aluminum, a good conductor; its surface is coated with a thin layer of selenium, which has the interesting propertyfl (called “photoconductivity”) of being an electrical nonconductor in the dark, but a conductor when exposed to light. In a photocopier, lenses and mirrors focus an image of the original sheet of

paper onto the drum, much like a camera lens focuses an image on film. Step 1 is

(2) Lens focuses image of original

FIGURE 21-49 Inside a photocopy machine: (1) the selenium drum is given a + charge; (2) the lens focuses image on drum—only dark spots stay charged; (3) toner particles (negatively charged) are attracted to positive areas on drum;

(4) the image is transferred to paper; (5) heat binds the image to the paper.

Charging rod

(5) Heater rollers

the placing of a uniform positive charge on the drum’s selenium layer by a charged rod or roller, done in the dark. In step 2, the image to be copied or printed is projected 0111t0 the drum. For simplicity, let us assume the image is a dark letter A on a white background (as on the page of a book) as shown in Fig. 21-49. The letter A on the drum is dark, but all around it is light. At all these light places, the selenium bfi:cornes conducting and electrons flow in from the aluminum beneath,

PHYSICS

APPLIED

Photocopy machines

neutralizing those positive areas. In the dark areas of the letter A, the selenium is

nonconducting and so retains a positive charge, Fig. 21-49. In step 3, a fine dark powder known as foner is given a negative charge, and brushed on the drum as it rotates. The negatively charged toner particles are attracted to the positive areas on the drum (the A in our case) and stick only there. In step 4, the rotating drum presses against a piece of paper which has been positively charged more strongly than the selenium, so the toner particles are transferred to the paper, forming the final image. Finally, step 5, the paper is heated to fix the toner particles firmly on the paper. In a color copier (or printer), this process is repeated for each color—black, cyan (blue), magenta (red), and yellow. Combining these four colors in different woportions produces any desired color. A laser printer, on the other hand, uses a computer output to program the intensity of a laser beam onto the selenium-coated drum. The thin beam of light from the laser is scanned (by a movable mirror) from side to side across the drum in a series of horizontal lines, each line just below the previous line. As the beam swee ps across the drum, the intensity of the beam is varied by the computer output, bei ng strong for a point that is meant to be white or bright, and weak or zero for points that are meant to come out dark. After each sweep, the drum rotates ver y slightly for additional sweeps, Fig. 21-50, until a complete image is formed on it. The light parts of the selenium become conducting and lose their electric chjge, and the toner sticks only to the dark, electrically charged areas. The drum then transfers the image to paper, as in a photocopier. An inkjet printer does not use a drum. Instead nozzles spray tiny droplets of ink directly at the paper. The nozzles are swept across the paper, each sweep just above the previous one as the paper moves down. On each sweep, the ink makes dots on the paper, except for those points where no ink is desired, as directed by the comput er. The image consists of a huge number of very tiny dots. The quality or resoluti on of a printer is usually specified in dots per inch (dpi) in each (linear) dir ection. Laser beam scans

@

Toner hopper

APPLIED

@pnvsncs

APPLIED

Laser printer

Inkjet printer

Laser Movable mirror

T

@) rpHYSsics

et

FIGURE 21-50

Inside a laser

printer: A movable mirror sweeps the laser beam in horizontal lines

across the drum.,

Heater rollers

*SECTION 21-13

Photocopy Machines and Computer Printers Use Electrostatics

583

| Summary There are two kinds of electric charge, positive and negative. These designations are to be taken algebraically—that is, any charge is plus or minus so many coulombs (C), in SI units. Electric charge is conserved: if a certain amount of one type of charge is produced in a process, an equal amount of the opposite type is also produced; thus the ner charge produced is zero. According to atomic theory, electricity originates in the atom, each consisting of a positively charged nucleus surrounded by negatively charged electrons. Each electron has a charge

-e=-16 x 107°C.

Electric conductors are those materials in which many electrons are relatively free to move, whereas electric insulators are those in which very few electrons are free to move. An object is negatively charged when it has an excess of electrons, and positively charged when it has less than its normal amount of electrons. The charge on any object is thus a whole number times +e or —e. That is, charge is quantized. An object can become charged by rubbing (in which electrons are transferred from one material to another), by conduction (which is transfer of charge from one charged object to another by touching), or by induction (the separation of

charge

within

an object because

of the close approach

of

another charged object but without touching). Electric charges exert a force on each other. If two charges are of opposite types, one positive and one negative, they each exert an attractive force on the other. If the two charges are the same type, each repels the other.

The magnitude of the force one point charge exerts on another is proportional to the product of their charges, and inversely proportional to the square of the distance between them: F

-

& ZQZ -1 r

4meq

2 ZQZ; r

(21-1,21-2)

this is Coulomb’s law. In ST units, k is often written as 1/4meg.

We think of an electric field as existing in space around any charge or group of charges. The force on another charged object is then said to be due to the electric field present at its location. The electric field, E, at any point in space due to one or more charges, is defined as the force per unit charge that would act on a positive test charge g placed at that point:

£=X

21-3)

q

The magnitude charge Q is

of the electric field a distance r from a point E

=k % (21-4a) r The total electric field at a point in space is equal to the vector sum of the individual fields due to each contributing charge (principle of superposition). Electric fields are represented by electric field lines that start on positive charges and end on negative charges. Their direction indicates the direction the force would be on a tiny positive test charge placed at each point. The lines can be drawn so that the number per unit area is proportional to the magnitude of E. The static electric field inside a conductor is zero, and the

electric field lines just outside a charged conductor are perpendicular to its surface. An electric dipole is a combination of two equal but opposite charges, +Q and —Q, separated by a distance £. The dipole moment is p = Qf. A dipole placed in a uniform electric field feels no net force but does feel a net torque (unless p is parallel to E). The electric field produced by a dipole decreases

as the third power of the distance r from the dipole (E o 1 /r3},« for r large compared to £. [*In the replication of DNA, the electrostatic force plays a crucial role in selecting the proper molecules so the genetic information is passed on accurately from generation to generation. |

I Questions 1. If you charge a pocket comb by rubbing it with a silk scarf, how can you determine if the comb is positively or negatively charged? 2. Why does a shirt or blouse taken from a clothes dryer sometimes cling to your body? 3. Explain why fog or rain droplets tend to form around ions or electrons in the air. 4. A positively charged rod is brought close to a neutral piece of paper, which it attracts. Draw a diagram showing the separation of charge in the paper, and explain why attraction occurs. 5. Why does a plastic ruler that has been rubbed with a cloth have the ability to pick up small pieces of paper? Why is this difficult to do on a humid day? 6. Contrast the net charge on a conductor to the “free charges” in the conductor. " Figures 21-7 and 21-8 show how a charged rod placed near an uncharged metal object can attract (or repel) electrons. There are a great many electrons in the metal, yet only some of them move as shown. Why not all of them? 8. When an electroscope is charged, the two leaves repel each other and remain at an angle. What balances the electric force of repulsion so that the leaves don’t separate further?

584

CHAPTER 21

Electric Charge and Electric Field

9. The form of Coulomb’s law is very similar to that for Newton’s law of universal gravitation. What are the differences between these two laws? Compare also gravitational mass and electric charge. 10. We are not normally aware of the gravitational or electric force between two ordinary objects. What is the reason in each case? Give an example where we are aware of each one and why. 11. Is the electric force a conservative force? Why or why not? (See Chapter 8.) 12. What experimental observations mentioned in the text rule out

the

possibility

that

the

numerator

in Coulomb’s

law

contains the sum (Q; + Q,) rather than the product Q; Q,?

13. When a charged ruler attracts small pieces of paper, sometimes a piece jumps quickly away after touching the ruler. Explain. 14 Explain why the test charges we use when measuring electric fields must be small. 15. When

determining an electric field, must we use a positive

test charge, or would a negative one do as well? Explain. 16. Draw the electric field lines surrounding two negative electric charges a distance £ apart.

17. Assume that the two opposite charges in Fig. 21-34a are 12.0 cm apart. Consider the magnitude of the electric field 2.5cm from the positive charge. On which side of this charge—top, bottom, left, or right—is the electric field the

strongeAs&? The weakest?

rls

. Conside 1 the electric field at the three points indicated by the lette‘rs A, B, and C in Fig. 21-51. First draw an arrow at

each point indicating the direction of the net force that a positive test charge would experience if placed at that point, then list the letters in order

.B

@

of decreasing field strength (strongest first).

FIGURE 21-51

Question 18.

19. Why cainelectric field lines never cross? 20. Show, using the three rules for field lines given in Section 21-8, that the electric field lines starting or ending on a single point charge must be symmetrically spaced around the charge.

21. Given two point charges, Q and 2Q, a distance { apart, is there a point along the straight line that passes through them where £ = 0 when their signs are (a) opposite, (b) the same? If yes, state roughly where this point will be. 22. Suppose the ring of Fig. 21-28 has a uniformly distributed negative charge Q. What is the magnitude and direction of E at point P? 23 Consider a small positive test charge located on an electric field line at some point, such as point P in Fig. 21-34a. Is the direction of the velocity and/or acceleration of the test charge along this line? Discuss. 24 We wish to determine the electric field at a point near a positively charged metal sphere (a good conductor). We do so by bringing a small test charge, gy, to this point and measure the force Ky on it. Will Fy/qq be greater than, less than, or equal to the electric field E as it was at that point before the test charge was present? 25, In what ways does the electron motion in Example 21-16 resemble projectile motion (Section 3-7)? In which ways not? 26. Describe the motion of the dipole shown in Fig. 21-44 if it is released from rest at the position shown. 27. Explain why there can be a net force on an electric dipole placed in a nonuniform electric field.

| Problems 21-5 Coulomb’s Law [tmC=1073C, 1uC =107%C, 1nC =107°C.] r

11.

L (I) What is the magnitude of the electric force of attraction between an iron nucleus (g = +26e) and its innermost

electron if the distance between them is 1.5 X 1072 m?

2. (I) How many electrons make up a charge of —38.0 uC? 3. What is the magnitude of the force a +25 uC charge exerts on a +2.5 mC charge 28 cm away?

(II) Two positive point charges are a fixed distance apart. The sum of their charges is Qr. What charge must each have in order to (a) maximize the electric force between them, and (b) minimize it?

12. (II) Particles of charge +75, +48, and —85 uC are placed in a line (Fig. 21-52). The center one is 0.35m from each of

the others. Calculate the net force on each charge due to the

other two.

. (I) What is the repulsive electrical force between two protons

FIGURE 21-52

4.0 X 10~ m apart from each other in an atomic nucleus?

. (II) When an object such as rubbing it with a cloth, the microcoulombs. If that charge does the mass of a 35-g comb

. (II) Two

a plastic comb is charged by net charge is typically a few is 3.0 uC, by what percentage change during charging?

charged dust particles exert a force of 3.2 X 102N on each other. What will be the force if they are moved so they are only one-eighth as far apart?

. (II) Two charged spheres are 8.45 cm apart. They are moved, and the force on each of them is found to have been tripled. How far apart are they now? . (II) A person scuffing her feet on a wool rug on a dry day accumulates a net charge of —46 uC. How many excess electrons does she get, and by how much does her mass increase?

. (I) Wh

t is the total charge of all the electrons in a 15-kg bar of gold? What is the net charge of the bar? (Gold has 79 electrons per atom and an atomic mass of 197 u.)

10. (I

C ‘ pare

the

electric force

orbit (r = 0.53 x 107°m)

holding

the electron

in

around the proton nucleus of

the hydrogen atom, with the gravitational force between the same electron and proton. What is the ratio of these two forces?

Problem 12.

+75

pnC

S

+48 uC

nn

0.35m

—85 uC

13. (IT) Three charged particles are placed at the corners of an equilateral triangle of side 1.20m (Fig. 21-53). The charges are +7.0 uC, —8.0uC,

and

—6.0 uC. Calculate the

magnitude and direction of the net force on each due to the other two.

FIGURE 21-53 Problem 13.

0,=-8.0uC

03=-6.0 uC

14. (II) Two small nonconducting spheres have a total charge of 90.0 uC. (a) When placed 1.16 m apart, the force each exerts on the other is 12.0N and is repulsive. What is the charge on each? (b) What if the force were attractive? 15. (II) A charge of 4.15 mC is placed at each corner of a square 0.100m on a side. Determine the magnitude and direction of the force on each charge.

Problems

585

Problem 16.

415mC

0.100m _445mc

17. (II) A charge Q is transferred from an initially uncharged plastic ball to an identical ball 12cm away. The force of attraction is then 17 mN. How many electrons were transferred from one ball to the other? 18. (IIT) Two charges, —Qy and —4Q, are a distance { apart. These two charges are free to move but do not because there is a third charge nearby. What must be the magnitude of the third charge and its placement in order for the first two to be in equilibrium? 19. (IIT) Two positive charges +Q are affixed rigidly to the x axis, one at x = +d and the other at x = —d. A third charge +¢ of mass m, which is constrained to move only along the x axis, is displaced from the origin by a small distance s (4mr?) = Q. 3

This is Gauss’s law, with Q.. = Q, and we derived it for the special case of a spherical surface enclosing a point charge at its center. But what about some other surface, such as the irregular surface labeled A, in Fig. 22-8? The same number of field lines (due to our charge Q) pass through surface A,, as pass through the spherical surface, A, . Therefore, because the flux through a surface is proportional to the number of lines through it as we saw in Section 22-1, the flux through A, is the same as through A, :

3517:-(1; - jfii:-dx -2€9 A,

A

Hence, we can expect that

3€i: .4k = £€0 would be valid for any surface surrounding a single point charge Q. Finally, let us look at the case of more than one charge. For each charge, Q;, enclosed by the chosen surface,

3@17:,- -dk = 2,€0

where E; refers to the electric field produced by Q; alone. By the superposition principle for electric fields (Section 21-6), the total field E is equal to the sum of the fields due to each separate charge, E = ZE;. Hence

3Ei:-dx - %(EE,-)-dK = 3%€0 _ Qem, €0

FIGURE 22-9 Electric flux through a closed surface. (Same as Fig. 22-3.) No electric charge is enclosed by this

surface (Qenet = 0).

>r

where Qcnq = 20Q; is the total net charge enclosed within the surface. Thus we see, based on this simple argument, that Gauss’s law follows from Coulomb’s law for any distribution of static electric charge enclosed within a closed surface of any shape. The derivation of Gauss’s law from Coulomb’s law is valid for electric fields produced by static electric charges. We will see later that electric fields can also be produced by changing magnetic fields. Coulomb’s law cannot be used to describe such electric fields. But Gauss’s law is found to hold also for electric fields produced in any of these ways. Hence Gauss’s law is a more general law than Coulomb’s law. It holds for any electric field whatsoever. Even for the case of static electric fields that we are considering in this Chapter, it is important to recognize that E on the left side of Gauss’s law is not necessarily due only to the charge Q.. that appears on the right. For example, in Fig. 22-9 there is an electric field E at all points on the imaginary gaussian surface, but it is not due to the charge enclosed by the surface (which is Qg,q = 0 in this case). The electric field E which appears on the left side of Gauss’s law is the total electric field at each point, on the gaussian surface chosen, not just that due to the charge Qcpq,

which appears on the right side. Gauss’s law has been found to be valid for the total field at any surface. It tells us that any difference between the input and output flux of the electric field over any surface is due to charge within that surface. "Note that Gauss’s law would look more complicated in terms of the constant originally used in Coulomb’s law (Egs. 21-1 or 21-4a): Coulomb’s law Gauss’s law

E = k—Q;

%E-dfi

r

E

594

CHAPTER 22

=

1 471’&0

=Q0 r2

B=

dhv

k = 1/4me

that we

= 4mkQ =

Q = €

Gauss’s law has a simpler form using €,; Coulomb’s law is simpler using k. The normal convention is to use €( rather than k because Gauss’s law is considered more general and therefore it is preferable to have it in simpler form.

‘

CONCEPTUAL EXAMPLE 22-2 | Flux from Gauss's law. Consider the two gaussian surfaces, A; and A, , shown in Fig. 22-10.The only charge present is the charge Q at the center of surface A;. What is the net flux through each surface, A; and A,? éRESPONSE The surface A, encloses the charge +Q. By Gauss’s law, the net flux through A, is then Q/¢,. For surface A,, the charge +Q is outside the surface. Surface A, encloses zero net charge, so the net electric flux through A, is zero, by Gauss’s law. Note that all field lines that enter the volume enclosed by surface A, also leave it. EXERCISE d A point charge Q is at the center of a spherical gaussian surface A. When a second char&e Q is placed just outside A, the total flux through this spherical surface A is (a) unchanged, (b) doubled, (c) halved, (d) none of these.

EXERCISE C Three 2.95 uC charges are in a small box. What is the net flux leaving the

box?

(d) 1.0 X 10

(@) 33 x10”N-m?/C, (b) 33 X 10°N-m?/C, eN-m?/C, (e) 6.7 X 105N -m?/C.

(¢)

FIGURE 22-10 Example 22-2. Two gaussian surfaces.

1.0 x 10"N-m?/C,

We note that the integral in Gauss’s law is often rather difficult to carry out in practice. We rarely need to do it except for some fairly simple situations that we now discuss.

22-3 Applications of Gauss’s Law Gauss’s law is a very compact and elegant way to write the relation between electric charge and electric field. It also offers a simple way to determine the electric field when the charge distribution is simple and/or possesses a high degree of symmetry. In order to apply Gauss’s law, however, we must choose the “gaussian” surface very carefully (for the integral on the left side of Gauss’s law) so we can determine E. We normally try to think of a surface that has just the symmetry needed so that E will be constant on all or on parts of its surface. Sometimes we choose a surface so the flux through part of the surface is zero.

m

Spherical

conductor. A thin spherical shell of radius ry

possesses a total net charge Q that is uniformly distributed on it (Fig. 22-11). Determinfi the electric field at points (a) outside the shell, and (b) inside the shell. (c) What if the conductor were a solid sphere? APPROACH Because the charge is distributed symmetrically, the electric field must also be symmetric. Thus the field outside the sphere must be directed radially outward (inward if Q < 0) and must depend only on r, not on angle (spherical coordinates). SOLUTION (@) The electric field will have the same magnitude at all points on an imaginary gaussian surface, if we choose that surface as a sphere of radius r (r > ry) concentric with the shell, and shown in Fig. 22-11 as the dashed circle 4. Because E is perpendicular to this surface, the cosine of the angle between E and dA is always 1. Gauss’s law then gives (with O.,q = Q in Eq. 22-4)

FIGURE 22-11

Cross-sectional

drawing of a thin spherical shell of radius ry, carrying a net charge Q uniformly distributed. A and A, represent two gaussian surfaces we use to determine E. Example 22-3.

3@173 Cd& = E(4mr) = 2.

where 47

0

is the surface area of our sphere (gaussian surface) of radius r. Thus

£

-1 =

2

drreq r?

[r > nl

Thus the field outside a uniformly charged spherical shell is the same as if all the charge were concentrated at the center as a point charge. (b) Inside the shell, the electric field must also be symmetric. So £ must again have the same value at all points on a spherical gaussian surface (A, in Fig. 22-11) concentric with the shell. Thus E can be factored out of the integral and, with Qcna = 0 because the charge enclosed within the sphere A, is zero, we have jEE-dK AHence

E

=0

=

E(4nr?)

=

0. [r
R, this result reduces to (1/4me,)(Q/x), the potential of a point charge, as we should expect.

Potential due to a charged disk. A thin flat disk, of

cradius Ry, has a uniformly distributed charge Q, Fig. 23-15. Determine potential at a point P on the axis of the disk, a distance x from its center.

the

APPROACH We divide the disk into thin rings of radius R and thickness dR and use the result of Example 23—-8 to sum over the disk. SOLUTION The charge Q is distributed uniformly, so the charge contained in each ring is proportional to its area. The disk has area 7R} and each thin ring has area dA = (2wR)(dR). Hence

|

FIGURE 23-14 Example 23-8. Calculating the potential at point P, a distance x from the center of a uniform ring of charge. FIGURE 23-15 Example 23-9. Calculating the electric potential at point P on the axis of a uniformly charged thin disk.

dq _ 2wRdR Q

7R}

4y- o (2mR)(dR) 2TRIGR) __ 20RdR 20RdR

SO

mRy

R}

Then the potential at P, using Eq. 23-6b in which r is replaced by (x> + R%)z, is V

dq

1

=

41re J (x2 + RZ)% -

NOTE

~

s

20

JRO

RdR

4meqRS

|o

(x2 + Rz)%

-

0

-

2

2meo R} (x

22

&

1|R=Ro

) R=0

67 + R = ]

Q

2

2\1

_

ol 1)) - o

For x => R,, this formula reduces to

‘

V

=

2meo R}

*1

t5s =]~

2 x*

*

=

dmregx

:

| This is thL formula for a point charge, as we expect.

SECTION 23-4

Potential Due to Any Charge Distribution

615

a1 T—H*f*'

23-5

} T T [

Equipotential Surfaces quip

| } +—l_l..1-— |H B 20 Visv | sydV 16y FIGURE 23-16 Equipotential lines

The electric potential can be represented graphically by drawing equipotential lines or, in three dimensions, equipotential surfaces. An equipotential surface is one or*™ which all points are at the same potential. That is, the potential difference between any two points on the surface is zero, and no work is required to move a charge from one point to the other. An equipotential surface must be perpendicular to the electric field at any point. If this were not so—that is, if there were a component of E parallel to the surface—it would require work to move the charge along the surface against this component of E; and this would contradict the idea that it is an equipotential surface. This can also be seen from Eq. 23-4a, AV = — | fl: - dl. On a surface where V is constant, AV = 0, so we must have either E = 0, df = 0, or cosf = 0 where 6 is the angle between E and df. Thus in a region where E is not zero, the path df along an equipotential must have cos6# = 0, meaning # = 90° and E is perpendicular to the equipotential. The fact that the electric field lines and equipotential surfaces are mutually perpendicular helps us locate the equipotentials when the electric field lines are

oppositely charged parallel plates. Note that they are perpendicular to

which are the intersections of equipotential surfaces with the plane of the drawing. In Fig. 23-16, a few of the equipotential lines are drawn (dashed green lines)

!

|

T

| 1

Ll

(the green dashed lines) between two

known.

In

a

normal

two-dimensional

drawing,

the electric field lines (solid red lines). ~ for the electric field (red lines) between

we

show

equipotential

[lines,

two parallel plates at a potential

difference of 20 V. The negative plate is arbitrarily chosen to be zero volts and the potential of each equipotential line is indicated. Note that E points toward lower values of V.

FIGURE 23-17

Example 23-10.

:Eul:fcatzicsffigrlcilixzfilsltazl}cliaerq:ipotential P

ge-

ST

ERTN

Point charge equipotential surfaces. For a single point

charge with Q = 4.0 X 107°C, plane containing the charge) Vs =30V.

sketch the equipotential surfaces (or lines in a corresponding to V; =10V, V;, =20V, and*™

APPROACH The electric potential V depends on the distance r from the charge (Eq. 23-5). SOLUTION The electric field for a positive point charge is directed radially outward. Since the equipotential surfaces must be perpendicular to the lines of electric field, they will be spherical in shape, centered on

the point charge, Fig. 23-17. From Eq. 23-5 we have r = (1/4me,)(Q/V), so that for V, =10V, r, = (9.0 X 10°N-m?/C?)(4.0 X 10°C)/(10V) = 3.6 m, for V, =20V, r, = 1.8m, and for V3 = 30V, r; = 1.2m, as shown. NOTE The equipotential surface with the largest potential is closest to the positive charge. How would this change if Q were negative?

FIGURE 23-18

Equipotential lines

(green, dashed) are always

perpendicular to the electric field lines (solid red) shown here for two equal

but oppositely charged particles.

The

equipotential

lines for the case of two equal but oppositely charged

particles are shown in Fig. 23-18 as green dashed lines. Equipotential lines and

surfaces, unlike field lines, are always continuous and never end, and so continue

beyond the borders of Figs. 23-16 and 23-18.

We saw in Section 21-9 that there can be no electric field within a conductor in the static case, for otherwise the free electrons would feel a force and would move. Indeed, the entire volume of a conductor must be entirely at the same potential in the static case, and the surface of a conductor is then an equipotential surface. (If it weren’t, the free electrons at the surface would move, since whenever there is a potential difference between two points, free

charges will move.) This is fully consistent with our result, discussed earlier, ™ that the electric field at the surface of a conductor must be perpendicular to the surface.

616

CHAPTER 23

Electric Potential

A useful analogy for equipotential lines is a topographic map: the contour lines are essentially gravitational equipotential lines (Fig. 23-19).

FIGURE 23-19 A topographic map (here, a portion of the Sierra Nevada in California) shows continuous contour lines, each of which is at a fixed height above sea level. Here they are at 80 ft (25 m) intervals. If you walk along one contour line, you neither climb nor descend. If you cross lines, and especially if you climb perpendicular to the lines, you will be changing your gravitational potential (rapidly, if the lines are close together).

Dipole Potential Two equal point charges Q, of opposite sign, separated by a distance £, are called

an electric dipole, as we saw in Section 21-11. Also, the two charges we saw in

Figs. 23-12 and 23-18 constitute an electric dipole, and the latter shows the electric field lines and equipotential surfaces for a dipole. Because electric dipoles occur often in physics, as well as in other fields, it is useful to examine them more closély. The electric potential at an arbitrary point P due to a dipole, Fig. 23-20, is the sum of the potentials due to each of the two charges (we take V =0 at r = o0):

voolle. 1 (=9 _ e

r

dmey (r + Ar)

1 .f1_ _1}_ 0

dmeq

~\r

r + Ar

Ar

4mey r(r + Ar)

9

rwhere r is the distance from P to the positive charge and r + Ar is the distance to the negative charge. This equation becomes simpler if we consider points P whose distance from the dipole is much larger than the separation of the two charges— that is, for r >>f. From Fig. 23-20 we see that Ar ~ [Lcosf; since r >> Ar = fcosf, we can neglect Ar in the denominator as compared to r. Therefore, we obtain vV=

L2

C;) LA

pcc;s

0

[dipole; r >> €]

(23-7)

dmey 1 dmeg 1 where p = QI is called the dipole moment. We see that the potential decreases as the square of the d1sta.nce from the dipole, whereas. for a single point charge the potential decreases with the first power of the distance (Eq. 23-5). It is not surprising that the potential should fall off faster for a dipole; for when you are far from a dipole, the two equal but opposite charges appear so close together as to

FIGURE 23-20

TABLE 23-2 Dipole Moments of Selected Molecules

tend to neutralize each other. Table 23-2 gives the dipole moments for several molecules. The + and — signs

Molecule H,(Yo™)

unit called a debye is sometimes used: 1 debye = 3.33 X 107 C-m.

SCH=00)

indicate on which atoms these charges lie. The last two entries are a part of many organic molecules and play an important role in molecular biology. A dipole moment has units of coulomb-meters (C-m), although for molecules a smaller

23-7

~

We can

E Determined from V =

+

use Eq. 23-4a, V, -V, = — | : E - dl, to determine the difference in

potential between two points if the electric field is known in the region between those two points. By inverting Eq. 23—-4a, we can write the electric field in terms of the potential. Then the electric field can be determined from a knowledge of V. Let us see how to do this.

SECTION 23-7

Dipole Moment

(C:m) 61 %107

H® 1) NOH,M SNO—H®)

eimet S

.

Electric dipole.

Calculation of potential V at point P.

34 X 10730 5.0 X 1073 ~30t x 10730

~80' x 1079

L

L g

dipole moment will vary somewhat,

CRPETRIE Sliihe E & Mt nilinugle.

E Determined from VvV

617

We write Eq. 23-4a in differential form as

dv = -E-dl

= -E,dl,

where dV is the infinitesimal difference in potential between two points a distance dl apart, and E, is the component of the electric field in the direction of thfi infinitesimal displacement df. We can then write \ By

=

o

(23-8)

Thus the component of the electric field in any direction is equal to the negative of the rate of change of the electric potential with distance in that direction. The quantity dV//df is called the gradient of V in a particular direction. If the direction is not specified, the term gradient refers to that direction in which V changes most rapidly; this would be the direction of E at that point, so we can write F

av

)

e

=

-

[if d€| E]

If E is written as a function of x, y, and z, and we let £ refer to the x, y, and z axes, then Eq. 23-8 becomes v aV v

Ey=—>—

Here, 8V

=%

By

=

(23-9)

/dx is the “partial derivative” of V with respect to x, with y and z held constant.’

For example,if V(x,y,z) = (2V/m?)x* + (8 V/m?)y*z + (2V/m?)Z?, then E,= -38V/3x = —(4 V/m?)x, and

E, = —V/oy = —(16 V/m®)yz, E,= —3V/oz = —(8 V/m®)y* — (4 V/m?)z. E for ring and disk. Use electric potential to determine

the electric field at point P on the axis of (@) a circular ring of charge (Fig. 23—14)‘ and (b) a uniformly charged disk (Fig. 23-15). APPROACH We obtained V as a function of x in Examples 23-8 and 23-9, so we find E by taking derivatives (Eqgs. 23-9). SOLUTION (a) From Example 23-8, V o= Then

v

1

v

Ox

dmeo (2 + R

This is the same result we obtained in Example 21-9. (b) From Example 23-9, V

=

SO E

!

=

Q

2

27e R} [(x oV

—_——

ax

=

2\L

+

Rz )

—

o]

1

2meo RS [

x] —_

x|, X

.

(x* + R%)%]

For points very close to the disk, x R, or Ry) straight wire carrying a uniform charge per unit length A. 1% (IT) Suppose the end of your finger is charged. (a) Estimate the breakdown voltage in air for your finger. (b) About what surface charge density would have to be on your finger at this voltage? 18. (II) Estimate the electric field in the membrane wall of a living cell. Assume the wall is 10 nm thick and has a potential of 0.10 V across it. 19. (IT) A nonconducting sphere of radius ry carries a total charge Q distributed uniformly throughout its volume. Determine the electric potential as a function of the distance r from the center of the sphere for (a) r > ry and (b) r < ry. Take V =0 at r = co. (c) Plot V versus r and E versus r. 20. (III) Repeat Problem 19 assuming the charge density pg increases as the square of the distance from the center of the sphere, and pg = 0 at the center. 21. (IIT) The volume charge density pg within a sphere of radius r, is distributed in accordance with the following spherically symmetric relation

pe(r)

= Po[l - :—;]

where r is measured from the center of the sphere and py is a constant, For a point P inside the sphere (r < ry), determine the electric potential V. Let V' = 0 at infinity. 22, (III) A hollow spherical conductor, carrying a net charge +Q, has inner radius ry and outer radius r, = 2r; (Fig. 23-26). At the center of the sphere is a point charge +Q/2. (a) Write the electric field strength E in all three regions as a function of r. Then determine the potential as a function of r, the distance from the center, for (b) r >ry, (¢) nm?, is equivalent to a parallel-plate

capacitor with C ~ e, A/d = (8.85 x 10" C%/N- mz)(lo2 m?)/(10°m) ~ 1F.

One type of computer keyboard operates by capacitance. As shown in Fig. 24-5, each key is connected to the upper plate of a capacitor. The upper plate moves down when the key is pressed, reducing the spacing between the capacitor plates, and increasing the capacitance (Eq. 24-2: smaller d, larger C). The change in capacitance results in an electric signal that is detected by an electronic circuit. The proportionality, C' o A/d in Eq. 24-2, is valid also for a parallel-plate capacitor tha‘tt is rolled up into a spiral cylinder, as in Fig. 24—1b. However, the constant factor, €, must be replaced if an insulator such as paper separates the plates, as is usual, and this is discussed in Section 24-5. For a true cylindrical capacitor—consisting of two long coaxial cylinders—the result is somewhat different as the next Example shows.

Cylindrical

PHYSICS

APPLIED

Very high capacitance

Key Movable

plate

Capacitor

Insulator (flexible)

Fixed

plate

FIGURE 24-5 Key on a computer keyboard. Pressing the key reduces the capacitor spacing thus increasing the capacitance which can be detected electronically.

@)pHYSICS

APPLIED

Computer key

capacitor. A cylindrical capacitor consists of a

cylinder (or wire) of radius R}, surrounded by a coaxial cylindrical shell of inner radius R, , Fig. 24—6a. Both cylinders have length £ which we assume is much greater than the separation of the cylinders, R, — Ry, so we can neglect end effects. The capacitor is Acharged (by connecting it to a battery) so that one cylinder has a charge +Q (say, the inner one) ‘bnd the other one a charge —Q. Determine a formula for the capacitance. APPROACH To obtain C = Q/V, we need to determine the potential difference V' between the cylinders in terms of Q. We can use our earlier result (Example 21-11 or 22-6) that the electric field outside a long wire is directed radially outward and

FIGURE 24-6 (a) Cylindrical capacitor consists of two coaxial cylindrical conductors. (b) The electric field lines are shown in cross-sectional view.

has magnitude E = (1/2me,)(A/R), where R is the distance from the axis and A is the charge per unit length, Q/£. Then E = (1/2me,)(Q/ER) for points between the cylinders. SOLUTION To obtain the potential drfference V in terms of O, we use this result

for Ein

Eq.23-4a,

V

=V, - V, = f E-dl, and write the line integral from

the outer cylinder to the inner one (so V > 0) along a radial line:" V

=

Vb

-

Va

=

I

_faE'dl

|

=

Q

(®™dR

2

R _

ZWEOeJ

__

‘ 2meg! Q and V are proportional, and the capacitance C is _

NOTE If the

Q

=

_

277'60'@

.

R

R,

Q

2megl

R

lindrical capacitor

space between cylinders, R, — Ry, = AR is small, we have In(R,/R;) =

In[(Ry + AR)/Ry] = In[1 + AR/R,] ~ AR/R, ~ 2we0(Rb/ AR

is just Eq. ’24—2

= €y A/AR

(d = AR),

(see

Appendix

because the area of cylinder bis

a nice check.

A-3)

so

A = 2w RyL. This

EXERCISE C What is the capacitance per unit length of a cylindrical capacitor with radii ARa =2.5mm and Ry = 040mm? (a) 30 pF/m; (b) —30 pF/m; (c) 56 pF/m; (d) —56 pF/m; (e) 100 pF/m; (f) —100 pF/m. "Note that E points outward in Fig. 24-6b, but dt points inward for our chosen direction of integration; the angle between E and df is 180° and cos 180° = —1. Also, df = —dR because dR increases outward. These two minus signs cancel.

(b)

SECTION 24-2

631

Spherical capacitor. A spherical capacitor consists of two

thin concentric spherical conducting shells, of radius r, and r, as shown in Fig. 24-7. The inner shell carries a uniformly distributed charge Q on its surface, and the outer shell an equal but opposite charge —Q. Determine the capacitance‘\ of the two shells. APPROACH In Example 22-3 we used Gauss’s law to show that the electric field outside a uniformly charged conducting sphere is E = Q/4meor* as if all the charge were concentrated at the center. Now we use Eq. 23-4a, V = — [/ E-dl. SOLUTION We integrate Eq. 23—4a along a radial line to obtain the potential difference between the two conducting shells: Voo FIGURE 24-7

b_.

=

—[E-d! a

Cross section

through the center of a spherical capacitor. The thin inner shell has radius ry and the thin outer shell has radius r, .

SOLVING

Checking with a limiting case

=

-

Q f dmreg ),

0

s

lzdr 1

(1

1)

47rey \ 1y

Finally, C

Z PROBLEM

-

=

Q

—

=

Vba

4re

( 0

Talv ra

—

s

Q

47e

(ra—n Talh

) Iy

NOTE If the separation Ar = r, — n, is very small,then C = 4mer’/Ar =~ €q A/Ar

(since A = 4mr?), which is the parallel-plate formula, Eq. 24-2.

A single isolated conductor can also be said to have a capacitance, C. In this

case, C can still be defined as the ratio of the charge to absolute potential V on the conductor (relativeto V =0 at r = c0), so that the relation Q

=CV

remains valid. For example, the potential of a single conducting sphere of radius r,

can be obtained from our results in Example 24-3 by letting r, become infinitel ™ large. As r, = oo, then

V=_Q_=> R), Fig. 24-8.

A

2meg(d — x)’ + to —). Now

Capacitance, Dielectrics, Electric Energy Storage

S we find the potential difference

between the two wires using Eq. 23-4a and integrating along the straight line from the surface of the negative wire to the surface of the positive wire, noting

that E and df point in opposite directions (E-df < 0):

(

d—R

____b*.*_[ln(d) —In(R)], TE( because we are given that

The

d >> R.

So

il [ln

I Questions rl.

Suppose two nearby conductors carry the same negative charge. Can there be a potential difference between them? If so, can the definition of capacitance,

here?

C = Q/V,

be used

2. Suppose the separation of plates d in a parallel-plate capac-

itor is not very small compared to the dimensions of the plates. Would you expect Eq. 24-2 to give an overestimate or underestimate of the true capacitance? Explain. 3. Suppose one of the plates of a parallel-plate capacitor was moved so that the area of overlap was reduced by half, but they are still parallel. How would this affect the capacitance?

)

(&S

]

&

When a battery is connected to a capacitor, why do the two plates acquire charges of the same magnitude? Will this be true if the two conductors are different sizes or shapes? Describe a simple method of measuring e, using a capacitor. Suppose three identical capacitors are connected to a battery. Will they store more energy if connected in series or in parallel? A large copper sheet of thickness £ is placed between the parallel plates of a capacitor, but does not touch the plates. How will this affect the capacitance? 8. The parallel plates of an isolated capacitor carry opposite charges, Q. If the separation of the plates is increased, is a force required to do so? Is the potential difference changed? What happens to the work done in the pulling process? 9. How does the energy in a capacitor change if (@) the potenr‘ tial difference is doubled, (b) the charge on each plate is ' doubled, and (c) the separation of the plates is doubled, as the capacitor remains connected to a battery in each case?

10. If the voltage across a capacitor is doubled, the amount of energy it can store (a) doubles; (b) is halved; (c) is quadrupled; (d) is unaffected; (e) none of these. 11. Anisolated charged capacitor has horizontal plates. If a thin dielectric is inserted a short way between the plates, Fig. 24-19, will it move left or right when it is released?

. . Dielectric

FIGURE 24-19 Question 11.

s

12. Suppose a battery remains connected to the capacitor in Question 11. What then will happen when the dielectric is released? 13. How does the energy stored in a capacitor change when a dielectric is inserted if (a) the capacitor is isolated so Q does not change; (b) the capacitor remains connected to a battery so V does not change? 14. For dielectrics consisting of polar molecules, how would you expect the dielectric constant to change with temperature? 15. A dielectric is pulled out from between the plates of a capacitor which remains connected to a battery. What changes occur to the capacitance, charge on the plates, potential difference, energy stored in the capacitor, and electric field? 16. We have seen that the capacitance C depends on the size, shape, and position of the two conductors, as well as on the dielectric constant K. What then did we mean when we said that C is a constant in Eq. 24-1? 17. What value might we assign to the dielectric constant for a good conductor? Explain.

Questions

643

| Problems 24-1

Capacitors

1. (I) The two plates of a capacitor hold +2800uC and —2800 uC of charge, respectively, when the potential difference is 930 V. What is the capacitance? . (I) How much charge flows from a 12.0-V battery when it is connected to a 12.6-uF capacitor? . (I) The potential difference between two short sections of parallel wire in air is 24.0 V. They carry equal and opposite charge of magnitude 75 pC. What is the capacitance of the two wires? . (I) The charge on a capacitor increases by 26 wC when the voltage across it increases from 28 V to 78 V. What is the capacitance of the capacitor? . (II) A 7.7-uF capacitor is charged by a 125-V battery (Fig. 24-20a) and then is disconnected from the battery. When this capacitor (C,) is then connected (Fig. 24-20b) to a second (initially uncharged) capacitor, C,, the final voltage on each capacitor is 15 V. What is the value of C,? [Hint: Charge is conserved.] 1]

10

Cl

Cl

in i"

11 ]

v (a)

C (b)

FIGURE 24-20

Problems 5 and 48.

. (II) An isolated capacitor C, carries a charge Q. Its wires are then connected to those of a second capacitor C,, previously uncharged. What charge will each carry now? What will be the potential difference across each? . (IT) It takes 1517 of energy to move a 0.20-mC charge from one plate of a 15-uF capacitor to the other. How much charge is on each plate? . (IT) A 2.70-uF capacitor is charged to 475V and a 4.00-uF capacitor is charged to 525 V. (a) These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the negative plates are connected to each other. What will be the potential difference across each capacitor and the charge on each? (b) What is the voltage and charge for each capacitor if plates of opposite sign are connected? . (II) Compact “ultracapacitors” with capacitance values up to several thousand farads are now commercially available. One application for ultracapacitors is in providing power for electrical circuits when other sources (such as a battery) are turned off. To get an idea of how much charge can be stored in such a component, assume a 1200-F ultracapacitor is initially charged to 12.0V by a battery and is then disconnected from the battery. If charge is then drawn off the plates of this capacitor at a rate of 1.0 mC/s, say, to power the backup memory of some electrical gadget, how long (in days) will it take for the potential difference across this capacitor to drop to 6.0 V?

644

CHAPTER 24

10. (II) In a dynamic random access memory (DRAM) computer chip, each memory cell contains a capacito for charge storage. Each of these cells represents .. single binary-bit value of 1 when its 35-fF capacitor

(1LfF = 1075F) is charged at 1.5V, or 0 when uncharged at

0V. (a) When it is fully charged, how many excess electrons are on a cell capacitor’s negative plate? (b) After charge has been placed on a cell capacitor’s plate, it slowly “leaks” off (through a variety of mechanisms) at a constant rate of 0.30fC/s. How long does it take for the potential difference across this capacitor to decrease by 1.0% from its fully charged value? (Because of this leakage effect, the charge on a DRAM capacitor is “refreshed” many times per second.)

24-2

Determination of Capacitance

11. (I) To make a 0.40-uF capacitor, what area must the plates have if they are to be separated by a 2.8-mm air gap? 12. (I) What is the capacitance per unit length (F/m) of a coaxial cable whose inner conductor has a 1.0-mm diameter and the outer cylindrical sheath has a 5.0-mm diameter? Assume the space between is filled with air.

13. (I) Determine the capacitance of the Earth, assuming it to be a spherical conductor. 14. (IT) Use Gauss’s law to show that E = 0 inside the inner conductor of a cylindrical capacitor (see Fig. 24-6 and Example 24-2) as well as outside the outer cylinder. 15. (IT) Dry air will break down if the electric field exceeds about 3.0 X 10° V/m. What amount of charge can be placeh on a capacitor if the area of each plate is 6.8 cm?? 16. (I1) An electric field of 4.80 X 10°V/m is desired between two parallel plates, each of area 21.0cm? and separated by 0.250cm of air. What charge must be on each plate? 17. (IT) How strong is the electric field between the plates of a 0.80-uF air-gap capacitor if they are 2.0 mm apart and each has a charge of 92 uC? 18. (I) A large metal sheet of thickness ¢ is placed between, and parallel to, the plates of the parallel-plate capacitor of Fig. 24-4. It does not touch the plates, and extends beyond their edges. (a) What is now the net capacitance in terms of A, d, and £? (b) If £ = 0.40d,

by

what factor does the capacitance change when the sheet is inserted? 19. (III) Small distances are commonly measured capacitively. Consider an air-filled parallel-plate capacitor with fixed plate area A = 25mm’ and a variable plate-separation distance x. Assume this capacitor is attached to a capacitance-measuring instrument which can measure capacitance C in the range 1.0pF to 1000.0pF with an accuracy of AC = 0.1pF. (a) If C is measured while x is varied, over what range

(Xmin = X = Xpax) can the plate-separation distance (in um)

be determined by this setup? (b) Define Ax to be the accuracy (magnitude) to which x can be determined, and determine a formula for Ax. (c¢) Determine the percent accuracy to which xi, and x5 can be measured. «

Capacitance, Dielectrics, Electric Energy Storage

~

20. (IIT) In an electrostatic air cleaner (“precipitator”), the strong nonuniform electric field in the central region of a cylindrical capacitor (with outer and inner cylindrical radii R, and Ry,) is used to create ionized air molecules for use in charging dust and soot particles (Fig. 24-21). Under standard

atmospheric

conditions,

if air is subjected

26. (IT) Three conducting plates, each of area A, are connected as shown in Fig. 24-22. (a) Are the two capacitors thus formed connected in series or in parallel? (b) Determine C as a function of d, d,, and A. Assume d; + d, is much less

than the dimensions of the plates. (¢) The middle plate can be moved (changing the values of d; and d,), so as to vary the capacitance. What are the minimum and maximum values of the net capacitance?

to an

electric field magnitude that exceeds its dielectric strength Eg = 2.7 X 10°N/C,

air

molecules

will

dissociate

into

positively charged ions and free electrons. In a precipitator, the region within which air is ionized (the corona discharge region) occupies a cylindrical volume of radius R that is typically five times that of the inner cylinder. Assume a particular precipitator is constructed with R, = 0.10 mm and R, = 10.0cm. In order to create a corona discharge region with radius R = 5.0 Ry, what potential difference V should be applied between the precipitator’s inner and outer conducting cylinders? [Besides dissociating air, the charged inner cylinder repels the resulting positive ions from the corona discharge region, where they are put to use in charging dust particles, which are then “collected” on the

negatively charged outer cylinder.] Corona

discharge

FIGURE 24-22 Problem 26.

la b ¥

|

!

27. (IT) Consider three capacitors, of capacitance 3600 pF, 5800 pF, and 0.0100 «F. What maximum and minimum capacitance can you form from these? How do you make the connection in each case? 28. (IT) A 0.50-uF and a 0.80-uF capacitor are connected in series to a 9.0-V battery. Calculate (a) the potential difference across each capacitor and (b) the charge on each. (c) Repeat parts (a) and (b) assuming the two capacitors are in parallel. 29. (I1) In Fig. 24-23, suppose C;=C,=C3=C4=C. (a) Determine the equivalent capacitance between points a C o and b. (b) Determine the charge 1 n

region (ionized air)

on each capacitor and the poten-

=5.0R,

tial difference terms of V.

across

each

1

in

"

G

IL 11

Cy= FIGURE 24-21

24-3

FIGURE 24-23

Problem 20.

Problems 29 and 30.

Vv

a

b

Capacitors in Series and Parallel

21. (I) The capacitance of a portion of a circuit is to be reduced from 2900 pF to 1600 pF. What capacitance can be added to the circuit to produce this effect without removing existing circuit elements? Must any existing connections be broken to accomplish this? 22. (I) (a) Six 3.8-uF capacitors are connected in parallel. What is the equivalent capacitance? (b) What is their equivalent capacitance if connected in series? 23. (IT) Given three capacitors, C; = 2,0 uF, C; = 1.5 uF, and C; = 30uF, what arrangement of parallel and series connections with a 12-V battery will give the minimum voltage drop across the 2.0-uF capacitor? What is the minimum voltage drop? . (IT) Suppose three parallel-plate capacitors, whose plates have areas A, A,, and Aj and separations d;, d,, and d;, are connected in parallel. Show, using only Eq. 24-2, that Eq. 24-3 is valid. 25. (II) An electric circuit was accidentally constructed using a 5.0-uF capacitor instead of the required 16-uF value. Without removing the 5.0-uF capacitor, what can a technician add to correct this circuit?

30. (IT) Suppose in Fig. 24-23 that C; = C;, = C3 = 16.0 uF and C4 = 28.5 uF. If the charge on C, is Q; = 124 uC, determine

the charge

on each of the other capacitors, the

voltage across each capacitor, and the voltage V,}, across the entire combination. 31

(IT) The switch S in Fig. 24-24 is connected downward so that capacitor C, becomes fully charged by the battery of voltage V,. If the switch is then

connected upward, determine the charge on each capacitor after

¢ i

the switching.

G

FIGURE 24-24

Problem 31.

U

|

Vo

Problems

645

32. (II) (a) Determine the equivalent capacitance between points a and b for the combination of capacitors shown in Fig. 24-25. (b) Determine the charge on each capacitor and the voltage across each if V4, = V.

1

ey

—e b

11 H G,

il i Cy

FIGURE 24-25

FIGURE 24-28 Problem 39.

Problems 32 and 33.

(IT) Supposein Problem 32, Fig. 24-25, that C; = C3 = 8.0 uF,

3.

C,=C4=16uF, and Q;=23uC. Determine (a) the charge on each of the other capacitors, (b) the voltage across each capacitor, and (c) the voltage W,, across the combination. . (IT) Two capacitors connected in parallel produce an equivalent capacitance of 35.0 uF but when connected in series the equivalent capacitance is only 5.5 uF. What is the individual capacitance of each capacitor?

. (IIT) A voltage V is applied to the capacitor network shown in Fig. 24-29. (a) What is the equivalent capacitance? [Hint: Assume a potential difference Vj}, exists across the network as shown; write potential differences for various pathways through the network from a to b in terms of the charges on the

capacitors and the capac-

itances.] (b) Determine the equivalent capacitance if C;=Cy=80uF and C1

=

C3

35. (IT) In the capacitance bridge shown in Fig. 24-26, a voltage Vo is applied and the variable capacitor C; is adjusted until there is zero voltage between points a and b as measured on

the voltmeter («~—V)—). Determine the unknown capacitance

C, if

C, = 18.0uF

Assume

no

C; =89uF

and

charge

and

the fixed capacitors

C; = 4.8uF.

through the voltmeter.

flows

a

G

Cy

G

G

FIGURE 24-26

i}

Problem 35.

Vo

. (IT) Two capacitors, C; = 3200pF and connected in series to a 12.0-V battery. later disconnected from the battery and to each other, positive plate to positive plate to negative plate. What then will be capacitor?

C, = 1800 pF, are The capacitors are connected directly plate, and negative the charge on each

37. (II) (a) Determine the equivalent capacitance of the circuit shown in Fg 24-27. (b) If C, = C, = 2C; = 24.0 uF, how much charge is stored on each capacitor when V = 35.0V?

FIGURE 24-27

Problems 37, 38, and 45.

ae«——y—ob

C,=3.00uF, . (II) In Fg 24-27, let C; =200uF, C; =400uF, and V =240V. What is the potential difference across each capacitor?

646

CHAPTER 24

C5

=

45

[,LF

FIGURE 24-29

Problem 40.

have

b

=

24-4

@

1 I

20—

6, where A is the area of each plate and 6 is small. Assume the plates are square. [Hint: Imagine the capacitor as many infinitesimal capacitors in parallel.]

1

1l

the capacitance C in terms of A, d, and

.9

1L

i)

3

so it makes a small angle 6 with the other plate, as shown in Fig. 24-28. Determine a formula for


= 313

and

V? = 3V3. The square

root of each of these is the rms (root-mean-square) value of the current or voltage:

The

rms

Ly = VP= = —\—05 = 07071, =VV2 = =L = 0.707V,. Ve = V2 values

of

V

and

[ are

sometimes

(25-9a) (25-9b) called

the

effective

wvalues.

*A graph of cos® wr versus ¢ is identical to that for sin® wr in Fig. 25-22, except that the points are shifte(’fl (by £ cycle) on the time axis. Hence the average value of sin” and cos?, averaged over one or more full cycles,

will be the same: sin? wf = cos?wt. From the trigonometric identity sin?6 + cos’6 = 1, we can write 664

CHAPTER 25

(sinwt) + (cos?wt) = 2(sifwt) Hence the average value of sin® f is -

=

1.

They are useful because they can be substituted directly into the power formulas, Egs. 25-6 and 25-7, to get the average power:

I P

.

(25-10a)

P =38R

= B R

(25-10b)

uil = 1Yo 17 I & P = b _ Yms R

b (25-10c)

Thus, a direct current whose values of / and V equal the rms values of / and V for an alternating current will produce the same power. Hence it is usually the rms value of current and voltage that is specified or measured. For example, in the United States and Canada, standard line voltage' is 120-V ac. The 120 V is V,n; the peak voltage V; is

Vo = V2V

= 170V.

In much of the world (Europe, Australia, Asia) the rms voltage is 240V, so the peak voltage is 340V. DOV NP LEAEE Hair dryer. (a) Calculate the resistance and the peak current in a 1000-W hair dryer (Fig. 25-23) connected to a 120-V line. (b) What happens if it is connected to a 240-V line in Britain? APPROACH

We are given P and Vi, SO Iims = P/Vins

SOLUTION

(a) We solve Eq. 25-10a for the rms current:

and Iy = V2 Is. Then we find R from V = IR. Lms

AThen

=

P

v

T

I, = V21,

A " The resistance is

=

l

Vims Liss

_

1000 W oV

(Eq. 25-10a or 25-6), Heating

coils

8.33 A.

= 118A. 120V 833 A

=

j

Cord 1440

FIGURE 25-23 A hair dryer. Most of the current goes through the heating coils, a pure resistance; a all part to th tor to t

'

The resistance could equally well be calculated using peak values: R

Motor

:fi; faszxifigl: 253?.0 SR

Vo 170V =—=—— = 144Q. I 11.8A

(b) When connected to a 240-V line, more current would flow and the resistance would change with the increased temperature (Section 25-4). But let us make an estimate (?f the power transformed based on the same 14.4-Q) resistance. The average power would be

~P = Vi,R = (240V) (Tan) - ‘00w This is four times the dryer’s power rating and would undoubtedly melt the heating element or the wire coils of the motor. EXERCISE H Each channel of a stereo receiver is capable of an average power output of 100 W into an 8-Q) loudspeaker (see Fig. 25-14). What are the rms voltage and the rms current fed to the speaker (a) at the maximum power of 100 W, and (b) at 1.0 W when the volume is turned down?

«*

The line voltage can vary, depending on the total load; the frequency of 60 Hz or 50 Hz, however, remains extremely steady.

SECTION 25-7

Alternating Current

665

25-8

Microscopic View of Electric Current: Current Density and Drift Velocity

Up to now in this Chapter we have dealt mainly with a macroscopic view ™ electric current. We

y

A

_,_'

—

i

B =

FIGURE 25-24 Electric field E in a uniform wire of cross-sectional areaA carrying a current /. The

current density j = I/A.

(;_

saw, however,

that according

to atomic

theory, the electric

current in metal wires is carried by negatively charged electrons, and that in liquid solutions current can also be carried by positive and/or negative ions. Let us now look at this microscopic picture in more detail. When a potential difference is applied to the two ends of a wire of uniform cross section, the direction of the electric field E is parallel to the walls of the wire

(Fig. 25-24). The existence of E within the conducting wire does not contradict our earlier result that E = 0 inside a conductor in the electrostatic case, as we are no longer dealing with the static case. Charges are free to move in a conductor, and

hence can move under the action of the electric field. If all the charges are at rest, then E We as the current section,

must be zero (electrostatics). now define a new microscopic quantity, the current density, j. It is defined electric current per unit cross-sectional area at any point in space. If the density j in a wire of cross-sectional area A is uniform over the cross then j is related to the electric current by I j= " or I = jA. (25-11)

If the current density is not uniform, then the general relation is

I =

FIGURE 25-25

Electric field B in

a wire gives electrons in random

motion a drift velocity vg.

FIGURE 25-26

Electrons in the

volume A{ will all pass through the

cross section indicated in a time A, where £ = vqAt.

J j-dA,

(25-12)

where dA is an element of surface and I is the current through the surface over which the integration is taken. The direction of the current density at any point is the direction that a positive charge would move when placed at that point—that is, the direction of j at any point is generally the same as the direction of E Fig. 25-24. The current density exists for any point in space. The current 7, on ti other hand, refers to a conductor as a whole, and hence is a macroscopic quantity.’ The direction of j is chosen to represent the direction of net flow of positive charge. In a conductor, it is negatively charged electrons that move, so they move in the direction of —j, or —E (to the left in Fig, 25-24). We can imagine the free electrons as

moving about randomly at high speeds, bouncing off the atoms of the wire (somewhat

like the molecules of a gas—Chapter 18). When an electric field exists in the wire, Fig. 25-25, the electrons feel a force and initially begin to accelerate. But they soon reach a more or less steady average velocity in the direction of E, known as their drift velocity, U4 (collisions with atoms in the wire keep them from accelerating further). The drift velocity is normally very much smaller than the electrons’ average random speed. We can relate the drift velocity vy to the macroscopic current / in the wire. In a time At, the electrons will travel a distance £ = vgAt on average. Suppose the wire has cross-sectional area A. Then in time At, electrons in a volume V = Al = Awvy At will pass through the cross section A of wire, as shown in Fig. 25-26. If there are n free

electrons (each of charge —e) per unit volume (n = N/V), then the total charge AQ

that passes through the area A in a time At is AQ

(no.ofcharges, N) X (charge per particle) (nV)(—e) = —(nAvgAt)(e). The current / in the wire is thus I

= =

=

AA?_

=

—neAuvy.

The current density, j = I/A,

is

j =

—nevy.

(25-14)

j =

—nevy,

(25-15)

In vector form, this is written 666

CHAPTER 25

where

(25-13)

“

the minus sign indicates that the direction of (positive) current flow is

opposite to the drift velocity of electrons.

We can generalize Eq. an electrolyte. If there are each of density n; (number drift velocity v;, then the

(25-16)

E n;q;Vy;.

=

j

r,

25-15 to any type of charge flow, such as flow of ions in several types of ions (which can include free electrons), per unit volume), charge ¢; (¢; = —e for electrons) and net current density at any point is

1

The total current / passing through an area A perpendicular to a uniform j is then 2 n;q; vg; A ]

diameter and (b) electrons electron

Electron speeds in a wire. A copper wire 3.2mm in

carries a 5.0-A current. Determine (@) the drift velocity of the free electrons. assuming they behave like an ideal per Cu atom is free to move (the others

APPROACH

For (a) j = I/A = I/mr%

the current density in the wire, (¢) Estimate the rms speed of gas at 20°C. Assume that one remain bound to the atom).

For (b) we can apply Eq. 25-14 to find

vg if we can determine the number n of free electrons per unit volume. Since we

assume there is one free electron per atom, the density of free electrons, , is the

same as the density of Cu atoms. The atomic mass of Cu is 63.5u (see Periodic

Table inside the back cover), so 63.5 g of Cu contains one mole or 6.02 X 107 free

electrons. The mass density of copper (Table 13-1) is pp = 8.9 X 10°kg/m?, where pp = m/V. (We use pp to distinguish it here from p for resistivity.) In (c) we use K = 3kT, Eq. 18-4. (Do not confuse V for volume with V' for voltage.)

SOLUTION (a) The current density is (with » = §(3.2mm) = 1.6 X 10> m)

T

1

=

—

A7

I

= —

S.0A

=

w7 7 (6% 10°m)

=

62 %

10°

e

o

A(b) The number of free electrons per unit volume, n = N/V (where is

N

N

"ty =

" Then, by =

T

(

V = m/pp),

N(lmol)

m/pp

m(1mole)

6.02 X 10% electrons

63.5 x 10 kg

>(

Pl

8.9 x 10°k

3)

g/n’)

=

84 x10%m™3

o

Eq. 25-14, the drift velocity has magnitude L

ne

6.2 X 10°A Ty /m’

=

(8.4 x 10

=

)(1.6 x 107°C)

=

4.6 X 10°m/s

~ 0.05 mm/s.

(c) If we model the free electrons as an ideal gas (a rather rough approximation), we use Eq. 18-5 to estimate the random rms speed of an electron as it darts around:

T

[3kT

\Nm

J

3(1.38 x 102 J/K)(293K)

9.11 x 10~ kg

Bk

S

e

The drift velocity (average speed in the direction of the current) is very much less than the rms thermal speed of the electrons, by a factor of about 10°.

NOTE The result in (c) is an underestimate. Quantum theory calculations, and experiments, give the rms speed in copper to be about 1.6 X 10°m/s. The drift velocity of electrons in a wire is very slow, only about 0.05 mm/s (Example 25-14 above), which means it takes an electron 20 X 10°s, or 53h, to travel only 1 m. This is not, of course, how fast “electricity travels”: when you flip a light switch, the light—even if many meters away—goes on nearly instantaneously. Why? Because electric fields travel essentially at the speed of light (3 X 10°m/s). fVe can think of electrons in a wire as being like a pipe full of water: when a little .ater enters one end of the pipe, almost immediately some water comes out at the other end.

SECTION 25-8

Microscopic View of Electric Current: Current Density and Drift Velocity

667

*Electric Field Inside a Wire Equation 25-2b, V = IR, can be written in terms of microscopic quantities as follows. We write the resistance R in terms of the resistivity p: R

=

¥

pz;

fl

and we write V and [ as I = jA

and

V

=

EL

The last relation follows from Egs. 23-4, where we assume the electric field is uniform within the wire and £ is the length of the wire (or a portion of the wire) between whose ends the potential difference is V. Thus, from

V = IR,

we have

Et = (1'&(9%) = jol

$O

j=

1 ;E

=

oE,

(25-17)

where o = 1/p is the conductivity (Eq. 25-4). For a metal conductor, p and o do not depend on V (and hence noton E). Therefore the current density j is proportional to the electrical field E in the conductor. This is the “microscopic” statement of Ohm’s law. Equation 25-17, which can be written in vector form as

TR = cE

j

=

—E, p

is sometimes taken as the definition of conductivity o and resistivity p.

Electric field inside a wire. What is the electric field

inside the wire of Example 25-14?

APPROACH We use Eq.25-17 and p = 1.68 X 108Q-m

SOLUTION Example 25-14 gives j = 6.2 X 10° A/m?, so

E = pj = (1.68 X 108Q-m)(6.2 X 10°A/m?)

NOTE

for copper. = 1.0 X 1072V/m.

-

For comparison, the electric field between the plates of a capacitor is often

much larger; in Example 24-1, for example, E is on the order of 10*

V/m. Thus we

see that only a modest electric field is needed for current flow in practical cases.

*25-9

Superconductivity

A superconducting

At very low temperatures, well below 0°C, the resistivity (Section 25-4) of certain metals and certain compounds or alloys becomes zero as measured by the highestprecision techniques. Materials in such a state are said to be superconducting. It was first observed by H. K. Onnes (1853-1926) in 1911 when he cooled mercury below 4.2 K (—269°C) and found that the resistance of mercury suddenly droppefd

temperature is below T, its “critical temperature.”At Tc, the resistivity jumps to a “normal” nonzero value and increases with temperature as

transition temperature or critical temperature, T, which is usually within a few degrees of absolute zero. Current in a ring-shaped superconducting material has been observed to flow for years in the absence of a potential difference, with no measurable decrease. Measurements show that the resistivity p of superconductors

FIGURE 25-27

material has zero resistivity when its

Resistivity, p

most materials do (Eq. 25-5).

t 668

CHAPTER 25

T

to zero. In general, superconductors become superconducting only below a certain

is less than 4 X 1072 ()-m, which is over 10'® times smaller than that for copper,

and is considered to be zero in practice. See Fig. 25-27. Before 1986 the highest temperature at which a material was found to superconduct was 23 K, which required liquid helium to keep the material cold. In 1987, a compound of yttrium, barium, copper, and oxygen (YBCO) was developed that can be superconducting at 90 K. This was an important breakthrough since liquid nitrogen, which boils at 77 K (sufficiently cold to keep the material superconducting is more easily and cheaply obtained than the liquid helium needed for conventione superconductors. Superconductivity at temperatures as high as 160K has been reported, though in fragile compounds.

Most applications today use a bismuth-strontium-calcium-copper oxide, known (for short) as BSCCO. A major challenge is how to make a useable, bendable wire

Signal from another neuron \ Synapse

out of the BSCCO, which is very brittle. (One solution is to embed tiny filaments

of the high-T. superconductor in a metal alloy, which is not resistanceless, but the rfisistance is much less than that of a conventional copper cable.)

*95-10

Electrical Conduction

Nervous System

in the

The flow of electric charge in the human nervous system provides us the means for being aware of the world. Although the detailed functioning is not well understood, we do have a reasonable understanding of how messages are transmitted within the nervous system: they are electrical signals passing along the basic element of the nervous system, the neuron. Neurons are living cells of unusual shape (Fig. 25-28). Attached to the main cell body are several small appendages known as dendrites and a long tail called the axon. Signals are received by the dendrites and are propagated along the axon. When a signal reaches the nerve endings, it is transmitted to the next neuron or to a muscle at a connection called a synapse. A neuron, before transmitting an electrical signal, is in the so-called “resting state.” Like nearly all living cells, neurons have a net positive charge on the outer surface of the cell membrane and a negative charge on the inner surface. This difference in charge, or “dipole layer,” means that a potential difference exists across the cell membrane. When a neuron is not transmitting a signal, this “resting potential,” normally stated as | Vinside —

outside »

is typically —60mV to —90mV, depending on the type of organism. The most ommon ions in a cell are K*, Na*, and CI". There are large differences in the ncentrations of these ions inside and outside a cell, as indicated by the typical values givenLi(n Table 25-2. Other ions are also present, so the fluids both inside and outside the

Dendrites

axon are electrically neutral. Because of the differences in concentration,

there is a tendency for ions to diffuse across the membrane (see Section 18-7 on diffusion). However, in the resting state the cell membrane prevents any net flow of Na* (through a mechanism of “active pumping” of Na* out of the cell). But it does allow the flow of Cl ions, and less so of K*ions, and it is these two ions that produce thqodipole charge layer on the membrane. Because there is a greater concentration of K* inside the cell than outside, more K" ions tend to diffuse

Myelin sheath Node of Ranvier

Synapse [

Nerve endings

Another neuron or a muscle

FIGURE 25-28 A simplified sketch of a typical neuron. TABLE 25-2

Concentrations of lons Inside and Outside a Typical Axon Concentration

Concentration

(mol/m?)

(mol/m®)

inside axon

K+

Na* al=

outside axon

140

5

15 9

140 125

Extracellular fluid @ +

outward across the membrane than diffuse inward. A K" ion that passes through

the membrane becomes attached to the outer surface of the membrane, and leaves

behind an equal negative charge that lies on the inner surface of the membrane (Fig. 25-29). Independently, Cl ions tend to diffuse into the cell since their concentratioln outside is higher. Both K™ and CI™ diffusion tends to charge the interior surface of the membrane negatively and the outside positively. As charge accumulates on the membrane surface, it becomes increasingly difficult for more ions to diffuse: K* ions trying to move outward, for example, are repelled by the positive charge already there. Equilibrium is reached when the tendency to diffuse because of the concentration difference is just balanced by the electrical potential difference across the membrane. The greater the concentration difference, the

FIGURE 25-29 How a dipole layer of charge forms on a cell membrane. FIGURE 25-30 Measuring the potential difference between the inside and outside of a nerve cell.

greater the Hotential difference across the membrane (=60 mV to —90 mV).

The most important aspect of a neuron is not that it has a resting potential (most cells do), but rather that it can respond to a stimulus and conduct an electrical signal along its length. The stimulus could be thermal (when you touch a hot stove) or chemical (as in taste buds); it could be pressure (as on the skin or at the eardrum), or light fis in the eye); or it could be the electric stimulus of a signal coming from the brain or .other neuron. In the laboratory, the stimulus is usually electrical and is applied by a tiny probe at some point on the neuron. If the stimulus exceeds some threshold, a voltage pulse will travel down the axon. This voltage pulse can be detected at a point on the axon using a voltmeter or an oscilloscope connected as in Fig. 25-30.

*SECTION 25-10

669

This voltage pulse has the shape shown in Fig. 25-31, and is called an action

20T

/\

Time (ms)

|

-

[

o

Potential V (mV)

60+

Resting

3

:

potential

FIGURE 25-31

Action potential.

FIGURE 25-32 Propagation of an action potential along an axon

membrase,

Point of stimulation

R

Membrane At

Exterior

(a) + 4+ + 4+ 4+ + + +

:

(b)

:

(© + 4+ + A+t

Tt

o

:

s

Action

potential

potential. As can be seen, the potential increases from a resting potential of about —70mV and becomes a positive 30mV or 40mV. The action potential lasts for

about 1 ms and travels down an axon with a speed of 30 m/s to 150 m/s. When an action potential is stimulated, the nerve is said to have “fired.”

What causes the action potential? Apparently, the cell membrane has the ability to

alter its permeability properties. At the point suddenly becomes much more permeable Na” ions rush into the cell and the inner charged, and the potential difference quickly Just as suddenly, the membrane returns to its

meable to Na™ and in fact pumps out Na™ ions. The diffusion of CI” and K" ions again predominates and the original resting potential is restored (—70 mV in Fig. 25-31). What causes the action potential to travel along the axon? The action potential occurs at .the: poiqt_ of stimula.tio.n, as shown 'in Fig. 25-323.' The mgmbra}ne momentarily is positive on the inside and negative on the outside at this point.

Nearby charges are attracted toward this region, as shown in Fig. 25-32b. The

potential in these adjacent regions then drops, causing an action potential there. Thus, as the membrane returns to normal at the original point, nearby it experiences an action potential, so the action potential moves down the axon (Figs. 25-32c and d). You may wonder if the number of ions that pass through the membrane would significantly alter the concentrations. The answer is no; and we can show why by treating the axon as a capacitor in the following Example.

EXAMPLE 25-16

Capacitance of an axon. (¢) Do an order-of-

The

is about

magnitude estimate for the capacitance of an axon 10cm long of radius 10 wm. thickness of the membrane

(d)

10 ®m, and the dielectric constant

is

about 3. (b) By what factor does the concentration (number of ions per volume) of Na™ ions in the cell change as a result of one action potential?

APPROACH We model the membrane of an axon as a cylindrically shape« parallel-plate capacitor, with opposite charges on each side. The separation of th “plates” is the thickness of the membrane, d ~ 10~ m. We first calculate the area of the cylinder and then can use Eq.24-8, C = KeyA/d, to find the capacitance. In (b), we use the voltage change during one action potential to find the amount of charge moved across the membrane. SOLUTION (a) The area A is the area of a cylinder of radius r and length £:

Action potential moving to the right

where the stimulus occurs, the membrane to Na® than to K™ and Cl™ ions. Thus, surface of the wall becomes positively swings positive (=~ +35 mV in Fig. 25-31). original characteristics: it becomes imper-

A

= 2mrl

~

(628)(10°m)(0.1m)

~ 6 X 10°m?

From Eq. 24-8, we have

C = Keog

~ (3)(8.85 X 1072C?*/N-m?)

6 X 10°%m? ~ 1078F. 108 m

(b) Since the voltage changes from =70 mV to about +30 mV, the total change is about 100 mV. The amount of charge that moves is then

Q = CV

~ (10%F)(01V)

= 10°C.

Each ion carries a charge e = 1.6 X 107 C, so the number of ions that flow per

action potential is Q/e = (10 C)/(1.6 X 107 C) ~ 10'°. The volume of our

cylindrical axon is

V = o

and

the

concentration

~ (3)10°m)(0.1m) of

Na'ions

inside

= 3x10 " m’, the

15mol/m?® = 15 x 6.02 X 10* ions/m® ~ 10 ions/m>.

cell

(Table

25-2)

is

Thus, the cell contains

(10% fons/m?) X (3 x 107" m*) ~ 3 X 10" Na" ions. One action potential, then, will change the concentration of Na” ions by about 10'°/(3 X 10') = § X 1074, or 1 part in 30,000. This tiny change would not be measurable.

670

CHAPTER 25

Thus, even 1000 action potentials will not alter the concentration significant'._fl The sodium pump does not, therefore, have to remove Na” ions quickly after an action potential, but can operate slowly over time to maintain a relatively constant concentration.

| Summary An electric battery serves as a source of nearly constant potential difference by transforming chemical energy into lectric energy. A simple battery consists of two electrodes made of different metals immersed in a solution or paste known as an electrolyte. Electric current, /, refers to the rate of flow of electric charge and is measured in amperes (A): 1 A equals a flow of 1 C/s past a given point. The direction of conventional current is that of positive charge flow. In a wire, it is actually negatively charged electrons that move, so they flow in a direction opposite to the conventional current. A positive charge flow in one direction is almost always equivalent to a negative charge flow in the opposite direction. Positive conventional current always flows from a high potential to a low potential. The resistance R of a device is defined by the relation V

=

IR,

(25-2)

where [ is the current in the device when a potential difference V is applied across it. For materials such as metals, R is a constant independent of V (thus / « V), a result known as Ohm’s law. Thus, the current / coming from a battery of voltage V depends on the resistance R of the circuit connected to it. Voltage is applied across a device or between the ends of a wire. Current passes through a wire or device. Resistance is a property of the wire or device. The unit of resistance is the ohm (Q), where 1Q = 1V/A. See Table 25-3.

r.

TABLE 25-3 Summary of Units Current

Potential difference Power

RéTsistance

1A =1C/s

1V =1J/C

1W =11J/s

19 =1V/A

The resistance R of a wire is inversely proportional to its cross-sectional area A, and directly proportional to its length £ and to a property of the material called its resistivity: \ pl R = —. (25-3)

A

The resistivity, p, increases with temperature for metals, but for semiconductors it may decrease.

|

The rate from electric equal to the transformed,

at which energy is transformed in a resistance R to other forms of energy (such as heat and light) is product of current and voltage. That is, the power measured in watts, is given by P

=1V,

(25-6)

which for resistors can be written as 2

P =1R-=

The ST unit of The total the product of is operated. In but

electric

VF-

(25-7)

power is the watt (1 W = 1J/s). electric energy transformed in any device equals the power and the time during which the device SI units, energy is given in joules (1J = 1W:"s),

companies

use

a larger

the

current

unit,

the

kilowatt-hour

(1kWh = 3.6 X 10°7J). Electric current can be direct current (dc), in which the current is steady in one direction; or it can be alternating current

(ac),

in

which

I

=

reverses

direction

at

a

particular frequency f, typically 60 Hz. Alternating currents are typically sinusoidal in time, Iysinwt,

(25-8)

where @ = 27f, and are produced by an alternating voltage. The rms values of sinusoidally alternating currents and voltages are given by I rms

If

= = \/E

and

|Z

Vims

= —=> '\/5

(25-9)

respectively, where I and V;, are the peak values. The power rela-

tionship, P = IV = I’R = V?/R, is valid for the average power

in alternating currents when the rms values of V and [/ are used. Current density j is the current per cross-sectional area. From a microscopic point of view, the current density is related to

the

number

of

charge

carriers

per

charge, ¢, and their drift velocity, v4, by

unit

volume,

j = ngvq.

n, their

(25-16)

The electric field within a wire is related to j by j = oE where

o = 1/p is the conductivity. [*At very low temperatures certain materials become superconducting, which means their electrical resistance becomes zero. | [*The human nervous system operates via electrical

conduction: when a nerve “fires,” an electrical signal travels as a

voltage pulse known as an action potential.]

g

1. What quantity is measured by a battery rating given in ampere-hours (A-h)? 2. When an electric cell is connected to a circuit, electrons flow away from the negative terminal in the circuit. But within the cell, electrons flow ro the negative terminal. Explain. 3. When a flashlight is operated, what is being used up: battery current, battery voltage, battery energy, battery power, or battery resistance? Explain. r4. One terminal of a car battery is said to be connected to “ground.” Since it is not really connected to the ground, what is meant by this expression?

W

g Questions When you turn on a water faucet, immediately. You don’t have to wait the faucet valve to the spout. Why true when you connect a wire to the Can a copper wire and an aluminum have the same resistance? Explain.

7. The equation

P = V2/R

the water usually flows for water to flow from not? Is the same thing terminals of a battery? wire of the same length

indicates that the power dissi-

pated in a resistor decreases if the resistance is increased, whereas the equation P = I’R implies the opposite. Is there a contradiction here? Explain. 8. What happens when a lightbulb burns out? Questions

671

9. If the resistance of a small immersion heater (to heat water for tea or soup, Fig. 25-33)

was increased, would it speed up or slow down the heating process? Explain.

~ 14. A 15-A fuse blows repeatedly. Why replace this fuse with a 25-A fuse?

is it dangerous

to

15. When electric lights are operated on low-frequency ac (say, 5 Hz), they flicker noticeably. Why? 16 Driven by ac power, the same electrons pass back and for

through your reading lamp over and over again. Explai.. why the light stays lit instead of going out after the first pass of electrons. 17. The heating element in a toaster is made of Nichrome wire. Immediately after the toaster is turned on, is the current

(Iims) in the wire increasing, decreasing, or staying constant?

FIGURE: 25-33

Explain.

Question 2,

18. Is current used up in a resistor? Explain.

10. If a rectangular solid made of carbon has sides of lengths a,2a, and 3a, how would you connect the wires from a battery so as to obtain (a) the least resistance, (b) the greatest resistance?

19, Compare the drift velocities and electric currents in two wires that are geometrically identical and the density of atoms is similar, but the number of free electrons per atom

are turned on and not after they have been on for some time.

20, A voltage V is connected across a wire of length £ and

11. Explain why lightbulbs almost always burn out just as they 12, Which draws more current, a 100-W lightbulb or a 75-W bulb? Which has the higher resistance? 13. Electric power is transferred over large distances at very high voltages. Explain how the high voltage reduces power losses in the transmission lines.

in the material of one wire is twice that in the other.

radius . How is the electron drift velocity affected if (a) £ is doubled, (b) r is doubled, (c¢) V is doubled? 21. Why is it more dangerous to turn on an electric appliance when you are standing outside in bare feet than when you are inside wearing shoes with thick soles?

| Problems 25-2 and 25-3

Electric Current, Resistance, Ohm's Law

(Note: The charge on one electron is 1.60 X 1071° C.)

1. (I) A current of 1.30 A flows in a wire. How many electrons

are flowing past any point in the wire per second?

2. (I) A service station charges a battery using a current of 6.7-A for 5.0 h. How much charge passes through the battery? 3. (I) What is the current in amperes if 1200 Na* ions flow across a cell membrane in 3.5 us? The charge on the sodium is the same as on an electron, but positive. 4. (I) What is the resistance of a toaster if 120 V produces a current of 42 A? 5. (II) An electric clothes dryer has a heating element with a resistance of 8.6 (). (a) What is the current in the element when it is connected to 240 V? (b) How much charge passes through the element in 50 min? (Assume direct current.)

6. (II) A hair dryer draws 9.5A when plugged into a 120-V

line. (a) What is its resistance? (b) How much charge passes through it in 15 min? (Assume direct current.)

7. (II) A 4.5-V battery is connected to a bulb whose resistance

is 1.6 Q. How many electrons leave the battery per minute?

8. (IT) A bird stands on a dc electric transmission line carrying

resistor. (¢) What is its resistance, and (b) how many joules of energy does. Sas ba.lttery Jasie LA TS

10. (I) An electric device draws 6.50A

at 240V. (a) If the

voltage drops by 15%, what will be the current, assumi nothing else changes? (b) If the resistance of the device were reduced by 15%, what current would be drawn at 240 V?

25-4

Resistivity

11. (I) What is the diameter of a 1.00-m length of tungsten wire whose resistance is 0.32 2? 12. (I) What is the resistance of a 4.5-m length of copper wire 1.5 mm in diameter? 13. (II) Calculate the ratio of the resistance of 10.0m of aluminum wire 2.0 mm in diameter, to 20.0 m of copper wire

LB mmin diameter.

_

14. (H)_ Can a 2.2-mm-d1amete'r copper wire have the Samc resistance as a tungsten wire of the same length? Give

numerical details.

.

o

a

wir_e

(diameter = 0.32mm, length = 11cm)

resulting currents / are measured as follows:

potential differ-

1 (mA)

ence

between

¥

V (V)

0100

72

0200

144

0300

216

0400

288

and

the

0.500

360

(a) If this wire obeys Ohm’s law, graphing / vs. V will result in a straight-line plot. Explain why this is so and determine the theoretical predictions for the straight line’s slope and y-intercept. (b) Plot I vs. V. Based on this plot, can you conclude that the wire obeys Ohm’s law (i.e., did you obtain a straight line with the expected y-intercept)? If so, determine the wire’s resistance R. (c) Calculate the wire’s resistivi and use Table 25-1 to identify the solid material from whi it is composed.

the bird’s feet?

FIGURE 25-34

Problem 8.

CHAPTER 25

'

~ 15. (II) A sequence of potential differences V' is applied across

3100 A (Fig. 25-34). The line has 2.5 X 107°Q resistance

per meter, and the bird’s feet are 4.0 cm apart. What is the

672

9. (II) A 12-V battery causes a current of 0.60A through a

Electric Currents and Resistance

16. (II) How much would you have to raise the temperature of a copper wire (originally at 20°C) to increase its resistance by 15%? 17. (II) A certain copper wire has a resistance of 10.0 Q. At what point along its length must the wire be cut so that the

resistance of one piece is 4.0 times the resistance of the

other? What is the resistance of each piece? 18. (IT) Determine at what temperature aluminum will have the same resistivity as tungsten does at 20°C. 19, (IT) A 100-W lightbulb has a resistance of about 12 {) when cold (20°C) and 140 Q) when on (hot). Estimate the temperature of the filament when hot assuming an average temperature coefficient of resistivity a = 0.0045 (C°)~!. 20. (IT) Compute the voltage drop along a 26-m length of household no. 14 copper wire (used in 15-A circuits). The wire has diameter 1.628 mm and carries a 12-A current. 21. (IT) Two aluminum wires have the same resistance. If one has twice the length of the other, what is the ratio of the diameter of thé‘: longer wire to the diameter of the shorter wire? 22. (ImH A reqtangular solid made of carbon has sides of lengths 1.0cm, 2.0cm, and 4.0cm, lying y along the x, y, and z axes, respectively (Fig. 25-35). Determine the resistance for current that passes through the solid in (@) the x direction, (b) the y direction, and (c¢) the z direction. Assume the resistivity is p=30x10°Qm. FIGURE 25-35 Problem 22.

Lre

to be

p

B = omi

measured.

As

the

object

deforms,

the

length of the wire sensor changes by A{, and the resulting change AR in the sensor’s resistance is measured. Assuming that as the solid wire is deformed to a length £, its density (and volume) remains constant (only approximately valid),

show that the strain (= A£/f) of the wire sensor, and thus

of the object to which it is attached, is AR/2Ry. 25. (IT) A length of wire is cut in half and the two lengths are wrapped together side by side to make a thicker wire. How does the resistance of this new combination compare to the resistance of the original wire? 26. (IIT) For some applications, it is important that the value of a resistance not change with temperature. For example, suppose you made a 3.70-k() resistor from a carbon resistor and a Nichrome wire-wound resistor connected together so the total resistance is the sum of their separate resistances. What value should each of these resistors have (at 0°C) so that the combination is temperature independent?

n

for current that flows radially outward. [Hint: Divide the resistor into concentric cylindrical shells and integrate.] (b) Evaluate the resistance R for such a resistor made of carbon whose inner and outer radii are 1.0mm and 1.8 mm and whose length is 2.4 cm. (Choose p = 15 X 107 Q-m.) (c¢) What is the resistance in } ¢ } part (b) for current flowing . parallel to the axis?

4

23. (IT) A length of aluminum wire is connected to a precision 10.00-V power supply, and a current of 0.4212 A is precisely measured at 20.0°C. The wire is placed in a new environment of unknown temperature where the measured current is 0.3818 A. What is the unknown temperature? . (II) Small changes in the length of an object can be measured using a strain gauge sensor, which is a wire with undeformed length £, cross-sectional area A;, and resistance Rj. This sensor is rigidly affixed to the object’s surface, a}igning its length in the direction in which length changes

27. (III) Determine a formula for the total resistance of a spherical shell made of material whose conductivity is o and whose inner and outer radii are r; and r,. Assume the current flows radially outward. 28. (ITT) The filament of a lightbulb has a resistance of 12 ) at 20°C and 140Q when hot (as in Problem 19). (a) Calculate the temperature of the filament when it is hot, and take into account the change in length and area of the filament due to thermal expansion (assume tungsten for which the thermal expansion coefficient is ~5.5 X 1076 C°™!), (b) In this temperature range, what is the percentage change in resistance due to thermal expansion, and what is the percentage change in resistance due solely to the change in p? Use Eq. 25-5. 29, (IIT) A 10.0-m length of wire consists of 5.0m of copper followed by 5.0m of aluminum, both of diameter 1.4 mm. A voltage difference of 85 mV is placed across the composite wire. (a) What is the total resistance (sum) of the two wires? (b) What is the current through the wire? (c¢) What are the voltages across the aluminum part and across the copper part? 30. (ITI) A hollow cylindrical resistor with inner radius r; and outer radius r,, and length £, is made of a material whose resistivity is p (Fig. 25-36). (a) Show that the resistance is given by

FIGURE 25-36

Problem 30.

25-5 and 25-6 Electric Power 31 (I) What is the maximum power consumption of a 3.0-V portable CD player that draws a maximum of 270 mA of current? 32. (I) The heating element of an electric oven is designed to produce 3.3 kW of heat when connected to a 240-V source. What must be the resistance of the element? 33. (I) What is the maximum voltage that can be applied across a 3.3-k() resistor rated at § watt? 34. (I) (a) Determine the resistance of, and current through, a 75-W lightbulb connected to its proper source voltage of

110 V. (b) Repeat for a 440-W bulb.

3s. (II) An electric power plant can produce electricity at a fixed power P, but the plant operator is free to choose the voltage V at which it is produced. This electricity is carried as an electric current / through a transmission line (resistance R) from the plant to the user, where it provides the user with electric power P’. (a) Show that the reduction in power AP = P — P’ due to transmission losses is given by

AP = P?R/V?

(b) In order to reduce power losses during

transmission, should the operator choose V to be as large as small as possible? 36. (IT) A 120-V hair dryer has two settings: 850 W and 1250 (a) At which setting do you expect the resistance to higher? After making a guess, determine the resistance (b) the lower setting; and (c) the higher setting,

Problems

or

W, be at

673

37. (II) A 115-V fish-tank heater is rated at 95 W. Calculate (a) through the heater when it is operating, and )itthe current ” its resistance.

25-7 Alternating Current 49 . (I) Calculate the peak

38. (II) You buy a 75-W lightbulb in Europe, where electricity is

50. (I) An ac voltage, whose peak value is 180V, is across a 380-(),

delivered to homes at 240 V. If you use the lightbulb in the United

States at 120V

(assume

a

2.7-kQ

resistor

resistor. What are the rms and peak currents in the resistor? S

51, (II) Estimate the resistance of the 120-Vyy circuits in your

[Hinr: Assume roughly that brightness is proportional to

electrical is unplugged, and (b) there are two 75-W light-

120-V bulbs?

power consumed.] 39. (II) How many kWh of energy does a 550-W toaster use in the morning if it is in operation for a total of 6.0min? At a cost of 9.0 cents/kWh, estimate how much this would add to your monthly electric energy bill if you made toast four mornings per week.

40. (1) At $0.095/kWh, what does it cost to leave a 25-W porch light on day and night for a year?

41. (II) What is the total amount of energy stored in a 12-V, 75-A-h car battery when it is fully charged? 42. (II) An ordinary flashlight uses two D-cell 1.5-V batteries connected in series as in Fig. 25-4b (Fig. 25-37). The bulb draws 380 mA when turned on. (a) Calculate the resistance of the bulb and the power dissipated. (b) By what factor would the power increase if four D-cells in series were used with the same bulb? (Neglect heating effects of the filament.) Why shouldn’t you try this?

house

FIGURE 25-37 “ b3 (IT) How

many

75-W

Eroblemd2

power

company,

when

(a) everything

by a 2.5-hp pump connected to a 240-Vy ac power source?

(b) What is the maximum current passing through the pump? 55. (II) A heater coil connected to a 240-V;¢ ac line has a resistance of 44 (). (a) What is the average power used? (b) What are the maximum and minimum values of the instantaneous power? 56 . (IT) For a time-dependent voltage V/(¢), which is periodic with period 7, the rms voltage is defined to be

Vems = [# fUT v? dt]%. Use this definition to determine Vs

(in terms of the peak voltage Vp) for (a) a sinusoidal voltage, ie, V(t) = Vysin(2mt/T) for 0 = ¢ = T; and (b) a positive square-wave voltage, i.e.,

Microscopic View of Electric Current

.

to 120V

Fig. 25-20, can be used without blowing a 15-A fuse?

as in

44, (IT) An extension cord made of two wires of diameter 0.129cm (no. 16 copper wire) and of length 2.7m (9ft) is connected to an electric heater which draws 15.0A on a 120-V line. How much power is dissipated in the cord?

45. (I) A power station delivers 750kW of power at 12,000V to a factory through wires with total resistance 3.0 {). How

much less power is wasted if the electricity is delivered at

50,000 V rather than 12,000 V?

46. (III) A small immersion heater can be used in a car to heat a cup of water for coffee or tea. If the heater can heat 120 mL of water from 25°C to 95°C in 8.0 min, (a) approximately how much current does it draw from the car’s 12-V battery, and (b) what is its resistance? Assume the manufacturer’s claim of 75% efficiency.

47. (III) The current in an electromagnet connected to a 240-V line is 17.5 A. At what rate must cooling water pass over the coils if the water temperature is to rise by no more than 6.50 C°?

48. (II) A 1.0-m-long round tungsten wire is to reach a temperature of 3100K when a current of 15.0A flows through it. What diameter should the wire be? Assume the wire loses energy only by radiation (emissivity e = 1.0, Section 19-10) and the surrounding temperature is 20°C. CHAPTER 25

by the

57. (II) A 0.65-mm-diameter copper wire carries a tiny currer.fl of 2.3 uA. Estimate (a) the electron drift velocity, (b) the current density, and (c) the electric field in the wire. 58. (II) A 5.80-m length of 2.0-mm-diameter wire carries a : ; 750-mA current when 22.0mV is applied to its ends. If the

.

lightbulbs, connected

as seen

bulbs burning. 52. (IT) The peak value of an alternating current in a 1500-W device is 5.4 A. What is the rms voltage across it? 53. (II) An 1800-W arc welder is connected to a 660-V;ys ac line. Calculate (a) the peak voltage and (b) the peak current. 54. (II) (a) What is the maximum instantaneous power dissipated

25-8

674

in

its resistance does not

change), how bl'ight will it be relative to 75-W

43.

current

connected to a 220-V rms ac source.

Electric Currents and Resistance

.

.

drift velocity is 1.7 X 107>

_5

.

.

m/s, determine (a) the resistance R

of the wire, (b) the resistivity p, (c) the current density j, (d) the electric field inside the wire, and (e) the number n of free electrons per unit volume. 59 . (II) At a point high in the Earth’s atmosphere, He?" ions in a concentration of 2.8 X 10'2/m? are moving due north at a

speed of 2.0 X 10 m/s. Also,a 7.0 X 10''/m? concentration

of O3 ions is moving due south at a speed of 6.2 X 105 m/s. Determine

the magnitude

density j at this point.

and

direction

of the current

*25-10 Nerve Conduction *60. (I) What is the magnitude of the electric field across an axon membrane 1.0 X 108 m thick if the resting potential is =70mV? *61. (II) A neuron is stimulated with an electric pulse. The action potential is detected at a point 3.40cm down the axon 0.0052 s later. When the action potential is detected 7.20 cm

from the point of stimulation, the time required is 0.0063 s.

What

is the speed of the electric pulse along the axon?

(Why are two measurements needed instead of only one?)

*62. (III) During an action potential, Na™ ions move into the cell at a rate of about 3 X 107" mol/m?-s. How much power must be produced by the “active Na” pumping” system t produce this flow against a +30-mV potential difference? Assume that the axon is 10 cm long and 20 um in diameter.

| General Problems .

63. A person accidentally leaves a car with the lights on. If each of the two headlights uses 40 W and each of the two taillights 6 W, for a total of 92W, how long will a fresh 12-V battery last if it is rated at 85 A-h? Assume the full 12V appears across each bulb. 64. How many coulombs are there in 1.00 ampere-hour? 65. You want to design a portable electric blanket that runs on a 1.5-V battery. If you use copper wire with a 0.50-mm diameter as the heating element, how long should the wire be if you want to generate 15 W of heating power? What happens if you accidentally connect the blanket to a 9.0-V battery? 66. What is the average current drawn by a 1.0-hp 120-V motor? (1hp = 746 W.) 67. The conductance G of an object is defined as the reciprocal of the resistance R;thatis, G = 1/R. The unit of conductance

is a mho (= ohm™), which is also called the siemens (S).

68. 69.

.

71.

72.

73. 74.

75.

What is the conductance (in siemens) of an object that draws 480 mA of current at 3.0 V? The heating element of a 110-V, 1500-W heater is 3.5m long. If it is made of iron, what must its diameter be? (a) A particular household uses a 1.8-kW heater 2.0 h/day (“on” time), four 100-W lightbulbs 6.0 h/day, a 3.0-kW electric stove element for a total of 1.0 h/day, and miscellaneous power amounting to 2.0 kWh/day. If electricity costs $0.105 per kWh, what will be their monthly bill (30d)? (b) How much coal (which produces 7500 kcal/kg) must be burned by a 35%-efficient power plant to provide the yearly needs of this household? A small city requires about 15 MW of power. Suppose that instead of using high-voltage lines to supply the power, the power is delivered at 120V. Assuming a two-wire line of 0.50-cm-diameter copper wire, estimate the cost of the energy lost to heat per hour per meter. Assume the cost of electricity is about 9.0 cents per kWh. A 1400-W hair dryer is designed for 117V. (a) What will be the percfirtage change in power output if the voltage drops to 105 V? Assume no change in resistance. (b) How would the actual chdnge in resistivity with temperature affect your answer? The wiring in a house must be thick enough so it does not become so hot as to start a fire. What diameter must a copper wire be if it is to carry a maximum current of 35 A and produce no more than 1.5 W of heat per meter of length? Determine the resistance of the tungsten filament in a 75-W 120-V incandescent lightbulb (a) at its operating temperature of about 3000 K, () at room temperature. Suppose a current is given by the equation / = 1.80 sin 210¢, where 7 is in amperes and ¢ in seconds. () What is the frequency? (b) What is the rms value of the current? (c¢) If this is the current through a 24.0-Q resistor, write the equation that describes the voltage as a function of time. A microwave oven running at 65% efficiency delivers 950 W to the interior. Find (a) the power drawn

from the source,

and (b) the current drawn. Assume a source voltage of 120 V. 76. A 1.00-€) wire is stretched uniformly to 1.20 times its original length. What is its resistance now? . 220V is applied to two different conductors made of the same material. One conductor is twice as long and twice the diameter of the second. What is the ratio of the power transformed in the first relative to the second?

78. An electric heater is used to heat a room of volume 54 m®, Air is brought into the room at 5°C and is completely replaced twice per hour. Heat loss through the walls amounts to approximately 850kcal/h. If the air is to be maintained at 20°C, what minimum wattage must the heater have? (The specific heat of air is about 0.17 kcal/kg- C°.) . A 2800-W oven is connected to a 240-V source. (a) What is the resistance of the oven? (b) How long will it take to bring 120 mL of 15°C water to 100°C assuming 75% efficiency? (¢) How much will this cost at 11 cents/kWh? 80. A proposed electric vehicle makes use of storage batteries as its source of energy. Its mass is 1560 kg and it is powered by 24 batteries, each 12V, 95 A-h. Assume that the car is driven on level roads at an average speed of 45 km/h, and

the average friction force is 240 N. Assume 100% efficiency and neglect energy used for acceleration. No energy is consumed when the vehicle is stopped, since the engine doesn’t need to idle. (a) Determine the horsepower required. (b) After approximately how many kilometers must the batteries be recharged? 81. A 12.5-Q) resistor is made from a coil of copper wire whose total mass is 15.5 g. What is the diameter of the wire, and how long is it? 82. A fish-tank heater is rated at 95W when connected to 120 V. The heating element is a coil of Nichrome wire. When uncoiled, the wire has a total length of 3.8 m. What is the diameter of the wire? 83. A 100-W, 120-V lightbulb has a resistance of 12 () when cold (20°C) and 140Q when on (hot). Calculate its power consumption (a) at the instant it is turned on, and (b) after a few moments when it is hot. 84. In an automobile, the system voltage varies from about 12V when the car is off to about 13.8 V when the car is on and the charging system is in operation, a difference of 15%. By what percentage does the power delivered to the headlights vary as the voltage changes from 12V to 13.8 V? Assume the headlight resistance remains constant. 85. The Tevatron accelerator at Fermilab (Illinois) is designed to carry an 11-mA beam of protons traveling at very nearly

the speed of light (3.0 X 108 m/s) around a ring 6300m in

circumference. How many protons are in the beam? 86. Lightbulb A is rated at 120V and 40 W for household applications. Lightbulb B is rated at 12V and 40 W for automotive applications. (a) What is the current through each bulb? (b) What is the resistance of each bulb? (c¢) In one hour, how much charge passes through each bulb? (d) In one hour, how much energy does each bulb use? (e) Which bulb requires larger diameter wires to connect its power source and the bulb? 87. An air conditioner draws 14 A at 220-V ac. The connecting cord is copper wire with a diameter of 1.628 mm. (a) How much power does the air conditioner draw? (b) If the total length of wire is 15 m, how much power is dissipated in the wiring? (c) If no. 12 wire, with a diameter of 2.053 mm, was

used instead, how much power would be dissipated in the wiring? (d) Assuming that the air conditioner is run 12 h per day, how much money per month (30 days) would be saved by using no. 12 wire? Assume

12 cents per kWh.

that the cost of electricity is

General Problems

675

. Copper wire of diameter 0.259 cm is used to connect a set of appliances at 120V, which draw 1750 W of power total. (a) What power is wasted in 25.0 m of this wire? (b) What is your answer if wire of diameter 0.412 cm is used? 89. Battery-powered electricity is very expensive compared with that available from a wall receptacle. Estimate the cost per kWh of (a) an alkaline D-cell (cost $1.70) and (b) an alkaline AA-cell (cost $1.25). These batteries can provide a continuous current of 25mA

for 820h

95. The level of liquid helium (temperature = 4K) in its storage tank can be monitored using a vertically aligned niobium—-titanium (NbTi) wire, whose length f spans the height of the tank. In this level-sensing setup, an

electronic circuit maintains a constant electrical current / ¢

all times in the NbTi wire and a voltmeter monitors the voltage difference V across this wire. Since the superconducting transition temperature for NbTi is 10 K, the portion

of the wire immersed in the liquid helium is in the superconducting state, while the portion above the liquid (in helium vapor with temperature above 10K) is in the normal state. Define f = x/f to be the fraction of the tank filled with liquid helium (Fig. 25-40) and V, to be the value of V when the tank is empty (f =0). DeterHelium mine the relation vapor between f and V (>10K)

and 120 h, respectively, at 1.5V.

Compare to normal 120-V ac house current at $0.10/kWh. . How far does an average electron move along the wires of a 550-W toaster during an alternating current cycle? The power cord has copper wires of diameter 1.7mm and is plugged into a standard 60-Hz 120-V ac outlet. [Hint: The maximum current in the cycle is related to the maximum drift velocity. The maximum velocity in an oscillation is related to the maximum displacement; see Chapter 14.] 91. A copper pipe has an inside diameter of 3.00cm and an outside diameter of 5.00 cm (Fig. 25-38). What is the resistance of a 10.0-m length of this pipe? 3.00 cm

e

__”//

T

E

(in terms of Vj).

© e0 T -]

§|

3

y

Liquid

helium

1

| x

FIGURE 25-38 Problem 91. FIGURE 25-40 Problem 95.

92. For the wire in Fig. 25-39, whose diameter varies uniformly from a to b as shown, suppose a current / = 2.0 A enters ata.If

a =25mm

and

density (assume uniform) at each end?

FIGURE 25-39 Problems 92 and 93.

b = 4.0mm,

|

|

|

*Numerical/Computer *96. (II) The resistance, R, of a particular thermistor as a function of temperature 7 is shown in this Table:

TCC)

R(®)

T (C)

R (Q)

20 22 24 26 28 30 32 34

126,740 115,190 104,800 95,447 87,022 79,422 72,560 66,356

36 38 40 42 44 46 48 50

60,743 55,658 51,048 46,863 43,602 39,605 36,458 33,591

03 The cross section of a portion of wire increases uniformly as shown in Fig. 25-39 so it has the shape of a truncated cone. The diameter at one end is a and at the other it is b, and the total length along the axis is £. If the material has resistivity p,

determine the resistance R between the two ends in terms of a, b, L, and p. Assume that the current flows uniformly through each section, and that the taper is small, i.e., (b — a)

v

T

E

%

DC Circuit CHAPTER-OPENING

QUESTION—Guess

The automobile headlight bulbs

now!

CONTENTS

shown in the circuits here are

identical. The connection which produces more light is (a) circuit 1. (b) circuit 2. (¢) both the same.

(d) not enough information. Circuit 1

Circuit 2

26-1

EMF and Terminal Voltage

26-2

Resistors in Series and in Parallel

26-3

Kirchhoff’s Rules

26-4

Series and Parallel EMFs; Battery Charging

26-5

Circuits Containing Resistor and Capacitor (RC Circuits)

26-6

Electric Hazards

#26-7

Ammeters and Voltmeters

lectric circuits are basic parts of all electronic devices from radio and TV

sets to computers and automobiles. Scientific measurements, from physics

to biology and medicine, make use of electric circuits. In Chapter 25, we discussed the basic principles of electric current. Now we will apply these principles to analyze dc circuits involving combinations of batteries, resistors, and capacitors. We also study the operation of some useful instruments.” ’

AC circuits that contain only a voltage source and resistors can be analyzed like the dc circuits in this Chapter. However, ac circuits that contain capacitors and other circuit elements are more complicated, and we discuss them in Chapter 30.

677

TABLE 26-1 Symbols for Circuit Elements :

Symbol i

:

Device

When we draw a diagram for a circuit, we represent batteries, capacitors, and resistors by the symbols shown in Table 26—1. Wires whose resistance is negligible compared with other resistance in the circuit are drawn simply as straight lines.

Some circuit diagrams show a ground symbol (< or J,) which may mean a real

Battery

connection to the ground, perhaps via a metal pipe, or it may simply mean

“Wr

: Retmsto'r

For the most part in this Chapter, except in Section 26-5 on RC circuits, we will be interested in circuits operating in their steady state. That is, we won’t be looking at a circuit

=

Switch

I

or 46

S

Capacitor

common connection, such as the frame of a car.

o

W;rees);svtnéficx;eghgxble

Lol

Ground

o

.

at the moment a change is made in it, such as when a battery or resistor is connected

or disconnected, but rather later when the currents have reached their steady values.

26—-1

EMF and Terminal Voltage

To have current in an electric circuit, we need a device such as a battery or an electric generator that transforms one type of energy (chemical, mechanical, or light, for example) into electric energy. Such a device is called a source of electromotive force or of emf, (The term “electromotive force” is a misnomer since it does not refer to a “force” that is measured in newtons. Hence, to avoid confusion, we prefer to use the abbreviation, emf.) The potential difference between the terminals of such a source, when no current flows to an external circuit, is called the emf of the source. The symbol € is usually used for emf (don’t confuse it with E for electric field), and its unit is volts. A CAUTION A battery is not a source of constant current—the current out of a battery Why battery voliage isn’t perfecily ~ varies according to the resistance in the circuit. A battery is, however, a nearly constant— constant voltage source, but not perfectly constant as we now discuss. You may have noticed in your own experience that when a current is drawn from a battery, the potential difference (voltage) across its terminals drops below its rated emf. For example, if you start a car with the headlights on, you may notice the headlights dim. This happens because the starter draws a large current, and the battery voltage drops as a result. The voltage drop occurs because the chemical reactions in a battery (Section 25-1) cannot supply charge fast enough to maintain the full emf. For Onaem, thing, charge must move (within the electrolyte) between the electrodes of the battery, and there is always some hindrance to completely free flow. Thus, a battery itself has some resistance, which is called its internal resistance; it is usually designated r. A real battery is modeled as if it were a perfect emf € in series with a resistor r, FIGURE 26-1 Diagram for an as shown in Fig. 26-1. Since this resistance r is inside the battery, we can never electric cell or battery. separate it from the battery. The two points a and b in the diagram represent the two terminals of the battery. What we measure is the terminal voltage Vi = Vo — Vp. When no current is drawn from the battery, the terminal voltage equals the emf, which is determined by the chemical reactions in the battery: Voo = €. However, when a current / flows naturally from the battery there is an internal drop in voltage equal to Ir. Thus the terminal voltage (the actual voltage) is’ Vb = € —1In. (26-1) For example, if a 12-V battery has an internal resistance of 0.1 ), then when 10 A flows from the battery, the terminal voltage is 12V — (10A)(0.1Q) = 11 V. The internal resistance of a battery is usually small. For example, an ordinary flashlight battery when fresh may have an internal resistance of perhaps 0.05 ). (However, as it ages and the electrolyte dries out, the internal resistance increases to many ohms.) Car batteries have lower internal resistance.

FIGURE 26-2 T Il

Example 26-1.

R=6500 M

a —o—AAAA—

osa '

678

CHAPTER 26

8=

b ——

120V

Battery with

internal

resistance. A 65.0-Q resistor is

connected to the terminals of a battery whose emf is 12.0.V and v-vhos.e internal resistance is 0.5 (), Fig. 26-2. Calculate (a) the current in the circuit, (b) the terminal voltage of the battery, V,, and (c) the power dissipated in the resistor R and in the battery’s internal resistance r. APPROACH We first consider the battery as a whole, which is shown in Fig. 26-2 as an emf € and internal resistance r between points a and b. Then we appl‘fl

V = IR to the circuit itself.

"When a battery is being charged, a current is forced to pass through it; we then have to write Vap

=

€ +

I,

See Section 26—4 or Problem 28 and Fig. 26-46.

SOLUTION

(a) From Eq. 26-1, we have Vab

=

¢

—

I

We applyEOhm’s law (Egs. 25-2) to this battery and the resistance R of the fcircuit: Voo = IR. Hence IR =¢ —Ir or €= 1I(R+r), andso ¢ 120V 120V

I'=

57

~ &on+050

~ &sa

(b) The terminal voltage is Vip = € —Ir = 120V — (0183 A)(05Q) (c) The power dissipated (Eq. 25-7) in R is Py, = I’R = (0.183A)%(65.0Q) = 2.18W, and in r is e

P,

=

I*r

EXERCISE A Repeat

=

(0.183A)*(05Q)

Example

26-1

assuming

whereas ¢ and r remain as before. In much

of what

follows, unless

=

_ X8A =

119V.

0.02W.

now

that

the

resistance

stated otherwise, we assume

R = 10.0(),

that the battery’s

internal resistance is negligible, and that the battery voltage given is its terminal voltage, which we will usually write simply as V rather than V,;,. Be careful not to confuse V (italic) for voltage and V (not italic) for the volt unit.

26-2

Resistors in Series and in Parallel

When two or more resistors are connected end to end along a single path as shown in Fig. 26-3a, they are said to be connected in series. The resistors could be simple

resistors as were pictured in Fig. 25-12, or they could be lightbulbs (Fig. 26-3b), or

heating elements, or other resistive devices. Any charge that passes through R, in Fig. 26-3a will also pass through R, and then R;. Hence the same current / passes

’fihrough each resistor. (If it did not, this would imply that either charge was not onserved, or that charge was accumulating at some point in the circuit, which does not happen in the steady state.)

We let V represent the potential difference (voltage) across in Fig. 26-3a. We assume all other resistance in the circuit can : : equals the terminal voltage supplied by the battery. We let Vi, potential di‘fferences across each of the resistors, R;, R,, and

From Ohm’s law, V = IR, we can write V; = IR,,

V, =1IR,,

all three resistors be ignored, so V. V,, and V; be the R;, respectively.

and

V; = IR;.

Because the resistors are connected end to end, energy conservation tells us that the total voltage V is equal to the sum of the voltages' across each resistor:

V=V+V+V,

=IR

+ IR, + IR;.

[series]

=~ FIGURE 26-3

(a) Resistances

connected in series. (b) Resistances

couldbe lightbulbs, or any other type

of resistance. (c) Equivalent single

reSIStan_c‘;eReq_th;t dra;vs theRsame current:

Req = R,

Ry +

R,

Kz +

3.

R;

VWWW—AM—MVW ; , )

;

4

V2

V3

¥

(a)

(26-2)

Now let us determine the equivalent single resistance R4 that would draw the same current / as our combination of three resistors in series; see Fig. 26-3c. Such a single resistance R.q would be related to V by V

=

IR

eq-

We equate this expression with Eq. 26-2, V = I(R, + R, + R;), and find

J: Ryq

= R

[series]

+ R, + R;.

(26-3)

This is, in fact, what we expect. When we put several resistances in series, the total or equivalent rgsistance is the sum of the separate resistances. (Sometimes we may also call it the “net resistance.”) This sum applies to any number of resistances in series. Note that when you add more resistance to the circuit, the current through the circuit will decrease. For example, if a 12-V battery is connected to a 4-() resistor, the current

(b) r’

will be 3 A. But if the 12-V battery is connected to three 4-Q) resistors in series, the

!

total resistathe is 12 () and the current through the entire circuit will be only 1 A.

r.‘o see in more detail why this is true, note that an electric charge g passing through R; loses an amount of potential energy equal to ¢V;. In passing through R, and Rs, the potential energy U decreases by gV, and gV3, for a total AU = gV; + gV, + gV3; this sum must equal the energy given

to g by the battery, gV, so that energy

V =V, + V5 + V5, which is Eq. 26-2.

is conserved.

Hence

qV = g(V; + V5 + V3),

and

so

WV

(c)

+

= {,'

SECTION 26-2

679

Another simple way to connect resistors is in parallel so that the current from the source splits into separate branches or paths, as shown in Fig. 26-4a and b. The wiring in houses and buildings is arranged so all electric devices are in parallel, as we already saw in Chapter 25, Fig. 25-20. With parallel wiring, if youfl disconnect one device (say, R; in Fig. 26—4a), the current to the other devices it not interrupted. Compare to a series circuit, where if one device (say, R; in Fig. 26-3a) is disconnected, the current is stopped to all the others. In a parallel circuit, Fig. 26—4a, the total current / that leaves the battery splits into three separate paths. We let /;, I,, and I3 be the currents through each of the resistors, Ry, R,, and R;, respectively. Because electric charge is conserved,

the current / flowing into junction A (where the different wires or conductors meet, Fig. 26—4a) must equal the current flowing out of the junction. Thus I =1+ L+ L. [parallel] When resistors are connected in parallel, each has the same voltage across it. (Indeed, any two points in a circuit connected by a wire of negligible resistance are at the same potential.) Hence the full voltage of the battery is applied to each resistor in Fig. 26—-4a. Applying Ohm’s law to each resistor, we have R%’

11=

L,

=

Rlz’

and

L

=

%3

Let us now determine what single resistor R.q (Fig. 26-4c) will draw the same current / as these three resistances in parallel. This equivalent resistance R.q must satisfy Ohm’s law too:

(b)

I =

V

.

Ry We now combine the equations above: I =L+ L+ L Vv Vv Vv vV R

()

Lty

R

1

1%

R

R

R,

_

4+

R,

—.

R,

fl

CHAPTER 26

DC Circuits

=

1

1 + =+

1 =

%

[parallel]

(26-4)

For example, suppose you connect two 4-() loudspeakers to a single set of output terminals of your stereo amplifier or receiver. (Ignore the other channel for a moment—our two speakers are both connected to the left channel, say.) The equivalent resistance of the two 4-() “resistors” in parallel is

1

Rq

FIGURE 26-5 Water pipes in parallel—analogy to electric currents in parallel.

680

—

When we divide out the V from each term, we have

FIGURE 26-4 (a) Resistances connected in parallel. (b) The resistances could be lightbulbs. (c) The equivalent circuit with Req obtained from Eq. 26—4: 1 1 1 1 — T e— _t —

Rq

=

1

40

12 1

40

40

20

and s0 R.q = 2(). Thus the net (or equivalent) resistance is less than each single resistance. This may at first seem surprising. But remember that when you connect resistors in parallel, you are giving the current additional paths to follow. Hence the net resistance will be less. Equations 26-3 and 26-4 make good sense. Recalling Eq. 25-3 for resistivity, R = pl/A, we see that placing resistors in series increases the length and therefore the resistance; putting resistors in parallel increases the area through which current flows, thus reducing the overall resistance. An analogy may help here. Consider two identical pipes taking in water near the top of a dam and releasing it below as shown in Fig. 26-5. The gravitational potential difference, proportional to the height 4, is the same for both pipes, just as the voltage is the same for parallel resistors. If both pipes are open, rather than only one, twice as much water will flow through. That is, with two equal pipes open, the net resistance to the flow of water will be reduced, by half, just as for electrical resistors in parallel. Note that if both pipes are closed, the dam offers infinite resistance to the flow o water. This corresponds in the electrical case to an open circuit—when the path is not continuous and no current flows—so the electrical resistance is infinite. EXERCISE B You have a 10-Q and a 15-() resistor. What is the smallest equivalent resistance that you can make with these two resistors?

and largest

CONCEPTUAL

EXAMPLE 26-2 | Series ot parallel? (a) The lightbulbs in Fig. 26-6

are identical. Which configuration produces more light? (b) Which way do you think the headlights of a car are wired? Ignore change of filament resistance R with current. RESPONSE (a) The equivalent resistance of the parallel circuit is found from Eq. 264, 1/Req = 1/R + 1/R = 2/R. Thus R.q = R/2. The parallel combination then has lower resistance (= R/2) than the series combination (ch = R + R = 2R). There will be more total current in the parallel configuration (2),since I = V//R.q and V'is the same for both circuits. The total power transformed, which is related to the light produced, is P = IV, so the greater current in (2) means more light produced. (b) Headlights are wired in parallel (2), because if one bulb goes out, the other bulb can stay lit. If qhey were in series (1), when one bulb burned out (the filament broke), the circuit would be open and no current would flow, so neither bulb would light. NOTE When you answered the Chapter-Opening Question on page 677, was your answer circuit 2? Can you express any misconceptions you might have had? CONCEPTUAL EXAMPLE 26-3 | An illuminating surprise. A 100-W, 120-V lightbulb and a 60-W, 120-V lightbulb are connected in two different ways as shown in Fig. 26—b. In each case, which bulb glows more brightly? Ignore change of filament resistance with current (and temperature). RESPONSE (a) These are normal lightbulbs with their power rating given for 120 V. They both receive 120V, so the 100-W bulb is naturally brighter. (b) The resistance of the 100-W bulb is less than that of the 60-W bulb (calculated from P = V2/R at constant 120 V). Here they are connected in series and receive the same current. Hence, from P = I’R (I = constant) the higher-resistance “60-W” bulb will transform more power and thus be brighter. NOTE When connected in series as in (b), the two bulbs do not dissipate 60 W and 100 VY because neither bulb receives 120 V.

(1) Series

(2) Parallel

FIGURE 26-6

Example 26-2.

FIGURE 26-7

Example 26-3. 60 W

(CV

e

AU

ey

(b)

Note that whenever a group of resistors is replaced by the equivalent resistance, (-urrent and voltage and power in the rest of the circuit are unaffected.

series

and

parallel

resistors. How

much

FIGURE 26-8 (a) Circuit for Examples 26—4 and 26-5. (b) Equivalent circuit, showing the equivalent resistance of 290 () for the two parallel resistors in (a).

so0Q 1i

consider tfilis Rp to be in series with the 400-() resistor.

SOLUTION The equivalent resistance, Rp, of the 500-Q and 700-Q) resistors in parallel is glven by

take this reciprocal. Notice that the units of reciprocal ohms,

R

= Qoo

1

E

— 8

~J

S S S = 0.0020Q7" + 0.0014Q7' = 0.0034 Q7L \ RP 5000 7000 This is 1/Rp, so we take the reciprocal to find Rp. It is a common mistake to forget to ) !, are a reminder. Thus

T 0%

This 290 () is the equivalent resistance of the two parallel resistors, and is in series with the 400-() resistor as shown in the equivalent circuit of Fig. 26—8b. To find

c

T

|

a 400Q p

-—~

with

1

Circuit

S

58

o]

EXAMPLE yI

current is drawn from the battery shown in Fig. 26-8a? APPROACH The current 7 that flows out of the battery all passes through the 400-Q) refilstor but then it splits into /; and I, passing through the 500-Q) and 700-Q) resistors. The latter two resistors are in parallel with each other. We look for someLE-nng that we already know how to treat. So let’s start by finding the equivalent resistance, Rp, of the parallel resistors, 500 {) and 700 £). Then we can

a

400Q

b

the total equivalent resistance R.q, we add the 400-Q and 290-Q) resistances together, since they are in series, and find

Req = 400Q + 2900 = 690(). The total current flowing from the battery is then A

I

]

|4 R

=

120V

000

—_———

0.0174 A ~

(b) 17mA.

NOTE This / is also the current flowing through the 400-) resistor, but not through the 500-() and 700-(2 resistors (both currents are less —see the next Example). NOTE Complex resistor circuits can often be analyzed in this way, considering |the circuit as a combination of series and parallel resistances.

A CAUTION

Remember to take the reciprocal

SECTION 26-2

681

Current in one branch. What is the current through the

a

400Q

p

c

500-€) resistor in Fig. 26-8a? ’ APPROACH We need to find the voltage across the 500-Q) resistor, which is the voltage between points b and c in Fig. 26-8a, and we call it V;.. Once Vj is known, we can apply Ohm’s law, V = IR, to get the current. First we find th voltage across the 400-() resistor, V,,, since we know that 17.4 mA passes through it (Example 26-4). SOLUTION V,, can be found using V = IR: Vab

=

(0.0174 A)(400Q)

=

7.0V.

Since the total voltage across the network of resistors is V,. = 12.0V, then Vj, must be 12.0V — 7.0V = 5.0V. Then Ohm’s law applied to the 500-Q) resistor tells us that the current /; through that resistor is =

I FIGURE 26-8 (repeated) (a) Circuit for Examples 26—4 and 26-5. (b) Equivalent circuit, showing the equivalent resistance of 290 ) for the two parallel resistors in (a).

50V 5000

—_—

=

1.00

X

X 107° —2 A

=

10 mA

!

This is the answer we wanted. We can also calculate the current I, through the 700-() resistor since the voltage across it is also 5.0 V:

NOTE When /; combines with 7, to form the total current / (at point c in Fig. 26-8a), their sum is 10mA + 7mA = 17 mA. This equals the total current / as calculated in Example 26-4, as it should.

CONCEPTUAL

EXAMPLE 26-6 | Bulb brightness in a circuit. The circuit shown

in Fig. 26-9 has three identical lightbulbs, each of resistance R. (a) When switch S is closed, how will the brightness of bulbs A and B compare with that of bulb C? (b) What happens when switch S is opened? Use a minimum of mathematics in your answers.

I

1

FIGURE 26-9

Example 26-6,

three identical lightbulbs. Each yellow circle with -“AA\~ inside represents a lightbulb and its resistance.

RESPONSE (a) With switch S closed, the current that passes through bulb C must split into two equal parts when it reaches the junction leading to bulbs A and B. splits into equal parts because the resistance of bulb A equals that of B. Thus, bulb. A and B each receive half of C’s current; A and B will be equally bright, but they will be less bright than bulb C (P = I2R). (b) When the switch S is open, no current can flow through bulb A, so it will be dark. We now have a simple one-loop series circuit, and we expect bulbs B and C to be equally bright. However, the equivalent resistance of this circuit (=R + R) is greater than that of the circuit with the switch closed. When we open the switch, we increase the resistance and reduce the current leaving the battery. Thus, bulb C will be dimmer when we open the switch. Bulb B gets more current when the switch is open (you may have to use some mathematics here), and so it will be brighter than with the switch closed; and B will be as bright as C.

A two-speed

fan. One way a multiple-speed

ventilation fan for a car can be designed is to put resistors in series with the fan motor. The resistors reduce the current through the motor and make it run more slowly. Suppose the current in the motor is 5.0 A when it is connected directly across a 12-V battery. (a) What series resistor should be used to reduce the current to 2.0 A for low-speed operation? (b) What power rating should the resistor have? APPROACH

682

CHAPTER 26

DC Circuits

An

electric motor

in series with a resistor can be treated as two

resistors in series. The power comes from P = IV. SOLUTION (2) When the motor is connected to 12V and drawing 5.0 A, its resistance is R =V/I = (12V)/(5.0A) = 2.4Q. We will assume that this is the motor’s resistance for all speeds. (This is an approximation because the current through the motor depends on its speed.) Then, when a current of 2.0A is flowing, the voltage across the motor is (2.0 A)(2.4 Q) = 4.8 V. The remaining 120V — 48V = 72V must appear across the series resistor. When 2.0 A flows through the resistor, its resistance must be R = (7.2V)/(20A) = 3.6Q. (b) Tk power dissipated by the resistor is P = (7.2V)(2.0A) = 144 W. To be safe, a power rating of 20 W would be appropriate.

Analyzing a circuit. A 9.0-V battery whose internal resis-

tance r is 0.50£} is connected in the circuit shown in Fig. 26-10a. (¢) How much current is drawn from the battery? (b) What is the terminal voltage of the fbattery? (c) What is the current in the 6.0-() resistor? APPROACH To find the current out of the battery, we first need to determine the equivalent resistance R.q of the entire circuit, including r, which we do by identifying and isolating simple series or parallel combinations of resistors. Once we find

I

from

Ohm’s

law,

[ = ¢/ R.q,

we

get

the

terminal

voltage

using

Vap = € — Ir. For (c) we apply Ohm’s law to the 6.0-() resistor. SOLUTION (@) We want to determine the equivalent resistance of the circuit. But where do we start? We note that the 4.0-Q and 8.0-() resistors are in parallel, and so have anl‘ equivalent resistance R.q; given by

| 1

—

| Reg S0

Req

= 2.7Q.

1

800

This

1

-

400

2.7Q

3

=

8.00°

is in series

.

with

the

6.0-() resistor, as shown

in

the equivalent circuit of Fig. 26-10b. The net resistance of the lower arm of the circuit is t}Fen - Ry

=

600

+ 270

=

874,

as shown in Fig. 26-10c. The equivalent resistance R.y3 of the 8.7-() and 10.0-Q resistances in parallel is given by

1

Re |

$0 Reg3 = the 0.50-() resistance Athe curren‘

(

=

1

1000

-

1

10.0Q

Ty7a - 0HY =

3

VW

=

R.»=87Q

(1/021 Q') = 4.8 Q. This 4.8 Q is in series with the 5.0-0) resistor and internal resistance of the battery (Fig. 26-10d), so the total equivalent R.q of the circuit is R.q = 4.8 + 5.0Q + 0.50Q = 10.3€Q. Hence drawn is

% 9.0V | I = Rq = —— 1030 = 087A.

-

500

©)

(b) The terminal voltage of the battery is

Vip = €—Ir = 90V — (0.87A)(0.50Q) = 86V.

Y MWv

© &=90v

Reqmg

(c) Now we can work back and get the current in the 6.0-() resistor. It must be the same as the current through the 8.7() shown in Fig. 26-10c (why?). The voltage across that 8.7 () will be the emf of the battery minus the voltage drops

across

r and

the

5.0-Q

resistor:

500

Vg;=9.0V — (0.87A)(0.50Q + 5.0Q).

Applying Ohm’s law, we get the current (call it /')

9.0V — (0.87A)(0.50Q + 5.00) 870 _— ‘ ' ) This is the current through the 6.0-() resistor. I'

=

=

'

r=050Q (d)

{5’-—",;OV =2 FIGURE 26-10 Circuit for Example 26-8, where r is the

0.48A.

WW-

internal resistance of the battery.

26-3

Kirchhoff’s Rules

In the last kfew Examples we have been able to find the currents in circuits by combining resistances in series and parallel, and using Ohm’s law. This technique can be used for many circuits. However, some circuits are too complicated for that analysis. For example, we cannot find the currents in each part of the circuit shown in Fig. 26-11 simply by combining resistances as we did before. To deal

with such complicated

circuits, we

FIGURE 26-11 Currents can be calculated using Kirchhoff’s rules.

r»

300

h

use Kirchhoff’s rules, devised by

G. R. Kirchhoff (1824-1887) in the mid-nineteenth century. There are two rules, ra:d they are simply convenient applications of the laws of conservation of charge nd energy.

SECTION 26-3

Kitchhoff's Rules

683

Kirchhoff’s first rule or junction rule is based on the conservation of electric charge that we already used to derive the rule for parallel resistors. It states that at any junction point, the sum of all currents entering the junction must equal the sum of all currents leaving the junction.

Junction rule (conservation of charge)

That is, whatever charge goes in must come out. We already saw an instance of this in the NOTE at the end of Example 26-5. Kirchhoff’s second rule or loop rule is based on the conservation of energy. It states that the sum of the changes in potential around any closed loop of a circuit must be zero.

Loop rule (conservation of energy)

To see why this rule should hold, consider a rough analogy with the potential

A

400Q

p

energy

290Q

S

[ —{—e

1

PR

SR

120V (a)

12_

€

a

v

120V

0

!

J

70V J:——— | 50V 1l

|

n

oot 120V

400 Q

290 Q

—e

(b) FIGURE 26-12

Changes in

potential around the circuit in (a) are plotted in (b).

of a roller coaster on its track. When

the roller coaster starts from

the

station, it has a particular potential energy. As it climbs the first hill, its potential energy increases and reaches a maximum at the top. As it descends the other side, its potential energy decreases and reaches a local minimum at the bottom of the hill. As the roller coaster continues on its path, its potential energy goes through more changes. But when it arrives back at the starting point, it has exactly as much potential energy as it had when it started at this point. Another way of saying this is that there was as much uphill as there was downhill. Similar reasoning can be applied to an electric circuit. We will analyze the circuit of Fig. 26-11 shortly but first we consider the simpler circuit in Fig. 26-12. We have chosen it to be the same as the equivalent circuit of Fig. 26—8b already discussed. The current in this circuit is /7 = (12.0V)/(690Q) = 0.0174 A, as we calculated in Example 26-4. (We keep an extra digit in / to reduce rounding errors.) The positive side of the battery, point e in Fig. 26-12a, is at a high potential compared to point d at the negative side of the battery. That is, point e is like the top of a hill for a roller coaster. We follow the current around the circuit starting at any point. We choose to start at point d and follow a positiveq test charge completely around this circuit. As we go, we note all changes i potential. When the test charge returns to point d, the potential will be the same as when we started (total change in potential around the circuit is zero). We plot the changes in potential around the circuit in Fig. 26-12b; point d is arbitrarily taken as zero. As our positive test charge goes from point d, which is the negative or low potential side of the battery, to point e, which is the positive terminal (high potential side) of the battery, the potential increases by 12.0 V. (This is like the roller coaster being pulled up the first hill.) That is,

Vg = +120V.

“ PROBLEM

SOLVING

Be consistent with signs when applying the loop rule

When our test charge moves from point e to point a, there is no change in potential since there is no source of emf and we assume negligible resistance in the connecting wires. Next, as the charge passes through the 400-() resistor to get to point b, there is a decrease in potential of V = IR = (0.0174 A)(400(2) = 7.0V. The positive test charge is flowing “downhill” since it is heading toward the negative terminal of the battery, as indicated in the graph of Fig. 26-12b. Because this is a decrease in potential, we use a negative sign: Vba

=

Vb

=

Va

=

=70V.

As the charge proceeds from b to c there is another potential decrease (a “yoltage drop”) of (0.0174 A) X (290Q) = 5.0V, and this too is a decrease in potential: Vcb

=

=50V.

There is no change in potential as our test charge moves from c to d as we assume negligible resistance in the wires. The sum of all the changes in potential around the circuit of Fig. 26-12 is +120V 684

CHAPTER 26

- 70V

= 50V

=

0.

This is exactly what Kirchhoff’s loop rule said it would be.

sOLVs, Y

~_

¢

Kirchhoff's Rules

3 &% &

1. Label the current in each separate branch of the given circuit with a different subscript, such as L, L, I; (see Fig. 26-11 or 26-13). Each current refers 90 a segment between two junctions. Choose the direction of each current, using an arrow. The direction can be chosen arbitrarily: if the current is actuall§ in the opposite direction, it will come out with a minus sign in the solution. . Identify the unknowns. You will need as many independent equations as there are unknowns. You may write down more equations than this, but you will find that some of the equations will be redundant (that is, not be independent in the sense of providing new information). You may use V = IR for each resistor, which sometimes will reduce the number of unknowns. . Apply Kirchhoff’s junction rule at one or more junctions.

4. Apply Kirchhoff’s loop rule for one or more loops: follow each loop in one direction only. Pay careful attention to subscripts, and to signs: (a) For a resistor, apply Ohm'’s law; the potential difference is negative (a decrease) if your chosen loop direction is the same as the chosen current direction through that resistor; the potential difference is positive (an increase) if your chosen loop direction is opposite to the chosen current direction. (b) For a battery, the potential difference is positive if your chosen loop direction is from the negative terminal toward the positive terminal; the potential difference is negative if the loop direction is from the positive terminal toward the negative terminal. . Solve the equations algebraically for the unknowns. Be careful when manipulating equations not to err with signs. At the end, check your answers by plugging them into the original equations, or even by using any additional loop or junction rule equations not used previously.

SOV IETR Using Kirchhoff's rules. Calculate the currents /;, [, and I in the three branches of the circuit in Fig. 26-13 (which is the same as Fig. 26-11).

APPROACH AND SOLUTION

/|

L

>

~

1. Label the currents and their directions. Figure 26-13 uses the labels I, I,, and L; for the current in the three separate branches. Since (positive) current tends to move away from the positive terminal of a battery, we choose I, and I; to have the directions shown in Fig. 26—13. The direction of /; is not obvious in advance, so we arbitrarily chose the direction indicated. If the current actually flows in the opposite direction, our answer will have a negative sign. Identify the unknowns. We have three unknowns and therefore we need three equations, which we get by applying Kirchhoff’s junction and loop rules. Junction rule: We apply Kirchhoff’s junction rule to the currents at point a, where I; enters and I, and [ leave:

=1

+ 1.

This same equation holds at point d, so we get no new information by writing an equation for point d. Loop ruFe: We apply Kirchhoff’s loop rule to two different closed loops. First we apply it to the upper loop ahdcba. We start (and end) at point a. From a to h we have a potential decrease V;, = —(7;)(30Q). From h to d there is no change, but from d to c the potential increases by 45 V: that is, V4 = +45V. From c to a the potential decreases through the two resistances by an amount Vie = —(5)(40Q + 1Q) = —(41 Q)I;. Thus we have Vi, + Veg + Ve = 0, or

=30

+ 45 — 41,

20Q

(a)

= 0,

FIGURE 26-13 Currents can be calculated using Kirchhoff’s rules. See Example 26-9.

(b)

where we have omitted the units (volts and amps) so we can more easily do the algebra. For our second loop, we take the outer loop ahdefga. (We could have chosen the lower loop abcdefga instead.) Again we start at point a and have Vp, = —(L)(30Q), and Vg, = 0. But when we take our positive test charge from d to e, it actually is going uphill, against the current—or at least against

the assumed

direction

of the

current, which

is what

counts

in this

calculati(‘l)n. Thus Veq = ,(20Q) has a positive sign. Similarly, Vi, = L(1Q). From f to g there is a decrease in potential of 80 V since we go from the high potential terminal of the battery to the low. Thus V, = —80V. Finally, Vag = 0, and the sum of the potential changes around this loop is =301, + (20 + 1)I, — 80 = 0. (©) Our major work is done. The rest is algebra.

SECTION 26-3

685

5. Solve the equations. We have three equations—Ilabeled (a), (b), and (c¢)—and three unknowns. From Eq. (¢) we have

80 + 301,

ey

= 3.8 + 1.41,.

(d)

= 1.1 - 0.731,.

(e)

From Eq. (b) we have

45 — 301,

L = *41—1

We substitute Egs. (d) and (e) into Eq. (a): L

=L-L

=11-0734

— 38 — 141,.

We solve for I, collecting terms:

3L

I,

= =

=27 —-0.87A.

The negative sign indicates that the direction of /; is initially assumed and shown in Fig. 26-13. The answer in amperes because all values were in volts and ohms. L, = 38 + 141, = 38 + 1.4(-0.87) = and from Eq. (e) L

=

11-0734

=

1.1 - 0.73(-087)

actually opposite to that automatically comes out From Eq. (d) we have 2.6, =

1.7 A.

This completes the solution. NOTE The unknowns in different situations are not necessarily currents. It might be that the currents are given and we have to solve for unknown resistance or voltage. The variables are then different, but the technique is the same. EXERCISE C Write the equation for the lower loop abcdefga of Example 26-9 and show, assuming the currents calculated in this Example, that the potentials add to zero for this lower loop.

26-4

(a)

a

b

-”+

€

Il+

(b) io—{hvm—-bo—mjl—i—

AMWW—)

12V -+

I

(©) _.El L

12V FIGURE 26-14

Batteries in series

(a) and (b), and in parallel (c). 686

CHAPTER 26

DC Circuits

Series and Parallel EMFs;

-

Battery Charging

When two or more sources of emf, such as batteries, are arranged in series as in Fig. 26-14a, the total voltage is the algebraic sum of their respective voltages. On the other hand, when a 20-V and a 12-V battery are connected oppositely, as shown in Fig. 26-14b, the net voltage V,, is 8 V (ignoring voltage drop across internal resistances). That is, a positive test charge moved from a to b gains in potential by 20 V, but when it passes from b to c it drops by 12 V. So the net change is 20V — 12V = 8V. You might think that connecting batteries in reverse like this would be wasteful. For most purposes that would be true. But such a reverse arrangement is precisely how a battery charger works. In Fig. 26-14b, the 20-V source is charging up the 12-V battery. Because of its greater voltage, the 20-V source is forcing charge back into the 12-V battery: electrons are being forced into its negative terminal and removed from its positive terminal. An automobile alternator keeps the car battery charged in the same way. A voltmeter placed across the terminals of a (12-V) car battery with the engine running fairly fast can tell you whether or not the alternator is charging the battery. If it is, the voltmeter reads 13 or 14 V. If the battery is not being charged, the voltage will be 12V, or less if the battery is discharging. Car batteries can be recharged, but other batteries may not be rechargeable, since the chemical reactions in many cannot be reversed. In such cases, the arrangement of Fig. 26-14b would simply waste energy. Sources of emf can also be arranged in parallel, Fig. 26—-14c. With equal emfs, a parallel arrangement can provide more energy when large currents are needed. Each of the cells in parallel has to produce only a fraction of the total current, ss the energy loss due to internal resistance is less than for a single cell; and ti batteries will go dead less quickly.

SOV IEAN Jump starting a car. A good car battery is being used to jump start a car with a weak battery. The good battery has an emf of 12.5V and internal re‘sistance 0.020 ). Suppose the weak battery has an emf of 10.1 V and internal resistance 0.10 ). Each copper jumper cable is 3.0 m long and 0.50 cm in diameter, and can be attached as shown in Fig. 26-15. Assume the starter motor can be represented as a resistor Ry = 0.15 (). Determine the current through the starter motor (a) if only the weak battery is connected to it, and (b) if the good battery is also connected, as shown in Fig. 26-15. APPROAC}I We apply Kirchhoff’s rules, but in (b) we will first need to determine the resistance of the jumper cables using their dimensions and the resistivity (p = 1.68 X 108 Q-m for copper) as discussed in Section 25-4. SOLUTION (a) The circuit with only the weak battery and no jumper cables is simple: an emf of 10.1 V connected to two resistances in series, 0.10Q + 0.15€ =

1

0.25 Q. Hence the currentis I = V/R = (10.1V)/(025Q) = 40 A. (b) We need to find the the good battery. From

resistance of the jumper cables Eq. 25-3, each has resistance

that connect R, = pl/A =

(1.68 X 108 Q-m)(3.0m)/(7)(0.25 X 107> m)* = 0.0026 Q. Kirchhoff’slooprule

for the full outside loop gives [

12.5V_11(2RJ+"1)-I3RS

125V

= L(0.025Q)

= 0

101V = 5(015Q) ~ 5(0.100) = 0.

(i)

The junction rule at point B gives

= I

We have three equations in three unknowns. From Eq. (iii), [y =

substitute this into Eq. (i):

125V - (I — L)(0.025Q) — K(015Q) = L(0175Q)

+ L(0.025Q)

=

(i)

5 — I, and we

= 0,

of the weak 10.1-V battery is now Vg, = 101V — (=5A)(0.10Q) = 10.6 V.

The circuit shown in Fig. 26-15, without the starter motor, is how a battery

can be charged. The stronger battery pushes charge back into the weaker battery.

EXERCISED If the jumper cables of Example 26-10 were mistakenly connected in reverse, theLpositive terminal of each battery would be connected to the negative terminal of the other battery (Fig. 26-16). What would be the current / even before the starter motor is engaged (the switch S in Fig. 26-16 is open)? Why could this cause the batteries to explode?

26-5

r Weak battery 1 I I

lflllli”_:q.B

{O{;Q

&=

b o

Starter

101y

|

R,=0.15Q

switch | S

113

s

(closed)

@

U

e

FIGURE 26-15

Example 26-10

a jump start.

0.

Combining this last equation with Eq. (ii) gives 5 = 71 A, quite a bit better than in part (a). The other currentsare I, = =5 A and [, = 76 A. Notethat [, = =5 A is in the opposite direction from that assumed in Fig. 26-15. The terminal voltage NOTE

\§

__________

(i)

since (2R; + r) = (0.0052Q + 0.020Q) = 0.025 Q.

125V

§ n

The loop rule for the lower loop, including the weak battery and the starter, gives

L+

~

125V i 11

Jumper cables

g/

=0

— L(015Q)

&=

10020Q

Circuits Containing Resistor

“and Capacitor (RC Circuits)

FIGURE 26-16

Exercise D.

D 125v

Ry A

0.020Q DON’'T

TRY

THIS

0.10Q

|

Ry

10.1V

s N

Our study o‘f circuits in this Chapter has, until now, dealt with steady currents that do not change in time. Now we examine circuits that contain both resistance and capacitance. Such a circuit is called an RC circuit. RC circuits are common in everyday life: they are used to control the speed of a car’s windshield wiper, and the timing of the change of traffic lights. They are used in camera flashes, in heart fi acemakers, and in many other electronic devices. In RC circuits, we are not so terested in the final “steady state” voltage and charge on the capacitor, but rather in how these variables change in time.

SECTION 26-5

Circuits Containing Resistor and Capacitor (RC Circuits)

687

Let us now examine the RC circuit shown in Fig. 26—17a. When the switch S is closed, current immediately begins to flow through the circuit. Electrons will flow out from the negative terminal of the battery, through the resistor R, and accumulate on the upper plate of the capacitor. And electrons will flow into the positive terminal of the battery, leaving a positive charge on the other plate of the capacitc*™ As charge accumulates on the capacitor, the potential difference across it increases (Ve = Q/C), and the current is reduced until eventually the voltage across the capacitor equals the emf of the battery, €. There is then no potential difference across the resistor, and no further current flows. Potential difference V. across the capacitor thus increases in time as shown in Fig. 26—-17b. The mathematical form of this curve—that is, V- as a function of time—can be derived using conservation of energy (or Kirchhoff’s loop rule). The emf € of the battery will equal the sum of the voltage drops across the resistor (/R) and the capacitor (Q/C):

R

MW _-:.—6

C=

+

/

S

(a)

€ = IR + 0&=

| |

I t=0 t=RC !

(b)

|

2RC

Time

|

3RC

¢

8/R

The resistance R includes all resistance in the circuit, including the internal resistance of the battery; I is the current in the circuit at any instant, and Q is the charge on the capacitor at that same instant. Although ¢, R, and C are constants, both Q and I are functions of time. The rate at which charge flows through the resistor (/ = dQ/dt) is equal to the rate at which charge accumulates on the capacitor. Thus we can write

Current /

¢=

0

(26-5)

dQ 1 R—=+ —0Q. Rar *¢c@

This equation can be solved by rearranging it: g dt RC

(c)

1

2RC

C¢-Q

3RC

Time

FIGURE 26-17

!

RC

We now integrate from capacitor:

¢ = 0, when

| 0oi o LRC|,[[a Cé€ —Q

closes in the RC circuit shown in (a),

-In(C¢

- Q)

Q

_t

RC ~In(C% — Q) - (~InC¥) = Et'c"

increases with time as shown in (b),

and the current through the resistor decreases with time as shown in (c).

1 or

g %

-,

t

After the switch S

the voltage across the capacitor

Q = 0, to time ¢ when a charge Q is on the

=

o

t

0

e p—t/RC

Q = C¥1 - e/RO),

(26-6a)

The potential difference across the capacitor is V. = Q/C,

Ve = ¥(1 — e7/FC),

so

(26-6b)

From Egs. 26-6 we see that the charge Q on the capacitor, and the voltage V¢

across

it, increase

from

zero

at

t =0

to

maximum

values

Q, = Cé

and

Ve = € after a very long time. The quantity RC that appears in the exponent is called the time constant 7 of the circuit: T = RC. (26-7)

It represents the time* required for the capacitor to reach (1 — ') = 0.63 or 63% of

its full charge and voltage. Thus the product RC is a measure of how quickly t**

688

CHAPTER 26

"The constant e, known as the base for natural logarithms, has the value this e with e for the charge on the electron.

*The units of RC are Q-F = (V/A)(C/V) = C/(C/s) = s.

e = 2.718 ---. Do not confuse

capacitor

gets

C = 3.0uF,

charged.

In

a circuit,

for

example,

where

R = 200k{)

the time constant is (2.0 X 10°Q)(3.0 X 10°°F) = 0.60s.

and

If the

resistance is much lower, the time constant is much smaller. This makes sense, since lower resistance will retard the flow of charge less. All circuits contain some (A sistance (if only in the connecting wires), so a capacitor never can be charged mstantaneously when connected to a battery. From Eqgs. 26-6, it appears that Q and V. never quite reach their maximum values within a finite time. However, they reach 86% of maximum in 2RC, 95% in 3RC, 98% in 4RC, and so on. Q and V. approach their maximum values asymptotically. For example, if R =20k{) and C = 030uF, the time constant is

(2.0 x 10°Q)(3.0

X 1077 F) = 6.0 X 10s.

So the capacitor is more than 98%

charged in less than 45 of a second. The current / through the circuit of Fig. 26—17a at any time f can be obtained by differentiating Eq. 26—6a: { I

=

ig

=

dt

Ee#/RC

R

(26-8)

'

Thus, at ¢ = 0, the current is a maximum, / = I, = €/R, as expected for a circuit containing or}ly a resistor (there is not yet a potential difference across the capacitor). The current then drops exponentially in time with a time constant equal to RC, as the voltage across the capacitor increases. This is shown in Fig. 26-17c. The time constant RC represents the time required for the current to drop to 1/e ~ 0.37 of its initial value.

RC

circuit, with

emf. The capacitance in the circuit of

Fig. 26-17ais C = 0.30 uF, the total resistance is 20 k(), and the battery emf is 12 V. Determine (a) the time constant, (b) the maximum charge the capacitor could acquire, (c) the time it takes for the charge to reach 99% of this value, (d) the current / when the charge Q is half its maximum value, (e) the maximum current, and (f) the charge Q when the current / is 0.20 its

}v\aximum value. APPROACH

We use Fig. 26-17 and Eqgs. 26-5, 6, 7, and 8.

SOLUTIONT(a) The time constant is RC = (2.0 X 10*Q)(3.0 X 107F) = 6.0 X 103s. (b) The maximum charge would be Q = C% = (3.0 X 107 F)(12V) = 3.6 uC. (c) In Eq. 26—-6a, we set Q = 0.99C€:

0.99C% or

= C¥(1 — e/FC),

e /RC

= 1 - 0.99

= 0.01.

Then t E SO

=

—1n(0.01)

t = 46RC

or 28 ms (d) From value, 1.8 equation, ]

=

4.6

= 28 X 10735

(less than 35s). part (b) the maximum charge is 3.6 uC. When the charge is half this uC, the current / in the circuit can be found using the original differential or Eq. 26-5: =

1

—|¢

(

—

0

=

-

)

1

—

20 X 10°Q (12V

_—

1.8 x10°¢C

e

0.30 X 10‘6F)

=

300 uA

"

(e) The current is maximum at ¢ = 0, when there is no charge on the capacitor (Q = 0): I = € _ 12V

R

20x10'Q

= 600 pA.

;(f) Again using Eq. 26-5, now with 7 = 0.20/; = 120 nA, we have

|_

0 =C(¢-IR)

= (3.0 X 107F)[12V - (1.2 x 1074 A)(2.0 X 10*Q)] = 2.9 uC. SECTION 26-5

Circuits Containing Resistor and Capacitor (RC Circuits)

689

The circuit just discussed involved the charging of a capacitor by a battery through a resistance. Now let us look at another situation: a capacitor is already charged to a voltage V;, and charge Q,, and it is then allowed to discharge through a resistance R as shown in Fig. 26—18a. (In this case there is no battery.) When the switch S is closed, charge begins to flow through resistor R from one side of thAfl capacitor toward the other side, until the capacitor is fully discharged. The voltage across the resistor at any instant equals that across the capacitor: IR

QC

=

The rate at which charge leaves the capacitor equals the negative of the current in the resistor, / = —dQ/dt, because the capacitor is discharging (Q is decreasing). So we write the above equation as 0

t=RC

2RC

3RC

_A0L,_

2

dtR

!

C

Time

(b)

FIGURE 26-18 For the RC circuit shown in (a), the voltage V- across the capacitor decreases with time, as shown in (b), after the switch S is closed at ¢ = 0. The charge on the capacitor follows the same curve since Ve « Q.

and integrate it from ¢ = 0 when the charge on the capacitor is Q,, to some time ¢ later when the charge is O: In 9

or

e

Qo

Q

=

RC

Qe

EC,

(26-92)

The voltage across the capacitor

Vo = Ve

(Ve = Q/C)

as a function of time is

e,

(26-9bam,

where the initial voltage V; = Q,/C. Thus the charge on the capacitor, and the voltage across it, decrease exponentially in time with a time constant RC. This is shown in Fig. 26-18b. The current is

[ =

aQ

——=

dt

_

= = ¢/RC

RC

=

[ e/RC

0

(

26-10

)

and it too is seen to decrease exponentially in time with the same time constant RC. The charge on the capacitor, the voltage across it, and the current in the resistor all decrease to 37% of their original value in one time constant = 7 = RC. EXERCISE E In 10 time constants, the charge on the capacitor in Fig. 26-18 will be about

(@) Q0/20,000, (b) Qo/5000, (c) Qo/1000, (d) Qo/10, (e) Qo/3? FIGURE 26-19 a

gor [ oy

200V

- 'I'

I

Discharging

RC

circuit. In the RC

circuit shown

Fig. 26-19, the battery has fully charged the capacitor, so

S

capacitance C = 1.02 uF. The current [ is observed to decrease to 0.50 of its initial value in 40 us. (¢) What is the value of Q, the charge on the capacitor, at t = 0? (b) What is the value of R? (c¢) What is Q at ¢ = 60 us?

R

W

T1.02 uF

Q, = C¢é.

in

Example 26-12.

Then at

t = 0 the switch is thrown from position a to b. The battery emf is 20.0 V, and the

APPROACH At ¢t = 0, the capacitor has charge Q, = C%, and then the battery is removed from the circuit and the capacitor begins discharging through the resistor, as in Fig. 26-18. At any time ¢ later (Eq. 26-9a) we have Q

—

Qoe—t/RC

—

chefl/RC.

> 690

CHAPTER 26

DC Circuits

SOLUTION (a) At ¢ = 0, Q =0,

= Cé

= (102X 10°F)(20.0V)

= 2,04 X 10°C

= 204 uC.

f(b) To find R, we are given that at ¢ = 40 us, I = 0.50/,. Hence

0500,

= [e

¥,

Taking natural logs on both sides t — RC

0693

=

R

t (0.693)C

SO =

(In0.50 = —0.693):

=

(40 X 107°5) (0.693)(1.02 X 10°°F)

=

57Q.

(c) At 1 = 60 s, B

60x10~%s

Q = Que/’C = (20.4 X 105C)e (TONRXI0TF)

= 73,,C,

CONCEPTUAL EXAMPLE 26-13 | Bulbin RC circuit. In the circuit of Fig. 26-20, the capacitor is originally uncharged. Describe the behavior of the lightbulb from the instant switch S is closed until a long time later. RESPONSE When the switch is first closed, the current in the circuit is high and the lightbulb burns brightly. As the capacitor charges, the voltage across the capacitor increases causing the current to be reduced, and the lightbulb dims. As the potential difference across the capacitor approaches the same voltage as the Ettery, the current decreases toward zero and the lightbulb goes out.

A

l} FIGURE 26-20

Example 26-13.

- Applications of RC Circuits The charging and discharging in an RC circuit can be used to produce voltage pulses at a fiegular frequency. The charge on the capacitor increases to a particular voltage, and’ then discharges. One way of initiating the discharge of the capacitor is by the use of a gas- -filled tube which has an electrical breakdown when the voltage

across it regches a certain value Vy. After the discharge is finished, the tube no

longer conducts current and the recharging process repeats itself, starting at a lower voltage V;. Figure 26-21 shows a possible circuit, and the “sawtooth” voltage it produces. A simple blinking light can be an application of a sawtooth oscillator circuit. Here the erxflf is supplied by a battery; the neon bulb flashes on at a rate of perhaps 1 cycle per second. The main component of a “flasher unit” is a moderately large capacitor. The intermittent windshield wipers of a car can also use an RC circuit. The RC time constant, which can be changed using a multi-positioned switch for dlfferenf values of R with fixed C, determines the rate at which the wipers come

@ PHYSICS APPLIED Sawtooth, blinkers, windshield wipers

FIGURE 26-21

(a) An RC circuit,

coupled with a gas-filled tube as a switch, can produce a repeating “sawtooth” voltage, as shown in (b). AW R R +| 1 T S Gas-fitgfig

on.

SEVILE D ITALE

ESTIMATE | Resistor in a turn signal. Estimate the order

of magnitude of the resistor in a turn-signal circuit.

APPROACH A typical turn signal flashes perhaps twice per second, so the time constant is on the order of 0.5s. A moderate capacitor might have C = 1 uF. SOLUTION Setting 7 = RC = 0.5, we find |

i

R=2-_05 . sp0xq. C 1 X10™°F SECTION 26-5

Time (b) Circuits Containing Resistor and Capacitor (RC Circuits)

691

@)PHYSICS

APPLIED

Heart pacemaker

FIGURE 26-22 Electronic batterypowered pacemaker can be seen on the rib cage in this X-ray.

@PHYSlcs

APPLIED

Dangers of electricity

FIGURE 26-23 A person receives an electric shock when the circuit is completed.

120V

;

(a)

An interesting medical use of an RC circuit is the electronic heart pacemaker, which can make a stopped heart start beating again by applying an electric stimulus through electrodes attached to the chest. The stimulus can be repeated at the normal heartbeat rate if necessary. The heart itself contains pacemaker cells, which send out tiny electric pulses at a rate of 60 to 80 per minute. Thes signals induce the start of each heartbeat. In some forms of heart disease, the natural pacemaker fails to function properly, and the heart loses its beat. Such patients use electronic pacemakers which produce a regular voltage pulse that starts and controls the frequency of the heartbeat. The electrodes are implanted in or near the heart (Fig. 26-22), and the circuit contains a capacitor and a resistor. The charge on the capacitor increases to a certain point and then discharges a pulse to the heart. Then it starts charging again. The pulsing rate depends on the values of R and C.

26—-6

Electric Hazards

Excess electric current can heat wires in buildings and cause fires, as discussed in Section 25-6. Electric current can also damage the human body or even be fatal. Electric current through the human body can cause damage in two ways: (1) Electric current heats tissue and can cause burns; (2) electric current stimulates nerves and muscles, and we feel a “shock.” The severity of a shock depends on the magnitude of the current, how long it acts, and through what part of the body it passes. A current passing through vital organs such as the heart or brain is especially serious for it can interfere with their operation. Most people can “feel” a current of about 1 mA. Currents of a few mA cause pain but rarely cause much damage in a healthy person. Currents above 10 mA cause severe contraction of the muscles, and a person may not be able to let go of the source of the current (say, a faulty appliance or wire). Death from paralysis of the respiratory system can occur. Artificial respiratios however, can sometimes revive a victim. If a current above about 80 to 100 m. passes across the torso, so that a portion passes through the heart for more than a second or two, the heart muscles will begin to contract irregularly and blood will not be properly pumped. This condition is called ventricular fibrillation. If it lasts for long, death results. Strangely enough, if the current is much larger, on the order of 1 A, death by heart failure may be less likely,” but such currents can cause serious burns, especially if concentrated through a small area of the body. The seriousness of a shock depends on the applied voltage and on the effective resistance of the body. Living tissue has low resistance since the fluid of cells contains ions that can conduct quite well. However, the outer layer of skin, when dry, offers high resistance and is thus protective. The effective resistance between two points on opposite sides of the body when the skin is dry is in the range of 10* to 10° Q). But when the skin is wet, the resistance may be 10* Q) or less. A person who is barefoot or wearing thin-soled shoes will be in good contact with the ground, and touching a 120-V line with a wet hand can result in a current I =

120V 10000

120 mA.

As we saw, this could be lethal. A person who has received a shock has become part of a complete circuit. Figure 26-23 shows two ways the circuit might be completed when a person "Larger currents apparently bring the entire heart to a standstill. Upon release of the current, the heart returns to its normal rhythm. This may not happen when fibrillation occurs because, once started, it can be hard

(®) 692

CHAPTER 26

to stop. Fibrillation may

also occur as a result of a heart

attack

or during

heart surgery.

device known as a defibrillator (described in Section 24-4) can apply a brief high current to the hear., causing complete heart stoppage which is often followed by resumption of normal beating.

DC Circuits

(@

(b)

©

FIGURE 26-'24 (a) An electric oven operating normally with a 2-prong plug. (b) Short to the case with ungrounded case: shock. (c) Short to the case with the case grounded by a 3-prong plug.

acc1dentall touches a “hot” electric wire—“hot” meaning a high potentlal such as 120V (normal U.S. household voltage) relative to ground. The other wire of building wiring is connected to ground—either by a wire connected to a buried conductor, or via a metal water pipe into the ground. In Fig. 26-23a, the current passes from the high-voltage wire, through the person, to the ground through his bare feet, and back along the ground (a fair conductor) to the ground terminal of the source. If the person stands on a good insulator—thick rubber-soled shoes or a dry wood floor—there will be much more resistance in the circuit and consequently much less current through the person. If the person stands with bare feet on the ground, or is in a bathtub, there is lethal danger because the resistance is much less and the current greater. In a bathtub (or swimming pool), not only are ou wet, which reduces your resistance, but the water is in contact with the drain H ipe (typlcally metal) that leads to the ground. It is strongly recommended that you not touch anything electrical when wet or in bare feet. Building codes that require the use of non-metal pipes would be protective. In Fig. 26-23b, a person touches a faulty “hot” wire with one hand, and the other hand touches a sink faucet (connected to ground via the pipe). The current is partlcularlyLdangerous because it passes across the chest, through the heart and lungs. A useful rule: if one hand is touching something electrical, keep your other hand in yofir pocket (don’t use it!), and wear thick rubber-soled shoes. It is also a good idea to remove metal jewelry, especially rings (your finger is usually moist under a ring). You can come into contact with a hot wire by touching a bare wire whose insulation has worn off, or from a bare wire inside an appliance when you're tinkering with it. (Always unplug an electrical device before investigating’ its insides!) Another possibility is that a wire inside a device may break or lose its insulation and come in contact with the case. If the case is metal, it will conduct elqctricity A person could then suffer a severe shock merely by touching the case, as shown in Fig. 26-24b. To prevent an accident, metal cases are supposed to be connected directly to ground by a separate ground wire. Then if a “hot” wire touches the grounded case, a short circuit to ground immediately occurs internally, as shown in Fig. 26-24c, and most of the current passes through the low-resistance ground wire rather than through the person. Furthermore, the high current should open the fuse or circuit breaker. Grounding a metal case is done by a separate ground wire connected to the third (round) prong of a 3-prong plug. Never cut off the third prong of a plug—it could save your life.

Even then touch it.

you

can

get a bad

shock

from

a capacitor

that

hasn’t been

discharged

until

4\ CAUTION Keep one hand in your pocket when other touches electricity @ PHYSICS APPLIED Grounding and shocks

you

SECTION 26-6

Electric Hazards

693

A three-prong plug, and an adapter, are shown in Figs. 26-25a and b.

Why is a third wire needed? The 120 V is carried by the other two wires—one hot (120V ac), the other neutral, which is itself grounded. The third “dedicated” ground wire with the round prong may seem redundant. But it is protection for two reasons: (1) it protects against internal wiring that may have been don incorrectly; (2) the neutral wire carries normal current (“return” current from

the 120 V) and it does have resistance; so there can be a voltage drop along

(b)

FIGURE 26-25

(a) A 3-prongplug,

and (b) an adapter (gray) for

it—normally small, but if connections are poor or corroded, or the plug is loose, the resistance could be large enough that you might feel that voltage if you touched the neutral wire some distance from its grounding point. Some electrical devices come with only two wires, and the plug’s two prongs are of different widths; the plug can be inserted only one way into the outlet so that the intended neutral (wider prong) in the device is connected to neutral in the wiring (Fig. 26-25c). For example, the screw threads of a lightbulb are meant to be connected to neutral (and the base contact to hot), to avoid shocks when changing a bulb in a possibly protruding socket. Devices with 2-prong plugs do nor have their cases grounded; they are supposed to have double electric insulation. Take extra care anyway. The insulation on a wire may be color coded. Hand-held meters may have red (hot) and black (ground) lead wires. But in a house, black is usually hot (or it may be red), whereas white is neutral and green is the dedicated ground, Fig. 26-26. But beware: these color codes cannot always be trusted. [In the U.S., three wires normally enter a house: two kot wires at 120 V each (which add together to 240 V for appliances or devices that run on 240 V) plus the grounded neutral (carrying return current for the two hots). See Fig. 26-26. The “dedicated” ground wire (non-current carrying) is a fourth wire that does not come from the electric company but enters the house from a nearby heavy stake in the ground or a buried metal pipe. The two hot wires can feed separate 120-V circuits in the house, so each 120-V circuit inside the house has only three wires, including ground.] Normal circuit breakers (Sections 25-6 and 28-8) protect equipment an buildings from overload and fires. They protect humans only in some circumstances, such as the very high currents that result from a short, if they respond quickly enough. Ground fault circuit interrupters (GFCI), described in Section 29-8, are designed to protect people from the much lower currents (10 mA to 100 mA) that are lethal but would not throw a 15-A circuit breaker or blow a 20-A fuse. It is current that harms, but it is voltage that drives the current. 30 volts is sometimes said to be the threshhold for danger. But even a 12-V car battery (Which can supply large currents) can cause nasty burns and shock.

Another danger is leakage current, by which we mean a current along an

old-fashioned 2-prong outlets—

unintended path. Leakage currents are often “capacitively coupled.” For example,

FIGURE 26-26 Four wires entering a typical house. The color codes for wires are not always as

electrical codes limit leakage currents to 1 mA for any device. A 1-mA leakage current is usually harmless. It can be very dangerous, however, to a hospital patient with implanted electrodes connected to ground through the apparatus. This is due to the absence of the protective skin layer and because the current can pass

be sure to screw down the ground tab. (c) A polarized 2-prong plug.

shown here—be careful!

r120v

From

gffig‘a‘;\y _

20N

Black (orred)

120 Vg

o—White ¥ 120\*,

Black

_[ = S;gl:gg ;:)le

694

§

H

2£120V

240 Ve

Neutral yHot— 0

+120¥

I S = E;S::d g

CHAPTER 26

DC Circuits

a wire in a lamp forms a capacitor with the metal case; charges moving in one conductor attract or repel charge in the other, so there is a current. Typical

directly through the heart as compared to the usual situation where the current

enters at the hands and spreads out through the body. Although 100 mA may be

peeded to cause heart fibrillation when entering through the hands (very little

of it actually passes through the heart), as little as 0.02mA has been known to

cause fibrillation when passing directly to the heart. Thus, a “wired” patient

is in considerable danger from leakage current even from as simple an act as

touching a lamp.

Finally, don’t touch a downed power line (lethal!) or even get near it. A hot power line is at thousands of volts. A huge current can flow along the ground or pavement, from where the high-voltage wire touches the ground along its path to the groundin“™ point of the neutral line, enough that the voltage between your two feet could be large. Tip: stand on one foot or run (only one foot touching the ground at a time).

*26-7

Ammeters and Voltmeters

An ammeter is used to measure current, and a voltmeter measures potential differnce or voltage. Measurements of current and voltage are made with meters that

are of two types: (1) analog meters, which display numerical values by the position of a pointer that can move across a scale (Fig. 26-27a); and (2) digital meters, which display the numerical value in numbers (Fig. 26-27b). We now discuss the meters themselves and how they work, then how they are connected to circuits to make measurements. Finally we will discuss how using meters affects the circuit being measured, possibly causing erroneous results—and what to do about it.

@ PHYSICS DC meters

APPLIED

* Analog Ammeters and Voltmeters The crucial part of an analog ammeter or voltmeter, in which the reading is by a pointer on a scale (Fig. 26-27a), is a galvanometer. The galvanometer works on the principle of the force between a magnetic field and a current-carrying coil of wire, and will be discussed in Chapter 27. For now, we merely need to know that the deflection of the needle of a galvanometer is proportional to the current flowing through it. The full-scale current sensitivity of a galvanometer, I,,, is the electric current needed to make the needle deflect full scale. A galvanometer can be used directly to measure small dc currents. For example, a galvanometer whose sensitivity /,, is 50 A can measure currents from about 1 uA (currents smaller than this would be hard to read on the scale) up to 50 uA. To measure larger currents, a resistor is placed in parallel with the galvanometer. Thus, an analog ammeter, represented by the symbol +®-, consists of a galvanometer (+@~) in parallel with a resistor called the shunt resistor, as shown in Fig. 26-28. (“Shunt” is a synonym for “in parallel.”) The shunt resistance is Ry, and the resistance of the galvanometer coil, through which current passes, is r. The value of Ry, is chosen according to the full-scale deflection desired; Ry, is normally very small—giving an eter a very small net resistance—so most of the current passes through Ry, and ary little (= 50 uwA) passes through the galvanometer to deflect the needle.

1R

FIGURE 26-27

(a) An analog

1 (b)

multimeter being used as a voltmeter. (b) An electronic digital meter.

FIGURE 26-28 Anammeterisa galvanometer in parallel with a (shunt) resistor with low resistance, Ry}, .

A G =

Ammeter | [— —)—e = o—

)

Rsh

UL

ITAEN

Ammeter design. Design an ammeter to read 1.0 A at full

r = 30().

Check if the scale is linear.

scale using a galvanometer with a full-scale sensitivity of 50 uA and a resistance

APPROACH

Only

50 uA (=1Ig = 0.000050 A)

of the 1.0-A current must pass

through the galvanometer to give full-scale deflection. The rest of the current

(Ir = 0.999950 A) passes through the small shunt resistor, Ry,, Fig. 26-28. The

potential difference across the galvanometer equals that across the shunt resistor (they are in parallel). We apply Ohm’s law to find Ry, . SOLUTION Because [ = I + Iz, when I = 1.0A flows into the meter, we want I through the shunt resistor to be I = 0.999950 A. The potential difference across the shunt is the same as across the galvanometer, so Ohm’s law tells us then

IgRy Ry

= Igr; =

Ior

(50 x10°A)300Q)

= 1.5 X 1073%Q, Ix (0.999950 A) Aor 0.0015 Q. The shunt resistor must thus have a very low resistance and most of the current passes through it.

Because I = Ix(Rq,/r) and (Rg,/r) is constant, we see that the scale is linear.

*SECTION 26-7

Ammeters and Voltmeters

695

An analog voltmeter (~@-) also consists of a galvanometer and a resistor. But the resistor Ry is connected in series, Fig. 26-29, and it is usually large, giving a voltmeter a high internal resistance. FIGURE 26-29 A_Voltrpeter is a galvanometer in series with a resistor with high resistance, Rgc;.

Voltmeter @

R N

ser

-~

r

Sl

@

DGV TITALE Voltmeter design. Using a galvanometer resistance r = 30() and full-scale current sensitivity of 50 A, meter that reads from 0 to 15 V. Is the scale linear?

with internal design a volt-

APPROACH When a potential difference of 15V exists across the terminals of our voltmeter, we want 50 A to be passing through it so as to give a full-scale deflection.

SOLUTION From Ohm’s law, V = IR, we have (see Fig. 26-29) 0

15V

=

(50pA)(r + Re)s

15V r = 300kQ — 30Q = 300kQ. 50X 105 A Notice that r = 300 is so small compared to the value of Ry, that it doesn’t influence the calculation significantly. The scale will again be linear: if the voltage to be measured is 6.0V, the current passing through the voltmeter will be Ry

=

(6.0V)/(3.0 X 10°Q) = 2.0 X 10° A, or 20 pA. This will produce two-fifths of

full-scale deflection, as required

measured FIGURE 26-30

An ohmmeter.

FIGURE 26-31 and voltage.

Measuring current

4

b

(a) a

"

Vv

b

c

R,

R, (b)

o) o

%

N My

c

4

CHAPTER 26

Suppose you wish to determine the current / in the circuit shown in Fig. 26-31a, and the voltage V across the resistor R;. How exactly are ammeters and voltmeters connected to the circuit being measured? Because an ammeter is used to measure the current flowing in the circuit, it must be inserted directly into the circuit, in series with the other elements, as shown in Fig. 26-31b. The smaller its internal resistance, the less it affects the circuit.

A voltmeter, on the other hand, is connected “externally,” in parallel with the

circuit element across which the voltage is to be measured. It is used to measure the potential difference between two points. Its two wire leads (connecting wires) are connected to the two points, as shown in Fig. 26-31c, where the voltage acrossem, R, is being measured. The larger its internal resistance, (R, + r) in Fig. 26-29, the less it affects the circuit being measured.

(© 696

The meters just described are for direct current. A dc meter can be modified to measure ac (alternating current, Section 25-7) with the addition of diodes (Chapter 40), which allow current to flow in one direction only. An ac meter can be calibrated to read rms or peak values. Voltmeters and ammeters can have several series or shunt resistors to offer a choice of range. Multimeters can measure voltage, current, and resistance. Sometime =y a multimeter is called a VOM (Volt-Ohm-Meter or Volt-Ohm-Milliammeter). An ohmmeter measures resistance, and must contain a battery of known voltage connected in series to a resistor (Ry;) and to an ammeter (Fig. 26-30). The resistor whose resistance is to be measured completes the circuit. The needle deflection is inversely proportional to the resistance. The scale calibration depends on the value of the series resistor. Because an ohmmeter sends a current through the device whose resistance is to be measured, it should not be used on very delicate devices that could be damaged by the current. The sensitivity of a meter is generally specified on the face. It may be given as so many ohms per volt, which indicates how many ohms of resistance there are in the meter per volt of full-scale reading. For example, if the sensitivity is 30,000 /V, this means that on the 10-V scale the meter has a resistance of 300,000 (), whereas on a 100-V scale the meter resistance is 3 M(). The full-scale current sensitivity, /,, discussed earlier, is just the reciprocal of the sensitivity in (/V.

* How to Connect Meters

b

a +—A\WW\—

(6.0 V/15.0V = 2/5).

DC Circuits

*Effects of Meter Resistance It is important to know the sensitivity of a meter, for in many cases the resistance of the meter can seriously affect your results. Take the following Example. *

DOV

TRV

Voltage reading versus true voltage. Suppose you are testing

an electronic circuit which has two resistors, R, and R,, each 15 k), connected in series as shown in Fig. 26-32a. The battery maintains 8.0 V across them and has negligible internal resistance. A voltmeter whose sensitivity is 10,000 /V is put

@

PHYSICS APPLIED Correcting for meter resistance

&

b

a

—WW——AWA—2 R R,

on the 5.0-V scale. What voltage does the meter read when connected across Ry,

Fig. 26—Sjb, and what error is caused by the finite resistance of the meter? APPROACH The meter acts as a resistor in parallel with R,. We use parallel and series resistor analyses and Ohm’s law to find currents and voltages. SOLUTION On the 5.0-V scale, the voltmeter has an internal resistance of

(5.0 V)(10,000 /V) = 50,000 . When connected across R, , as in Fig. 26-32b, we

h;\; téns 50k€) in parallel with Ry = 15k{). The net resistance R.q of these two is

st

B

R,

SR

TR

S

S0kQ

-

15kQ

150kQ’

s0 Req =J11.5 k. This R, = 11.5kQ is in series with R, = 15k(), so the total resistance of the circuit is now 26.5k() (instead of the original 30k(2). Hence the current from the battery is =

80V 265KQ

=

303.0 X X 107* -4 A

=

(. 0.30 mA .

(a) N\ @

Y/

A

B

R, “ (b)

AAAA

C

Ry

L

FIGURE 26-32 E xample

26-17.

Then the voltage drop across R, which is the same as that across the voltmeter,

is (3.0x10*A)(11.5 X 10°Q) = 3.5V. [The voltage drop across R, is (3.0 x 107* A)(15 X 10°Q) = 4.5V, for a total of 8.0 V.] If we assume the meter is precise, it will read 3.5 V. In the original circuit, without the meter, R; = R, so the voltage across R, is half that of the battery, or 4.0V. Thus the voltmeter, because of its internal resistance, gives a low reading, In this case it is off by 0.5 V, or more than 10%.

Example 26-17 illustrates how seriously a meter can affect a circuit and give a misleading reading. If the resistance of a voltmeter is much higher than the resistance

of the circui{t, however, it will have little effect and its readings can be trusted, at least

to the manufactured precision of the meter, which for ordinary analog meters is typically 3% to 4% of full-scale deflection. An ammeter also can interfere with a circuit, but

the effect is minimal if its resistance is much less than that of the circuit as

a whole. For both voltmeters and ammeters, the more sensitive the galvanometer, the less effect it will have. A 50,000-)/V meter is far better than a 1000-Q)/V meter.

*Digital Meters Digital meters (see Fig. 26-27b) are used in the same way as analog meters: they are inserted directly into the circuit, in series, to measure current (Fig. 26-31b), and connected “outside,” in parallel with the circuit, to measure voltage (Fig. 26-31c). The internal construction of digital meters, however, is different from that of analog meters in that digital meters do not use a galvanometer. The electronic circuitry and digital readout are more sensitive than a galvanometer, and have less effect on the circuit to be measured. When we measure dc voltages, a digital meter’s resistance is

very high, commonly on the order of 10 to 100 MQ (10’-10% Q2), and doesn’t change

significantly when different voltage scales are selected. A 100-M(Q digital meter draws off very little current when connected across even a 1-M() resistance. The precision of digital meters is exceptional, often one part in 10* (=0.01%) or better. This precision is not the same as accuracy, however. A precise meter of internal

resistance 10® () will not give accurate results if used to measure a voltage across a 108-Q) resistor—in which case it is necessary to do a calculation like that in Example 26-17.

Whenever we make a measurement on a circuit, to some degree we affect that

’;ircuit (Example 26-17). This is true for ve make a measurement on a system, temperature measurement, for example, the system, thus altering its temperature.

other types of measurement as well: when we affect that system in some way. On a the thermometer can exchange heat with It is important to be able to make needed

corrections, as we saw in Example 26-17.

*SECTION 26-7

697

| Summary A device that transforms another type of energy into electrical energy is called a source of emf. A battery behaves like a source of emf in series with an internal resistance. The emf is the potential difference determined by the chemical reactions in the battery and equals the terminal voltage when no current is drawn. When a current is drawn, the voltage at the battery’s terminals is less than its emf by an amount equal to the potential decrease [r across the internal resistance. When resistances are connected in series (end to end in a single linear path), the equivalent resistance is the sum of the individual resistances:

Reqg = Ry + Ry + . In a series combination, Rq

resistance.

(26-3)

is greater than any component

When resistors are connected in parallel, the reciprocal of the equivalent resistance equals the sum of the reciprocals of

the individual resistances: 1

R

—_—

=

1

R

—

1

R

—_— 4 e,

26-4 (26-4)

In a parallel connection, the net resistance is less than any of the individual resistances. Kirchhoff’s rules are helpful in determining the currents and voltages in circuits. Kirchhoff’s junction rule is based on conservation of electric charge and states that the sum of all currents entering any junction equals the sum of all currents leaving that junction. The second, or loop rule, is based on conservation of energy and states that the algebraic sum of the

changes in potential around any closed path of the circuit must be zero.

When an RC circuit containing a resistor R in series with “™)

capacitance C is connected to a dc source of emf, the voltage across the capacitor rises gradually in time characterized by an

exponential of the form (1 — e T =

*/KC), where the time constant,

RC,

(26-7)

is the time it takes for the voltage to reach 63 percent of its maximum value. The current through the resistor decreases as e */RC, A capacitor discharging through a resistor is characterized by the same time constant: in a time 7 = RC, the voltage across the capacitor drops to 37 percent of its initial value. The charge on the capacitor, and voltage across it, decreases

as e

"/RC as does the current.

Electric shocks are caused by current passing through the body. To avoid shocks, the body must not become part of a complete circuit by allowing different parts of the body to touch objects at different potentials. Commonly, shocks are caused by one part of the body touching ground and another part touching a high electric potential. [*An ammeter measures current. An analog ammeter consists of a galvanometer and a parallel shunt resistor that carries most of the current. An analog voltmeter consists of a galvanometer and a series resistor. An ammeter is inserted into the circuit whose current is to be measured. A voltmeter is external,

being

connected

in

parallel

to

the

element

whose

voltage is to be measured. Digital voltmeters have greater internal resistance and affect the circuit to be measured less than do analog meters.] -

] Questions 1. Explain why birds can sit on power lines safely, whereas leaning a metal ladder up against a power line to fetch a stuck kite is extremely dangerous. 2. Discuss the advantages and disadvantages of Christmas tree lights connected in parallel versus those connected

9. When applying Kirchhoff’s loop rule (such as in Fig. 26-33), does the sign (or direction) of a battery’s emf depend on the direction of current through the battery? What about the terminal voltage? r=10Q

in series.

=18V

3 If all you have is a 120-V line, would it be possible to light several 6-V lamps without burning them out? How?

4. Two lightbulbs of resistance R; and R, (R, > R;) and a battery are all connected in series. Which bulb is brighter? What if they are connected in parallel? Explain. 5. Household outlets are often double outlets. Are connected in series or parallel? How do you know?

these

6. With two identical lightbulbs and two identical batteries, how would you arrange the bulbs and batteries in a circuit to get the maximum possible total power to the lightbulbs? (Assume the batteries have negligible internal resistance.) 7. If two identical resistors are connected in series to a battery, does the battery have to supply more power or less power than when only one of the resistors is connected? Explain. 8. You have a single 60-W bulb on in your room. How does the overall resistance of your room’s electric circuit change when you turn on an additional 100-W bulb? 698

CHAPTER 26

DC Circuits

FIGURE 26-33

Question 9.

R=66Q

r=20Q 6=12V

10. Compare and discuss the formulas for resistors and capacitors when connected in series and in parallel.

for

11. For what use are batteries connected in series? For what use are they connected in parallel? Does it matter if the batteries are nearly identical or not in either case? 12. Can the terminal voltage of a battery ever exceed its emf? Explain. 13. Explain in detail how you could measure the internal resistance of a battery. 14. In an RC circuit, current flows from the battery until the capacitor is completely charged. Is the total energy supplieh by the battery equal to the total energy stored by the capa itor? If not, where does the extra energy go?

15. Given

the circuit shown in Fig. 26-34, use the words “increases,” “decreases,” or “stays the same” to complete the

(

following statements: (a) If R; increases, the potential difference between A and E . Assume no resistance in ® and €. (b) If Ry increases, the potential difference between A and (c) (d) (e) (f)

E

If If If If

R, R, R, R,

through R

the the the the

Wire

Wire

voltage drop across Ry current through R, current through Ry current R

.

FIGURE 26-36

Rs M—«

C

D

Rg

drop across Ry If Ry, Rs, and R; increase, €(r=0) .

*20, Explain why an ideal ammeter would have zero resistance

u

FIGURE 26-34

%ERa

Ry

B )4

Question 15. R,, Rs,

and‘R7 are variable

Ry

resiftors (you can

chajnge their resistance), giv ,‘n the symbol AW-.

Fass

A

8

E

16. Figure 26-35 is a diagram of a capacitor (or condenser) microphone. The changing air pressure in a sound wave causes one plate of the capacitor C to move back ) and fortT. Explain how a

of

the

frequency as the . wave is produced.

same

*22. A small battery-operated flashlight requires a single 1.5-V battery. The bulb is barely glowing, but when you take the battery out and check it with a voltmeter, it registers 1.5 V. How would you explain this? 23. Different lamps might have batteries connected in either of the two arrangements shown in Fig. 26-37. What would be the advantages of each scheme?

Sound "Iy pressure =

li

Question 16.

and an ideal voltmeter infinite resistance.

*21, A voltmeter connected across a resistor always reads less than the actual voltage across the resistor when the meter is not present. Explain.

C

sound

FIGURE 26-35 Diagram of a capacitor microphone.

*18. What is the main difference between an analog voltmeter and an analog ammeter? *19, What would happen if you mistakenly used an ammeter where you needed to use a voltmeter?

MW-

RZZg

|Wire

Question 17.

‘v‘v‘é‘v

(h) If Rs increases, the voltage

current

shown in Fig. 26-36 can be used to operate the same lightbulb from opposite sides of a room.

. Assume ® and € have resistance.

increases, decreases, decreases, decreases,

(g) If Rsincreases, the voltage drop across R, _____.

(i)

17. Design a circuit in which two different switches of the type

pj,vable plate ~ F——

Voutput

v FIGURE 26-37

Question23.

(diaphragm)

(@)

(b)

| Problems 26-1

Emf and Terminal Voltage

1. (I) Calculate the terminal voltage for a battery with an internal resistance of 0.900 ) and an emf of 6.00 V when the battery is connected in series with (a) an 81.0-() resistor, and (b) an 810-) resistor. . (I) Four

1.50-V cells are connected in series to a 12-() light-

bulb. If the resulting current is 0.45 A, what is the internal

resistance of each cell, assuming they neglecting the resistance of the wires?

are identical

and

. (II) A 1.5-V dry cell can be tested by connecting it to a lowresistance ammeter. It should be able to supply at least 25 A. What is the internal resistance of the cell in this case, assuming ?t is much greater than that of the ammeter? (II) What is the internal resistance of a 12.0-V car battery whose terminal voltage drops to 8.4V when the starter draws 95 A? What is the resistance of the starter?

26-2 Resistors in Series and Parallel In these Problems neglect the internal resistance of a battery unless the Problem refers to it.

S. (I) A 650-Q and a 2200-() resistor are connected in series with a 12-V battery. What is the voltage across the 2200-Q) resistor? 6. (I) Three 45-Q lightbulbs and three 65-Q lightbulbs are connected in series. (¢) What is the total resistance of the circuit? (b) What is the total resistance if all six are wired in parallel? . (I) Suppose that you have a 680-(}, a 720-(), and a 1.20-kQ resistor. What is (a) the maximum, and (b) the minimum resistance you can obtain by combining these? . (I) How many 10-{) resistors must be connected in series to give an equivalent resistance to five 100-{) resistors connected in parallel?

Problems

699

(II) Suppose that you have a 9.0-V battery and you wish to apply a voltage of only 4.0 V. Given an unlimited supply of

1.0-Q) resistors, how could you connect them so as to make a

“voltage divider” that produces a 4.0-V output for a 9.0-V input? 10. (IT) Three 1.70-k() resistors can be connected together in four different ways, making combinations of series and/or parallel circuits. What are these four ways, and what is the net resistance in each case? 11. (II) A battery with an emf of 12.0V shows a terminal voltage of 11.8 V when operating in a circuit with two lightbulbs, each rated at 4.0W

(at 12.0 V), which are connected

in parallel. What is the battery’s internal resistance?

12. (II) Eight identical bulbs are connected in series across a 110-V line. (@) What is the voltage across each bulb? (b) If the current is 0.42 A, what is the resistance of each bulb, and

what is the power dissipated in each? 13. (II) Eight bulbs are connected in parallel to a 110-V source by two long leads of total resistance 1.4 Q). If 240 mA flows through each bulb, what is the resistance of each, and what fraction of the total power is wasted in the leads? 14. (IT) The performance of the starter circuit in an automobile can be significantly degraded by a small amount of corrosion on a battery terminal. Figure 26-38a depicts a properly functioning circuit with a battery (12.5-V emf, 0.02-Q internal resistance) attached via corrosion-free cables to a starter motor of resistance Rs = 0.15Q. Suppose that later, corrosion between a battery terminal and a starter cable introduces an extra series resistance of just Rc=010Q into the circuit as suggested in Fig. 26-38b. Let P, be the power delivered to the starter in the circuit free of corrosion, and let P be the

power delivered to the circuit with corrosion. Determine the ratio P/P,,.

FIGURE 26-38

VWW\ Rg=0.15Q

Problem 14.

(b)

15. (IT) A close inspection of an electric circuit reveals that a 480-() resistor was inadvertently soldered in the place where a 370-Q resistor is needed. How can this be fixed without removing anything from the existing circuit? 16. (IT) Determine (a) the equivalent resistance of the circuit shown in Fig. 26-39, and (b) the voltage across each resistor. 820Q FIGURE 26-39

Problem 16.

700

CHAPTER 26

DC Circuits

680 Q

960 Q

17. (II) A 75-W, 110-V bulb is connected in parallel with a 25-W, 110-V bulb. What is the net resistance? 18. (II) (a) Determine the equivalent resistance of the “ladder” of equal 125-Q) resistors shown in Fig. 26-40. In other words, what resistance would an ohmmeter read i connected between points A and B? (b) What is the current through each of the three resistors on the left if a 50.0-V battery is connected between points A and B? R

R

R

R

:

:

A

FIGURE 26-40 Problem 18.

B

19; (II) What is the net resistance of the circuit connected to the battery in Fig. 26-41? R B

R % A §

FIGURE 26-41

R

R R

pC

§R

%

Problems 19 and 20.

20. (IT) Calculate the current through each resistor in Fig. 26-41 if each resistance R = 1.20k{) and V = 12.0V. What 1‘ the potential difference between points A and B? 21. (IT) The two terminals of a voltage source with emf € and internal resistance r are connected to the two sides of a load

resistance R. For what value of R will the maximum power be delivered from the source to the load?

22, (IT) Two resistors when connected in series to a 110-V line use one-fourth the power that is used when they are connected in parallel. If one resistor is 3.8 k(}, what is the resistance of the other? 23. (IIT) Three equal resistors (R) are connected to a battery as shown in Fig. 26-42. Qualitatively, what happens to (a) the voltage drop across each of these resistors, (b) the current flow through each, and (c) the terminal voltage of the battery, when the switch S is opened, after having been closed for a long time? (d) If the emf of the battery is 9.0V, what is its terminal voltage when the switch is closed if the internal resistance r is 0.50Q and R = 5.50(2? (e) What is the terminal R S voltage when the switch is open? é 4

R

R

FIGURE 26-42 Problem 23.

. (III) A 2.8-kQ and a 3.7-kQ) resistor are connected in parallel; this combination is connected in series with a 1.8-k€ resistor. If each resistor is rated at ;W (maximum\ without overheating), what is the maximum voltage that ca. be applied across the whole network?

25. (III) Consider the network of resistors shown in Fig. 26-43. Answer qualitatively: (a) What happens to the voltage across each resistor when the switch S is closed? (b) What happens to the current through each when the switch is closed? r‘ (c) What happens to the power output of the battery when the switch is closed? (d) Let Ry = R, = Ry = Ry = 125Q and V =22.0V. Determine the current through each resistor before and after closing the switch. Are your qualitative predictions confirmed?

PM FIGURE 26-43

is15V.

1.5V

R

R R

FIGURE 26-47

Problem 29.

1.5V

b

30. (1) (a) A network of five equal resistors R is connected to a battery € as shown in Fig. 26-48. Determine the current / that flows out of the battery. (b) Use the value determined for I to find the single resistor Ry that is equivalent to the five-resistor network.

$n

Ry

Problem 25.

29. (II) For the circuit shown in Fig. 26—47, find the potential difference between points a and b. Each resistor has R & R = 130Q and each battery

R

MW

26. (III) You are designing a wire resistance heater to heat an enclosed volume of gas. For the apparatus to function properly, this i’heater must transfer heat to the gas at a very constant

rate.

While

in

operation,

the

resistance

of

the

heater will always be close to the value R = R;, but may fluctuate slightly causing its resistance to vary a small

amount

AR (

emf to be measured, €y, is placed into

this situation we call R,. Next, a standard emf, €, which is

Bas

(@)

the unknown

the circuit as shown, the movable contact C is moved until the galvanometer G gives a null reading (i.e., zero) when the switch S is closed. The resistance between A and C for

;\,—J

Power supply

o

T

b

o |

.

Electronic

device

%—/

Power outage protector

FIGURE 26-75 Problem 86.

General Problems

705

87. The circuit shown in Fig. 26-76 is a primitive 4-bit digitalto-analog converter (DAC). In this circuit, to represent each digit (2") of a binary number, a “1” has the n'* switch closed whereas zero (“0”) has the switch open. For example, 0010 is represented by closing switch n =1, while all other switches are open. Show that the voltage V across the 1.0-Q resistor for the binary numbers 0001, 0010, 0100, and 1010 (which represent

1,2, 4, 10)

follows the pattern that

=0

I0XE

n=2

4.0 kQ

= n=3

20 kQ

I _L—-——O/

you expect

for a 4-bit FIGURE 26-76

Problem 87.

unknown

10.00 V== FIGURE 26-77 Problem 88.

+“’1
‘ 9

(27-1)

If the direction of the current is perpendicular to the field B (6 = 90°), then the force is

" Fpux = IB.

[current 1 B] (27-2)

If the current is parallel to the field (6 = 0°), the force is zero. The magnitude of B can be defined using Eq. 27-2 as B = Frax/ 1€, where F, is the magnitude of the force on a straight length £ of wire carrying a current / when the wire is perpendicular to B. The

relation

between

the

force

magnetic field B that causes the force, so, we recall that the direction of F is and the ma gnitude by Eq. 27-1. This is Cross produ ct (see Section 11-2), so we

F

on

a wire

carrying

current

/, and

the

can be written as a vector equation. To do given by the right-hand rule (Fig. 27-11c), consistent with the definition of the vector can write

F = [I X B

(27-3)

ector whose magnitude is the length of the wire and its direction is along the wire (assumed straight) in the direction of the conventional (positive) current. aere, fis a

SECTION 27-3

Force on an Electric Current in a Magnetic Field; Definition of B

711

Equation 27-3 applies if the magnetic field is uniform and the wire is straight. If B is not uniform, or if the wire does not everywhere make the same angle 6 with B, then Eq. 27-3 can be written more generally as

dF = Idi x B,

27-4"

where dF is the infinitesimal force acting on a differential length d€ of the wire. The total force on the wire is then found by integrating. Equation 27-4 can serve (just as well as Eq. 27-2 or 27-3) as a practical definition of B. An equivalent way to define B, in terms of the force on a moving electric charge, is discussed in the next Section. EXERCISE Assuming the force the angle magnetic

! /4

-

B

/6 / ¢

fi

/

VA

FIGURE 27-12

(Repeated for

C A wire carrying current / is perpendicular to a magnetic field of strength B. a fixed length of wire, which of the following changes will result in decreasing on the wire by a factor of 2? (a) Decrease the angle from 90° to 45°; (b) decrease from 90° to 30° (c) decrease the current in the wire to //2; (d) decrease the field strength to B/2; (e) none of these will do it.

The ST unit for magnetic field B is the tesla (T). From Eqgs. 27-1,2, 3, or 4, we see that 1T = 1 N/A-m. An older name for the tesla is the “weber per meter squared” (1Wb/m? = 1T). Another unit sometimes used to specify magnetic field is a cgs unit, the gauss (G): 1G = 10*T. A field given in gauss should always be changed to teslas before using with other SI units. To get a “feel” for these units, we note that the magnetic field of the Earth at its surface is about 3 G or 0.5 X 107*T. On the other hand, the field near a small magnet attached to your refrigerator may be 100 G (0.01 T) whereas strong electromagnets can produce fields on the order of 2T and superconducting magnets can produce over 10T.

DGV

AR YES N Magnetic

carrying a magnet at uniform at magnitude

on

a

current-carrying

wire. A

wire

APPROACH We use Eq.27-1, F = I€Bsin. SOLUTION

Example 27-1.) Current-carrying

wire in a magnetic field. Force on the wire is directed into the page.

force

30-A current has a length £ = 12cm between the pole faces of a an angle 6 = 60° (Fig. 27-12). The magnetic field is approximatel;‘q 0.90T. We ignore the field beyond the pole pieces. What is the of the force on the wire?

|

The force F on the 12-cm length of wire within the uniform field B is F

=

IfBsinf

=

(30A)(0.12m)(0.90T)(0.866)

=

2.8N.

EXERCISE D A straight power line carries 30 A and is perpendicular to the Earth’s magnetic field of 0.50 x 10*T. What magnitude force is exerted on 100m of this

power line?

On a diagram, when we want to represent an electric current or a magnetic field that is pointing out of the page (toward us) or into the page, we use © or X, respectively. The © is meant to resemble the tip of an arrow pointing directly toward the reader, whereas the X or ) resembles the tail of an arrow moving away. (See Figs. 27-13 and 27-14.)

Measuring a magnetic field. A rectangular loop of wire

hangs vertically as shown in Fig. 27-13. A magnetic field B is directed horizontally, perpendicular to the wire, and points out of the page at all points as represented by the symbol ©. The magnetic field B is very nearly uniform along the horizontal portion of wire ab (length £ = 10.0cm) which is near the center of the gap of a large magnet producing the field. The top portion of the wire loop is free of the field. The loop hangs from a balance which measures a downward magnetic force (in addition to the gravitational force) of F = 3.48 X 1072 N when the wire carries a current / = 0.245 A. What is the magnitude of the magnetic field B?

712

CHAPTER 27

Magnetism

APPROACH Three straight sections of the wire loop are in the magnetic field: a horizontal section and two vertical sections. We apply Eq. 27-1 to each section and use the right-hand rule. SOLUTION The magnetic force on the left vertical section of wire points to the left; the force on the vertical section on the right points to the right. These two forces are equal and in opposite directions and so add up to zero. Hence, the net magnetic force on the loop is thaton the horizontal section ab, whose length is

A

£ =0.100 én The angle 0 between B and the wire is § = 90°, so sin6 = 1. Thus

¢

Eq.27-1

NOTE

gives B

-

-

.

E

T 7T

This technique

3.48 X 102N

(0245A)(0.100m)

1T

can be a precise means

| strength.

DOV e carrying aLcurrent as shown in Fig. magnetic field B,. length £ within the field By.

e

o

of determining

magnetic

field

Magnetic force on a semicircular wire. A rigid wire, I, consists of a semicircle of radius R and two straight portions 27-14. The wire lies in a plane perpendicular to a uniform Note choice of x and y axis. The straight portions each have field. Determine the net force on the wire due to the magnetic

-

11 e s ©; 0|6 B e

10.0 cm

IT

A s S 0 6 0|6 © (toward viewer) A

(® O} [OJCONOJIO;

®'

'®

QI

@

©

|® ®© O

G)l@

OFO

©

©

@

O

@'

FIGURE 27-13 Measuring a magnetic field B. Example 27-2.

APPROACH The forces on the two straight sections are equal (= I£B,) and in opposite directions, so they cancel. Hence the net force is that on the semicircular portion.

ot

H‘

|

: X

Eag

|= "

5

|| % I |x

]2;(0 (aw)e 0,b is true if g < 0; is true if ¢ < 0, b is true if g > 0.

SECTION 27-4

F.=oF E=4 (b)

Force on an Electric Charge Moving in a Magnetic Field

717

Axis of otiion P

6'0‘

1

(')F‘ .2

Fid

~

axis

~

-1}

27-5

;"2 (b) #

L Il

0

&

FIGURE 27-22

_NIA w to coil face)

©

Axis

!

Torque on a Current Loop;

Magnetic Dipole Moment

When an electric current flows in a closed loop of wire placed in an extern:™ magnetic field, as shown in Fig. 27-22, the magnetic force on the current can produce a torque. This is the principle behind a number of important practical devices, including motors and analog voltmeters and ammeters, which we discuss in the next Section. The interaction between a current and a magnetic field is important in other areas as well, including atomic physics. Current flows through the rectangular loop in Fig. 27-22a, whose face we assume is parallel to B. B exerts no force and no torque on the horizontal segments of wire because they are parallel to the field and sin® = 0 in Eq.27-1. But the magnetic field does exert a force on each of the vertical sections of wire as shown, F, and F, (see also top view, Fig. 27-22b). By right-hand-rule2 (Fig. 27-11c or Table 27-1) the direction of the force on the upward current on the left is in the opposite direction from the equal magnitude force F, on the downward current on the right. These forces give rise to a net torque that acts to rotate the coil about its vertical axis. Let us calculate the magnitude of this torque. From Eq. 27-2 (current L B), the force F = IaB, where a is the length of the vertical arm of the coil. The lever arm for each force is b/2, where b is the width of the coil and the “axis” is at the midpoint. The torques produced by F; and F, act in the same direction, so the total torque is the sum of the two torques:

B :

vfiz

Calculating the

torque on a current loop in a

T

=

IaBé

+ IaBé

2

2

=

JabB

=

IAB,

where A = ab is the area of the coil. If the coil consists of N loops of wire, the current is then NI, so the torque becomes

— NIAB.

-~

magnetic field B. (a) Loop face

If the coil makes an angle 6 with the magnetic field, as shown m Fig. 27-22c, the

reducing the torque since the lever

the coil, Fig. 27-22c. So the torque becomes

parallel to B field lines; (b) top view; (c) loop makes an angle to B, arm is reduced.

forces are unchanged but each lever arm is reduced from 1b to 1bsin 6. Note that the angle 6 is taken to be the angle between B and the perpendicular to the face of T =

.

NIABsin6.

(27-9)

This formula, derived here for a rectangular coil, is valid for any shape of flat coil. The quantity NIA is called the magnetic dipole moment of the coil and is considered a vector:

i = NIA,

(27-10)

where the direction of A (and therefore of ji) is perpendicular to the plane of the coil (the green arrow in Fig. 27-22c) consistent with the right-hand rule (cup your right hand so your fmgers wrap around the loop in the direction of current flow, then your thumb pomts in the direction ofji and A). With this definition ofji, we can rewrite Eq. 27-9 in vector form:

F# = NIAXB or

7 =

2

ji X B,

(27-11)

which gives the correct magnitude and direction for the torque 7. Equation 27-11 has the same form as Eq. 21-9b for an electric dipole (with electric dipole moment p) in an electric field E, which is 7 = p X E. And

just

as

an

electric

dipole

has

potential

energy

given

by

U = —p- ‘E

when in an electric field, we expect a similar form for a magnetic dlpolfi in a magnetic field. In order to rotate a current loop (Fig. 27-22) so as . increase 6, we must do work against the torque due to the magnetic field.

718

CHAPTER 27

Magnetism

Hence the potential energy depends on angle (see Eq. 10-22, the work-energy principle for rotational motion) as U

=

ere

=

fNIABsianB

rf we choose U = 0 at 6 = /2, potential energy is

U =

—pBcosf

=

—pBcosh

+ C.

then the arbitrary constant C is zero and the

= —fi-B,

27-12)

as expected Fcompare Eq. 21-10). Bar magnets and compass needles, as well as current loops, can be considered as magnetic dipoles. Note the striking similarities of the fields produced by a bar magnet and a current loop, Figs. 27-4b and 27-9. Torque on a coil. A circular coil of wire has a diameter of 20.0 cm and contains 10 loops. The current in each loop is 3.00 A, and the coil is placed in a 2.00-T external magnetic field. Determine the maximum and minimum torque exerted on the coil by the field. APPROAC Equation 27-9 is valid for any shape of coil, including circular loops. Maximum and minimum torque are determined by the angle 6 the coil makes with the magnetic field. SOLUTION The area of one loop of the coil is A

=

7r?

=

7(0.100m)®>

=

3.14 X 107 m%

The maximum torque occurs when the coil’s face is parallel to the magnetic field, so 6 = 90°

in Fig. 27-22c, and

sinf = 1 in Eq.27-9:

7 = NIABsin# = (10)(3.00 A)(3.14 x 102 m?)(2.00T)(1) = 1.88 N-m. The minimum torque occurs if sinf = 0, for which from Eq. 27-9. NOTE

S

6 = 0°,

and then

If the coil is free to turn, it will rotate toward the orientation with

m

7 =0 6 = 0°.

Magnetic moment of a hydrogen atom. Determine the

magnetic dipole moment of the electron orbiting the proton of a hydrogen atom at a given instant, assuming (in the Bohr model) it is in its ground state with a

circular orbit of radius 0.529 X 107'°m. [This is a very rough picture of atomic structure,

but nonetheless gives an accurate result.]

APPROACH We proton equal to SOLUTION The second law, F =

start by setting the electrostatic force on the electron due to the ma = mv?/r since the electron’s acceleration is centripetal. electron is held in its orbit by the coulomb force, so Newton’s ma, gives

darer?

o)

oo &2

v

4qregmr

\/(

8.99 x 10°N-m?/C?)(1.60 x 107° C)* /CX ) = 2.19 X 10°m/s. (9.11 x 1073 kg)(0.529 X 107°m)

Since current is the electric charge that passes a given point per unit time, the revolving electron is equivalent to a current ]

=

e—

=

ev

—,

2mr where T = 2mrr/v is the time required for one orbit. Since the area of the orbit is A = mr?, the magnetic dipole moment is »

p=1IA

T

Sy (mr?) = Jevr

= 3(1.60 x 107 C)(2.19 X 10°m/s)(0.529 X 10°m) = 9.27 X 10

A-m?,

or 9.27 X }0'2“ J/T.

i

SECTION 27-5

Torque on a Current Loop; Magnetic Dipole Moment

719

*27—-6 Applications: Motors, Loudspeakers, Galvanometers *Electric Motors @

PHYSICS st

APPLIED DC motor ,

ature

_

S

An electric motor changes electric energy into (rotational) mechanical energy. A motor works on the principle that a torque is exerted on a coil of current-carrying wire suspended in the magnetic field of a magnet, described in Section 27-5. The coil i . . is mounted on a large cylinder called the rotor or armature, Fig. 27-23. The curved pole pieces and iron armature tend to concentrate the magnetic field lines so the field lines are parallel to the face of the wire (see Fig. 27-28 which shows the field lines). Actually, there are several coils, although only one is indicated in Fig. 27-23. The armature is mounted on a shaft or axle. When the armature is in the position shown in Fig. 27-23, the magnetic field exerts forces on the current in the loop as shown (perpendicular to B and to the current direction). However, when the coil, which is rotating clockwise in Fig. 27-23, passes beyond the vertical position, the forces would then act to return the coil back to vertical if the current remained the same. But if the current could somehow be reversed at that critical moment, the forces would reverse, and the coil would continue rotating in the same direction. Thus, alternation

Voltage source

of the current is necessary if a motor is to turn continuously in one direction. This can be achieved in a dc motor with the use of commutators and brushes: as shown in Fig. 27-24, input current passes through stationary brushes that rub against the

FIGURE 27-23 Diagram of a simple dc motor.

conducting commutators mounted on the motor shaft. At every half revolution, each

commutator changes its connection over to the other brush. Thus the current in the coil reverses every half revolution as required for continuous rotation. Lead wires to armature coil

Bss Battery (dc) o=’

FIGURE 27-26

FIGURE 27-24 The commutator-brush arrangement in a dc motor ensures alternation of the current in the armature to keep rotation continuous. The commutators

Loudspeaker.

FIGURE 27-25 many windings.

Motor with

are attached to the motor shaft and turn with it, whereas

the brushes remain stationary.

Rigid

Elc:til ohf \gitre attachea

to

.

mlegtlal frame

.

g

.

.

Most motors contain several coils, called windings, each located in a different place on the armature, Fig. 27-25. Current flows through each coil only during a small part of a revolution, at the time when its orientation results in the maximum torque. In this way, a motor produces a much steadier torque than can be obtained from a single coil.

speaker cone)

An ac motor, with ac current as input, can work without commutators since the

current itself alternates. Many motors use wire coils to produce the magnetic field (electromagnets) instead of a permanent magnet. Indeed the design of most motors is more complex than described here, but the general principles remain the same.

* Loudspeakers

U

720

CHAPTER 27

Magnetism

A loudspeaker also works on the principle that a magnet exerts a force on a current-carrying wire. The electrical output of a stereo or TV set is connected to the wire leads of the speaker. The speaker leads are connected internally to a coil of wire, which is itself attached to the speaker cone, Fig. 27-26. The speaker cone is usually made of stiffened cardboard and is mounted so that it can move back apfi forth freely. A permanent magnet is mounted directly in line with the coil of wi. When the alternating current of an audio signal flows through the wire coil, which is free to move within the magnet, the coil experiences a force due to the magnetic

field of the magnet. (The force is to the right at the instant shown in Fig. 27-26.)

As the current alternates at the frequency of the audio signal, the coil and attached speaker cone move back and forth at the same frequency, causing alternate compres-

sions and rarefactions of the adjacent air, and sound waves are produced. A speaker

us changes electrical energy into sound energy, and the frequencies and intensities "the emitted sound waves can be an accurate reproduction of the electrical input.

*Galvanometer The basic component of analog meters (those with pointer and dial), including analog ammeters, voltmeters, and ohmmeters, is a galvanometer. We have already seen how these meters are designed (Section 26-7), and now we can examine how the

crucial

element,

a

galvanometer,

works.

As

shown

in

Fig.

27-27,

a

galvanometer consists of a coil of wire (with attached pointer) suspended in the magnetic field of a permanent magnet. When current flows through the loop of wire, the magnetic field exerts a torque on the loop, as given by Eq. 27-9, T

This torque proportional

=

NIABsin6.

is opposed by a spring which exerts a torque 7, approximately to the angle ¢ through which it is turned (Hooke’s law). That is, s

=

ko,

where k is the stiffness constant of the spring. The coil and attached pointer rotate to the angle where the torques balance. When the needle is in equilibrium at rest, the torques are equal: k¢ = NIABsin#, or

|

NIABsin 0

¢=——"0

The deflection of the pointer, ¢, is directly proportional to the current / flowing in the coil, but also depends on the angle 0 the coil makes with B. For a useful meter e need ¢ to depend only on the current /, independent of 6. To solve this problem, A agnets with curved pole pieces are used and the galvanometer coil is wrapped around a cylindrical iron core as shown in Fig. 27-28. The iron tends to concentrate the magnetié field lines so that B always points parallel to the face of the coil at the wire outside the core. The force is then always perpendicular to the face of the coil, and the torq‘ue will not vary with angle. Thus ¢ will be proportional to / as required.

27-7

y

Pivot \1‘

FIGURE 27-27

Galvanometer.

FIGURE 27-28

Galvanometer coil

wrapped on an iron core. Pointer

Iron core

Discovery and Properties of the Electron

The electro | plays a basic role in our understanding of electricity and magnetism today. But its existence was not suggested until the 1890s. We discuss it here because magnetic fields were crucial for measuring its properties. Toward the end of the nineteenth century, studies were being done on the discharge of electricity through rarefied gases. One apparatus, diagrammed in Fig. 27-29, was a glass tube fitted with electrodes and evacuated so only a small

amount of gas remained inside. When a very high voltage was applied to the electrodes, a dark space seemed to extend outward from the cathode (negative electrode) toward the opposite end of the tube; and that far end of the tube would glow. If one}or more

~ FIGURE 27-29

Screens

screens containing a small hole was inserted as shown, the

glow was restricted to a tiny spot on the end of the tube. It seemed as though something being emitted by the cathode traveled to the opposite end of the tube. These “somethings” were named cathode rays. There was much discussion at the time about what these rays might be. Some scientists thought they might resemble light. But the observation that the bright ot at the end of the tube could be deflected to one side by an electric or _iagnetic fie[ld suggested that cathode rays could be charged particles; and the direction of the deflection was consistent with a negative charge. Furthermore, if the tube contained certain types of rarefied gas, the path of the cathode rays was

made visible’ by a slight glow.

Discharge tube.

Insome models, one of the screens is the anode (positive plate).

voltage

SECTION 27-7

721

FIGURE 27-30 Cathode rays deflected by electric and magnetic fields. High e voltage

Electric field plates

Coils to produce magnetic field

Estimates of the charge e of the (assumed) cathode-ray particles, as well as of their charge-to-mass ratio e/m, had been made by 1897. But in that year, J. J. Thomson (1856-1940) was able to measure e/m directly, using the apparatus shown in Fig. 27-30. Cathode rays are accelerated by a high voltage and then pass between a pair of parallel plates built into the tube. The voltage applied to the plates produces an electric field, and a pair of coils produces a magnetic field. When only the electric field is present, say with the upper plate positive, the cathode rays are deflected upward as in path a in Fig. 27-30. If only a magnetic field exists, say inward, the rays are deflected downward along path c. These observations are just what is expected for a negatively charged particle. The force on the rays due to the magnetic field is F = evB, where e is the charge and v is the velocity of the cathode rays. In the absence of an electric field, the rays are bent into a curved path, so we have, from F = ma, evB

=

v2 m—,r

and thus

LN m

722

CHAPTER 27

Br

M

The radius of curvature r can be measured and so can B. The velocity v can be found by applying an electric field in addition to the magnetic field. The electric field E is adjusted so that the cathode rays are undeflected and follow path b in Fig. 27-30. This is just like the velocity selector of Example 27-10 where the force due to the electric field, F = eE, is balanced by the force due to the magnetic field, F = evB. Thus eE = evB and v = E/B. Combining this with the above equation we have e E (27-13) m B The quantities on the right side can all be measured so that although e and m could not be determined separately, the ratio e/m could be determined. The accepted value today is e/m = 1.76 X 10'' C/kg. Cathode rays soon came to be called electrons. It is worth noting that the “discovery” of the electron, like many others in science, is not quite so obvious as discovering gold or oil. Should the discovery of the electron be credited to the person who first saw a glow in the tube? Or to the person who first called them cathode rays? Perhaps neither one, for they had no conception of the electron as we know it today. In fact, the credit for the discovery is generally given to Thomson, but not because he was the first to see the glow in the tube. Rather it is because he believed that this phenomenon was due to tiny negatively charged particles and made careful measurements on them. Furthermore he argued that these particles were constituents of atoms, and not ions or atoms themselves as many thought, and he developed an electron theory of matter. His view is close to what we accept today, and this is why Thomson is credited with th “discovery.” Note, however, that neither he nor anyone else ever actually saw ar. electron itself. We discuss this briefly, for it illustrates the fact that discovery in science is not always a clear-cut matter. In fact some philosophers of science think the word “discovery” is often not appropriate, such as in this case.

Thomson believed that an electron was not an atom, but rather a constituent, or part, of an atom. Convincing evidence for this came soon with the determination

of the charge and the mass of the cathode rays. Thomson’s student J. S. ade the first direct (but rough) measurements of e in 1897. But it was >fined oil-drop experiment of Robert A. Millikan (1868-1953) that precise value for the charge on the electron and showed that charge discrete amounts. In this experiment, tiny droplets of mineral oil carrying

Atomizer

Townsend the more yielded a comes in %Ej{t*—fi}’{“ an electric ~ Droplets - — =

charge were allowed to fall under gravity between two parallel plates, Fig. 27-31.

The electric field E between the plates was adjusted until the drop was suspended in midair. The downwgrd. pull of gravity, mg, was then just balanced by the upward force due to the electric field. Thus gE = mg, so the charge g = mg/E. The mass of the droplet was determined by measuring its terminal velocity in the absence of the electric field. Sometimes the drop was charged negatively, and sometimes positively, s uggesting that the drop had acquired or lost electrons (by friction, leaving the atomizer). Millikan’s painstaking observations and analysis presented convincing evidence that any charge was an integral multiple of a smallest charge, e, that was ascribed to the electron, and that the value of e was 1.6 X 107"C. This value o f e, combined with the measurement of e/m, gives the mass of the electron

fi;\

-_

-

\_1

Telescope

e m=e=E FIGURE 27-31 Millikan’s oil-drop experiment.

to be (1.6 X 107 C)/(1.76 x 10" C/kg) = 9.1 X 107*' kg. This mass is less than a the mass of the smallest atom, and thus confirmed the idea that the electron is only a part of an atom. The accepted value today for the mass of the thousandth

electron is m, = 9.11 X 1073 kg. CRT, Revisited

r

The cathode ray tube (CRT), which can serve as the picture tube of TV sets, oscilloscopes, and computer monitors, was discussed in Chapter 23. There, in Fig. 23-22, we saw a design using electric deflection plates to maneuver the electron beam. Many CRTs, however, make use of the magnetic field produced by coils to maneuver the electron beam. They operate much like the coils shown in Fig. 27-30.

27-8

The Hall Effect

When a current-carrying conductor is held fixed in a magnetic field, the field exerts a sideways force on the charges moving in the conductor. For example, if electrons move to the right in the rectangular conductor shown in Fig. 27-32a, the inward magnetic field will exert a downward force on the electrons Fy = —evy X B, where V4 is the drift velocity of the electrons (Section 25-8). Thus the electrons will tend to move nearer to face D than face C. There will thus be a potential difference between faces C and D of the conductor. This potential difference builds up until the electric field Ey tha it produces exerts a force, eEy, on the moving charges that is equal and opposite to the magnetic force. This effect is called the Hall effect after E. H. Hall, who discovered it in 1879. The difference of potential produced is called the Hall emf. The ele ctric field due to the separation of charge is called the Hall field, Ey;, and points downward in Fig. 27-32a, as shown. In equilibrium, the force due to this electric fiel is balanced by the magnetic force evy B, so eEy Hence

=

FIGURE 27-32 The Hall effect. (a) Negative charges moving to the right as the current. (b) Positive charges moving to the left as the current.

evyB.

Ey = vy B. The Hall emf is then (Eq. 23-4b, assuming the conductor is long 1 is uniform)

and thin so

%y = Eud = v4Bd,

(27-14)

where d is the width of the conductor. A current of negative charges moving to the right is equivalent to positive charges moving to the left, at least for most purposes. But the Hall effect can distinguish these two. As can be seen in Fig. 27-32b, positive particles moving to the ft are deflected downward, so that the bottom surface is positive relative to the top surface. This is the reverse of part (a). Indeed, the direction of the emf in the Hall effect first revealed that it is negative particles that move in metal conductors.

e

SECTION 27-8

The Hall Effect

723

The magnitude of the Hall emf is proportional to the strength of the magnetic field. The Hall effect can thus be used to measure magnetic field strengths. First the conductor, called a Hall probe, is calibrated with known magnetic fields. Then, for the same current, its emf output will be a measure of B. Hall probes can be made very small and are convenient and accurate to use. S The Hall effect can also be used to measure the drift velocity of charge carriers when the external magnetic field B is known. Such a measurement also allows us to determine the density of charge carriers in the material.

DIy

EREN

Drift velocity using the Hall effect. A long copper strip

1.8 cm wide and 1.0 mm thick is placed in a 1.2-T magnetic field as in Fig. 27-32a. When a steady current of 15 A passes through it, the Hall emf is measured to be 1.02 uV. Determine the drift velocity of the electrons and the density of free (conducting) electrons (number per unit volume) in the copper. APPROACH We use Eq. 27-14 to obtain the drift velocity, and Eq. 25-13 Chapter 25 to find the density of conducting electrons. SOLUTION The drift velocity (Eq. 27-14) is

(a) FIGURE 27-32a (Repeated here for Example 27-13.)

Vg —

1.02 X 105V

&

(1.2T)(1.8 X 102m)

Bd

= 47 X 10 m/s.

The density of charge carriers # is obtained from Eq.25-13, I = nevq A, A is the cross-sectional area through which the current 7 flows. Then

"o

I

evg A

of

where

15A "~ (1.6 X 10°C)(4.7 X 10°m/s)(1.8 X 102m)(1.0 X 103 m) 11 X 10%¥m™3,

This value for the density of free electrons in copper, n = 11 X 10% per m?,

3

the experimentally measured value. It represents more than one free electron pe.

atom, which as we saw in Example 25-14 is 8.4 X 10¥ m™.

*27-9 Mass Spectrometer @) PHYSICS

APPLIED

The mass spectrometer

FIGURE 27-33 Bainbridge-type mass spectrometer. The magnetic fields B and B’ point out of the paper (indicated by the dots), for positive ions. s

=l

S L B

I &

s;+:EB'-?‘-\ PN ™~ B

and E

2r

R

4-—/01

Detector}. or film (.

« .

o

.

0

/,,

A mass spectrometer is a device to measure masses of atoms. It is used today not only in physics but also in chemistry, geology, and medicine, often to identify atoms (and their concentration) in given samples. As shown in Fig. 27-33, ions are produced by heating, or by an electric current, in the source or sample S. The particles, of mass m and electric charge g, pass through slit S, and enter crossed electric and magnetic fields. Ions follow a straight-line path in this “velocity selector” (as in Example 27-10) if the electric force gF is balanced by the magnetic force quB: that is, if gE = quvB, or v = E/B. Thus only those ions whose speed is v = E/B will pass through undeflected and emerge through slit S,. In the semicircular region, after S,, there is only a magnetic field, B’, so the ions follow a circular path. The radius of the circular path is found from their mark on film (or detectors) if B’ is fixed; or else r is fixed by the position of a detector and B’ is varied until detection occurs. Newton’s second law, 2F = ma, applied to an ion moving in a circle under the influence only of the magnetic field B’ gives quvB' = mv*/r. Since v = E/B, we have

4

o« / 7

_ v

gBB'r E

All the quantities on the right side are known or can be measured, and thus m ca. be determined.

724

CHAPTER 27

Magnetism

Historically, the masses of many atoms were measured this way. When a pure substance was used, it was sometimes found that two or more closely spaced marks would appear on the film. For example, neon produced two marks whose radii corresponded to atoms of mass 20 and 22 atomic mass units (u). r ‘mpurities were ruled out and it was concluded that there must be two types of neon with different masses. These different forms were called isotopes. It was soon found that most elements are mixtures of isotopes, and the difference in mass is due to different numbers of neutrons (discussed in Chapter 41). Mass spectrometry. Carbon atoms of atomic mass 12.0u are found to be mixed with another, unknown, element. In a mass spectrometer with fixed B, the carbon traverses a path of radius 22.4cm and the unknown’s path has a 26.2-cm radius. What is the unknown element? Assume the ions of both elements have the same charge. |

APPROACH The carbon and unknown atoms pass through the same electric and magnetic fields. Hence their masses are proportional to the radius of their respective paths (see equation on previous page). SOLUTIOITI We write a ratio for the masses, using the equation at the bottom of the previous page: my gBB'ry/E 262cm 117

mec

qBB'rc/E

~

24cm

Thus m, = 1.17 X 12.0u = 14.0u. The other (see the ‘eriodic Table, inside the back cover).

element

is probably

nitrogen

NOTE The unknown could also be an isotope such as carbon-14 ('{C). See

Appendix F. Further physical or chemical analysis would be needed.

b

Summary A magnet hz‘as two poles, north and south. The north pole is that end which points toward geographic north when the magnet is freely suspended. Like poles of two magnets repel each other, whereas unlike poles attract. We can picture that a magnetic field surrounds every magnet. TheII unit for magnetic field e is the tesla (T). . Electric currents produce magnetic fields. For example, the lines of ma ‘ etic field due to a current in a straight wire form circles around the wire, and the field exerts a force on magnets

(or cunentsj near it. A magfietic field exerts a force on an electric current. The force on an ijnfini_t’esimal length of wire df carrying a current / in a magnetic field _1.3 i

-

dF = Idl X B.

(27-49)

If the field B is uniform over a straight length € of wire, then the force is ) which has

F = [IxB magnitude F = I{Bsin®

@7-3) (27-1)

where 6 is the angle between magnetic field B and the current direction. Tde direction of the force is perpendicular to the wire and to the rpagnetic field, and is given by the right-hand rule. This relatioq serves as the definition of magnetic field B. Similarly, a magnetic field B exerts a force on a charge ¢ moving withrvelocity Vv given by

~ F = gv x B.

T'he magnitude of the force is F

=

quBsiné,

where 6 is the angle between ¥ and B.

(27-5a) (27-5b)

The path of a charged particle moving perpendicular to a uniform magnetic field is a circle. ~ If both electric and magnetic fields (E and B) are present, the force on a charge g moving with velocity V is F ==

qfi

+

27-7)

gv X B.

The torque on a current loop in a magnetic field B is 7

=

jixB,

(27-11)

where ji is the magnetic dipole moment of the loop: i

=

NIA.

(27-10)

HereN is the number of coils carrying current / in the loop and A is a vector perpendicular to the plane of the loop (use right-hand rule, fingers along current in loop) and has magnitude squal o the ares bf.the loop, The measurement of the charge-to-mass ratio (e/m) of the electron was done using magnetic and electric fields. The charge e on the electron was first measured in the Millikan oil-drop experiment and then its mass was obtained from the measured value of the e/m ratio. In the Hall effect, moving charges in a conductor placed in a magnetic field are forced to one side, producing an emf between the two sides of the conductor. [*A mass spectrometer uses magnetic and electric fields to measure the mass of ions.]

Summary

725

| Questions 13. A charged particle is moving in a circle under the influence of a uniform magnetic field. If an electric field that points ilh the same direction as the magnetic field is turned oz describe the path the charged particle will take. 14 The force on a particle in a magnetic field is the idea behind electromagnetic pumping. It is used to pump metallic fluids (such as sodium) and to pump blood in artificial heart machines. The basic design is shown in Fig. 27-36. An electric field is applied perpendicular to a blood vessel and to a magnetic field. Explain how ions are caused to move. Do positive and negative ions feel a force in the same direction?

FIGURE 27-36 Electromagnetic pumping in a blood vessel.

Question 14.

15. A beam of electrons is directed toward a horizontal wire carrying a current from left to right (Fig. 27-37). In what direction is the beam deflected? _yl

FIGURE 27-37 Ouestion 15

' Electron direction

A charged particle moves in a straight line through a particular region of space. Could there be a nonzero rnagnetifl field in this region? If so, give two possible situations. 17 If a moving charged particle is deflected sideways in some region of space, can we conclude, for certain, that B # 0in that region? Explain. 18 How could you tell whether moving electrons in a certain region of space are being deflected by an electric field or by a magnetic field (or by both)? 19. How can you make a compass without using iron or other ferromagnetic material? 20 Describe how you could determine the dipole moment of a bar magnet or compass needle. 21 In what positions (if any) will a current loop placed in a uniform magnetic field be in (a) stable equilibrium, and (b) unstable equilibrium? A rectangular piece of semiconductor is inserted in a *22, magnetic field and a battery is connected to its ends as shown in Fig. 27-38. When a sensitive voltmeter is connected between points a and b, o it is found that point a is at a B* higher potential than b. What is a 16


N.

The magnetic force on wire 2 must be upward, and Eq. 28-2 gives

1B, F = Ko LI where d = 020m and [, = 80 A. We solve this for I, and set the two force magnitudes equal (letting £ = 1.0 m): 2

27d (F) i ol

28-3

B

27(0.20m) (118 X 10 N/m) (47 x 107 T-m/A)(80 A) (1.0m)

=

15 A

Definitions of the Ampere and the Coulomb

You may have wondered how the constant u, in Eq. 28-1 could be exactly 4 X 107 T-m/A. Here is how it happened. With an older definition of the ampere, po was measured experimentally to be very close to this value. Today,

Mo

is defined

to

be

exactly

47 X 107 T-m/A.

if the ampere were defined independently. The now defined in terms of the magnetic field B value of ;. In particular, we use the force between two Eq. 28-2, to define the ampere precisely. If I; = wires are exactly 1 m apart, then

This

could

be

done

parallel current-carrying wires, I, = 1 A exactly, and the two

) _ pohlL _ (47 x107TFT ) m/A) (1A)(1A (im) =

not

ampere, the unit of current, is it produces using the defined

—

2 X 107" =7 N/m.

Thus, one ampere is defined as that current flowing in each of two long parall wires 1 m apart, which results in a force of exactly 2 X 1077N per meter of leng of each wire.

736

CHAPTER 28

Sources of Magnetic Field

'

This is the precise definition of the ampere. The coulomb is then defined as being exactly one ampere-second: 1 C = 1 A-s. The value of k or €, in Coulomb’s law (Section 21-5) is obtained from experiment. This may seem a rather roundabout way of defining quantities. The reason r. *hind it is the desire for operational definitions of quantities—that is, definitions of quantities that can actually be measured given a definite set of operations to carry out. For example, the unit of charge, the coulomb, could be defined in terms of the force between two equal charges after defining a value for €y or k in Egs. 21-1 or 21-2. However, to carry out an actual experiment to measure the force between two charges is very difficult. For one thing, any desired amount of charge is not easily obtained precisely; and charge tends to leak from objects into the air. The amount of current in a wire, on the other hand, can b e varied accurately and continuously (by putting a variable resistor in a circuit . Thus the force between two current-carrying conductors is far easier to measure precisely. This is why we first define the ampere, and then define the coulomb in terms of the ampere. At the National Institute of Standards and Technology in Maryland, precise measurement of current is made using circula r coils of wire rather than straight lengths because it is more convenient and accurate

Electric and magnetic field strengths are also defined operationally: the elecof the measurable force on a charge, via Eq. 21-3; and the magnetic fie Id in terms of the force per unit length on a current-carrying wire, via Eq.27-2. tric field in terms

28-4 Ampere’s Law In Section 2 8—1 we saw that Eq. 28-1 gives the relation between the current in a

»ng

straight wire and the magnetic field it produces. This equation is valid only

ra long st raight wire. Is there a general relation between a current in a wire of

any shape and the magnetic field around it? The answer is yes: the French scientist André Marie Ampere (1775-1836) proposed such a relation shortly after Oersted’s di scovery. Consider an arbitrary closed path around a current as shown in Fig. 28-8 and imagine this path as being made up of short segments each of length AL First, we take the product of the length of each segment times the component of B parallel to that segment (call this component B)). If we now sum all these ter ms, according to Ampere, the result will be equal to u, times the net current I, that passes through the surface enclosed by the path: EBH

Al

=

FIGURE 28-8 Arbitrary path enclosing a current, for Ampere’s law. The path is broken down into segments of equal length AL

pglend&

The lengths Al are chosen so that B is essentially constant along each length. The sum must be made over a closed path; and I, is the net current passing through the surface bounded by this closed path (orange in Fig. 28-8). In the limit AL — 0, this relation becomes

:fifi-di = polenals where dl is that the p Ampere’s 1 and I.,q is ;ath or loop.

an infinitesimal length vector and rallel component of B is taken. w. The integrand in Eq. 28-3 is the current passing through the

(28-3)

Closed path made up of segments of

Area enclosed

length A{

by the path I

| AMPERE'S LAW |

the vector dot product assures Equation 28-3 is known as taken around a closed path, space enclosed by the chosen

SECTION 28-4

Ampere's Law

737

To understand Ampere’s law better, let us apply it to the simple case of a single long straight wire carrying a current / which we’ve already examined, and which served as an inspiration for Ampere himself. Suppose we want to find the magnitude of B at some point A which is a distance r from the wireq (Fig. 28-9). We know the magnetic field lines are circles with the wire at thei. center. So to apply Eq. 28-3 we choose as our path of integration a circle of radius r. The choice of path is ours, so we choose one that will be convenient: at any point on this circular path, B will be tangent to the circle. Furthermore, since all points on the path are the same distance from the wire, by symmetry we expect B to have the same magnitude at each point. Thus for any short segment of the circle (Fig. 28-9), B will be parallel to that segment, and (setting I, = I)

Tl

FIGURE 28-9

radius r.

Circular path of

wol

jgfi - dl

#QBdfl = Bigdf = B(2mr), where

¢ dl = 27r, the circumference of the circle. We solve for B and obtain

_ kol

B_27rr

This is just Eq. 28-1 for the field near a long straight wire as discussed earlier. Ampere’s law thus works for this simple case. A great many experiments indicate that Ampere’s law is valid in general. However, as with Gauss’s law for the electric field, its practical value as a means to calculate the magnetic field is limited mainly to simple or symmetric situations. Its importance is that it relates the magnetic field to the current in a direct and mathematically elegant waya

Ampere’s

FIGURE 28-10 Magnetic field lines around two long parallel wires whose equal currents, /; and I, are coming out of the paper toward the viewer.

law

is thus considered

one

of the basic laws

of electricity

an.

magnetism. It is valid for any situation where the currents and fields are steady and not changing in time, and no magnetic materials are present.

We now can see why the constant in Eq. 28-1 is written uo/27. This is done so that only p, appears in Eq. 28-3, rather than, say, 27k if we had used k in Eq. 28-1. In this way, the more fundamental equation, Ampere’s law, has the simpler form. It should be noted that the B in Ampere’s law is not necessarily due only to

the current /,. Ampere’s law, like Gauss’s law for the electric field, is valid in

general. B is the field at each point in space along the chosen path due to all sources—including the current / enclosed by the path, but also due to any other sources. For example, the field surrounding two parallel current-carrying wires is

the vector sum of the fields produced by each, and the field lines are shown in Fig. 28-10. If the path chosen for the integral (Eq. 28-3) is a circle centered on one of the wires with radius less than the distance between the wires (the dashed line in Fig. 28-10), only the current (/;) in the encircled wire is included on the right side of Eq. 28-3. B on the left side of the equation must be the

total B at each point due to both wires. Note also that ¢B-dl for the path

shown in Fig. 28-10 is the same whether the second wire is present or not (in both cases, it equals u, I; according to Ampere’s law). How can this be? It can be so because the fields due to the two wires partially cancel one another at some points between them, such as point C in the diagram (B = 0 at a point midway between the wires if I, = L,); at other points, such as D in Fig. 28-10, the fields add together to produce a larger field. In the sum, gifi-di, these effects just balance so that $§B-dl = pyl;, whether the second wire is there or not. The

integral $B-df will be the same in each case, even though B will not be thom,

same at every point for each of the two cases.

738

CHAPTER 28

Sources of Magnetic Field

Field inside and outside a wire. A long straight cylindrical

wire conductor of radius R carries a current / of uniform current density in the conductor. Determine the magnetic field due to this current at (a) points (.outside the conductor (r > R), and (b) points inside the conductor (r < R). See Fig. 28-11. Assume that r, the radial distance from the axis, is much less than the length of the wire. (¢) If R =2.0mm and [ = 60A, what is B at

r=10 rnn;1, r=20mm,

and

r = 3.0mm?

APPROACH We can use symmetry: Because the wire is long, straight, and cylindrical, we Expect from symmetry that the magnetic field must be the same at all points that are the same distance from the center of the conductor. There is no reason why any such point should have preference over others at the same distance from the wire (they are physically equivalent). So B must have the same value at all points the same distance from the center. We also expect B to be tangent to‘ circles around the wire (Fig. 28-1), so we choose a circular path of integration as we did in Fig. 28-9. SOLUTION (@) We apply Ampere’s law, integrating around a circle (r > R) centered on the wire (Fig. 28-11a), and then /., = I -

#fi.di

or

B

B(Zfl.r)

0

/"LOIencl

=

:

=

B

ML,

[r>

2ar

r=R

FIGURE 28-11

r

(b)

Magnetic field

inside and outside a cylindrical

R]

conductor (Example 28-6).

which is the same result as for a thin wire.

(b) Inside the wire (r < R), we again choose a circular path concentric with the cylinder; we expect B to be tangential to this path, and again, because of the symmetry, it will have the same magnitude at all points on the circle. The current Aenclosed in this case is less than / by a factor of the ratio of the areas: Ing

=

I—0;

So Ampere’s law gives %fi vdl

=

Mo Lena

B(2mr) = uol(f’—) SO

[r

< R]

The field is zero at the center of the conductor and increases linearly with r until r = R; beyond r = R, B decreases as 1/r. This is shown in Fig. 28-11b. Note that these results are valid only for points close to the center of the conductor as compared to its length. For a current to flow, there must be connecting wires (to a battery, say), and the field due to these conducting wires, if not very far away, will destroy the assumed symmetry. (c) At r = 2.0mm, the surface of the wire, r = R, so B

=

pol

2mR

(4w X 107 T-m/A)(60A)

(27)(2.0 X 103 m)

=

We saw in (b) that inside the wire B is linear in r. So at

Abe

half what itis at

» = 20mm

or

A CAUTION Connecting wires can destroy assumed symmetry

6.0 X 107°T. r = 1.0mm,

B will

3.0 X 107 T. Outside the wire, B falls off

as 1/r, so at r = 3.0mm it will be two-thirds as great as at r = 2.0mm, or B = 4.0 X 107 T. To check, we use our result in (a), B = pol/27r, which gives the same result.

SECTION 28-4

Ampére's Law

739

@ansms

APPLIED Coaxial cable (shielding)

Insulating sleeve. _g&=

Cylindrical braid 7

Solid wire

FIGURE 28-12

Coaxial cable.

FIGURE 28-13

Exercise C.

Example 28-7.

CONCEPTUAL

EXAMPLE

28-7 | Coaxial cable. A coaxial cable is a single wire

surrounded by a cylindrical metallic braid, as shown in Fig. 28-12. The two conductors are separated by an insulator. The central wire carries current to the other end of the cable, and the outer braid carries the return current and is usually considered ground. Describe the magnetic field (@) in the space between th conductors, and (b) outside the cable.

RESPONSE (a) In the space between the conductors, we can apply Ampere’s law for a circular path around the center wire, just as we did for the case shown in Figs. 28-9 and 28-11. The magnetic field lines will be concentric circles centered on the center of the wire, and the magnitude is given by Eq. 28-1. The current in the outer conductor has no bearing on this result. (Ampere’s law uses only the current enclosed inside the path; as long as the currents outside the path don’t affect the symmetry of the field, they do not contribute to the field along the path at all). (b) Outside the cable, we can draw a similar circular path, for we expect the field to have the same cylindrical symmetry. Now, however, there are two currents enclosed by the path, and they add up to zero. The field outside the cable is zero. The nice feature of coaxial cables is that they are self-shielding: no stray magnetic fields exist outside the cable. The outer cylindrical conductor also shields external electric fields from coming in (see also Example 21-14). This makes them ideal for carrying signals near sensitive equipment. Audiophiles use coaxial cables between stereo equipment components and even to the loudspeakers. EXERCISE C In Fig. 28-13, A and B are wires each carrying a 3.0-A current but in opposite directions. On the circle C, which statement is true? (a) B = 0; (b) §B- dt = 0;

FIGURE 28-14

Example 28-8.

Al

b

. B

0 I

1

l

i =l

a

(a)

| d (impossible)

The integral along sections ab and cd is zero, since B L d€. Thus

351*3 *dl = Byl — Byl

/——_-.\

B

i

» @

::7

(possible)

5O

LV, & N

Ampere

7

A nice use for Ampére’s law. Use Ampere’s law to show

3@1*; dl = 0.

/_._\

(b)

LN

APPROACH The wider spacing of lines near the top of Fig. 28-14a indicates t field B has a smaller magnitude at the top than it does lower down. We apps, Ampere’s law to the rectangular path abcd shown dashed in Fig. 28-14a. SOLUTION Because no current is enclosed by the chosen path, Ampere’s law gives

i !

T

POV

(d) B= —3u; () §B-dl = 6p.

that in any region of space where there are no currents the magnetic field cannot be both unidirectional and nonuniform as shown in Fig. 28-14a.

C

1

(¢) B =3u;

¢

which is not zero since the field By, along the path bc is less than the field By, along path da. Hence we have a contradiction: 4513 - df cannot be both zero (since I = 0) and nonzero. Thus we have shown that a nonuniform unidirectional field is not consistent with Ampere’s law. A nonuniform field whose direction also changes, as in Fig. 28-14b, is consistent with Ampere’s law (convince yourself this is so), and possible. The fringing of a permanent magnet’s field (Fig. 27-7) has this shape. .

s Law

; : : 1. Ampere’s law, like Gaus ’s law, is alwaysa valid statem:;l:. I;llst :swa lcaelc uliltliSSn t?)vovl 1ist ?szkfl ite dvmlainly to

a =

systems with a high degree of symmetry. The first step

in applying Ampere’s law is to identify useful symmetry. . Choose an integration path that reflects the symmetry (see the Examples). Search for paths where B has constant magnitude along the entire path or along segments of the path. Make sure your integration path passes through the point where you wish to evaluate the magnetic field.

740

CHAPTER 28

= (Bp. — Ban)t,

Sources of Magnetic Field

.

.

=

3. Use symmetry to determ;lne the dlrelcltlon off B al}i)n]%

the integration path. With a smart choice of path, ; . . will be either parallel or perpendicular to the path..

Determine the enclosed current, I.,q. Be careful with

signs. Let the fingers of your right hand curl along the direction of B so that your thumb shows the direction of positive current. If you have a solid conductor and your integration path does not enclose the full current, you can use the current density (current per unit area) multiplied by the enclosed area (as in Example 28-6). |

28-5

Magnetic Field of a Solenoid

f\ long coil of wire consisting of many loops is called a solenoid. Each loop produces a magnetic field as was shown in Fig. 27-9. In Fig. 28-15a, we see the field due to a solenoid when the coils are far apart. Near each wire, the field lines are very nearly circles as for a straight wire (that is, at distances that are small compared to the curvature of the wire). Between any two wires, the fields due to each loop tend to cancel. Toward the center of the solenoid, the fields add up to give a field that can be fairly large and fairly uniform. For a long solenoid with closely packed coils, the field is nearly uniform and parallel to the solenoid axis within the entire cross section, as shown in Fig. 28-15b. The field outside the solenoid is very small compared to the field inside, except near the ends. Note that the same number of field lines that are concentrated inside the solenoid, spread out into the vast open space outside. We now use Ampere’s law to determine the magnetic field inside a very long (ideally, infinitely long) closely packed solenoid. We choose the path abed shown in Fig. 28-16, far from either end, for applying Ampere’s law. We will consider this path as made up of four segments, the sides of the rectangle: ab, bc, cd, da. Then the left side of Eq. 28-3, Ampere’s law, becomes |

3@1‘;‘{[

b

=

[ B-di a

+

.

J B-di b

+

d

f B-di c

+

a

j B-di.

~ -

-1}

and a Toroid

1

/

@)

©9000000000000000000

0!'

(b) FIGURE 28-15

Magnetic field due

to a solenoid: (a) loosely spaced

d

turns, (b) closely spaced turns.

The field outside the solenoid is so small as to be negligible compared to the field inside. Thus the fif‘st term in this sum will be zero. Furthermore, B is perpendicular to the segments bj and da inside the solenoid, and is nearly zero between and outside the coils,

f

Y

| ‘

e >

OCOOOOOOOOO

O

—

= B Crm——— = o= = = — — -d ]

T

BI®R®®B®R® |of | L ok o it

@R

I

|

®

———

@

'Currefnt gg;g

Current into

page

FIGURE 28-16 Cross-sectional view into a solenoid. The magnetic

field inside is straight except at the ends. Red dashed lines indicate the S path chosen for use in. Ampére’s law.

©® and ® are electric current

direction (in the wire loops) out of

the page and into the page.

so these terms too are zero. Therefore we have reduced the integral to the segment cd where B is the nearly uniform field inside the solenoid, and is parallel to @€, so d

%fi-di = [[Bedt = B c

where £ is the length cd. Now we determine the current enclosed by this loop for

the right sifie of Ampere’s law, Eq. 28-3. If a current / flows in the wire of the sole-

noid, the total current enclosed by our path abcd is NI where N is the number of loops our path encircles (five in Fig. 28-16). Thus Ampere’s law gives us Bl

=

Mo NI.

If we let n = N/{ be the number of loops per unit length, then B = puqgnl. [solenoid]

(28-4)

This is the magnitude of the magnetic field within a solenoid. Note that B depends only on the number of loops per unit length, n, and the current /. The field does not depend on position within the solenoid, so B is uniform. This is strictly true only for an infinite solenoid, but it is a good approximation for real ones for points not close to the ends.

DGV

E R

Field inside a solenoid. A thin 10-cm-long solenoid used

for fast el?ctromechanical switching has a total of 400 turns of wire and carries a current oi} 2.0 A. Calculate the field inside near the center. APPROACH We use Eq. 28-4, where the number of turns per unit length is n = 400/0.10m = 4.0 X 10°m™.

SOLUTION

B = uonl = (47 X107 T+m/A)4.0x10°m*)(2.0A) = 1.0 X 102 T.

SECTION 28-5

741

A close look at Fig. 28-15 shows that the field outside of a solenoid is much like that of a bar magnet (Fig. 27-4). Indeed, a solenoid acts like a magnet, with one end acting as a north pole and the other as south pole, depending on the direction of the current in the loops. Since magnetic field lines leave thefl north pole of a magnet, the north poles of the solenoids in Fig. 28-15 are o the right. Solenoids have many practical applications, and we discuss some of them later in the Chapter, in Section 28-8.

GNP

EEALN

Toroid. Use Ampere’s law to determine the magnetic field

(a) inside and (b) outside a toroid, which is like a solenoid bent into the shape of a circle as shown in Fig. 28—-17a. APPROACH The magnetic field lines inside the toroid will be circles concentric with the toroid. (If you think of the toroid as a solenoid bent into a circle, the field lines bend along with the solenoid.) The direction of B is clockwise. We choose as our path of integration one of these field lines of radius r inside the toroid as shown by the dashed line labeled “path 1” in Fig. 28-17a. We make this choice to use the symmetry of the situation, so B will be tangent to the path and will have the same magnitude at all points along the path (although it is not necessarily the same across the whole cross section of the toroid). This chosen path encloses all the coils; if there are N coils, each carrying current /, then /., = NI.

SOLUTION

(a) Ampere’s law applied along this path gives %E'di

=

.u“Olenc]

B(2@r)

=

poNI,

where N is the total number Thus -

)

of coils and 7 is the current in each of the coils.

B = BN 27r

FIGURE 28-17 (a) A toroid. (b) A section of the toroid showing direction of the current for three

loops: ® means current toward you, &) means current away from you.

The magnetic field B is not uniform within the toroid: it is largest along the inner edge (where r is smallest) and smallest at the outer edge. However, if the toroid is large, but thin (so that the difference between the inner and outer radii is small compared to the average radius), the field will be essentially uniform

within the toroid. In this case, the formula

for B reduces

to that for a straight solenoid B = ponl/ where n = N/(2zr) is the number of coils per unit length. (b) Outside the toroid, we choose as our path of integration a circle concentric with the toroid, “path 2” in Fig. 28-17a. This path encloses N loops carrying current / in one direction and N loops carrying the same current in the opposite direction. (Figure 28-17b shows the directions of the current for the parts of the loop on the inside and outside of the toroid.) Thus the net current enclosed by path 2 is zero. For a very tightly packed toroid, all points on path 2 are equidistant from the toroid and equivalent, so we expect B to be the same at all points along the path. Hence, Ampere’s law gives (fifi dl

or

B(27r)

=

o Lencl

0

B = 0. The same is true for a path taken at a radius smaller than that of the toroid. So there is no field exterior to a very tightly wound toroid. It is all inside th loops.

742

CHAPTER 28

-

Sources of Magnetic Field

28-6

Biot-Savart Law

The usefulness of Ampere’s law for determining the magnetic field B due to r articular electric currents is restricted to situations where the symmetry of the given currents allows us to evaluate gfifi-d? readily. This does not, of course, invalidate Ampere’s law nor does it reduce its fundamental importance. Recall the electric case, where Gauss’s law is considered fundamental but is limited in its use for actually calculating E. We must often determine the electric field E by another method su ‘ming over contributions due to infinitesimal charge elements dg via

Coulomb’s law: dE = (1/4me,)(dq/r?). A magnetic equivalent to this infinitesimal

form of Coulomb’s law would be helpful for currents that do not have great symmetry. Such a law was developed by Jean Baptiste Biot (1774-1862) and Felix Savart (1791-1841) shortly after Oersted’s discovery in 1820 that a current produces a magnetic field. According to Biot and Savart, a current / flowing in any path can be considered as many tiny (infinitesimal) current elements, such as in the wire of Fig. 28-18. If dl represents any infinitesimal length along which the current is flowing, then the magnetic field, 4B, at any point P in space, due to this element of current, is given by dB

=

fio_l_ at >2< L 4 r

(28-5)

Biot-Savart law

where F is the displacement vector from the element df to the point P, and I = F/r is the unit vector (magnitude = 1) in the direction of ¥ (see Fig. 28-18). dB

(out)

OP

r FIGURE 28-18

Biot-Savart law:

thg field at P due to current element

Idlis dB = (uol/4m)(dl X t/r?).

Equation 28-5 is known as the Biot-Savart law. The magnitude of dB is ‘

dB =

o ldl sin 6 darr

;

(28-6)

|

where 6 is l‘Ihe angle between df and ¥ (Fig. 28-18). The total magnetic field at point P is then found by summing (integrating) over all current elements:

e .

s

Mol

dl X © r

Note that this is a vector sum. The Biot-Savart law is the magnetic equivalent of Coulomb’s law in its infinitesimal form. It is even an inverse square law, like Coulomb’s law. An important difference between the Biot-Savart law and Ampere’s law (Eq. 28-3) is that in Ampere’s law [fl_i-di - ,uolencl],fi is not necessarily due only to the current enclosed by the path of integration. But in the iot-Savart law the field dB in Eq. 28-5 is due only, and entirely, to the current " clement Idl. To find the total B at any point in space, it is necessary to include all currents. SECTION 28-6

Biot-Savart Law

743

ST

dB (inward)

TIIN

B due to current 7 in straight wire. For the ficld near a

long straight wire carrying a current /, show that the Biot-Savart law gives the same result as Eq. 28-1, B = pyI/27r.

S

-

APPROACH We calculate the magnetic field in Fig. 28-19 at point P, which is perpendicular distance R from an infinitely long wire. The current is moving upwards, and both df and , which appear in the cross product of Eq. 28-5, are in the plane of the page. Hence the direction of the field @B due to each element of current must be directed into the plane of the page as shown (right-hand rule for the cross product

dl X 1).Thus all the B have the same direction at point P, and add up to give B the same direction consistent with our previous results (Figs. 28-1 and 28-11).

FIGURE 28-19

Determining B

SOLUTION The magnitude of B will be ol B = K4

due to a long straight wire using the Biot-Savart law.

where dy = dl

f+°°

dysin 6

oo

and

1P

’

r* = R*+ y’. Note that we are integrating over y (the

length of the wire) so R is considered constant. Both y and 6 are variables, but they are not independent. In fact, y = —R/tan 6. Note that we measure y as positive upward from point 0, so for the current element we are considering y < 0. Then

dy

Y

+Rcsc*0do

=

e

Rd§ _sinf0

=

Rd§ _= r’de (R/r)} R

From Fig. 28-19 we can see that y = —oco corresponds to 6 = 0 y = +oo corresponds to 6 = 7 radians. So our integral becomes

Mol 1

(7 sinfdd .

5 ===

ol

=

47 R L=0 '

and that

N = — Mol

—-——cosb|

4nR

%, T 2R

This is just Eq. 28-1 for the field near a long wire, where R has been used instead of .

FIGURE 28-20 Determining B due to a current loop.

FUIITFTEFE Current loop. Determine B for points on the axis of afl circular loop of wire of radius R carrying a current /, Fig. 28—-20. APPROACH For an element of current at the top of the loop, the magnetic field 4B at point P on the axis is perpendicular to ¥ as shown, and has magnitude (Eq. 28-5) [.Lol d[

dB =

4arr?

since df is perpendicular to F so |df X f| = dl. We can break dB down into

components dB; and dB, , which are parallel and perpendicular to the axis as shown.

SOLUTION When we sum over all the elements of the loop, symmetry tells us that the perpendicular components will cancel on opposite sides, so Hence, the total B will point along the axis, and will have magnitude B

=B

:

=

j

|dBcos¢p

¢

=

J

R r

|dB—

=

B, = 0.

R (R* + X2y

| dB-———7’

J

where x is the distance of P from the center of the ring, and r*> = R* + x% Now we put in dB from the equation above and integrate around the current loop, noting that all segments

point P:

_ml__ 4m

dl of current are the same

R

(R* + xz)%

f

__

mlIR

2(R* + xz)%

since [dl = 2mR, the circumference of the loop. NOTE At the very center of the loop (where x = 0) B

744

CHAPTER 28

Sources of Magnetic Field

ol

= 5K

distance, (R2 + xz)%, from

the field has its maximum value [at center of current loop]

fl

Recall from Section 27-5 that a current loop, such as that just discussed (Fig. 28-20), is considered a magnetic dipole. We saw there that a current loop has a magnetic dipole moment r

p

=

NIA,

where A is‘ the area of the loop and N is the number of coils in the loop, each carrying current /. We also saw in Chapter 27 that a magnetic dipole placed in an external magnetic field experiences a torque and possesses potential energy, just like an elé‘ctric dipole. In Example 28-12, we looked at another aspect of a magnetic dipole: the magnetic field produced by a magnetic dipole has magnitude, along the dipole axis, of

po IR?

2(R2 + xz)% We can write this in terms of the magnetic dipole moment single loop‘ N =1):

S

S

w = IA = I7R?* (for a

[magnetic dipole]

27 (R* + x2)2

(28-7a)

(Be careful to distinguish p for dipole moment from pu,, the magnetic permeability

constant.)

For distances far from the loop, x >> R, this becomes

G

on axis,

53

[magnetic dipole, x >> R]

-

B

(28-7b)

The magneftic field on the axis of a magnetic dipole decreases with the cube if the distance, just as the electric field does for an electric dipole. B decreases as

the

cube

of

the

distance

also

for

points

not

on

the

axis,

although

the

multiplying factor is not the same. The magnetic field due to a current loop can be determined at various points using the Biot-Savart law and the results are in accord with experiment. The field lines around a current loop are shown in Fig. 28-21.

B due to a wire segment. One quarter of a circular loop

of wire carries a current / as shown in Fig. 28-22. The current / enters and leaves on straight segments of wire, as shown; the straight wires are along the radial direction from the center C of the circular portion. Find the magnetic field at point C.

FIGURE 28-21 Magnetic field due to a circular loop of wire. (Same as Fig.27-9.) ~ FIGURE 28-22

Example 28-13.

APPROACH The current in the straight sections produces no magnetic field at point C because df and ¥ in the Biot-Savart law (Eq. 28-5) are parallel and therefore? dl X t = 0. Each piece df of the curved section of wire produces a

field @B that points into the page at C (right-hand rule).

SOLUTION The magnitude of each dB due to each df of the circular portion of wire is (Eq. 28-6) dB =

Mo

1

df

47 R?

where r = R is the radius of the curved section, and sinf in Eq. 28-6 is sin 90° = 1. With r = R for all pieces df, we integrate over a quarter of a circle.

' &

L

B=fd3=fljde=“° 417 R?

1

=E‘(L{"

47R? \ 4

8R

SECTION

28-6

Biot-Savart Law

745

28—7 Magnetic Materials—Ferromagnetism

(b) FIGURE 28-23 (a) An unmagnetized piece of iron is made up of domains that are randomly arranged. Each domain is like a tiny magnet; the arrows represent the magnetization direction, with the

arrowhead being the N pole. (b) In a magnet, the domains are preferentially aligned in one direction (down in this case), and may be altered in size by the magnetization process.

FIGURE 28-24 Iron filings line up along magnetic field lines due to a

permanent magnet.

A

CAUTION

B B lines form closed loops, E start on @ and end on =

746

CHAPTER 28

Magnetic fields can be produced (1) by magnetic materials (magnets) and (2) by electric currents. Common magnetic materials include ordinary magnets, iron core, in motors and electromagnets, recording tape, computer hard drives and magnet. stripes on credit cards. We saw in Section 27-1 that iron (and a few other materials) can be made into strong magnets. These materials are said to be ferromagnetic. We now look into the sources of ferromagnetism. A bar magnet, with its two opposite poles near either end, resembles an electric dipole (equal-magnitude positive and negative charges separated by a distance). Indeed, a bar magnet is sometimes referred to as a “magnetic dipole.” There are opposite “poles” separated by a distance. And the magnetic field lines of a bar magnet form a pattern much like that for the electric field of an electric dipole: compare Fig. 21-34a with Fig. 27-4 (or 28-24). Microscopic examination reveals that a piece of iron is made up of tiny regions known as domains, less than 1 mm in length or width. Each domain behaves like a tiny magnet with a north and a south pole. In an unmagnetized piece of iron, these domains are arranged randomly, as shown in Fig. 28-23a. The magnetic effects of the domains cancel each other out, so this piece of iron is not a magnet. In a magnet, the domains are preferentially aligned in one direction as shown in Fig, 28-23b (downward in this case). A magnet can be made from an unmagnetized piece of iron by placing it in a strong magnetic field. (You can make a needle magnetic, for example, by stroking it with one pole of a strong magnet.) The magnetization direction of domains may actually rotate slightly to be more nearly parallel to the external field, and the borders of domains may move so domains with magnetic orientation parallel to the external field grow larger (compare Figs. 28—-23a and b). We can now explain how a magnet can pick up unmagnetized pieces of iron like paper clips. The field of the magnet’s south pole (say) causes a slight realignment of the domains in the unmagnetized object, which then becomes a temporary magnet with its north pole facing the south pole of the permanent magnet; thus, attractic*™ results. Similarly, elongated iron filings in a magnetic field acquire aligned domains ana align themselves to reveal the shape of the magnetic field, Fig. 28-24. An iron magnet can remain magnetized for a long time, and is referred to as a “permanent magnet.” But if you drop a magnet on the floor or strike it with a hammer, you can jar the domains into randomness and the magnet loses some or all of its magnetism. Heating a permanent magnet can also cause loss of magnetism, for raising the temperature increases the random thermal motion of atoms, which tends to randomize the domains. Above a certain temperature known as the Curie temperature (1043 K for iron), a magnet cannot be made at all. Iron, nickel, cobalt, gadolinium, and certain alloys are ferromagnetic at room temperature; several other elements and alloys have low Curie temperature and thus are ferromagnetic only at low temperatures. Most other metals, such as aluminum and copper, do not show any noticeable magnetic effect (but see Section 28-10). The striking similarity between the fields produced by a bar magnet and by a loop of electric current (Figs. 27-4b and 28-21) offers a clue that perhaps magnetic fields produced by electric currents may have something to do with ferromagnetism. According to modern atomic theory, atoms can be visualized as having electrons that orbit around a central nucleus. The electrons are charged, and so constitute an electric current and therefore produce a magnetic field; but the fields due to orbiting electrons generally all add up to zero. Electrons themselves produce an additional magnetic field, as if they and their electric charge were spinning about their own axes. It is the magnetic field due to electron spin’ that is believed to produce ferromagnetism in most ferromagnetic materials. It is believed today that all magnetic fields are caused by electric currents. This means that magnetic field lines always form closed loops, unlike electric field lines which begin on positive charges and end on negative charges. “The name “spin” comes from an early suggestion that this intrinsic magnetic moment arises from the electron “spinning” on its axis (as well as “orbiting” the nucleus) to produce the extra field. However this view of a spinning electron is oversimplified and not valid.

EXERCISE D Return to the Chapter-Opening Question, page 733, and answer it again now. Try to explain why you may have answered differently the first time.

~28-8

Electromagnets and Solenoids—Applications

A long coil ‘of wire consisting of many loops of wire, as discussed in Section 28-5, is called a s#lenoid. The magnetic field within a solenoid can be fairly large since it will be the sum of the fields due to the current in each loop (see Fig. 28-25). The solenoid acts like a magnet; one end can be considered the north pole and the other the sq[uth pole, depending on the direction of the current in the loops (use the right-hand rule). Since the magnetic field lines leave the north pole of a magnet, the north pole of the solenoid in Fig. 28-25 is on the right. If a piece of iron is placed inside a solenoid, the magnetic field is increased greatly becafixse the domains of the iron are aligned by the magnetic field produced by the current. The resulting magnetic field is the sum of that due to the current and that due to the iron, and can be hundreds or thousands of times larger than that due to the current alone (see Section 28-9). This arrangement is called an electromagnet. The alloys of iron used in electromagnets acquire and lose their magnetism quite readily when the current is turned on or off, and so are referred to as “soft iron.” (It is “soft” only in a magnetic sense.) Iron that holds its magnetism even when there is no externally applied field is called “hard iron.” Hard iron is used in permanent magnets. Soft iron is usually used in electromagnets so that the field can be turned on and off éadily. Whether iron is hard or soft depends on heat treatment, type of alloy, and other factors. Electromagnets have many practical applications, from use in motors and generators to producing large magnetic fields for research. Sometimes an iron core is not present—the magnetic field comes only from the current in the wire coils. hen the current flows continuously in a normal electromagnet, a great deal of .aste heat (I°R power) can be produced. Cooling coils, which are tubes carrying

FIGURE 28-25 Magnetic field of a solenoid. The north pole of this solenoid, thought of as a magnet, is

on the right, and the south pole is on the left.

@pHYsics

APPLIED

@pHysics

aAPPLIED

Electromagnets and solenoids

Doorbell, car starter Iron rod

water, are needed to absorb the heat in larger installations.

For some applications, the current-carrying wires are made of superconducting material kept below the transition temperature (Section 25-9). Very high fields

can be produced with superconducting wire without an iron core. No electric

power is needed to maintain large current in the superconducting coils, which means large* savings of electricity; nor must huge amounts of heat be dissipated. It is not a free ride, though, because energy is needed to keep the superconducting coils at the necessary low temperature. Another useful device consists of a solenoid into which a rod of iron is partially inserted. This combination is also referred to as a solenoid. One simple use is as a doorbell (Fflg. 28-26). When the circuit is closed by pushing the button, the coil effectively becomes a magnet and exerts a force on the iron rod. The rod is pulled into the coil and strikes the bell. A large solenoid is used in the starters of cars; when you erfigage the starter, you are closing a circuit that not only turns the starter motor, but activates a solenoid that first moves the starter into direct contact with the gears oj the engine’s flywheel. Solenoids are used as switches in many devices. They have the advantage of moving mechanical parts quickly and accurately.

* Magnetic k:ircuit Breakers Modern circuit breakers that protect houses and buildings from overload and fire contain not only a “thermal” part (bimetallic strip as described in Section 25-6, Fig. 25-19) but also a magnetic sensor. If the current is above a certain level, the magnetic field it produces pulls an iron plate that breaks the same contact points as in Fig. 25-19b and c. In more sophisticated circuit breakers, including rround fault circuit interrupters (GFCIs—discussed in Section 29-8), a solenoid is .sed. The iron rod of Fig. 28-26, instead of striking a bell, strikes one side of a pair of points, opening them and opening the circuit. Magnetic circuit breakers react quickly ( By.) Initially, point a, the domains (Section 28-7) are randomly oriented. As B, increases, the domains become more and more aligned until at point b, nearly all are aligned. The iron is said to be approaching saturation. Point b is typically 70% of full saturation. (If B, is increased further, the curve continues to rise very slowly, and reaches 98% saturation only when B, reaches a value about a thousandfold above that at point b; the last few domains are very difficult to align.) Next, suppose the external field B, is reduced by decreasing the current in the toroid coils. As the current is reduced to zero, shown as point c in Fig. 28-29, the domains do not become completely random. Some permanent magnetism remains. If the current is then reversed in direction, enough domains can be turned around so B = 0 (point d). As the reverse current is increased further, the iron approaches saturation in the opposite direction (point e). Finally, if the current is again reduced to zero and then increased in the original direction, the total field follows the path efgb, again

approaching saturation at point b.

Notice that the field did not pass through the origin (point a) in this cycle. The fact that the curves do not retrace themselves on the same path is called hysteresis. The curve bedefgb is called a hysteresis loop. In such a cycle, much energy is transformed to thermal energy (friction) due to realigning of the domains. It can\ be shown that the energy dissipated in this way is proportional to the area of th. hysteresis loop.

Sources of Magnetic Field

At points c and f, the iron core is magnetized even though there is no current in the coils. These points correspond to a permanent magnet. For a permanent magnet, it is desired that ac and af be as large as possible. Materials for which this is true are said to have high retentivity. é Materials with a broad hysteresis curve as in Fig. 28-29 are said to be magnetically “hard” and make good permanent magnets. On the other hand, a hysteresis curve such as that in Fig. 28-30 occurs for “soft” iron, which is preferred for electromagnets and transformers (Section 29-6) since the field can be more readily switched off, and the field can be reversed with less loss of energy. A ferro agnetic material can be demagnetized—that is, made unmagnetized. This can be done by reversing the magnetizing current repeatedly while decreasing its magnitude. This results in the curve of Fig. 28-31. The heads of a tape recorder are demagnetized in this way. The alternating magnetic field acting at the heads due to a hahdheld demagnetlzer is strong when the demagnetizer is placed near the heads and decreases as it is moved slowly away. Video and audio tapes themselves can be erased and ruined by a magnetic field, as can computer hard disks, other magnetic storage devices, and the magnetic stripes on credit cards.

*28-10

Paramagnetism and Diamagnetism

B

/ /

FIGURE 28-30 for soft iron.

By

Hysteresis curve

FIGURE 28-31 Successive hysteresis loops during demagnetization. B

All material’s are magnetic to at least a tiny extent. Nonferromagnetic materials fall into two principal classes: paramagnetic, in which the magnetic permeability w is very slightly greater than ug; and diamagnetic, in which u is very slightly less than Mo The ratio of u to p, for any material is called the relative permeability K, :

B,

Mo

Another useful parameter is the magnetic susceptibility X, defined as Xoo

=

Kin — L.

~aramagnetic substances substances have K, < 1 effect is.

TABLE 28-1

have and

K, >1 and X, > 0, whereas diamagnetic X, < 0. See Table 28-1, and note how small the

Paramagnetism and Diamagnetism: Magnetic Susceptibilities

Parnmagnet#c substance

Xm

Aluminum

231079

Magnesium Oxygen (STP) Platinum Tungsten

1.2 % 107° 3.4 1078 29 x 1074 6.8 X 1073

Calcium

b

Diamagnetic substance

Xm

Copper

-9.8 X 107°

Gold Lead Nitrogen (STP) Silicon

-3.6 -1.7 -5.0 -42

Diamond

-22 %1073

X x x x

1073 1073 107° 1076

The difference between paramagnetic and diamagnetic materials can be understood theoretically at the molecular level on the basis of whether or not the molecules have a permanent magnetic dipole moment. One type of paramagnetism occurs in materials whose molecules (or ions) have a permanent magnetic dipole moment.” In the absence of an external field, the molecules are randomly oriented and no magnetic effects are observed. However, when an external magnetic field is applied, say, by putting the material in a solenoid, the applied field exerts a torque on the magnetic dipoles (Section 27-5), tending to align them parallel to the field. The total magnetic field (external plus that due to aligned magnetic dipoles) will be slightly greater than B,,. The thermal motion of the molecules reduces the alignment, however. Other types of paramagnetism also occur whose origin is different from that described here, such as in metals where free electrons can contribute.

*SECTION 28-10

Paramagnetism and Diamagnetism

749

A useful quantity is the magnetization vector, M, defined as the magnetic dipole moment per unit volume,

> where fi is the magnetic dipole moment of the sample and V its volume. It is found experimentally that M is directly proportional to the external magnetic field (tending to align the dipoles) and inversely proportional to the kelvin temperature 7 (tending to randomize dipole directions). This is called Curie’s law, after Pierre Curie (1859-1906), who first noted it: B C T

M

where C is a constant. If the ratio B/T is very large (B very large or T very small) Curie’s law is no longer accurate; as B is increased (or 7 decreased), the magnetization approaches some maximum value, M,,,,. This makes sense, of course, since M.« corresponds to complete alignment of all the permanent magnetic dipoles. However, even for very large magnetic fields, 2.0 T, deviations from Curie’s law are normally noted only at very low temperatures, on the order of a few kelvins. Ferromagnetic materials, as mentioned in Section 28-7, are no longer ferromagnetic above a characteristic temperature called the Curie temperature (1043 K for iron). Above this Curie temperature, they generally are paramagnetic. Diamagnetic materials (for which u,, is slightly less than w,) are made up of molecules that have no permanent magnetic dipole moment. When an external magnetic field is applied, magnetic dipoles are induced, but the induced magnetic dipole moment is in the direction opposite to that of the field. Hence the total field will be slightly less than the external field. The effect of the external field—in the crude model of electrons orbiting nuclei—is to increase the “orbital” speed of electrons revolving in one direction, and to decrease the speed of electrong revolving in the other direction; the net result is a net dipole moment opposing tl:xA external field. Diamagnetism is present in all materials, but is weaker even than paramagnetism and so is overwhelmed by paramagnetic and ferromagnetic effects in materials that display these other forms of magnetism.

| Summary The magnetic field B at a distance r from a long straight wire is directly proportional to the current / in the wire and inversely proportional to r:

The magnetic field lines are circles centered at the wire. The force that one long current-carrying wire exerts on a second parallel current-carrying wire 1 m away serves as the definition of the ampere unit, and ultimately of the coulomb as well. Ampere’s law states that the line integral of the magnetic field B around any closed loop is equal to g times the total net current /¢ enclosed by the loop: Ho Lencl -

(28-3)

The magnetic field inside a long tightly wound solenoid is B

=

(284)

[.L()fl[

where n is the number current in each coil.

of coils per unit length and [/ is the

750

Sources of Magnetic Field

CHAPTER 28

dB

(28-1)

B=27rr

=

The Biot-Savart law is useful for determining the magnetic due to a known arrangement of currents. It states =

Kol

%fi dl

field that

=

[..L()I d?

y—

X

F

’

(28‘-5)

where dB is the contribution to the total field at some point P due to a current / along an infinitesimal length df of its path, and # is the unit vector along the direction of the displacement vector ¥

from df to P. The total field B will be the integral over all dB.

Iron and a few other materials can be made into strong permanent magnets. They are said to be ferromagnetic. Ferromagnetic materials are made up of tiny domains—each a tiny magnet—which are preferentially aligned in a permanent magnet, but randomly aligned in a nonmagnetized sample. [*When a ferromagnetic material is placed in a magnetic field By due to a current, say inside a solenoid or toroid, the material becomes magnetized. When the current is turned off, however, the material remains magnetized, and when the current

is increased in the opposite direction (and then again reversed), a graph of the total field B versus By is a hysteresis loop, and th fact that the curves do not retrace themselves is called hysteresis., [*All materials exhibit some magnetic effects. Nonferromagnetic materials have much smaller paramagnetic or diamagnetic

properties.]

| Questions

¢

1. The magnetic field due to current in wires in your home can affect a compass. Discuss the problem in terms of currents, depending on whether they are ac or dc, and their distance away. Compare and contrast the magnetic field due to a long straight current and the electric field due to a long straight line of electric charge at rest (Section 21-7). i

.

P

Two 1n51}lated long wires carrying equal currents / cross at right angles to each other. Describe the magnetic force one

16. A type of magnetic switch similar to a solenoid is a relay (Fig. 28-32). A relay is an electromagnet (the iron rod inside the coil does not move) which, when activated, attracts a piece

of iron on a pivot. Design a relay to close an electrical switch. A relay is used when you need to switch on a circuit carrying a very large current but you do not want that large current flowing through the main switch. For example, the starter switch of a car is connected to a relay so that the large current needed for the starter doesn’t pass to the dashboard switch.

exerts on the other.

. A horizontal wire carries a large current. A second wire carrying a current in the same direction is suspended below it. Can the current in the upper wire hold the lower wire in suspension against gravity? Under what conditions will the lower wire be in equilibrium? 5 A horizontal current-carrying wire, free to move in Earth’s gravitational

field, is suspended

directly

above

a second,

parallel, current-carrying wire. (¢) In what direction is the current in the lower wire? (b) Can the upper wire be held in stable eciuilibrium due to the magnetic force of the lower g

wire? Explain.

(a) Write Ampere’s law for a path that surrounds both conductors in Fig. 28-10. (b) Repeat, assuming the lower

current, I, is in the opposite direction (I, = —1).

. Suppose the cylindrical conductor of Fig. 28-11a has a concentric cylindrical hollow cavity inside it (so it looks like a pipe). What can you say about B in the cavity? . Explain why a field such as that shown in Fig. 28-14b is consistent with Ampere’s law. Could the lines curve upward instead of downward? . What would be the effect (a) the diameter of all the spacing between loops was length was doubled along number of loops.

on B inside a long solenoid if loops was doubled, or (b) the doubled, or (c) the solenoid’s with a doubling in the total

10. Use the Biot-Savart law to show that the field of the current loop in Fig. 28-21 is correct as shown for points off the axis. 11. Do you think B will be the same for all points in the plane of the current loop of Fig. 28-21? Explain. 12. Why does twisting the lead-in wires to electrical devices reduce the magnetic effects of the leads? 13. Compare the Biot-Savart law with Coulomb’s law. What are the similarities and differences? 14. How might you define or determine the magnetic pole strength (the magnetic equivalent of a single electric charge) for (a) a bar magnet, (b) a current loop? 15. How might you measure the magnetic dipole moment of the Earth?

FIGURE 28-32

Question 16.

17. A heavy magnet attracts, from rest, a heavy block of iron. Before striking the magnet the block has acquired considerable kinetic energy. () What is the source of this kinetic energy? (b) When the block strikes the magnet, some of the latter’s domains may be jarred into randomness; describe the energy transformations. 18. Will a magnet attract any metallic object, such as those made of aluminum, or only those made of iron? (Try it and see.) Why is this so? 19. An unmagnetized nail will not attract an unmagnetized paper clip. However, if one end of the nail is in contact with a magnet, the other end will attract a paper clip. Explain. 20. Can an iron rod attract a magnet? Can a magnet attract an iron rod? What must you consider to answer these questions? 21. How do you suppose the first magnets found in Magnesia were formed? 22 Why will either pole of a magnet attract an unmagnetized piece of iron? 23 Suppose you have three iron rods, two of which are magnetized but the third is not. How would you determine which two are the magnets without using any additional objects? 24, Two iron bars attract each other no matter which ends are placed close together. Are both magnets? Explain. *285. Describe the magnetization curve for (a) a paramagnetic substance and (b) a diamagnetic substance, and compare to that for a ferromagnetic substance (Fig. 28-29). *26. Can all materials be considered (a) diamagnetic, (b) paramagnetic, (¢) ferromagnetic? Explain.

i Prob{lems 28-1and 28-2 Straight Wires, Magnetic Field, and Force r

1 (I) Jumper cables used to start a stalled vehicle often carry a 65-A current. How strong is the magnetic field 3.5cm from one cable? Compare to the Earth’s magnetic

field (5.0

X 107°7T).

. (I) If an electric wire is allowed to produce a magnetic field

no larger than that of the Earth (0.50 X 107*T) at a

distance of 15cm from the wire, what current the wire can carry?

is the maximum

Problems

751

3. (I) Determine the magnitude and direction of the force between two parallel wires 25 m long and 4.0 cm apart, each carrying 35 A in the same direction. (I) A vertical straight wire carrying an upward 28-A current exerts an attractive force per unit length of 7.8 X 107*N/m on a second parallel wire 7.0 cm away. What current (magnitude and direction) flows in the second wire? . (I) In Fig. 28-33, a long straight wire carries current / out of the page toward the viewer. Indicate, with appropriate arrows,

the

direction

and E in the plane of the page.

of

B

at

each

of

the

points

C,

priate arrows, the direction of B at each

of the points 1 to 6 in the plane of the page. State if the field is zero at any of the points.

b4 @

e 4

o1 ¢ 2

D,

e 5

*:3

FIGURE 28-35

Problem13.

1©

6

14. (II) A long pair of insulated wires serves to conduct 28.0 A of dc current to and from an instrument. If the wires are of negligible diameter but are 2.8mm apart, what is the magnetic field 10.0cm from their midpoint, in their plane (Fig. 28-36)? Compare to the magnetic field of the Earth.

FIGURE 28-33 Problem 5.

=

13. (IT) Two long straight wires each carry a current / out of the page toward the viewer, Fig. 28-35. Indicate, with appro-

(II) An experiment on the Earth’s magnetic field is being carried out 1.00m from an electric cable. What is the maximum allowable current in the cable if the experiment is to be accurate to + 2.0%? (II) Two long thin parallel wires 13.0cm apart carry 35-A currents in the same direction. Determine the magnetic field vector at a point 10.0cm from one wire and 6.0cm from the other (Fig. 28-34).

10.0icm, ===+

M

2.8 mm

FIGURE 28-36

Problems 14 and 15.

-fl

15, (II) A third wire is placed in the plane of the two wires shown in Fig. 28-36 parallel and just to the right. If it carries 25.0 A upward, what force per meter of length does it exert on each of the other two wires? Assume it is 2.8 mm from

FIGURE 28-34 Problem 7.

the nearest wire, center to center.

. (II) A horizontal compass is placed 18 cm due south from a straight vertical wire carrying a 43-A current downward. In what direction does the compass needle point at this location? Assume the horizontal component of the Earth’s

16. (I) A power line carries a current of 95 A west along the

declination is 0°. . (II) A long horizontal wire carries 24.0 A of current due north. What is the net magnetic field 20.0 cm due west of the

cancel the Earth’s? 17. (II) A compass needle

field at this point is 0.45 X 10T

and the magnetic

wire if the Earth’s field there points downward, 44° below

the horizontal, and has magnitude 5.0 X 1075 T?

10. (I) A straight stream of protons passes a given point in space at a rate of 2.5 X 10° protons/s. What magnetic field do they produce 2.0 m from the beam? 11. (II) Determine the magnetic field midway between two long straight wires 2.0 cm apart in terms of the current / in one when the other carries 25 A. Assume these currents are (a) in the same direction, and (b) in opposite directions. 12 (II) Two straight parallel wires are separated by 6.0cm. There is a 2.0-A current flowing in the first wire. If the magnetic field strength is found to be zero between the two wires at a distance of 2.2 cm from the first wire, what is the magnitude and direction of the current in the second wire?

752

CHAPTER 28

Sources of Magnetic Field

tops of 8.5-m-high poles. (a) What is the magnitude and

direction of the magnetic field produced by this wire at the ground directly below? How does this compare with the

Earth’s field of about : G? (b) Where would the line’s field points

28°

E

of

N

outdoors.

However, when it is placed 12.0 cm to the east of a vertical

wire inside a building, it points 55° E of N. What is the magnitude and direction of the current in the wire? The

Earth’s field there is 0.50 X 107*T and is horizontal.

18. (II) A rectangular loop of wire is placed next to a straight

wire, as shown in Fig. 28-37. There is a current of 3.5 A in

both wires. Determine the magnitude and direction of the net

354

force on the loop.

35A 5.0cm

FIGURE 28-37

Problem 18.

l

10.0 cm

t

3.0‘cm

19. (II) Let two long parallel wires, a distance d apart, carry

equal currents / in the same direction. One wire is at x = 0,

r

the other at x = d, Fig. 28-38. Determine x axis between the wires as a function of x.

B along the

24. (ITI) A triangular loop of side length a carries a current [ (Fig. 28-41). If this loop is placed a distance d away from a very long straight wire carrying a current ', determine the force

on the loop.

{/

T

FIGURE 28-41

Problem 24.

FIGURE 28-38

28-4 and 28-5

Problems 19 and 20.

20. (IT) Repeat Problem 19 if the wire at x = 0 carries twice the current (2/) as the other wire, and in the opposite direction. long wires are oriented so that they are perpendic21. (IT) Two ular to each other. At their closest, they are 20.0 cm apart (Fig. 28-39). What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 20.0 A and the bottom one carries 12.0A? I;=200A 10.0 cm

B=7 10.0 cm

FIGURE 28-39

Problem 21.

e

=B

Bottom wire

Iy = 120A

(IT) Two long parallel wires 8.20cm apart carry 16.5-A currents in the same direction. Determine the magnetic field vector at a point P, 12.0cm from one wire and 13.0cm from the Ether. See Fig. 28-40.

[Hint: Use the law of cosines.]

\I

—

]/

|

Ampére’s Law, Solenoids and Toroids

25. (I) A 40.0-cm-long solenoid 1.35 cm in diameter is to produce a field of 0.385 mT at its center. How much current should the solenoid carry if it has 765 turns of wire? 26. (I) A 32-cm-long solenoid, 1.8 cm in diameter, is to produce a 0.30-T magnetic field at its center. If the maximum current is 4.5 A, how many turns must the solenoid have? 27. (I) A 2.5-mm-diameter copper wire carries a 33-A current (uniform across its cross section). Determine the magnetic field: (a) at the surface of the wire; (b) inside the wire, 0.50 mm below the surface; (c) outside the wire 2.5 mm from the surface. 28. (IT) A toroid (Fig. 28-17) has a 50.0-cm inner diameter and a 54.0-cm outer diameter. It carries a 25.0 A current in its 687 coils. Determine the range of values for B inside the toroid. 29. (II) A 20.0-m-long copper wire, 2.00 mm in diameter including insulation, is tightly wrapped in a single layer with adjacent

coils touching, to form a solenoid of diameter 2.50 cm (outer

edge). What is (a) the length of the solenoid and (b) the field at the center when the current in the wire is 16.7 A? 30. (II) (a) Use Eq. 28-1, and the vector nature of B, to show that the magnetic field lines around two long parallel wires carrying equal currents I; = I, are as shown in Fig. 28-10.

(b) Draw the equipotential lines around two stationary posi-

tive electric charges. (c¢) Are these two diagrams similar? Identical? Why or why not? 31 (II) A coaxial cable consists of a solid inner conductor of radius R;, surrounded by a concentric cylindrical tube of inner radius R, and outer radius R; (Fig. 28-42). The conductors carry equal and opposite currents /; distributed uniformly across their cross sections. Determine the magnetic field at a distance R from the axis for:

(@) R d? Does this make sense? Explain.

R,=2.00cm,

Graph

Bfrom

and

R; = 2.50cm.

R=0to R =3.00cm.

|

FIGURE 28-42

Problems 31 and 32.

32. (III) Suppose the current in the coaxial cable of Problem 31, Fig. 28-42, is not uniformly distributed, but instead the current density j varies linearly with distance from the center: j; = C; R for the inner conductor and j, = ;R for the outer conductor. Each conductor still carries the same total current Iy, in opposite directions. Determine the magnetic field in terms of I; in the same four regions of space as in Problem 31.

Problems

753

28-6 Biot-Savart Law 33. (I) The Earth’s magnetic field is essentially that of a magnetic dipole. If the field near the North Pole is about

1.0 X 107*T,

what will it be (approximately)

13,000 km

above the surface at the North Pole? (IT) A wire, in a plane, has the shape shown in Fig. 28-43, two arcs of a circle connected by radial lengths of wire. Determine B at point C in terms of Ry, R,, 6, and the current /.

39. (IT) A nonconducting circular disk, of radius R, carries a uniformly distributed electric charge Q. The plate is set spinning with angular velocity w about an axis perpendicular to the plate through its center (Fig. 28-47). Determine (a) its magnetic dipole moment and (b) the magnetiq field at points on its axis a distance x from its center; (c) does Eq. 28-7b apply in this case for

x > R?

FIGURE 28-47 Problem 39.

3s. (IT) A circular conducting ring of radius R is connected to two exterior straight wires at two ends of a diameter (Fig. 28-44). The current / splits into

unequal

portions

(as

shown) while passing through the ring. What is B at the center of the ring?

0.651

1,

kol

d

2R (d* + 4R%):

.

R

(b) Show that this is consistent with Example 28-11 for an infinite wire. FIGURE 28-48

A

Problem 40.

41. (IT) A segment of wire of length d carries a current / as shown in Fig. 28-49. (a) Show that for points along the posi-

—

FIGURE 28-44 Problem 35.

_

tive x axis (the axis of the wire),

03517

. (IT) A small loop of wire of radius 1.8cm is placed at the center of a wire loop with radius 25.0 cm. The planes of the loops are perpendicular to each other, and a 7.0-A current flows in each. Estimate the torque the large loop exerts on the smaller one. What simplifying assumption did you make? 37. (IT) A wire is formed into the shape of two half circles connected by equal-length straight sections as shown in Fig. 28-45. A current / flows in the circuit clockwise as shown. Determine (a) the magnitude and direction of the magnetic field at the center, C, and (b) the magnetic dipole moment of the circuit.

such as point Q, the magnetic field B is zero. (b) Determine a formula for the field at points along the y axis, such as point P.

|

.

38. (IT) A single point charge g is moving with velocity V. Use the Biot-Savart law to show that the magnetic field B it produces at a point P, whose position vector relative to the charge is ¥ (Fig. 28—-46), is given by

(Assume

v is much less than the

[} FIGURE 28-46

Problem 38.

754

CHAPTER 28

v

Sources of Magnetic Field

I *—

X

Q

FIGURE 28-49 Problem 41. 42. (III) Use the result of Problem 41 to find the magnetic field at point P in Fig. 28-50 due to the current in the square loop.

T

d—-e

el

mE——

43. (III) A wire is bent into the shape of a regular polygon with n sides whose vertices are a distance R from the center. (See Fig. 28-51, which shows the special case of n = 6.) If the wire carries a current I, I(y——> (a) determine the magnetic field at the center; (b) if n is allowed

to become show that reduces to (Example

e~

speed of light.)

P

—d—

Problem 42. Problem 37.

2

—

FIGURE 28-50 FIGURE 28-45

[STINW

FIGURE 28-43 Problem 34.

(SIS

. (IT) Consider a straight section of wire of length d, as in Fig. 28-48, which carries a current /. (a) Show that the magnetic field at a point P a distance R from the wire along its perpendicular bisector is TI

. (III)

Start

very large (n — o), the formula in part (a) that for a circular loop 28-12). FIGURE 28-51 Problem 43. with

the

result

of

Example

28-12

for

the

magnetic field along the axis of a single loop to obtair the field inside a very long solenoid with n turns per meter (Eq. 28-4) that stretches from +oo to —oo.

f‘

45. (III) A single rectangular loop of wire, with sides a and b, carries a current /. An xy coordinate system has its origin at the lower left corner of the rectangle with the x axis Y

parallel to side b (Fig. 28-52)

and the y axis parallel to side a. Determine the magnetic field B

T

loop.

J

at all points (v, y) within the FIGURE 28-52

Problem 45.

I

@

11 0

:

IT

f——b—

46. (III) A square loop of wire, of side d, carries a current /. (a) Determine the magnetic field B at points on a line perpen-

dicular to the plane of the square which passes through the

center of the square (Fig. 28-53). Express B as a function of x, the dista?ce along the line from the center of the square. (b) For Ir x >> d, does the square appear T to be a magnetic

dipole?

what is its dipole moment?

If so,

d

X

*28-9

Magnetic Materials; Hysteresis

*48. (I) The following are some values of B and B, for a piece of annealed iron as it is being magnetized: B,(10™*T)

00

013

025

050

063

10

13

B(T)

0.0

0.0042

0010

0.028

0.043 0.095

045

0.67

By(107*T)

19

25

63

130

1300

10,000

B(T)

101

118

144

172

226

315

130 158

078

Determine the magnetic permeability u for each value and plot a graph of u versus By. *49. (I) A large thin toroid has 285 loops of wire per meter, and a 3.0-A current flows through the wire. If the relative permeability of the iron is u/p, = 2200, what is the total field B inside the toroid? #50. (II) An iron-core solenoid is 38cm long and 1.8cm in diameter, and has 640 turns of wire. The magnetic field inside the solenoid is 22T when 48 A flows in the wire. What is the permeability u at this high field strength?

FIGURE 28-53 Problem 46.

—d—1

28-7 Magnetic Materials—Ferromagnetism 47. (II) An iron atom has a magnetic dipole moment of about

1.8 X 1002 A-m?

(a) Determine the dipole moment of an

iron bar 9.0 cm long, 1.2 cm wide, and

’\

1.0 cm thick, if it is

100 percent saturated. (b) What torque would be exerted on this bar when placed in a 0.80-T field acting at right angles to the bar?

| General Problems 51. Three long parallel wires are 3.5cm from one another. (Looking along them, they are at three corners of an equilateffsl triangle.) The current in each wire is 8.00 A, but its direction in wire M is opposite to that in wires N and P (Fig. 28-54). Determine the magnetic force per unit length on each wire due to the other two.

55. Two long straight parallel wires are 15cm apart. Wire A carries 2.0-A current. Wire B’s current is 4.0A in the same direction. (a¢) Determine the magnetic field due to wire A at the position of wire B. (b) Determine the magnetic field due to wire B at the position of wire A. (c) Are these two magnetic fields equal and opposite? Why or why not? (d) Determine the force per unit length on wire A due to wire B, and that on wire B due

to wire A. Are these two forces equal and opposite? Why or why not?

FIGURE 28-54 ProblerTs 51,52,and 53. 52. In Fig. 28-54, determine the magnitude and direction of the magnetic field midway between points M and N.

56. A rectangular loop of wire carries a 2.0-A current and lies in a plane which also contains a very long straight wire carrying a 10.0-A current as shown in Fig. 28-55. Determine (a) the net force and (b) the net torque on the 50cm loop due to the straight == wire.

\

100 A

53. In Fig. 2§—54 the top wire is 1.00-mm-diameter copper wire and is suspended in air due to the two magnetic forces from the bottom two wires. The current is 40.0A in each of the two bottom wires. Calculate the required current flow in the suspended wire. r

54. An electron enters a large solenoid at a 7.0° angle to the . axis. If the field is a uniform 3.3 X 1072 T, determine the radius and pitch (distance between loops) of the electron’s

helical path if its speed is 1.3 X 107 m/s.

2,0Ai1

FIGURE 28-55 Problem 56.

20 Alll26 em

L | T

e 0 cm

General Problems

755

57. A very large flat conducting sheet of thickness ¢ carries a

uniform current density j throughout (Fig. 28-56). Determine the distance (Assume long and

magnetic field (magnitude y above the plane. the plane is infinitely wide.) FIGURE 28-56 Problem 57.

and direction)

at a

58. A long horizontal wire carries a current of 48 A. A second wire, made of 1.00-mm-diameter copper wire and parallel to the first, is kept in suspension magnetically 5.0cm below (Fig. 28-57). (a) Determine the magnitude and direction of the current in the lower wire. (b) Is the lower wire in stable equilibrium? (c¢) Repeat parts (a) and (b) if the second wire is suspended 5.0 cm above 1=48 A the first due to the first’s — magnetic field.

FIGURE 28-57 Problem 58.

5.0cm

F=

63. Near

the Earth’s

poles

the magnetic

field is about

1G

(1 X 107*T). Imagine a simple model in which the Earth’s

field is produced by a single current loop around the equator. Estimate roughly the current this loop would carry. . A 175-g model airplane charged to 18.0 mC and traveling : 2.8 m/s passes within 8.6 cm of a wire, nearly parallel to its path, carrying a 25-A current. What acceleration (in g’s) does this interaction give the airplane? 65. Suppose that an electromagnet uses a coil 2.0 m in diameter made from square copper wire 2.0 mm on a side; the power supply produces 35V at a maximum power output of 1.0kW. (a) How many turns are needed to run the power supply at maximum power? (b) What is the magnetic field strength at the center of the coil? (¢) If you use a greater number of turns and this same power supply, will a greater magnetic field result? Explain. . Four long straight parallel wires located at the corners of a square of side d carry equal currents /y perpendicular to the page as shown in Fig. 28-59. Determine the magnitude and direction of B at the center C of the square.

59. A square loop of wire, of side d, carries a current /. Show

that the magnetic field at the center of the square is 2\/5

Mo I

md [Hint: Determine B for each segment of length d.] 60. In Problem 59, if you reshaped the square wire into a circle, would B increase or decrease at the center? Explain. 61. Helmholtz coils are two identical circular coils having the same radius R and the same number of turns N, separated by a distance equal to the radius R and carrying the same current / in the same direction. (See Fig. 28-58.) They are used in scientific instruments to generate nearly uniform magnetic fields.(They can be seen in the photo, Fig. 27-18.) (a) Determine the magnetic field B at points x along the line joining their centers. Let x = 0 at the center of one coil, and x = R at the center of the other. (b) Show that the field midway between the coils is particularly uniform by 2

showing that Z—f =0 and % the coils. (¢) If

R = 10.0cm,

y

A\

756

CHAPTER 28

=

71l a um—

\[) |

, ‘(RW\T\_" /A

——

"\ “J l

FIGURE 28-60

Problem 67.

___:L

[

13

a —

. A thin 12-cm-long solenoid has a total of 420 turns of wire M‘I i i

X

\

62.For two long parallel wires separated by a distance d, carrying currents /; and [, as in Fig. 28-10, show directly (Eq. 28-1) that Ampere’s law is valid (but do not use Ampere’s law) for a circular path of radius r (r < d) centered on /y: %fi’di

—a—

= 0 at the midpoint between

S

FIGURE 28-58

67. Determine the magnetic field at the point P due to a very long wire with a square bend as shown in Fig. 28-60. The point P is halfway between the two corners. [Hint: You can use the results of Problems 40 and 41.]

N = 250 turns and / = 2.0 A,

what is the field at the midpoint between the coils, x = R/2?

Problem 61.

FIGURE 28-59 Problem 66.

[1-011.

Sources of Magnetic Field

and carries a current of 2.0 A. Calculate the field inside the solenoid near the center.

69. A 550-turn solenoid is 15cm long. The current into it is 33 A. A 3.0-cm-long straight wire cuts through the center of the solenoid, along a diameter. This wire carries a 22-A current downward (and is connected by other wires that don’t concern us). What is the force on this wire assuming the solenoid’s field points due east? 70. You have 1.0kg of copper and want to make a practical solenoid that produces the greatest possible magnetic field for a given voltage. Should you make your copper

wire long and thin, short and fat, or something else“™ Consider other variables, such as solenoid diameter, length, and so on.

~

71. A small solenoid (radius r,) is inside a larger solenoid (radius r, > r,). They are coaxial with n, and ny turns per unit length, respectively. The solenoids carry the same current, but in opposite directions. Let r be the radial distance from the common axis of the solenoids. If the

magnetic field inside the inner solenoid (r < r,) is to be in the oppfisne

direction as the field between the solenoids

(ra < r < n), but have half the magnitude, determine the required ratio ny/n,.

72. Find B at the center of the 4.0-cm-radius semicircle in Fig. 28-61. The straight wires extend a great distance outward to the left and carry a current / = 6.0A.

T

FIGUBE 28-61 Problem 72.

)

e

73. The design of a magneto-optical atom trap requires a magnetic field B that is directly proportional to position x along an axis. Such a field perturbs the absorption of laser light by atoms in the manner needed to spatially confine atoms in the trap. Let us dem&mstrate that “anti-Helmholtz” coils will provide the required

field

B = Cx,

where

C

is

a

constant.

Anti-

Helmbholtz coils consist of two identical circular wire coils, each with radius R and N turns, carrying current / in opposite directions (Fig. 28-62). The coils share a common axis (defined as the x axis with x = 0 at the midpoint (0) between the coils). Assum# that the centers of the coils are separated by a distance equal to the radius R of the coils. (a) Show that the magnetic field at position x along the x axis is given by

74. You want to get an idea of the magnitude of magnetic fields produced by overhead power lines. You estimate that a transmission wire is about 12 m above the ground. The local power company tells you that the line operates at 15kV and provide a maximum of 45 MW to the local area. Estimate the maximum magnetic field you might experience walking under such a power line, and compare to the Earth’s field. [For an ac current, values

will be changing,]

are rms, and

the magnetic

field

. Numencal/ Computer *75. (IT) A circular current loop of radius 15 cm containing 250 turns carries a current of 2.0 A. Its center is at the origin and its axis lies along the x axis. Calculate the magnetic field B at a point x on the x axis for x = —40cm to +40cm in steps of 2cm and make a graph of B as a function of x. *76. (II1) A set of Helmholtz coils (see Problem 61, Fig. 28-58) have a radius R = 10.0 cm and are separated by a distance R = 10.0 cm. Each coil has 250 loops carrying a current I =2.0A. (a) Determine the total magnetic field B along the x axis (the center line for the two coils) in steps of 0.2 cm from the center of one coil (x = 0) to the center of the other (x = R). (b) Graph B as a function of x. (c) By what % does B vary from x = 5.0cm to x = 6.0 cm?

=229 o o 2 oo 2

(b) For small excursions from the origin where |x|

| ¢

?TEQ

02

9

Electromagnetic Induction~ and Faraday’'s Law CHAPTER-OPENING

CONTENTS 29-1

Induced EMF

29-2

Faraday’s Law of Induction; Lenz’s Law

29-3

EMF Induced in a Moving Conductor

#29-4

Electric Generators

29-5

Back EMF and Counter Torque; Eddy Currents

29-6

Transformers and

29-7

A Changing Magnetic Flux Produces an Electric Field

*29-8

Applications of Induction: Sound Systems, Computer Memory, Seismograph, GFCI

758

Transmission of Power

QUESTION—Guess

now!

In the photograph above, the bar magnet is inserted into the coil of wire, and is left there for 1minute; then it is removed from the coil. What would an observer watching the galvanometer see? (a) No change; without a battery there is no current to detect. (b) A small current flows while the magnet is inside the coil of wire. (¢) A current spike as the magnet enters the coil, and then nothing,. (d) A current spike as the magnet enters the coil, and then a steady small current. (e) A current spike as the magnet enters the coil, then nothing, and then a current spike in the opposite direction as the magnet leaves the coil. n Chapter 27, we discussed two ways in which electricity and magnetism are related: (1) an electric current produces a magnetic field; and (2) a magnetic field exerts a force on an electric current or moving electric charge. These discoveries were made in 1820-1821. Scientists then began to wonder: if electric currents produce a magnetic field, is it possible that a magnetic field can produce an electric current? Ten years later the American Joseph Henry (1797-1878) and the Englishman Michael Faraday (1791-1867) independently found that it was possible. Henry actually made the discovery first. But Faraday published his results earlier and«™, investigated the subject in more detail. We now discuss this phenomenon and some of its world-changing applications including the electric generator.

29-1

Induced EMF

In his attempt to produce an electric current from a magnetic field, Faraday used an paratus like that shown in Fig. 29-1. A coil of wire, X, was connected to a battery. (he current that flowed through X produced a magnetic field that was intensified by the ring«‘Fhaped iron core around which the wire was wrapped. Faraday hoped that a strong steady current in X would produce a great enough magnetic field to produce a current in a second coil Y wrapped on the same iron ring. This second circuit, Y, contained a galvanometer to detect any current but contained no battery. Galvanometer

FIGURE 29-1

to induce an emf.

Faraday’s experiment

He met no success with constant currents. But the long-sought effect was finally observed when Faraday noticed the galvanometer in circuit Y deflect strongly at the moment he closed the switch in circuit X. And the galvanometer deflected strongly in the oppc’)Fte direction when he opened the switch in X. A constant current in X produced a constant magnetic field which produced no current in Y. Only when the current in X was starting or stopping was a current produced in Y. Faraday concluded that although a constant magnetic field produces no current in a conductor, a changing magnetic field can produce an electric current. Such a current is called anT induced current. When the magnetic field through coil Y changes, a current occurs in Y as if there were a source of emf in circuit Y. We therefore say that a changing magnetic field induces an emf. . Faraday did further experiments on electromagnetic induction, as this phenomenon is called. For example, Fig. 29-2 shows that if a magnet is moved quickly into a coil of wire, a current is induced in the wire. If the magnet is quickly removed, a current is induced in the opposite direction (B through the coil decreases). Furthermore, if the magnet is held steady and the coil of wire is moved toward or away from the magnet, again an emf is induced and a current flows. Motion or change is required to induce an emf. It doesn’t matter whether the magnet or the coil moves. It is their relative motion that counts.

A CAUTION

Changing B, not B itself, induces current

A

CAUTION

Relative motion—magnet or coil moving induces current

FIGURE 29-2 (a) A current is induced when a magnet is moved toward a coil, momentarily increasing the magnetic field through the coil. (b) The induced current is opposite when the magnet is moved away from the coil (B decreases). Note that the galvanometer zero is at the center of the scale and the needle deflects left or right, depending on the direction of the current. In (c), no current is induced if the magnet does not move relative to the coil. It is the relative motion that counts here: the magnet can be held steady and the coil moved, which also induces an emf.

A

|

/=2

J} S N

(B in coil increasing)

4

-

(b)

'S

|

decreasing)

()

'E§ g

~

™

No

movement

(Bincoil

constant)

SECTION 29-1

759

EXERCISE A Return to the Chapter-Opening Question, page 758, and answer it again now. Try to explain why you may have answered differently the first time.

=

Faraday's Law of Induction; Lenz's Law

29-2

Faraday investigated quantitatively what factors influence the magnitude of the emf induced. He found first of all that the more rapidly the magnetic field changes, the greater the induced emf. He also found that the induced emf depends on the area of the circuit loop. Thus we say that the emf is proportional to the rate of change of the magnetic flux, @, passing through the circuit or loop of area A. Magnetic flux for a uniform magnetic field is defined in the same way we did for electric flux in Chapter 22, namely as

FIGURE 29-3

Determining the flux

through a flat loop of wire. This loop is square, of side £ and area A = 2,

FIGURE 29-4

Magnetic flux @

is proportional to the number of lines of B that pass through the loop.

¢y = BjA

4

ety @p=BAcosds® (b)

FIGURE 29-5 Wire logr

d=(F Bz=BA ©

Example 29-1. & ® ®

|

Arga

I~

(29-1a)

Here B,

loop, and 6 is the angle between B and the vector A (representing the area)

whose direction is perpendicular to the face of the loop. These quantities are

shown in Fig. 29-3 for a square loop of side £ whose areais A = . If the area is of some other shape, or B is not uniform, the magnetic flux can be written"

(29-1b)

As we saw in Chapter 27, the lines of B (like lines of E) can be drawn such that the number of lines per unit area is proportional to the field strength. Then the flux @5 can be thought of as being proportional to the total number of lines

.

passing

o

el

throug

&

.the.area enclosed

&

by.the

prop loop.

by

s

This is 1llusirateod in Fig. 29—?, -

where the loop is viewed from the side (on edge). For 6 = 90°, no magnetic field lines pass through the loop and ®; = 0, whereas ®5 is a maximum when 6 = 0°. The unit of magnetic flux is the tesla-meter?; this is called a weber:

1Wb =1T-m%

B

-

Arja

[B uniform]

Dy = J B-dA.

>/a ®=0 (@)

= B-A.

is the component of the mggnetic field B perpgndicular to the face of the

B2

il

= BAcost

.

CONCEPTUAL

EXAMPLE

29-1 | Determining

flux. A square loop of wire

encloses area A, as shown in Fig. 29-5. A uniform magnetic field B perpendicular to the loop extends over the area A,. What is the magnetic flux through the loop A,? RESPONSE

We assume that the magnetic field is zero outside the area A,. The total

magnetic flux through area A, is the flux through area A,, which by Eq.29-1a for a uniform field is BA,, plus the flux through the remaining area (= A; — A,), which

is zero because B = 0. So the total flux is ®, = BA, + 0(A;, — A,) = BA,. It is not equal to BA; because B is not uniform over A, .

With our definition of flux, Eqs. 29-1, we can now write down the results of Faraday’s investigations: The emf induced in a circuit is equal to the rate of change of magnetic flux through the circuit:

FARADAY’S LAW OF INDUCTION

g =

%3 dt

(29-2a)

This fundamental result is known as Faraday’s law of induction, and is one of the basic laws of electromagnetism. "The integral is taken over an open surface—that is, one bounded by a closed curve such as a circle or square. In the present discussion, the area is that enclosed by the loop under discussion. The area is not an enclosed surface as we used in Gauss’s law, Chapter 22.

760

CHAPTER 29

Electromagnetic Induction and Faraday's Law

fl

J

If the circuit contains N loops that are closely wrapped so the same flux passes through each, the emfs induced in each loop add together, so

€=

‘&

-

_

"Na

)

[Nloops] 29-2b)

FARADAY'S LAW

| oF NDUCTION

A loop of wire in a magnetic field. A square loop of

wire of side £ =5.0cm is in a is the mégnetic flux in the loop of the I(Jop and (b) when B is loop? (c) What is the magnitude has a resistance of 0.012Q) and it in 0.14s?

uniform magnetic field B = 0.16 T. What (a) when B is perpendicular to the face at an angle of 30° to the area A of the of the average current in the loop if it is rotated from position (b) to position (a)

APPROACH We use the definition ®; = B-A to calculate the magnetic flux. Then we use Faraday’s law of induction to find the induced emf in the coil, and from that the induced current (I = ¢/R).

SOLUTION

The area of the coil is A = £ = (5.0 X 10?m)’ = 2.5 X 107 m’,

and the direction of A is perpendicular to the face of the loop (Fig. 29-3). (a) B is perpendicular to the coil’s face, and thus parallel to A (Fig. 29-3), so s

ol

®, = B-A

= BAcos0°

= (0.16T)(2.5 X 10°m?)(1)

= 4.0 x 10°* Wb.

(b) The angle between B and A is 30°, so

b, = B-A fl

=

BAcosf#

= (016T)(2.5 x 10°m?)cos30°

= 3.5 x 10 Whb.

(c) The magnitude of the induced emf is

o

APy

_ (4.0 X 107*Wb) — (3.5 X 107*Wb)

At

_ ag

g

0.14s

The current is then I

=

€

'E

=

3.6 X 104V

OO].T

=

0.030 A

=

30mA.

The m‘illnus signs in Eqs. 29-2a and b are there to remind us in which direction the induce ‘ emf acts. Experiments show that a current produced by an induced emf moves in the magnetic field created by that current opposes in flux.

a direction so that the original change

This is known as Lenz’s law. Be aware that we are now discussing two distinct magnetic fields: (1) the changing magnetic field or flux that induces the current, and (2) the magnetic field produced by the induced current (all currents produce a field). The second field opposes the change in the first. Lenz’s law can be said another way, valid even if no current can flow (as when a circuit is not complete): r

A CAUTION Distinguish two different magnetic fields

An induced emf is always in a direction that opposes the original change in flux that caused it.

SECTION 29-2

Faraday's Law of Induction; Lenz's Law

761

(@)

*

" '

Magnet moves up

}

(B in coil in,

LS

-~ toward coil increasing)

FIGURE 29-2

- Magnet

IN

l

)

A

§ |}

g

o

Bincoil

decreasing)

(repeated).

FIGURE 29-6 A current can be induced by changing the area of the coil, even though B doesn’t change. Here the area is reduced by pulling on its sides: the flux through the coil is reduced as we go from (a) to (b). Here the brief induced current acts in the direction shown so as to try to maintain the original flux (® = BA) by producing its own magnetic field into the page.

Let us apply Lenz’s law to the relative motion between a magnet and a coil, Fig. 29-2. The changing flux through the coil induces an emf in the coil, producing a current. This induced current produces its own magnetic field. In Fig. 29-2a the distance between the coil and the magnet decreases. The magnet’s magnetic field (and number of field lines) through the coil increases, and therefore th flux increases. The magnetic field of the magnet points upward. To oppose the upward increase, the magnetic field inside the coil produced by the induced current needs to point downward. Thus, Lenz’s law tells us that the current moves as shown (use the right-hand rule). In Fig. 29-2b, the flux decreases (because the magnet is moved away and B decreases), so the induced current in the coil produces an upward magnetic field through the coil that is “trying” to maintain the status quo. Thus the current in Fig. 29-2b is in the opposite direction from Fig. 29-2a. It is important to note that an emf is induced whenever there is a change in flux through the coil, and we now consider some more possibilities. XXX

XXXXXXX

XX XXX

XXX

XX RXK XXX

-

i

X

XX XX XXX RXRERXNX

dnward)

XXXXHHXKXKK

FIGURE 29-7 A current can be induced by rotating a coil in a magnetic field. The flux through the coil changes from (a) to (b) because 0 (in Eq.29-1a, ® = BA cos 6) went from 0° (cos@ = 1) to 90° (cos @ = 0).

X

X X

x x

X X X X X X

we see that an emf can be

changing magnetic field B; (2) by changing thefl or (3) by changing the loop’s orientation 6 wit. and 29-2 illustrated case 1. Examples of cases 2 and 29-7, respectively.

X X D % x X X

A e

—

B

X X X X

% X X X x Flux X X decreasingX X X X X

X XX

induction stove.

EXAMPLE

XX X X X X

(inward) x x X X X|x X X X X X X

29-3 | Induction

X X X X X X

X X X X X X

X XX X X X

(a) Maximum flux CONCEPTUAL

- eaced

X X X X X X X X X X X X XXX XXX X X b o db & R

Since magnetic flux ®; = [B-dA = [BcosfdA,

induced in three ways: (1) by a area A of the loop in the field; respect to the field. Figures 29-1 and 3 are illustrated in Figs. 29-6

X X X % X X

Flux - through coilis decreased _ because A

(b)

% %

Example 29-3: An

XX

o

(a)

XK

FIGURE 29-8

XXX

XXX

X X X = —

That is, as the area A decreases, the

current acts to increase B in the original (inward) direction.

XX XX

X X X X

X X % X X X

X X X X X X

XX X X X X X X X X X X

X X% X X X % XR (b) Zero flux

stove.

In

an

induction

stove

(Fig. 29-8), an ac current exists in a coil that is the “burner” (a burner that never gets hot). Why will it heat a metal pan but not a glass container?

RESPONSE The ac current sets up a changing magnetic field that passes through the pan bottom. This changing magnetic field induces a current in the pan bottom, and since the pan offers resistance, electric energy is transformed to thermal energy which heats the pot and its contents. A glass container offers such high resistance that little current is induced and little energy is transferrecem,

(P =V?R).

762

CHAPTER 29

Electromagnetic Induction and Faraday's Law

=2 “

$0 by Lenz's Law to determine the direction electric current induced in a loop due to a change in magnetic flux inside the loop. o produce an induced current you need (a) a clos ed conducting loop, and (b) an ex ternal magnetic flux through the loop that is changing in time. 1. Determine whether the magnetic flux (d, =BAcos#) Lenz’s

of the

inside

is

law

the

unchanged. 2. The magnetic

(a) po

used

( conventional)

ints

loop

in

field

the

is

decreasing,

due

same

Example 29-4.

to

the

direction

increasing,

induced as

the

or

current:

external

field if the flux is decreasing; (b) points in the opposite direction from the external field if the flux is increasing; or (c) is zero if the flux is not changing. . Once you know the direction of the induced magnetic field, use the right-hand rule to find the direction of the induced current. . Always keep in mind that there are two magnetic fields: (1) an external field whose flux must be changing if it is to induce an electric current, and (2) a magnetic field produced by the induced current.

3 @ =0

@

®

C‘)

~

»

Pulling a round loop to

he right out of a magnetic

field which points out of the page

CONCEPTUAL

(b)

Shrinking a loop in a magnetic field pointing into the page

()

S magnetic pole moving from below, up toward the loop

(d)

N magnetic pole

Rotating the loop by pulling

in the plane of the page

us and pushing the right

the left side toward

moving toward loop

side in; the magnetic field points from right to left

EXAMPLE 29-4 | Practice with Lenz’s law. In which direction is

the current induced in the circular loop for each situation in Fig. 29-9?

RESPONS E (a) Initially, the magnetic field pointing out of the page passes through the loop. 1 f you pull the loop out of the field, magnetic flux through the loop decreases; so the induced current will be in a direction to maintain the decreasing flux through the loop: the current will be counterclockwise to produce a magnetic field outward (toward the reader). (b) The ex ternal field is into the page. The coil area gets smaller, so the flux will decrease; | hence the induced current will be clockwise, producing its own field into the page to make up for the flux decrease. (c) Magnetic field lines point into the S pole of a magnet, so as the magnet moves tow ard us and the loop, the magnet’s field points into the page and is getting e stronger. The current in the loop will be induced in the counterclockwise direction i n order to produce a field B out of the page. (d) The fijld is in the plane of the loop, so no magnetic field lines pass through the loop and the flux through the loop is zero throughout the process; hence there is no change in external magnetic flux with time, and there will be no induced emf or current in the loop. (e) In1t1ally there is no flux through the loop When you start to rotate the loop, the extern al field through the loop begins increasing to the left. To counteract ‘this change in flux, the loop will have current induced in a counterclockwise l direction so as to produce its own field to the right.

SECTION 29-2

A

CAUTION

Magnetic field created by induced

current opposes change in external flux, not necessarily opposing the external field

Faraday’s Law of Induction; Lenz's Law

763

|

X

X

X

X

X

X

X

X

X

X

MX

X

X

X

X

X

Pulling a coil from a magnetic field. A 100-loop square

X X

XX

X

X

X

X

X

X|

|

FIGURE 29-10 Example 29-5. The square coil in a magnetic field B = 0.600T is pulled abruptly to the right to a region where B = 0.

coil of wire, with side £ = 5.00cm and total resistance 100 (), is positioned perpendicular to a uniform 0.600-T magnetic field, as shown in Fig. 29-10. It is quickly pulled from the field at constant speed (moving perpendicular to B) to aq region where B drops abruptly to zero. At ¢t = 0, the right edge of the coil is a the edge of the field. It takes 0.100s for the whole coil to reach the field-free region. Find (a) the rate of change in flux through the coil, and (b) the emf and current induced. (c) How much energy is dissipated in the coil? (d) What was the average force required (Fyy)? APPROACH We start by finding how the magnetic flux, ®z = BA, changes during the time interval Af = 0.100s. Faraday’s law then gives the induced emf and Ohm’s law gives the current. SOLUTION (a) The area of the coilis A = £ = (5.00 X 10‘2m)2 =2.50 X 103 m?

The flux through one loop is initially ®; = BA = (0.600 T)(2.50 X 107 m?) = 1.50 X 103 Wb.

After

0.100s, the flux is zero. The

rate

constant (because the coil is square), equal to

A®,

At

0-— (150 X 107 Wb)

0.100's

=

of change

in flux is

—1.50 X 1072 Whb/s.

(b) The emf induced (Eq. 29-2) in the 100-loop coil during this 0.100-s interval is

=

=R

Ady

= —(100)(—1.50 X 102

3

Wb/s)

= 1.50V.

The current is found by applying Ohm'’s law to the 100-) coil: FIGURE 29-11

O

Exercise B.

———e——————

[ increasing

-

| decreasing

O

(a)

(b)

I =

—=———=150XxX10%2A

=

15.0mA.

By Lenz’s law, the current must be clockwise to produce more B into the pag™ and thus oppose the decreasing flux into the page. >~ (c) The total energy dissipated in the coil is the product of the power (= I’R) and the time:

E = Pt = I*Rt = (150 x 102 AR(100Q)(0.100s) = 225 X 107J. (d) We can use the result of part (c) and apply the work-energy principle: the energy dissipated E is equal to the work W needed to pull the coil out of the field (Chapters 7 and 8). Because W = Fd where d = 5.00cm, then w 225 %X 10737 F=—=——"——— d 5.00 X 102 m

=

0.0450N.

Alternate Solution (d) We can also calculate the force directly using F = 7€ x B, Q

‘ I constant

(©

I increasing {

Q

(d) 764

CHAPTER 29

Eq. 27-3, which here for constant B is F = I£B. The force the magnetic field exerts on the top and bottom sections of the square coil of Fig. 29-10 are in opposite directions and cancel each other. The magnetic force Fy exerted on the left vertical section of the square coil acts to the left as shown because the current is up (clockwise). The right side of the loop is in the region where B=0 Hence the external force, to the right, needed to just overcome the magnetic force to the left (on N = 100 loops) is

F.. = NIB = (100)(0.0150 A)(0.0500m)(0.600T) = 0.0450N, which is the same answer, confirming our use of energy conservation above. EXERCISE B What is the direction of the induced current in the circular loop due to thf‘fl current shown in each part of Fig. 29-11?

Electromagnetic Induction and Faraday's Law

|

29-3

®© 0 0 0

EMF Induced in a Moving Conductor

Another way to induce an emf is shown in Fig. 29-12a, and this situation helps f']lluminate the nature of the induced emf. Assume that a uniform magnetic field B is arpendicular to the area bounded by the U-shaped conductor and the movable rod resting on it. If the rod is made to move at a speed v, it travels a distance dx = v dt in a time dt. Therefore, the area of the loop increases by an amount dA = £dx = fv dt in a time dt. By Faraday’s law there is an induced emf & whose magnitude is given by ddy B dA Biv dt ¢ = 5 a g - Blw. (29-3)

ofo

olo ©

©

B (outward)

ololleiio ¢ ;1»::

@l@a’;«@:;@ ©

EXERCISE AL In what direction will the electrons flow in Fig. 29-12 if the rod moves to the eft, decreasing the area of the current loop?

GNP

IR

ESTIMATE | Does a moving airplane develop a large emf?

An airplane travels 1000 km/h in a region where the Earth’s magnetic field is about 5 X 107°T and is nearly vertical (Fig. 29-13). What is the potential difference induced between the wing tips that are 70 m apart?

(0|0 vdt

(b

in Fig. 29-1 |b. If the rod was not in contact with the U-shaped conductor, electrons

result” as from Faraday’s law above, Eq. 29-3.

© (a)

Equation 29-3 is valid as long as B, £, and v are mutually perpendicular. (If they are not, we use only the components of each that are mutually perpendicular.) An emf induced on a conductor moving in a magnetic field is sometimes called motional emf. We can also obtain Eq. 29-3 without using Faraday’s law. We saw in Chapter 27 that a charged particle moving perpendicular to a magnetic field B with speed v experiencesj: force F = gv X B (Eq.27-5a). When the rod of Fig. 29-12a moves to the right with speed v, the electrons in the rod also move with this speed. Therefore, since v L fi,%each electron feels a force F = quvB, which acts up the page as shown would collect at the upper end of the rod, leaving the lower end positive (see signs in Fig. 29-12b). There must thus be an induced emf. If the rod is in contact with the U-shaped conductor (Fig. 29-12a), the electrons will flow into the U. There will then be a clockwise (conventional) current in the loop. To calculate the emf, we determine the work W needed to move a charge ¢ from one end of the rod to the other agains| this potential difference: W = force X distance = (qvB)({). The emf equals the work done per unit charge, so €= W/q = quB{/q = Blv, the same

0 0

FIGURE 29-12 (a) A conducting rod is moved to the right on a U-shaped conductor in a uniform magnetic field B that points out of the page. The induced current is clockwise. (b) Upward force on an electron in the metal rod (moving to the right) due to B pointing out of the page; hence electrons can collect at top of rod, leaving + charge at bottom. FIGURE 29-13

Example 29-6.

APPROACH We consider the wings to be a 70-m-long conductor moving through the Earth’s magnetic field. We use Eq. 29-3 to get the emf. SOLUTION

NOTE

S

Since

v = 1000 km/h

= 280 m/s,

and v L B, we have

€ = Blv = (5x107°T)(70m)(280m/s)

~ 1V.

Not much to worry about.

Electromagnetic

blood-flow

measurement. The rate of

blood flow in our body’s vessels can be measured using the apparatus shown in Fig. 29-14, since blood contains charged ions. Suppose that the blood vessel is 2.0mm

in diameter,

the

magnetic

field

is 0.080T,

0.10 mV. What is the flow velocity v of the blood?

APPROACH The magnetic field B points horizontally toward S pole). The induced emf acts over the width vessel, perpendicular to B and v (Fig. 29-14), just as use Eq. 29-3 to get v. (V in Fig. 29-14 corresponds to SOLUTION We solve for v in Eq. 29-3:

€&

Bl

(1.0 x 107*V)

(0.080T)(2.0 X 10> m)

=

and

the

measured

emf

@)PHYSICS

APPLIED

Blood-flow measurement

is

from left to right (N pole £ = 2.0 mm of the blood in Fig. 29-12. We can then v in Fig. 29-12.)

FIGURE 29-14 Measurement of blood velocity from the induced emf. Example 29-7.

0.63m/s.

OTE In actual practice, an alternating current is used to produce an alternating magnetic fileld. The induced emf is then alternating.

e

"This force argument, which is basically the same as for the Hall effect (Section 27-8), explains this one way of inducing an emf. It does not explain the general case of electromagnetic induction.

SECTION 29-3

765

Force on the rod. To make the rod of Fig. 29-12a move to

the right at constant speed », you need to apply an external force on the rod to the right. (a) Explain and determine the magnitude of the required force. (b) What external power is needed to move the rod? (Do not confuse this external force on the rod with the upward force on the electrons shown in Fig. 29-12b.)

®©

0

0

0

O

©

0

0

0

Of

O

60

60

6

@

0

©

©

0

60

0

0

FIGURE 29-12

(b)

06 ’

APPROACH When the rod moves to the right, electrons flow upward in the rod according to the right-hand rule. So the conventional current is downward in the rod. We can see this also from Lenz’s law: the outward magnetic flux through the loop is increasing, so the induced current must oppose the increase. Thus the current is clockwise so as to produce a magnetlc field into the page (right-hand rule). The magnetic force on the moving rod is F=1I1[{xB fora constant B (Eq. 27-3). The right-hand rule tells us this magnetic force is :getlrxie left, and is thus a “drag force” opposing our effort to move the rod to ght.

SOLUTION .(a) The magnitude the magnetic force F = I{B, Eq. 29-3), and the resistance U-shaped conductor. The force

right on a U-shaped conductor in a

uniform magnetic field B that points out of the page. The induced current

is clockwise. (b) Upward force on an

electron in the metal rod (moving to

If B, £, and

bottom.

2

R are constant, then

a constant

speed

.

v is produced

by a

constant external force. (Constant R implies that the parallel rails have negligible resistance.)

(b) The external power needed to move the rod for constant R is

the right) due to B pointing out of the page; hence electrons can collect at top of rod, leaving + charge at

the external force, to the right, needs to balance the left. The current [ = &/R = Blv/R (see is that of the whole circuit: the rod and the required to move the rod is thus

F= 1B - (’ifl)w - EL, R R

(repeated)

(a) A conducting rod is moved to the

of to R F

Pt

=

Fv

=

B*? R

The power dissipated in the resistance is P = I’R. With

I = &/R = Blv/R,

i

*,

202,,2

P = PR=2 fe” , so the power input equals the power dissipated in the resistance at any moment.

29-4

FIGURE 29-15 QR

An ac generator.

Axle rotated mechanically

Sliprings

766

CHAPTER 29

Electric Generators

We discussed alternating currents (ac) briefly in Section 25-7. Now we examine how ac is generated: by an electric generator or dynamo, one of the most important practical results of Faraday’s great discovery. A generator transforms mechanical energy into electric energy, just the opposite of what a motor does. A simplified diagram of an ac generator is shown in Fig. 29-15. A generator consists of many loops of wire (only one is shown) wound on an armature that can rotate in a magnetic field. The axle is turned by some mechanical means (falling water, steam turbine, car motor belt), and an emf is induced in the rotating coil. An electric current is thus the output of a generator Suppose in Fig. 29-15 that the armature is rotating clockwise; then F = gv x B applied to charged partlcles in the wire (or Lenz’s law) tells us that the (conventional) current in the wire labeled b on the armature is ~ outward, toward us; therefore the current is outward from brush b. (Each brush is fixed and presses against a continuous slip ring that rotates with the armature.) After one-half revolution, wire b will be where wire a is now in the drawink and the current then at brush b will be inward. Thus the current produced alternating.

Electromagnetic Induction and Faraday's Law

Let us assume the loop is being made to rotate in a uniform magnetic field B with constant angular velocity w. From Faraday’s law (Eq. 29-2a), the induced emf is

r

where A is the area of the loop and 6 is the angle between w = db/dt, then 6 = 6, + wt. We arbitrarily take 6, = 0, so

€ =

—BA % (coswt)

=

B and A. Since

BAwsin ot.

If the rotati ng coil contains N loops,

€ = =

NB/.4w sin wt

(29-4)

%sinwt.

the o utput emf is sinusoidal (Fig. 29-16) with amplitude &, = NBAw. Such a rot ating coil in a magnetic field is the basic operating principle of an

Thus

ac generat

The frequency f (=w/27) is 60 Hz for general use in the United States and Canada, whereas 50 Hz is used in many countries. Most of the power generated in the United States is done at steam plants, where the burning of fossil fuels (coal, oil, natural gas) boils water to produce high-pressure steam that turns a turbine connected t o the generator axle. Falling water from the top of a dam (hydroelectric) is also common (Fig. 29-17). At nuclear power plants, the nuclear energy released is used to roduce steam to turn turbines. Indeed, a heat engine (Chapter 20) a generator is the principal means of generating electric power. The connected frequency f 60 Hz or 50 Hz is maintained very premsely by power compames, and in doing Problems, we will assume it is at least as precise as other numbers given.

emf

VANWAN/ VvVV

Time

FIGURE 29-16 An ac generator produces an alternating current. The output emf € = % sinwf, where % = NABw (Eq.29-4).

FrpHysics

APPLIED

@PHYSlcs

APPLIED

Power plants FIGURE 29-17 Water-driven generators at the base of Bonneville Dam, Oregon.

|

r

W Eap Bl An ac generator. The armature rotates in a 0.15-T magnetic field. If the area of the many loops must the coil contain if the peak output is APPROACH From Eq. 29-4 we see that the maximum SOLUTIO : We solve Eq. 29-4 for N with o = 27f =

|

%

BAw

of a 60-Hz ac generator coil is 2.0 X 1072 m?, how to be &, = 170 V? emf is % = NBAw.

(628)(60s7!) = 377s7%:

170V

(015T)(2.0 X 102m?)(377s7")

e

Adcg erator is much like an ac generator, except the slip rings are replaced by split-rin commutators, Fig. 29-18a, just as in a dc motor (Section 27-6). The output of such a generator is as shown and can be smoothed out by placing a capacitor in parallel with the output (Section 26-5). More common is the use of many armature windings, as in Fig. 29-18b, which produces a smoother

output.

DC generator

FIGURE 29-18

(a) A dc generator

with one set of commutators, and

(b) a dc generator with many sets of commutators and windings.

(a)

t

(b)

t

SECTION 29-4

Electric Generators

767

FIGURE 29-19 (a) Simplified schematic diagram of an alternator. The input current to the rotor from the battery is connected through continuous slip rings. Sometimes the rotor electromagnet is replaced by a permanent magnet. (b) Actual shape of an alternator. The rotor is made to turn by a belt from the engine. The current in the wire coil of the rotor produces a magnetic field inside it on its axis that points horizontally from left to right, thus making north and south poles of the plates attached at either end. These end plates are made with triangular fingers that are bent over the coil—hence there are alternating N and S poles quite close to one another, with magnetic field lines between them as shown by the blue lines. As the rotor turns, these field lines pass through the fixed stator coils (shown on the right for clarity, but in operation the rotor rotates within the stator), inducing a current in them, which is the output.

-

Loops of wire (in which current is induced)

Stator coil (emf induced in)

Current to Bproduce field

@ N

Input current

Ou[put

current (induced)

South pole1

Z: (Rotating

electromagnet)

Stator coil (emf induced in) (a)

Rotor

Stator assembly

(b) @

PHYSICS

APPLIED Alternators~

Automobiles used to use dc generators. Today they mainly use altemato,q which avoid the problems of wear and electrical arcing (sparks) across the split-ring commutators of dc generators. Alternators differ from generators in that an electromagnet, called the rotor, is fed by current from the battery and is made to rotate by a belt from the engine. The magnetic field of the turning rotor passes through a surrounding set of stationary coils called the stator (Fig. 29-19), inducing an alternating current in the stator coils, which is the output. This ac output is changed to dc for charging the battery by the use of semiconductor diodes, which allow current flow in one direction only.

*29-5

Back EMF and Counter Torque; Eddy Currents

*Back EMF, in a Motor A motor turns and produces mechanical energy when a current is made to flow in it. From our description in Section 27-6 of a simple dc motor, you might expect that the armature would accelerate indefinitely due to the torque on it. However, as the armature of the motor turns, the magnetic flux through the coil changes and an emf is generated. This induced emf acts to oppose the motion (Lenz’s law) and is called the back emf or counter emf. The greater the speed of the motor, the greater the back emf. A motor normally turns and does work on something, but if there were no load, the motor’s speed would increase until the back emf equaled the input voltage. When there is a mechanical load, the speed of the motor may be limited also by the load. The back emf will then be less than the external applig‘ voltage. The greater the mechanical load, the slower the motor rotates and t lower is the back emf (€ o« w, Eq.29-4).

768

CHAPTER 29

Electromagnetic Induction and Faraday’s Law

DOV

EES [

Back emf in @ motor. The armature windings of a dc motor

have a resistance of 5.0 (). The motor is connected to a 120-V line, and when the motor reaches full speed against its normal load, the back emf is 108 V. Calculate #(a) the current into the motor when it is just starting up, and (b) the current when .he motor reaches full speed. APPROACH As the motor is just starting up, it is turning very slowly, so there is no induced back emf. The only voltage is the 120-V line. The current is given

by Ohm’s law with

R = 5.0().

At full speed, we must include as emfs both

the 120-V applied emf and the opposing back emf.

SOLUTION

(a) At start up, the current is controlled by the 120 V applied to the

coil’s 5.0-() resistance. By Ohm’s law,

[ = L g 280% R

500

=

120V

—

fi%cul;eeg‘fn

Ofllr;wtr:)grs

armature winding |

50Q

8induced = 108V

24A.

»

(b) When the motor is at full speed, the back emf must be included in the equivalent circuit shown in Fig. 29-20. In this case, Ohm’s law (or Kirchhoff’s rule) gives _- 108V

irl

1(5.00Q).

= I1|20V FIGURE 29-20

Circuit of a motor

showing Example induced 29-10, back emf.

Therefore I

=

12V 500

24 A

NOTE This result shows that the current can be very high when a motor first starts up. This is why the lights in your house may dim when the motor of the refrigerator (or other large motor) starts up. The large initial current causes the voltage to the lights and at the outlets to drop, since the house wiring has resistance and there is some voltage drop across it when large currents are drawn. \ONCEPT

AL EXAMPLE

29-11 | Motor overload. When using an appliance

such as a bl ender, electric drill, or sewing machine, if the appliance is overloaded or jammed so that the motor slows appreciably or stops while the power is still connected, the device can burn out and be ruined. Explain why this happens.

@)PHYSIcs

APPLIED

Burning out a motor

RESPONSE The motors are designed to run at a certain speed for a given applied voltage, an d the designer must take the expected back emf into account. If the rotation sp eed is reduced, the back emf will not be as high as expected (€ x w, Eq. 29-4), and the current will increase, and may become large enough that the windings of the motor heat up to the point of ruining the motor.

* Counter Torque In a generator, the situation is the reverse of that for a motor. As we saw, the mechanical turning of the armature induces an emf in the loops, which is the output. If the generator is not connected to an external circuit, the emf exists at the terminals but there is no current. In this case, it takes little effort to turn the armature. But if the generator is connected to a device that draws current, then a current flows in the coils of the armature. Because this current-carrying coil is in an external magnetic field, there will be a torque exerted on it (as in a motor), and this torque opposes the motion (use the right-hand rule for the force on a wire in Fig. 29-15). This is called a counter torque. The greater the electrical load—that is, the more current that is drawn—the greater will be the counter torque. Hence the external applied torque will have to be greater to keep the generator turning. This makes sense from the conservation of energy principle. More mechanical-energy input is needed to produce more electrical-energy output. "XERCISE D A bicycle headlight is powered by a generator that is turned by the bicycle ~heel. (a) Ifi you pedal faster, how does the power to the light change? (b) Does the generator refist being turned as the bicycle’s speed increases, and if so how?

*SECTION 29-5

Back EMF and Counter Torque; Eddy Currents

769

*Eddy Currents

#i Gaward)

(a)

(b) FIGURE 29-21 Production of eddy currents in a rotating wheel. The grey ~

lines in (b) indicate induced current. FIGURE 29-22 Pt —

@

E

Airport metal

PHYSICS APPLIED Airport metal detector

FGURE 29-23 Reparingasep

down transformer on a utility pole.

ke

Induced currents are not always confined to well-defined paths such as in wires. Consider, for example, the rotating metal wheel in Fig. 29-21a. An external magnetic field is applied to a limited area of the wheel as shown and points into thesm, page. The section of wheel in the magnetic field has an emf induced in it because the conductor is moving, carrying electrons with it. The flow of induced (conventional) current in the wheel is upward in the region of the magnetic field (Fig. 29-21b), and the current follows a downward return path outside that region. Why? According to Lenz’s law, the induced currents oppose the change that causes them. Consider the part of the wheel labeled c in Fig. 29-21b, where the magnetic field is zero but is just about to enter a region where B points into the page. To oppose this inward increase in magnetic field, the induced current is counterclockwise to produce a field pointing out of the page (right-hand-rule 1). Similarly, region d is about to move to e, where B is zero; hence the current is clockwise to produce an inward field opposed to this decreasing flux inward. These currents are referred to as eddy currents. They can be present in any conductor that is moving across a magnetic field or through which the magnetic flux is changing. In Fig. 29-21b, the magnetic field exerts a force F on the induced currents it has created, and that force opposes the rotational motion. Eddy currents can be used in this way as a smooth braking device on, say, a rapid-transit car. In order to stop the car, an electromagnet can be turned on that applies its field either to the wheels or to the moving steel rail below. Eddy currents can also be used to dampen (reduce) the oscillation of a vibrating system. Eddy currents, however, can be a problem. For

example, eddy currents induced in the armature of a motor or generator produce

?eat. (P = I ig) and waste energy. To re.duce the ec!dy currents, the grmatures are aminated, that is, they are made of very thin sheets of iron that. are well insulated frgm one another. The total path length of the eddy currents is confined to each slab, which increases the total resistance; hence the current is less and there is less wasted energy. Walk-through metal detectors at airports (Fig. 29-22) detect metal objects usin electromagnetic induction and eddy currents. Several coils are situated in the wal. of the walk-through at different heights. In a technique called “pulse induction,”" the coils are given repeated brief pulses of current (on the order of microseconds), hundreds or thousands of times a second. Each pulse in a coil produces a magnetic field for a very brief period of time. When a passenger passes through the walk-through, any metal object being carried will have eddy currents induced in it. The eddy currents persist briefly after each input pulse, and the small magnetic field produced by the persisting eddy current (before the next external pulse) can be detected, setting off an alert or alarm. Stores and libraries sometimes use similar systems to discourage theft.

) )—(

Transformers and Transmission of Power

A transformer is a device for increasing or decreasing an ac voltage. Transformers are found everywhere: on utility poles (Fig. 29-23) to reduce the high voltage from

the electric company to a usable voltage in houses (120 V or 240 V), in chargers for

cell phones, laptops, and other electronic devices, in CRT monitors and in your car to give the needed high voltage (to the spark plugs), and in many other applications. A transformer consists of two coils of wire known as the primary and secondary coils. The two coils can be interwoven (with insulated wire); or they can be linked by an iron core which is laminated to minimize eddy-current losses (Section 29-5), as shown in Fig. 29-24. Transformers are designed so that (nearly) all the magnetic flux produced by the current in the primary coil also passes through the secondary

coil, and we assume this is true in what follows. We also assume that energy losse r,), we assume B =

at all times. Determine the electric field E at any point P a distance r from the center of the circular area due to the changing B.

APPROACH

The changing magnetic flux through a circle of radius r, shown

dashed in Fig. 29-27b, will produce an emf around this circle. Because all points on the dashed circle are equivalent physically, the electric field too will show this symmetry and will be in the plane perpendicular to B. Thus we can expect E to be perpendicular to B and to be tangent to the circle of radius r. The direction of E will be as shown in Fig. 29-27b and c, since by Lenz’s law the induced E needs to be capable of producing a current that generates a magnetic field opposing the original change in B. By symmetry, we also expect E to have the same magnitude at all points on the circle of radius r.

SOLUTION We take the circle shown in Fig 29-27b as our path of integration in Eq. 29-8. We ignore the minus sign so we can concentrate on magnitude since we already found the direction of E from Lenz’s law, and obtain

E{25r) = lar )Zf

[r


r,, the flux through a circle of radius ris ®z = wr§B. Then Eq. 29-8 gives -~ E(2@r) or

=

nr%d—B dt

= Indb 2

T ordt

[r > r

[ > ro]

Thus the magnitude of the induced electric field increases linearly from zero at the

center of the magnet to E = (dB/dt)(ry/2) at the edge, and then decreases inversely

\/

with distance in the region beyond the edge of the magnetic field. The electric field lines are circles as shown in Fig. 29-27c. A graph of E vs. r is shown in Fig, 29-27d.

(c)

EXERCISE E Consider the magnet shown in Fig. 29-27 with a radius ry = 6.0cm. If the magnetic field changes uniformly from 0.040T to 0.090T in 0.18 s, what is the magnitude of the resulting electric field at (a) » = 3.0cm and (b) r = 9.0cm?

| |

i

*Forces Due to Changing B are Nonconservative

|

I

@

ro

r

Example produced charges at static case field lines loops. This static case,

29-14 illustrates an important difference between electric fields by changing magnetic fields and electric fields produced by electric rest (electrostatic fields). Electric field lines produced in the electro(Chapters 21 to 24) start and stop on electric charges. But the electric produced by a changing magnetic field are continuous; they form closed distinction goes even further and is an important one. In the electrothe potential difference between two points is given by (Eq. 23-4a) b

Voo =

o —V,

= —J E-dl.

-

If the integral is around a closed loop, so points a and b are the same, then V;, = 0.

774

CHAPTER 29

Electromagnetic Induction and Faraday's Law

Hence the integral of E-df around a closed path is zero:

$E-at = o

[electrostatic field]

rhis followed from the fact that the electrostatic force (Coulomb’s law) is a conservative force, and so a potential energy function could be defined. Indeed, the relation above, $E-df = 0, tells us that the work done per unit charge around any closed path is zero (or the work done between any two points is independent of path—see Chapter 8), which is a property only of a conservative force. But in the nonelectrostatic case, when the electric field is produced by a changing magnetic field, the integral around a closed path is not zero, but is given by Eq. 29-8:

-

ngdf—

_

Small coil of wire

dd,

Magnet

=

We thus come to the conclusion that the forces due to changing magnetic fields are nonconservative. We are not able therefore to define a potential energy, or potential function, at a given point in space for the nonelectrostatic case. Although static electric fields are conservative fields, the electric field produced by a changing magnetic field is a nonconservative field.

*29—8‘ Applications of Induction: Sound Systems, Computer

Memory, Seismograph, GFCI

*Microgthne There are various types of microphones, and many operate on the principle of inducsion. In one form, a microphone is just the inverse of a loudspeaker (Section 27-6). . small coil connected to a membrane is suspended close to a small permanent magnet, as shown in Fig. 29-28. The coil moves in the magnetic field when sound waves strike the membrane and this motion induces an emf. The frequency of the induced emf will be just that of the impinging sound waves, and this emf is the “signal” that can be amplified and sent to loudspeakers, or sent to a recorder.

*Read/Write on Tape and Disks Recording and playback on tape or disks is done by magnetic heads. Recording tapes for use in audio and video tape recorders contain a thin layer of magnetic oxide on a it‘[;in plastic tape. During recording, the audio and/or video signal voltage is sent to the recording head, which acts as a tiny electromagnet (Fig. 29-29) that magnetizes the tiny section of tape passing over the narrow gap in the head at each instant. In playback, the changing magnetism of the moving tape at the gap causes corresponding changes in the magnetic field within the soft-iron head, which in turn induces an emf in the coil (Faraday’s law). This induced emf is the output signal that can be amplified and sent to a loudspeaker (audio) or to the picture tube (video). In audio and video recorders, the signals may be analog—they vary continuously in amplitude over time. The variation in degree of magnetization of the tape at any point reflects the variation in amplitude and frequency of the audio or video signal. Digital information, such as used on computer hard drives or on magnetic computer tape and some types of digital tape recorders, is read and written using heads‘that are basically the same as just described (Fig. 29-29). The essential difference is in the signals, which are not analog, but are digital, and in particular binary, meaning that only two values are possible for each of the extremely high number of predetermined spaces on the tape or disk. The two possible values are usually referred as 1 and 0. The signal voltage does not vary continuously but rather takes on only .wo values, +5 V and 0V, for example, corresponding to the 1 or 0. Thus, information is carried as a Teries of bits, each of which can have only one of two values, 1 or 0.

*SECTION 29-8

Membrane

To recorder or amplifier FIGURE 29-28 Diagramofa microphone that works by induction. FIGURE 29-29 (a) Read/Write (playback/recording) head for tape or disk. In writing or recording, the electric input signal to the head, which acts as an electromagnet, magnetizes the passing tape or disk. In reading or playback, the changing magnetic field of the passing tape or disk induces a changing magnetic field in the head, which in turn induces in the coil an emf that is the output signal. (b) Photo of a hard drive showing several platters and read/write heads that can quickly move from the edge of the disk to the center. Electric signal input (or output)

Moving magnetic tape or disk

(b)

Applications of Induction: Sound Systems, Computer Memory, Seismograph, GFCI

775

Suspension

Coil

*Credit Card Swipe

springs

Permanent magnet

When you swipe your credit card at a store or gas station, the magnetic stripe on the back of the card passes over a read head just as in a tape recorder or ~ computer. The magnetic stripe contains personal information about your accountem, and connects by telephone line for approval if your account is in order.

*Seismograph FIGURE 29-30 One type of seismograph, in which the coil is fixed to the case and moves with the Earth, The magnet, suspended by springs,

i

Aol

e

instantaneously with the coil (and case), so there is relative motion between magnet and coil.

In geophysics, a seismograph measures the intensity of earthquake waves using a magnet and a coil of wire. Either the magnet or the coil is fixed to the case, and the other is inertial (suspended by a spring; Fig. 29-30). The relative motion of magnet and coil when the Earth shakes induces an emf output. .

.

* Ground Fault Circuit Interrupter (GFCI) Fuses and circuit breakers (Sections 25-6 and 28-8) protect buildings from fire, and apparatus from damage, due to undesired high currents. But they do not turn off the current until it is very much greater than that which causes permanent damage to humans or death (=100 mA). If fast enough, they may protect in case of a short. A ground fault circuit interrupter (GFCI) is meant to protect humans; GFClIs can react to currents as small as 5 mA.

Simple electronic circuit ]

Sensin

FIGURE 29-31 A ground fault circuit interrupter (GFCI). Hot

Neutral

FIGURE 29-32

(a) A GFCI wall

outlet. GFCIs can be recognized

because they have “test” and “reset”

buttons. (b) Add-on GFCI that plugs into outlet.

Solenoid b circuit breaker

g 1

120V

;

>

I Iron ring

S

—%

®

—

.

1

k

4 Electric circuit with one or more devices

(possible sources of trouble)

Electromagnetic induction is the physical basis of a GFCI. As shown in Fig. 29-31, the two conductors of a power line leading to an electrical circuit or device

(red) pass through a small iron ring. Around the ring are many loops of thin wire

that serve as a sensing coil. Under normal conditions (no ground fault), the current

moving in the hot wire is exactly balanced by the returning current in the neutral wire. If something goes wrong and the hot wire touches the ungrounded metal case of the device or appliance, some of the entering current can pass through a person who touches the case and then to ground (a ground fault). Then the return current in the neutral wire will be less than the entering current in the hot wire, so there is a net current passing through the GFCI’s iron ring. Because the current is ac, it is changing and the current difference produces a changing magnetic field in the iron, thus inducing an emf in the sensing coil wrapped around the iron. For example, if a device draws 8.0 A, and there is a ground fault through a person of 100 mA (= 0.1 A), then 7.9 A will appear in the neutral wire. The emf induced in the sensing coil by this 100-mA difference is amplified by a simple transistor circuit and sent to its own solenoid circuit breaker that opens the circuit at the switch S, thus protecting your life. If the case of the faulty device is grounded, the current difference is even higher when there is a fault, and the GFCI trips immediately.

GFCIs can sense currents as low as 5mA and react in 1 msec, saving lives. They can be small enough to fit as a wall outlet (Fig. 29-32a), or as a plug-in unit into which you plug a hair dryer or toaster (Fig. 29-32b). It is especially important to have GFCls installed in kitchens, bathrooms, outdoors, and near swimming pools X, the circuit is said to be predominantly “inductive.” And if Xo > X, the circuit is said to be predominantly “capacitive.” Discuss the reasons for these terms. In particular, do they say anything about the relative values of L and C at a given frequency?

13. Do the results of Section 30-8 approach the proper expected results when w approaches zero? What are the expected results?

14. Under what conditions is the impedance in an LRC circuit a minimum? 15. Is it possible for the instantaneous power output of an ac generator connected to an LRC circuit ever to be negative? Explain. 16. In an ac LRC circuit, does the power factor, cos ¢, depend on frequency? Does the power dissipated depend on frequency? 17. Describe briefly how the frequency of the source emf affects the impedance of (a) a pure resistance, (b) a pure capacitance, (c) a pure inductance, (d) an LRC circuit near resonance (R small), (e) an LRC circuit far from resonance (R small). 18. Discuss the response of an LRC circuit as R — 0 when the frequency is (a) at resonance, (b) near resonance, (c) far from resonance. Is there energy dissipation in each case? Discuss the transformations of energy that occur in each case. 19. An LRC resonant circuit is often called an oscillator circuit. What is it that oscillates? 20, Compare the oscillations of an LRC circuit to the vibration of a mass m on a spring. What do L and C correspond to in the mechanical system?

| Problems 30-1 r\

30-2

Mutual Inductance

1 (II) A 2.44-m-long coil containing 225 loops is wound on an iron core (average w = 1850ug ) along with a second coil of 115 loops. The loops of each coil have a radius of 2.00cm. If the current in the first coil drops uniformly from 12.0A to zero in 98.0ms, determine: (a) the mutual inductance M; (b) the emf induced in the second coil. . (II) Determine the mutual inductance per unit length between

two

long

solenoids,

one

inside

the

5 (I) If the current in a 280-mH 25.0A to 10.0A induced emf?

FIGURE 30-25 Problem 4.

in 360 ms, what

coil changes steadily from is the magnitude

of the

. (I) How many turns of wire would be required to make a 130-mH inductance out of a 30.0-cm-long air-filled coil with a diameter of 4.2 cm? . (I) What is the inductance of a coil if the coil produces an emf of 2.50 V when the current in it changes from —28.0 mA to +25.0 mA in 12.0 ms?

other, whose

radii are r; and r, (r, < r;) and whose turns per unit length

are ny and n,. . (I) A small thin coil with N, loops, each of area A,, is placed inside a long solenoid, near its center. The solenoid has N; loops in its length £ and has area A;. Determine the mutual inductance as a function of 6, the angle between the plane of the small coil and the axis of the solenoid. . (IIT) A long straight wire and a small rectangular wire loop lie in the same plane, Fig. 30-25. Determine the mutual inductance in terms "_fl_’{ of {1, ¢, and w. Assume the wire is very long compared to £;,4, and w, and that the rest of its circuit is very far away compared to £y, ¢,, T! and w.

Self-Inductance

. (IT) An

air-filled

cylindrical

inductor

has

2800

turns, and

it is 2.5cm in diameter and 21.7 cm long. (a) What is its inductance? (b) How many turns would you need to generate the same inductance if the core were filled with iron of magnetic permeability 1200 times that of free space? . (I) A coil has 3.25-Q) resistance and 440-mH inductance. If the current is 3.00A and is increasing at a rate of 3.60 A/s, what is the potential difference across the coil at this moment? 10. (IT) If the outer conductor of a coaxial cable has radius 3.0 mm, what should be the radius of the inner conductor so that the inductance per unit length does not exceed 55 nH per meter? 11.

(IT) To demonstrate the large size of the henry unit, a physics professor wants to wind an air-filled solenoid with self-inductance of 1.0 H on the outside of a 12-cm diameter plastic hollow tube using copper wire with a 0.81-mm diameter. The solenoid is to be tightly wound with each turn touching its neighbor (the wire has a thin insulating layer on its surface so the neighboring turns are not in electrical contact). How long will the plastic tube need to be and how many kilometers of copper wire will be required? What will be the resistance of this solenoid?

Problems

805

12. (IT) The wire of a tightly wound solenoid is unwound and used to make another tightly wound solenoid of 2.5 times the diameter. By what factor does the inductance change? 13. (IT) A toroid has a rectangular cross section as shown in Fig. 30-26. Show that the self-inductance is [.LoNzh n ——In—

L=

27

r

where N is the total number of turns and r{, r,, and h are the dimensions shown in Fig. 30-26. [Hint: Use Ampere’s law to get B as a function of r inside the toroid, and integrate.]

23. (IT) How many time constants does it take for the potential difference across the resistor in an LR circuit like that in Fig. 30-7 to drop to 3.0% of its original value? . (IT) Tt takes 2.56ms for the current in an LR circuit to increase from zero to 0.75 its maximum

25. (IT) (a) Determine the energy stored in the inductor L as a function of time for the LR circuit of Fig. 30-6a. (b) After how many time constants does the stored energy reach 99.9% of its maximum value? 26. (IT) In the circuit of Fig. 30-27, determine the current in

each resistor (I;, I, I;) at the moment closed, (b) a long time

FIGURE 30-26

Problems 13 and 19. cross section, with

N turns carrying a current /.

14. (IT) Ignoring any mutual inductance, what is the equivalent inductance of two inductors connected (a) in series, (b) in parallel?

30-3

Magnetic Energy Storage

15. (I) The magnetic field inside an air-filled solenoid 38.0 cm long and 2.10 cm in diameter is 0.600 T. Approximately how much energy is stored in this field? 16. (I) Typical large

values

after

the

after

it

switch

is

for electric

and

magnetic

fields

attained in laboratories are about 1.0 X 10*V/m and 2.0 T. (a) Determine the energy density for each field and compare. (b) What magnitude electric field would be needed to produce the same energy density as the 2.0-T magnetic field?

17. (IT) What is the energy density at the center of a circular loop of wire carrying a 23.0-A current if the radius of the loop is 28.0 cm? 18. (IT) Calculate the magnetic and electric energy densities at the surface of a 3.0-mm-diameter copper wire carrying a 15-A current. 19. (IT) For the toroid of Fig. 30-26, determine the energy

is

opened,

FIGURE 30-27

— s

Problem 26.

resistor R. At time

¢ = 0,

= Ipe /",

Show

volume

to obtain

the

total

of its N loops.

20. (IT) Determine the total energy stored per unit length in the magnetic field between the coaxial cylinders of a coaxial cable (Fig. 30-5) by using Eq. 30-7 for the energy density and integrating over the volume. 21. (II) A long straight wire of radius R carries current / uniformly distributed across its cross-sectional area. Find the magnetic energy stored per unit length in the interior of this wire. 30-4 LR Circuits (IT) After how many time constants does the current in Fig. 30-6 reach within (a) 5.0%, (b) 1.0%, and (c) 0.10% of its maximum value?

and

Vp = 120V,

&pnax.

mechanical

decays

through L AEEE

that

mine

the

the switch is quickly switched to

position B and the current (which is much greater than R) according to

energy stored in the toroid, which carries a current / in each

this over

|

é

27. (II) (a) In Fig. 30-28a, assume that the switch S has been in position A for sufficient time so that a steady current Iy = Vo/R flows through the resistor R. At time ¢ = 0, the switch is quickly switched to position B and the current through R decays according to I = Iye™"/". Show that th maximum emf &, induced in the inductor during this time period equals the battery voltage V;. (b) In Fig. 30-28b, assume that the switch has been in position A for sufficient time so that a steady current Iy = V;/R flows through the

density in the magnetic field as a function of r (r; < r < ry) integrate

Ry

(d) after a long time?

the maximum emf &,.x induced in the inductor during this time period is (R'/R)Vy. If R' = 55R

and

(a) the switch is

L

closed. After the switch has been closed for a long time, and then reopened, what is each current (c) just

A toroid of rectangular

value. Determin

(a) the time constant of the circuit, (b) the resistance of the circuit if L = 31.0 mH.

R

deter-

[When

switch

®

a

B

is

opened, a high-resistance air gap is created, which is modeled as R’ here. This Problem illustrates why high-voltage sparking can occur if a currentcarrying inductor is

suddenly cut off from its

power source. ]

Vo L A

R

R’

22.

806

CHAPTER 30

FIGURE 30-28 Problem 27.

Inductance, Electromagnetic Oscillations, and AC Circuits

0

resistor

R’

28. (IT) You want to turn on the current through a coil of selfinductance L in a controlled manner, so you place it in series r

with a resistor R = 2200(), a switch, and a dc voltage source Vy = 240V. After closing the switch, you find that

the current through the coil builds up to its steady-state value with a time constant 7. You are pleased with the current’s steady-state value, but want 7 to be half as long. What new values should you use for R and V;? 29. (IT) A 12-V battery has been connected to an LR circuit for sufficient time so that a steady current flows through the resistor

R = 22k}

and inductor

L = 18 mH.

At

t =0,

the battery is removed from the circuit and the current decays exponentially through R. Determine the emf & across the inductor as time f increases. At what time is & greatest and what is this maximum value (V)? 30. (IIT) Two tightly wound solenoids have the same length and circular cross-sectional area. But solenoid 1 uses wire that is 1.5 times as thick as solenoid 2. (¢) What is the ratio of their inductances? (b) What is the ratio of their inductive time constants (assuming no other resistance in the circuits)? 30-5 LC Circuits and Oscillations 31 (I) The variable capacitor in the tuner of an AM radio has a capacitance of 1350 pF when the radio is tuned to a station at 550 kHz. () What must be the capacitance for a station at 1600 kHz? (b) What is the inductance (assumed constant)? Ignore resistance. 32. (I) (a) If the initial conditions of an LC circuit were [ = [ and Q=0 at t=0, write Q as a function of time. (b) Practically, how could you set up these initial conditions? . (II) In some experiments, short distances are measured by using capacitance. Consider forming an LC circuit using a parallel-plate capacitor with plate area A, and a known inductance L. (a) If charge is found to oscillate in this circuit at frequency f = w/2m when the capacitor plates are

separated

by

distance

x, show

that

x = 4m’Ae(f°L.

(b) When the plate separation is changed by Ax, the circuit’s oscillation frequency will change by Af. Show that Ax/x =~ 2(Af/f). (c) If f is on the order of 1 MHz and can be measured to a precision of Af = 1Hz, with what percent accuracy can x be determined? Assume fringing effects at the capacitor’s edges can be neglected. . (II) A 425-pF capacitor is charged to 135V and then quickly connected to a 175-mH inductor. Determine (a) the frequency of oscillation, (b) the peak value of the current, and (c) the maximum energy stored in the magnetic field of the inductor. 35. (I) At t =0, let Q =0Qpy, and [ =0 in an LC circuit. (a) At the first moment when the energy is shared equally by the inductor and the capacitor, what is the charge on the capacitor? (b) How much time has elapsed (in terms of the period T)?

30-6 36.

37

LC Oscillations with Resistance

(IT) A damped

LC circuit loses 3.5% of its electromagnetic

energy per cycle to thermal energy. If C = 1.00 uF, what is the value of R?

L = 65mH

and

(IT) In an oscillating LRC circuit, how much time does it take for the energy stored in the fields of the capacitor and inductor to fall to 75% of the initial value? (See Fig. 30-13; assume R Electromagnetic Waves CONTENTS 31-1

Changing Electric Fields

Produce Magnetic Fields; Ampere’s Law and

Displacement Current Gauss’s Law for Magnetism Maxwell’s Equations Production of Electromagnetic Waves

Electromagnetic Waves, and

Their Speed, Derived from Maxwell’s Equations 31-6

31-7 31-8 31-9 31-10

812

CHAPTER-OPENING

QUESTION—Guess

now!

Which of the following best describes the difference between radio waves and X-rays? (a) X-rays are radiation while radio waves are electromagnetic waves. (b) Both can be thought of as electromagnetic waves. They differ only in wavelength and frequency. (¢) X-rays are pure energy. Radio waves are made of fields, not energy. (d) Radio waves come from electric currents in an antenna. X-rays are not related to electric charge. (e)

The fact that X-rays can expose film, and radio waves cannot, means they are

fundamentally different.

Light as an Electromagnetic

Wave and the Electromagnetic Spectrum Measuring the Speed of Light Energy in EM Waves; the Poynting Vector Radiation Pressure Radio and Television; Wireless Communication

he culmination of electromagnetic theory in the nineteenth century was the prediction, and the experimental verification, that waves of electromagnetic fields could travel through space. This achievement opened a whole new world of communication: first the wireless telegraph, then radio and television, and more recently cell phones, remote-control devices, wi-fi, and Bluetooth. Most important was the spectacular prediction that visible light is an electromagnetic wave.

“

The theoretical prediction of electromagnetic waves was the work of the Scottish physicist James Clerk Maxwell (1831-1879; Fig. 31-1), who unified, in one magnificent theory, all the phenomena of electricity and magnetism. The development of electromagnetic theory in the early part of the nineteenth f“ ntury by Oersted, Ampere, and others was not actually done in terms of electric and magnetic fields. The idea of the field was introduced somewhat later by Faraday, and was not generally used until Maxwell showed that all electric and magnetic phenomena could be described using only four equations involving electric and magnetic fields. These equations, known as Maxwell’s equations, are the basic equations for all electromagnetism. They are fundamental in the same sense that Newton’s three laws of motion and the law of universal gravitation are for mechanics. In a sense, they are even more fundamental, since they are consistent with the theory of relativity (Chapter 36), whereas Newton’s laws are

not. Because all of electromagnetism is contained in this set of four equations,

Maxwell’s equations are considered one of the great triumphs of human intellect. Before we discuss Maxwell’s equations and electromagnetic waves, we first

=~ FIGURE 31-1

James Clerk Maxwell

(1831-1879).

need to discuss a major new prediction of Maxwell’s, and, in addition, Gauss’s law

for magnetism.

31-1

Changing Electric Fields Produce Magnetic Fields; Ampere’s Law and Displacement Current

Ampere’s Law That a magnetic field is produced by an electric current was discovered Oersted, and the mathematic relation is given by Ampere’s law (Eq. 28-3): r

%Bdl

=

by

Mo lendl-

Is it possible that magnetic fields could be produced in another way as well? For if a changing magnetic field produces an electric field, as discussed in Section 29-7, then perhaps the reverse might be true as well: that a changing electric field will produce a magnetic field. If this were true, it would signify a beautiful symmetry in nature. To back up this idea that a changing electric field might produce a magnetic field, we use an indirect argument that goes something like this. According to Ampere’s law, we divide any chosen closed path into short segments df, take the

dot product of each df with the magnetic field B at that segment, and sum

FIGURE 31-2 Ampere’s law applied to two different surfaces

bounded by the same closed path.

(integrate) all these products over the chosen closed path. That sum will equal times the total current [ that passes through a surface bounded by the path of the

=~ FIGURE 31-3

(Section 28-4), we imagined the current as passing through the circular area enclosed by our circular loop, and that area is the flat surface 1 shown in Fig. 31-2.

passes through surface 1, but no ~ conduction current passes through

line integral. When we applied Ampere’s law to the field around a straight wire

However, we could just as well use the sackshaped surface 2 in Fig. 31-2 as the surface for Ampere’s law, since the same current / passes through it.

Now consider the closed circular path for the situation of Fig. 31-3, where a

capacitor is being discharged. Ampere’s law works for surface 1 (current I passes through surface 1), but it does not work for surface 2, since no current passes through surface 2_. Tllere is a magnetic field around

the wire, so the left side of

Ampere’s law ([B-dl) is not zero; yet no current flows through surface 2, so the right side of Ampere’s law is zero. We seem to have a contradiction of Ampere’s law. There is a magnetic field present in Fig. 31-3, however, only if charge is flowing to or away from the capacitor plates. The changing charge on the plates means that the electric field between the plates is changing in time. Maxwell resolved the rroblem of no current through surface 2 in Fig. 31-3 by proposing that there needs 0 be an extra term on the right in Ampere’s law involving the changing electric field.

SECTION 31-1

A capacitor

discharging. A conduction current

i‘:rfa?e 2,‘?“ extra term is needed in [peres taw.

_

Closed

'

Lc: Surface 1 Surface 2

Changing Electric Fields Produce Magnetic Fields; Ampére’s Law and Displacement Current

813

Closed

_

Su tface ]

Surface 2

Let us see what this term should be by determining it for the changing electric

field between the capacitor plates in Fig. 31-3. The charge Q on a capacitor of capacitance C is Q = CV where V is the potential difference between the plates (Eq. 24-1). Also recall that V = Ed (Eq. 23-4) where d is the (small) separatio of the plates and E is the (uniform) electric field strength between them, if w_ ignore any fringing of the field. Also, for a parallel-plate capacitor, C = €, Ald, where A is the area of each plate (Eq. 24-2). We combine these to obtain: Q

FIGURE 31-3

(repeated) See text.

=

CV

A

=

(Ed)

=

EoAE.

If the charge on each plate changes at a rate dQ/dt, the electric field changes at a proportional rate. That is, by differentiating this expression for Q, we have: dQ

dE

dt

Ay

Now dQ/dt is also the current / flowing into or out of the capacitor: [ = where

@

= EA

dQ

a4

dE

ddy,

T

ar

is the electric flux through the closed path (surface 2 in Fig. 31-3).

In order to make Ampere’s

law work

for surface 2 in Fig. 31-3, as well as for

surface 1 (where current / flows), we therefore write: Ampere’s law

(genel;al form)

-

%B'di

dd

= polena + Mo€o Et—fi'

(31-1)

This equation represents the general form of Ampere’s law,” and embodies Maxwell’s idea that a magnetic field can be caused not only by an ordinary electric current, but also by a changing electric field or changing electric flux. Although we arrived at it for a special case, Eq. 31-1 has proved valid in general. The last term on the right in Eq. 31-1 is usually very small, and not easy to measure experimentally. -

Charging capacitor. A 30-pF air-gap capacitor has circular

plates of area A = 100cm? It is charged by a 70-V battery through a 2.0-Q resistor. At the instant the battery is connected, the electric field between the plates is changing most rapidly. At this instant, calculate (a) the current into the plates, and (b) the rate of change of electric field between the plates. (c) Determine the magnetic field induced between the plates. Assume E is uniform between the plates at any instant and is zero at all points beyond the edges of the plates. APPROACH In Section 26-5 we discussed RC circuits and saw that the charge on a capacitor being charged, as a function of time, is

Q = CVy(1 — ek9),

where Vj is the voltage of the battery. To find the current at ¢ = 0, we differentiate this and substitute the values V;, = 70V, C = 30pF, R = 2.0Q. SOLUTION (@) We take the derivative of Q and evaluate it at t = 0:

a0

dt t=0 =0 This is the rate at which charge accumulates on the capacitor and equals the current flowing in the circuit at ¢t = 0. (b) The electric field between two closely spaced conductors is given by (Eq. 21-8)

E=92 .94

as we saw in Chapter 21 (see Example 21-13). "Actually, there is a third term on the right for the case when a magnetic field is produced b

magnetized materials. This can be accounted for by changing u to u, but we will mainly be interestew in cases where no magnetic material is present. In the presence of a dielectric, € is replaced by e = Ke (see Section 24-5).

814

CHAPTER 31

Maxwell’s Equations and Electromagnetic Waves

Hence

A (face of

o

SASIAE

dt

f

A

€A

(885 % 1072C%*N-m?)(1.0

=4.0 X 104 V/m-s

X 102m?)

‘C’?‘igimr

'

(c) Although we will not prove it, we might expect the lines of B, because of symmetry, to be circles, and to be perpendicular to E, as shown in Fig. 31-4; this is the same symmetry we saw for the inverse situation of a changing magnetic field producing an electric field (Section 29-7, see Fig. 29-27). To determine the magnitude of B between the plates we apply Ampere’s law, Eq. 31-1, with the current /,,q = 0: =

éBdi

=

)

Mp€p

gt—g"

E (out of papér)

We choose our path to be a circle of radius r, centered at the center of the plate, and thus following a magnetic field line such as the one shown in Fig. 31-4. For

r = ry (the radius of plate) the flux through a circle of radius r is E (7rr?) since £ .

.

.

.

2

.

is assumed uniform between the plates at any moment. So from Ampere’s law we have d

B(2mr)

= wpeg 5 (mr®E) Mo

Hence B

We assume

=

Mo€o

dE

[rSrO]

E = 0 for r > r,, so for points beyond the edge of the plates all the

within

Ampere’s law gives B(2mr)

=

the plates

B

=

Mo€oty

and

®

= Emr].

Thus

dE

Hvoeor(z) _KE

2r dt B has its maximum value at rp="VA/m =56cm),is =

(area = 7rd)

Mofio%(fl"%E)

Ho€omrG

or

Bmax

opnciior. Bheisshplaterymiol

;i (oward viewer: lines of B are circles. (Example 31-1.)

€ €o Trrzd—Edt

_OzordT'

flux is contained A

FIGURE 31-4 Frontal view of a circular plate of a parallel-plate

[rZrO] r = r, which, from either relation above (using

dE

_q%d_t

=} (4m X 107 T-m/A)(8.85 x 1072CYN-m?)(5.6 X 10 2m)(4.0 X 10" V/m-s) =12

X 107*T.

This is a very small field and lasts only briefly (the time constant RC = 6.0 X 107'!s)

|and so would be very difficult to measure.

Let us write the magnetic field B outside the capacitor plates of Example 31-1 in terms of the current 7 that leaves the plates. The electric field between the plates is

E =0/eg = Q/(egA),

as we saw in part b, so dE/dt = I/(epA). Hence B for

r > ryis,

B

-

Mo€ors dE 2r

dt

_

po€ors 2r

I

egmr}

_ Mol 2mr

This is the same formula for the field that surrounds a wire (Eq. 28-1). Thus the B ield outside the capacitor is the same as that outside the wire. In other words, che magnetic field produced by the changing electric fiecld between the plates is the same as that produced by the current in the wire.

SECTION 31-1

Changing Electric Fields Produce Magnetic Fields; Ampére’s Law and Displacement Current

815

Displacement Current Maxwell interpreted the second term on the right in Eq. 31-1 as being equivalent to an electric current. He called it a displacement current, /. An ordinary current / is then called a conduction current. Ampere’s law can then be written -

$ib-dl = pall + Ip)e

(31-2)

Iy

(31-3)

where Displacement current

=

€n——"

O dt

The term “displacement current” was based on an old discarded theory. Don’t let it confuse you: I, does not represent a flow of electric charge’, nor is there a displacement.

31-2

Gauss’s Law for Magnetism

We are almost in a position to state Maxwell’s equations, but first we need to discuss the magnetic equivalent of Gauss’s law. As we saw in Chapter 29, for a magnetic field B the magnetic flux ® 5 through a surface is defined as

b, = fB-dK =

where the integral is over the area of either an open or a closed surface. The magnetic flux through a closed surface—that is, a surface which completely encloses a volume—is written

, = jgfi-d&

-

In the electric case, we saw in Section 22-2 that the electric flux ®, through & closed surface is equal to the total net charge Q enclosed by the surface, divided by €, (Eq. 22-4):

%E'dfi - £€p This relation is Gauss’s law for electricity.

We can write a similar relation for the magnetic flux. We have seen, however,

that in spite of intense searches, no isolated magnetic poles (monopoles)—the magnetic equivalent of single electric charges—have ever been observed. Hence, Gauss’s law for magnetism is

3@3@2& = 0. FIGURE 31-5 Magnetic field lines for a bar magnet.

(31-4)

In terms of magnetic field lines, this relation tells us that as many lines enter the

enclosed

volume

as leave it. If, indeed, magnetic

monopoles

do not exist, then

there are no “sources” or “sinks” for magnetic field lines to start or stop on, corresponding to electric field lines starting on positive charges and ending on negative charges. Magnetic field lines must then be continuous. Even for a bar magnet, a magnetic field B exists inside as well as outside the magnetic material, and the lines of B are closed loops as shown in Fig. 31-5. "The interpretation of the changing electric field as a current does fit in well with our discussion in Chapter 30 where we saw that an alternating current can be said to pass through a capacitor (althoug] charge doesn’t). It also means that Kirchhoff’s junction rule will be valid even at a capacitor pla® conduction current flows into the plate, but no conduction current flows out of the plate—instead « “displacement current” flows out of one plate (toward the other plate).

816

CHAPTER 31

Maxwell's Equations and Electromagnetic Waves

™

31-3 Maxwell’s Equations With the extension of Ampere’s law given by Eq. 31-1, plus Gauss’s law for agnetism (Eq. 31-4), we are now ready to state all four of Maxwell’s equations. we have seen them all before in the past ten Chapters. In the absence of dielectric or magnetic materials, Maxwell’s equations are:

SEE-dK 48

(31-5a)

3§fi-d;&

(31-5b) b

|

~i

< fi;B-di

I

= Il

e

jflE-d

o

€9

%

ao wol + [.L()eoTE'

(31-5¢)

MAXWELL’S

EQUATIONS

_

(31-5d)

The first two of Maxwell’s equations are the same as Gauss’s law for electricity

(Chapter 22, Eq. 22-4) and Gauss’s law for magnetism (Section 31-2, Eq. 31-4).

The third is Faraday’s law (Chapter 29, Eq. 29-8) and the fourth is Ampere’s law as modified by Maxwell (Eq. 31-1). (We dropped the subscripts on Q.4 and 1, for simplicity.) They can be summarized in words: (1) a generalized form of Coulomb’s law relating electric field to its sources, electric charges; (2) the same for the magnetic field, except that if there are no magnetic monopoles, magnetic field lines are continuous—they do not begin or end (as electric field lines do on charges); (3) an electric field is produced by a changing magnetic field; (4) a magnetic field is produced by an electric current or by a changing electric field. r Maxwell’s equations are the basic equations for all electromagnetism, and e as fundamental as Newton’s three laws of motion and the law of universal gravitation. Maxwell’s equations can also be written in differential form; see Appendix E. In earlier Chapters, we have seen that we can treat electric and magnetic fields separately if they do not vary in time. But we cannot treat them independently if they do change in time. For a changing magnetic field produces an electric field; and a changing electric field produces a magnetic field. An important outcome of these relations is the production of electromagnetic waves.

31-4 Production of Electromagnetic Waves A magnetic field will be produced in empty space if there is a changing electric field. A changing magnetic field produces an electric field that is itself changing. This changing electric field will, in turn, produce a magnetic field, which will be changing, and so it too will produce a changing electric field; and so on. Maxwell found that the net result of these interacting changing fields was a wave of electric and magnetic fields that can propagate (travel) through space! We now examine, in a simplified way, how such electromagnetic waves can be produced. Consider two conducting rods that will serve as an “antenna” (Fig. 31-6a). Suppose these two rods are connected by a switch to the opposite terminals of a battery. When the switch is closed, the upper rod quickly becomes positively charged and the lower one negatively charged. Electric field lines are formed as indicated by the lines in Fig. 31-6b. While the charges are flowing, a current exists whose direction is indicated by the black arrows. A magnetic field is therefore

produced near the antenna. The magnetic field lines encircle the rod-like antenna

# -

S J _[/

@)

FIGURE 31-6 Fields produced by charge flowing into conductors. It

(1o time for the F and B fields to

(;avel outward to distant points. The

and therefore, in Fig. 31-6, B points into the page (®) on the right and out of the fjelds are shown to the right of the age (O) on the left. How far out do these electric and magnetic fields extend? In antenna, but they move out in all ae static case, the fields extend outward indefinitely far, However, when the directions, symmetrically about the switch in Fig. 31-6 is closed, the fields quickly appear nearby, but it takes time for ~ (vertical) antenna. them to reach distant points. Both electric and magnetic fields store energy, and this energy cannot be transferred to distant points at infinite speed. SECTION 31-4 817

charges on two conductors (the

Now we look at the situation of Fig. 31-7 where our antenna is connected to an ac generator. In Fig. 31-7a, the connection has just been completed. Charge starts building up and fields form just as in Fig. 31-6. The + and — signs in Fig. 31-7a indicate the net charge on each rod at a given instant. The black arrowsem, indicate the direction of the current. The electric field is represented by the rec. lines in the plane of the page; and the magnetic field, according to the right-hand rule, is into (®) or out of (©) the page, in blue. In Fig. 31-7b, the voltage of the ac generator has reversed in direction; the current is reversed and the new magnetic field is in the opposite direction. Because the new fields have changed direction, the old lines fold back to connect up to some of the new lines and form closed loops as shown.” The old fields, however, don’t suddenly disappear; they are on their way to distant points. Indeed, because a changing magnetic field produces an electric field, and a changing electric field produces a magnetic field, this combination of changing electric and magnetic fields moving outward is self-supporting, no longer depending on the antenna charges. The fields not far from the antenna, referred to as the near field, become quite complicated, but we are not so interested in them. We are instead mainly interested in the fields far from the antenna (they are generally what we detect), which we refer to as the radiation field, or far field. The electric field lines form loops, as shown in Fig. 31-8, and continue moving outward. The magnetic field lines also form closed loops, but are not shown since they are perpendicular to the page. Although the lines are shown only on the right of the source, fields also travel in other directions. The field strengths are greatest in directions perpendicular to the e i oscillating charges; and Fhey. drop to zero along the direction of oscillation—above

antenna) connected to an ac source

and below the antenna in Fig, 31-8.

11

J

T

FIGURE 31-7 Sequence showing electric and magnetic fields that pigln spread outward from oscillating (see the text).

FIGURE 31-8 (a) The radiation fields (far from the antenna) produced by a sinusoidal signal on the antenna. The red closed loops represent electric field lines. The magnetic field lines, perpendicular to the page and represented by blue %) and ©, also form closed loops. (b) Very far from the antenna the wave fronts (field lines) are essentially flat over a fairly large area, and are referred to as plane waves.

(b)

The magnitudes of both E and B in the radiation field are found to decrease with distance as 1/r. (Compare this to the static electric field given by Coulomb’s law where E decreases as 1/r%) The energy carried by the electromagnetic wave is proportional (as for any wave, Chapter 15) to the square of the amplitude, E? or B?, as will be discussed further in Section 31-8. So the intensity of the wave decreases as 1/r%, which we call the inverse square law. Several things about the radiation field can be noted from Fig. 31-8. First, the electric and magnetic fields at any point are perpendicular to each other, and to the direction of wave travel. Second, we can see that the fields alternate in direction (B is into the page at some points and out of the page at others; E points up at some points and down at others). Thus, the field strengths vary from a maximum in one direction, to zero, to a maximum in the other direction. The electric and magnetic fields are “in phase”: that is, they each are zero at the same points and reach their maxima at the same points in space. Finally, very far from the antenna (Fig. 31-8b) the field lines are quite flat over a reasonably large area, and the waves are referred to as plane waves. If the source voltage varies sinusoidally, then the electric and magnetic field strengths in the radiation field will also vary sinusoidally. The sinusoidal character of the waves is diagrammed in Fig. 31-9, which shows the field directions and magnitudes plotted as a function of position. Notice that B and E are perpendicular to each other and to the direction of travel (= the direction of the wave velocity V) “ The direction of ¥ can be had from a right-hand rule using E X B: fingers along E, then along B, gives v along thumb. 818

CHAPTER 31

"We are considering waves traveling through empty space. There are no charges for lines of E to start

or stop on, so they form closed loops. Magnetic field lines always form closed loops.

E Direction of motion / of wave

We call these waves electromagnetic (EM) waves. They are transverse waves

FIGURE 31-9

Electric and

because the amplitude is perpendicular to the direction of wave travel. However, = magnetic field strengths in an_ EM waves are always waves of fields, not of matter (like waves on water or a ¢lectromagnetic wave. E and B are

rope). Because they are fields, EM waves can propagate in empty space.

As we have seen, EM waves are produced by electric charges that are oscillating,

and hence are undergoing acceleration. In fact, we can say in general that

atright angles to each other. The

~ €Ptire patternmoves in a direction pexpendiulax to bath K and B.

accelerating electric charges give rise to electromagnetic waves. Electromagnetic waves can be produced in other ways as well, requiring description at the atomic and nuclear levels, as we will discuss later. EXERCISE A At a particular instant in time, a wave has its electric field pointing north and its magnetic field pointing up. In which direction is the wave traveling? (a) South, (b) west, (c) east, (d) down, (e) not enough information.

-

31-5

Electromagnetic Waves, and

Their Speed, Derived from Maxwell’s Equations

Let us now examine how the existence of EM waves follows from Maxwell’s equations. We will see that Maxwell’s prediction of the existence of EM waves was startling. Equally startling was the speed at which they were predicted to travel. We begin by considering a region of free space, where there are no charges or conduction currents—that is, far from the source so that the wave fronts (the field lines in Fig. 31-8) are essentially flat over a reasonable area. We call them plane waves, as we saw, because at any instant E and B are uniform over a reasonably large plane perpendicular to the direction of propagation. We choose a coordinate system, so that the wave is traveling in the x direction with velocity v = vi, with E parallel to the y axis and B parallel to the z axis, as in Fig. 31-9. Maxwell’s equations, with Q = [/ = 0, become

%F:-d&

= 0

(31-6a)

%fi-d[fi

=0

(31-6b)

Lo E-di = e

%B'dl

ddy g dd

(31-6¢)

= “OEOETE”

(31-64)

otice the beautiful symmetry of these equations. The term on the right in the last cquation, conceived by Maxwell, is essential for this symmetry. It is also essential if electromagnetic waves are to be produced, as we will now see.

SECTION 31-5

Electromagnetic Waves, and Their Speed, Derived from Maxwell's Equations

819

If the wave is sinusoidal with wavelength A and frequency f, then, as we saw in Chapter 15, Section 15-4, such a traveling wave can be written as

E = E, = Eysin(kx — of) B = B, = Bysin(kx — t)

Al

-7

A

where

k = 27'” E FIGURE 31-10

Applying Faraday’s

law to the rectangle (Ay)(dx).

© = 2xf,

and

fA = f;(’- = v,

(31-8)

with v being the speed of the wave. Although visualizing the wave as sinusoidal is helpful, we will not have to assume so in most of what follows. Consider now a small rectangle in the plane of the electric field as shown in Fig. 31-10. This rectangle has a finite height Ay, and a very thin width which we

take to be the infinitesimal distance dx. First we show that E, B, and ¥ are in the

orientation shown by applying Lenz’s law to this rectangular loop. The changing magnetic flux through this loop is related to the electric field around the loop by Faraday’s law (Maxwell’s third equation, Eq. 31-6c). For the case shown, B through the loop is decreasing in time (the wave is moving to the right). So the electric field must be in a direction to oppose this change, meaning E must be greater on the right side of the loop than on the left, as shown (so it could produce a counterclockwise current whose magnetic field would act to oppose the change in ® ;—but of course there is no current). This brief argument shows that the orientation of E, B, and ¥ are in the correct relation as shown. That is, v is in the direction of E X B. Now let us apply Faraday’s law, which is Maxwell’s third equation (Eq. 31-6c),

to the rectangle of height Ay and width dx shown in Fig. 31-10. First we consider

$E-dl. Along the short top and bottom sections of length dx, E is perpendicular

to d, so E-df = 0. Along the vertical sides, we let E be the electric field along the left side, and on the right side where it will be slightly larger, it is E + P Thus, if we take our loop counterclockwise,

3@17:-(12 = (E + dE) Ay — E Ay = dE Ay. For the right side of Faraday’s law, the magnetic flux through the loop changes as ddy dB i T dx Ay, since the area of the loop, (dx)(Ay), is not changing. Thus, Faraday’s law gives us dE Ay or

=

— Z—tB dx Ay

dEdx _ _dBdt

Actually, both £ and B are functions of position x and time . We should therefore use partial derivatives:

i

(31-9)

where dE/dx means the derivative of E with respect to x while ¢ is held fixed, and dB/at is the derivative of B with respect to ¢ while x is kept fixed. We can obtain another important relation between E and B in addition to Eq. 31-9. To do so, we consider now a small rectangle in the plane of B, whose length and width are Az and dx as shown in Fig. 31-11. To this rectangular loop we apply Maxwell’s fourth equation (the extension of Ampere’s law), Eq. 31-6d: FIGURE 31-11 Applying Maxwell’s fourth equation to the rectangle

(Az)(dx).

820

CHAPTER 31

= %B-d_l'

where we have taken

=

ad ,quO—E di

I = 0 since we assume the absence of conduction currents.

Maxwell’s Equations and Electromagnetic Waves

fl

Along the short sides (dx), B-dl is zero since B is perpendicular to df. Along the longer sides (Az), we let B be the magnetic field along the left side of length Az, and B + dB be the field along the right side. We again integrate counterclockwise, so

= BAz - (B + dB) Az = —dB Az

%fi-di

(.

The right side of Maxwell’s fourth equation is

dd

dE

,I.L()E()d_t' dx Az.

=

/.L()Eod_t'g

Equating the two expressions, we obtain _dB

AZ

=

M€

dE dt

dx

AZ

or

dB

oE

ax

ety

(31-10)

where we have replaced dB/dx and dE/dt by the proper partial derivatives as before. We can use Egs. 31-9 and 31-10 to obtain a relation between the magnitudes of E and B, and the speed v. Let E and B be given by Eqs. 31-7 as a function of x and t. When we apply Eq. 31-9, taking the derivatives of E and B as given by Egs. 31-7, we obtain kEycos(kx — ot)

or

Eo

=

wBycos(kx — wt)

w

—=—='v’

B,

since v = w/k

k

(see Eq.31-8 or 15-12). Since E and B are in phase, we see that E

rand B are related by

% —

(31-11)

at any point in space, where v is the velocity of the wave. Now we apply Eq. 31-10 to the sinusoidal fields (Egs. 31-7) and we obtain kBycos(kx — wt)

or

=

pgeqwEycos(kx — wt)

By _ Pofge . o E()

k

/"LOO'

We just saw that By/E, = 1/v, so

N

Mo €o

-

Solving for » we find v

=c¢

=

1 V € Mo

)

(31-12)

where c is the special symbol for the speed of electromagnetic waves in free space. We see that c is a constant, independent of the wavelength or frequency. If we put in values for €, and uy we find C

(8

=

1

Veoro

—i

1

/(885 X 1072C%/N-m?)(4m x 107 T-m/A)

= 3.00 X 108m/s. This is a remarkable result. For this is precisely equal to the measured speed of light!

SECTION 31-5

Electromagnetic Waves, and Their Speed, Derived from Maxwell’s Equations

821

Determining E and B in EM waves. Assume a 60.0-Hz EM

wave is a sinusoidal wave propagating in the z direction with E pointing in the

x direction,

and

E; = 2.00 V/m.

Write

APPROACH

We find A from Af = v = c. Then we use Fig. 31-9 and Egs. 31—’

functions of position and time.

vector

expressions

for

E

and

B

as

and 31-8 for the mathematical form of traveling electric and magnetic fields of an EM wave, SOLUTION The wavelength is

¢

3.00 X 10m/s

===

—————— =500 X

f

From Eq. 31-8 we have

60.0s !

2w

2

m

105m.

k=—=_———72— =126 o

X T Smx10m |

=

27f

=

2w(60.0Hz)

=

From Eq. 31-11 with v = ¢, we find that E 2,00V

B, = =2¢ =

/M

xX10"°m™! exiTm

377rad/s.

_ 667 % 10°T.

3.00 % 108m/s The direction of propagation is that of E X B, as in Fig. 31-9. With E pointing in the x direction, and the wave propagating in the z direction, the y direction. Using Eqgs. 31-7 we find:

B must point in

E = i(2.00V/m)sin[(1.26 X 10°m™)z — (377rad/s)] — A . B = j(6.67 X 10°T)sin[(1.26 X 10°m™)z — (377rad/s)t] * Derivation of Speed of Light (General) We can derive the speed of EM waves without having to assume sinusoidal waves by combining Eqs. 31-9 and 31-10 as follows. We take the derivative, with respec« to ¢ of Eq. 31-10

*B

&’E

stox

002

We next take the derivative of Eq. 31-9 with respect to x:

#E

B

ox?

atox

Since 4*B/dt dx appears in both relations, we obtain

PE o

_o e 1 PE

(31-13a)

By taking other derivatives of Egs. 31-9 and 31-10 we obtain the same relation for B:

7B _ 1 7B at?

B

(31-13b)

Mo€g ax?

Both of Egs. 31-13 have the form of the wave equation for a plane wave traveling in the x direction, as discussed in Section 15-5 (Eq. 15-16):

#Dot _ LoDe

where D stands for any type of displacement. We Eqgs. 31-13 is given by v

=

see that the velocity v for

!

Ho€o

in agreement with Eq. 31-12. Thus we see that a natural outcome of Maxwell’s equations is that E and B obey the wave equation for waves traveling with speec*™ v =1/\/po€o.

It was

on

this

basis

that

Maxwell

electromagnetic waves and predicted their speed.

822

CHAPTER 31

Maxwell's Equations and Electromagnetic Waves

predicted

the

existence

of

31-6

Light as an Electromagnetic Wave and the Electromagnetic Spectrum

.he calculations in Section 31-5 gave the result that Maxwell himself determined: that the speed of EM waves in empty space is given by c =

£

L

B

V €0 to

=

3.00 X 108 m/s,

the same as the measured speed of light in vacuum. Light had been shown some 60 years previously to behave like a wave (we’ll discuss this in Chapter 34). But nobody knew what kind of wave it was. What is it that is oscillating in a light wave? Maxwell, on the basis of the calculated speed of EM waves, argued that light must be an electromagnetic wave. This idea soon came to be generally accepted by scientists, but not fully until after EM waves were experimentally detected. EM waves were first generated and detected experimentally by Heinrich Hertz (1857-1894) in 1887, eight years after Maxwell’s death. Hertz used a spark-gap apparatus in which charge was made to rush back and forth for a short time, generating waves whose frequency was about 10° Hz. He detected them some distance away using a loop of wire in which an emf was produced when a changing magnetic field passed through. These waves were later shown to travel at the speed of light, 3.00 X 10®m/s, and to exhibit all the characteristics of light such as reflection,

refraction, and interference. The only difference was that they were not visible. Hertz’s experiment was a strong confirmation of Maxwell’s theory. The wavelengths of visible light were measured in the first decade of the nineteenth century, long before anyone imagined that light was an electromagnetic

wave. The wavelengths were found to lie between 4.0 X 1077 m and 7.5 X 10" m,

or 400nm to 750nm (1 nm = 10~ m). The frequencies of visible light can be found ’Qsing Eq. 15-1 or 31-8, which we rewrite here: c = Af, (31-14) where f and A are the frequency and wavelength, respectively, of the wave. Here, ¢ is the speed of light, 3.00 X 108 m/s; it gets the special symbol ¢ because of its universality for all EM waves in free space. Equation 31-14 tells us that the

frequencies of visible light are between 4.0 X 10" Hz and 7.5 X 10" Hz. (Recall that 1 Hz = 1cycle per second = 1s7'))

But visible light is only one kind of EM wave. As we have seen, Hertz produced EM waves of much lower frequency, about 10° Hz. These are now called radio waves, because frequencies in this range are used to transmit radio and TV signals. Electromagnetic waves, or EM radiation as we sometimes call it, have been produced or detected over a wide range of frequencies. They are usually categorized as shown in Fig. 31-12, which is known as the electromagnetic spectrum. FIGURE 31-12 Electromagnetic spectrum. 104m

104

Wavelength (m) 3m

106

108

3x108m

1010

1012

Frequency (Hz)

1016

A=7.5x10"m

f

SECTION 31-6

1014

=4x104H

“

3x10712m

-

1018

1020

4.0%x10"7m «

Visible light

7.5x1014

i

-

Light as an Electromagnetic Wave and the Electromagnetic Spectrum

823

Radio waves and microwaves can be produced in the laboratory using electronic equipment (Fig. 31-7). Higher-frequency waves are very difficult to produce electronically. These and other types of EM waves are produced in natural processes,

as emission

from

atoms,

molecules,

and

nuclei

(more

on

this later),.‘

EM waves can be produced by the acceleration of electrons or other charged particles, such as electrons in the antenna of Fig. 31-7. Another example is X-rays, which are produced (Chapter 35) when fast-moving electrons are rapidly decelerated upon striking a metal target. Even the visible light emitted by an ordinary incandescent light is due to electrons undergoing acceleration within the hot filament. We will meet various types of EM waves later. However, it is worth mentioning here that infrared (IR) radiation (EM waves whose frequency is just less than that of visible light) is mainly responsible for the heating effect of the Sun. The Sun emits not only visible light but substantial amounts of IR and UV (ultraviolet) as well. The molecules of our skin tend to “resonate” at infrared frequencies, so it is these that are preferentially absorbed and thus warm us up. We humans experience EM waves differently, depending on their wavelengths: Our eyes detect wavelengths between about 4 X 107" m and 7.5 X 1077 m (visible light), whereas our skin detects longer wavelengths (IR). Many EM wavelengths we don’t detect directly at all.

A

CAUTION

Sound and EM waves

are different

EXERCISE B Return to the Chapter-Opening Question, page 812, and answer it again now. Try to explain why you may have answered differently the first time.

Light and other electromagnetic waves travel at a speed of 3 X 10%m/s. Compare this to sound, which travels (see Chapter 16) at a speed of about 300 m/s in air, a million times slower; or to typical freeway speeds of a car, 30m/s (100 km/h, or 60 mi/h), 10 million times slower than light. EM waves differ from sound waves in another big way: sound waves travel in a medium such as air, and involve motion of air molecules; EM waves do not involve any material—only fields, and they can travel in empty space.

me

Wavelengths of EM waves. Calculate the wavelength (a) of

a 60-Hz EM wave, (b) of a 93.3-MHz FM radio wave, and (c) of a beam of visible*™

red light from a laser at frequency 4.74 X 10" Hz. APPROACH

All of these waves

are electromagnetic waves, so their speed is

¢ = 3.00 X 108 m/s. We solve for A in Eq. 31-14: A = ¢/f.

3.00 X 10°m/s SOLUTION N@r(a)A = =7=c = ————" = 50 10°m, e 50 Xx 10°m or 5000 km. 60 Hz is the frequency of ac current in the United States, and, as we see here, one wavelength stretches all the way across the continental USA. b)A

®)

=

3.00 X 10°m/s

——————

%33 x 10551

=

3. o2

.

The length of an FM antenna is about half this (1), or 13m. 3.00 X 10°m/s )

_(___)

= ———————

4.74 X 10Ms71

= 6.33 X 107"m (=633 nm). (

)

EXERCISE C What are the frequencies of (a) an 80-m-wavelength radio wave, and (b) an

X-ray of wavelength 5.5 X 107! m?

DG

IGNEIET N ESTIMATE | Cell phone

antenna. The antenna of a cell

phone is often ; wavelength long. A particular cell phone has an 8.5-cm-long straight rod for its antenna. Estimate the operating frequency of this phone. APPROACH The basic equation relating wave speed, wavelength, and frequency is ¢ = Af; the wavelength A equals four times the antenna’s length. SOLUTION The antenna is 5A long, so A = 4(8.5cm) = 34cm = 0.34 m. Then

f=c/A=(3.0x10°m/s)/(0.34m) = 8.8 X 10°Hz = 880 MHz.

NOTE Radio antennas are not always straight conductors. The conductor may be a round loop to save space. See Fig. 31-21b.

| EXERCISE D How long should a } A antenna be for an aircraft radio operating at 165 MHz?

824

CHAPTER 31

Maxwell’'s Equations and Electromagnetic Waves

fl

Electromagnetic waves can travel along transmission lines as well as in empty ~ FIGURE 31-13 space. When a source of emf is connected to a transmission line—be it two parallel wires or a coaxial cable (Fig. 31-13)—the electric field within the wire is not set up immediately at all points along the wires. This is based on the same argument f‘ ve used in Section 31-4 with reference to Fig. 31-7. Indeed, it can be shown that if the wires are separated by empty space or air, the electrical signal travels along the wires at the speed ¢ = 3.0 X 10°m/s. For example, when you flip a light switch, the light actually goes on a tiny fraction of a second later. If the wires are in a medium whose electric permittivity is € and magnetic permeability is u (Sections 24-5 and 28-9, respectively), the speed is not given by Eq. 31-12, but by v

=

EXAMPLE 31-5

Coaxial cable.

1 ’\/G[L.

Phone

call time lag. You make a telephone

call from New York to a friend in London. Estimate how long it will take the electrical signal generated by your voice to reach London, assuming the signal is (a) carried on a telephone cable under the Atlantic Ocean, and (b) sent via satellite 36,000 km above the ocean. Would this cause a noticeable delay in either case? APPROACH The signal is carried on a telephone wire or in the air via satellite. In either case it is an electromagnetic wave. Electronics as well as the wire or cable slow things down, but as a rough estimate we take the speed to be ¢ = 3.0 X 108 m/s. SOLUTION The distance from New York to London is about 5000 km.

(a) The time delay via the cable is ¢ = d/c ~ (5 x 10°m)/(3.0 X 108 m/s) = 0.017s.

(b) Via satellite the time would be longer because communications satellites, which are usually geosynchronous (Example 6-6), move at a height of 36,000 km. The signal would have to go up to the satellite and back down, or about 72,000 km. The actual distance the signal would travel would be a little more than this as the signal would go up and down on a diagonal. Thus

(k t=d/c~72%x10m/(3 x 10°m/s) = 0.24s.

NOTE When the signal travels via the underwater cable, there is only a hint of a delay and conversations are fairly normal. When the signal is sent via satellite, the delay is noticeable. The length of time between the end of when you speak and your friend receives it and replies, and then you hear the reply, is about a half second beyond the normal time in a conversation. This is enough to be noticeable, and you have to

adjust for it so you don’t start talking again while your friend’s reply is on the way Eck to you.

EXERCISE E If you are on the phone via satellite to someone only 100 km away, would you hear the same effect? EXERCISE F If your voice traveled as a sound wave, how long would it take to go from New York to London?

31-7 Measuring the Speed of Light Galileo attempted to measure the speed of light by trying to measure the time required for light to travel a known distance between two hilltops. He stationed an

assistant on one hilltop and himself on another, and ordered the assistant to lift the

cover from a lamp the instant he saw a flash from Galileo’s lamp. Galileo measured the time between the flash of his lamp and when he received the light from his assistant’s lamp. The time was so short that Galileo concluded it merely represented human reaction time, and that the speed of light must be extremely high. The first successful determination that the speed of light is finite was made by the Danish astronomer Ole Roemer (1644-1710). Roemer had noted that the carefully measured orbital period of Io, a moon of Jupiter with an average period rof 42.5 h, varied slightly, depending on the relative position of Earth and Jupiter. * He attributed this variation in the apparent period to the change in distance between the Earth and Jupiter during one of Io’s periods, and the time it took light to travel the extra distance. Roemer concluded that the speed of light—though great—is finite.

SECTION 31-7

825

S Fltgl:itl-lsmed # Observer

o mirror

)

Since then a number of techniques have been used to measure the speed of light. Among the most important were those carried out by the American Albert A.

(dEBaly

the in Fig. 31-14 for a series of high-precision experiments carried out from 1880 to 1920s. Light from a source would hit one face of a rotating eight-sided mirrc The reflected light traveled to a stationary mirror a large distance away and back

Stationary \fichelson (1852-1931). Michelson used the rotating mirror apparatus diagrammed i

r L 1ight

source

(Mt. Wilson) ke 35km—» FIGURE 31-14 Michelson’s speedof-light apparatus (not to scale).

again

as shown.

If the

rotating

mirror

was

turning

at just

the

right

rate, the

returning beam of light would reflect from one of the eight mirrors into a small telescope through which the observer looked. If the speed of rotation was only slightly different, the beam would be deflected to one side and would not be seen by the observer. From the required speed of the rotating mirror and the known distance to the stationary mirror, the speed of light could be calculated. In the 1920s, Michelson set up the rotating mirror on the top of Mt. Wilson in southern California and the stationary mirror on Mt. Baldy (Mt. San Antonio) 35 km away. He later measured the speed of light in vacuum using a long evacuated tube. Today the speed of light, ¢, in vacuum is taken as c =

299792458 x 10°m/s,

and is defined to be this value. This means that the standard for length, the meter, is no longer defined separately. Instead, as we noted in Section 1-4, the meter is now formally defined as the distance light travels in vacuum in 1/299,792,458 of a second. We usually round off ¢ to

c = 3.00 X 10°m/s when extremely precise results are not required. In air, the speed is only slightly less.

31—-8 Energy in EM Waves; the Poynting Vector Electromagnetic waves carry energy from one region of space to another. This energy is associated with the moving electric and magnetic fields. In Section 24-4

we saw that the energy

density uy (J/m’)

stored in an electric field E

-

up = g E? (Eq. 24-6). The energy density stored in a magnetic field B, as wé discussed in Section 30-3, is given by up = 3 B*/uo (Eq.30-7). Thus, the total energy stored per unit volume in a region of space where there is an electromagnetic wave is =

lozoE2 o = =

(31-15)

In this equation, E and B represent the electric and magnetic field strengths of the wave at any instant in a small region of space. We can write Eq. 31-15 in terms

of the E field alone, using Egs. 31-11 to obtain

u

L1 = >

€

gy

E° +

>

,

ekE”

(B = E/c)

and 31-12

= g E%

(c = 1/\/ep)

(31-16a)

Note here that the energy density associated with the B field equals that due to the E field, and each contributes half to the total energy. We can also write the energy density in terms of the B field only: BZ

u

=

EOE2

=

EOCZB2

=

p—a

(31—16b)

0

or in one term containing both E and B,

4 = eE Equations Now unit area. The units energy is 826

CHAPTER 31

EB

= egEcB = —2== =

,|2EB.

(31-16¢)

31-16 give the energy density in any region of space at any instant. let us determine the energy the wave transports per unit time per This is given by a vector S, which is called the Poynting vector.] of S are W/m2 The direction of § is the direction in which th transported, which is the direction in which the wave is moving.

" After J. H. Poynting (1852-1914).

Let us imagine the wave is passing through an area A perpendicular to the x axis as shown in Fig. 31-15. In a short time df, the wave moves to the right a distance

dx = cdt

ime dt is ensity u instant. So times the

where c is the wave speed. The energy that passes through A in the

the energy that occupies the volume dV = A dx = Acdt. The energy is u = ey E* where E is the electric field in this volume at the given the total energy dU contained in this volume dV is the energy density u volume: dU = u dV = (€gE*)(Ac dt). Therefore the energy crossing

the area A per time df is S

=

Z;l_—

=

egck?

(31-17)

FIGURE 31-15

t

Since E = cB and

Electromagnetic

wave carrying energy through area A.

¢ = 1/\/eyuyg,

this can also be written:

B? EB S=e0cE2=C—=—Mo Mo

The direction of § is along ¥, perpendicular to E and B, so the Poynting vector § can be written

§ =

(E x B).

Mo

(31-18)

Equation 31-17 or 31-18 gives the energy transported per unit area per unit time at any instant. We often want to know the average over an extended period of time since the frequencies are usually so_high we don’t detect the rapid time variation. If E and B are sinusoidal, then E? = E}/2, just as for electric currents and voltages (Section 25-7), where E, is the maximum value of E. Thus we can write for the magnitude of the Poynting vector, on the average, 3

=

_Eoch

=

__BO

=

2—%’

(31—198)

where B, is the maximum value of B. This time averaged value of § is the intensity, defined as the average power transferred across unit area (Section 15-3). We can also write for the average value of S: —S

E;:s rms B-~

SR

rms

Mo

(

31-19b

)

where Ejyg and By are the rms values (Eps = VE, By = \/?) Fm E and B from the Sun. Radiation from the Sun reaches the Earth (above the atmosphere) at a rate of about 1350J/s-m? (= 1350 W/m?).

Assume that this is a single EM wave, and calculate the maximum values of E and B.

APPROACH We solve Eq. 31-19a

5 = 13501/s-m>

[25

E,

From Eq.31-11,

B,

\/

€oc

B = E/c,

=

for E, in terms of S using

2(13503/s - m?)

V(885 x 10°2C%/N-m?)(3.00 X 10°m/s)

1.01 X 10’ V/m.

Il

SOLUTION

(S = Se,cE})

fl

g

so

101

x 10°V/m

3.00 X 108m/s

=

337 X 1075T.

HNOTE Although B has a small numerical value compared to E (because of the © way the different units for £ and B are defined), B contributes the same energy lt_othe wave as E does, as we saw earlier (Egs. 31-15 and 16).

SECTION 31-8

Energy in EM Waves; the Poynting Vector

827

31-9

Radiation Pressure

If electromagnetic waves carry energy, then we might expect them to also carry

linear momentum. When an electromagnetic wave encounters the surface of ar™ object, a force will be exerted on the surface as a result of the momentum transfe: (F = dp/dt), just as when a moving object strikes a surface. The force per unit area exerted by the waves is called radiation pressure, and its existence was predicted by Maxwell. He showed that if a beam of EM radiation (light, for example) is completely absorbed by an object, then the momentum transferred is Ap

=

radiation fully absorbed

e ¢

(31-20a)

where AU is the energy absorbed by the object in a time At, and c is the speed of light.” If instead, the radiation is fully reflected (suppose the object is a mirror), then the momentum transferred is twice as great, just as when a ball bounces elastically off a surface: radiation

b = mitl,

fully

c

If a surface where a is Using exerted by

reflected

| (31-20b)

absorbs some of the energy, and reflects some of it, then Ap = a AU/c, a factor between 1 and 2. Newton’s second law we can calculate the force and the pressure radiation on the object. The force F is given by

dp

ke dt The average rate that energy is delivered to the object is related to the Poynting vector by dU where A is the cross-sectional area of the object which intercepts the radiation. The radiation pressure P (assuming full absorption) is given by (see Eq. 31-20a)

il

_F

_1dp

Adt

A

1dU

Acdt

S

¢

[ fully

| absorbed |

(3-2la)

If the light is fully reflected, the pressure is twice as great (Eq. 31— 20b):

. DGV

[ fully

25

.

| reflected |

(121l

RIEYE ESTIMATE | Solar pressure. Radiation from the Sun that

reaches the Earth’s surface (after passing through the atmosphere) transports energy at a rate of about 1000 W/m? Estimate the pressure and force exerted by the Sun on your outstretched hand.

APPROACH The radiation is partially reflected and partially absorbed, so let us estimate simply P = S/c. SOLUTION

P

=

5

1000 W/m?

————— =~ 3 X 10°N/m2 C 3 X 108m/s An estimate of the area of your outstretched hand might be about 10cm 20cm, so A = 0.02m? Then the force is

F = PA

NOTE

=

~ (3x10°N/m?)(0.02m?)

by

~ 6 X 10®N.

These numbers are tiny. The force of gravity on your hand, for comparison,

is maybe a half pound, or with m = 0.2kg, mg ~ (0.2kg)(9.8 m/s?) ~ 2 N. The

radiation pressure on your hand is imperceptible compared to gravity.

828

CHAPTER 31

"Very roughly, if we think of light as particles (and we do—see Chapter 37), the force that would be neede(fl to bring such a particle moving at speed c to “rest” (i.e. absorption) is F = Ap/At. But F is also related to energy by Eq. 87, F = AU/Ax, so Ap = F At = AU/(Ax/At) = AU/c where we identify (Ax/At) with the speed of light c.

EXAMPLE 31-8

A solar sail. Proposals have been made to use

the radiation pressure from the Sun to help propel spacecraft around the solar system. (a) About how much force would be applied on a 1km X 1km highly reflective sail, and (b) by how much would this increase the speed of a 5000-kg é spacecraft in one year? (c) If the spacecraft started from rest, about how far would it travel in a year? APPROACH of Example

Pressure P is force per unit area,so F = PA. We use the estimate 31-7, doubling it for a reflecting surface P = 2S/c. We find the

acceleration from Newton’s second law, and assume it is constant, and then find

the speed from v = v, + af. The distance traveled is given by x = lat?, SOLUTION (a) Doubling the result of Example 31-7, the solar pressure is 2S/c =

6 X 10°°N/m% Then the forceis F ~ PA = (6 X 10°°N/m?)(10°m?) ~ 6N.

(b) The acceleration is a ~ F/m ~ (6N)/(5000kg) ~ 1.2 X 10*m/s’. The speed

increase is v — vy = at = (1.2 X 1073 m/s?)(365 days)(24 hr/day)(3600s/hr) ~

4 X 10*m/s (~ 150,000 km/h!).

(c) Starting from rest, this acceleration would

result in a distance of about ar?> ~ 6 X 10'" m in a year, about four times the

Sun-Earth distance. The starting point should be far from the Earth so the Earth’s gravitational force is small compared to 6 N. NOTE A large sail providing a small force over a long time can result in a lot of

motion.

Although you cannot directly feel the effects of radiation pressure, the phenomenon is quite dramatic when applied to atoms irradiated by a finely

focused laser beam. An atom has a mass on the order of 107?" kg, and a laser beam can deliver energy at a rate of 1000 W/m?

@ PHYSICS Optical tweezers

APPLIED

This is the same intensity used in

Example 31-7, but here a radiation pressure of 10°° N/m? would be very significant on a molecule whose mass might be 107 to 10"* kg, It is possible to move atoms and

molecules

around

by steering

them

with

a laser beam,

in a device

called

~optical tweezers.” Optical tweezers have some remarkable applications. They are r' great interest to biologists, especially since optical tweezers can manipulate live microorganisms, and components within a cell, without damaging them. Optical tweezers have been used to measure the elastic properties of DNA by pulling each end of the molecule with such a laser “tweezers.”

31-10

Radio and Television; Wireless Communication L]

L]

Electromagn‘etic waves offer the possibility of transmitting information over long distances. Among the first to realize this and put it into practice was Guglielmo

Marconi (1874-1937) who, in the 1890s, invented and developed wireless communication. With it, messages could be sent at the speed of light without the use of wires. The first signals were merely long and short pulses that could be translated into words by a code, such as the “dots” and “dashes” of the Morse code: they were digital wireless, believe it or not. In 1895 Marconi sent wireless signals a kilometer or two in Italy. By 1901 he had sent test signals 3000 km

@ PHYSICS APPLIED Wireless transmission

across the ocean from Newfoundland, Canada, to Cornwall, England. In 1903 he

sent the first practical commercial messages from Cape Cod, Massachusetts, to England: the London Times printed news items sent from its New York correspondent. 1903 was also the year of the first powered airplane flight by the Wright brothers. The hallmarks of the modern age—wireless communication and flight—date from the same year. Our modern world of wireless communication, including radio, television, cordless phones, cell phones, Bluetooth, wi-fi, and satellite

communication, are simply modern applications of Marconi’s pioneering work. The next decade saw the development of vacuum tubes. Out of this early work rdio and television were born. We now discuss briefly (1) how radio and TV signals are transmitted, and (2) how they are received at home. SECTION 31-10

Radio and Television; Wireless Communication

829

Transmitting Audio

(electrical)

(amplified)

signal

.

Sound

Audio

signal

g Microphone

waves

n

antenna

Modulated

xer

signal

M

RF signal = carrier FIGURE 31-16

e

W

.

FIGURE 31-17

W

Block diagram of a radio transmitter.

Program (audio)

In amplitude

osclizlgtor

The process by which a radio station transmits information (words and music) is outlined in Fig. 31-16. The audio (sound) information is changed into an electrical signal of the same frequencies by, say, a microphone or magnetic read/write head. This electrical signal is called an audiofrequency (AF) signal, since the frequencies are in the audio range (20 to 20,000 Hz). The signal is amplified electronically and is then mixed with a radio-frequency (RF) signal called its carrier frequency, which represents that station. AM radio stations have carrier frequencies from about 530 kHz to 1700 kHz. For example, “710 on your dial” means a station whose carrier frequency is 710 kHz. FM radio stations have much

higher

carrier

frequencies,

between

88 MHz

and

108 MHz.

The

carrier

modulation (AM), the amplitude of

frequencies for broadcast TV stations in the United States lie between 54 MHz

proportion to the audio signal’s

between 470 MHz and 698 MHz.

the carrier signal is made to vary in

amplitude.

F)PHYSICS

APPLIED AM

FIGURE 31-18

and

FM

In frequency

modulation (FM), the frequency of the carrier signal is made to change in proportion to the audio signal’s amplitude. This method is used by FM radio and television.

/‘\/\/\ Program (audio)

and 72 MHz, between 76 MHz and 88 MHz, between 174 MHz and 216 MHz, and

The mixing of the audio and carrier frequencies is done in two ways. In amplitude modulation (AM), the amplitude of the high-frequency carrier wave is made to vary in proportion to the amplitude of the audio signal, as shown in Fig. 31-17. It is called “amplitude modulation” because the amplituc of the carrier is altered (“modulate” means to change or alter). In frequene,

modulation (FM), the frequency of the carrier wave is made to change in proportion to the audio signal’s amplitude, as shown in Fig. 31-18. The mixed signal is amplified further and sent to the transmitting antenna, where the complex mixture of frequencies is sent out in the form of EM waves. In digital communication, the signal is put into a digital form (Section 29-8) which modulates the carrier. A television transmitter works in a similar way, using FM

for audio and AM

for video; both audio and video signals (see Section 23-9) are mixed with carrier

Carrier

Now let us look at the other end of the process, the reception of radio and TV programs at home. A simple radio receiver is diagrammed in Fig. 31-19. M/\N\/\M/\/V\/\W The EM waves sent out by all stations are received by the antenna. The signals the antenna detects and sends to the receiver are very small and Total signal (FM) ~ contain frequencies from many different stations. The receiver selects out a particular RF frequency (actually a narrow range of frequencies) corresponding PHYSICS APPLIED 4 5 particular station using a resonant LC circuit (Sections 30-6 and 30-9). Radio and TV receivers

FIGURE 31-19

Block diagram of a simple radio receiver.

Receiving

antenna

RF al sigh

830

CHAPTER 31

Maxwell's Equations and Electromagnetic Waves

A_“di‘f signa

Loudspeaker

-

A simple way of tuning a station is shown in Fig. 31-20. A particular station is

Antenna

“tuned in” by adjusting C and/or L so that the resonant frequency of the circuit equals that of the station’s carrier frequency. The signal, containing both audio and

carrier frequencies, next goes to the demodulator, or detector (Fig. 31-19), where r‘demodulation” takes place—that is, the RF carrier frequency is separated from the audio signal. The audio signal is amplified and sent to a loudspeaker or headphones. Modern receivers have more stages than those shown. Various means are used to increase the sensitivity and selectivity (ability to detect weak signals and distinguish them from other stations), and to minimize distortion of the original signal.” A television receiver does similar things to both the audio and the video signals. The audio signal goes finally to the loudspeaker, and the video signal to the monitor, such as a cathode ray tube (CRT) or LCD screen (Sections 23-9 and 35-12). One kind of antenna consists of one or more conducting rods; the electric field in the EM

waves

LU

Transistor aniphifies

Tuning circuit

L = FIGURE 31-20 stage of a radio.

i 7]

Simple tuning

-

exerts a force on the electrons in the conductor, causing

them to move back and forth at the frequencies of the waves (Fig. 31-21a). A second type of antenna consists of a tubular coil of wire which detects the magnetic field of the wave: the changing B field induces an emf in the coil (Fig. 31-21b). Antenna rod

A\

2 T

L a&zflna

47 Current produced by electric field

lnduced/'

B N

current

— :

B \

‘

BM wave

To receiver

B

(a)

(b)

To receiver

(TV set)

~IGURE 31-21 Antennas. (a) Electric field of EM wave produces a current in an antenna consisting of straight wire or rods. (b) Changing magnetic field induces an emf and current in a loop antenna. |

A satellite dish (Fig. 31-22) consists of a parabolic reflector that focuses the EM waves onto a “horn,” similar to a concave mirror telescope (Fig. 33-38).

FIGURE 31-22

A satellite dish.

DEUILIREIEEE Tuning a station. Calculate the transmitting wavelength of an FM radio station that transmits at 100 MHz. APPROACH

Radio is transmitted as an EM wave, so the speed is ¢ = 3.0 X 108m/s.

The wavelength is found from Eq. 31-14, A = ¢/f.

SOLUTION The carrier frequency is f = 100 MHz = 1.0 X 10%s7!, so

T

c _ (30x10°m/s)

T (10 x10°5))

=

3.0m

NOTE The wavelengths of other FM signals (88 MHz to 108 MHz) are close to the 3.0-m wavelength of this station. FM antennas are typically 1.5 m long, or about a half wavelength. This length is chosen so that the antenna reacts in a resonant fashion and thus is more sensitive to FM frequencies. AM radio antennas would have to be very long to be either 1A or 1 A. "For FM stereo broadcasting, two signals are carried by the carrier wave. One signal contains frequencies up to about 15 kHz, which includes most audio frequencies. The other signal includes the same range of frequencies, but 19 kHz is added to it. A stereo receiver subtracts this 19,000-Hz signal and distributes the two signals to the left and right channels. The first signal consists of the sum of left and right r"\annels (L + R), so mono radios detect all the sound. The second signal is the difference between left ad right (L — R). Hence the receiver must add and subtract the two signals to get pure left and right signals for each channel.

SECTION 31-10

Radio and Television; Wireless Communication

831

Other EM Wave Communications PHYSICS

APPLIED

Cell phones, radio control, remote control, cable TV, and satellite TV and radio

| Summary

The various regions of the radio-wave spectrum are assigned by governmental agencies for various purposes. Besides those mentioned above, there are “bands” assigned for use by ships, airplanes, police, military, amateurs, satellites and space, and radar. Ce& phones, for example, are complete radio transmitters and receivers. In the U. CDMA cell phones function on two different bands: 800 MHz and 1900 MHz (=1.9 GHz). Europe, Asia, and much of the rest of the world use a different system: the international standard called GSM (Global System for Mobile Communication), on 900-MHz and 1800-MHz bands. The U.S. now also has the GSM option (at 850 MHz and 1.9 GHz), as does much of the rest of the Americas. A 700-MHz band is now being made available for cell phones (it used to carry TV broadcast channels 52-69, now no longer used). Radio-controlled toys (cars, sailboats, robotic animals, etc.) can use various frequencies from 27 MHz to 75 MHz. Automobile remote (keyless) entry may operate around 300 MHz or 400 MHz. Cable TV channels are carried as electromagnetic waves along a coaxial cable (see Fig. 31-13) rather than being broadcast and received through the “air.” The channels are in the same part of the EM spectrum, hundreds of MHz, but some are at frequencies not available for TV broadcast. Digital satellite TV and radio are carried in the microwave portion of the spectrum (12 to 14 GHz and 2.3 GHz, respectively).

James Clerk Maxwell synthesized an elegant theory in which all electric and magnetic phenomena could be described using four equations, now called Maxwell’s equations. They are based on earlier ideas, but Maxwell added one more—that a changing electric field produces a magnetic field. Maxwell’s equations are L 0

%E'dA

e

jgii-d[i

=0

éfz dl

(31-5a)

0

(31-5h)

_ & dt

S éB'dl

= ol

(31-5¢)

+ Ko€o

ddg

(31-5d)

The first two are Gauss’s laws for electricity and for magnetism; the other two are Faraday’s law and Ampere’s law (as extended by Maxwell), respectively. Maxwell’s theory predicted that transverse electromagnetic (EM) waves would be produced by accelerating electric charges, and these waves would propagate through space at the speed of

light c, given by

c =

I Questions

1

V € Mo

__ 3.00 X 108m/s.

(31-12)

1. An electric field E points away from you, and its magnitude is increasing. Will the induced magnetic field be clockwise or counterclockwise? What if E points toward you and is decreasing? 2. What is the direction of the displacement current in Fig. 31-3? (Note: The capacitor is discharging.) 3. Why is it that the magnetic field of a displacement current in a capacitor is so much harder to detect than the magnetic field of a conduction current? 4. Are there any good reasons for calling the term Ho€o dPg/dt in Eq.31-1 as being due to a “current”? Explain.

832

CHAPTER 31

The wavelength A and frequency f of EM waves are related to their speed ¢ by c = Af, (31-14) just as for other waves. The oscillating electric and magnetic fields in an EM wave are perpendicular to each other and to the direction of propagation.

EM waves are waves of fields, not matter, and can propagate in

empty space.

After EM waves were experimentally detected in the lefl 1800s, the idea that light is an EM wave (although of muc.. higher frequency than those detected directly) became generally accepted. The electromagnetic spectrum includes EM waves of a

wide variety of wavelengths, from microwaves and radio waves {0 visible light to X-rays and gamma rays, all of which travel

through space at a speed ¢ = 3.00 X 108 m/s.

The energy carried by EM waves can be described by the

Poynting vector

$

{ [P (31-18) —EXB . which gives the rate energy is carried across unit area per unit ;o when the electric and magnetic fields in an EM wave in =

free space are E and B.

EM waves carry momentum and exert a radiation pressure proportional to the intensity S of the wave.

5. The electric field in an EM wave traveling north oscillates in an east—west plane. Describe the direction of the magnetic field vector in this wave. 6 Is sound an electromagnetic wave? If not, what kind of wave is it? 7 Can EM waves travel through a perfect vacuum? Can sound waves? 8 When you flip a light switch, does the overhead light go on immediately? Explain. 9. Are the wavelengths of radio and television signals long or shorter than those detectable by the human eye?

Maxwell’'s Equations and Electromagnetic Waves

10. What does the wavelength calculated in Example 31-2 tell you about the phase of a 60-Hz ac current that starts at a power plant as compared to its phase at a house 200 km away? 1. When you connect two loudspeakers to the output of a fi stereo amplifier, should you be sure the lead wires are equal in length so that there will not be a time lag between speakers? Explain. 12 In the electromagnetic spectrum, what type of EM wave would hz?ve a wavelength of 10°km; 1km; 1m; 1cm; 1mm; 1 um? 13. Can radio waves have the same frequencies as sound waves (20 Hz-20,000 Hz)? 14. Discuss

how

cordless

telephones

What about cellular telephones?

make

use of EM

waves.

15. Can two radio or TV stations broadcast on the same carrier frequency? Explain. 16. If a radio transmitter has a vertical antenna, should a receiver’s antenna (rod type) be vertical or horizontal to obtain best reception? 17. The carrier frequencies of FM broadcasts are much higher than for AM broadcasts. On the basis of what you learned about diffraction in Chapter 15, explain why AM signals can be detected more readily than FM signals behind low hills or buildings. 18. A lost person may signal by flashing a flashlight on and off using Morse code. This is actually a modulated EM wave. Is it AM or FM? What is the frequency of the carrier, approximately?

f Problems 31-1

B Produced by Changing E

1. (I) Determine the rate at which the electric field changes between the round plates of a capacitor, 6.0 cm in diameter, if the plates are spaced 1.1 mm apart and the voltage across them is ch'anging at a rate of 120 V/s, 2. (I) Calculate the displacement current I, between the square plates, 5.8 cm on a side, of a capacitor if the electric

field is changing at a rate of 2.0 X 10°V/m-s.

3. (IT) At a given instant, a 2.8-A current flows in the wires connected to a parallel-plate capacitor. What is the rate at which the electric field is changing between the plates if the square plates are 1.60 cm on a side? 4. (IT) A 1500-nF capacitor with circular parallel plates 2.0 cm r‘ in diameter is accumulating charge at the rate of 38.0mC/s at some instant in time. What will be the induced magnetic

field strength 10.0 cm radially outward from the center of

the plates? What will be the value of the field strength after the capacitor is fully charged? 5. (II) Show that the displacement current through a parallelplate capacitor can be written I = C dV/dt, where V is the voltage across the capacitor at any instant. 6. (IT) Suppose an air-gap capacitor has circular plates of radius R =25cm

and

separation

d = 1.6mm.

A

76.0-Hz

emf,

€= %coswt, is applied to the capacitor. The maximum displacement current is 35 wA. Determine (@) the maximum conduction current /, (b) the value of &), (¢) the maximum value of d®g/dt between the plates. Neglect fringing. 7. (IIT) Suppose that a circular parallel-plate capacitor has radius Ry, = 3.0cm and plate separation d = 5.0mm. A sinusoidal potential difference V = Vsin(2xft) is applied across the plates, where Vy = 150V and f = 60Hz. (a) In the region between the plates, show that the magnitude of the induced magnetic field is given by B = By(R) cos(2mft), where R is the radial distance from the capacitor’s central axis. (b) Determine the expression for the amplitude By(R) of this time-dependent (sinusoidal) field when R = R, and when R > R;. (c) Plot By(R) in tesla for the range 0=R=10cm. 31-5

EM Waves

8. (I) If the electric field in an EM wave has a peak magnitude

of 0.57 X 1074V/m, what is the peak magnitude of the magnetic field strength?

9. (I) If the magnetic field in a traveling EM wave has a peak magnitude of 12.5nT, what is the peak magnitude of the electric field?

10. (I) In an EM wave traveling west, the B field oscillates vertically and has a frequency of 80.0kHz and an rms strength of 7.75 X 107°T. Determine the frequency and rms strength of the electric field. What is its direction? 11. (IT) The electric field of a plane EM wave is given by Ey = Egcos(kz + wt), Ey = E; = 0. Determine (a) the direction of propagation and (b) the magnitude and direction of B. 12. (IIT) Consider two possible candidates E(x, ) as solutions of the wave equation for an EM wave’s electric field. Let A and o

be constants. Show that (a) E(x,t) = Ae wave equation, and that (b) satisfy the wave equation.

31-6

)" satisfies the

E(x,t) = Ae

@)

does not

Electromagnetic Spectrum

13. (I) What is the frequency of a microwave whose wavelength is 1.50 cm? 14. (I) (a) What is the wavelength of a 25.75 X 10°Hz radar signal? (b) What is the frequency of an X-ray with wavelength 0.12 nm? 15. (I) How long does it take light to reach us from the Sun,

1.50 x 10® km away?

16. (I) An EM wave has frequency 8.56 X 10'* Hz. What is its wavelength, and how would we classify it? 17. (I) Electromagnetic waves and sound waves can have the same frequency. (a) What is the wavelength of a 1.00-kHz electromagnetic wave? (b) What is the wavelength of a 1.00-kHz sound wave? (The speed of sound in air is 341 m/s.) (¢) Can you hear a 1.00-kHz electromagnetic wave? 18. (IT) Pulsed lasers used for science and medicine produce very brief bursts of electromagnetic energy. If the laser light wavelength is 1062 nm (Neodymium-YAG laser), and the pulse lasts for 38 picoseconds, how many wavelengths are found within the laser pulse? How brief would the pulse need to be to fit only one wavelength? 19. (IT) How long would it take a message sent as radio waves from Earth to reach Mars when Mars is (a) nearest Earth,

(b) farthest from Earth? Assume that Mars and Earth are in the same plane and that their orbits around the Sun are

circles (Mars is 230 X 10°km from the Sun).

20. (IT) An electromagnetic wave has an electric field given by E

=

i(225V/m) sin[(0.077m™)z — (2.3 X 107 rad/s)t].

(a) What are the wavelength and frequency of the wave? (b) Write down an expression for the magnetic field. Problems

833

31-7

Speed of Light

21. (IT) What Michelson’s reflect light faces (1/8 of

31-8

33. (IIT) The Arecibo radio telescope in Puerto Rico can detect

is the minimum angular speed at which eight-sided mirror would have had to rotate to into an observer’s eye by succeeding mirror a revolution, Fig. 31-14)?

EM Wave Energy; Poynting Vector

22. (I) The E field in an EM wave has a peak of 26.5mV/m. What is the average rate at which this wave carries energy across unit area per unit time? 23. (IT) The magnetic field in a traveling EM wave has an rms strength of 22.5 nT. How long does it take to deliver 3357J of energy to 1.00 cm? of a wall that it hits perpendicularly? . (IT) How much energy is transported across a 1.00 cm? area per hour by an EM wave whose E field has an rms strength of 32.8 mV/m? 25, (IT) A spherically spreading EM wave comes from a 1500-W source. At a distance of 5.0 m, what is the intensity, and what is the rms value of the electric field? . (IT) If the amplitude of the B field of an EM wave is 2.5 x 1077T, (a) what is the amplitude of the E field? (b) What is the average power per unit area of the EM wave? 27. (IT) What is the energy contained in a 1.00-m? volume near the Earth’s surface due to radiant energy from the Sun? See Example 31-6. . (IT) A 15.8-mW laser puts out a narrow beam 2.00 mm in diameter. What are the rms values of E and B in the beam?

29. (IT) Estimate the average power output of the Sun, given that about 1350 W/m? reaches the upper atmosphere of the Earth. 30. (II) A high-energy pulsed laser emits a 1.0-ns-long pulse of

a radio wave with an intensity as low as 1 X 1072 W/m?

As a “best-case” scenario for communication with extraterrestrials, consider the following: suppose an advanced civilization located at point A, a distance x away fror Earth, is somehow able to harness the entire power output of a Sun-like star, converting that power completely into a radio-wave signal which is transmitted uniformly in all directions from A. (a) In order for Arecibo to detect this radio signal, what is the maximum value for x in light-years

(1ly ~ 10'*m)?

834

CHAPTER 31

How

does

this

maximum

value

Sun is 1350 W/m?.

31-9 Radiation Pressure 34. (IT) Estimate the radiation pressure due to a 75-W bulb at a distance of 8.0cm from the center of the bulb. Estimate the force exerted on your fingertip if you place it at this point. 3s. (IT) Laser light can be focused (at best) to a spot with a radius r equal to its wavelength A. Suppose that a 1.0-W

beam of green laser light (A = 5 X 107 m) is used to form

such a spot and that a cylindrical particle of about that size (let the radius and height equal r) is illuminated by the laser as shown in Fig. 31-23. Estimate the acceleration of the particle, if its density equals that of water and it absorbs the radiation. [This order-of-magnitude calculation convinced researchers of the feasibility of “optical tweezers,” p. 829.] fl

average power 1.8 X 10'' W, The beam is 2.2 X 103m in

radius. Determine (a) the energy delivered in each pulse, and (b) the rms value of the electric field. 3L (I) How practical is solar power for various devices? Assume that on a sunny day, sunlight has an intensity of 1000 W/m? at the surface of Earth and that, when illuminated by that sunlight, a solar-cell panel can convert 10% of the sunlight’s energy into electric power. For each device given below, calculate the area A of solar panel needed to power it. (a) A calculator consumes 50 mW. Find A in cm? Is A small enough so that the solar panel can be mounted directly on the calculator that it is powering? (b) A hair dryer consumes 1500 W. Find A in m?. Assuming no other electronic devices are operating within a house at the same time, is A small enough so that the hair dryer can be powered by a solar panel mounted on the house’s roof? (c) A car requires 20hp for highway driving at constant velocity (this car would perform poorly in situations requiring acceleration). Find A4 in m% Is A small enough so that this solar panel can be mounted directly on the car and power it in “real time”? 32. (II1) (a) Show that the Poynting vector S points radially inward toward the center of a circular parallel-plate capacitor when it is being charged as in Example 31-1. (b) Integrate S over the cylindrical boundary of the capacitor gap to show that the rate at which energy enters the capacitor is equal to the rate at which electrostatic energy is being stored in the electric field of the capacitor (Section 24-4). Ignore fringing of E.

(b)

compare with the 100,000-ly size of our Milky Way galaxy? The intensity of sunlight at Earth’s orbital distance from the

FIGURE 31-23

—r—

Problem 35.

T

e e

!

—

.

A=5X%X10""m

. (IT) The powerful laser used in a laser light show provides a 3-mm diameter beam of green light with a power of 3 W. When a space-walking astronaut is outside the Space Shuttle, her colleague inside the Shuttle playfully aims such a laser beam at the astronaut’s space suit. The masses of the suited astronaut and the Space Shuttle are 120kg and 103,000 kg, respectively. (a) Assuming the suit is perfectly reflecting, determine the “radiation-pressure” force exerted on the astronaut by the laser beam. (b) Assuming the astronaut is separated from the Shuttle’s center of mass by 20 m, model the Shuttle as a sphere in order to estimate the gravitation force it exerts on the astronaut. (¢) Which of the two forces is larger, and by what factor? 37. (II) What size should the solar panel on a satellite orbiting Jupite)\ be if it is to collect the same amount of radiation from th. Sun as a 1.0-m? solar panel on a satellite orbiting Earth?

Maxwell's Equations and Electromagnetic Waves

31-10

Radio, TV, Wireless

38. (I) What is the range of wavelengths for (a) FM radio (88 MHz to 108 MHz) and (b) AM radio (535kHz to 1700 kHz)? Y. (I) Estimate the wavelength for 1.9-GHz cell phone reception. 40. (I) The variable capacitor in the tuner of an AM radio has a capacitance of 2200 pF when the radio is tuned to a station at 550 kHz. What must the capacitance be for a station near the other end of the dial, 1610 kHz?

41. (IT) A certain FM radio tuning circuit has a fixed capacitor C = 620 pF. Tuning is done by a variable inductance. What range of values must the inductance have to tune stations from 88 MHz to 108 MHz? 42. (IT) A satellite beams microwave radiation with a power of 12kW toward the Earth’s surface, S50 km away. When the beam strikes Earth, its circular diameter is about 1500 m.

Find the rms electric field strength surface of the Earth.

of the beam

at the

| General Problems 43. A 1.60-m-long FM antenna is oriented parallel to the electric field of an EM wave. How large must the electric field

be to produce a 1.00-mV (rms) voltage between the ends of

the antenna? What square meter?

is the

rate

of energy

transport

per

44. Who will hear the voice of a singer first: a person in the

balcony 50.0m away from the stage (see Fig. 31-24), or a person 1500 km away at home whose ear is next to the radio listening to a live broadcast? Roughly how much sooner? Assume the microphone is a few centimeters from the singer and the temperature is 20°C. e

T

FIGURE 31-24 Problem 44. 45. Light is emitted from an ordinary lightbulb filament in r wave-train bursts about 107®s in duration. What is the length in space of such wave trains? 46. Radio-controlled clocks throughout the United States receive a radio signal from a transmitter in Fort Collins, Colorado, that accurately (within a microsecond) marks the beginning of each minute. A slight delay, however, is introduced because this signal must travel from the transmitter to the clocks. Assuming Fort Collins is no more than 3000 km from any point in the US., what is the longest travel-time delay? 47. The voice from an astronaut on the Moon (Fig. 31-25) was beamed to a listening crowd on Earth. If you were standing 25 m from the loudspeaker on Earth, what was the total time lag between when you heard the sound and when the sound entered a microphone on the Moon? Explain whether the microphone was inside the space helmet, or outside, and why.

. Cosmic microwave background radiation fills all space with

an average energy density of 4 X 107#J/m?>, (a) Find the

rms value of the electric field associated with this radiation. (b) How far from a 7.5-kW radio transmitter emitting uniformly in all directions would you find a comparable value?

49. What are E; and By 2.00m from a 75-W light source? Assume the bulb emits radiation of a single frequency uniformly in all directions. . Estimate the rms electric field in the sunlight that hits Mars, knowing that the Earth receives about 1350 W/m? and that Mars is 1.52 times farther from the Sun (on average) than is the Earth. 51. At a given instant in time, a traveling EM wave is noted to have its maximum magnetic field pointing west and its maximum electric field pointing south. In which direction is the wave traveling? If the rate of energy flow is 560 W/m?, what are the maximum values for the two fields? 52. How large an emf (rms) will be generated in an antenna that consists of a circular coil 2.2 cm in diameter having 320 turns of wire, when an EM wave of frequency 810 kHz transporting energy at an average rate of 1.0 X 10~*W/m? passes through

it? [Hint: you be applied to magnetic field 53. The average

1.0 X 107" W/m? when it arrives at a 33-cm-diameter satel-

lite TV antenna. (a) Calculate the total energy received by the antenna during 6.0 hours of viewing this station’s programs. (b) What are the amplitudes of the E and B fields of the EM wave? 54. A radio station is allowed to broadcast at an average power not to exceed 25kW. If an electric field amplitude of 0.020 V/m is considered to be acceptable for receiving the radio transmission, estimate how many kilometers away you might be able to hear this station. . A point source emits light energy uniformly in all directions at an average rate F, with a single frequency f. Show that the peak electric field in the wave is given by E()

r

FIGURE 31-25 Problem 47,

can use Eq. 29-4 for a generator, since it could an observer moving with the coil so that the is oscillating with the frequency f = w/27.] intensity of a particular TV station’s signal is

=

[ocPy 2mr

9

:

56. Suppose a 35-kW radio station emits EM waves uniformly in all directions. (¢) How much energy per second crosses a 1.0-m? area 1.0km from the transmitting antenna? (b) What is the rms magnitude of the E field at this point, assuming the station is operating at full power? What is the rms voltage induced in a 1.0-m-long vertical car antenna (c) 1.0 km away, (d) 50 km away?

General Problems

835

57. What avoid from point

is the maximum power level of a radio station so as to electrical breakdown of air at a distance of 0.50m the transmitting antenna? Assume the antenna is a source. Air breaks down in an electric field of about

61. The electric field of an EM along the «x ax15 in free

spectrum (near 5 X 10'* Hz), the dielectric constant of water

is 1.77. Predict the speed of light in water and compare this value (as a percentage) with the speed of light in a vacuum. 59. The metal walls of a microwave oven form a cavity of dimensions 37cm X 37cm X 20 cm. When 2.45-GHz microwaves are continuously introduced into this cavity, reflection of incident waves from the walls set up standing waves with nodes at the walls. Along the 37-cm dimension of the oven, how many nodes exist (excluding the nodes at the wall) and what is the distance between adjacent nodes? [Because no heating occurs at these nodes, most microwaves rotate food while operating,] . Imagine that a steady current / flows in a straight cylindrical wire of radius Rj and resistivity p. (a) If the current is then changed at a rate d//dt, show that a displacement current /p, exists in the wire of magnitude ey p(dl/dt). (b) If the current in a copper wire is changed at the rate of 1.0 A/ms, determine the magnitude of /. (c) Determine the magnitude of the magnetic field Bp (T) created by /p at the surface of a copper wire with Ry = 1.0mm. Compare (as a ratio) Bp with the field created at the surface of the wire by a steady current of 10A.

pulse traveling is given by

Ey,= Egexp[~a’x*> — B** + 2apxt], where Ey, a, and B are

positive constants. (a) Is the pulse moving in the +x or —x direction? (b) Express B in terms of a and ¢ (speed of ligh™™

3 x 10°V/m.

. In free space (“vacuum”), where the net charge and current flow is zero, the speed of an EM wave is given by v = 1/\/€gug. If, instead, an EM wave travels in a nonconducting (“dielectric”) material with dielectric constant K, then v = 1/\/Kegpg. For frequencies corresponding to the visible

wave space

in free

space).

(¢)

Determine

the

expression

for

the

magnetic field of this EM wave. 62. Suppose that a right-moving EM wave overlaps with a leftmoving EM wave so that, in a certain region of space, the total electric field in the y direction and magnetic field in the z direction are given by Ey = Epsin(kx — of) + Egsin(kx + ot) and B; = Bysin(kx — wt) — Bysin(kx + wt). (a) Find the mathematical expression that represents the standing electric and magnetic waves in the y and z directions, respectively. (b) Determine the Poynting vector and find the x locations at which it is zero at all times. 63. The electric and magnetic fields of a certain EM wave in free space are given by E=E, sin(kx — wt)J + Eqcos(kx — wt)k and B By cos(kx — wt)j — Bysin(kx — wt)k. (a) Show that E and B are perpendlcular to each other at all times. () For this wave, E and B are in a plane parallel to the yz plane. Show _that the wave moves in a direction perpendicular to both E and B. (c) At any arbitrary choice of position x and time ¢, show that the magnitudes of E and B always equal E, and By, respectively. (d) At x =0, draw the orientation of E and B in the yz plane at ¢ = 0. Then qualitatively describe the motion of these vectors in the yz plane as time increases. [Note: The EM wave in this Problem is “circularly polarized.”]

S

Answers to Exercises A: (o).

D: 45cm.

B: (b).

E: Yes; the signal still travels 72,000 km.

C: (a) 3.8 X 10°Hz; (b) 5.5 X 10"* Hz 836

CHAPTER 31

F:

Over 4 hours.

Maxwell's Equations and Electromagnetic Waves

Reflection from still water, as from a glass mirror, can be analyzed using the ray model of light. Is this picture right side up? How can you tell? What are the clues? Notice the people and position of the Sun. Ray diagrams, which we will learn to draw in this Chapter, can provide the answer. See Example 32-3. In this first Chapter on light and optics, we use the ray model of light to understand the formation of images by mirrors, both plane and curved (spherical). We also begin our study of refraction—how light rays bend when they go from one medium to another—which prepares us for our study in the next Chapter of lenses, which are the crucial part of so many optical instruments.

~

Light: Reflection and Refraction

Do Q

—Guess now! A 2.0-m-tall person is standing 2.0 m from a flat vertical mirror staring at her image. What minimum height must the mirror have if the person is to see her entire body, from the top of her head to her feet? 0.50 m. 1.0m. 1.5m. 2.0m. 2.5m.

he sense of sight is extremely important to us, for it provides us with a large part of our information about the world. How do we see? What is the something called /ight that enters our eyes and causes the sensation of sight? How does light behave so that we can see everything that we do? We saw in Chapter 31 that light can be considered a form of electromagnetic radiation. We now examine the subject of light in detail in the next four Chapters. We see an object in one of two ways: (1) the object may be a source of light, such as a lightbulb, a flame, or a star, in which case we see the light emitted directly from the source; or, more commonly, (2) we see an object by light reflected from it. In the latter case, the light may have originated from the Sun, artificial lights, or a -ampfire. An understanding of how objects emit light was not achieved until the 1920s, and will be discussed in Chapter 37. How light is reflected from objects was understood earlier, and will be discussed here, in Section 32-2.

CONTENTS 32-1 The Ray Model of Light 32-2

Reflection; Image Formation by a Plane Mirror

32-3

Formation of Images by Spherical Mirrors

32-4

Index of Refraction

32-5

Refraction: Snell’s Law

32-6

Visible Spectrum and Dispersion

32-7

Total Internal Reflection; Fiber Optics

#32-8

Refraction at a Spherical Surface

837

32—1

The Ray Model of Light

A great deal of evidence suggests that light travels in straight lines under a wide variety of circumstances. For example, a source of light like the Sun casts distinofl shadows, and the light from a laser pointer appears to be a straight line. In fact, wc infer the positions of objects in our environment by assuming that light moves from the object to our eyes in straight-line paths. Our orientation to the physical world is based on this assumption.

This reasonable assumption is the basis of the ray model of light. This model

assumes that light travels in straight-line paths called light rays. Actually, a ray is an idealization; it is meant to represent an extremely narrow beam of light. When we see an object, according to the ray model, light reaches our eyes from each point on the object. Although light rays leave each point in many different directions, normally only a small bundle of these rays can enter an observer’s eye, as shown in

This bundle enters the eye

Fig. 32-1. If the person’s head moves to one side, a different bundle of rays will

FIGURE 32-1 Light rays come from each single point on an object. A small bundle of rays leaving one point is shown entering a person’s eye.

enter the eye from each point. We saw in Chapter 31 that light can be considered as an electromagnetic wave. Although the ray model of light does not deal with this aspect of light (we discuss the wave nature of light in Chapters 34 and 35), the ray model has been very successful in describing many aspects of light such as reflection, refraction, and the formation of images by mirrors and lenses.” Because these explanations involve straight-line rays at various angles, this subject is referred to as geometric optics.

32-2

Reflection; Image Formation by a Plane Mirror

When light strikes the surface of an object, some of the light is reflected. The rest can be absorbed by the object (and transformed to thermal energy) or, if th.« object is transparent like glass or water, part can be transmitted through. For avery smooth shiny object such as a silvered mirror, over 95% of the light may be reflected. When a narrow beam of light strikes a flat surface (Fig. 32-2), we define the angle of incidence, 6, to be the angle an incident ray makes with the normal (perpendicular) to the surface, and the angle of reflection, 6,, to be the angle the reflected ray makes with the normal. It is found that the incident and reflected rays lie in the same plane with the normal to the surface, and that the angle of reflection equals the angle of incidence, 6, = 0;. This is the law of reflection, and it is depicted in Fig. 32-2. It was known to the ancient Greeks, and you can confirm it yourself by shining a narrow flashlight beam or a laser pointer at a mirror in a darkened room. Normal to surface

Normal to surface

|

FIGURE 32-2 Law of reflection: (a) Shows a 3-D view of an incident

ray being reflected at the top of a flat surface; (b) shows a side or “end-on” view, which we will usually use because of its clarity.

} I Incident light ray

Angle of | Angle of incidence : reflection 6

R

Light: Reflection and Refraction

|

Reflected light ray

Angle of | Angle of incidence : reflection

SO

In ignoring the wave properties through apertures, these must phenomena of interference and what happens to the light at the

CHAPTER 32

I

|

(a)

838

:

(b)

of light we must be careful that when the light rays pass by objects 0‘ be large compared to the wavelength of the light (so the wav diffraction, as discussed in Chapter 15, can be ignored), and we ignore edges of objects until we get to Chapters 34 and 35.

When light is incident upon a rough surface, even microscopically rough such as this page, it is reflected in many directions, as shown in Fig. 32-3. This is called diffuse reflection. The law of reflection still holds, however, at each small section of fhe surface. Because of diffuse reflection in all directions, an ordinary object can be cen at many different angles by the light reflected from it. When you move your head to the side, different reflected rays reach your eye from each point on the = FIGURE 32-3 Diffuse reflection object (such as this page), Fig. 32—4a. Let us compare diffuse reflection to reflection ~ from a rough surface. from a mirror, which is known as specular reflection. (“Speculum” is Latin for mirror.) When a narrow beam of light shines on a mirror, the light will not reach your eye unless your eye is positioned at just the right place where the law of reflection is satisfied, as shown in Fig. 32—4b. This is what gives rise to the special image-forming properties of mirrors. 'é’he eye here Eye at both reflected light

(a)

$

&

w

(4]

does see

reflected light

)

(b)

FIGURE 32-4 A narrow beam of light shines on (a) white paper, and (b) a mirror. In part (a), you can see with your eye the white light reflected at various positions because of diffuse reflection. But in part (b), you see the reflected light only when your eye is placed correctly (6; = 6;); mirror reflection is also known as specular reflection. (Galileo, using similar arguments, showed that the Moon must have a rough surface rather than a highly polished surface like a mirror, as some people thought.)

m

Reflection from flat mirrors. Two flat mirrors are perpen-

licular to each other. An incoming beam of light makes an angle of 15° with the | first mirror as shown in Fig. 32—-5a. What angle will the outgoing beam make with the second mirror?

APPROACH We sketch the path of the beam as it reflects off the two mirrors, and draw the two normals to the mirrors for the two reflections. We use geometry and the law of reflection to find the various angles. FIGURE 32-5

Example 32-1.

(a)

(b)

SOLUTION In Fig. 32-5b, 6, + 15° = 90°, so 6, = 75° by the law of reflection 6, = 6; = 75° too. The two normals to the two mirrors are perpendicular to each other, so 6, + 6; + 90° = 180° as for any triangle. Thus 0; = 180° — 90° — 75° = 15°. By the law of reflection, 6, = 6; = 15°, so 6s = 75° is the angle the reflected ray makes with the second mirror surface. NOTE The outgoing ray is parallel to the incoming ray. Red reflectors on bicycles and cars use this principle.

When you look straight into a mirror, you see what appears to be yourself

as well as various objects around

and behind you, Fig. 32-6. Your face and the

ther objects look as if they are in front of you, beyond the mirror. But what ,Ou see in the mirror is an image of the objects, including yourself, that are in front of the mirror. SECTION 32-2

FIGURE 32-6

When you lookin a

mirror, you see an image of yourself

and objects around you. You don’t see yourself as others see you, because left and right appear reversedin the image.

Reflection; Image Formation by a Plane Mirror

839

A “plane” mirror is one with a smooth flat reflecting surface. Figure 32-7 shows how an image is formed by a plane mirror according to the ray model. We are viewing the mirror, on edge, in the diagram of Fig. 32-7, and the rays are shown reflecting from the front surface. (Good mirrors are generally made by putting a highly reflective metallic coating on one surface of a very flaq piece of glass.) Rays from two different points on an object (the bottle on the left in Fig. 32-7) are shown: two rays are shown leaving from a point on the top of the bottle, and two more from a point on the bottom. Rays leave each point on the object going in many directions, but only those that enclose the bundle of rays that enter the eye from each of the two points are shown. Each set of diverging rays that reflect from the mirror and enter the eye appear to come from a single point (called the image point) behind the mirror, as shown by the dashed lines. That is, our eyes and brain interpret any rays that enter an eye as having traveled straightline paths. The point from which each bundle of rays seems to come is one point on the image. For each point on the object, there is a corresponding image point. Plane mirror Reflecting surface

FIGURE 32-7

Formation of a virtual

image by a plane mirror.

B

T\\\ —~

=

1

A }

S Ssg

B ~a=i

\ 3 A

S~

S

o

~~"s

/ &3,

&

\~

.|Sy dy

d,

&~ i l/l

1 ] 1 |

C |

Let us concentrate on the two rays that leave point A on the object i Fig. 32-7, and strike the mirror at points B and B’. We use geometry for the rays at B. The angles ADB and CDB are right angles; and because of the law of reflection, 6; = 6, at point B. Therefore, angles ABD and CBD are also equal. The two triangles ABD and CBD are thus congruent, and the length AD = CD. That is, the image appears as far behind the mirror as the object is in front. The image distance, d; (perpendicular distance from mirror to image, Fig. 32-7), equals the object distance, d,, (perpendicular distance from object to mirror). From the geometry, we can also see that the height of the image is the same as that of the object. The light rays do not actually pass through the image location itself in Fig. 32-7. (Note where the red lines are dashed to show they are our projections, not rays.) The image would not appear on paper or film placed at the location of the image. Therefore, it is called a virtual image. This is to distinguish it from a real image in which the light does pass through the image and which therefore could appear on film or in an electronic sensor, and even on a white sheet of paper or screen placed at the position of the image. Our eyes can see both real and virtual images, as long as the diverging rays enter our pupils. We will see that curved mirrors and lenses can form real images, as well as virtual. A movie projector lens, for example, produces a real image that is visible on the screen.

PHYSICS

APPLIED

How tall a mirror do you need to see a reflection of your entire self?

840

CHAPTER 32

How tall must a full-length mirror be? A woman 1.60m

tall stands in front of a vertical plane mirror. What is the minimum height of the mirror, and how close must its lower edge be to the floor, if she is to be able to see her whole body? Assume her eyes are 10 cm below the top of her head. APPROACH For her to see her whole body, light rays from the top of her head and from the bottom of her foot must reflect from the mirror and enter her eye: see Fig. 32-8. We don’t show two rays diverging from each point as we did i Fig. 32-7, where we wanted to find where the image is. Now that we know the image is the same distance behind a plane mirror as the object is in front, we only need to show one ray leaving point G (top of head) and one ray leaving point A (her toe), and then use simple geometry.

FIGURE 32-8 Seeing oneselfin a mirror. Example 32-2.

A

SOLUTION

D

C

First consider the ray that leaves her foot at A, reflects at B, and enters the

eye at E. The mirror needs to extend no lower than B. The angle of reflection equals the angle of incidence, so the height BD is half of the height AE. Because AE = 1.60m — 0.10m = 1.50m, then BD = 0.75m. Similarly, if the woman is to see the top of her head, the top edge of the mirror only needs to reach point F which is 5cm below the top of her head (half of GE = 10 cm). Thus, DF = 1.55m, and the mirror needs to have a vertical height of only (1.55m — 0.75m) = 0.80 m. The mirror’s bottom edge must be 0.75 m above the floor. NOTE We see that a mirror, if positioned well, need be only half as tall as a person for that person to see all of himself or herself. | EXERCISE A Does the result of Example 32-2 depend on your distance from the mirror? (Try it.) EXERCISE B Return to the Chapter-Opening Question, page 837, and answer it again now. Try to explain why you may have answered differently the first time. EXERCISE C Suppose you are standing about 3 m in front of a mirror in a hair salon. You can see yourself from your head to your waist, where the end of the mirror cuts off the rest of your image. If you walk closer to the mirror () you will not be able to see any more of your image; (b) you will be able to see more of your image, below your waist; (c) you will see less of your image, with the cutoff rising to be above your waist.

CONCEPTUAL

EXAMPLE 32-3 | Is the photo upside down? Close examination

of the photograph on the first page of this Chapter reveals that in the top portion, the image of the Sun is seen clearly, whereas in the lower portion, the image of the Sun is partially blocked by the tree branches. Show why the reflection is not the same as the real scene by drawing a sketch of this situation, showing the Sun, the camera, the branch, and two rays going from the Sun to the camera (one direct and one reflected). TIs the photograph right side up? RESPONSE We need to draw two diagrams, one assuming the photo on p. 837 is right side up, and another assuming it is upside down. Figure 32-9 is drawn assuming the photo is upside down. In this case, the Sun blocked by the tree would be the direct view, and the full view of the Sun the reflection: the ray which reflects off the

water and into the camera travels at an angle below the branch, whereas the ray that travels directly to the camera passes through the branches. This works. Try to draw a diagram assuming the photo is right side up (thus assuming that the image of the Sun in the reflection is higher above the horizon than it is as viewed directly). It won’t work. The photo on p. 837 is upside down. Also, what about the people in the photo? Try to draw a diagram showing why they don’t appear in the reflection. [ Hint: Assume they are not sitting on the edge of poolside, but back from the edge a bit.] Then try to draw a diagram of the reverse (i.e., assume the photo is right side up so the people are visible only in the reflection). Reflected images |_are not perfect replicas when different planes (distances) are involved. Branches Direct ect ray ———————————————————————

r Sun

Reflected ray

Camera or

eye

FIGURE 32-9

Example 32-3.

SECTION 32-2

841

Normal —~ to surface

~

Nomal_ g

Rays

0.

i skt

o surface

-

source

FIGURE 32-10 Mirrors with convex and concave spherical surfaces. Note that 6, = 6; for each ray.

(@)

32-3

Eomvak mirror

(b)

Coticave mirror

Formation of Images by Spherical

Mirrors

Reflecting surfaces do not have to be flat. The most common curved mirrors are spherical, which means they form a section of a sphere. A spherical mirror is called convex if the reflection takes place on the outer surface of the spherical shape so that the center of the mirror surface bulges out toward the viewer (Fig. 32—-10a). A mirror is called concave if the reflecting surface is on the inner surface of the sphere so that the mirror surface sinks away from the viewer (like a “cave”), Fig. 32-10b. Concave mirrors are used as shaving or cosmetic mirrors (Fig. 32-11a) because they magnify, and convex mirrors are sometimes used on cars and trucks (rearview mirrors) and in shops (to watch for thieves), because they take in a wide field of view (Fig. 32-11b).

FIGURE 32-11 (a) A concave cosmetic mirror gives a magnified image. (b) A convex mirror in a store reduces image size and so includes a wide field of view. Note the extreme distortion—this mirror has a large curved surface and does not fit the “paraxial ray” approximation discussed on the next page.

(a)

(b)

Focal Point and Focal Length To see how spherical mirrors form images, we first consider an object that is very far from a concave mirror. For a distant object, as shown in Fig, 32-12, the rays from each point on the object that strike the mirror will be nearly parallel. For an object infinitely far away (the Sun and stars approach this), the rays would be precisely parallel.

FIGURE 32-12 If the object’s distance is large compared to the size of the mirror (or lens), the rays are nearly parallel. They are parallel for an object at infinity (co).

842

CHAPTER 32

Light: Reflection and Refraction

These rays strike the mirror, and they are essentially p parallel.

-

Now consider such parallel rays falling on a concave mirror as in Fig. 32-13. The law of reflection holds for each of these rays at the point each strikes the mirror. As can be seen, they are not all brought to a single point. In order to form a sharp image, the rays must come to a point. Thus a spherical mirror ill not make as sharp an image as a plane mirror will. However, as we show oelow, if the mirror is small compared to its radius of curvature, so that a reflected ray makes only a small angle with the incident ray (26 in Fig. 32-14), then the rays will cross each other at very nearly a single point, or focus. In the case shown in Fig. 32-14, the incoming rays are parallel to the principal axis, which is defined as the straight line perpendicular to the curved surface

in Fig. 32-14). The point F, where

rays come to a focus after The distance between F and focal length, f, of the mirror. infinitely far away along the would be at F.

C

i

Principal | axis. |

incident parallel

reflection, is called the focal point of the mirror. the center of the mirror, length FA, is called the The focal point is also the image point for an object principal axis. The image of the Sun, for example, \ B

Cl—

.

r

AV / riGURE 32-13

Parallel rays

giking a concave spherical mirror do

ot intersect (or focus) at precisely a single point. (This “defect” is referred to as “spherical aberration.”)

FIGURE 32-14 Rays parallel to the principal axis of a concave spherical mirror come to a focus at F, the focal point, as long as the mirror is small in width as compared to its radius of

A

F

curvature, r, so that the rays are “paraxial”— ‘\N

at its center (line CA

\

that is, make only small angles with the horizontal axis.

ll

Now we will show, for a mirror whose reflecting surface is small compared to its radius of curvature, that the rays very nearly meet at a common point, F, and we will also calculate the focal length f. In this approximation, we consider only rays that make a small angle with the principal axis; such rays are called axial rays, and their angles are exaggerated in Fig. 32-14 to make the labels ear. First we consider a ray that strikes the mirror at B in Fig. 32-14. The point C is the center of curvature of the mirror (the center of the sphere of which the mirror is a part). So the dashed line CB is equal to r, the radius of curvature, and CB is norm?l to the mirror’s surface at B. The incoming ray that hits the mirror at B makes an angle 6 with this normal, and hence the reflected ray, BF, also makes an angle 6 with the normal (law of reflection). Note that angle BCF is also 0 as shown. The triangle CBF is isosceles because two of its angles are equal. Thus we have length CF = BF. We assume the mirror surface is small compared

to the mirror’s radius of curvature, so the angles

are small, and the

length FB is nearly equal to length FA. In this approximation, FA = FC. But FA = f, the focal length, and CA = 2 X FA = r. Thus the focal length is half the radius of curvature:

[spherical mirror]

F = =

(32-1)

We assumed only that the angle 6 was small, so this result applies for all other incident paraxial rays. Thus all paraxial rays pass through the same point F, the focal point. Since it is only approximately true that the rays come to a perfect focus at F, the more curved the mirror, the worse the approximation (Fig. 32-13) and the more blurred the image. This “defect” of spherical mirrors is called spherical aberration; we will discuss it more with regard to lenses in Chapter 33. A parabolic reflector, on the

other

hand, will reflect

the rays to a perfect

focus. However,

because parabolic shapes are much harder to make and thus much more expensive, spherical mirrors are used for most purposes. (Many astronomical telescopes use farabolic reflectors.) We consider here only spherical mirrors and we will assume iat they are small compared to their radius of curvature so that the image is sharp and Eq. 32-1 holds.

SECTION 32-3

Formation of Images by Spherical Mirrors

843

Image Formation—Ray Diagrams We saw that for an object at infinity, the image is located at the focal point of a concave spherical mirror, where f = r/2. But where does the image lie for an object not at infinity? First consider the object shown as an arrow iqN Fig. 32—-15a, which is placed between F and C at point O (O for object). Let u_ determine where the image will be for a given point O’ at the top of the object. (a)

o’

Ray 1 goes out from

C

O’ parallel to the axis

1

F

v __] A

/I‘———

and reflects through F.

O’

(b)

FIGURE 32-15

Rays leave point O’

Ray 2 goes through F and then reflects back

parallel to the axis.

on the object (an arrow). Shown are the three most useful rays for determining where the image I is formed. [Note that our mirror is not small compared to f, so our diagram will not give the precise position of the image.]

1 —

Ray 3 is chosen perpendicular to mirror, and so must reflect

*

through C (center of curvature).

1]

844

CHAPTER 32

2

A /

c

o’

1

.

=

PROBLEM SOLVING Image point is where reflected rays intersect

F

D)

I

back on itself and go

= “

0

3

(c)

RAY DIAGRAM Finding the image position for a curved mirror

g

1

Diverging rays heading toward eye

q

To do this we can draw several rays and make sure these reflect from the mirror such that the angle of reflection equals the angle of incidence. Many rays could be drawn leaving any point on an object, but determining the image position is simplified if we deal with three particularly simple rays. These are the rays labeled 1, 2, and 3 in Fig. 32-15 and we draw them leaving object point O’ as follows: Ray 1 is drawn parallel to the axis; therefore after reflection it must pass along a line through F (Fig. 32-15a). Ray 2 leaves O’ and is made to pass through F (Fig. 32-15b); therefore it must reflect so it is parallel to the axis. Ray 3 passes through C, the center of curvature (Fig. 32-15¢); it is along a radius of the spherical surface and is perpendicular to the mirror, so it is reflected back on itself. All three rays leave a single point O’ on the object. After reflection from a (small) mirror, the point at which these rays cross is the image point I'. All other rays from the same object point will also pass through this image point. To find the image point for any object point, only these three types of rays need to be drawn. Only two of these rays are needed, but the third serves as a check.

We have shown the image point in Fig. 32-15 only for a single point on the object. Other points on the object are imaged nearby, so a complete image of the object is formed, as shown by the dashed arrow in Fig. 32—15c. Because the light actually passes through the image itself, this is a real image that will appear on a piece of paper or film placed there. This can be compared to the virtual image formed by a plane mirror (the light does not actually pass through that image, Fig. 32-7). The image in Fig. 32-15 can be seen by the eye when the eye is placed t“™ the left of the image, so that some of the rays diverging from each point on the image (as point I') can enter the eye as shown in Fig. 32-15c. (See also Figs. 32-1 and 32-7.)

Mirror Equation and Magnification Image points can be determined, roughly, by described, Fig. 32-15. But it is difficult to draw ys as we assumed. For more accurate results, _wves the image distance if the object distance

drawing the three rays as just small angles for the “paraxial” we now derive an equation that and radius of curvature of the

mirror are known. To do this, we refer to Fig. 32-16. The object distance, d,,, is

the distance of the object (point O) from the center of the mirror. The image distance, d;, is the distance of the image (point I) from the center of the mirror. The height of the object OO’ is called 4, and the height of the image, I'I, is ;.

FIGURE 32-16 Diagram for deriving the mirror equation. For the derivation, we assume the mirror

size is small compared to its radius of curvature.

Two rays leaving O’ are shown: O'FBI’ (same as ray 2 in Fig. 32-15) and O'ATl’, which is a fourth type of ray that reflects at the center of the mirror and can also be used to find an image point. The ray O’ Al’ obeys the law of reflection, so the two right triangles

' &

O’ AO

hi

and I' Al are similar. Therefore, we have

d

For the other ray shown, O'FBI’, the triangles O'FO and AFB are also similar because the angles are equal and we use the approximation AB = h; (mirror small compared to its radius). Furthermore FA = f, the focal length of the mirror, o

ho

OF

h,

do_f.

FA

f

The left sides of the two preceding expressions are the same, so we can equate the right sides:

Sl

e

=

i

(32-2)

Mirror equation

This is the equation we were seeking. It is called the mirror equation and relates the object and image distances to the focal length f (where f = r/2). The lateral magnification, 7, of a mirror is defined as the height of the image divided by the height of the object. From our first set of similar triangles above, or the first equation on this page, we can write:

" Wy ke TQeT d,

32-3 ey

fihe minus sign in Eq. 32-3 is inserted as a convention. Indeed, we must be careful sout the signs of all quantities in Eqs. 32-2 and 32-3. Sign conventions are chosen so as to give the correct locations and orientations of images, as predicted by ray diagrams.

SECTION 32-3

Formation of Images by Spherical Mirrors

845

FIGURE 32-16

previous page.)

ZJPROBLEM Sign conventions

SOLVING

I

T

(Repeated from

h;

;

;

'

| | | | | | | | | 1 | 1

B

The sign conventions we use are: the image height A; is positive if the image is upright, and negative if inverted, relative to the object (assuming #, is taken as positive); d; or d, is positive if image or object is in front of the mirror (as in Fig. 32-16); if either image or object is behind the mirror, the corresponding distance is negative (an example can be seen in Fig. 32-17, Example 32-6). Thus the magnification (Eq. 32-3) is positive for an upright image and negative for an inverted image (upside down). We summarize sign conventions more fully in the Problem Solving Strategy following our discussion of convex mirrors later in this Section.

Concave Mirror Examples Image in a concave mirror. A 1.50-cm-high diamond ring is placed 20.0cm from a concave mirror with radius of curvature 30.0cm. Determine (a) the position of the image, and (b) its size. APPROACH We determine the focal length from the radius of curvature (Eq. 32-1), f = r/2 = 15.0cm. The ray diagram is basically like that shown in Fig. 32-16 (repeated here on this page), since the object is between F and C. Tl"fl position and size of the image are found from Egs. 32-2 and 32-3. SOLUTION Referring to Fig. 32-16,we have CA = r = 30.0cm, FA = f =15.0cm, and OA = d, = 20.0cm. (a) From Eq. 32-2,

_1

dy

=

A CAUTION

1

—_

1

150cm _ 200em

So d; = 1/(0.0167cm™!) = 60.0cm.

Remember to take the reciprocal

=

-1

. »0167cm.

Because d, is positive, the image is 60.0 cm

in front of the mirror, on the same side as the object. (b) From Eq. 32-3, the magnification is m

=

===

The image is 3.0 times larger than the object, and its height is h;

=

mh,

=

(-=3.00)(1.5cm)

=

—4.5cm.

The minus sign reminds us that the image is inverted, as in Fig. 32-16.

NOTE When an object is further from a concave mirror than the focal point, w™ can see from Fig. 32-15 or 32-16 that the image is always inverted and real.

846

CHAPTER 32

Light: Reflection and Refraction

CONCEPTUAL EXAMPLE 32-5 | Reversible rays. If the object in Example 32—4 is placed instead where the image is (see Fig. 32-16), where will the new image be? RESPONSE The mirror equation is symmetric in d, and d;. Thus the new image will be where the old object was. Indeed, in Fig. 32-16 we need only reverse the &irection of the rays to get our new situation.

Object closer to concave mirror. A 1.00-cm-high object

is placed 10.0cm from a concave mirror whose radius of curvature is 30.0cm. (a¢) Draw a ray diagram to locate (approximately) the position of the image. (b) Determine the position of the image and the magnification analytically. APPROACH

We draw the ray diagram using the rays as in Fig. 32-15, page 844.

An analytic solution uses Eqgs. 32-1, 32-2, and 32-3.

SOLUTION (a) Since f = r/2 = 15.0cm, the object is between the mirror and the focal point. We draw the three rays as described earlier (Fig. 32-15); they are shown leaving the tip of the object in Fig. 32-17. Ray 1 leaves the tip of our object heading toward the mirror parallel to the axis, and reflects through F. Ray 2 cannot head toward F because it would not strike the mirror; so ray 2 must point as if it started at F (dashed line) and heads to the mirror, and then is reflected parallel to the principal axis. Ray 3 is perpendicular to the mirror, as before. The rays reflected from the mirror diverge and so never meet at a point. They appear, however, to be coming from a point behind the mirror. This point locates the image of the tip of the arrow. The image is thus behind the mirror and is virtual. (Why?) (b) We use Eq. 32-2 to find d; when d, = 10.0 cm: 1r d; Therefore,

1 f

1 d,

d; = —30.0cm.

_

1 150cm The

B

minus

1 10.0cm sign

_2-3 30.0cm

means

the

1 ) 30.0cm image

is

behind

the mirror, which our diagram also told us. The magnification is m = —di/d, = —(=30.0cm)/(10.0cm) = +3.00. So the image is 3.00 times larger than the object. The plus sign indicates that the image is upright (same as object), which is consistent with the ray diagram, Fig. 32-17. NOTE The image distance cannot be obtained accurately by measuring on Fig. 32-17, because our diagram violates the paraxial ray assumption (we draw rays at steeper angles to make them clearly visible). NOTE When the object is located inside the focal point of a concave mirror

(dy < f), the image

is always upright and virtual. And

if the object O in

PHYSICS

APPLIED

Seeing yourself upright and magnified in a concave mirror

—

e

o

-]

Fig. 32-17 is you, you see yourself clearly, because the reflected rays at point O are diverging. Your image is upright and enlarged.

r

FIGURE 32-17 Object placed within the focal point F. The image is behind the mirror and is virtual, Example 32-6. [Note that the vertical scale (height of object = 1.0cm) is different from the horizontal (OA = 10.0cm) for ease of drawing, and reduces the precision of the drawing.]

SECTION 32-3

Formation of Images by Spherical Mirrors

847

@

PHYSICS

APPLIED

Magnifying mirror

(shaving/cosmetic)

It is useful to compare Figs. 32-15 and 32-17. We can see that if the object is

~ Within the focal point (d0 < f), as in Fig. 32-17, the image is virtual, upright, and

~ magnified. This is how a shaving or cosmetic mirror is used—you must place your head closer to the mirror than the focal point if you are to see yourself right-sid% up (see the photograph of Fig. 32-11a). If the object is beyond the focal point, as Fig. 32-15, the image is real and inverted (upside down—and hard to use!). Whether the magnification has magnitude greater or less than 1.0 in the latter case depends on the position of the object relative to the center of curvature, point C. Practice making ray diagrams with various object distances. The mirror equation also holds for a plane mirror: the focal length is f =r/2 = oo, and Eq. 32-2 gives d; = —d,.

Seeing the Image For a person’s eye to see a sharp image, the eye must be at a place where it intercepts diverging rays from points on the image, as is the case for the eye’s position in Figs. 32-15 and 32-16. Our eyes are made to see normal objects, which always means the rays are diverging toward the eye as shown in Fig. 32-1. (Or, for very distant objects like stars, the rays become essentially parallel, as in Fig. 32-12.) If you placed your eye between points O and I in Fig. 32-16, for example, converging rays from the object OO’ would enter your eye and the lens of your eye could not bring them to a focus; you would see a blurry image. [We will discuss the eye more in Chapter 33.]

)

FIGURE 32-18 You can see a clear inverted image of your face when you are beyond C (d, > 2f), because the rays that arrive at your eye are diverging. Standard rays 2 and 3 are shown leaving point O on your nose. Ray 2 (and other nearby rays) enters your eye. Notice that rays are diverging as they move to the left of image point I.

FIGURE 32-19

Convex mirror: () the:focal poisitis at Fbehind tiiemitros (b) the imge Tof the obj ect at O is virtual, upright, and smaller than the object. [Not to scale for Example 32-7.]

~ §;-(5 -

4 T (a) é

\.fi,W‘ 1 ‘lg\,“ == 2 \ | ‘ '\ | (b) r—do—t+d;~ 0

848

1 3

At

CHAPTER 32

o

)

£ . ' in Fig. 32-16, si ’ If you are the object OO’ in Fig. 32-16, situated between F and C, and are trying to see your'self .in the mirror, you would see a blur; but the person whose eye is sh.own in Flg. 32-16 can see you clearly. You can see yourself clearly, but upside down, if you are to t‘he left of C in Fig. 32-16, where d, > 2f. Why? Because then the rays arriving from the image will be diverging at your position as demonstrated in Fig. 32-18, and your eye can then focus them. You can also see yourself clearly, and right-side up, if you are closer

to the mirror than its focal point (d, < f), as we saw in Example

Fig. 32-17.

Convex Mirrors

k= “’l

Image

of your nose

32-6,

The analysis used for concave mirrors can be applied to convex mirrors. Even the mirror equation (Eq. 32-2) holds for a convex mirror, although the quantities involved must be carefully defined. Figure 32—-19a shows parallel rays falling on a convex mirror. Again spherical aberration is significant (Fig. 32-13), unless we assume the mirror’s size is very small compared to its radius of curvature. The reflected rays diverge, but seem to come from point F behind the mirror. This is the focal point, and its distance from the center of the mirror (point A) is the fo length, f. It is easy to show that again f = r/2. We see that an object at infini_ produces a virtual image in a convex mirror. Indeed, no matter where the object is

Light: Reflection and Refraction

placed on the reflecting side of a convex mirror, the image will be virtual and upright, as indicated in Fig. 32-19b. To find the image we draw rays 1 and 3 according to the rules used before on the concave mirror, as shown in Fig. 32-19b. Aote that although rays 1 and 3 don’t actually pass through points F and C, the 1e along which each is drawn does (shown dashed). The mirror equation, Eq. 32-2, holds for convex mirrors but the focal length f must be considered negative, as must the radius of curvature. The proof is left as a Problem. It is also left as a Problem to show that Eq. 32-3 for the magnification is also valid.

s0L¥z,

= ©

Q)

-2 o

¢

Spherical Mirrors 1. Always draw a ray diagram even though you are going to make an analytic calculation—the diagram serves as a check, even if not precise. From one point on the object, draw at least two, preferably three, of the easy-to-draw

rays

using

the rules

described

in

3. Sign Conventions (a) When the object, image, or focal point is on the reflecting side of the mirror (on the left in our drawings), the corresponding distance is positive.

If any of these points is behind the mirror (on the

Fig. 32-15. The image point is where the reflected

rays intersect or appear to intersect.

2. Apply the mirror equation, Eq. 32-2, and the magnification equation, Eq. 32-3. It is crucially important to follow the sign conventions—see the next point.

right) the corresponding distance is negative."

(b) The image height &; is positive if the image is

upright, and negative if inverted, relative to the object (h, is always taken as positive). 4. Check that the analytical solution is consistent with the ray diagram.

"Object distances are positive for material objects, but can be negative in systems with more than one mirror or lens—see Section 33-3.

Convex rearview mirror. An external rearview car mirror is

convex with a radius of curvature of 16.0 m (Fig. 32-20). Determine the location f the image and its magnification for an object 10.0 m from the mirror.

@

PHYSICS

APPLIED

Convex rearview mirror

APPROACH We follow the steps of the Problem Solving Strategy explicitly. SOLUTION 1. Draw a ray diagram. The ray diagram will be like Fig. 32-19b, but the large object distance (d, = 10.0m) makes a precise drawing difficult. We have a convex mirror, so r is negative by convention. 2. Mirror and magnification equations. The center of curvature of a convex mirror is behind the mirror, as is its focal point, so we set r = —16.0m so that the focal length is f = r/2 = —8.0m. The object is in front of the mirror, d, = 10.0m. Solving the mirror equation, Eq. 32-2, for 1/d; gives 1

—_—

=

1

=

-

1

—_—

4 . F . 6.

=

1

1

~Bim.

10Om

=

—10.0 — 8.0

Sim

=

18

8.0m

=

L\GURE32-20

Example 32-7.

3. Sign conventions. The image distance is negative, —4.4m, so the image is behind the mirror. The magnification is m = +0.44, so the image is upright (same orientation as object) and less than half as tall as the object. | 4. Check. Our results are consistent with Fig. 32-19b. Convex rearview mirrors on vehicles sometimes come with a warning that objects are closer than they appear in the mirror. The fact that d; may be smaller than d, (as in Example 32-7) seems to contradict this observation. The real reason fwe object seems farther away is that its image in the convex mirror is smaller than « would be in a plane mirror, and we judge distance of ordinary objects such as other cars mostly by their size.

SECTION 32-3

Formation of Images by Spherical Mirrors

849

TABLE 32-1

Refraction’

32—-4

Indices of

Material

n= £ v

Vacuum Air (at STP) Water Ethyl alcohol Glass Fused quartz Crown glass Light flint Lucite or Plexiglas Sodium chloride Diamond

1.0000 1.0003 1.33 1.36 1.46 1.52 1.58 1.51 1.53 242

fA = 589 nm.

Index of Refraction

We saw in Chapter 31 that the speed of light in vacuum is

¢ = 299792458 X 108 m/s, which is usually rounded off to

3.00 X 108 m/s when extremely precise results are not required. In air, the speed is only slightly less. In other transparent materials such as glass and water, the speed is always less than that in vacuum. For example, in water light travels at about 3c. The ratio of the speed of light in vacuum to the speed v in a given material is called the index of refraction, , of that material:

n =
ny.

(b) Light refracted when passing from water (n) into air (ny): ny > n,.

ray

' )

o } 2 | i n2 > n]

Reflected

Air (ny) Water (1)

CHAPTER 32

Light: Reflection and Refraction

:

|

' Reflocted

ray

Refracted ray

(a) Ray bends toward L

850

IRaeyfraCted

n] > n2

Air (n,) Water (1)

:

Incident ray

:

%Source

(b) Ray bends away from L

-

FIGURE 32-22 Ray diagram showing why a person’s legs look shorter when standing in waist-deep water: the path of light traveling from the bather’s foot to the observer’s eye bends at the water’s surface, and our brain interprets the light as having traveled in a straight line, from higher up (dashed line).

Foot appeart,

FIGURE 32-23

A straw in water

looks bent even when it isn’t.

to be here

Refraction is responsible for a number of common optical illusions. For example, a person standing in waist-deep water appears to have shortened legs. As shown in Fig. 32-22, the rays leaving the person’s foot are bent at the surface. The observer’s brain assumes the rays to have traveled a straight-line path (dashed red line), and so the feet appear to be higher than they really are. Similarly, when you put a straw in water, it appears to be bent (Fig. 32-23).

Snell’s Law The angle of refraction depends on the speed of light in the two media and on the incident angle. An analytical relation between 6, and 6, in Fig. 32-21 was arrived at experimentally about 1621 by Willebrord Snell (1591-1626). It is known as Snell’s law and is written: ny sin 01

=y

sin 92 '

(32—5)

6 is the angle of incidence and 6, is the angle of refraction; n,; and n, are the respective indices of refraction in the materials. See Fig. 32-21. The incident and refracted rays lie in the same plane, which also includes the perpendicular to the surface. Snell’s law is the law of refraction. (Snell’s law was derived in Section 15-10 where Eq. 15-19 is just a combination of Eqs. 32-5 and 32-4. We also derive it in Chapter 34 using the wave theory of light.)

SNELL’S LAW (LAW OF REFRACTION)

It is clear from Snell’s law that if n, > ny, then 6, < 6,. That is, if light enters

medium where # is greater (and its speed is less), then the ray is bent toward the normal. And if n, < ny, then 6, > 6, so the ray bends away from the normal. This is what we saw in Fig. 32-21. EXERCISE D Light passes from a medium with » = 1.3 into a medium with the light bent toward or away from the perpendicular to the interface?

D EVILEY LI

n = 1.5. Is

Refraction through flat glass. Light traveling in air strikes a

flat piece of uniformly thick glass at an incident angle of 60°, as shown in Fig. 32-24. If the index of refraction of the glass is 1.50, (¢) what is the angle of refraction 6, in the glass; (b) what is the angle 65 at which the ray emerges from the glass?

r ; ; ;:foll’l:fi : ;iezcz OfL gllg;tspassmg (Example 32-8).

APPROACH We apply Snell’s law at the first surface, where the light enters the glass, and again at the second surface where it leaves the glass and enters the air. SOLUTION (a) The incident ray is in air,so n; = 1.00 and n, = 1.50. Applying

Air

Glass

Air

Snell’s law where the light enters the glass (8, = 60°) gives sinf, SO

GA

.

=

1'—28sin60°

=

0.5774,

35.30-

(b) Since the faces of the glass are parallel, the incident angle at the second surface is just 6, (simple geometry),so sinf, = 0.5774. At this second interface,

ny =150 given by

A

and

n, = 1.00. Thus the ray re-enters the air at an angle 65 (=6,)

. sinflg

1.50 = ——sinf, = 0.866, B 1.00 = and 6 = 60°. The direction of a light ray is thus unchanged by passing through a flat piece of glass of uniform thickness. NOTE This result is valid for any angle of incidence. The ray is displaced slightly to one side, however. You can observe this by looking through a piece of glass (near its edge) at some object and then moving your head to the side slightly so

that you see the object directly. It “jumps.”

Raoy i)rt%r:cl

/“Image” (where object 3%5238&33;};” 31? glass

SECTION 32-5

851

Apparent depth of a pool. A swimmer has dropped her

goggles to the bottom of a pool at the shallow end, marked as 1.0 m deep. But the goggles don’t look that deep. Why? How deep do the goggles appear to be when you look straight down into the water? -

FIGURE 32-25

APPROACH We draw a ray diagram showing two rays going upward from . point on the goggles at a small angle, and being refracted at the water’s (flat) surface, Fig. 32-25. The two rays traveling upward from the goggles are refracted away from the normal as they exit the water, and so appear to be diverging from a point above the goggles (dashed lines), which is why the water seems less deep than it actually is. SOLUTION To calculate the apparent depth d’' (Fig. 32-25), given a real depth d = 1.0m, we use Snell’s law with n; = 1.33 for water and n, = 1.0 for air:

Example 32-9.

sin 92

=

m

sin 01 .

We are considering only small angles, so sin 6 ~ tan6 ~ 6, with @ in radians. So Snell’s law becomes 92

~

n101.

From Fig. 32-25, we see that =7

X

and

0,

~

tanf;

=

x 7

Putting these into Snell’s law, 6, ~ n, 6, , we get p

or

T~ d

,

=

d

n

X

_10m

133

_

0.75 m. fl

The pool seems only three-fourths as deep as it actually is.

32-6 Visible Spectrum and Dispersion An obvious property of visible light is its color. Color is related to the wavelengths or frequencies of the light. (How this was discovered will be discussed in Chapter 34.) Visible light—that to which our eyes are sensitive—has wavelengths in air in the range of about 400 nm to 750 nm.” This is known as the visible spectrum, and within it lie the different colors from violet to red, as shown in Fig. 32-26. Light

FIGURE 32-26 The spectrum of visible light, showing thfe range of wavelengths for the various colors as seen in air. Many colors, such as brown, do not appear in the

with wavelength shorter than 400 nm (= violet) is called ultraviolet (UV), with wavelength greater than 750 nm (= red) is called infrared (IR). human eyes are not sensitive to UV or IR, some types of photographic digital cameras do respond to them. A prism can separate white light into a rainbow of colors, as Fig. 32-27. This happens because the index of refraction of a material on the wavelength, as shown for several materials in Fig. 32-28. White

v

400 nm F

;

500 nm }

:

600 nm ;

:

700 nm f

;

and light Although film and

shown in depends light is a

R

spectrum; they are made from a

mixture of wavelengths.

*Sometimes the angstrom (A) unit is used when referring to light: 1A = 1 X 107°m. light falls in the wavelength range of 4000 A to 7500 A.

*The complete electromagnetic spectrum is illustrated in Fig. 31-12.

852

CHAPTER 32

Light: Reflection and Refraction

Then visiblemiy

Refractive index

w

\E|ilic¢tefllint glasI

|

Borate flint glas Qulartz

.

|

Silicate crown glass

1.5

f

Fused quartz

1.4

400

500

600

700

Wavelength (nm)

Violet

FIGURE 32-27 White light passing through a prism is broken down into its constituent colors.

|

Blue

Green

Yellow

Orange

Red

FIGURE 32-28 Index of refraction as a function of wavelength for various transparent solids.

mixture of all visible wavelengths, and when incident on a prism, as in Fig. 32-29, the different wavelengths are bent to varying degrees. Because the index of refraction is greater for the shorter wavelengths, violet light is bent the most and red the least, as indicated. This spreading of white light into the full spectrum is

called dispersion.

o

Rainbows are a spectacular example of dispersion — by drops of water. You

can see rainbows when you look at falling water droplets with the Sun behind

you. Figure 32-30 shows how red droplets and are reflected off the least and so reaches the observer’s ’jn the diagram. Thus the top of the FIGURE 32-30

and violet rays are bent by spherical water back surface of the droplet. Red is bent the eyes from droplets higher in the sky, as shown rainbow is red.

screen

FIGURE 32-29 White light

dispersed by a prism into the visible

spectrum.

@ PHYSICS ReIRBOT

APPLIED

(a) Ray diagram explaining how a rainbow (b) is formed.

These two rays

Sunligh,

are seen by

i

observer (not to scale)

Red

Orange i

Green

‘

Sun”ght

&

E Violet

Blue

Violet

red (b)

(a) The visible spectrum, Fig. 32-26, does not show all the colors seen in nature. For example, there is no brown in Fig. 32-26. Many of the colors we see are a mixture of wavelengths. For practical purposes, most natural colors can be reproduced using three primary colors. They are red, green, and blue for direct source viewing such as TV and computer monitors. For inks used in printing, the primary colors are cyan (the color of the margin notes in this book), yellow, and magenta (the color we use for light rays in diagrams). For any wave, its velocity v is related to its wavelength A and frequency f by v = fA (Eq. 15-1 or 31-14). When a wave travels from one material into another, the frequency of the wave does not change across the boundary since a point (an atom) at the boundary oscillates at that frequency. Thus if light goes from air into P

a material with index of refraction n, the wavelength becomes (recall Eq. 32-4): v c A

/\n=?=fi=;

(32-6)

where A is the wavelength in vacuum or air and A,, is the wavelength in the material with index of refraction n.

SECTION 32-6

853

CONCEPTUAL

EXAMPLE

32-10

| Observed

color

of light

under

water.

We said that color depends on wavelength. For example, for an object emitting 650 nm light in air, we see red. But this is true only in air. If we observe this same object when under water, it still looks red. But the wavelength in water A, i\ (Eq. 32-6) 650nm/1.33 = 489 nm. Light with wavelength 489 nm would appes blue in air. Can you explain why the light appears red rather than blue when observed under water? RESPONSE It must be that it is not the wavelength that the eye responds to, but rather the frequency. For example, the frequency of 650nm red light in air

is f=c/A=(30x10"m/s)/(650 X 10°m) = 4.6 X 10“Hz,

and does not

change when the light travels from one medium to another. Only A changes.

NOTE If we classified colors by frequency, the color assignments would be valid for any material. We typically specify colors by wavelength in air (even if less general) not just because we usually see objects in air, but because wavelength is what is commonly measured (it is easier to measure than frequency).

el

32-7

Total Internal Reflection;

Fiber Optics

When light passes from one material into a second material where the index of refraction is less (say, from water into air), the light bends away from the normal, as for rays I and J in Fig. 32-31. At a particular incident angle, the angle of refraction will be 90°, and the refracted ray would skim the surface (ray K) in this

FIGURE 32-31 Since n, < nq, light rays internally reflected if the incident angle 6; as for ray L. If 6; < 6c, as for rays I and J, part of the light is reflected, and the rest is

are totally > 6, only a refracted.

|

ny (< ny) /g n A

|

|LK 6

fl

lL l |

Source

case. The incident angle at which this occurs is called the critical angle, 6. From Snell’s law, 6 is given by

sinfe = 2sin90° = 2. n

n

(32-7)

For any incident angle less than 6., there will be a refracted ray, although part of the light will also be reflected at the boundary. However, for incident angles A

CAUTION Total internal reflection (occurs only if refractive index is smaller beyond boundary)

greater than 6., Snell’s law would tell us that sin 6, is greater than 1.00. Yet the

sine of an angle can never be greater than 1.00. In thi§ case there is no refracte'd ray at all, and all of the light is reflected, as for ray L in Fig. 32-31. This effec.t is called total internal reflection. Total internal reflection can occur only when light strikes a boundary where the medium beyond has a lower index of refraction. EXERCISE E Fill a sink with water. Place a waterproof watch just below the surface with the watch’s flat crystal parallel to the water surface. From above you can still see the watch reading. As you move your head to one side far enough what do you see™™y and why?

854

CHAPTER 32

Light: Reflection and Refraction

CONCEPTUAL

EXAMPLE 32-11|

View up from under water. Describe what a

person would see who looked up at the world from beneath the perfectly smooth surface of a lake or swimming pool. f\RESPONSE

For an air—water interface, the critical angle is given by 1.00

sinfc

=

133

=

0.750.

Therefore, 6 = 49°. Thus the person would see the outside world compressed into a circle whose edge makes a 49° angle with the vertical. Beyond this angle, the person would see reflections from the sides and bottom of the lake or pool (Fig. 32-32).

FIGURE 32-32 (a) Light rays, and (b) view looking upward from beneath the water (the surface of the water must be very smooth). Example 32-11.

(b) Diamonds achieve their brilliance from a combination of dispersion and total internal reflection. Because diamonds have a very high index of refraction of about 2.4, the critical angle for total internal reflection is only 25°. The light dispersed into a spectrum inside the diamond therefore strikes many of the internal surfaces of the diamond before it strikes one at less than 25° and emerges. After many such reflections, the light has traveled far enough that the colors have become sufficiently separated to be seen individually and brilliantly by the eye after leaving the diamond. r Many optical instruments, such as binoculars, use total internal reflection jdthin a prism to reflect light. The advantage is that very nearly 100% of the light is reflected, whereas even the best mirrors reflect somewhat less than 100%. Thus the image is brighter, especially after several reflections. For glass with n = 1.50, 6c = 41.8°. Therefore, 45° prisms will reflect all the light internally, if oriented as shown in the binoculars of Fig, 32-33.

FIGURE 32-33 Total internal reflection of light by prisms in binoculars.

EXERCISE F If 45.0° plastic prisms are used in binoculars, what minimum index of refraction

must the plastic have?

Fiber Optics Total internal reflection is the principle behind fiber optics. Glass and plastic fibers as thin as a few micrometers in diameter are common. A bundle of such tiny fibers is called a light pipe or cable, and light" can be transmitted along it with almost no loss because of total internal reflection. Figure 32-34 shows how light traveling down a thin fiber makes only glancing collisions with the walls so that

total internal reflection occurs. Even if the light pipe is bent into a complicated

shape, the critical angle still won’t be exceeded, so light is transmitted practically undiminished to the other end. Very small losses do occur, mainly by reflection at the ends and absorption within the fiber. Important applications of fiber-optic cables are in communications and medicine. They are used in place of wire to carry telephone calls, video signals, and computer data. The signal is a modulated light beam (a light beam whose intensity can be varied) and data is transmitted at a much higher rate and with less loss and less interference than an electrical signal in a copper wire. Fibers have been developed that can support over one hundred separate wavelengths, each modulated

=~ FIGURE 32-34 Light reflected totally at the interior surface of a

glass or transparent plastic fiber. %

@ PHYSICS APPLIED Fiber optics in communications

to carry up to 10 gigabits (10' bits) of information per second. That amounts to a

™ -ranit (10' bits) per second for the full one hundred wavelengths.

"Fiber optic devices can use not only visible light but also infrared light, ultraviolet light, and microwaves.

SECTION 32-7

Total Internal Reflection; Fiber Optics

855

PHYSICS APPLIED Medicine—endoscopes

The sophisticated use of fiber optics to transmit a clear picture is particularly useful in medicine, Fig. 32-35. For example, a patient’s lungs can be examined by inserting a light pipe known as a bronchoscope through the mouth and down the bronchial tube. Light is sent down an outer set of fibers to illuminate the lungs. The« reflected light returns up a central core set of fibers. Light directly in front of eacl fiber travels up that fiber. At the opposite end, a viewer sees a series of bright and dark spots, much like a TV screen—that is, a picture of what lies at the opposite end. Lenses are used at each end. The image may be viewed directly or on a monitor screen or film. The fibers must be optically insulated from one another, usually by a thin coating of material with index of refraction less than that of the fiber. The more fibers there are, and the smaller they are, the more detailed the picture. Such instruments, including bronchoscopes, colonoscopes (for viewing the colon), and endoscopes (stomach or other organs), are extremely useful for examining hard-to-reach places.

*32-8 *

(b) FIGURE 32-35 (a) How a fiber-optic image is made. (b) Example of a fiber-optic device inserted through the mouth to view the vocal cords, with image on screen.

Refraction at a Spherical Surface

We now examine the refraction of rays at the spherical surface of a transparent material. Such a surface could be one face of a lens or the cornea of the eye. To be general, let us consider an object which is located in a medium whose index of refraction is n;, and rays from each point on the object can enter a medium whose index of refraction is n,. The radius of curvature of the spherical boundary is R, and its center of curvature is at point C, Fig. 32-36. We now show that all rays leaving a point O on the object will be focused at a single point I, the image point, if we consider only paraxial rays: rays that make a small angle with the axis.

FIGURE 32-36 Rays from a point O on an object will be focused at a single image point I by a spherical boundary between two transparent materials (ny > n,), as long as the rays make small angles with the axis.

To do so, we consider a single ray that leaves point O as shown in Fig. 32-37. From Snell’s law, Eq. 32-5, we have n;sinf;

=

n,sinb,.

We are assuming that angles 6,, 0,, «, 8, and Y are small, so sin # ~ 6 (in radians), and Snell’s law becomes, approximately, n101

Also, B+

=

n262.

¢ = 180° and 6, + ¥ + ¢ = 180°, so fi

=

7

+

02.

Similarly for triangle OPC, 6[

FIGURE 32-37 Diagram for showing that all paraxial rays from

O focus at the same point I (n, > ny).

=

a

+

fi.

3

0

o n

856

CHAPTER 32

Light: Reflection and Refraction

These three relations can be combined to yield nya

+

flzy

=

(nz

=

fll)B.

Since we are considering only the case of small angles, we can write, approximately,

where d, and d; are the object and image distances and 4 is the height as shown in Fig. 32-37. We substitute these into the previous equation, divide through by 4, and obtain ny

2. +

ny

ny,

a

—

R

ny

(32-8)

For a given object distance d,,, this equation tells us d;, the image distance, does not depend on the angle of a ray. Hence all paraxial rays meet at the same point I. This is true only for rays that make small angles with the axis and with each other, and is equivalent to assuming that the width of the refracting spherical surface is small compared to its radius of curvature. If this assumption is not true, the rays will not converge to a point; there will be spherical aberration, just as for a mirror (see Fig. 32-13), and the image will be blurry. (Spherical aberration will be discussed further in Section 33-10.) We derived Eq. 32-8 using Fig. 32-37 for which the spherical surface is convex (as viewed by the incoming ray). It is also valid for a concave surface—as can be seen using Fig. 32-38—if we use the following conventions: 1. If the surface is convex (so the center of curvature C is on the side of the surface opposite to that from which the light comes), R is positive; if the surface is concave (C on the same side from which the light comes) R is negative. 2. The image distance, d;, follows the same convention: positive if on the opposite side from where the light comes, negative if on the same side. r 3. The object distance is positive if on the same side from which the light comes (this is the normal case, although when several surfaces bend the light it may not be so0), otherwise it is negative. For the case shown in Fig. 32-38 with a concave surface, both R and d; are negative when used in Eq. 32-8. Note, in this case, that the image is virtual.

FIGURE 32-38

Rays from O

refracted by a concave surface form

a virtual image (n, > ny). Per our

conventions, R < 0,d; < 0,d, > 0.

DOV

ER P

Apparent depth 1l. A person looks vertically down into a

1.0-m-deep pool. How deep does the water appear to be? APPROACH

Eg.32-8.

Example 32-9 solved this problem using Snell’s law. Here we use

SOLUTION A ray diagram is shown in Fig. 32-39. Point O represents a point on the pool’s bottom. The rays diverge and appear to come from point I, the image.

We have

r

Hence

d, = 1.0m

and, for a flat surface,

1.33

1.00

1.0m

di

=

d; = —(1.0m)/(1.33)

R = co. Then Eq. 32-8 becomes

(1.00 — 1.33) o0

= —0.75m.

=

FIGURE 32-39

Example 32-12.

Air: ny = 1.00

Water: ny = 1.33

0.

So the pool appears to be only three-

fourths as deep as it actually is, the same result we found in Example 32-9. The

* | minus sign tells us the image point I is on the same side of the surface as O, and the image is virtual. At angles other than vertical, this conclusion must be modified.

*SECTION 32-8

Refraction at a Spherical Surface

857

A spherical “lens” A point source of light is placed at a

distance of 25.0 cm from the center of a glass sphere Fig. 32-40. Find the image of the source.

(n = 1.5) of radius 10.0cm,

APPROACH As shown in Fig. 32-40, there are two refractions, and we treat the successively, one at a time. The light rays from the source first refract from th._ convex glass surface facing the source. We analyze this first refraction, treating it as in Fig. 32-36, ignoring the back side of the sphere. SOLUTION Using Eq. 32-8 (assuming paraxial rays) with n; = 1.0, n, = 1.5, R =10.0cm,

and

d, = 25.0cm

— 10.0cm

= 15.0cm,

distance as formed at surface 1, d;; :

_L_LCM‘W)_fl)hL 1. This occurs if 6; exceeds the critical angle 6 given by ;

sinfc

=

| Questions 1. What would be the appearance of the Moon if it had (a) a rough surface; (b) a polished mirrorlike surface? 2. Archimedes

3. 4.

5. 6. 7. 8. 9.

is said to have burned the whole Roman

fleet

in the harbor of Syracuse by focusing the rays of the Sun with a huge spherical mirror. Is this' reasonable? What is the focal length of a plane mirror? What is the magnification of a plane mirror? An object is placed along the principal axis of a spherical mirror. The magnification of the object is —3.0. Is the image real or virtual, inverted or upright? Is the mirror concave or convex? On which side of the mirror is the image located? Using the rules for the three rays discussed with reference to Fig. 32-15, draw ray 2 for Fig. 32-19b. Does the mirror equation, Eq. 32-2, hold for a plane mirror? Explain. If a concave mirror produces a real image, is the image necessarily inverted? How might you determine the speed of light in a solid, rectangular, transparent object? When you look at the Moon’s reflection from a ripply sea, it appears elongated (Fig. 32-41). Explain.

ny

—ny

(32-7)

14. When a wide beam of parallel light enters water at an angle, the beam broadens. Explain. 15. You look into an aquarium and view a fish inside. One ray of light from the fish as it emerges from the tank is shown in Fig. 32-42. The apparent position of the fish is also shown. In the drawing, indicate the approximate position of the actual fish. Briefly justify your answer.

| i, FIGURE 32-42 Question 15.

fi

|

:

)

16. How can you “see” a round drop of water on a table even though the water is transparent and colorless? 17. A ray of light is refracted through three different materials (Fig. 32-43). Rank the materials according to their index of refraction, least to greatest.

FIGURE 32-43

1

/

/

Question 17.

18.

Can a light ray traveling in air be totally reflected when it strikes a smooth water surface if the incident angle is chosen correctly? Explain.

19. When you look up at an object in air from beneath the surface in a swimming pool, does the object appear to be the same size as when you see it directly in air? Explain. 20. What type of mirror is shown in Fig. 32-44?

FIGURE 32-41

Question 9.

10. How can a spherical mirror have a negative object distance? 11. What is the angle of refraction when a light ray is incident perpendicular to the boundary between two transparent materials? 12. When you look down into a swimming pool or a lake, are you likely to overestimate or underestimate its depth? Explain. How does the apparent depth vary with the viewing angle? (Use ray diagrams.) 13. Draw a ray diagram to show why a stick or straw looks bent when part of it is under water (Fig. 32-23). f "Students at MIT did a feasibility study. See www.mit.edu.

FIGURE 32-44

Question 20.

21. Light rays from stars (including our Sun) always bend toward the vertical direction as they pass through the Earth’s atmosphere. (a) Why does this make sense? (b) What can you conclude about the apparent positions of stars as viewed from Earth?

Questions

859

| Problems 32-2

Reflection; Plane Mirrors

1. (I) When you see close to diagrams

you look at yourself in a 60-cm-tall plane mirror, the same amount of your body whether you are the mirror or far away. (Try it and see.) Use ray to show why this should be true.

32-3 9.

Spherical Mirrors

(I) A solar cooker, really a concave mirror pointed at the«

Sun, focuses the Sun’s rays 18.8 cm in front of the mirro.

What is the radius of the spherical surface from which the mirror was made?

. (I) Suppose that you want to take a photograph of yourself as you look at your image in a mirror 2.8 m away. For what

~ 10. (I) How far from a concave mirror (radius 24.0 cm) must an object be placed if its image is to be at infinity?

. (II) Two plane mirrors meet at a 135° angle, Fig. 32-45. If light rays strike one mirror at 38° as shown, at what angle ¢

the image flips at a distance of 0.50 m. What is the radius of curvature of the mirror?

distance should the camera lens be focused?

do they leave the second mirror?

FIGURE 32-45 Problem 3.

¢

389

. (II) A person whose eyes are 1.64 m above the floor stands

2.30m in front of a vertical plane mirror whose bottom edge is 38 cm above the floor, Fig. 32-46. What is the horizontal

11. (I) When walking toward a concave mirror you notice that

12. (II) A small candle is 35 cm from a concave mirror having a radius of curvature of 24 cm. (a) What is the focal length of the mirror? (b) Where will the image of the candle be located? (c) Will the image be upright or inverted? 13. (II) You look at yourself in a shiny 9.2-cm-diameter Christmas tree ball. If your face is 25.0 cm away from the ball’s front surface, where is your image? Is it real or virtual? Is it upright or inverted?

14, (II) A mirror at an amusement park shows an upright image

of any person who stands 1.7 m in front of it. If the image is three times the person’s height, what is the radius of

distance x to the base of the wall supporting the mirror of

the nearest point on the flogr that can be seen reflected in

the mirror?

FIGURE 32-46

Problem 4.

&

=230

T %

1.64 mFL

J’ =

m—]

-

curvature of the mirror? (See Fig. 32-44.)

15. (II) A dentist wants a small mirror that, when 2.00 cm from a tooth, will produce a 4.0% upright image. What kind of mirror

must be used and what must its radius of curvature be?

138cm

16. (IT) Some rearview mirrors produce images of cars to your

rear that are smaller than they would be if the mirror were

=

. (II) Show that if two plane mirrors meet at an angle ¢, a single ray reflected successively from both mirrors is deflected through an angle of 2¢ independent

of the inci-

dent angle. Assume ¢ < 90° and that only two reflections,

flat. Are the mirrors concave or convex? What is a mirror’s

17.

radius of curvature if cars 180m away appear 0.33 the My normal size? (II) You are standing 3.0m from a convex security mirror in a

store. You estimate the height of your image to be half of your

one from each mirror, take place. . (II) Suppose you are 88 cm from a plane mirror. What area of the mirror is used to reflect the rays entering one eye from a point on the tip of your nose if your pupil diameter is 4.5 mm?

actual height. Estimate the radius of curvature of the mirror. 18. (II) An object 3.0 mm high is placed 18 cm from a convex mirror of radius of curvature 18 cm. (a) Show by ray tracing that the image is virtua}, aqd estim‘ate the image distance.

. (II) Stand up two plane mirrors so they form a 90.0° angle

g’gnsllgw ;g:itztizignegz}zz:; ;:lagt; (::fStiI;cg gzln l()ce) ng:gu:letg

as in Fig. 32-47. When you look into this double mirror, you see you, in a ray this

yourself as others see instead of reversed as single mirror. Make a diagram to show how occurs.

b

i

s

RIS s

&

Y

P

tae nnage. Sizg; UAIAE q e 19. (I1) Th(? image of a dlstant.tree is v1{tua1 and very small when v1ewe(_1 in a cur\"ed mirror. Tl_le image appears to be 16.0 cm.behlnq the mirror. What kind of mirror is it, and what is its radius of curvature? 20. (II) Use two techniques, (a) a ray diagram, and (b) the mitror equation, to show that the magnitude of the magnification of a concave mirror is less than 1 if the object is

beyond the center of curvature C(d, > r), and is greater

than 1 if the object is within C (d, < r).

FIGURE 32-47 Problems 7 and 8.

21. (II) Show, using a ray diagram, that the magnification m of a convex mirror is m = —d;/d,, just as for a concave mirror. [Hint: Consider a ray from the top of the object that reflects at the center of the mirror.] 22. (II) Use ray diagrams to show that the mirror equation,

. (III) Suppose a third mirror is placed beneath the two shown Eq. 32-2, is val_1d for a convex mirror as long as f is in Fig. 3247, so that all three are perpendicular to each other. considered negative. (a) Show that for such a “corner reflector,” any incident ray ~ 23. (II) The magnification of a convex mirror is +0.55X fo will return in its original direction after three reflections. (b) What happens if it makes only two reflections?

860

CHAPTER 32

Light: Reflection and Refraction

objects 3.2m from the mirror. What is the focal length o. this mirror?

. (II) (a) Where should an object be placed in front of a concave mirror so that it produces an image at the same location as the object? (b) Is the image real or virtual? (c) Is the image inverted or upright? (d) What is the magnification of the image? . (II) A 4.5-cm tall object is placed 26cm in front of a spherical mirror. It is desired to produce a virtual image that is upright and 3.5 cm tall. (@) What type of mirror should be used? (b) Where is the image located? (c¢) What is the focal length of the mirror? (d) What is the radius of curvature of the mirror? 26. (II) A shaving or makeup mirror is designed to magnify your face by a factor of 1.35 when your face is placed 20.0cm in front of it. (a) What type of mirror is it? (b) Describe the type of image that it makes of your face. (c) Calculate the required radius of curvature for the mirror. 27. (IT) A concave mirror has focal length f. When an object is placed a distance d, > f from this mirror, a real image with

magnification m is formed. (a) Show that m = f/(f — d,).

(b) Sketch m vs. d, over the range f < d, < +oo where f =045 m. (c) For what value of d, will the real image have the same (lateral) size as the object? (d) To obtain a real image that is much larger than the object, in what general region should the object be placed relative to the mirror? 28. (IT) Let the focal length of a convex mirror be written as f = —|f|. Show that the magnification m of an object a

distance d,, from this mirror is given by m = |f|/(do + |f]).

Based on this relation, explain why your nose looks than the rest of your face when looking into a mirror. . (I) A spherical mirror of focal length f produces an of an object with magnification m. (a) Show that the is a distance

d, = f (1 = %)

bigger convex

—m?*

where

m

is the

from the reflecting side of

normal

“lateral”

magnification,

Eq. 32-3. Why the minus sign? [Hint: Find the image positions for both ends of the wire, and assume £ is very small.]

32-4

Index of Refraction

(I) The speed of light in ice is 2.29 X 10%m/s. What is the index of refraction of ice? 33. (I) What is the speed of light in (a) ethyl alcohol, (b) lucite, (c) crown glass? 34. (I) Our nearest star (other than the Sun) is 4.2 light years away. That is, it takes 4.2 years for the light to reach Earth. r How far away is it in meters? . (I) How long does it take light to reach us from the Sun, 32.

1.50 x 108 km away?

7 5 (IT) Light is emitted from an ordinary lightbulb filament in wave-train bursts of about 10®s in duration. What is the length in space of such wave trains?

32-5

Snell's Law

38. (I) A diver shines a flashlight upward from beneath the water at a 38.5° angle to the vertical. At what angle does the light leave the water? 39. (I) A flashlight beam strikes the surface of a pane of glass (n = 1.56) at a 63° angle to the normal. What is the angle of refraction? 40. (I) Rays of the Sun are seen to make a 33.0° angle to the vertical beneath the water. At what angle above the horizon is the Sun? 41. (I) A light beam coming from an underwater spotlight exits the water at an angle of 56.0°. At what angle of incidence did it hit the air-water interface from below the surface? 42. (IT) A beam of light in air strikes a slab of glass (n = 1.56) and is partially reflected and partially refracted. Determine the angle of incidence if the angle of reflection is twice the angle of refraction. 43. (II) A light beam strikes a 2.0-cm-thick piece of plastic with a refractive index of 1.62 at a 45° angle. The plastic is on top of a 3.0-cm45°) thick piece of glass for which n = 1.47. What is the distance D in Fig. 32-48?

image object

the mirror. (b) Use the relation in part (@) to show that, no matter where an object is placed in front of a convex mirror, its image will have a magnification in the range 0 = m = +1. 30. (IIT) An object is placed a distance r in front of a wall, where r exactly equals the radius of curvature of a certain concave mirror. At what distance from the wall should this mirror be placed so that a real image of the object is formed on the wall? What is the magnification of the image? 31 (III) A short thin object (like a short length of wire) of length £ is placed along the axis of a spherical mirror (perpendicular to the glass surface). Show that its image has length ¢ = m*0 so the longitudinal magnification is equal to

36. (IT) The speed of light in a certain substance is 88% of its value in water. What is the index of refraction of that substance?

FIGURE 32-48

D=1

Problem 43.

. (I) An aquarium filled with water has flat glass sides whose index of refraction is 1.56. A beam of light from outside the aquarium strikes the glass at a 43.5° angle to the perpendicular (Fig. 32-49). What is the angle of this light ray when it enters (a) the glass, and then Glass (b) the water? (c) What would Air be the refracted angle if the ray Water entered the water directly?

FIGURE 32-49

Problem 44.

43

45. (IT) In searching the bottom of a pool at night, a watchman shines a narrow beam of light from his flashlight, 1.3m above the water level, onto the surface of the water at a point

2.5m from his foot at the edge of the pool (Fig. 32-50). Where does the spot of light hit the bottom

of

the

beneath

his foot, if the

measured from bottom of the

pool,

the wall

pool is 2.1 m deep?

FIGURE 32-50 Problem 45.

25m

-

2.1m J.

Problems

861

46. (II) The block of glass (n = Fig. 32-51 is surrounded by block at its left-hand face reemerges into the air from parallel to the block’s base. Determine 6, .

1.5) shown in cross section in air. A ray of light enters the with incident angle 6; and the right-hand face directed

54. (II) A parallel beam of light containing two wavelengths, A

= 465nm

and

A, = 652nm,

glass of an equilateral At what angle does each beam leave the prism (give angle with normal to the face)?

prism

as

enters

shown

the

silicate

in Fig.

flint

32-54.

See Fig. 32-28.

FIGURE 32-54

FIGURE 32-51

Problem 54,

Problem 46.

47. (II) A laser beam of diameter d; = 3.0mm in air has an incident angle 6; = 25° at a flat air-glass surface. If the index of refraction of the glass is n = 1.5, determine the diameter d, of the beam after it enters the glass. 48. (II) Light is incident on an equilateral glass prism at a 45.0° angle to one face, Fig. 32-52. Calculate the angle at which light emerges from the opposite face. Assume that n = 1.54.

55. (III) A ray of light with wavelength A is incident from air at precisely 60° ( = 6) on a spherical water drop of radius r and index of refraction n (which depends on A). When the ray reemerges into the air from the far side of the drop, it has been deflected an angle ¢ from its original direction as shown in Fig. 32-55. By how much does the value of ¢ for violet light (n = 1.341) differ from the value for red light

(n = 1.330)?

FIGURE 32-52

FIGURE 32-55

Problems 48 and 65.

Problem 55.

49. (II) A triangular prism made of crown glass (n = 1.52) with base angles of 30.0° is surrounded by air. If parallel rays are incident normally on its base as shown in Fig. 32-53, what is the angle ¢ between the two emerging rays?

56. (III) For visible light, the index of refraction n of glass is roughly 1.5, although this value varies by about 1% across the visible range. Consider a ray of white light incident from air at angle 6 onto a flat piece of glass. (@) Show that, upon entering the glass, the visible colors contained in this incig dent ray will be dispersed over a range A6, of refract«‘ angles given approximately by ABZ

FIGURE 32-53

50. (IT) Show in general that for a light beam incident on a uniform layer of transparent material, as in Fig. 32-24, the

direction of the emerging beam is parallel to the incident beam, independent of the incident angle 6. Assume the air on the two sides of the transparent material is the same. 51. (III) A light ray is incident on a flat piece of glass with index of refraction n as in Fig. 32-24. Show that if the incident angle 6 is small, the emerging ray is displaced a distance d = t0(n — 1)/n, where ¢ is the thickness of the glass, 6 is in radians, and d is the perpendicular distance between the incident ray and the (dashed) line of the emerging ray (Fig. 32-24).

52. (I) By what percent is the speed of blue light (450 nm) less than the speed of red light (680 nm), in silicate flint glass (see Fig. 32-28)? 53. (I) A light beam strikes a piece of glass at a 60.00° incident angle. The beam contains two wavelengths, 450.0 nm and

CHAPTER 32

"

(d/dx)(sin"'x)=1/0/1 — x%.]

32-7

in degrees?

Light: Reflection and Refraction

(c) If 6; = 90°,

Total Internal Reflection

57. (I) What is the critical angle for the interface between water

and diamond? To be internally reflected, the light must start in

which material? 58. (I) The critical angle for a certain liquid—air surface is 49.6°. What is the index of refraction of the liquid? 59. (II) A beam of light is emitted in a pool of water from a depth of 72.0 cm. Where must it strike the air—water interface, relative to the spot directly above it, in order that the

light does not exit the water? 60. (II) A ray of light, after entering a light fiber, reflects at an of 14.5°

with

the

long

axis

of the

fiber, as in Fig.

32-56. Calculate the distance along the axis of the fiber that the light ray travels between successive reflections off the sides of the fiber. Assume that the fiber has an index of refraction of 1.55 and is 1.40 X 10~ m in diameter.

of the glass is

1.4831 and 1.4754, respectively. What is the angle between the two refracted beams?

862

An

(b) If 6 =0° what is A, what is Af, in degrees?

angle

32-6 Visible Spectrum; Dispersion

the index of refraction

sin 6,

\/n? — sin’6,

[Hint: For x in radians,

Problem 49.

700.0 nm, for which

i

FIGURE 32-56

Problem 60.

61. (II) A beam of light is emitted 8.0 cm beneath the surface of a liquid and strikes the surface 7.6cm from the point directly above the source. If total internal reflection occurs,

o

what can you say about the index of refraction of the liquid?

62. (II) Figure 32-57 shows a liquid-detecting prism device that might be used inside a washing machine or other liquid-containing appliance. If no liquid covers the prism’s hypotenuse, total internal reflection of the beam from the

light source produces a large signal in the light sensor. If liquid covers the hypotenuse, some light escapes from the prism into the liquid and the light sensor’s signal decreases. Thus a large signal from the light sensor indicates the absence of liquid in the reservoir. If this device is designed to detect the presence of water, determine the allowable range for the prism’s index of refraction n. Will the device work properly if the prism is constructed from (inexpensive) lucite? For lucite, =1.5. FIGURE 32-57

Problem 62.

64. (II) (a) What is the minimum index of refraction for a glass or plastic prism to be used in binoculars (Fig. 32-33) so that total internal reflection occurs at 45°? (b) Will binoculars work if their prisms (assume n = 1.58) are immersed in water? (c) What minimum » is needed if the prisms are immersed in water? 65. (III) Suppose a ray strikes the left face of the prism in Fig. 32-52 at 45.0° as shown, but is totally internally reflected at the opposite side. If the apex angle (at the top) is @ = 60.0°, what can you say about the index of refraction of the prism? 66. (III) A beam of light enters the end of an optic fiber as shown in Fig. 32-59. (a) Show that we can guarantee total internal reflection at the side surface of the material (at point A), if the index of refraction is greater than about 1.42. In other words, regardless of the angle «, the light beam reflects back into the material at point A, assuming air outside. FIGURE 32-59

Problem 66.

Liquid reservoir L [ |

~ ,

| |

| | | | Light | source e

Air

| |

Transparent material

| l

o

e

e

o e

Light sensor e

J

| | | | |

*32-8

Refraction at Spherical Surface

#67. (II) A 13.0-cm-thick plane piece of glass (n = 1.58) lies on the surface of a 12.0-cm-deep pool of water. How far below the top of the glass does the bottom of the pool seem, as viewed from directly above?

Enclosure

*68. (II) A fish is swimming in water inside a thin spherical glass bowl of uniform thickness. Assuming the radius of curvature 63. (II) Two rays A and B travel down a cylindrical optical fiber of diameter

d = 75.0 um,

length

£ = 1.0 km,

and index of

refraction ny; = 1.465. Ray A travels a straight path down the fiber’s axis, whereas ray B propagates down the fiber by repeated reflections at the critical angle each time it impinges on the fiber’s boundary. Determine the extra time At it takes for ray B to travel down the entire fiber in comparison with ray A (Fig. 32-58), assuming (a) the fiber is surrounded by air, (b) the fiber is surrounded by a cylindrical glass “cladding” with index of refraction n, = 1.460.

(.

I

750 wm | 4= k

Air or glass

=

: 1.00 km

FIGURE 32-58

of the bowl is 28.0 cm, locate the image of the fish if the fish

is located: (a) at the center of the bowl; (b) 20.0 cm from the side of the bowl between the observer and the center of the bowl. Assume the fish is small.

*69. (III) In Section 32-8, we derived Eq. 32-8 for a convex spherical surface with n, > n;. Using the same conventions and using diagrams similar to Fig. 32-37, show that Eq. 32-8 is valid also for (a) a convex spherical surface with n, < n,, (b) a concave spherical surface with n, > ny, and (c) a concave spherical surface with n, < n,.

B

. = 1.465 l‘—

Problem 63.

.

!

*70. (III) A viewer relative back to

coin lies at the bottom of a 0.75-m-deep pool. If a sees it at a 45° angle, where is the image of the coin, to the coin? [Hint: The image is found by tracing the intersection of two rays.]

Problems

863

E General Problems 71. Two identical concave mirrors are set facing each other 1.0m apart. A small lightbulb is placed halfway between the mirrors. A small piece of paper placed just to the left of the bulb prevents light from the bulb from directly shining on the left mirror, but light reflected from the right mirror still reaches the left mirror. A good image of the bulb appears on the left side of the piece of paper. What is the focal length of the mirrors? 72. A

slab

of thickness

D, whose

two

faces

are parallel, has

index of refraction n. A ray of light incident from air onto one face of the slab at incident angle 6, splits into two rays A and B. Ray A reflects directly back into the air, while B travels a total distance { within the slab before reemerging from the slab’s face a distance d from its point of entry (Fig. 32-60). (a) Derive expressions for £ and d in terms of D, n, and 6. (b) For normal incidence (i.e., #; = 0°) show that your expressions yield the expected values for £ and d.

76. The critical angle of a certain piece of plastic in air is 6c = 39.3°. What is the critical angle of the same plastic if

it is immersed in water?

77. Each student in a physics lab is assigned to find the location where a bright object may be placed in order that a concave mirror, with radius of curvature

r = 46 cm,

will produce an

image three times the size of the object. Two students complete the assignment at different times using identical equipment, but when they compare notes later, they discover that their answers for the object distance are not the same. Explain why they do not necessarily need to repeat the lab, and justify your response with a calculation. 78. A kaleidoscope makes symmetric patterns with two plane mirrors having a 60° angle between them as shown in Fig. 32-63. Draw the location of the images (some of them images of images) of the object placed between 60° the mirrors. | FIGURE 32-63 Problem 78.

FIGURE 32-60 Problem 72.

73. Two plane mirrors are facing each other 2.2m apart as in Fig. 32-61. You stand 1.5 m away from one of these mirrors and look into it. You will see multiple images of yourself. (a) How far away from you are the first three images of yourself in the mirror in 1.5m front of you? (b) Are these first three images facing toward you or away from you?

FIGURE 32-61

79. When light passes through a prism, the angle that the refracted ray makes relative to the incident ray is called the deviation angle 8, Fig. 32-64. Show that this angle is a minimum when the ray passes through the prism symmetrically, perpendicular to the bisector of the apex angle ¢, and show

that the minimum

deviation

the prism’s index of refraction n by

angle, 8,,, is related to

q

sin}(¢ + 6) sin ¢/2

[Hint: For 6 in radians, (d/d8)(sin"'0) = 1/V1 — 6%]

22m

Problem 73.

FIGURE 32-64

Problems 79 and 80.

74. We wish to determine the depth of with water by measuring the width noting that the bottom edge of the angle of 13.0° above the horizontal Calculate the depth of the pool.

r 0 and d; < 0), and for an inverted image the magnification is negative (h; < 0 and d; > 0). From sign convention 1, it follows that the power (Eq. 33-1) of a converging lens, in diopters, is positive, whereas the power of a diverging lens is negative. A converging lens is sometimes referred to as a positive lens, and a diverging lens as a negative lens. %w

50L¥r,

o

Thin Lenses

i

1. Draw

o

¢

a ray diagram, as precise as possible, but

even a rough one can serve as confirmation of analytic results. Choose one point on the object and draw at least two, or preferably three, of

the

|

and

easy-to-draw rays 33-8.

mntersect.

The

image

described

point

in Figs.

is where

the

33-6

rays

2. For analytic solutions, solve for unknowns in the thin lens equation (Eq. 33-2) and the magnification equation (Eq. 33-3). The thin lens equation involves reciprocals —don’t forget to take the reciprocal.

3. Follow the sign conventions listed just above.

4 Check that your analytic answers are consistent with

your ray diagram.

SECTION 33-2

The Thin Lens Equation; Magnification

871

Axis

Leaf '

3

F 4

:';

magesm,

1

100 cm

T

FIGURE 33-11

)

F

Example 33-2. (Not to scale.)

DG

TR

Image

formed

by

converging

lens. What

is (a)

the

position, and (b) the size, of the image of a 7.6-cm-high leaf placed 1.00m from a +50.0-mm-focal-length camera lens? APPROACH SOLUTION

We follow the steps of the Problem Solving Strategy explicitly.

1. Ray diagram. Figure 33-11 is an approximate ray diagram, showing only rays 1 and 3 for a single point on the leaf. We see that the image ought to be a little behind the focal point F, to the right of the lens. 2. Thin lens and magnification equations. (a) We find the image position analytically using the thin lens equation, Eq. 33-2. The camera lens is converging, with f = +5.00cm, and d, = 100cm, and so the thin lens equation gives

rF_1 d;

1 _

f

-d,

1

500cm

1

100 cm

_200-10 100 cm

190

.

100 cm

Then, taking the reciprocal, d;,

=

100 cm 00

-

) 5.26 cm,

or 52.6 mm behind the lens.

(b) The magnification is m

=

h;

=

d d—o

=

526cm Doem

0.0526,

SO mh,

=

(—0.0526)(7.6cm)

=

—0.40cm.

The image is 4.0 mm high. 3. Sign conventions. The image distance d; came out positive, so the image is behind the lens. The image height is #; = —0.40 cm; the minus sign means the image is inverted. 4. Consistency. The analytic results of steps (2) and (3) are consistent with the ray diagram, Fig. 33-11: the image is behind the lens and inverted. NOTE Part (a) tells us that the image is 2.6 mm farther from the lens than the image for an object at infinity, which would equal the focal length, 50.0 mm. Indeed, when focusing a camera lens, the closer the object is to the camera, the farther the lens must be from the sensor or film. EXERCISE A If the leaf (object) of Example 33-2 is moved farther from the lens, does the

image move closer to or farther from the lens? (Don’t calculate!)

872

CHAPTER 33

Lenses and Optical Instruments

-

Object close to converging lens. An object is placed 10 cm

from a 15-cm-focal-length converging lens. Determine the image position and size (a) analytically, and (b) using a ray diagram. fi

APPROACH We first use Egs. 33-2 and 33-3 to obtain an analytic solution, and then confirm with a ray diagram using the special rays 1, 2, and 3 for a single object point. SOLUTIOlTl (a) Given f = 15cm and d, = 10cm, then i d;

1 15cm

1 10cm

1 30cm

ACAUTION

and d; = —30cm. (Remember to take the reciprocal!) Because d; is negative, the image must be virtual and on the same side of the lens as the object. The magnification d;

=

’ i

—-— d,

=

-

—-30cm

10 cm

=

Don'’t forget to take the reciprocal

3.0

The imagé is three times as large as the object and is upright. This lens is being used as a simple magnifying glass, which we discuss in more detail in Section 33-7. (b) The ray diagram is shown in Fig. 3312 and confirms the result in part (a). We choose point O’ on the top of the object and draw ray 1, which is easy. But ray 2 may take some thought: if we draw it heading toward F’, it is going the wrong way—so we have to draw it as if coming from F’ (and so dashed), striking the lens, and then going out parallel to the lens axis. We project it back parallel, with a dashed line, as we must do also for ray 1, in order to find where they cross. Ray 3 is drawn through the lens center, and it crosses the other two rays at the image point, I'. NOTE From Fig. 33-12 we can see that, whenever an object is placed between a converging lens and its focal point, the image is virtual.

FIGURE 33-12 An object placed within the focal point of a converging lens produces a virtual image. Example 33-3.

DOV

T

W

Diverging lens. Where must a small insect be placed if a

25-cm-focal-length diverging lens is to form a virtual image 20 cm from the lens,

~ FIGURE 33-13

Exercise C.

on the same side as the object? |

APPROACH The ray diagram is basically that of Fig. 33-10 because our lens here is diverging and our image is in front of the lens within the focal distance. (It would be a valuable exercise to draw the ray diagram to scale, precisely, now.) The insect’s distance, d,, can be calculated using the thin lens equation.

SOLUTION The lens is diverging, so f is negative: f = —25cm. The image distance must be negative too because the image is in front of the lens (sign conventions), so d; = —20cm. Equation 33-2 gives i

d,

i

f

1T

4

1

25cm

+1

20cm

_

—4+5

100cm

1

100cm

So the objtjact must be 100 cm in front of the lens. |

EXERCISE B Return to the Chapter-Opening Question, page 866, and answer it again now. Try to explain why you may have answered differently the first time. f

I

EXERCISE C Figure 33-13 shows a converging lens held above three equal-sized letters A. In (a) the lens is 5 cm from the paper, and in (b) the lens is 15 cm from the paper. Estimate the focal length of the lens. What is the image position for each case?

SECTION 33-2

The Thin Lens Equation; Magnification

873

33-3

Combinations of Lenses

Optical instruments typically use lenses in combination. When light passes through more than one lens, we find the image formed by the first lens as if it were alone‘ This image becomes the object for the second lens, and we find the image the. formed by this second lens, which is the final image if there are only two lenses. The total magnification will be the product of the separate magnifications of each lens, as we shall see. Even if the second lens intercepts the light from the first lens before it forms an image, this technique still works.

A two-lens system. Two converging lenses, A and B, with

focal lengths f, =20.0cm and fz = 25.0cm, are placed 80.0cm apart, as shown in Fig. 33-14a. An object is placed 60.0cm in front of the first lens as shown in Fig. 33-14b. Determine (a) the position, and (b) the magnification, of the final image formed by the combination of the two lenses. APPROACH Starting at the tip of our object O, we draw rays 1, 2, and 3 for the first lens, A, and also a ray 4 which, after passing through lens A, acts as “ray 3” (through the center) for the second lens, B. Ray 2 for lens A exits parallel, and so is ray 1 for lens B. To determine the position of the image I, formed by lens A, we use Eq.33-2 with f, = 20.0cm and d,, = 60.0cm. The distance of I, (lens A’s image) from lens B is the object distance d,g for lens B. The final image is found using the thin lens equation, this time with all distances relative to lens B. For (b) the magnifications are found from Eq. 33-3 for each lens in turn. SOLUTION (@) The object is a distance d,, = +60.0cm from the first lens, A, and this lens forms an image whose position can be calculated using the thin lens equation:

1 dix A

CAUTION

Object distance for second lens is not equal to the image

distance for first lens

1

fs

s

1

dog

250cm

1

500cm

_2-=-1

500cm

(a) A

(b)

CHAPTER 33

1 _3-1 600cm 60.0cm

_

1 30.0cm

d;z from the lens B:

and B, used in combination, Example 33-5. The small numbers refer to the easily drawn rays.

874

1 200cm

-

The image formed by lens B, again using the thin lens equation, is at a distance

dy

Two lenses,A

1 don

So the first image I, is at dj, = 30.0cm behind the first lens. Th. image becomes the object for the second lens, B. It is a distance dog = 80.0cm — 30.0cm = 50.0cm in front of lens B, as shown in Fig. 33-14b.

1

FIGURE 33-14

1 fa

Lenses and Optical Instruments

doa

B /

1

50.0cm

(b) Lens A has a magnification (Eq. 33-3)

fikh

=

—

dip

don

=

-

30.0cm

60.0 cm

=

—(.

500

1

fThus, the first image is inverted and is half as high as the object (again Eq. 33-3): hiA

=

My

hoA

=

_O.SOOhOA.

Lens B takes this image as object and changes its height by a factor _

diB

M8 =

_

50.0 cm

TG T

S00em

_

0%

The second lens reinverts the image (the minus sign) but doesn’t change its size. The final image height is (remember h,p is the same as h;,) hiB

=

mBhoB

o

mBhiA

=

mBmAhoA

=

(mtotal)hoA~

The total magnification is the product of m, and mg, which here equals Myoral = Mamg = (—1.000)(—0.500) = +0.500, or half the original height, and the final ir?age is upright.

m

Measuring f for a diverging lens. To measure the focal

length of a diverging lens, a converging lens is placed in contact with it, as shown in Fig. 33-15. The Sun’s rays are focused by this combination at a point 28.5 cm, behind the lenses as shown. If the converging lens has a focal length f- of 16.0 cm, what is the focal length f;, of the diverging lens? Assume both lenses are thin and the space between them is negligible.

APPROACH The image distance for the first lens equals its focal length (16.0 cm) since the object distance is infinity (co). The position of this image, even though it is never actually formed, acts as the object for the second (diverging) lens. We apply the thin lens equation to the diverging lens to find its focal length, given Athat the final image is at d; = 28.5 cm. SOLUTION If the diverging lens was absent, the converging lens would form the image at its focal point—that is, at a distance f- = 16.0cm behind it (dashed lines in Fig.| 33-15). When the diverging lens is placed next to the converging lens, we treat the image formed by the first lens as the object for the second lens. Since this object lies to the right of the diverging lens, this is a situation where d,, is negative (see the sign conventions, page 871). Thus, for the diverging lens, the object is virtual and d, = —16.0cm. The diverging lens forms the image of this virtual object at a distance d; = 28.5 cm away (given). Thus, 1

—_

fo

=

1

—

dy

4+

1

—

di

=

1

—160cm

+

1

285cm

=

—=(.

R

_1'

We take the reciprocal to find f, = —1/(0.0274cm™!) = —36.5 cm.

NOTE If tflis technique is to work, the converging lens must be “stronger” than the diverging lens—that is, it must have a focal length whose magnitude is less than that of the diverging lens. (Rays from the Sun are focused 28.5 cm behind the combination, so the focal length of the total combination is f; = 28.5 cm.) Image point

made by first lens

(object point

for second lens)

:

T

FIGURE 33-15

Image made by

second lens

(final image)

.

Determining the

focal length of a diverging lens. Example 33-6.

|

=285cm———

SECTION 33-3

Combinations of Lenses

875

*33-4 Lensmaker’s Equation In this Section, we will show that parallel rays are brought to a focus at a single

point for a thin lens. At the same time, we will also derive an equation that relatess

the focal length of a lens to the radii of curvature of its two surfaces and its inde. of refraction, which is known as the lensmaker’s equation. In Fig. 33-16, a ray parallel to the axis of a lens is refracted at the front surface of the lens at point A; and is refracted at the back surface at point A,. This ray then passes through point F, which we call the focal point for this ray. Point A, is a height A, above the axis, and point A, is height &, above the axis. C; and C, are the centers of curvature of the two lens surfaces; so the length C; A; = R,, the radius of curvature of the front surface, and C,A, = R, is the radius of the second surface. We consider a double convex lens and by convention choose the radii of both lens surfaces as positive. The thickness of the lens has been grossly exaggerated so that the various angles would be clear. But we will assume that the lens is actually very thin and that angles between the rays and the axis are small. In this approximation, the sines and tangents of all the angles will be equal to the angles themselves in radians. For example, sin §; ~ tan 6, ~ 6, (radians).

FIGURE 33-16 Diagram of a ray passing through a lens for derivation of the lensmaker’s equation.

To this approximation, then, Snell’s law tells us that

where

01

=

n02

94

=

n03

n is the index of refraction of the glass, and we

surrounded by air (n = 1). Notice also in Fig. 33-16 that 0,

=

sinf,

‘

=

assume

that the lens is

—

:

The last follows because the distance from F to the lens (assumed very thin) is f. From the diagram, the angle 7 is Y

=

0]

==

02 .

A careful examination of Fig. 33-16 shows also that a

876

CHAPTER 33

=

03

-

Y.

This can be seen by drawing a horizontal line to the left from point A,, which divides the angle 6, into two parts. The upper part equals ¥ and the lower palfi equals a. (The opposite angles between an oblique line and two parallel lines are equal.) Thus, 6; = ¥ + a. Furthermore, by drawing a horizontal line to the right from point A,, we divide 6, into two parts. The upper part is @ and the lower is 3.

Thus 04

=

+

B.

We now combine all these equations:

or

Because the lens is thin, #; =~ h, and all A’s can be canceled from all the numerators. We then multiply through by n and rearrange to find that

= = fi= 1)(—1- - —1—).

(33-4)

This is called the lensmaker’s equation. It relates the focal length of a thin lens to the radii of curvature of its two surfaces and its index of refraction. Notice that f for a thin lens does not depend on A, or h,. Thus the position of the point F does not depend on where the ray strikes the lens. Hence, all rays parallel to the axis of a thin lens will pass through the same point F, which we wished to prove. In our derivation, both surfaces are convex and R, and R, are considered positive." Equation 33-4 also works for lenses with one or both surfaces concave; but for a concave surface, the radius must be considered negative. Notice in Eq. 33-4 that the equation is symmetrical in Ry and R,. Thus, if a lens is turned around so that light impinges on the other surface, the focal length is the same even if the two lens surfaces are different.

ACAUTION Sign conventions

Calculating f for a converging lens. A convex meniscus

lens (Figs. 33-2a and 33-17) is made from glass with n = 1.50. The radius of wurvature of the convex surface is 22.4cm and that of the concave surface is 46.2 cm. (a) What is the focal length? (b) Where will the image be for an object 2.00 m away? APPROACFI We use Eq. 33-4, noting that R, is negative because it refers to the concave surface. SOLUTION (a¢) Ry = 224cm and R, = —46.2cm. Then

1 f

(1.50—1.00)(

L

462cm

FIGURE 33-17

Example 33-7.

)

0.0115cm™.

So

1

I == — Soi5em and

-

2.4cm

G,

Ry =-46.2 cm—-**lA

the lens is converging.

Ry = —46.2cm

and

=

§7,

Notice

R, = +22.4cm,

oEmr that if we

(b) From the lens equation, with f = 0.870m 1

d

1

f

1

d,

turn the lens around

we get the same result.

il

080m

and

d, = 2.00m,

so that

we have

1

200m

0.649 m™!,

so d; =1/0.649m™! = 1.54m. EXERCISE D A Lucite planoconcave lens (see Fig. 33-2b) has one flat surface and the ther has R = —18.4cm. What is the focal length? Is the lens converging or diverging? fSome books use a different convention—for example, R; and R, are considered positive if their centers of curvature are to the right of the lens, in which case a minus sign appears in their equivalent of Eq. 33-4.

*SECTION 33-4

877

@PHYSlcs

APPLIED The camera

Viewfinder

Iris > diaphragm O dpe]

or

Shutter

“stop

FIGURE 33-18

Sensor or film

A simple camera.

@PHYSlcs

APPLIED

Digital cameras

Color pixel

Electrodes

FIGURE 33-19 Portion of a typical CCD sensor. A square group of four

pixels 8 is sometimes called a “color pixel.”

FIGURE 33-20 Suppose we take a picture that includes a thin black line (our object) on a white background. The image of this black line has a colored halo (red above, blue below) due to the mosaic arrangement of color filter pixels, as shown by the colors transmitted. Computer averaging can minimize color problems such as this (the green at top and bottom of image can be averaged with nearby pixels to give white or nearly so) but the image is consequently “softened” or blurred. The layered color pixel described in the text would avoid this artifact. Object

Pixels

Image

33—-5 Cameras: Film and Digital The basic elements of a camera are a lens, a light-tight box, a shutter to let light pass through the lens only briefly, and in a digital camera an electronic sensor o, in a traditional camera a piece of film (Fig. 33-18). When the shutter is opene. for a brief “exposure,” light from external objects in the field of view is focused by the lens as an image on the film or sensor. Film contains light-sensitive chemicals that change when light strikes them. In the development process, chemical reactions cause the changed areas to turn opaque, so the image is recorded on

the film."

You can see the image yourself if you remove the back of a conventional camera and view through a piece of tissue paper (on which an image can form) placed where the film should be with the shutter open.

Digital Cameras, Electronic Sensors (CCD, CMOS) In a digital camera, the film is replaced by a semiconductor sensor. Two types are in common use: CCD (charge-coupled device) and CMOS (complementary metal oxide semiconductor). A CCD sensor is made up of millions of tiny pixels (“picture elements”)—see Figs. 35-42 and 33-19. A 6-MP (6-megapixel) sensor would contain about 2000 pixels vertically by 3000 pixels horizontally over an area of perhaps 4 X 6 mm or even 24 X 36 mm. Light reaching any pixel liberates electrons from the semiconductor. The more intense the light, the more charge accumulates during the brief exposure time. Conducting electrodes carry each pixel’s charge (serially in time, row by row—hence the name “chargecoupled™) to a central processor that stores the relative brightness of pixels, and allows reformation of the image later onto a computer screen or printer. A CCD is fully reusable. Once the pixel charges are transferred to memory, a new picture can be taken.

A CMOS sensor also uses a silicon semiconductor, and incorporates som% electronics within each pixel, allowing parallel readout. Color is achieved by red, green, and blue filters over alternating pixels as

shown in Fig. 33-19, similar to a color CRT or LCD screen. The sensor type shown in Fig. 33-19 contains twice as many green pixels as red or blue (because green is claimed to have a stronger influence on the sensation of sharpness). The computeranalyzed color at each pixel is that pixel’s intensity averaged with the intensities of the nearest-neighbor colors. To reduce the amount of memory for each picture, compression programs average over pixels, but with a consequent loss of sharpness, or “resolution.”

*Digital Artifacts Digital cameras can produce image artifacts (artificial effects in the image not present in the original) resulting from the imaging process. One example using the “mosaic” of pixels (Fig. 33-19) is described in Fig. 33-20. An alternative technology uses a semitransparent silicon semiconductor layer system wherein different wavelengths of light penetrate silicon to different depths: each pixel is a sandwich of partly transparent layers, one for each color. The top layer can absorb blue light, allowing green and red light to pass through. The second layer absorbs green and the bottom layer detects the red. All three colors are detected by each pixel, resulting in better color resolution and fewer artifacts. "This is called a negative, because the black areas correspond to bright objects and vice versa. The same process occurs during printing to produce a black-and-white “positive” picture from the negative. Color film has three emulsion layers (or dyes) corresponding to the three primary colors. “Each different color of pixel in a CCD is counted as a separate pixel. In contrast, in an LCD screc (Section 35-12), a group of three subpixels is counted as one pixel, a more conservative count.

878

CHAPTER 33

Lenses and Optical Instruments

Camera Adjustments There are three main adjustments on good-quality cameras: shutter speed, f-stop, and focusing. Although most cameras today make these adjustments automatically, r\t is valuable to understand these adjustments to use a camera effectively. For pecial or top-quality work, a manual camera is indispensable (Fig. 33-21).

Exposure

time

or

shutter

speed

This refers to how quickly the digital

sensor can make an accurate reading, or how long the shutter of a camera is open and the film or sensor exposed. It could vary from a second or more (“time exposures”), to S or less. To avoid blurring from camera movement, exposure times shorter than 1i5s are normally needed. If the object is moving, even shorter exposure times are needed to “stop” the action. If the exposure (or sampling) time is not fast enough, the image will be blurred by camera shake no matter how many pixels a digital camera claims. Blurring in low light conditions is more of a problem with cell-phone cameras whose sensors are not usually the most sophisticated. Digital still cameras or cell phones that take short videos must have a fast enough “sampling” time and fast “clearing” (of the charge) time so as to take pictures at least 15 frames per second, and preferably 30 fps.

f-stop

The amount of light reaching the film must be carefully controlled to avoid ~ FIGURE 33-21

underexposure (too little light so the picture is dark and only the brightest objects show up) or overexposure (too much light, so that all bright objects look the same, with a consequent lack of contrast and a “washed-out” appearance). Most cameras these days make f-stop and shutter speed adjustments automatically. A high quality camera controls the exposure with a “stop” or iris diaphragm, whose opening is of variable diameter, placed behind the lens (Fig. 33-18). The size of the opening is controlled (automatically or manually) to compensate for bright or dark lighting

On this camera,

thet{l-stops andl Uie g;lcutstmg rng dare ::e S;g;?j?n :}I::smzll (:hstiieos top of the camera body.

conditions, the sensitivity of the sensor or film," and for different shutter speeds. The

size of the opening is specified by the f-number or f-stop, defined as f-stop

. D’ f

=

where f is the focal length of the lens and D is the diameter of the lens opening (Fig. 33-18). For example, when a S50-mm-focal-length lens has an opening D = 25mm, we say it is set at f/2. When this lens is set at f/8, the opening is

only 63 mm (50/6} = 8). For faster shutter speeds, or low light conditions, a wider

lens opening must be used to get a proper exposure, which corresponds to a smaller f-stop number. The smaller the f-stop number, the larger the opening and the more light passes through the lens to the sensor or film. The smallest f-number of a lens (largest opening) is referred to as the speed of the lens. The best lenses may have a speed of f/2.0, or even faster. The advantage of a fast lens is that it allows pictures to be taken under poor lighting conditions. Good quality lenses consist of several elements to reduce the defects present in simple thin lenses (Section 33-10). Standard f-stops are 1.0, 1.4, 2.0, 2.8, 40, 5.6, 8 11, 16, 22, and 32 (Fig. 33-21). Each of these stops corresponds to a diameter reduction by a factor of about V2 ~ 1.4. Because the amount of light reaching the film is proportional to the area of the opening, and therefore proportional to the diameter squared, each standard f-stop corresponds to a factor of 2 in light intensity reaching the film. Focusing Focusing is the operation of placing the lens at the correct position relative to the sensor or film for the sharpest image. The image distance is smallest for objects at infinity (the symbol oo is used for infinity) and is equal to the focal length. For closer objects, the image distance is greater than the focal length, as can be seen from the lens equation, 1/f = 1/d, + 1/d, (Eq. 33-2). To focus on nearby objects, the lens must therefore be moved away from the sensor or film, and this is usually done on a manual camera by turning a ring on the lens. fiifferent films have different sensitivities to light, referred to as the “film speed” and specified as an [SO (or ASA) number.” A “faster” film is more sensitive and needs less light to produce a good image. Faster films are grainier so offer less sharpness (resolution) when enlarged. Digital cameras may have a “gain” or “ISO” adjustment for sensitivity. Adjusting a CCD to be “faster” for low light conditions results in “noise,” the digital equivalent of graininess.

SECTION

33-5

879

4 FIGURE 33-22 Photos taken with a camera lens (a) focused on a nearby object with distant object blurry, and (b) focused on a more distant object with nearby object blurry.

‘

(a)

(b)

If the lens is focused on a nearby object, a sharp image of it will be formed, but the image of distant objects may be blurry (Fig. 33-22). The rays from a point on the distant object will be out of focus—they will form a circle on the sensor or film as shown (exaggerated) in Fig. 33-23. The distant object will thus produce an image consisting of overlapping circles and will be blurred. These circles are called circles of confusion. To include near and distant objects in the same photo, you (or the camera) can try setting the lens focus at an intermediate position. For a given distance setting, there is a range of distances over which the circles of confusion will be small enough that the images will be reasonably sharp. This is called the depth of field. The depth of field varies with the lens opening. If the lens opening is smaller, only rays through the central part of the lens are accepted, and these form smaller circles of confusion for a given object distance. Hence, at smaller lens openings, a greater range of object distances will fit within the circle of confusion criterion, so the depth of field is greater.” For sensor or film width of 36 mm (including 35-mm film cameras), the depth of fie. is usually based on a maximum circle of confusion diameter of 0.003 mm. FIGURE 33-23 When the lens is positioned to focus on a nearby object, points on a distant object produce circles and are therefore blurred. (The effect is shown greatly exaggerated.)

Rays from nearby object (in focus)

T |

Rays from

distant object

“Circle of confusion” for distant object (greatly exaggerated)

DOV E T Camera focus. How far must a 50.0-mm-focal-length camera lens be moved from its infinity setting to sharply focus an object 3.00 m away? APPROACH For an object at infinity, the image is at the focal point, by definition. For an object distance of 3.00 m, we use the thin lens equation, Eq. 33-2, to find the image distance (distance of lens to film or sensor). SOLUTION When focused at infinity, the lens is 50.0 mm from the film. When focused at d, = 3.00 m, the image distance is given by the lens equation,

1

d;

1

1

fds

I 500mm

1

__3000-50

3000mm

We solve for d; and find d; = 50.8 mm, from the film or digital sensor.

(3000)(50.0) mm

2950 150,000 mm

so the lens needs to move 0.8 mm away

*Smaller lens openings, however, result in reduced resolution due to diffraction (discussed in Chapter . Best resolution is typically found around f/8.

880

CHAPTER 33

Lenses and Optical Instruments

fl

CONCEPTUAL

EXAMPLE

33-9 | Shutter speed. To improve the depth of field,

you “stop down” your camera lens by two f-stops from f/4 to f/8. What should you do to the shutter speed to maintain the same exposure?

AESPONSE The amount of light admitted by the lens is proportional to the area " of the lens opening. Reducing the lens opening by two f-stops reduces the diameter by a factor of 2, and the area by a factor of 4. To maintain the same exposure, the shutter must be open four times as long. If the shutter speed had been 355, you would have to increase the exposure time to 3s.

*Picture Sharpness The sharpness of a picture depends not only on accurate focusing and short exposure times, but also on the graininess of the film, or the number of pixels for a digital camera. Fine-grained films are “slower,” meaning they require longer exposures for a given light level. Digital cameras have averaging (or “compression”) programs, such as JPEG, which reduce memory size by averaging over pixels where little contrast is detected. Hence it is unusual to use all pixels available. They also average over pixels in low light conditions, resulting in a less sharp photo. The quality of the lens strongly affects the image quality, and we discuss lens resolution and diffraction effects in Sections 33-10 and 35-4. The sharpness, or resolution, of a lens is often given as so many lines per millimeter, measured by photographing a standard set of parallel lines on fine-grain film or high quality sensor, or as so many dots per inch (dpi). The minimum spacing of distinguishable lines or dots gives the resolution; 50lines/mm is reasonable, 100lines/mm is quite good (=100 dots/mm ~2500 dpi on the sensor).

DGV

EAN

Pixels

and

resolution.

A 6-MP

(6-megapixel)

digital

camera offers a maximum resolution of 2000 X 3000 pixels on a 16-mm X 24-mm CCD sensor. How sharp should the lens be to make use of this resolution?

APPROACH We find the number of pixels per millimeter and require the lens to Je at least that good. SOLUTION We can either take the image height (2000 pixels in 16 mm) or the width (3000 pixels in 24 mm): 3000 pixels amm

125 pixels/mm.

We would want the lens to be able to resolve at least 125 lines or dots per mm as well, which would be a very good lens. If the lens is not this good, fewer pixels and less memory could be used. NOTE Increasing lens resolution is a tougher problem today than is squeezing more pixels on a CCD or CMOS. The sensor for high MP cameras must also be physically larger for greater light sensitivity (low light conditions).

Blown-up photograph. An enlarged photograph looks sharp

at normal viewing distances if the dots or lines are resolved to about 10 dots/mm. Would an 8 X 10-inch enlargement of a photo taken by the camera in Example 33-10 seem sharp? To what maximum size could you enlarge this 2000 X 3000-pixel image?

APPROACH We assume the image is 2000 X 3000 pixels on a 16 X 24-mm CCD

as in Example 33-10, or 125 pixels/mm. We make an enlarged photo 8 X 10in. = 20cm X 25cm. SOLUTION The short side of the CCD is 16 mm = 1.6cm long, and that side of the photograph is 8 inches or 20 cm. Thus the size is increased by a factor of 20cm/1.6cm = 12.5X (or25cm/2.4cm ~ 10x). To fill the 8 X 10-in. paper, we assume the enlargement is 12.5X. The pixels are thus enlarged 12.5X; so the ixel count of 125/mm on the CCD becomes 10 per mm on the print. Hence an 8 X 10-inch print is just about the maximum possible for a sharp photograph with 6 megaPixels. If you feel 7 dots per mm is good enough, you can enlarge to

mybe 11 X 14 inches.

SECTION 33-5

Cameras:

Film and Digital

881

*Telephotos and Wide-angles

FIGURE 33-24 Single-lens reflex (SLR) camera, showing how the image is viewed through the lens with the help of a movable mirror and prism.

@PHYSICS FIGURE 33-25 human eye.

APPLIED

The eye

Diagram of a

Ciliary

muscles

Iris Aqueous humor

Fovea

Pupil Cornea

Optic Ciliary muscles

nerve

Camera lenses are categorized into normal, telephoto, and wide angle, according to focal length and film size. A normal lens covers the sensor or film with a field of view that corresponds approximately to that of normal vision. A normal lens fo 35-mm film has a focal length in the vicinity of 50 mm. The best digital cameras aim for a sensor of the same size' (24 mm X 36mm). (If the sensor is smaller, digital cameras sometimes specify focal lengths to correspond with classic 35-mm cameras.) Telephoto lenses act like telescopes to magnify images. They have longer focal lengths than a normal lens: as we saw in Section 33-2 (Eq. 33-3), the height of the image for a given object distance is proportional to the image distance, and the image distance will be greater for a lens with longer focal length. For distant objects, the image height is very nearly proportional to the focal length. Thus a 200-mm telephoto lens for use with a 35-mm camera gives a 4X magnification over the normal 50-mm lens. A wide-angle lens has a shorter focal length than normal: a wider field of view is included, and objects appear smaller. A zoom lens is one whose focal length can be changed so that you seem to zoom up to, or away from, the subject as you change the focal length. Digital cameras may have an optical zoom meaning the lens can change focal length and maintain resolution. But an “electronic” or digital zoom just enlarges the dots (pixels) with loss of sharpness. Different types of viewing systems are used in cameras. In some cameras, you view through a small window just above the lens as in Fig. 33-18. In a single-lens reflex camera (SLR), you actually view through the lens with the use of prisms and mirrors (Fig. 33-24). A mirror hangs at a 45° angle behind the lens and flips up out of the way just before the shutter opens. SLRs have the advantage that you can see almost exactly what you will get. Digital cameras use an LCD display, and it too can show what you will get on the photo if it is carefully designed.

33—-6 The Human Eye; Corrective Lenses " The human eye resembles a camera in its basic structure (Fig. 33-25), but is far more sophisticated. The interior of the eye is filled with a transparent gel-like substance called the vitreous humor with index of refraction n = 1.337. Light enters this enclosed volume through the cornea and lens. Between the cornea and lens is a watery fluid, the aqueous humor (aqua is “water” in Latin) with n = 1.336. A diaphragm, called the iris (the colored part of your eye), adjusts automatically to control the amount of light entering the eye, similar to a camera. The hole in the iris through which light passes (the pupil) is black because no light is reflected from it (it’s a hole), and very little light is reflected back out from the interior of the eye. The retina, which plays the role of the film or sensor in a camera, is on the curved rear surface of the eye. The retina consists of a complex array of nerves and receptors known as rods and cones which act to change light energy into electrical signals that travel along the nerves. The reconstruction of the image from all these tiny receptors is done mainly in the brain, although some analysis may also be done in the complex interconnected nerve network at the retina itself. At the center of the retina is a small area called the fovea, about 0.25 mm in diameter, where the cones are very closely packed and the sharpest image and best color discrimination are found. Unlike a camera, the eye contains no shutter. The equivalent operation is carried out by the nervous system, which analyzes the signals to form images at the rate of about 30 per second. This can be compared to motion picture or television cameras, which operate by taking a series of still pictures at a rate of 24 (movies) or 30 (U.S. television) per second; their rapid projection on the screen gives the appearance of motion. TA “35-mm camera” uses film that is physically 35 mm wide; that 35 mm is not to be confused with . focal length. 35-mm film has sprocket holes, so only 24 mm of its height is used for the photo; the width is usually 36 mm for stills. Thus one frame is 24 mm X 36 mm. Movie frames are 18 mm X 24 mm.

882

CHAPTER 33

Lenses and Optical Instruments

The lens of the eye (n = 1386

to 1.406) does little of the bending of

the light rays. Most of the refraction is done at the front surface of the cornea (n = 1.376) at its interface with air (n = 1.0). The lens acts as a fine djustment for focusing at different distances. This is accomplished by the rfi iliary muscles (Fig. 33-25), which change the curvature of the lens so that its focal length is changed. To focus on a distant object, the ciliary muscles of the eye are relaxed and the lens is thin, as shown in Fig. 33-26a, and parallel rays focus at the focal point (on the retina). To focus on a nearby object, the muscles contract, causing the center of the lens to thicken, Fig. 33-26b, thus shortening the focal length so that images of nearby objects can be focused on the retina, behind the new focal point. This focusing adjustment is called accommodation. The closest distance at which the eye can focus clearly is called the near point of the eye. For young adults it is typically 25 cm, although younger children can often focus on objects as close as 10cm. As people grow older, the ability to accommodate is reduced and the near point increases. A given person’s far point is the farthest distance at which an object can be seen clearly. For some purposes it is useful to speak of a normal eye (a sort of average over the population), defined as

an eye having a near point of 25 cm and a far point of infinity. To check your own

near point, blace this book close to your eye and slowly move it away until the type 1s shar}?.

;

:

The “normal” eye is sort of an ideal. Many people have eyes that do not accommodate within the “normal” range of 25cm to infinity, or have some other defect. Two common defects are nearsightedness and farsightedness. Both can be corrected to a large extent with lenses—either eyeglasses or contact lenses. In nearsightedness, or myopia, the eye can focus only on nearby objects. The far point is not infinity but some shorter distance, so distant objects are not seen clearly. It is usually caused by an eyeball that is too long, although sometimes it is e curvature of the cornea that is too great. In either case, images of distant sbjects are focused in front of the retina. A diverging lens, because it causes parallel rays to diverge, allows the rays to be focused at the retina (Fig. 33-27a) and thus coflrects this defect. In farsightedness, or hyperopia, the eye cannot focus on nearby objects. Although distant objects are usually seen clearly, the near point is somewhat greater than the “normal” 25cm, which makes reading difficult. This defect is caused by an eyeball that is too short or (less often) by a cornea that is not sufficiently curved. It is corrected by a converging lens, Fig. 33-27b. Similar to hyperopia is presbyopia, which refers to the lessening ability of the eye to accommodate as a person ages, and the near point moves out. Converging lenses also compensate for this.

Focal point of lens and cornea -

Focal point of lens and cornea

~ FIGURE 33-26

'

Accommodation

lf)gcisr:a%?a}ln?:izt}(fz;‘)aT:rf:ltfifl?éne d focused on a nearl;y object.

@ PHYSICS Corrective lenses

APPLIED

i

(a) Nearsighted eye

N

FIGURE 33-27 Correcting eye defects with lenses: (a) a nearsighted eye, which cannot focus clearly on distant objects, can be corrected by use of a diverging lens; (b) a farsighted eye, which cannot focus clearly on nearby objects, can be corrected by use of a converging lens.

(b) Farsighted eye

SECTION 33-6

The Human Eye; Corrective Lenses

883

Image (line)

(point) FIGURE 33-28 A cylindrical lens forms a line image of a point object because it is converging in one plane only.

Astigmatism is usually caused by an out-of-round cornea or lens so that point objects are focused as short lines, which blurs the image. It is as if the cornea were spherical with a cylindrical section superimposed. As shown in Fig. 33-28, a cylindrical lens focuses a point into a line parallel to its axis. An astigmatic eye ma}'q focus rays in one plane, such as the vertical plane, at a shorter distance than does for rays in a horizontal plane. Astigmatism is corrected with the use of a compensating cylindrical lens. Lenses for eyes that are nearsighted or farsighted as well as astigmatic are ground with superimposed spherical and cylindrical surfaces, so that the radius of curvature of the correcting lens is different in different planes.

DEVILZNR

AP

Farsighted eye. Sue is farsighted with a near point of 100 cm.

Reading glasses must have what lens power so that she can read a newspaper at a distance of 25 cm? Assume the lens is very close to the eye. APPROACH When the object is placed 25 cm from the lens, we want the image to be 100 cm away on the same side of the lens (so the eye can focus it), and so the image is virtual, as shown in Fig, 33-29, and d; = —100 cm will be negative. We use the thin lens equation (Eq. 33-2) to determine the needed focal length. Optometrists’ prescriptions specify the power (P = 1/f, Eq. 33-1) given in

diopters (1D =1m™). SOLUTION gives

Given that

d, = 25cm

l__1_+_1__ f

d,

d,

and

1+

25cm

d; = —100cm,

1

—100 cm

.

the thin lens equation

4=3 .

100 cm

1

33 cm

So f =33cm = 0.33m. The power P of the lens is P =1/f = +3.0D. The plus sign indicates that it is a converging lens. NOTE We chose the image position to be where the eye can actually focus. The lens needs to put the image there, given the desired placement of the object (newspaper).

o

-

FIGURE 33-29 Lens of reading glasses (Example 33-12).

FIGURE 33-30

Nearsighted

Example 33-13.

Object

”

at oo

2cm

:

eye. A nearsighted

eye has near and far

points of 12cm and 17 cm, respectively. () What lens power is needed for this person to see distant objects clearly, and (b) what then will be the near point? Assume that the lens is 2.0 cm from the eye (typical for eyeglasses).

APPROACH For a distant object (d, = o), the lens must put the image at the 17 cm (Far point)

(a)

|

far point of the eye as shown in Fig. 33-30a, 17 cm in front of the eye. We can use the thin lens equation to find the focal length of the lens, and from this its lens power. The new near point (as shown in Fig. 33-30b) can be calculated for the lens by again using the thin lens equation.

SOLUTION

(a) For an object at infinity (d, = co), the image must be in front of

the lens 17cm from the eye or (17cm — 2cm) = 15cm from the lens; hence d; = —15cm. We use the thin lens equation to solve for the focal length of the needed lens:

1

12 cm (Near point)

(b) 884

CHAPTER 33

f

_1

do

1 di

1

oo

.

1

—15cm

T

So f=-15ecm = —-0.15m or P =1/f = —6.7D. that it must be a diverging lens for the myopic eye.

Lenses and Optical Instruments

1

15em The minus sign indicate

(b) The near point when glasses are worn is where an object is placed (d,) so that the lens forms an image at the “near point of the naked eye,” namely 12 cm from the eye. That image point is (12cm — 2cm) = 10cm in front of the lens, so r‘ d; = —0.10 m and the thin lens equation gives 1 do

1 f

1 4

e

1 0.15m

+

1 0.10m

_=2+3 = 0.30m

=

1 0.30m

.

So d, = 30cm, which means the near point when the person is wearing glasses is 30 cm in front of the lens, or 32 cm from the eye. Suppose contact lenses are used to correct the eye in Example 33-13. Since contacts are placed directly on the cornea, we would not subtract out the 2.0 cm so for the image distances. That is, for distant objects d;=f = —17cm, P =1/f = =59D. The new near point would be 41 cm. Thus we see that a contact lens and an eyeglass lens will require slightly different powers, or focal lengths, for the same eye because of their different placements relative to the eye. We also see that glasses in this case give a better near point than contacts.

@ PHYSICS Contact lenses

APPLIED

EXERCISE E What power contact lens is needed for an eye to see distant objects if its far point is 25 cm? |

When your eyes are under water, distant underwater objects look blurry because at the water—cornea interface, the difference in indices of refraction is very small: n = 1.33 for water, 1.376 for the cornea. Hence light rays are bent very little and are focused far behind the retina, Fig. 33-31a. If you wear goggles or a face mask, you restore an air—cornea interface (n = 1.0 and 1.376, respectively) and the rays can be focused, Fig. 33-31b. f

FIGURE 33-31 (a) Under water, we see a blurry image because light rays are bent much less than in air. (b) If we wear goggles, we again have an air—cornea interface and can see clearly.

Object

Object \‘:’.atcr f), and (b) is virtual and

upright if the object is within the focal point (d, < f).

Next, describe the image if the object is itself an image (formed by another lens), and its position is on the opposil\ side of the lens from the incoming light, (c¢) for —d, > and (d) for 0 < —d, < f.

16. (IT) A converging lens has focal length f. When an object is placed a distance d, > f from this lens, a real image with

magnification m is formed. (a) Show that m = f/(f — d,).

(.

(b) Sketch m vs. d, over the range f < d, < oo where f = 0.45 cm. (c) For what value of d, will the real image have the same (lateral) size as the object? (d) To obtain a real image that is much larger than the object, in what general region should the object be placed relative to the lens? 17. (II) In a slide or movie projector, the film acts as the object whose image is projected on a screen (Fig. 33—-46). If a 105-mm-focallength lens is to project an image on a screen 6.50m away, how far from the lens should the slide be? If the slide is 36 mm wide, how wide will the picture be on the screen?

FIGURE 33-46 Slide projector, Problem 17.

f\L |-

.

S

]lg

Screen

18. (III) A bright object is placed on one side of a converging lens of focal length f, and a white screen for viewing the image is on the opposite side. The distance dr = d; + d,

25. (II) Two lenses, one converging with focal length 20.0 cm and one diverging with focal length —10.0 cm, are placed 25.0cm apart. An object is placed 60.0cm in front of the converging lens. Determine (a) the position and (b) the magnification of the final image formed. (c¢) Sketch a ray diagram for this system. 26. (IT) A diverging lens is placed next to a converging lens of focal length f-, as in Fig. 33-15. If f; represents the focal length of the combination, show that the focal length of the diverging lens, fp, is given by

27. (IT) A lighted candle is placed 36 cm in front of a converging lens of focal length f; = 13 cm, which in turn is 56 cm in front of another converging lens of focal length f, = 16 cm (see Fig. 33-47). (a) Draw a ray diagram and estimate the location and the relative size of the final image. (b) Calculate the position and relative size of the final image.

between the object and the screen is kept fixed, but the lens

xx' = f 2

&

where x is the distance of the object from the focal point on the front side of the lens, and x' is the distance of the image to the focal point on the other side of the lens. Calculate the location of an image if the object is placed 48.0 cm in front of a convex lens with a focal length of 38.0 cm using (b) the standard form of the thin lens formula, and (c¢) the Newtonian form, derived above.

33-3 Lens Combinations 20. (IT) A diverging lens with f = —33.5cm is placed 14.0 cm behind a converging lens with f = 20.0cm. Where will an object at infinity be focused? 21. (IT) Two 25.0-cm-focal-length converging lenses are placed 16.5 cm apart. An object is placed 35.0cm in front of one lens. Where will the final image formed by the second lens be located? What is the total magnification? 22. (II) A 34.0-cm-focal-length converging lens is 24.0cm behind a diverging lens. Parallel light strikes the diverging lens. After passing through the converging lens, the light is again parallel. What is the focal length of the diverging lens? [Hint: first draw a ray diagram.] 23. (II) The two converging lenses of Example 33-5 are now placed only 20.0 cm apart. The object is still 60.0 cm in front of the first lens

as in Fig. 33-14.

In this case, determine

(a) the position of the final image, and (b) the overall magnification. (c) Sketch the ray diagram for this system. 24, (IT) A diverging lens with a focal length of —14 cm is placed 12 cm to the right of a converging lens with a focal length of 18cm. An object is placed 33cm to the left of the converging lens. (a) Where will the final image be located? (b) Where will the image be if the diverging lens is 38 cm from the converging lens?

fi=13cm

FIGURE 33-47 Problem 27.

*33-4

8 soem

fr=16cm 56 cm

Lensmaker’s Equation

*28. (I) A double concave lens has surface radii of 33.4 cm and 28.8 cm. What is the focal length if n = 1.58? *29. (I) Both surfaces of a double convex lens have radii of 31.4 cm. If the focal length is 28.9 cm, what is the index of refraction of the lens material? . (I) Show that if the lens of Example 33-7 is reversed, the focal length is unchanged. . (I) A planoconvex lens (Fig. 33-2a) is to have a focal length of 18.7cm. If made from fused quartz, what must be the radius of curvature of the convex surface? *32. (II) An object is placed 90.0cm from a glass lens (n = 1.52) with one concave surface of radius 22.0 cm and one convex surface of radius 18.5 cm. Where is the final image? What is the magnification? *33. (II) A prescription for a corrective lens calls for +3.50 diopters. The lensmaker grinds the lens from a “blank” with n = 1.56 and convex front surface of radius of curvature of 30.0cm. What should be the radius of curvature of the other surface? 33-5 Camera 34. (I) A properly exposed photograph is taken at f/16 and 3;5. What lens opening is required if the shutter speed is 1a555? . (I) A television camera lens has a 17-cm focal length and a lens diameter of 6.0 cm. What is its f-number? . (II) A “pinhole” camera uses a tiny pinhole instead of a lens. Show, using ray diagrams, how reasonably sharp images can be formed using such a pinhole camera. In particular, consider two point objects 2.0 cm apart that are 1.0m from a 1.0-mm-diameter pinhole. Show that on a piece of film 7.0cm behind the pinhole the two objects produce two separate circles that do not overlap. 37 (II) Suppose that a correct exposure is 5i;s at f/11. Under the same conditions, what exposure time would be needed for a pinhole camera (Problem 36) if the pinhole diameter is 1.0mm and the film is 7.0 cm from the hole?

w W

can be moved. (a) Show that if dp > 4f, there will be two positions where the lens can be placed and a sharp image will be produced on the screen. (b) If dr < 4f, show that there will be no lens position where a sharp image is formed. (c) Determine a formula for the distance between the two lens positions in part (a), and the ratio of the image sizes. 19. (III) (a) Show that the lens equation can be written in the Newtonian form:

Problems

895

38. (II) Human vision normally covers an angle of about 40° horizontally. A “normal” camera lens then is defined as follows: When focused on a distant horizontal object which subtends an angle of 40°, the lens produces an image that extends across the full horizontal extent of the camera’s light-recording medium (film or electronic sensor). Determine the focal length f of the “normal” lens for the following types of cameras: (a) a 35-mm camera that records images on film 36 mm wide; (b) a digital camera that records images on a charge-coupled device (CCD) 1.00 cm wide. 39. (II) A nature photographer wishes to photograph a 38-m tall tree from a distance of 65 m. What focal-length lens should be used if the image is to fill the 24-mm height of the sensor?

33-6

Eye and Corrective Lenses

40. (I) A human eyeball is about 2.0 cm long and the pupil has a maximum diameter of about 8.0 mm. What is the “speed” of this lens? 41. (IT) A person struggles to read by holding a book at arm’s length, a distance of 55cm away. What power of reading glasses should be prescribed for her, assuming they will be placed 2.0cm from the eye and she wants to read at the “normal” near point of 25 cm? 42. (IT) Reading glasses of what power are needed for a person whose near point is 105 cm, so that he can read a computer

screen at 55 cm? Assume a lens—eye distance of 1.8 cm.

43. (II) If the nearsighted person in Example 33-13 wore contact lenses corrected for the far point (= 0o), what would be the near point? Would glasses be better in this case? . (II) An eye is corrected by a —4.50-D lens, 2.0 cm from the eye. (a) Is this eye near- or farsighted? (b) What is this eye’s far point without glasses?

33-7 51

47. (II) One lens of a nearsighted person’s eyeglasses has a focal length of —23.0 cm and the lens is 1.8 cm from the eye. If the person switches to contact lenses placed directly on

the

eye,

what

should

corresponding contact lens?

be

the

focal

length

of

the

. (II) What is the focal length of the eye lens system when viewing an object (a) at infinity, and (b) 38 cm from the eye? Assume that the lens—retina distance is 2.0 cm. 49. (I) A nearsighted person has near and far points of 10.6 and 20.0 cm respectively. If she puts on contact lenses with power P = —4.00D, what are her new near and far points? . (II) The closely packed cones in the fovea of the eye have a diameter of about 2 um. For the eye to discern two images on the fovea as distinct, assume that the images must be separated by at least one cone that is not excited. If these images are of two point-like objects at the eye’s 25-cm near point, how far apart are these barely resolvable objects? Assume the diameter of the eye (cornea-to-fovea distance) is 2.0 cm.

CHAPTER 33

Lenses and Optical Instruments

of

53. (I) A magnifier is rated at 3.0X for a normal eye focusing on an image at the near point. (a) What is its focal length? (b) What is its focal length if the 3.0X refers to a relaxed eye? 54. (II) Sherlock Holmes is using an 8.80-cm-focal-length lens as his magnifying glass. To obtain maximum magnification, where must the object be placed (assume a normal eye), and what will be the magnification? 55. (IT) A small insect is placed 5.85 cm from a +6.00-cm-focallength lens. Calculate (a) the position of the image, and (b) the angular magnification. 56. (II) A 3.40-mm-wide bolt is viewed with length lens. A normal eye views the image Calculate (@) the angular magnification, (b) image, and (c) the object distance from the

a 9.60-cm-focalat its near point. the width of the lens.

57. (IT) A magnifying glass with a focal length of 9.5 cm is used to read print placed at a distance of 8.3 cm. Calculate (a) the position of the image; (b) the angular magnification. 58. (II) A magnifying glass is rated at 3.0X for a normal eye that is relaxed. What would be the magnification for a relaxed eye whose near point is (a) 65 cm, and (b) 17 cm? Explain the differences. 59. (II) A converging lens of focal length f = 12cm is being used by a writer as a magnifying glass to read some fine print on his book contract. Initially, the writer holds the le

above the fine print so that its image is at infinity. To gex better look, he then moves the lens so that the image is at his 25-cm near point. How far, and in what direction

(toward or away from the fine print) did the writer move the lens? Assume the writer’s eye is adjusted to remain always very near the magnifying glass.

33-8 Telescopes 60. (I) What is the magnification of an astronomical telescope whose objective lens has a focal length of 78 cm, and whose eyepiece has a focal length of 2.8 cm? What is the overall length of the telescope when adjusted for a relaxed eye? 61. (I) The overall magnification of an astronomical telescope is desired to be 35X. If an objective of 88 cm focal length is used, what must be the focal length of the eyepiece? What is the overall length of the telescope when adjusted for use by the relaxed eye? 62. (II) A 7.0X binocular has 3.0-cm-focal-length What is the focal length of the objective lenses?

eyepieces.

63. (IT) An astronomical telescope has an objective with focal length 75cm and a +35D eyepiece. What is the total magnification? (II) An astronomical telescope has its two lenses spaced 78.0cm apart. If the objective lens has a focal length of 75.5 cm, what is the magnification of this telescope? Assume a relaxed eye. 65.

(I) A Galilean telescope adjusted for a relaxed eye 33.8cm long. If the objective lens has a focal length 36.0 cm, what is the magnification?

896

glass

S2. (I) What is the magnification of a lens used with a relaxe& eye if its focal length is 13 cm?

45. (IT) A person’s right eye can see objects clearly only if they are between 25cm and 78 cm away. (a) What power of contact lens is required so that objects far away are sharp? (b) What will be the near point with the lens in place? . (II) A person has a far point of 14 cm. What power glasses would correct this vision if the glasses were placed 2.0 cm from the eye? What power contact lenses, placed on the eye, would the person need?

Magnifying Glass

(I) What is the focal length of a magnifying 3.8X magnification for a relaxed normal eye?

66. (IT) What is the magnifying power of an astronomical telescope using a reflecting mirror whose radius of curvature is 6.4m and an eyepiece whose focal length is 2.8 cm?

¢ 68.

(IT) The Moon’s image appears to be magnified 120X by a reflecting astronomical telescope with an eyepiece having a focal length of 3.1 cm. What are the focal length and radius of curvature of the main (objective) mirror? (IT) A 120X astronomical telescope is adjusted for a relaxed eye when the two lenses are 1.25 m apart. What is the focal length of each lens?

69. (IT) An astronomical telescope longer than about 50 cm is not easy to hold by hand. Based on this fact, estimate the maximum angular magnification achievable for a telescope designed to be handheld. Assume its eyepiece lens, if used as a magnifying glass, provides a magnification of 5X for a relaxed eye with near point N = 25 cm. 70. (III) A reflecting telescope (Fig. 33-38b) has a radius of curvature of 3.00 m for its objective mirror and a radius of curvature of —1.50 m for its eyepiece mirror. If the distance between the two mirrors is 0.90m, how far in front of the eyepiece should you place the electronic sensor to record the image of a star? 71. (III) A 7.5X pair of binoculars has an objective focal length of 26 cm. If the binoculars are focused on an object 4.0m away (from the objective), what is the magnification? (The 7.5X% refers to objects at infinity; Eq. 33-7 holds only for objects at infinity and not for nearby ones.)

*33~-9

Microscopes

*T72, (I) A microscope uses an eyepiece with a focal length of 1.50 cm. Using a normal eye with a final image at infinity, the barrel length is 17.5 cm and the focal length of the objective lens is 0.65 cm. What is the magnification of the microscope? . (I) A 680X microscope uses a 0.40-cm-focal-length objective lens. If the barrel length is 17.5 cm, what is the focal length of the eyepiece? Assume a normal eye and that the final image is at infinity. *74. (I) A 17-cm-long microscope has an eyepiece with a focal length of 2.5cm and an objective with a focal length of 0.28 cm. What is the approximate magnification? *75. (IT) A microscope has a 13.0X eyepiece and a 58.0X objective lens 20.0 cm apart. Calculate (a) the total magnification, (b) the focal length of each lens, and (c) where the object must be for a normal relaxed eye to see it in focus.

*76 .(II) Repeat Problem 75 assuming that the final image is located 25 cm from the eyepiece (near point of a normal eye). * 77 (IT) A microscope has a 1.8-cm-focal-length eyepiece and a 0.80-cm objective. Assuming a relaxed normal eye, calculate (a) the position of the object if the distance between the lenses is 16.8 cm, and (b) the total magnification. *78. (IT) The eyepiece of a compound microscope has a focal length of 2.80 cm and the objective lens has f = 0.740 cm. If an object is placed 0.790cm from the objective lens, calculate

(a)

the

distance

between

the

lenses

when

the

microscope is adjusted for a relaxed eye, and (b) the total magnification. *79. (IT) An inexpensive instructional lab microscope allows the user to select its objective lens to have a focal length of 32 mm, 15 mm, or 3.9 mm. It also has two possible eyepieces

with magnifications 5X and 10X. Each objective forms a real image 160 mm beyond its focal point. What are the largest and smallest overall magnifications obtainable with this instrument? *80. (IIT) Given two 12-cm-focal-length lenses, you attempt to make a crude microscope using them. While holding these lenses a distance 55 cm apart, you position your microscope so that its objective lens is distance d, from a small object. Assume your eye’s near point N = 25cm. (a) For your microscope to function properly, what should d, be? (b) Assuming your eye is relaxed when using it, what magnification M does your microscope achieve? (c) Since the length of your microscope is not much greater than the focal lengths of its lenses, the approximation M ~ N{/f. f, is not valid. If you apply this approximation to your microscope, what % error do you make in your microscope’s true magnification?

*33-10

Lens Aberrations

*81. (IT) A planoconvex lens (Fig. 33-2a) has one flat surface and the other has R = 153 cm. This lens is used to view a red and yellow object which is 66.0 cm away from the lens. The index of refraction of the glass is 1.5106 for red light and 1.5226 for yellow light. What are the locations of the red and yellow images formed by the lens? *82. (IT) An achromatic lens is made of two very thin lenses, placed in contact, that have focal lengths f; = —28cm and f»=+25cm. (a) Is the combination converging or diverging? (b) What is the net focal length?

| General Problems 83. A 200-mm-focal-length lens can be adjusted so that it is 200.0mm to 206.4mm from the film. For what range of object distances can it be adjusted? . If a 135-mm distances

from

telephoto 1.30m

lens is designed

to oo, over what

to cover

distance

object

must

lens move relative to the plane of the sensor or film?

the

85. For a camera equipped with a 58-mm-focal-length lens, what is the object distance if the image height equals the object height? How far is the object from the image on the film? . Show that for objects very far away (assume infinity), the magnification of any camera lens is proportional to its focal length.

87. A small object is 25.0 cm from a diverging lens as shown in Fig. 33-48. A converging lens with a focal length of 12.0 cm is 30.0 cm to the right of the diverging lens. The two-lens system forms a real inverted image 17.0 cm to the right of the converging lens. What is the focal length of the diverging lens?

i—25.0

e

30.0 cm

cm—A

FIGURE 33-48

I 17.0 cm

Av

|

i

Problem 87.

General Problems

897

88. A converging lens with focal length of 13.0cm is placed in contact with a diverging lens with a focal length of —20.0 cm. What is the focal length of the combination, and is the combination converging or diverging?

97. A child has a near point of 15 cm. What is the maximum magnification the child can obtain using an 8.5-cm-focallength magnifier? What magnification can a normal eye obtain with the same lens? Which person sees more

the two lenses are 28 cm apart, determine the focal length of each lens.

98. A woman can see clearly with her right eye only when objects are between 45 cm and 155 cm away. Prescription

combination is given by fr = fif2/(fi + f2). (b) Show that

book 25 cm away (lower part) with her right eye? Assume

89. An astronomical telescope has a magnification of 8.0X. If

90. (a) Show that if two thin lenses of focal lengths f; and f; are placed in contact with each other, the focal length of the

the power P of the combination of two lenses is the sum of their separate powers,

P = P, + P,.

91. How large is the image of the Sun on film used in a camera

detail?

bifocals should have what powers so that she can see distant objects clearly (upper part) and be able to read a that the glasses will be 2.0 cm from the eye.

99. What is the magnifying power of a +4.0-D lens used as a

magnifier? Assume a relaxed normal eye.

with (a) a 28-mm-focal-length lens, (b) a 50-mm-focal-length lens, and (c) a 135-mm-focal-length lens? (d) If the 50-mm lens is considered normal for this camera, what relative magnification does each of the other two lenses provide?

100. A physicist lost in the mountains tries to make a telescope using the lenses from his reading glasses. They have powers of +2.0D and +4.5D, respectively. (¢) What maximum magnification telescope is possible? (b) Which lens should

away. 92. Two converging lenses are placed 30.0 cm apart. The focal length of the lens on the right is 20.0cm, and the focal length of the lens on the left is 15.0cm. An object is placed to the left of the 15.0-cm-focal-length lens. A final image

101. A 50-year-old man uses +2.5-D lenses to read a newspaper 25 cm away. Ten years later, he must hold the paper 32 cm away to see clearly with the same lenses. What power lenses does he need now in order to hold the paper 25 cm away? (Distances are measured from the lens.)

The Sun has diameter 1.4 X 10°km, and it is 1.5 X 10%km

from both lenses is inverted and located halfway between the two lenses. How

far to the left of the 15.0-cm-focal-

length lens is the original object?

93. When an object is placed 60.0 cm from a certain converging lens, it forms a real image. When the object is moved to 40.0 cm from the lens, the image moves 10.0 cm farther from the lens. Find the focal length of this lens. 94. Figure 33-49 was taken from the NIST Laboratory (National Institute of Standards and Technology) in Boulder, CO, 2 km from the hiker in the photo. The Sun’s image was 15mm across on the film. Estimate the focal length of the camera

lens

(actually a telescope). The

has diameter 1.4 X 10°km, and it is 1.5 X 10°km away.

Sun

.

be used as the eyepiece?

102. An object is moving toward a converging lens of focal

length f with constant speed v, such that its distance d,,

from the lens is always greater than f. (a) Determine the

velocity v; of the image as a function of d,. (b) Which direction (toward or away from the lens) does the image move? (c) For what d, does the image’s speed equal the object’s speed? 103. The objective lens and the eyepiece of a telescope are spaced 85 cm apart. If the eyepiece is +23 D, what is they total magnification of the telescope? *104. Two converging lenses, one with f = 4.0cm and the other with

f = 44 cm,

are made

into a telescope. (¢) What

are

the length and magnification? Which lens should be the

eyepiece? (b) Assume these lenses are now combined to make a microscope; if the magnification needs to be 25X,

FIGURE 33-49 Problem 94. 95. A movie star catches a reporter shooting pictures of her at home. She claims the reporter was trespassing. To prove her point, she gives as evidence the film she seized. Her 1.75-m height is 8.25 mm high on the film, and the focal length of the camera lens was 220 mm. How far away from the subject was the reporter standing? 96. As early morning passed toward midday, and the sunlight

got more intense, a photographer noted that, if she kept her shutter speed constant, she had to change the f-number from f/5.6 to f/16. By what factor had the sunlight intensity increased during that time?

898

CHAPTER 33

Lenses and Optical Instruments

how long would the microscope be? 105. Sam purchases +3.50-D eyeglasses which correct his faulty vision to put his near point at 25 cm. (Assume he wears the lenses 2.0 cm from his eyes.) (a) Calculate the focal length of Sam’s glasses. (b) Calculate Sam’s near point without glasses. (c) Pam, who has normal eyes with near point at 25 cm, puts on Sam’s glasses. Calculate Pam’s near point with Sam’s glasses on. 106. The proper functioning of certain optical devices (e.g., optical fibers and spectrometers) requires that the input light be a collection of diverging rays within a cone of halfangle 6 (Fig. 33-50). If the light initially exists as a collimated beam (i.e., parallel rays), show that a single lens of focal length f and diameter D can be used to create the required input light if D/f = 2tan6. If 6 = 3.5° for a certain spectrometer, what focal length lens should be used if the lens diameter is 5.0 cm? =

D 1

: FIGURE 33-50

6

Input light

for device

Problem 106.

“

107. In a science-fiction novel, an intelligent ocean-dwelling creature’s eye functions underwater with a near point of 25cm. This creature would like to create an underwater magnifier out of a thin plastic container filled with air. r. What shape should the air-filled plastic container have (i.e., determine radii of curvature of its surfaces) in order for it to be used by the creature as a 3.0X magnifier? Assume the eye is focused at its near point, 108. A telephoto lens system obtains a large magnification in a compact package. A simple such system can be constructed out of two lenses, one converging and one diverging, of focal lengths f; and f, = —1f;, respectively, separated

by

a

distance

£=3f;

as

shown

in

Fig.

*Numerical/Computer *109.

di(cm) m

point. Go on to show that the total magnification for the

two-lens system is m ~ —2f;/d,. (b) For an object located at infinity, show that the two-lens system forms an image that is a distance 3 f; behind the first lens. (c) A single 250mm-focal-length lens would have to be mounted about 250 mm from a camera’s film in order to produce an image of a distant object at d, with magnification —(250mm)/d,. To produce an image of this object with the same magnification using the two-lens system, what value of f; should be used and how far in front of the film should the first lens be placed? How much smaller is the “focusing length” (i.e., first lens-to-final image distance) of this two-lens system in comparison with the 250-mm “focusing length” of the equivalent single lens?

0=3 4f 1

20 043

25 -079

30 -114

35 -150

40 —1.89

(a) Show analytically that a graph of m vs. d; should produce a straight line. What are the theoretically expected values for the slope and the y-intercept of this line? [Hint: d, is not constant.] (b) Using the data above, graph m vs. d; and show that a straight line does indeed result. Use the slope of this line to determine the focal length of the lens. Does the y-intercept of your plot have the expected value? (¢) In performing such an experiment, one has the practical problem of locating the exact center of the lens since dj

33-51.

(a) For a distant object located at distance d, from the first lens, show that the first lens forms an image with magnification my ~ —fi/d, located very close to its focal

|

(III) In the “magnification” method, the focal length f of a converging lens is found by placing an object of known size at various locations in front of the lens and measuring the resulting real-image distances d; and their associated magnifications m (minus sign indicates that image is inverted). The data taken in such an experiment are given here:

must be measured

from this point. Imagine, instead, that

one measures the image distance d} from the back surface of the lens, which is a distance ¢ from the lens’s center. Then, d; = d{ + £. Show that, when implementing the magnification method in this fashion, a plot of m vs. d! will still result in a straight line. How can f be determined from this straight line?

: /

|

Light from distant

| | | Image

object

: |

“Focusing length” FIGURE 33-51

Problem 108.

Answers to Exercises A: Closer to it.

r‘:

D: —36 cm; diverging.

(b) and (d) are true.

C: f=10cm

for both cases; d;i = —10cm

E: in (a) and 30cm in (b).

F:

P=—40D. 48cm.

General Problems

899

The beautiful colors from the surface of this soap bubble can be nicely explained by the wave theory of light. A soap bubble is a very thin spherical film filled with air. Light reflected from the outer and inner surfaces of this thin film of soapy water interferes constructively to produce the bright colors. Which color we see at any point depends on the thickness of the soapy water film at that point and also on the viewing angle. Near the top of the bubble, we see a small black area surrounded by a silver or white area. The bubble’s thickness is smallest at that black spot, perhaps only about 30 nm thick, and is fully transparent (we see the black background). We cover fundamental aspects of the wave nature of light, including two-slit interference and interference in thin films.

®

The Wave Nature of Light; Interference CHAPTER-OPENING

CONTENTS 34-1

Waves versus Particles; Huygens’ Principle and Diffraction

34-2

Huygens’ Principle and the Law of Refraction

34-3

Interference— Young’s Double-Slit Experiment

*34-4 Intensity in the Double-Slit Interference Pattern

34-5

Interference in Thin Films

#34-6

Michelson Interferometer

#34-7

Luminous Intensity

QUESTION—Guess

now!

When a thin layer of oil lies on top of water or wet pavement, you can often see swirls of color. We also see swirls of color on the soap bubble shown above. What causes these colors? (a) Additives in the oil or soap reflect various colors. (b) Chemicals in the oil or soap absorb various colors. (¢) Dispersion due to differences in index of refraction in the oil or soap. (d) The interactions of the light with a thin boundary layer where the oil (or soap) and the water have mixed irregularly. (e) Light waves reflected from the top and bottom surfaces of the thin oil or soap film can add up constructively for particular wavelengths. hat light carries energy is obvious to anyone who has focused the Sun’s rays with a magnifying glass on a piece of paper and burned a hole in it. But how does light travel, and in what form is this energy carried? In our discussion of waves in Chapter 15, we noted that energy can be carried from place to place in basically two ways: by particles or by waves. In the first case, material objects or particles can carry energy, such as an avalanche of rocks or rushing water. In the second case, water waves and sound waves, for example, can

carry energy over long distances even though the oscillating particles of the medium do not travel these distances. In view of this, what can we say abou.m, the nature of light: does light travel as a stream of particles away from its source, ¢ does light travel in the form of waves that spread outward from the source?

Historically, this question has turned out to be a difficult one. For one thing,

light does not reveal itself in any obvious way as being made up of tiny particles; nor do we see tiny light waves passing by as we do water waves. The evidence seemed to favor first one side and then the other until about 1830, when most )hysicists had accepted the wave theory. By the end of the nineteenth century, light was considered to be an electromagnetic wave (Chapter 31). In the early twentieth century, light was shown to have a particle nature as well, as we shall discuss in Chapter 37. We now speak of the wave-particle duality of light. The wave theory of light remains valid and has proved very successful. We now

investigate the evidence for the wave theory and how it has been used to explain a wide range of phenomena.

34—-1

g S

Waves versus Particles;

Huygens' Principle and Diffraction

The Dutch scientist Christian Huygens (1629-1695), a contemporary of Newton, proposed a wave theory of light that had much merit. Still useful today is a technique Huygens developed for predicting the future position of a wave front

when an earlier position is known. By a wave front, we mean all the points along

a two- or three-dimensional wave that form a wave crest—what we simply call a

“wave” as seen on the ocean. Wave

fronts are perpendicular to rays as we

FIGURE 34-1 Huygens’ principle, used to determine wave front CD

~ when wave front AB is given. FIGURE 34-2

Huygens’ principle

already discussed in Chapter 15 (Fig. 15-20). Huygens’ principle can be stated as follows: Every point on a wave front can be considered as a source of tiny

is consistent with diffraction (a)around the edge of an obstacle,

The new wave front is the envelope of all the wavelets—that is, the tangent to all of them.

2 small hole whose size is on the order of the wavelength of the wave.

wavelets that spread out in the forward direction at the speed of the wave itself. ~ (b) through alarge hole, (c) through

As a simple example of the use of Huygens’ principle, consider the wave rfiront AB in Fig. 34-1, which is traveling away from a source S. We assume the nedium is isotropic—that is, the speed v of the waves is the same in all directions. To find the wave front a short time ¢ after it is at AB, tiny circles are drawn with radius r = of. The centers of these tiny circles are blue dots on the original wave front AB, and the circles represent Huygens’ (imaginary) wavelets. The tangent to all these wavelets, the curved line CD, is the new position of the wave front. Huygens’ principle is particularly useful for analyzing what happens when waves impinge on an obstacle and the wave fronts are partially interrupted. Huygens’ principle predicts that waves bend in behind an obstacle, as shown in Fig. 34-2. This is just what water waves do, as we saw in Chapter 15 (Figs. 15-31 and 15-32). The bending of waves behind obstacles into the “shadow region” is known as diffraction. Since diffraction occurs for waves, but not for particles, it can serve as one means for distinguishing the nature of light. Note, as shown in Fig. 34-2, that diffraction is most prominent when the size of the opening is on the order of the wavelength of the wave. If the opening is much larger than the wavelength, diffraction goes unnoticed. Does light exhibit diffraction? In the mid-seventeenth century, the Jesuit priest Francesco Grimaldi (1618-1663) had observed that when sunlight entered a darkened room through a tiny hole in a screen, the spot on the opposite wall was larger than would be expected from geometric rays. He also observed that the border of the image was not clear but was surrounded by colored fringes. Grimaldi attributed this to the diffraction of light. The wave model of light nicely accounts for diffraction, and we discuss diffraction in detail in the next Chapter. But the ray model (Chapter 32) cannot account for diffraction, and it is important to be aware of such limitations to the ray model. rGeometric optics using rays is successful in a wide range of situations only because aormal openings and obstacles are much larger than the wavelength of the light, and so relatively little diffraction or bending occurs.

SECTION 34-1

| (@)

(b)

)

(©

Waves versus Particles; Huygens’ Principle and Diffraction

901

34-2

A

g

L

|

L\ FIGURE 34-3

.

/MEdfum 1 %}"2‘1‘:‘;}3

Ray Refraction

: S explained, using Huygens’ principle. Wave fronts are perpendicular to the

rays.

Huygens' Principle and the Law of Refraction

The laws of reflection and refraction were well known in Newton’s time. The la'\)\ of reflection could not distinguish between the two theories we just discussed: waves versus particles. For when waves reflect from an obstacle, the angle of incidence equals the angle of reflection (Fig. 15-21). The same is true of particles—think of a tennis ball without spin striking a flat surface. The law of refraction is another matter. Consider a ray of light entering a medium where it is bent toward the normal, as when traveling from air into water.

As shown in Fig. 34-3, this bending can be constructed using Huygens’ principle if we assume the speed of light is less in the second medium (v, < v;). In time ¢, point B on wave front AB (perpendicular to the incoming ray) travels a distance v, ¢ to reach point D. Point A on the wave front, traveling in the second medium, goes a » ) . . . . distance v, to rt?ach point C, and v,¢ < v . Huygens’ principle is apphed‘ to points

A and B to obtain the curved wavelets shown at C and D. The wave front is tangent

to these two wavelets, so the new wave front is the line CD. Hence the rays, which are perpendicular to the wave fronts, bend toward the normal if v, < v, as drawn. Newton favored a particle theory of light which predicted the opposite result, that the speed of light would be greater in the second medium (v, > v;). Thus the wave theory predicts that the speed of light in water, for example, is less than in air; and Newton’s particle theory predicts the reverse. An experiment to actually measure the speed of light in water was performed in 1850 by the French physicist Jean Foucault, and it confirmed the wave-theory prediction. By then, however, the wave theory was already fully accepted, as we shall see in the next Section. Snell’s law of refraction follows directly from Huygens’ principle, given that the speed of light » in any medium is related to the speed in a vacuum, ¢, and the index of refraction, n, by Eq. 32-4: that is, » = ¢/n. From the Huygens’ construction of Fig, 34-3, angle ADC is equal to 6, and angle BAD is equal to 6,fl Then for the two triangles that have the common side AD, we have sinf;,

=

it’

sin92=£-

AD

AD

We divide these two equations and obtain sinf, v

sing,

Then, by Eq.32-4 ny

v

v; = ¢/n;

sin 01

=

n

and

v, = ¢/n,, so we have

sin 02,

which is Snell’s law of refraction, Eq. 32-5. (The law of reflection can be derived from Huygens’ principle in a similar way: see Problem 1 at the end of this Chapter.) When a light wave travels from one medium to another, its frequency does not change, but its wavelength does. This can be seen from Fig. 34-3, where each of the blue lines representing a wave front corresponds to a crest (peak) of the wave. Then A )\_1

o

nt

oy

n

m

;1

na ’

where, in the last step, we used Eq. 32-4, v = ¢/n. If medium 1 is a vacuum (or air), so n; = 1, v, = ¢, and we call A; simply A, then the wavelength in another

medium of index of refraction n (= n,) will be

by = =

(34-1)

This result is consistent with the frequency f being unchanged no matter what medium the wave is traveling in, since ¢ = fA. EXERCISE A A light beam in air with wavelength = 500 nm, frequency = 6.0 X 10" Hz,

902

CHAPTER 34

and speed = 3.0 X 108m/s goes into glass which has an index of refraction = 1.5. What are the wavelength, frequency, and speed of the light in the glass?

Direct ra

Ray

A/////,

Observer

Wave fronts can be used to explain how mirages are produced by refraction of light. For example, on a hot day motorists sometimes see a mirage of water on the highway ahead of them, with distant vehicles seemingly reflected in it (Fig. 34—4a). On a hot day, there can be a layer of very hot air next to the roadway (made hot by the Sun beating on the road). Hot air is less dense than cooler air, so the index of refraction is slightly lower in the hot air. In Fig. 34—4b, we see a diagram of light coming from one point on a distant car (on the right) heading left toward the observer. Wave fronts and two rays (perpendicular to the wave fronts) are shown. Ray A heads directly at the observer and follows a straight-line path, and represents the normal view of the distant car. Ray B is a ray initially directed slightly downward, but it bends slightly as it moves through layers of air of different index of refraction. The wave fronts, shown in blue in Fig. 34—4b, move slightly faster in the layers of air nearer the ground. Thus ray B is bent as shown, and seems to the observer to be coming from below (dashed line) as if reflected off the road. Hence the mirage.

34-3

¥

directed

slightly downward

FIGURE 34-4 (a) A highway mirage. (b) Drawing (greatly exaggerated) showing wave fronts and rays to explain highway mirages. Note how sections of the wave fronts near the ground move faster and so are farther apart.

@)PHYsSics

Highway mirages

APPLIED

Interference—Young's

Double-Slit Experiment

fl 1801, the Englishman Thomas Young (1773-1829) obtained convincing evidence for the wave nature of light and was even able to measure wavelengths for visible light. Figure 34-5a shows a schematic diagram of Young’s famous double-slit experiment. FIGURE 34-5 (a) Young’s double-slit experiment. (b) If light consists of particles, we would expect to see two bright lines on the screen behind the slits. (c) In fact, many lines are observed. The slits and their separation need to be very thin.

Sun’s rays

(particle theory

prediction)

(c)

Viewing screen (actual)

To have light from a single source, Young used the Sun passing through a very narrow slit ‘in a window covering. This beam of parallel rays falls on a screen containing two closely spaced slits, S; and S,. (The slits and their separation are very narrow, not much larger than the wavelength of the light.) If light consists of tiny particles, we might expect to see two bright lines on a screen placed behind the slits as in (b). But instead a series of bright lines are seen, as in (c). Young was able to explain this result as a wave-interference phenomenon. To understand why, we consider the simple situation of plane waves' of light of a single wavelength — called monochromatic, meaning “one color” — falling on the two slits as shown in Fig. 34—6. Because of diffraction, the waves leaving the two all slits spread out as shown. This is equivalent to the interference pattern oroduced when two rocks are thrown into a lake (Fig. 15-23), or when sound from two loudspeakers interferes (Fig. 16-15). Recall Section 15-8 on wave interference. See pages 410 and 818.

FIGURE 34-6 If light is a wave, light passing through one of two slits should interfere with light passing through the other slit.

Rays

S >

Screen

Viewing screen

o2

(b)

~_"

Viewing screen

Wave| fronts

(a)

SECTION 34-3

903

Bright (constructive interference)

Bright (constructive interference)

Dark (destructive interference)

Extra distance

\—)Exltra distance

=A

I

¢

Screen

=-=A

2

|

¢

Screen

|

(b)

(a) 8 =00

(c)

¢

Screen

(d)

FIGURE 34-7 How the wave theory explains the pattern of lines seen in the double-slit experiment. (a) At the center of the screen the waves from each slit travel the same distance and are in phase. (b) At this angle 6, the lower wave travels an extra distance of one whole wavelength, and the

waves are in phase; note from the shaded triangle that the path difference equals d sin 6. (c) For this angle 6, the lower wave travels an extra distance equal to one-half wavelength, so the two waves arrive at the screen fully out of phase. (d) A more detailed diagram showing the geometry for parts (b) and (c).

FIGURE 34-8 Two traveling waves are shown undergoing (a) constructive interference, (b) destructive interference. (See also Section 15-8.)

S

+

(a)

To see how an interference pattern is produced on the screen, we make use of Fig. 34-7. Waves of wavelength A are shown entering the slits S; and S,, which are a distance d apart. The waves spread out in all directions after passing through the slits (Fig. 34-6), but they are shown only for three different angles 6. In Fig. 34-7a, the waves reaching the center of the screen are shown (6 = 0°). The waves from the two slits travel the same distance, so they are in phase: a crest of one wave arrives at the same time as a crest of the other wave. Hence the amplitudes of the two waves add to form a larger amplitude as shown in Fig. 34-8a. This is constructive interference, and there is a bright area at the center of the screen. Constructive interference also occurs when the paths of the two rays differ by one wavelength (or any whole number of wavelengths), as shown in Fig, 34-7b; also hercq there will be brightness on the screen. But if one ray travels an extra distance of one-half wavelength (or 34,3 A, and so on), the two waves are exactly out of phase when they reach the screen: the crests of one wave arrive at the same time as the troughs of the other wave, and so they add to produce zero amplitude (Fig. 34-8b). This is destructive interference, and the screen is dark, Fig. 34-7c. Thus, there will be a series of bright and dark lines (or fringes) on the viewing screen. To determine exactly where the bright lines fall, first note that Fig. 34-7 is somewhat exaggerated; in real situations, the distance d between the slits is very small compared to the distance £ to the screen. The rays from each slit for each case will therefore be essentially parallel, and 6 is the angle they make with the horizontal as shown in Fig. 34-7d. From the shaded right triangles shown in Figs. 34—7b and c, we can see that the extra distance traveled by the lower ray is d sin 6 (seen more clearly in Fig. 34-7d). Constructive interference will occur, and a bright fringe will appear on the screen, when the path difference, d sin 6, equals a whole number of wavelengths: dsinf

(b)

=

mA,

m

=

0,1,2,-.

constructive interference (bright)

(34-2a)

The value of m is called the order of the interference fringe. The first order (m = 1), for example, is the first fringe on each side of the central fringe (which is at & = 0, m = 0). Destructive interference occurs when the path difference d sin 6

is$A,3 1, and so on: dsin@

= (m +3)A,

destructive

m = 0,1,2,---. | interference | (34-2b) (dark) -

The bright fringes are peaks or maxima of light intensity, the dark fringes are minima.

904

CHAPTER 34

The Wave Nature of Light; Interference

The intensity of the bright fringes is greatest for the central fringe (m = 0) and decreases for higher orders, as shown in Fig. 34-9. How much the intensity decreases with increasing order depends on the width of the two slits. f‘ CONCEPTUAL EXAMPLE 34-1 | Interference pattern lines. (a) Will there be an infinite number of points on the viewing screen where constructive and destructive interference occur, or only a finite number of points? (b) Are neighboring points of constructive interference uniformly spaced, or is the spacing between neighboring points of constructive interference not uniform? RESPONSE (a) When you look at Egs. 34-2a and b you might be tempted to say, given the statement m = 0,1,2,--- beside the equations, that there are an infinite number of points of constructive and destructive interference. However, recall that sin @ cannot exceed 1. Thus, there is an upper limit to the values of m that can be used in these equations. For Eq. 34-2a, the maximum value of m is the integer closest in value but smaller than d/\. So there are a finite number of points of constructive and destructive interference no matter how large the screen. (b) The spacing between neighboring points of constructive or destructive interference is not uniform: The spacing gets larger as 6 gets larger, and you can verify this statement mathematically. For small values of 6 the spacing is nearly uniform as you will see in Example 34-2.

Line

spacing

for

double-slit

interference. A

screen

containing two slits 0.100 mm apart is 1.20m from the viewing screen. Light of wavelength A = 500 nm falls on the slits from a distant source. Approximately how far apart will adjacent bright interference fringes be on the screen?

’

APPROACH The angular position of bright (constructive interference) fringes is found using Eq. 34-2a. The distance between the first two fringes (say) can be found using right triangles as shown in Fig. 34-10.

SOLUTION

¢ = 1.20m,

Given

d = 0.100mm = 1.00 X 10#m,

the first-order fringe

sinf;,

=

mA

d

A = 500 X 10~ m,

(m = 1) occurs at an angle 6 given by

(1)(500 x 10~m) 1.00 X 10* m

=

and

fringe will occur a distance x; above the center of the screen (see Fig. 34-10), given by xi/f = tan6; = 6, so

=

The second-order fringe

(1.20m)(5.00 X 107%)

Constructive interference 2 10 1 2 3

m=3

]

\‘ s | Y

.

m=2 1 0 0 1 2 Destructive interference

|

||

3

(b)

FIGURE 34-9 (a) Interference fringes produced by a double-slit experiment and detected by photographic film placed on the viewing screen. The arrow marks the central fringe. (b) Graph of the intensity of light in the interference pattern. Also shown are values of m for Eq. 34-2a (constructive interference) and Eq. 34-2b (destructive interference).

5.00 X 1073,

This is a very small angle, so we can take sin § ~ 6, with 0 in radians. The first-order

x, ~ 06,

(a)

= 6.00 mm.

(m = 2) will occur at

FIGURE 34-10 Examples 34-2 and 34-3. For small angles 0 (give 6 in radians), the interference fringes occur at distance x = 6f above the center fringe (m = 0); 0y and x are for the first-order fringe (m = 1);60, and x; are for m = 2.

)

Sl! above the center, and so on. Thus the lower order fringes are 6.00 mm apart. S, NOTE The spacing between fringes is essentially uniform until the approximation sin@ ~ 6 is no longer valid.

)

CONCEPTUAL

EXAMPLE

D]

a===— "

9'1

x

\

:¢

f=1.20m‘——“*J

34-3 | Changing the wavelength. (a) What happens

to the interference pattern shown in Fig. 34-10, Example 34-2, if the incident light (500 nm) is replaced by light of wavelength 700 nm? (b) What happens instead if the wavelength stays at 500 nm but the slits are moved farther apart? RESPONSE (a) When A increases in Eq. 34-2a but d stays the same, then the angle 6 for bright fringes increases and the interference pattern spreads out. (b) Increasing the slit spacing d reduces 6 for each order, so the lines are closer

ligether.

SECTION 34-3

Interference—Young's Double-Slit Experiment

905

White

~-2.0 mm-> 3.5 mm—> FIGURE 34-11 First-order fringes are a full spectrum, like a rainbow.

From Egs. 34-2 we can see that, except for the zeroth-order fringe at the center, the position of the fringes depends on wavelength. Consequently, when white light falls on the two slits, as Young found in his experiments, the central fringe is white, but the first- (and higher-) order fringes contain a spectrum of colors like a rainbow; 6 was found to be smallest for violet light and largest for re™™ (Fig. 34-11). By measuring the position of these fringes, Young was the first to determine the wave':le.ngth.s of v?sible light (using Eqs. 3‘4—2).. In doing so, he showed that wl.xat distinguishes different colors phy§1cally is their wavelength (or

FIGURE 34-10 (Repeated.) For small angles 6 (give 6 in radians),

passes through two slits 0.50 mm apart, and an interference pattern is observed on a screen 2.5 m away. The first-order fringe resembles a rainbow with violet and red light at opposite ends. The violet light is about 2.0 mm and the red 3.5 mm from the center of the central white fringe (Fig. 34—-11). Estimate the wavelengths

~

Also Example 34-4.

frequency), an idea put forward earlier by Grimaldi in 1665. DGV

the interference fringes occur at distance

x = 6

:1! anEE o °l

x)

’? : !

Lhmm_fiJ

from

double-slit

interference. White light

APPROACH We find the angles for violet and red light from the distances given and the diagram of Fig. 34-10. Then we use Eq. 34-2a to obtain the wavelengths. Because 3.5 mm is much less than 2.5 m, we can use the small-angle approximation. SOLUTION We use Eq.34-2a with m = 1 andsiné ~ tan @ ~ 6. Then for violet light, x = 2.0 mm, so (see also Fig. 34-10)

o, she=T]

Wavelengths

for the violet and red light.

above the center

fringe (m = 0); 6, and x, are for the first-order fringe (m = 1), 6, and x, are for m = 2.

e

L S

_dsin6

dsi

AT

"

d6

dx

S mi

—4 -3 _(50X107"m)(20x10"m)

or 400 nm. For red light, x = 3.5 mm,

Lo== dx

1

25m

>

40 x107"m,

so

(5.0X10’4m>< 10°m 1

_

25m

)

=

70x107m

=

Coherence

700nm.

«

The two slits in Fig. 34-7 act as if they were two sources of radiation. They are called coherent sources because the waves leaving them have the same wavelength and frequency, and bear the same phase relationship to each other at all times. This happens because the waves come from a single source to the left of the two slits in Fig. 34-7, splitting the original beam into two. An interference pattern is observed only when the sources are coherent. If two tiny lightbulbs replaced the two slits, an interference pattern would not be seen. The light emitted by one lightbulb would have a random phase with respect to the second bulb, and the screen would be more or less uniformly illuminated. Two such sources, whose output waves have phases that bear no fixed relationship to each other over time, are called incoherent sources.

*34—-4

Intensity in the Double-Slit Interference Pattern

We saw in Section 34-3 that the interference pattern produced by the coherent light from two slits, S; and S, (Figs. 34-7 and 34-9), produces a series of bright and dark fringes. If the two monochromatic waves of wavelength A are in phase at the slits, the maxima (brightest points) occur at angles 6 given by (Egs. 34-2) dsinf

=

mA,

and the minima (darkest points) when

dsind where m is an integer

906

CHAPTER 34

= (m + A, (m = 0,1,2,---).

The Wave Nature of Light; Interference

-

E9=E1

+

Ez.

Both E; and E, vary sinusoidally with frequency

f = c/A,

but they differ in

& L —

We now determine the intensity of the light at all points in the pattern, assuming that if either slit were covered, the light passing through the other would diffract sufficiently to illuminate a large portion of the screen uniformly. The intensity / of the light at any point is proportional to the square of its wave mplitude (Section 15-3). Treating light as an electromagnetic wave, I is proportional to the square of the electric field E (or to the magnetic field B, Section 31-8): I o E? The electric field E at any point P (see Fig. 34-12) will be the sum of the electric field vectors of the waves coming from each of the two slits, E; and E,. Since E, and E, are essentially parallel (on a screen far away compared to the slit separation), the magnitude of the electric field at angle 6 (that is, at point P) will be

Screen

FIGURE 34-12

Determining the

intensity in a double-slit interference

phase, depending on their different travel distances from the slits. The electric field ~ pattern. Not to scale:in fact £ > d, at P can then be written for the light from each of the two slits, using w = 27 f, as E]

=

E2

Ew

.

(34-3)

sin wt

E20 Sln(wt

+

8)

a“dafl“i two rays become essentially

parallel.

where Ejq and E,, are their respective amplitudes and 8§ is the phase difference. The value of & depends on the angle 6, so let us now determine & as a function of 6. At the center of the screen (point 0), § = 0. If the difference in path length from P to S; and S, is dsin@ = A/2, the two waves are exactly out of phase so 6 = (or 180°). If dsinf = A, the two waves differ in phase by 6 = 2. In general, then, 8 is related to 6 by

or

P

6

dsin @

2m

A

6

27 Td

=

. sin 6.

(34-4)

10 determine Ey = E; + E,, we add the two scalars E; and E, which are sine functions differing by the phase 8. One way to determine the sum of E; and E, is to use a phasor diagram. (We used this technique before, in Chapter 30.) As shown in Fig. 34-13, we draw an arrow of length Ej, to represent the amplitude of E; (Eq. 34-3); and the arrow of length E,,, which we draw to make a fixed angle & with Ej,, represents the amplitude of E,. When the diagram rotates at angular frequency w about the origin, the projections of E, and E,, on the vertical axis represent E; and E, as a function of time (see Eq. 34-3). We let Ey be the “vector” sum’ of E; and E,; Eg is the amplitude of the sum E, = E; + E,, and the projection of Ey, on the vertical axis is just Ey. If the two slits provide equal illumination, so that E, = E,; = E,, then from symmetry in Fig. 34-13, the angle ¢ = §/2, and we can write Eg

=

Egg sin(wt

+

g).

FIGURE 34-13 Phasor diagram for double-slit interference pattern.

(34—53)

From Fig. 34-13 we can also see that 5 Eg = 2E,cos ¢ = 2E, cos >

(34-5b)

Combining Egs. 34-5a and b, we obtain E, r

=

2E, cos%sin( wt + —;—),

(34-5c¢)

where 6 is given by Eq. 34-4. We are not adding the actual electric field vectors; instead we are using the “phasor” technique to add the amplitudes, taking into account the phase difference of the two waves.

*SECTION 34-4

Intensity in the Double-Slit Interference Pattern

907

We are not really interested in Ey as a function of time, since for visible light

the frequency (10" to 10" Hz) is much too high to be noticeable. We are interested in the average intensity, which is proportional to the amplitude squared, E3,. We

now drop the word “average,” and we let I, (I

< E3) be the intensity at any

point P at an angle 6 to the horizontal. We let [; be the intensity at point O, th ™

center of the screen, where

§ =6 =0,

so

Iy « (Ey + Ey) = (2E,)>.

Then

the ratio 7/, is equal to the ratio of the squares of the electric-field amplitudes at these two points, so

where we used Eq. 34-5b. Thus the intensity /, at any point is related to the maximum intensity at the center of the screen by

Iy = Iycos’ g

=

(34-6)

cosz< rd sin 0)

&

A

where 8 was given by Eq. 34—4. This is the relation we sought. From Eq. 34-6 we see that maxima occur where cosé/2 = 1, corresponds to 8 = 0,27, 47, -+, From Eq. 344, § has these values when dsinf

=

Minima occur where

mA,

which

m=20,1,2,--.

8 = ar, 37, Sar, -+, which corresponds to

dsin

= (m + 3\,

m = 01,2,

o |

These are the same results we obtained in Section 34-3. But now we know notonly the position of maxima and minima, but from Eq. 34-6 we can determine the intensity at all points. In the usual situation where the distance £ to the screen from the slits is large compared to the slit separation d (£ >> d), if we consider only points P whose distance y from the center (point O) is small compared to £ (y ]

2

.

[y 1.22) RP =5 = f6 = D f» (35-11)

where f is the objective lens’ focal length (not frequency). This distance s is called the resolving power of the lens because it is the minimum separation of two object points that can just be resolved — assuming the highest quality lens since this limit is imposed by the wave nature of light. A smaller RP means better resolution, better detail. EXERCISE C What is the resolving power of a microscope with a 5-mm-diameter objective which has f = 9mm? (a) 550nm, (b) 750nm, (c) 1200nm, (d) 0.05nm, (e) 0.005 nm.

Diffraction sets an ultimate limit on the detail that can be seen on any object. In Eq. 35-11 for resolving power of a microscope, the focal length of the lens cannot practically be made less than (approximately) the radius of the lens, and even that is very difficult (see the lensmaker’s equation, Eq. 33-4). In this best case, Eq. 35-11 gives, with f = D/2, RP

A =~ 3

(35-12)

Thus we can say, to within a factor of 2 or so, that A limits resolution

it is not possible to resolve detail of objects smaller than the wavelength of the eadiation being used.

This is an important and useful rule of thumb.

Compound lenses in microscopes are now designed so well that the actual limit

on resolution is often set by diffraction—that is, by the wavelength c& the light used. To obtain greater detail, one must use radiation of short: wavelength. The use of UV radiation can increase the resolution by a factor of perhaps 2. Far more important, however, was the discovery in the early twentieth century that electrons have wave properties (Chapter 37) and that their wavelengths can be very small. The wave nature of electrons is utilized in the electron microscope (Section 37-8), which can magnify 100 to 1000 times more than a visiblelight microscope because of the much shorter wavelengths. X-rays, too, have very short wavelengths and are often used to study objects in great detail (Section 35-10).

*35—6

Resolution of the Human Eye and Useful Magnification

The resolution of the human eye is limited by several factors, all of roughly the same order of magnitude. The resolution is best at the fovea, where the cone spacing is smallest, about 3 um (=3000nm). The diameter of the pupil varies from about 0.1cm to about 0.8cm. So for A = 550nm (where the eye’s sensitivity is greatest), the diffraction limit is about # ~ 1.22A/D ~ 8 X 10> rad to 6 X 10~*rad. The eye is about 2 cm long, giving a resolving power

(Eq. 35-11) of

s~ (2 X 1072m)(8 x 10°rad) ~ 2 um at best, to about 10 um at worst (pupil small). Spherical and chromatic aberration also limit the resolution to about 10 um.

The net result is that the eye can just resolve objects whose angular separation is around 4 best eye

5 X 107

rad.

resolution

This corresponds to objects separated by 1 cm at a distance of about 20 m. The typical near point of a human eye is about 25 cm. At this distance, the ey

can just resolve objects that are 932

CHAPTER 35

(25cm)(5 X 10#rad) ~ 10*m = {ymm

apart.

Since the best light microscopes can resolve objects no smaller than about 200 nm at best (Eq. 35-12 for violet light, A = 400nm ), the useful magnification

[= (resolution by naked eye)/(resolution by microscope)] is limited to about

10*m

-

maximum useful

Pl : R 500X e 200 X 107 m microscope magnification rln practice, magnifications of about 1000X are often used to minimize eyestrain. Any greater magnification would simply make visible the diffraction pattern produced by the microscope objective lens. Now you have the answers to the Chapter-Opening Question: (c), by the equation above, and (c¢) by the A rule.

35-7 Diffraction Grating

A large number

of equally spaced parallel slits is called a diffraction grating,

although the term “interference grating” might be as appropriate. Gratings can be

made by precision machining of very fine parallel lines on a glass plate. The untouched spaces between the lines serve as the slits. Photographic transparencies of an original grating serve as inexpensive gratings. Gratings containing 10,000 lines per centimeter are common, and are very useful for precise measurements of wavelengths. A diffraction grating containing slits is called a transmission grating. Another type of diffraction grating is the reflection grating, made by ruling fine lines on a metallic or glass surface from which light is reflected and analyzed. The analysis is basically the same as for a transmission grating, which we now discuss. The analysis of a diffraction grating is much like that of Young’s double-slit experiment. We assume parallel rays of light are incident on the grating as shown in Fig. 35-17. We also assume that the slits are narrow enough so that diffraction by each of them spreads light over a very wide angle on a distant screen beyond the grating, and interference can occur with light from all the other slits. Light rays that pass through each slit without deviation (6 = 0°) interfere constructively to produce a bright line at the center of the screen. Constructive interference also occurs at an angle 6 such that rays from adjacent slits travel an extra distance of ¢ = mA, where m is an integer. If d is the distance between slits, then we see rom Fig, 35-17 that Af = dsin#6, and

ginh

o= P

A

=01 2

[dlffragtlon grapng,] .

.

.

/\7Af =dsin@

FIGURE 35-17

Diffraction grating.

(35-13)

d principal maxima is the criterion to have a brightness maximum. This is the same equation as for the double-slit situation, and again m is called the order of the pattern. There is an important difference between a double-slit and a multiple-slit pattern. The bright maxima are much sharper and narrower for a grating. Why? Suppose that the angle 6 is increased just slightly beyond that required for a maximum. In the case of only two slits, the two waves will be only slightly out of phase, so nearly full constructive interference occurs. This means the maxima are wide (see Fig. 34-9). For a grating, the waves from two adjacent slits will also not be significantly out of phase. But waves from one slit and those from a second one a few hundred slits away may be exactly out of phase; all or nearly all the light can cancel in pairs in this way. For example, suppose the angle 6 is very slightly different from its first-order maximum, so that the extra path length for a pair of adjacent slits is not exactly A but rather 1.0010A. The wave through one slit and another one 500 slits below will have a path difference of 1A + (500)(0.0010A) = 1.5000A, or 13 wavelengths, so the two will cancel. A pair of slits, one below each of these, will also cancel. That is, the light from slit 1 cancels with that from slit 501; light from slit 2 cancels with that from slit 502, and so on. Thus even for a tiny angle’ corresponding to an extra path length of 1A, there is much destructive interference, and so the maxima are very narrow. The more lines there are in a grating, the sharper will be the peaks (see Fig. 35-18). Because a grating produces much sharper lines than two slits ’alone can (and much brighter lines because there are many more slits), a grating s a far more precise device for measuring wavelengths. "Depending on the total number of slits, there may or may not be complete cancellation for such an angle, so there will be very tiny peaks between the main maxima (see Fig. 35-18b), but they are usually much too small to be seen.

A

CAUTION

Diffraction grating is analyzed using interference formulas, not diffraction formulas

FIGURE function position slits, (b) grating,

35-18 Intensityasa of viewing angle 6 (or on the screen) for (a) two six slits. For a diffraction the number of slits is very

large (~10*) and the peaks are narrower still.

m=1

m=0

(b)

m=1

SECTION 35-7

933

Suppose the light striking a diffraction grating is not monochromatic, but consists of two or more distinct wavelengths. Then for all orders other than m = 0, each wavelength will produce a maximum at a different angle (Fig. 35-19a), just as for a double slit. If white light strikes a grating, the central (m = 0) maximum will be a sharp white peak. But for all other orders, there will be a distinct spectrum of colors spread out ovc‘fl a certain angular width, Fig. 35-19b. Because a diffraction grating spreads out light into’ its component wavelengths, the resulting pattern is called a spectrum. m=2

FIGURE 35-19

m=2m=1

Spectra produced by a

grating: (a) two wavelengths, 400 nm and

700 nm; (b) white light. The second order

400 700

nm

will normally be dimmer than the first

am

m=1

o

nm

m=0

DOV

nm

700 400

nm

(a)

Rainbow

RN

m=1m=2

:

400

order. (Higher orders are not shown.) If grating spacing is small enough, the second and higher orders will be missing. Rainbow (fainter)

m=1

White

m=2

o

nm nm

Rainbow

nm

Rainbow (fainter)

Diffraction grating: lines. Determine the angular positions

of the first- and second-order maxima for light of wavelength 400 nm and 700 nm incident on a grating containing 10,000 lines/cm.

APPROACH First we find the distance d between grating lines: if the grating has N lines in 1m, then the distance between lines must be d = 1/N meters. Then we use Eq. 35-13,sin & = mA/d, to find the angles for the two wavelengths for m =1 and 2.

SOLUTION The

which

grating

means

the

contains

distance

1.00 X 10°°m = 1.00 um. Sinf,y,

=

WS

o

M

SO 040 = 23.6°

SR 1

= and

0

oo

S

=

In first order

mr

=g

=

lines

=

g

—4

d = (1/1.00 X 10°)m =

0.400

o

Tl x10°m

22

=

1.00 X 10 °m

(1)(7.00 X 107m)

——

is

(m = 1), the angles are

_ (1)(4.00 X 107 m)

6,y = 44.4°. ]

1.00 X 10*lines/cm = 1.00 X 10° lines/m,

between

In second order,

(2)(4.00 X 107 m)

100 X 10°m

(2)(7.00 X 107m)

THox10°m

r

=

0.800 &

S0 B4 = 53.1°. But the second order does not exist for sin 0 cannot exceed 1. No higher orders will appear.

m

A = 700nm

because

Spectra overlap. White light containing wavelengths from

400 nm to 750 nm strikes a grating containing 4000 lines/cm. Show that the blue at A = 450nm of the third-order spectrum overlaps the red at 700nm of the second order.

APPROACH We use sin & = mA/d to calculate the angular positions of the m = blue maximum and the m = 2 red one.

934

CHAPTER 35

Diffraction and Polarization

fl

SOLUTION

The grating spacing is d = (1/4000) cm = 2.50 X 10°®m. The blue

of the third order occurs at an angle 6 given by

sing = T

d

3)(4.50 x 1077

= GX

:

(2.50 % 10°%m)

m) _ 0.540.

Red in second order occurs at

.

sinf

(2)(7.00 X 107" m) = 0.560, (2.50 X 10°m)

=

which is a greater angle; so the second order overlaps into the beginning of the third-order spectrum. CONCEPTUAL EXAMPLE 35-10 | Compact disk. When you look at the surface of a music CD (Fig. 35-20), you see the colors of a rainbow. () Estimate the distance between the curved lines (they are read by a laser). (b) Estimate the distance between lines, noting that a CD contains at most 80 min of music, that it rotates at speeds from 200 to 500 rev/min, and that 3 of its 6 cm radius contains the lines.

FIGURE 35-20 ~ Example 35-10.

A compact disk,

RESPONSE (a¢) The CD acts like a reflection diffraction grating. To satisfy Eq. 35-13, we might estimate the line spacing as one or a few (2 or 3) wavelengths (A = 550nm) or 0.5 to 1.5 um. (b) Average rotation speed of 350 rev/min times 80 min gives 28,000 total rotations or 28,000 lines, which are spread over

(3)(6cm) = 4cm. So we have a sort of reflection diffraction grating with about

(28,000 lines)/(4 cm) = 7000 lines/cm. The distance d between lines is roughly |1cm/7000lines ~ 1.4 X 10°m = 1.4 um. Our results in (a) and (b) agree.

35-8 The Spectrometer and Spectroscopy

A spectrometer or spectroscope, Fig. 35-21, is a device to measure wavelengths

curately using a diffraction grating (or a prism) to separate different wavelengths

Jf light. Light from a source passes through a narrow slit S in the “collimator.” The slit is at the focal point of the lens L, so parallel light falls on the grating. The movable telescope can bring the rays to a focus. Nothing will be seen in the

viewing telescope unless it is positioned at an angle 6 that corresponds to a diffraction

peak (first order is usually used) of a wavelength emitted by the source. The angle 6 can be measured to very high accuracy, so the wavelength of a line can be determined to high accuracy using Eq. 35-13: A

=

d

.

7n-sxn0,

where m is an integer lines. The line you see ally an image of the measure the angular

~ FIGURE 35-21

~ SPectroscope. Source

/ =

.

Spectrometer or

Grating

Collimatar )~ S

L

6y

Telescope

representing the order, and d is the distance between grating in a spectrometer corresponding to each wavelength is actuslit S. A narrower slit results in dimmer light but we can positions more precisely. If the light contains a continuous

,

A

Eye

range of wavelengths, then a continuous spectrum is seen in the spectroscope.

The spectrometer in Fig. 35-21 uses a transmission grating. Others may use a reflection grating, or sometimes a prism. A prism works because of dispersion (Section 32-6), bending light of different wavelengths into different angles. (A prism is not a linear device and must be calibrated.) An important use of a spectrometer is for the identification of atoms or molecules. When a gas is heated or an electric current is passed through it, the gas emits a characteristic line spectrum. That is, only certain discrete wavelengths of light are emitted, and these are different for different elements and compounds.” Figure 35-22 shows the line spectra for a number of elements in the gas state. Line spectra occur only for gases at high temperatures and low pressure and density. The light from heated solids, such as a lightbulb filament, and even from a dense gaseous object such s the Sun, produces a continuous spectrum including a wide range of wavelengths. "Why atoms and molecules emit line spectra was a great mystery for many years and played a central role in the development of modern quantum theory, as we shall see in Chapter 37.

SECTION 35-8

The Spectrometer and Spectroscopy

935

Atomic hydrogen

-

Mercury

T Sodium

Solar absorption spectrum

FIGURE 35-22 Line spectra for the gases indicated, and the

Figure 35-22 also shows the Sun’s “continuous spectrum,” which contains a number of dark lines (only the most prominent are shown), called absorption lines.

absorption lines.

The Sun’s absorption lines are due to absorption by atoms and molecules in the cooler outer atmosphere of the Sun, as well as by atoms and molecules in the Earth’s atmosphere. A careful analysis of all these thousands of lines reveals that at least two-thirds of all elements are present in the Sun’s atmosphere. The presence of elements in the atmosphere of other planets, in interstellar space, and in stars, is also determined by spectroscopy. Spectroscopy is useful for determining the presence of certain types of molecules in laboratory specimens where chemical analysis would be difficult. For example, biological DNA and different types of protein absorb light in particular regions of the spectrum (such as in the UV). The material to be examined, which is often in solution, is placed in a monochromatic light beam whose wavelength i selected by the placement angle of a diffraction grating or prism. The amount of absorption, as compared to a standard solution without the specimen, can reveal not only the presence of a particular type of molecule, but also its concentration. Light emission and absorption also occur outside the visible part of the spectrum, such as in the UV and IR regions. Glass absorbs light in these regions, so reflection gratings and mirrors (in place of lenses) are used. Special types of film or sensors are used for detection.

spectrum from the Sun showing

PHYSICS APPLIED Chemical and biochemical analysis by spectroscopy

Atoms and molecules can absorb light at the same wavelengths at which they emit light.

m Hydrogen spectrum. Light emitted by hot hydrogen gas is observed with a spectroscope using a diffraction grating having 1.00 X 10*lines/cm. The spectral lines nearest to the center (0°) are a violet line at 24.2°, a blue line at 25.7°, a blue-green line at 29.1°, and a red line at 41.0° from the center. What are the wavelengths of these spectral lines of hydrogen? APPROACH The wavelengths can be determined from the angles by using A = (d/m)sin@ where d is the spacing between slits, and m is the order of the spectrum (Eq. 35-13). SOLUTION Since these are the closest lines to 6 = 0°, this is the first-order

spectrum

(m = 1). The slit spacing is d = 1/(1.00 X 10*cm™) =1.00 X 10 *m.

The wavelength of the violet line is .0 X 1076 A= (—%) sinf = (M)

sin24.2° = 410 X 10" m = 410 nm.

The other wavelengths are:

blue: blue-green: red:

NOTE

936

CHAPTER 35

A = (1.00 X 10°m)sin25.7° = 434 nm, A = (1.00 X 10°m)sin29.1° = 486 nm, -6 : A = (1.00 X 10°°m)sin 41.0° o = 656 nm.

In an unknown mixture of gases, these four spectral lines need to be seen

to identify that the mixture contains hydrogen.

fl

“35-9 Peak Widths and Resolving

Power for a Diffraction Grating

f‘ Ve now look at the pattern of maxima produced by a multiple-slit grating using phasor diagrams. We can determine a formula for the width of each peak, and we will see why there are tiny maxima between the principal maxima, as indicated in Fig. 35-18b. First of all, it should be noted that the two-slit and six-slit patterns shown in Fig. 35-18 were drawn assuming very narrow slits so that diffraction does not limit the height of the peaks. For real diffraction gratings, this is not normally the case: the slit width D is often not much smaller than the slit separation d, and

diffraction thus limits the intensity of the peaks so the central peak (m = 0) is brighter than the side peaks. We won’t worry about this effect on intensity except to note that if a diffraction minimum coincides with a particular order of the interference pattern, that order will not appear. (For example, if d = 2D, all the even orders, m = 2,4,---, will be missing. Can you see why? Hint: See Example 35-4.) Figures 35-23 and 35-24 show phasor diagrams for a two-slit and a six-slit grating, respectively. Each short arrow represents the amplitude of a wave from a single slit, and their vector sum (as phasors) represents the total amplitude for a given viewing angle 6. Part (a) of each Figure shows the phasor diagram at 6 = 0°, at the center of the pattern, which is the central maximum (m = 0). Part (b) of each Figure shows the condition for the adjacent minimum: where the arrows first close on themselves (add to zero) so the amplitude Ej is zero. For two slits, this occurs when the two separate amplitudes are 180° out of phase. For six slits, it occurs when each amplitude makes a 60° angle with its neighbor. For two slits, the minimum occurs when the phase between slits is 8 = 27r/2 (in radians); for six slits it occurs when the phase 8 is 277/6; and in the general case of N slits, the minimum occurs for a phase difference between adjacent slits of 2w -14 o = B N (35-14) l

#hat does this correspond to in 6? First note that & is related to 6 by (Eq. 34-4) & dsin 6 2T, . A or é= A d sin 6. (35-15) Let us call A6, the angular position of the minimum next to the peak at 6 = 0. Then 1) d sin A6, 2

Ey

— e

(a) Central maximum: 6= 0, =0 E

=0

6= 180°

=£x.

(b) Minimum: 6= 180° FIGURE 35-23 Phasor diagram for two slits (a) at the central maximum, (b) at the nearest minimum. FIGURE 35-24 Phasor diagram for six slits (a) at the central maximum, (b) at the nearest minimum,

Ey IS 1>

IS 1%

IS >

S 1>

(a) Central maximum: =0, 6=0

A

We insert Eq. 35-14 for 6 and find sin Af,

A

=

/

=

A

Nd

0 =60°

(35-16a)

Nd Since A6, is usually small (N is usually very large for a grating), sin A6, ~ A6, so in the small angle limit we can write Aby,

> 1>

6=60° (b) Minimum: §=60°, E; =0

(35-16b)

It is clear from either of the last two relations that the larger N is, the narrower will be the central peak. (For N = 2,sin A§, = A/2d, which is what we obtained

earlier for the double slit, Eq. 34-2b with m = 0.) Either of Eqs. 35-16 shows why the peaks become narrower for larger N. The origin of the small secondary maxima between the principal peaks (see Fig. 35-18b) can be deduced from the diagram of Fig. 35-25. This is just a continuation of Fig. 35-24b (where 6 = 60° ); but now the phase & has been increased to almost 90°, where E, is a relative maximum. Note that Ej is much less than E, (Fig. 35-24a), so the intensity in this secondary maximum is much smaller than in a principal peak. rAs o (and 0) is increased further, Ey again decreases to zero (a “double circle”), then -eaches another tiny maximum, and so on. Eventually the diagram unfolds again and when & = 360°, all the amplitudes again lie in a straight line (as in Fig. 35-24a) corresponding to the next principal maximum (m = 1 in Eq. 35-13).

FIGURE 35-25 Phasor diagram for the secondary peak.

=i

*SECTION 35-9

937

Equation 35-16b gives the half-width of the central (m = 0) peak. To determine the half-width of higher order peaks, A8, for order m, we differentiate Eq. 35-15 so as to relate the change Ad in 8, to the change A# in the angle 6: Ad

~

S%AB

=

McosBAB.

-

If A6,, represents the half-width of a peak of order m (m = 1,2, -..)—that is, the angle between the peak maximum and the minimum to either side—then Aé = 2m/N as given by Eq. 35-14. We insert this into the above relation and find

A6, = A

(35-17)

Nd cos 6,

where 6,, is the angular position of the m™ peak as given by Eq.35-13. This derivation is valid, of course, only for small Aé (=27/N) which is indeed the case for real gratings since N is on the order of 10* or more. An important property of any diffraction grating used in a spectrometer is its ability to resolve two very closely spaced wavelengths (wavelength difference = AA). The resolving power R of a grating is defined as A R = rvy (35-18) With a little work, using Eq. 35-17, we can show that AA = A/Nm total number of grating lines and m is the order. Then we have R

=

fi

=

where N is the

Nm.

(35-19)

The larger the value of R, the closer two wavelengths can be resolvable. If R is given, the minimum separation AA between two wavelengths near A, is (by Eq. 35-18) AN

=

—

Resolving two close lines. Yellow sodium light, which consists

FIGURE 35-26

X-ray tube.

Electrons emitted by a heated

filament in a vacuum tube are accelerated by a high voltage. When they strike the surface of the anode, the “target,” X-rays are emitted.

of two wavelengths, A; = 589.00 nm and A, = 589.59 nm, falls on a 7500-line/cm diffraction grating. Determine () the maximum order m that will be present fomy sodium light, (b) the width of grating necessary to resolve the two sodium lines. APPROACH We first find d = 1cm/7500 = 1.33 X 10°m, and then use Eq.35-13 to find m. For (b) we use Egs. 35-18 and 35-19. SOLUTION (@) The maximum value of m at A = 589 nm, using Eq. 35-13 with sinf = 1, is B

—_

d ding = d_133X10°m A

T A

589x107m

_ 2.26

’

so m = 2 is the maximum order present. (b) The resolving power needed is

From

AA

059nm

Eq. 35-19, the total number

N = R/m = 1000/2 = 500,

N of lines needed

so the grating

need

for the

only

be

m = 2

order is

500/7500cm™! =

0.0667 cm

wide. A typical grating is a few centimeters wide, and so will easily

35-10

X-Rays and X-Ray Diffraction

|resolve the two lines.

High voltage

938

CHAPTER 35

In 1895, W. C. Roentgen (1845-1923) discovered that when electrons were accelerated by a high voltage in a vacuum tube and allowed to strike a glass or metal surface inside the tube, fluorescent minerals some distance away would glow, and photographic film would become exposed. Roentgen attributed these effects to a new type of radiation (different from cathode rays). They were given the name X-rays after the algebraic symbol x, meaning an unknown quantity. He soon found that X-rays penetrated through some materials better than throug'™ others, and within a few weeks he presented the first X-ray photograph (o his wife’s hand). The production of X-rays today is usually done in a tube (Fig. 35-26) similar to Roentgen’s, using voltages of typically 30 kV to 150 kV.

Investigations into the nature of X-rays indicated they were not charged particles (such as electrons) since they could not be deflected by electric or magnetic fields. It was suggested that they might be a form of invisible light. However, they showed no diffraction or interference effects using ordinary gratings. Indeed, if their wavelengths

ere much smaller than the typical grating spacing of 10 m (=10° nm), no effects

would be expected. Around 1912, Max von Laue (1879-1960) suggested that if the atoms in a crystal were arranged in a regular array (see Fig. 17-2a), such a crystal might serve as a diffraction grating for very short wavelengths on the order of the

spacing between atoms, estimated to be about 107°m (=107 nm). Experiments

soon showed that X-rays scattered from a crystal did indeed show the peaks and valleys of a diffraction pattern (Fig. 35-27). Thus it was shown, in a single blow, that X-rays have a wave nature and that atoms are arranged in a regular way in crystals. Today, X-rays are recognized as electromagnetic radiation with wavelengths in the range of about 1072 nm to 10 nm, the range readily produced in an X-ray tube. We saw in Section 35-5 that light of shorter wavelength provides greater resolution when we are examining an object microscopically. Since X-rays have much shorter wavelengths than visible light, they should in principle offer much greater resolution. However, there seems to be no effective material to use as lenses for the very short wavelengths of X-rays. Instead, the clever but complicated technique of X-ray diffraction (or crystallography) has proved very effective for examining the microscopic world of atoms and molecules. In a simple crystal such as NaCl, the atoms are arranged in an orderly cubical fashion, Fig. 35-28, with atoms spaced a distance d apart. Suppose that a beam of X-rays is incident on the crystal at an angle ¢ to the surface, and that the two rays shown are reflected from two subsequent planes of atoms as shown. The two rays will constructively interfere if the extra distance ray I travels is a whole number of wavelengths farther than the distance ray 1I travels. This extra distance is 2d sin ¢. Therefore, constructive interference will occur when mA

=

2dsin¢,

m=1,2,3,

here m can be any integer. (Notice that ¢ is not © wormal to the surface.) This is called the Bragg (1890-1971), who derived it and who, together (1862-1942), developed the theory and technique of 1912-1913. If the X-ray wavelength is known and distance d between atoms can be obtained. This is the

FIGURE 35-27 This X-ray diffraction pattern is one of the first observed by Max von Laue in 1912 when he aimed a beam of X-rays at a zinc sulfide crystal. The diffraction pattern was detected directly on a photographic plate. FIGURE 35-28 by a crystal.

X-ray diffraction

(35-20) the angle with respect to the equation after W. L. Bragg with his father W. H. Bragg X-ray diffraction by crystals in the angle ¢ is measured, the basis for X-ray crystallography.

EXERCISE D When X-rays of wavelength 0.10 X 10~m are scattered from a sodium chloride crystal, a second-order diffraction peak is observed at 21°. What is the spacing between the planes of atoms for this scattering?

. ;/;//;’///\\\ .

T

//‘/. ®

o

[ ]

//

//

.//

e

ONNN

®

o

o

o

\\

®

&

&

&

o

o

\

&

o

o

o

o

o

o

FIGURE 35-29

X-rays can be

diffracted from many possible

planes within a crystal.

Actual X-ray diffraction patterns are quite complicated. First of all, a crystal is a three-dimensional object, and X-rays can be diffracted from different planes at different angles within the crystal, as shown in Fig. 35-29. Although the analysis is complex, a great deal can be learned about any substance that can be put in crystalline form. X-ray diffraction has also been very useful in determining the structure of biologically important molecules, such as the double helix structure of DNA, worked out by James Watson and Francis Crick in 1953. See Fig. 35-30, and for

FIGURE 35-30 X-ray diffraction photo of DNA molecules taken by Rosalind Franklin in the early 1950s. The cross of spots suggested that DNA is a helix. B

models of the double helix, Figs. 21-47a and 21-48. Around 1960, the first detailed

structure of a protein molecule, myoglobin, was elucidated with the aid of X-ray rdiffraction. Soon the structure of an important constituent of blood, hemoglobin, /as worked out, and since then the structures of a great many molecules have been determined with the help of X-rays.

*SECTION 35-10

X-Rays and X-Ray Diffraction

939

35—-11

FIGURE 35-31 Transverse waves on a rope polarized (a) in a vertical plane and (b) in a horizontal plane.

FIGURE 35-32

(a) Vertically

polarized wave passes through a vertical slit, but (b) a horizontally polarized wave sHill not,

Polarization

An important and useful property of light is that it can be polarized. To see what this means, let us examine waves traveling on a rope. A rope can be sexfl into oscillation in a vertical plane as in Fig. 35-31a, or in a horizontal planc as in Fig. 35-31b. In either case, the wave is said to be linearly polarized or plane-polarized—that is, the oscillations are in a plane. If we now place an obstacle containing a vertical slit in the path of the wave, Fig. 35-32, a vertically polarized wave passes through the vertical slit, but a horizontally polarized wave will not. If a horizontal slit were used, the vertically polarized wave would be stopped. If both types of slit were used, both types of wave would be stopped by one slit or the other. Note that polarization can exist only for transverse waves, and not for longitudinal waves such as sound. The latter oscillate only along

the direction of motion, and neither orientation of slit would stop them. Light is not necessarily polarized. It can also be unpolarized, which means that the source has oscillations in many planes at once, as shown in Fig. 35-33. An ordinary incandescent lightbulb emits unpolarized light, as does the Sun.

_

e

Polaroids (Polarization by Absorption) Plane-polarized light can be obtained from unpolarized light using certain crystals

such as tourmaline. Or, more commonly, we use a Polaroid sheet. (Polaroid materials were invented in 1929 by Edwin Land.) A Polaroid sheet consists of long complex molecules arranged parallel to one another. Such a Polaroid acts like a series of parallel slits to allow one orientation of polarization to pass through nearly undiminished. This direction is called the transmission axis of the Polaroid. Polarization perpendicular to this direction is absorbed almost completely by the Polaroid. Absorption by a Polaroid can be explained at the molecular level. An electric field E that oscillates parallel to the long molecules can set electrons into motion along the molecules, thus doing work on them and transferring energy. Hence, if E is parallel to the molecules, it gets absorbed. An electric field E perpendicular t

75. At what angle above the horizon is the Sun when light reflecting off a smooth lake is polarized most strongly? 76. Unpolarized light falls on two polarizer sheets whose axes are at right angles. (a) What fraction of the incident light intensity is transmitted? (b) What fraction is transmitted if a third polarizer is placed between the first two so that its axis makes a 66° angle with the axis of the first polarizer? (c) What if the third polarizer is in front of the other two? . At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light by an additional factor (after the first Polaroid cuts it in

half) of (a) 4, (b) 10, (c) 100? 78. Four polarizers are placed in succession with their axes vertical, at 30.0° to the vertical, at 60.0° to the vertical, and

at 90.0° to the vertical. (a) Calculate what fraction of the incident unpolarized light is transmitted by the four polarizers. (b) Can the transmitted light be decreased by removing one of the polarizers? If so, which one? (c¢) Can the transmitted light intensity be extinguished by removing polarizers? If so, which one(s)? 79. Spy planes fly at extremely high altitudes (25 km) to avoid interception. Their cameras are reportedly able to discern features as small as 5 cm. What must be the minimum aperture of the camera lens to afford this resolution? (Use A = 580 nm.) 80. Two polarizers are oriented at 48° to each other and planepolarized light is incident on them. If only 25% of the light gets through both of them, what was the initial polarization direction of the incident light?

Answers to Exercises

(c).

CHAPTER 35

is

8 X 10°°m

from

Earth,

could

a

person

changed by amount A7 = 55C°. [Hint: Since first minima occur at a small angle.]

A

0.01c 0.50¢

0.90¢ 0.99¢

1.000 115

23 74

= 4.5 yr. Thus time dilation allows

such a trip, but the enormous practical problems of achieving such speeds may not be possible to overcome, certainly not in the near future. In this example, 100 years would pass on Earth, whereas only 4.5 years would pass for the astronaut on the trip. Is it just the clocks that would slow down for the astronaut? No. All processes, including aging and other life processes, run more slowly for the astronaut according to the Earth observer. But to the astronaut, time would pass in a normal rvay. The astronaut would experience 4.5 years of normal sleeping, eating, reading, .nd so on. And people on Earth would experience 100 years of ordinary activity. Twin Paradox Not long after Einstein proposed the special theory of relativity, an apparent paradox was pointed out. According to this twin paradox, suppose one of a pair of 20-year-old twins takes off in a spaceship traveling at very high speed to a distant star and back again, while the other twin remains on Earth. According to the Earth twin, the astronaut twin will age less. Whereas 20 years might pass for the Earth twin, perhaps only 1 year (depending on the spacecraft’s speed) would pass for the traveler. Thus, when the traveler returns, the earthbound twin could expect to be 40 years old whereas the traveling twin would be only 21. This is the viewpoint of the twin on the Earth. But what about the traveling twin? If all inertial reference frames are equally good, won’t the traveling twin make all the claims the Earth twin does, only in reverse? Can’t the astronaut twin claim that since the Earth is moving away at high speed, time passes more slowly on Earth and the twin on Earth will age less? This is the opposite of what the Earth twin predicts. They cannot both be right, for after all the spacecraft returns to Earth and a direct comparison of ages and clocks can be made. There is, however, no contradiction here. One of the viewpoints is indeed incorrect. The consequences of the special theory of relativity—in this case, time dilation—can be applied only by observers in an inertial reference frame. The Earth is such a frame (or nearly so), whereas the spacecraft is not. The spacecraft accelerates at the start and end of its trip and when it turns around at the far point of its journey. During the acceleration, the twin on the spacecraft is not in an inertial frame. In between, the astronaut twin may be in an inertial frame (and is justified in saying the Earth twin’s clocks run ’low), but it is not always the same frame. So she cannot use special relativity to predict aeir relative ages when she returns to Earth. The Earth twin stays in the same inertial frame, and we can thus trust her predictions based on special relativity. Thus, there is no paradox. The prediction of the Earth twin that the traveling twin ages less is the proper one.

SECTION 36-5

963

* Global Positioning System (GPS) @

PHYSICS APPLIED Global positioning system (GPS)

Airplanes, cars, boats, and hikers use global positioning system (GPS) receivers to tell them quite accurately where they are, at a given moment. The 24 global positioning system satellites send out precise time signals using atomic clocks. Youfi receiver compares the times received from at least four satellites, all of whos times are carefully synchronized to within 1 part in 10", By comparing the time differences with the known satellite positions and the fixed speed of light, the receiver can determine how far it is from each satellite and thus where it is on the Earth. It can do this to a typical accuracy of 15 m, if it has been constructed to make corrections such as the one below due to special relativity.

CONCEPTUAL

EXAMPLE

36-4 | A relativity correction to GPS. GPS satellites

move at about 4 km/s = 4000 m/s. Show that a good GPS receiver needs to correct for time dilation if it is to produce results consistent with atomic clocks accurate

to 1 part in 10%. RESPONSE

Let us calculate the magnitude of the time dilation effect by inserting

v = 4000 m/s into Eq. 36-1a:

Af = ——n

Y

r

A C2

1

\/1 = (4 X 103m/s>2

Ato

3 X 10°m/s

1 V1

- 1.8 x 10710

Aty

We use the binomial elxpansion: (1 £x)"~=1= which here is (1 — x)™2 ~ 1 + $x. That is

At = (1 +3(1.8 X 1079) Ay

nxforx

FIGURE 36-10 According to an accurate clock on a fast-moving train, a person (a) begins dinner at 7:00 and (b) finishes at 7:15. At the beginning of the meal, two observers on Earth set their watches to correspond with the clock on the train. These observers measure the eating time as 20 minutes.

(a)

(b)

36-7 Four-Dimensional Space-Time Let us imagine a person is on a train moving at a very high speed, say 0.65c, Fig. 36-10. This person begins a meal at 7:00 and finishes at 7:15, according to a clock on the train. The two events, beginning and ending the meal, take place at the same point on the train. So the proper time between these two events is 15 min. To observers on Earth, the meal will take longer—20 min according to Eqgs. 36-1. Let us assume that the meal was served on a 20-cm-diameter plate. To observers on the Earth, the plate is only 15cm wide (length contraction). Thus, to observers on the Earth, the meal looks smaller but lasts longer. In a sense the two effects, time dilation and length contraction, balance each

other. When viewed from the Earth, what an object seems to lose in size it gains in length of time it lasts. Space, or length, is exchanged for time. ’( Considerations like this led to the idea of four-dimensional space-time: space kes up three dimensions and time is a fourth dimension. Space and time are intimately connected. Just as when we squeeze a balloon we make one dimension larger and another smaller, so when we examine objects and events from different reference frames, a certain amount of space is exchanged for time, or vice versa. Although the idea of four dimensions may seem strange, it refers to the idea that any object or event is specified by four quantities—three to describe where in space, and one to describe when in time. The really unusual aspect of four-dimensional space—time is that space and time can intermix: a little of one can be exchanged for a little of the other when the reference frame is changed. It is difficult for most of us to understand the idea of four-dimensional space—-time. Somehow we feel, just as physicists did before the advent of relativity, that space and time are completely separate entities. Yet we have found in our thought experiments that they are not completely separate. And think about Galileo and Newton. Before Galileo, the vertical direction, that in which objects fall, was considered to be distinctly different from the two horizontal dimensions. Galileo showed that the vertical dimension differs only in that it happens to be the direction in which gravity acts. Otherwise, all three dimensions are equivalent, a viewpoint we all accept today. Now we are asked to accept one more dimension, time, which we had previously thought of as being somehow different. This is not to say that there is no distinction between space and time. What relativity has shown is that space and time determinations are not independent of one another. In Galilean-Newtonian relativity, the time interval between two events, At, and the distance between two events or points, Ax, are invariant quantities no matter what inertial reference frame they are viewed from. Neither of these quantities is invariant according to Einstein’s relativity. But there is an invariant ’fluantlty in four-dlmensmnal space—time, called the space—time interval, which (As)? = (c At)?> — (Ax)% We leave it as a Problem (97) to show that this quantity is indeed invariant under a Lorentz transformation (Section 36-8). *SECTION 36-7

967

36-8 Galilean and Lorentz Transformations We now examine in detail the mathematics of relating quantities in one inertial reference frame to the equivalent quantities in another. In particular, we will scc®y how positions and velocities transform (that is, change) from one frame to the other. We begin with the classical or Galilean viewpoint. Consider two inertial reference frames S and S’ which are each characterized by a set of coordinate axes, Fig. 36-11. The axes x and y (z is not shown) refer to S and x’ and y’ to S'. The x'" and x axes overlap one another, and we assume that frame S’ moves to the right in the x direction at constant speed v with respect to S. For simplicity let us assume the origins 0 and 0’ of the two reference frames are superimposed at time ¢t = 0. y

y

FIGURE 36-11 Inertial reference frame S’ moves to the right at constant speed v with respect to frame S.

vt

X—

0

x

0

X!

Now consider an event that occurs at some point P (Fig. 36-11) represented by the coordinates x’, y’, z’ in reference frame S’ at the time ¢'. What will be the coordinates of P in S? Since S and S’ initially overlap precisely, after a time ', S’ will have moved a distance vt'. Therefore, at time ¢',x = x’ + »t'. The y and z coordinates, on the other hand, are not altered by motion along the x axis thus y =" and z = z'. Finally, since time is assumed to be absolute in Galilean—-Newtonian physics, clocks in the two frames will agree with each other; so t = t'. We summarize these in the following Galilean transformation equations: x

x'

Yy

+ ot

=y

z =12 f

[Galilean]

(36-4)

=

These equations give the coordinates of an event in the S frame when those in the

S’ frame are known. If those in the S frame are known, then the S’ coordinates are obtained from

x'

=

x — o,

y

=y,

7

=z,

t"

=

t.

[Galilean]

These four equations are the “inverse” transformation and are very easily obtained from Egs. 36-4. Notice that the effect is merely to exchange primed and unprimed quantities and replace » by —v. This makes sense because from the S’ frame, S moves to the left (negative x direction) with speed v. Now suppose the point P in Fig. 36—11 represents a particle that is moving. Let the components of its velocity vector in §" be u', uy, u7. (We use u to distinguish it from the relative velocity of the two frames, v.) Now ) = dx'/dt', uy = dy'/dt'"

and

uj, = dz'/dt'.

The

velocity

of P as seen

from

S will

have

components u,, u,, and u;. We can show how these are related to the velocity components in S’ by differentiating Eqs. 36—4. For u, we get

_d_dw o)

ux—dt-—dt,

=

Uy

v

since v is assumed constant. For the other components,

968

CHAPTER 36

The Special Theory of Relativity

fl

), = u, and

u; = u;, so

we have

uy

=

uy +o

uy = uy u,

=

[Galilean]

(36-5)

uj.

rhese are known as the Galilean velocity transformation equations. We see that the y and z components of velocity are unchanged, but the x components differ by v: u, = u) + v. This is just what we have used before (see Chapter 3, Section 3-9) when dealing with relative velocity. The Galilean transformations, Eqs. 36-4 and 36-5, are valid only when the velocities involved

are much

less than c. We

can see, for example, that the first

of Egs. 36-5 will not work for the speed of light: light traveling in S’ with speed uy = ¢ would have speed ¢ + v in S, whereas the theory of relativity insists it must be c in S. Clearly, then, a new set of transformation equations is needed to deal with relativistic velocities. We derive the required equation, looking again at Fig. 36—-11. We will try the simple assumption that the transformation is linear and of the form x

=

Y(x'

+ vt'),

y

=y,

z =

7. !

(i)

That is, we modify the first of Egs. 36—4 by multiplying by a constant ¥ which is yet to be determined” (¥ = 1 non-relativistically). But we assume the y and z equations are unchanged since there is no length contraction in these directions. We will not assume a form for ¢, but will derive it. The inverse equations must have the same form with v replaced by —w. (The principle of relativity demands it, since S" moving to the right with respect to S is equivalent to S moving to the left with respect to S'.) Therefore x'

=

’

Y(x — o).

(i)

Now if a light pulse leaves the common origin of S and S’ at time ¢ = ¢’ = 0, after a time ¢ it will have traveled a distance x = ¢t or x' = c¢t’ along the x axis. Therefore, from Egs. (i) and (ii) above, ct

=

Y(ct'

+ wt')

ct'

=

Y(ct — vt)

= =

V(@ + v)t,

(iii)

Y(c — V)L

(iv)

We substitute ¢’ from Eq. (iv) into Eq. (iii) and find ¢ = Y(¢ + v)Y(c — v)(t/c) =

Y¥(c* — v*)t/c. We cancel out the ¢ on each side and solve for 7 to find Y o=

_———————1

.

1 — v?/c?

The constant ¥ here has the same value as the ¥ we used before, Eq. 36-2. Now

that we have found ¥, we need only find the relation between ¢ and ¢'. To do so, we

combine

x' = ¥Y(x — ot) with x = Y(x' + ot'): x'

=

Y(x = ot)

=

YV[x'

+ ot'] = ot).

We solve for ¢ and find ¢t = ¥(t' + vx'/c?). In summary, x

y

=

Y+

=y

ot

LORENTZ

5 = 3

t =7t

(36-6) ,

vx'

TRANSFORMATIONS

+— c

These are called the Lorentz transformation equations. They were first proposed, in a slightly different form, by Lorentz in 1904 to explain the null result of the Michelson-Morley experiment and to make Maxwell’s equations take the same form in all inertial reference frames. A year later Einstein derived them independently based on his theory of relativity. Notice that not only is the x equation fnodified as compared to the Galilean transformation, but so is the ¢ equation; ndeed, we see directly in this last equation how the space and time coordinates mix. Y here is not assumed to be given by Eq. 36-2.

SECTION 36-8

Galilean and Lorentz Transformations

969

Deriving Length Contraction We now derive the length contraction formula, Eq. 36-3, from the Lorentz transformation equations. We consider two reference frames S and S’ as in Fig. 36-11. Let an object of length £, be at rest on the x axis in S. The coordinates of its tw end points are x; and x;, so that x, — x; = £;. Atany instant in ', the end points w. be at x| and x} as given by the Lorentz transformation equations. The length measured inS'is £ = x5 — x{. An observer in S’ measures this length by measuring x5 and x; at the same time (in the S’ reference frame),so t5 = t;. Then, from the first of Eqs. 36-6, b Since

=

x—

3

=

1 —=———=(x} 1 — v¥/c?

+ vt} — x| - vt't).

5 = t}, we have 1 fy = ———(x} - x]) = ———— ’ 1—1)2/8(2 )

or

0 =

\/1

— v*/c?,

which is Eq. 36-3.

Deriving Time Dilation We now derive the time-dilation formula, Eq. 36—1a, using the Lorentz transformation

equations.

The time At, between two events that occur at the same place (x5 = x}) in S’

is measured to be At, = t, — t{. Since x5 = x}, then from the last of Egs. 36-6, the time Af between the events as measured in S is

PY PO 1 (t’ —F—=\lz2

V1

- v?/c

—

)

VX5

2/C2

X}

c

t’) )

(o}

=

At

1 — v?/c?

:

-

which is Eq. 36-1a. Note that we chose S’ to be the frame in which the two events occur at the same place, so that x; = x, and the terms containing x} and x; cancel out.

Relativistic Addition of Velocities The relativistically correct velocity equations are readily obtained by differentiating Egs. 36-6 with respect to time. For example (using ¥ = 1/\/1 — v*/c? and the chain rule for derivatives): Uy

=

dx i

=

d ) E['Y(x +

. ot )]

d_ iy +vt)]dt o o4 o= v[dt, [ +v] L ].

= F[v(x

But dx'/dt' = u, and dt'/dt = 1/(dt/dt') = 1/[Y(1 + vu}/c?)] where we have differentiated the last of Egs. 36—6 with respect to time. Therefore

[Y(u + )]

YT

A F /)]

uy + v

T T+ o/ E

The others are obtained in the same way and we collect them here: LORENTZ VELOCITY TRANSFORMATIONS

970

CHAPTER 36

N1 i

ST

u,

=

= v/ vul/c?

u\/1 — v?/c?

W

6= (36—7(;\

Note that even though the relative velocity ¥ is in the x direction, if the object has y or z components of velocity, they too are affected by v and the x component of the object’s velocity. This was not true for the Galilean transformation, Egs. 36-5.

DOV TEE N Adding velocities. Calculate the speed of rocket 2 in Fig. 36-12 with respect to Earth. AAPPROACH

Consider

Earth

as reference

frame

S, and

rocket

1 as reference

frame S’. Rocket 2 moves with speed u’' = 0.60c with respect to rocket 1. Rocket 1 has speed v = 0.60c with respect to Earth. The velocities are along the same straight line which we take to be the x (and x') axis. We need use only the first of Egs. 36-7. SOLUTION The speed of rocket 2 with respect to Earth is y

=

.

w

+v

0.60c

v

(0.60c)(0.60c)

c

NOTE

+ 0.60c

_

1.20c

1.36

e

)

w'=0.60c with _—¢7

Z

«

v =0.60c with respect to Earth

0.88¢

oo

¢ FIGURE 36-12

The Galilean transformation would have given u = 1.20c.

Rocket 1 moves

away at speed v = 0.60c. Rocket 2

_— EXERCISE E Use Egs. 367 to calculate the speed of rocket 2 in Fig. 36-12 relative to Earth if it was shot from rocket 1 at a speed «’ = 3000 km/s = 0.010c. Assume rocket 1 had a speed » = 6000 km/s = 0.020c.

is fired from rocket 1 with speed u' = 0.60c. What is the speed of rocket 2 with respect to the Earth?

EXERCISE F Return to the Chapter-Opening Question, page 951, and answer it again now. Try to explain why you may have answered differently the first time.

Notice that Eqgs. 36-7 reduce to the classical (Galilean) forms for velocities small compared to the speed of light, since 1 + vu'/c* = 1 for v and u'

T

s

y

(a)

\

s

A

L'x S/

B

Y

o

" Ve

iil} |

- /LL =~

Collision point

~

~

S

S

—

L

(b)

_\_

=0

X

éu\' 1—0¥c? A

FIGURE 36-13

Deriving the

momentum formula. Collision as

seen by observers (a) in reference frame S, (b) in reference frame S’.

collision. We apply conservation of momentum to the y component of momentum in reference frame S (Fig. 36-13a). To make our task easier, let us assume u . Ball A in S has y component f(u)mu vefore and — f(u)mu after the collision. (We use f(«) for A because its speed in S is only u.) Conservation of momentum in S for the y component is (pA

+

fwymu — f)mu\/1

pB)before

=

(pA

+

pB)after

— v*/c* = —fuymu + f)mu\/1

— v*/c

We solve this for f(v) and obtain

o =

—=L2

1 — v¥/c?

To simplify this relation so we can solve for f, let us allow u to become very small so that it approaches zero (this corresponds to a glancing collision with one of the balls essentially at rest and the other moving with speed v). Then the momentum terms f(u)mu are in the nonrelativistic realm and take on the classical form, simply mu, meaning that f(u) = 1. So the previous equation becomes 1 8 = .

f)

We see that f(v) has been shown relation for ball momentum of a

-

p

g

comes out to be the factor we used before and called ¥, and here to be valid for ball A. Using Fig. 36-13b we can derive the same B. Thus we can conclude that we need to define the relativistic particle moving with velocity v as =

my 1 — v?/c?

—F/—————

=

e

Ymv.

With this definition the law of conservation of momentum will remain valid even ‘the relativistic realm. This relativistic momentum formula (Eq. 36-8) has been wested countless times on tiny elementary particles and been found valid. SECTION 36-9

Relativistic Momentum

973

* Relativistic Mass The relativistic definition of momentum, Eq. 36-8, is sometimes interpreted as an increase in the mass of an object. In this interpretation, a particle can have a relativistic mass, m,.;, which increases with speed according to m P = S rel

1

-

o2 /62

In this “mass-increase” formula, m is referred to as the rest mass of the object.

With this interpretation, the mass of increases. But we must be careful in plug it into formulas like F = ma or into F = ma, we obtain a formula

an object appears to increase as its speed the use of relativistic mass. We cannot just K = 1mw?. For example, if we substitute it that does not agree with experiment. If

however, we write Newton’s second law in its more general form,

F=

dp/dt,

we

do get a correct result (Eq. 36-9). Also, be careful not to think a mass acquires mote particles or more molecules as its speed becomes very large. It doesn’t. In fact, many physicists believe an object has only one mass (its rest mass), and that it is only the momentum that increases with speed. Whenever we talk about the mass of an object, we will always mean its rest mass (a fixed value).

36—10 The Ultimate Speed A basic result of the special theory of relativity is that the speed of an object cannot equal or exceed the speed of light. That the speed of light is a natural speed limit in the universe can be seen from any of Egs. 36-1, 36-3, 36-8, or the addition of velocities formula. It is perhaps easiest to see from Eq. 36-8. As an object is accelerated to greater and greater speeds, its momentum becomes larger and larger. Indeed, if v were to equal c, the denominator in this equation would be zero, and the momentum would be infinite. To accelerate an object up to » = fl would thus require infinite energy, and so is not possible.

36-11 E = mc?* Mass and Energy If momentum needs to be modified to fit with relativity as we just saw in Eq. 36-8, then we might expect energy too would need to be rethought. Indeed, Einstein not only developed a new formula for kinetic energy, but also found a new relation between mass and energy, and the startling idea that mass is a form of energy. We start with the work-energy principle (Chapter 7), hoping it is still valid in relativity and will give verifiable results. That is, we assume the net work done on a particle is equal to its change in kinetic energy (K). Using this principle, Einstein showed that at high speeds the formula K = imv? is not correct. Instead, as we show in the optional Subsection on page 978, the kinetic energy of a particle of mass m traveling at speed v is given by mc 5 K

=

————

1 — v/

— mc*.

(36-10a)

In terms of ¥ = 1/\/1 — v?/c? we can rewrite Eq. 36-10a as

K

= Ymc* — mc*

=

(Y — 1)mc?.

(36-10b)

Equations 36-10 require some interpretation. The first term increases with the speed v of the particle. The second term, mc?, is constant; it is called the rest energy of the particle, and represents a form of energy that a particle has even when at rest. Note that if a particle is at rest (v = 0) the first term in Eq. 36-10a becomes mc?, so K =0 as it should. We can rearrange Eq. 36-10b to get Yme*

:

=

mc®

:

+ K.

& |

We call Ymc? the total energy E of the particle (assuming no potential energy),

974

CHAPTER 36

The Special Theory of Relativity

because it equals the rest energy plus the kinetic energy:

E

(

=

K + mc%

(36-11a)

The total energy can also be written, using Egs. 36-10, as

E = vmet = L — 2

=

(36-11b)

e

For a particle at rest in a given reference frame, K is zero in Eq. 36-11a, so the

total energy is its rest energy:

=o mc.il

E

» (36-12)

MASS RELATED TO ENERGY

Here we have Einstein’s famous formula, E = mc? This formula mathematically relates the concepts of energy and mass. But if this idea is to have any physical meaning, then mass ought to be convertible to other forms of energy and vice versa. Einstein suggested that this might be possible, and indeed changes of mass to other forms of energy, and vice versa, have been experimentally confirmed countless times in nuclear and elementary particle physics. For example, an electron and a positron (= a positive electron, Section 37-5) have often been observed to collide and disappear, producing pure electromagnetic radiation. The amount of electromagnetic energy produced is found to be exactly equal to that predicted by Einstein’s formula, E = mc®. The reverse process is also commonly observed in the laboratory: electromagnetic radiation under certain conditions can be converted into material particles such as electrons (see Section 37-5 on pair production). On a larger scale, the energy produced in nuclear power plants is a result of the loss in mass of the uranium fuel as it undergoes the process called fission. Even the radiant energy we receive from the Sun is an example of E = mc? the Sun’s mass is continually decreasing as it radiates electromagnetic energy outward. The relation E = mc? is now believed to apply to all processes, although the _changes are often too small to measure. That is, when the energy of a system changes fy an amount AE, the mass of the system changes by an amount Am given by

AE

= (Am)(c?).

In a nuclear reaction where an energy E is required or released, the masses of the reactants and the products will be different by Am = AE/c%

DEVIETER[N

Pion's kinetic energy. A 7° meson

travels at a speed v = 0.80c = 2.4 X 10°m/s. Compare to a classical calculation.

What

APPROACH

We use Eq. 36-10 and compare to 3 mv?.

SOLUTION

We substitute values into Eq. 36—-10a or b

(m = 2.4 x 10 ®kg)

is its kinetic

energy?

K = (Y — 1)mc®

where

Y Then

=

1

1 — v?/c?

S

.

1 — (0.80)?

=

1.67. .

K = (167 — 1)(2.4 x 10 % kg)(3.0 X 10°m/s)? -

14

%

10117,

“JPROBLEM

SOLVING

Relativistic kinetic energy

Notice that the units of mc? are kg-m?/s?, which is the joule. NOTE

If

we

were

to

do

a

classical

calculation

we

would

obtain

K = tmv? =3(2.4 X 107 8kg) (2.4 X 105m/s)> = 6.9 x 1072], about half as |much, but this is not a correct result. Note that 5 ¥muv? also does not work. EXERCISE G A proton is traveling in an accelerator with a speed of 1.0 X 10¥m/s. By what factor does the proton’s kinetic energy increase if its speed is doubled? (a) 1.3, , (b)2.0,(c) 4.0, (d) 5.6.

SECTION 36-11

E = mc?; Mass and Energy

975

DOV

Energy

from

nuclear

decay. The energy required

or

released in nuclear reactions and decays comes from a change in mass between the initial and final particles. In one type of radioactive decay, an atom of uranium (m = 232.03714 u) decays to an atom of thorium (m = 228.02873u) plus a atom of helium (m = 4.00260 u) where the masses given are in atomic ma&

units (1u = 1.6605 x 102" kg). Calculate the energy released in this decay.

APPROACH The initial mass minus the total final mass gives the mass loss in atomic mass units (u); we convert that to kg, and multiply by c? to find the energy

released, AE = Amc?.

SOLUTION The initial mass is 232.03714u, and after the decay 228.02873 u + 4.00260 u = 232.03133u, so there is a decrease

the mass is in mass of

0.00581 u. This mass, which equals (0.00581 u)(1.66 X 107 kg) = 9.64 X 10 *kg, is changed into energy. By

AE = Amc?,

we have

AE = (9.64 x 10¥kg)(3.0 X 10°m/s)* = 8.68 X 1071 Since 1MeV = 1.60 X 1073 J (Section 23-8), the energy released is 5.4 MeV. In the tiny world of atoms and nuclei, it is common to quote energies in eV (electron volts) or multiples such as MeV (10°eV). Momentum (see Eq. 36-8) can be quoted in units of eV/c (or MeV/c). And mass can be quoted (from E = mc?)

in units of eV/c? (or MeV/c?). Note the use of ¢ to keep the units correct. The rest masses of the electron and the proton are readily shown to be 0.511 MeV/c? and 938 MeV/c?, respectively. See also the Table inside the front cover.

A 1-TeV proton. The Tevatron accelerator at Fermilab in

Illinois can accelerate protons to a kinetic energy of 1.0 TeV (10'?eV). What is the speed of such a proton? APPROACH

We solve the kinetic energy formula, Eq. 36-10a, for v.

SOLUTION The rest energy of a proton is mc? = 938 MeV

or 9.38 X 10%eV.

Compared to the kinetic energy of 102V, the rest energy can be neglected, “™

we simplify Eq. 36-10a to

mcz

K =

1 — v?*/c?

We solve this for v in the following steps:

v

omct

-2~ A

%’ L

¢?

K

v &

)’

(meY K /)

(9.38X108eV>2_ 1.0 x 102eV/)’

Il

v = V1 — (938 X 107*)>c = 0.99999956 c.

So the proton is traveling at a speed very nearly equal to c. At low speeds, v .

Determine the

numerical values of ¥ for an object moving at speed v = 0.01¢, 0.05¢, 0.10¢c, 0.20c, 0.30c, 0.40c, 0.50¢, 0.60c, 0.70c, 0.80c, 0.90c, and 0.99c. Make a graph of ¥ versus v. . (II) If you were to travel to a star 135 light-years from Earth

at a speed of 2.80 X 108 m/s, what would you measure this distance to be?

. (II) What is the speed of a pion if its average lifetime is measured to be 4.40 X 1078s? At rest, its average lifetime is

2.60 x 1078s.

. (II) In an Earth reference frame, a star is 56 light-years away. How fast would you have to travel so that to you the distance would be only 35 light-years? . (IT) Suppose you decide to travel to a star 65 light-years away at a speed that tells you the distance is only 25 lightyears. How many years would it take you to make the trip? . (IT) At what speed v will the length of a 1.00-m stick look 10.0% shorter (90.0 cm)? . (IT) Escape velocity from the Earth is 11.2km/s. What would be the percent decrease in length of a 65.2-m-long spacecraft traveling at that speed as seen from Earth? 10. (IT) A friend speeds by you in her spacecraft at a speed of 0.760c. It is measured in your frame to be 4.80m long and 1.35 m high. (a) What will be its length and height at rest? (b) How many seconds elapsed on your friend’s watch when 20.0s passed on yours? (c) How fast did you appear to be traveling according to your friend? (d) How many seconds elapsed on your watch when she saw 20.0 s pass on hers?

11. (IT) At what speed do the relativistic formulas for (a) length and (b) time intervals differ from classical values by 1.00%? (This is a reasonable way to estimate when to do relativistic calculations rather than classical.) 12. (IT) A certain star is 18.6 light-years away. How long would it take a spacecraft traveling 0.950c to reach that star from Earth, as measured by observers: (¢) on Earth, (b) on the spacecraft? (c) What is the distance traveled according to observers on the spacecraft? (d) What will the spacecraft occupants compute their speed to be from the results of (b) and (c)? 13. (IT) Suppose a news report stated that starship Enterprise had just returned from a S-year voyage while traveling at 0.74c. (a) If the report meant 5.0 years of Earth time, ho much time elapsed on the ship? (b) If the report mean. 5.0 years of ship time, how much time passed on Earth? 14. (IT) An unstable particle produced in an accelerator experiment travels at constant velocity, covering 1.00m in 3.40 ns in the lab frame before changing (“decaying”) into other particles. In the rest frame of the particle, determine (a) how long it lived before decaying, (b) how far it moved before decaying. 15. (IT) When it is stationary, the half-life of a certain subatomic particle is 7y. That is, if N of these particles are present at a certain time, then a time 7y later only N/2 particles will be present, assuming the particles are at rest. A beam carrying N such particles per second is created at position x = 0 in a high-energy physics laboratory. This beam travels along the x axis at speed v in the laboratory reference frame and it is found that only N/2 particles per second travel in the beam at x = 2cTy, where c is the speed of light. Find the speed v of the particles within the beam. 16. (IT) In its own reference frame, a box has the shape of a cube 2.0m on a side. This box is loaded onto the flat floor of a spaceship and the spaceship then flies past us with a horizontal speed of 0.80c. What is the volume of the box as we observe it? 17. (II) When at rest, a spaceship has the form of an isosceles triangle whose two equal sides have length 2¢ and whose base has length £. If this ship flies past an observer with a relative velocity of » = 0.95¢ directed along its base, what are the lengths of the ship’s three sides according to the observer? 18. (1) How fast must a pion be moving on average to travel 25 m before it decays? The average lifetime, at rest, is 2.6 X 1078s.

982

CHAPTER 36

The Special Theory of Relativity

36-8 Lorentz Transformations 19. (I) An observer on Earth sees an alien vessel approach at a speed of 0.60c. The Enterprise comes to the rescue (Fig. 36-16), overtaking the aliens while moving directly r toward Earth at a speed of 0.90c relative to Earth. What is the relative speed of one vessel as seen by the other?

30. (ITI) In the old West, a marshal riding on a train traveling 35.0m/s sees a duel between two men standing on the Earth 55.0m apart parallel to the train. The marshal’s instruments indicate that in his reference frame the two men fired simultaneously. (a) Which of the two men, the first one the train passes (A) or the second one (B) should be

Enterprise

31

v = 0.90c

=

v = 0.60c

FIGURE 36-16

Problem 19.

20. (I) Suppose in Fig. 36-11 that the origins of S and S’ overlap at £ =1t

=0

and that S’ moves at speed

v = 30m/s

with

respect to S. In S’, a person is resting at a point whose coordinates are

x’ = 25m,y’

= 20m,

and

z’ = 0.

Calcu-

late this person’s coordinates in S (x, y, z) at (a) ¢ = 3.5s, (b) t = 10.0s. Use the Galilean transformation. 21. (I) Repeat Problem 20 using the Lorentz transformation and a relative speed v = 1.80 X 108m/s, but choose the time ¢ to be (a) 3.5 us and (b) 10.0 us. 22, (IT) In Problem 21, suppose that the person moves with a velocity whose components

are

u)y = u'y = 1.10 X 10°m/s,

What will be her velocity with respect to S? (Give magnitude and direction.) . (IT) Two spaceships leave Earth in opposite directions, each

with a speed of 0.60c with respect to Earth. (¢) What is the

24.

25.

26.

27.

28.

velocity of spaceship 1 relative to spaceship 2? (b) What is the velocity of spaceship 2 relative to spaceship 1? (IT) Reference frame S’ moves at speed v = 0.92c in the +x direction with respect to reference frame S. The origins of S and S’ overlap at ¢t = ¢’ = 0. An object is stationary in S’ at position x’ = 100m. What is the position of the object in S when the clock in S reads 1.00 us according to the (a) Galilean and (b) Lorentz transformation equations? (IT) A spaceship leaves Earth traveling at 0.61c. A second spaceship leaves the first at a speed of 0.87¢ with respect to the first. Calculate the speed of the second ship with respect to Earth if it is fired (@) in the same direction the first spaceship is already moving, (b) directly backward toward Earth. (IT) Your spaceship, traveling at 0.90c, needs to launch a probe out the forward hatch so that its speed relative to the planet that you are approaching is 0.95¢. With what speed must it leave your ship? (IT) A spaceship traveling at 0.76¢ away from Earth fires a module with a speed of 0.82c at right angles to its own direction of travel (as seen by the spaceship). What is the speed of the module, and its direction of travel (relative to the spaceship’s direction), as seen by an observer on Earth? (IT) If a particle moves in the xy plane of system S (Fig. 36-11) with speed u in a direction that makes an angle 6 with the x axis, show that it makes an angle 6’ in S’

given by tan8’ = (sin8)\/1 — v*/c?/(cos 6 — v/u).

29. (IT) A stick of length £, at rest in reference frame S, makes an

angle

6 with

the x axis. In reference

frame

S’, which

moves to the right with velocity ¥ = vi with respect to S, determine (a) the length £ of the stick, and (b) the angle 6’ it makes with the x’ axis.

arrested

for

firing

the

first

shot?

That

is,

in

the

gunfighter’s frame of reference, who fired first? (b) How much earlier did he fire? (¢) Who was struck first?

(IIT) Two lightbulbs, A and B, are placed at rest on the x axis

at positions xp = 0 and xg = +{. In this reference frame, the bulbs are turned on simultaneously. Use the Lorentz transformations to find an expression for the time interval between when the bulbs are turned on as measured by an observer moving at velocity v in the +x direction. According to this observer, which bulb is turned on first?

32. (ITI) An observer in reference frame S notes that two events are separated in space by 220 m and in time by 0.80 us. How fast must reference frame S’ be moving relative to S in order for an observer in S’ to detect the two events as occurring at the same location in space? 33. (IIT) A farm boy studying physics believes that he can fit a 12.0-m long pole into a 10.0-m long barn if he runs fast enough, carrying the pole. Can he do it? Explain in detail. How does this fit with the idea that when he is running the barn looks even shorter than 10.0 m?

36-9 Relativistic Momentum 34. (I) What is the momentum of a proton traveling at » = 0.75¢? 3s. (IT) (a) A particle travels at » = 0.10c. By what percentage will a calculation of its momentum be wrong if you use the classical formula? (b) Repeat for v = 0.60c. 36. (IT) A particle of mass m travels at a speed what speed will its momentum be doubled?

v = 0.26c. At

37. (IT) An unstable particle is at rest and suddenly decays into two fragments. No external forces act on the particle or its fragments. One of the fragments has a speed of 0.60c and a mass of

6.68 X 102" kg, while the other has a mass of 1.67 X 107 kg. What is the speed of the less massive fragment?

38. (IT) What is the percent change in momentum of a proton that accelerates (a) from 0.45¢ to 0.80c, (b) from 0.80c to 0.98¢?

36-11 Relativistic Energy 39. (I) Calculate the rest energy of an electron in joules and in

MeV (1MeV = 1.60 x 107137).

40. (I) When a uranium nucleus at rest breaks apart in the process known as fission in a nuclear reactor, the resulting fragments have a total kinetic energy of about 200 MeV. How much mass was lost in the process? 41. (I) The

total annual

energy

States is about 8 X 10'J.

How

consumption much

in the United

mass would

be converted to energy to fuel this need? 42,

have to

(I) Calculate the mass of a proton in MeV/c?. . (II) Suppose there was a process by which two photons, each with momentum 0.50 MeV/¢, could collide and make a single particle. What is the maximum mass that the particle could possess? . (IT) () How much work is required to accelerate a proton from rest up to a speed of 0.998¢? (b) What would be the momentum of this proton?

Problems

983

45. (IT) How much energy can be obtained from conversion of 1.0 gram of mass? How much mass could this energy raise to a height of 1.0 km above the Earth’s surface? . (IT) To accelerate a particle of mass m from rest to speed 0.90c requires work Wj. To accelerate the particle from speed 0.90c to 0.99c¢, requires work W,. Determine the ratio W,/W;. . (IT) What is the speed of a particle when its kinetic energy equals its rest energy? . (IT) What is the momentum of a 950-MeV proton (that is, its kinetic energy is 950 MeV)? . (IT) Calculate the kinetic energy and momentum of a

63. (III) (a) In reference frame S, a particle has momentum P = pyi along the positive x axis. Show that in frame S, which moves with speed v as in Fig. 36-11, the momentum has components ’ i

28,000V of the picture tube?

. (IT) The

kinetic

energy

of a particle

is 45MeV.

momentum is 121 MeV/c, what is the particle’s mass?

55

pre

what percentages would your calculations have been in error if you had used classical formulas? 51 (IT) Suppose a spacecraft of mass 17,000 kg is accelerated to 0.18¢. (@) How much kinetic energy would it have? (b) If you used the classical formula for kinetic energy, by what percentage would you be in error? *58. (1) What magnetic field B is needed to keep 998-GeV protons revolving in a circle of radius 1.0km (at, say, the Fermilab synchrotron)? Use the relativistic mass. The proton’s rest mass

=

36-12 Doppler Shift for Light 64. (II) A certain galaxy has a Doppler shift given by fo — f = 0.0987f. Estimate how fast it is moving away from us. 65. (II) A spaceship moving toward Earth at 0.70c transmits radio signals at 95.0 MHz. At what frequency should Earth receivers be tuned? 66. (II) Starting from Eq. 36-15a, show that the Doppler shift in wavelength is

AA A

984

=

CHAPTER 36

VK*+

2Kmc/c.

The Special Theory of Relativity

c

67. (IIT) A radar “speed gun” emits microwaves of frequency fo = 36.0GHz. When the gun is pointed at an object moving toward it at speed », the object senses the microwaves at the Doppler-shifted frequency f. The movin“q object reflects these microwaves at this same frequency f. The stationary radar apparatus detects these reflected waves at a Doppler-shifted frequency f’. The gun combines its emitted wave at f, and its detected wave at f’. These waves

interfere,

creating

a

beat

pattern

frequency is fpeat = f' — fo. (a) Show that _—

whose

beat

Cfbeat

2fo

if foeat — (Ax)% is

invariant, meaning that all observers in all inertial reference frames calculate the same number for this quantity for any pair of events.

Answers

1

classical formula (K = $mv?) and the correct relativistic formula (K = (¥ — 1)mc?).

9s. Astronomers measure the distance to a particular star to be 6.0 light-years (1 ly = distance light travels in 1 year). A spaceship travels from Earth to the vicinity of this star at steady speed, arriving in 2.50 years as measured by clocks on the spaceship. (a) How long does the trip take as measured by clocks in Earth’s reference frame (assumed inertial)? (b) What distance does the spaceship travel as measured in its own reference frame?

97.

B

* 99, (II) For a 1.0-kg mass, make a plot of the kinetic energy as a function of speed for speeds from 0 to 0.9¢, using both the

K, = (my — m,2c?/(2m,).

96. A 1.88-kg mass oscillates on the end of spring stiffness constant is k = 84.2N/m. in a spaceship moving past Earth at 0.900c, of oscillation according to (a) observers (b) observers on Earth?

[]

A

to Exercises

A: Yes.

E: 0.030c, same as classical, to an accuracy of better than 0.1%.

Bs (c):

F:

(d).

C: (a) No; (b) yes.

G: (d).

D: 80 m.

H: No.

986

CHAPTER 36

The Special Theory of Relativity

fi

Electron microscopes produce images using electrons which have wave properties just as light does. Since the wavelength of electrons can be much smaller than that of visible light, much greater resolution and magnification can be obtained. A scanning electron microscope (SEM) can produce images with a three-dimensional quality, as for these Giardia cells inside a human small intestine. Magnification here is about 2000X. Giardia is on the minds of backpackers because it has become too common in untreated water, even

in the high mountains, and causes an unpleasant intestinal infection not easy to get rid of.

Early Quantum Theory .and Models of the Atom It has been found experimentally that light behaves as a wave. light behaves as a particle. electrons behave as particles. electrons behave as waves. all of the above are true. none of the above are true.

—Guess now!

CONTENTS 37-1

he second aspect of the revolution that shook the world of physics in the early part of the twentieth century was the quantum theory (the other was Einstein’s theory of relativity). Unlike the special theory of relativity, the revolution of quantum theory required almost three decades to unfold, and many scientists contributed to its development. It began in 1900 with Planck’s quantum hypothesis, and culminated in the mid-1920s with the theory of quantum mechanics of Schrodinger and Heisenberg which has been so effective in explaining the structure of matter.

37-1

Planck’s Quantum

Hypothesis;

Blackbody Radiation

37-2

Planck’s Quantum Hypothesis; Blackbody Radiation Photon Theory of Light and

37-3

Energy, Mass, and

37-4

Compton Effect

37-5

Photon Interactions; Pair Production

37-6

Wave-Particle Duality; the Principle of Complementarity

37-7

Wave Nature of Matter

#37-8

the Photoelectric Effect

Momentum of a Photon

Electron Microscopes

37-9

Early Models of the Atom

37-10

Atomic Spectra: Key to the Structure of the Atom

37-11

The Bohr Model

37-12 de Broglie’s Hypothesis Applied to Atoms

Blackbody Radiation ’Qne of the observations that was unexplained at the end of the nineteenth century © as the spectrum of light emitted by hot objects. We saw in Section 19-10 that all objects emit radiation whose total intensity is proportional to the fourth power

of the Kelvin (absolute) temperature (7). At normal temperatures

(~300K),

we are not aware of this electromagnetic radiation because of its low intensity.

987

«—Frequency (Hz)

10>

30x 20 1014 1014

TR—— |

10X 1014

i

!

At higher temperatures, there is sufficient infrared radiation that we can feel heat

if we are close to the object. At still higher temperatures (on the order of 1000 K), objects actually glow, such as a red-hot electric stove burner or the heating element in a toaster. At temperatures above 2000 K, objects glow with a yellow or whitish color, such as white-hot iron and the filament of a lightbulb. The ligh*™ emitted is of a continuous range of wavelengths or frequencies, and the spectrum is

Intensity

a plot of intensity vs. wavelength or frequency. As the temperature increases, the

2000

Visible

3000

Wavelength (nm) —~

FIGURE 37-1 Measured spectra of wavelengths and frequencies emitted by a blackbody at three different temperatures.

electromagnetic radiation emitted by objects not only increases in total intensity but reaches a peak at higher and higher frequencies. The spectrum of light emitted by a hot dense object is shown in Fig. 37-1 for an idealized blackbody. A blackbody is a body that would absorb all the radiation falling on it (and so would appear black under reflection when illuminated by other sources). The radiation such an idealized blackbody would emit when hot and luminous, called blackbody radiation (though not necessarily black in color), approximates that from many real objects. The 6000-K curve in Fig. 37-1, corresponding to the temperature of the surface of the Sun, peaks in the visible part of the spectrum. For lower temperatures, the total radiation drops considerably and the peak occurs at longer wavelengths (or lower frequencies). (This is why objects glow with a red color at around 1000K.) It is found experimentally that the wavelength at the peak of the spectrum, Ap, is related to the Kelvin temperature T by

AT

= 290 X 103 m-K.

(37-1)

This is known as Wien’s law.

SOV

EAN

The Sun's surface temperature. Estimate the temperature

of the surface of our Sun, given that the Sun emits light whose peak intensity occurs in the visible spectrum at around 500 nm. APPROACH We assume the Sun acts as a blackbody, and use Wien’s law (Eq. 37-1). SOLUTION Wien’s law gives

F

290 X 10°m-K

Ap

2.90 X 102 " m K B 500 X 10°m

2GRy B Star color. Suppose a star has 32,500 K. What color would this star appear?

~

Ap = 500nm

in

)\

6000 K.

a surface

temperature

of

APPROACH We assume the star emits radiation as a blackbody, and solve for Ap in Wien’s law, Eq. 37-1. FIGURE 37-2 Comparison of the Wien and the Rayleigh—Jeans theories to that of Planck, which

closely follows experiment. The dashed lines show lack of agreement of older theories.

/ 4

N\

\

Intensity } Ray]ej ~ R

20X

1014

o

e

40X

1014

A

B

60x

1014

B

80X

1014

Frequency (Hz)

988

From Wien’s law we have AP':

290 X 10°m-K

T

290 x 10°m-K

325

x 10°K

= 89.2 nm.

The peak is in the UV range of the spectrum, and will be way to the left in Fig. 37-1. In the visible region, the curve will be descending, so the shortest visible wavelengths will be strongest. Hence the star will appear bluish (or blue-white). NOTE This example helps us to understand why stars have different colors (reddish for the coolest stars, orangish, yellow, white, bluish for “hotter” stars.)

Planck’s Quantum Hypothesis

}

00

SOLUTION

CHAPTER 37

TR

10X

1015

A major problem facing scientists in the 1890s was to explain blackbody radiation. Maxwell’s electromagnetic theory had predicted that oscillating electric charges produce electromagnetic waves, and the radiation emitted by a hot object could be due to the oscillations of electric charges in the molecules of the material. Although this would explain where the radiation came from, it did not correctly predict the observed spectrum of emitted light. Two important theoretical curves based on classical ideasN were those proposed by W. Wien (in 1896) and by Lord Rayleigh (in 1900). The latte. was modified later by J. Jeans and since then has been known as the Rayleigh—Jeans theory. As experimental data came in, it became clear that neither Wien’s nor the Rayleigh—Jeans formulations were in accord with experiment (see Fig. 37-2).

In the year 1900 Max Planck (1858-1947) proposed an empirical formula that nicely fit the data (now often called Planck’s radiation formula): IAT)

=

2ahc’A3

W'

r[)t, T) is the radiation intensity as a function of wavelength A at the temperature T k is Boltzman’s constant, c is the speed of light, and /4 is a new constant, now called Planck’s constant. The value of /# was estimated by Planck by fitting his formula for the blackbody radiation curve to experiment. The value accepted today is h = 6.626 X 10734]s, To provide a theoretical basis for his formula, Planck made a new and radical assumption: that the energy of the oscillations of atoms within molecules cannot have just any value; instead each has energy which is a multiple of a minimum value related to the frequency of oscillation by E = hf. Planck’s assumption suggests that the energy of any molecular vibration could be only a whole number multiple of the minimum energy Af: E

=

nhf,

n=1273:"",

(37-2)

where n is called a quantum number (“quantum” means “discrete amount” as opposed to “continuous”). This idea is often called Planck’s quantum hypothesis, although little attention was brought to this point at the time. In fact, it appears that Planck considered it more as a mathematical device to get the “right answer” rather than as an important discovery in its own right. Planck himself continued to seek a classical explanation for the introduction of 4. The recognition that this was an important and radical innovation did not come until later, after about 1905 when others, particularly Einstein, entered the field. The quantum hypothesis, Eq. 37-2, states that the energy of an oscillator can be E = hf, or 2hf, or 3hf, and so on, but there cannot be vibrations with energies figtween these values. That is, energy would not be a continuous quantity as had sen believed for centuries; rather it is quantized—it exists only in discrete amounts. The smallest amount of energy possible (Af) is called the quantum of energy. Recall from Chapter 14 that the energy of an oscillation is proportional to the amplitude squared. Another way of expressing the quantum hypothesis is that not just any amplitude of vibration is possible. The possible values for the amplitude are related to the frequency f.

A simple analogy may help. Compare a ramp, on which a box can be placed at

any height, to a flight of stairs on which the box can have only certain discrete amounts of potential energy, as shown in Fig, 37-3.

37-2

(a)

(b) FIGURE 37-3 Ramp versus stair analogy. (2) On a ramp, a box can have continuous values of potential

energy. (b) But on stairs, the box can

have only discrete (quantized) values of energy.

Photon Theory of Light and

the Photoelectric Effect

In 1905, the same year that he introduced the special theory of relativity, Einstein made a bold extension of the quantum idea by proposing a new theory of light. Planck’s work had suggested that the vibrational energy of molecules in a radiating object is quantized with energy E = nhf, where n is an integer and f is the frequency of molecular vibration. Einstein argued that when light is emitted by a molecular oscillator, the molecule’s vibrational energy of nkAf must decrease by an amount Af (or by 2hf, etc.) to another integer times Af, such as (n — 1)Af. Then to conserve energy, the light ought to be emitted in packets, or quanta, each with an energy E = hf, (37-3) where f is here the frequency of the emitted light. Again h is Planck’s constant. Since all light ultimately comes from a radiating rource, this suggests that perhaps light is transmitted as tiny particles, or photons, as ey are now called, as well as via waves predicted by Maxwell’s electromagnetic theory. The photon theory of light was a radical departure from classical ideas. Einstein proposed a test of the quantum theory of light: quantitative measurements on the photoelectric effect.

SECTION 37-2

989

Ph

oloce

11

@)

14 I FIGURE 37-4 effect.

The photoelectric

When light shines on a metal surface, electrons are found to be emitted from the surface. This effect is called the photoelectric effect and it occurs in many materials, but is most easily observed with metals. It can be observed using the apparatus shown in Fig. 37-4. A metal plate P and a smaller electrode C are placed inside an evacuated glass tube, called a photocell. The two electrodes ar connected to an ammeter and a source of emf, as shown. When the photocell is in the dark, the ammeter reads zero. But when light of sufficiently high frequency illuminates the plate, the ammeter indicates a current flowing in the circuit. We explain completion of the circuit by imagining that electrons, ejected from the plate by the impinging radiation, flow across the tube from the plate to the “collector” C as indicated in Fig. 37-4. That electrons should be emitted when light shines on a metal is consistent . . : o) o with the electromagnetic (EM) wave theory of light: the electric field of an EM wave could exert a force on electrons in the metal and eject some of them. Einstein pointed out, however, that the wave theory and the photon theory of light give very different predictions on the details of the photoelectric effect. For example, one thing that can be measured with the apparatus of Fig. 37-4 is the

maximum kinetic energy (Kp.) of the emitted electrons. This can be done by

using a variable voltage source and reversing the terminals so that electrode C is negative and P is positive. The electrons emitted from P will be repelled by the negative electrode, but if this reverse voltage is small enough, the fastest electrons will still reach C and there will be a current in the circuit. If the reversed voltage is increased, a point is reached where

the current reaches zero—no

electrons have

sufficient kinetic energy to reach C. This is called the stopping potential, or stopping wvoltage, V;, and from its measurement, K, ., can be determined using conservation of energy (loss of kinetic energy = gain in potential energy): Kmax

=

eVO .

Now let us examine the details of the photoelectric effect from the point of view of the wave theory versus Einstein’s particle theory. First the wave theory, assuming monochromatic light. The two important properties of a light wave are its intensity and its frequency (or wavelength). When these two quantities are varied, the wave theory makes the following predictions: Wave theory predictions

1. If the light intensity is increased, the number of electrons ejected and their maximum Kkinetic energy should be increased because the higher intensity means a greater electric field amplitude, and the greater electric field should eject electrons with higher speed. 2. The frequency of the light should not affect the kinetic energy of the ejected electrons. Only the intensity should affect K,y . The photon theory makes completely different predictions. First we note that in a monochromatic beam, all photons have the same energy (= Af). Increasing the intensity of the light beam means increasing the number of photons in the beam, but does not affect the energy of each photon as long as the frequency is not changed. According to Einstein’s theory, an electron is ejected from the metal by a collision with a single photon. In the process, all the photon energy is transferred to the electron and the photon ceases to exist. Since electrons are held in the metal by attractive forces, some minimum energy W, is required just to get an electron out through the surface. W, is called the work function, and is a few electron volts

(1eV = 1.6 X 107°J) for most metals. If the frequency f of the incoming light is

so low that Af is less than W, then the photons will not have enough energy to eject any electrons at all. If Af > W, then electrons will be ejected and energy will be conserved in the process. That is, the input energy (of the photon), Af, will equal the outgoing kinetic energy K of the electron plus the energy required to get it out of the metal, W: -

hf = K + W. 990

CHAPTER 37

The

(37-4a)

least tightly held electrons will be emitted

(Kmax),

in which

case

W in this equation

with the most

becomes

the work

kinetic energy

function

W,

and K becomes K ax: hf

=

Kpax + W.

[least bound electrons]

(37-4b)

Many electrons will require more energy than the bare minimum (W) to get out of the 1etal, and thus the kinetic energy of such electrons will be less than the maximum. From these considerations, the photon theory makes the following predictions: 1. An increase in intensity of the light beam means more photons are incident, so more electrons will be ejected; but since the energy of each photon is not changed, the maximum kinetic energy of electrons is not changed by an increase in intensity. 2. If the frequency of the light is increased, the maximum kinetic energy of the electrons increases linearly, according to Eq. 37-4b. That is, Kmax

=

hf

-

no

These predictions of the photon theory are clearly very different from the predictions of the wave theory. In 1913-1914, careful experiments were carried out by R. A. Millikan. The results were fully in agreement with Einstein’s photon theory. One other aspect of the photoelectric effect also confirmed the photon theory. If extremely low light intensity is used, the wave theory predicts a time delay before electron emission so that an electron can absorb enough energy to exceed the work function. The photon theory predicts no such delay—it only takes one

photon (if its frequency is high enough) to eject an electron—and experiments showed no delay. This too confirmed Einstein’s photon theory. light, A = 450 nm

Photon

predictions

g ;‘é% ) P4

Frequency of light f

=~ FIGURE 37-5 Photoelectric effect: ~ the maximum kinetic energy of €jected electrons increases linearly

Wit the frequency of incident light. Noelectrons ate emtitted if f < fp.

energy. Calculate the energy of a photon of blue

in air (or vacuum).

APPROACH The photon has energy E = hf (Eq.37-3) where SOLUTION Since f = c¢/A, we have

he E=hf=—"-= =3 or

theory

WO'

This relationship is plotted in Fig. 37-5. 3. If the frequency f is less than the “cutoff” frequency f,, where hf, = W;, electrons will be ejected, no matter how great the intensity of the light.

m

Photon

(6.63 X 10737:5)(3.0 X 108m/s) (4.5 x 107 m)

(4.4 x1077J)/(1.60 X 107J/eV) = 2.8eV.

Section 23-8, 1eV = 1.60 x 107].)

(See

f = c¢/A.

= 44 x 10777,

definition

of eV

in

DGVIRREYEL S ESTIMATE | Photons from a lightbulb. Estimate how many

visible light photons a 100-W lightbulb emits per second. Assume the bulb has a typical efficiency of about 3% (that is, 97% of the energy goes to heat). APPROACH Let’s assume an average wavelength in the middle of the visible spectrum, A ~ 500 nm. The energy of each photon is £ = hf = hc/A. Only 3% of the 100-W power is emitted as light,or 3 W = 3J/s. The number of photons emitted per second equals the light output of 3 J/s divided by the energy of each photon. SOLUTION The energy emitted in one second (=3J) is E = Nhf where N is the number of photons emitted per second and f = ¢/A. Hence

N E_EA_ hf hc

(6.63

37)(500 x 10°° G ™ X 10**J-5)(3.0 X 10°m/s)

gx 108

per second, or almost 10" photons emitted per second, an enormous number. rEXERCISE A Compare a light beam that contains infrared light of a single wavelength, 1000 nm, with a beam of monochromatic UV at 100 nm, both of the same intensity. Are there more 100-nm photons or more 1000-nm photons?

SECTION 37-2

991

Photoelectron

speed

and

energy. What is the kinetic

energy and the speed of an electron ejected from a sodium surface whose work function is W, = 2.28eV when illuminated by light of wavelength (a) 410 nm, (b) 550 nm?

APPROACH We first find the energy of the photons (E = Af = hc/A). If the energy is greater than W, then electrons will be ejected with varying amounts of kinetic energy, with a maximum of K., = Af — W,.

SOLUTION (a) For A = 410 nm, hf

=

%C- =

485 X 107°7J

or

3.03eV.

The maximum kinetic energy an electron can have is given by Eq. 37-4b, Kipax = 3.03eV — 228eV = 0.75¢eV, or (0.75eV)(1.60 x 1077J/eV) =

1.2 x 107 J. Since K = jmv® where m = 9.11 X 10 kg, Umax

2K

=

";_n—

=

5.1 X 10°m/s.

Most ejected electrons will have less kinetic energy and less speed than these maximum values.

(b) For

A =550nm, hf = hc/A = 3.61 X 107°J = 2.26eV.

Since this photon

energy is less than the work function, no electrons are ejected. NOTE In (a) we used the nonrelativistic equation for kinetic energy. If v had turned out to be more than about 0.1c, our calculation would have been inaccurate by at least a percent or so, and we would probably prefer to redo it using the relativistic form (Eq. 36-10).

FIGURE 37-6 Optical sound track on movie film. In the projector, light from a small source (different from

that for the picture) passes through the sound track on the moving film. The light and dark areas on the sound track vary the intensity of the transmitted light which reaches the photocell, whose output current is then a replica of the original sound. This output is amplified and sent to the loudspeakers. High-quality projectors can show movies containing several parallel sound tracks to go to different speakers around the theater. Picture

Sound track Photocell

Small

light

source

i

:

&

%

992

QU

ANGE

U

CHAPTER 37

EXERCISE B Determine the lowest frequency and the longest wavelength needed to e electrons from sodium.

It is easy to show that the energy of a photon in electron volts, when given the wavelength A in nm, is E (eV)

=

1.240 x 10° eV-nm A (nm)

[photon energy in eV]

Applications of the Photoelectric Effect The photoelectric effect, besides playing an important historical role in confirming the photon theory of light, also has many practical applications. Burglar alarms and automatic door openers often make use of the photocell circuit of Fig. 37-4. When a person interrupts the beam of light, the sudden drop in current in the circuit activates a switch—often a solenoid—which operates a bell or opens the door. UV or IR light is sometimes used in burglar alarms because of its invisibility. Many smoke detectors use the photoelectric effect to detect tiny amounts of smoke that interrupt the flow of light and so alter the electric current. Photographic light meters use this circuit as well. Photocells are used in many other devices, such as absorption spectrophotometers, to measure light intensity. One type of film sound track is a variably shaded narrow section at the side of the film. Light passing through the film is thus “modulated,” and the output electrical signal of the photocell detector follows the frequencies on the sound track. See Fig. 37-6. For many applications today, the vacuum-tube photocell of Fig. 37-4 has been replaced by a semiconductor device known as a photodiode (Section 40-9). In these semiconductors, the absorption of a photon liberates a bound electro which changes the conductivity of the material, so the current through a photodioc is altered.

Early Quantum Theory and Models of the Atom

37-3

Energy, Mass, and Momentum of a Photon

r 'e have just seen (Eq. by E = hf. Because a relativistic particle. Thus energy and momentum. p =mv/\/1

37-3) that the total energy of a single photon is given photon always travels at the speed of light, it is truly a we must use relativistic formulas for dealing with its The momentum of any particle of mass m is given by

— v*/c*.

Since

K

= hf.

v = ¢

for a photon,

the denominator

is zero. To

avoid having an infinite momentum, we conclude that the photon’s mass must be zero: m = 0. This makes sense too because a photon can never be at rest (it always moves at the speed of light). A photon’s kinetic energy is its total energy: The

=

momentum

E

of a photon

[photon] can

be

obtained

from

the

relativistic

formula

(Eq. 36-13) E? = p*?* + m*c* where we set m = 0, so E* = p** or

E

¢= Since

A CAUTION

[photon]

Momentum of photon is not mv

E = hf for a photon, its momentum is related to its wavelength by P

EXAMPLE

=

E

hf

h

c

¢

A

37-6

(37=5)

Photon momentum and force. Suppose the 10"

photons emitted per second from the 100-W lightbulb in Example 37-4 were all focused onto a piece of black paper and absorbed. () Calculate the momentum of one photon and (b) estimate the force all these photons could exert on the paper. APPROACH Each photon’s momentum is obtained from Eq. 37-5, p = h/A. »Next, each absorbed photon’s momentum changes from p = h/A to zero. We use Newton’s second law, F = Ap/At, to get the force. Let A = 500 nm. SOLUTION (a) Each photon has a momentum 6.63 X 10747 = -h = ————

Py

T

T500x107m

s

= 13x10%kg'm/s.

g-m/s

(b) Using Newton’s second law for N = 10" momentum changes from //A to 0, we obtain

F=

Ap

=

Nh/A-0

Ts

_

= H—

_h

_

photons (Example 37-4) whose P

T

—

~ (10¥s7')(10 7 kg m/s) ~ 107°N.

This is a tiny force, but we can see that a very strong light source could exert a measurable force, and near the Sun or a star the force due to photons in electromagnetic radiation could be considerable. See Section 31-9.

m

Photosynthesis. In photosynthesis, pigments such as chlorophyll

in plants capture the energy of sunlight to change CO, to useful carbohydrate. About nine photons are needed to transform one molecule of CO, to carbohydrate and O,. Assuming light of wavelength A = 670 nm (chlorophyll absorbs most strongly in the range 650nm to 700nm), how efficient is the photosynthetic process? The reverse chemical reaction releases an energy of 4.9 ¢V/molecule of C02 .

APPROACH The efficiency is the minimum energy required (4.9 eV) divided by the actual energy absorbed, nine times the energy (Af) of one photon. SOLUTION The energy of nine photons, each of energy hf = hc/A is

7779)(6.63 x 1047-5)(3.0 X 10°m/s)/(6.7 X 107m) = 2.7 X 107 l_’IEus the process is (4.9eV/17eV)

= 29%

efficient.

SECTION 37-3

or

17eV.

Energy, Mass, and Momentum of a Photon

993

BEFORE COLLISION

AFTER COLLISION ¥4

Scattered photon (4)

Incident photon (1) NAANAN

Electron at rest initially

FIGURE 37-7 The Compton effect. A single photon of wavelength A strikes an electron in some material, knocking it out of its atom. The scattered photon has less energy (some energy is given to the electron) and hence has a longer wavelength A'.

Intensity

FIGURE 37-8 Plots of intensity of radiation scattered from a target such as graphite (carbon), for three different angles. The values for A’ match Eq.37-6. For (a) ¢ = 0°, A" = Ay. In(b) and (c) a peak is found not only at A’ due to photons scattered from free electrons (or very nearly free), but also a peak at almost precisely Ag. The latter is due to scattering from electrons very tightly bound to their atoms so the mass in Eq. 37-6 becomes very large (mass of the atom) and A A becomes very small.

(a)o =0°

37-4

Compton Effect

Besides the photoelectric effect, a number of other experiments were carried out in the early twentieth century which also supported the photon theory. One o#y these was the Compton effect (1923) named after its discoverer, A. H. Comptown (1892-1962). Compton scattered short-wavelength light (actually X-rays) from various materials. He found that the scattered light had a slightly longer wavelength than did the incident light, and therefore a slightly lower frequency indicating a loss of energy. He explained this result on the basis of the photon theory as incident photons colliding with electrons of the material, Fig. 37-7. Using Eq. 37-5 for momentum of a photon, Compton applied the laws of conservation of momentum and energy to the collision of Fig. 37-7 and derived the following equation for the wavelength of the scattered photons: mec

(1 = cos¢),

(37-6a)

where m, is the mass of the electron. For ¢ = 0, the wavelength is unchanged (there is no collision for this case of the photon passing straight through). At any other angle, A’ is longer than A. The difference in wavelength, AL

=

A

— A

h mec

=

(37-6b)

(1 = cos ),

is called the Compton shift. The quantity #/m.c, which has the dimensions of length, is called the Compton wavelength A of a free electron, Ac

=

h g

= 243 243 xX 10 1073 nm

= 2.43 2 pm :

[electron]

Equations 37-6 predict that A’ depends on the angle ¢ at which the photons are detected. Compton’s measurements of 1923 were consistent with thi# formula, confirming the value of A- and the dependence of A’ on ¢. See Fig. 37-8. The wave theory of light predicts no wavelength shift: an incoming electromagnetic wave of frequency f should set electrons into oscillation at the same frequency f, and such oscillating electrons should reemit EM waves of this same frequency f (Chapter 31), and would not change with the angle ¢. Hence the Compton effect adds to the firm experimental foundation for the photon theory of light. EXERCISE C When a photon scatters off an electron by the Compton effect, which of the following increase: its energy, frequency, or wavelength?

Intensity

(b)o =45°

DUyl X-ray scattering. X-rays of wavelength 0.140 nm are scattered from a very thin slice of carbon. What will be the wavelengths of X-rays scattered at (a) 0°, (b) 90°, (c) 180°? APPROACH This is an example of the Compton effect, and we use Eq. 37-6a to find the wavelengths. SOLUTION

(a) For

¢ = 0°,cos¢p =1

and

1 — cos¢ = 0. Then Eq. 37-6 gives

A = A = 0.140nm. This makes sense since for ¢ = 0°, there really isn’t any collision as the photon goes straight through without interacting. (b) For ¢ = 90° cos¢p =0, and 1 — cos¢ = 1. So Intensity

6.63 X 10°#]s A = A+ mec ——h = 0140nm T + 911 x 10 kg)(3.00 X 10°m/s) =

0.140nm

+ 2.4 X 107?m

=

0.142 nm;

that is, the wavelength is longer by one Compton wavelength ( = 0.0024 nm

an electron).

994

CHAPTER 37

fo:

(c) For

¢ = 180°,

which means

the photon is scattered backward, returning in

the direction from which it came (a direct “head-on” collision), cos ¢ = —1, and 1 —-cos¢p =2. So

fk:

A= A+2 mh

= 0.140nm + 2(0.0024nm) = 0.145nm.

(3

NOTE The maximum shift in wavelength occurs for backward scattering, and it is twice the Compton wavelength. The Compton effect has been used to diagnose bone disease such as osteoporosis. Gamma rays, which are photons of even shorter wavelength than X-rays, coming from a radioactive source are scattered off bone material. The total intensity of the scattered radiation is proportional to the density of electrons, which is in turn proportional to the bone density. Changes in the density of bone can indicate the onset of osteoporosis.

@) PHYSICS

APPLIED

Measuring bone density

*Derivation of Compton Shift If the incoming photon in Fig. 37-7 has wavelength A, then its total energy and momentum

are

E=hf=%c-

and

p

==

After the collision of Fig. 37-7, the photon scattered at the angle ¢ has a wavelength which we call A’. Its energy and momentum are E'

fi

=

L

and

A

The electron, assumed

p

=

%

at rest before the collision but free to move when

struck,

scattered at an angle 6 as shown in Fig. 37-8. The electron’s kinetic energy is see Eq. 36-10):

K.

=

(—1 - 1>mec2 1 — v?/c?

where m, is the mass of the electron and v is its speed. The electron’s momentum is P

e

T

/5 T !

m

1 — v¥/c?

M.

We apply conservation of energy to the collision (see Fig. 37-7): incoming photon —

g5

scattered photon

o e

FUY

+ electron

me & 1 — v*/c?

We apply conservation of momentum to the x and y components of momentum: h =

=

h =08

0

=

fi,sincfi—

A

.

A

A

-

mgv cos 6

1 - v¥/c? mg v sinin6

.

1 — v¥/c?

We can combine these three equations to eliminate » and 6, and we obtain, as Compton did, an equation for the wavelength of the scattered photon in terms of its scattering angle ¢:

No= A+ which is Eq. 37-6a.

h

mecC

(1 = cos ), SECTION 37-4

Compton Effect

995

37-5

Photon Interactions; Pair Production

When a photon passes through matter, it interacts with the atoms and electrons. There are four important types of interactions that a photon can undergo: -~

et Photon

® Nucleus

o FIGURE 37-9 Pair production: a photon disappears and produces an electron and a

¥

positron.

1. The photoelectric effect: A photon may knock an electron out of an atom and in the process the photon disappears. 2. The photon may knock an atomic electron to a higher energy state in the atom if its energy is not sufficient to knock the electron out altogether. In this process the photon also disappears, and all its energy is given to the atom. Such an atom is then said to be in an excited state, and we shall discuss it more later. 3. The photon can be scattered from an electron (or a nucleus) and in the process lose some energy; this is the Compton effect (Fig. 37-7). But notice that the photon is not slowed i down. It still travels with speed c, but its N frequency will be lower because it has lost some energy. 4. Pair production: A photon can actually create matter, such as the production of an electron and a positron, Fig. 37-9. (A positron has the same mass as an electron, but the opposite charge, +e.) .

.

;

;

;

In process 4, pair production, the photon disappears in the process of creating the electron—positron pair. This is an example of mass being created from pure energy, and it occurs in accord with Einstein’s equation E = mc® Notice that a photon cannot create an electron alone since electric charge would not then be conserved. The inverse of pair production also occurs: if an electron collides with a positron, the

two

annihilate

each

other

and

their

energy, including

their mass,

appears as electromagnetic energy of photons. Because of this process, positrons usually do not last long in nature.

DGR YL Pair production. (¢) What is the minimum energy of a photon that can produce an electron-positron pair? (b) What is this photon’s wavelengtham, APPROACH The minimum photon energy E equals the rest energy two particles created, via Einstein’s famous equation E = mc® There is no energy left over, so the particles produced will have energy. The wavelength is A = ¢/f where E = hf for the original SOLUTION (a) Because E = mc? and the mass created is equal to masses, the photon must have energy

E = 2(911 x 103 kg)(3.0 X 108 m/s)* = 1.64 x 1073] (1MeV

= 10°eV = 1.60 X 10°J).

A

photon

with

less

= 1.02MeV energy

undergo pair production. (b) Since E = hf = hc/A, the wavelength of a 1.02-MeV photon is A=

L

E

(

6.63 X 107*J-5)(3.0 X 10°m/s X —2

(1.64 x 10717)

/s)

=

(mc?) of the (Eq. 36-12). zero kinetic photon. two electron

cannot

1.2 X 107% m,

which is 0.0012 nm. Such photons are in the gamma-ray (or very short X-ray) region of the electromagnetic spectrum (Fig. 31-12). NOTE Photons of higher energy (shorter wavelength) can also create an electron— positron pair, with the excess energy becoming kinetic energy of the particles. Pair production cannot occur in empty space, for momentum could not be conserved. In Example 37-9, for instance, energy is conserved, but only enough energy was provided to create the electron—positron pair at rest, and thus with zero momentum, which could not equal the initial momentum of the photon. Indeed, it can be shown that at any energy, an additional massive object, such as an atomic nucleus\ must take part in the interaction to carry off some of the momentum.

996

CHAPTER 37

Early Quantum Theory and Models of the Atom

37-6 Wave-Particle Duality; the Principle of Complementarity r The photoelectric effect, the Compton effect, and other experiments have placed the particle theory of light on a firm experimental basis. But what about the classic experiments of Young and others (Chapters 34 and 35) on interference and diffraction which showed that the wave theory of light also rests on a firm experimental basis? We seem to be in a dilemma. Some experiments indicate that light behaves like a wave; others indicate that it behaves like a stream of particles. These two theories seem to be incompatible, but both have been shown to have validity. Physicists finally came to the conclusion that this duality of light must be accepted as a fact of life. It is referred to as the wave-particle duality. Apparently, light is a more complex phenomenon than just a simple wave or a simple beam of particles. To clarify the situation, the great Danish physicist Niels Bohr (1885-1962, Fig. 37-10) proposed his famous principle of complementarity. It states that to understand an experiment, sometimes we find an explanation using wave theory

and sometimes using particle theory. Yet we must be aware of both the wave and

FIGURE 37-10

Niels Bohr (right),

particle aspects of light if we are to have a full understanding of light. Therefore these two aspects of light complement one another. It is not easy to “visualize” this duality. We cannot readily picture a combination

= walking with Enrico Fermi along the Appian Way outside Rome. This ~ photo shows one important way

different “faces” that light shows to experimenters. Part of the difficulty stems from how we think. Visual pictures (or models) in our minds are based on what we see in the everyday world. We apply the concepts of waves and particles to light because in the macroscopic world we see that energy is transferred from place to place by these two methods. We cannot see directly whether light is a wave or particle, so we do indirect experiments. To explain the experiments, we apply the models of waves or of particles to the nature of light. But these dre abstractions of the human mind. When we try to conceive of what light really “is,” we insist on a visual picture. Yet there is no reason why light should conform to these models (or visual images) taken from the macroscopic world. The “true” nature of

A CAUTION Not correct to say light is a wave and/or

of wave and particle. Instead, we must recognize that the two aspects of light are ~ physics is done.

light—if that means

anything—is

not possible to visualize. The best we can do is

recognize that our knowledge is limited to the indirect experiments, and that in terms of everyday language and images, light reveals both wave and particle properties. It is worth noting that Einstein’s equation E = hf itself links the particle and wave properties of a light beam. In this equation, E refers to the energy of a particle; and on the other side of the equation, we have the frequency f of the corresponding wave.

37-7

a particle. Light can act like a wave or

like a particle.

Wave Nature of Matter

In 1923, Louis de Broglie (1892-1987) extended the idea of the wave-particle duality. He appreciated the symmetry in nature, and argued that if light sometimes behaves like a wave and sometimes like a particle, then perhaps those things in nature thought to be particles—such as electrons and other material objects—might also have wave properties. De Broglie proposed that the wavelength of a material particle would be related to its momentum in the same way as for a photon, Eq. 37-5, p = h/A. That is, for a particle having linear momentum p = mw, the wavelength A is given by A

h ;-

37-7)

nd is valid classically (p = mv for v =100eV/(0.511x10%eV) ~ 10°%,

so relativity is not needed. Thus 1 Emv2

and

v=,/e—=\/() m A=

| or 0.12nm. Diffraction pattern

of electrons scattered from aluminum foil, as recorded on film.

9

eV 2eV

Then

FIGURE 37-11

=

.

mv

(

2)(1.6

X 1077

(9.1 x 6.63 X 107" ( =

(9.1 X 107"

50)

C)(100 V

73]( 10 kg) ad

J-s

)5

kg)(5.9 % 10°m/s)

) = 59 % 10°m/s. =

12X

107 %m,

EXERCISE D As a particle travels faster, does its de Broglie wavelength decrease, increase, or remain the same? From

Example

37-11, we see that electrons can have wavelengths

order of 107'°m, and even smaller. Although

on the

small, this wavelength can be

detected: the spacing of atoms in a crystal is on the order of 10"°m and the orderly array of atoms in a crystal could be used as a type of diffraction grating, as was done earlier for X-rays (see Section 35-10). C. J. Davisson and L. H. Germer

performed the crucial experiment; they scattered electrons from the surface of a metal crystal and, in early 1927, observed that the electrons were scattered into a pattern of regular peaks. When they interpreted these peaks as a diffraction pattern, the wavelength of the diffracted electron wave was found to be just that

predicted by de Broglie, Eq. 37-7. In the same year, G. P. Thomson (son of J. J. Thomson) used a different experimental arrangement and also detected diffraction of electrons. (See Fig. 37-11. Compare it to X-ray diffraction, Section 35-10. Later experiments showed that protons, neutrons, and other particles also have wave properties.

998

CHAPTER 37

Early Quantum Theory and Models of the Atom

Thus the wave-particle duality applies to material objects as well as to light. The principle of complementarity applies to matter as well. That is, we must be aware of both the particle and wave aspects in order to have an understanding of filatter, including electrons. But again we must recognize that a visual picture of a wave—particle” is not possible.

the surface of a solid. By studying the angular distribution of the diffracted electrons, one can indirectly measure the geometrical arrangement of atoms. Assume that the electrons strike perpendicular to the surface of a solid (see Fig. 37-12), and that their energy is low, K = 100eV, so that they interact only with the surface layer of atoms. If the smallest angle at which a diffraction maximum occurs is at 24°, what is the separation d between the atoms on the surface? SOLUTION Treating the electrons as waves, we need to determine the condition where the difference in path traveled by the wave diffracted from adjacent atoms is an integer multiple of the de Broglie wavelength, so that constructive interference occurs. The path length difference is d sin 6; so for the smallest value of 6 we must have

dsinf = A

L

-

APPLIED

i

"

FIGURE 37-12

¢

Example 37-12.

ety

$ia Agid

__F

2me,

Thus

Electron diffraction

The red dots represent atoms in an

However, A is related to the (non-relativistic) kinetic energy K by

_r

(K)PHYSICS

N

diffraction. The wave nature of electrons is

weaq o1 JuapoOUl

Electron

manifested in experiments where an electron beam interacts with the atoms on

2m \?

h (6.63 X 10775) = = 0.123 nm. V2m. K \/2(9.11 X 10 kg)(100eV)(1.6 x 107°J/eV)

The surface inter-atomic spacing is @

=

A sin

=

0.123 nm sin 24°

=

0.30

fm

nm.

EXERCISE E Return to the Chapter-Opening Question, page 987, and answer it again now. Try to explain why you may have answered differently the first time.

What Is an Electron?

*

We might ask ourselves: “What is an electron?” The early experiments of J. J. Thomson (Section 27-7) indicated a glow in a tube, and that glow moved when a magnetic field was applied. The results of these and other experiments were best interpreted as being caused by tiny negatively charged particles which we now call electrons. No one, however, has actually seen an electron directly. The drawings we sometimes make of electrons as tiny spheres with a negative charge on them are merely convenient pictures (now recognized to be inaccurate). Again we must rely on experimental results, some of which are best interpreted using the particle model and others using the wave model. These models are mere pictures that we use to extrapolate from the macroscopic world to the tiny microscopic world of the atom. And there is no reason to expect that these models somehow reflect the reality of an electron. We thus use a wave or a particle model (whichever works best in a situation) so that we can talk about what is happening, But we should not be led to believe that an electron is a wave or a particle. nstead we could say that an electron is the set of its properties that we can .neasure. Bertrand Russell said it well when he wrote that an electron is “a logical

construction.”

SECTION 37-7

Wave Nature of Matter

999

*37-8

Hot filament (source of electrons)

Projection

|

“lens” (eyepiece)

|

3

Image (on screen, film, or semiconductor detector)

FIGURE 37-13 Transmission electron microscope. The magnetic field coils are designed to be “magnetic lenses,” which bend the electron paths and bring them to a focus, as shown.

@

pHYsics

APPLIED

Electron microscope

Electron Microscopes

The idea that electrons have wave properties led to the development of the electron microscope, which can produce images of much greater magnificationm than does a light microscope. Figures 37-13 and 37-14 are diagrams of two types, developed around the middle of the twentieth century: the transmission electron microscope, which produces a two-dimensional image, and the scanning electron microscope (SEM), which produces images with a three-dimensional quality. In both types, the objective and eyepiece lenses are actually magnetic fields that exert forces on the electrons to bring them to a focus. The fields are produced by carefully designed current-carrying coils of wire. Photographs using each type are shown in Fig. 37-15. As discussed in Section 35-5, the maximum resolution of details on an object is about the size of the wavelength of the radiation used to view it. Electrons accelerated by voltages on the order of 10°V have wavelengths of about 0.004 nm. The maximum resolution obtainable would be on this order, but in practice, aberrations in the magnetic lenses limit the resolution in transmission electron microscopes to at best about 0.1to 0.5nm. This is still 10° times better than that attainable with a visible-light microscope, and corresponds to a useful magnification of about a million. Such magnifications are difficult to attain, and more common magnifications are 10* to 10° The maximum resolution attainable with a scanning electron microscope is somewhat less, typically 5 to 10nm although new high-resolution SEMs approach 1 nm. We discuss other sophisticated electron microscopes in the next Chapter, Section 38-10. FIGURE 37-15 Electron micrographs (in false color) of viruses attacking a cell of the bacterium Escherichia coli: (a) transmission electron micrograph (~ 50,000); (b) scanning electron micrograph (~ 35,000%).

FIGURE 37-14 Scanning electron microscope. Scanning coils move an electron beam back and forth across the specimen. Secondary electrons produced when the beam strikes the specimen are collected and modulate the intensity of the beam in the CRT to produce a picture.

vrv;

-

Electron source

Magnetic lens

(a)

Central electronics

37-9

Sweep

Scanning|l~ h f coils ™

1]

N

>

Specimen

1000

Electron collector

Secondary

electrons

CHAPTER 37

C

i Grid

(b)

Early Models of the Atom

The idea that matter is made up of atoms was accepted by most scientists by 1900. With the discovery of the electron in the 1890s, scientists began to think of the atom itself as having a structure with electrons as part of that structure. We now introduce our modern approach to the atom and the quantum theory with which it

is intertwined.”

‘Some readers may say: “Tell us the facts as we know them today, and don’t bother us with the historical background and its outmoded theories.” Such an approach would ignore the creative aspect of science and thus give a false impression of how science develops. Moreover, it is not really possible to understand today’s view of the atom without insight into the concepts that led to it.

Early Quantum Theory and Models of the Atom

-

A typical model of the atom in the 1890s visualized the atom as a homogeneous sphere of positive charge inside of which there were tiny negatively charged electrons, a little like plums in a pudding, Fig. 37-16. Around 1911, Ernest Rutherford (1871-1937) and his colleagues performed xperiments whose results contradicted the plum-pudding model of the atom. In these experiments a beam of positively charged alpha (a) particles was directed at a thin sheet of metal foil such as gold, Fig. 37-17a. (These newly discovered a particles were emitted by certain radioactive materials and were soon shown to be doubly ionized helium atoms—that is, having a charge of +2e.) It was expected from the plum-pudding model that the alpha particles would not be deflected significantly because electrons are so much lighter than alpha particles, and the alpha particles should not have encountered any massive concentration of positive charge to strongly repel them. The experimental results completely contradicted these predictions. It was found that most of the alpha particles passed through the foil unaffected, as if the foil were mostly empty space. And of those deflected, a few were deflected at very large angles—some even backward, nearly in the direction from which they had come. This could happen, Rutherford reasoned, only if the positively charged alpha particles were being repelled by a massive positive charge concentrated in a very small region of space (see Fig. 37-17b).

Positively

charged material FIGURE 37-16

Plum-pudding

model of the atom.

Viewing screen

Source

FIGURE 37-17 (a) Experimental setup for Rutherford’s experiment: « particles emitted by radon are deflected by a thin

containing radon

metallic foil and a few rebound backward;

r

(a)

X

’

\B

a;micle

(b) backward rebound of « particles explained as the repulsion from a heavy positively charged nucleus.

-

. Nucleus

(b)

He hypothesized that the atom must consist of a tiny but massive positively charged nucleus, containing over 99.9% of the mass of the atom, surrounded by electrons some distance away. The electrons would be moving in orbits about the nucleus—much as the planets move around the Sun—because if they were at rest, they would fall into the nucleus due to electrical attraction, Fig. 37-18. Rutherford’s experiments suggested that the nucleus must have a radius of about

10

to 107“m.

From

kinetic

theory, and

especially

Einstein’s

analysis

of

FIGURE 37-18

Rutherford’s

model of the atom, in which

electrons orbit a tiny positive nucleus (not to scale). The atom is visualized as mostly empty space.

Brownian motion (see Section 17-1), the radius of atoms was estimated to be about 107 m. Thus the electrons would seem to be at a distance from the nucleus of about 10,000 to 100,000 times the radius of the nucleus itself. (If the nucleus were the size of a baseball, the atom would have the diameter of a big city several kilometers across.) So an atom would be mostly empty space. Rutherford’s “planetary” model of the atom (also called the “nuclear model of the atom”) was a major step toward how we view the atom today. It was not, however, a complete model and presented some major problems, as we shall see.

37-10

Atomic Spectra: Key to the Structure of the Atom

’.aarlier in this Chapter we saw that heated solids (as well as liquids and dense gases) emit light with a continuous spectrum of wavelengths. This radiation is assumed to be due to oscillations of atoms and molecules, which are largely governed by the interaction of each atom or molecule with its neighbors.

SECTION 37-10

1001

W e /_[\—

4_9(.',,

4

&

High T voltage

4 =

- |Cathode

(a)

Rarefied gases can also be excited to emit light. This is done by intense heating, or more commonly by applying a high voltage to a “discharge tube” containing the gas at low pressure, Fig. 37-19. The radiation from excited gases had been observed early in the nineteenth century, and it was found that the spectrum was not continuous, but discrete. Since excited gases emit light of only certain wavelengths, when th™ light is analyzed through the slit of a spectroscope or spectrometer, a line spectrum is seen rather than a continuous spectrum. The line spectra in the visible region emitted by a number of elements are shown below in Fig. 37-20, and in Chapter 35, Fig. 35-22. The emission spectrum is characteristic of the material and can serve as a type of “fingerprint” for identification of the gas. We also saw (Chapter 35) that if a continuous spectrum passes through a rarefied gas, dark lines are observed in the emerging spectrum, at wavelengths corresponding to lines normally emitted by the gas. This is called an absorption spectrum (Fig. 37-20c), and it became clear that gases can absorb light at the same frequencies at which they emit. Using film sensitive to ultraviolet and to infrared light, it was found that gases emit and absorb discrete frequencies in these regions as well as in the visible.

(®) FIGURE 37-19 Gas-discharge tube: (a) diagram; (b) photo of an actual discharge tube for hydrogen. FIGURE 37-21

Balmer series of

lines for hydrogen. 365 Em— =

em—

}UV

410 =———Violet 434

Blue

FIGURE 37-20 Emission spectra of the gases (a) atomic hydrogen, (b) helium, and (c) the solar absorption spectrum. In low-density gases, the atoms

are far apart on the average

e

green

the

Hydrogen is the simplest atom—it has only one electron. It also has the

simplest spectrum. The spectrum of most atoms shows little apparent regularity. But the spacing between lines in the hydrogen spectrum decreases in a regular (1825-1898)

showed

that the four

lines in the visible portion of the hydrogen spectrum (with measured wavelengths

656 nm, 486 nm, 434 nm, and 410 nm) have wavelengths that fit the formula 1

X

(fim)

and hence

light emitted or absorbed is assumed to be by individual atoms rather than through interactions between atoms, as in a solid, liquid, or dense gas. Thus the line spectra serve as a key to the structure of the atom: any theory of atomic structure must be able to explain why atoms emit light only of discrete wavelengths, and it should be able to predict what these wavelengths are.

way, Fig. 37-21. Indeed, in 1885, J. J. Balmer

-

-

=

1

R(p

-

1

?>’

n

=

3,4,“'.

(37-—8)

Here n takes on the values 3, 4, 5, 6 for the four visible lines, and R, called the

Rydberg constant, has the value

R = 1.0974 X 10’m™".

Later it was found that

this Balmer series of lines extended into the UV region, ending at A = 365 nm,

as

shown in Fig. 37-21. Balmer’s formula, Eq. 37-8, also worked for these lines with

656 ———— 1002

CHAPTER 37

Red

higher integer values of n. The lines near 365 nm became too close together to distinguish, but the limit of the series at 365nm corresponds to n = oo (so 1/n* = 0 in Eq. 37-8). Later experiments on hydrogen showed that there were similar series of lines in the UV and IR regions, and each series had a pattern just like the Balmer seriesem, but at different wavelengths, Fig. 37-22. Each of these series was found to fit « formula with the same form as Eq. 37-8 but with the 1/2% replaced by 1/12,1/3% 1/4%, and so on. For example, the so-called Lyman series contains lines

-

SN

QN

£

=

E

(=]

=

\n

o

O o

—

Wavelength, A

B

=

v

al o0

o0 —

=)

Na) \O

~

FIGURE 37-22 Line spectrum of atomic hydrogen. Each series fits the 1 1 1 formulax n' =1

=

R

)

- ?)where

for the Lyman series,

n' = 2 for the Balmer series, v

W

Lyman series

J

-

~—

Balmer series =

Yy

uv

Paschen series S

-

~N"

Visible light

IR

n' = 3 for the Paschen series, and so on; n can take on all integer

values from n=n"+ 1 upto infinity. The only lines in the visible region of the electromagnetic spectrum are part of the Balmer series.

with wavelengths from 91 nm to 122 nm (in the UV region) and fits the formula 1

1

1

The wavelengths of the Paschen series (in the IR region) fit 1 A

1 R(32

1 B

nz)’

n o= 45

The Rutherford model was unable to explain why atoms emit line spectra. It

had other difficulties as well. According to the Rutherford model, electrons orbit

the nucleus, and since their paths are curved the electrons are accelerating. Hence they should give off light like any other accelerating electric charge (Chapter 31), with a frequency equal to its orbital frequency. Since light carries off energy and energy is conserved, the electron’s own energy must decrease to compensate. Hence electrons would be expected to spiral into the nucleus. As they spiraled inward, their frequency would increase in a short time and so too would the frequency of the light emitted. Thus the two main difficulties with the Rutherford odel are these: (1) it predicts that light of a continuous range of frequencies will fil: emitted, whereas experiment shows line spectra; (2) it predicts that atoms are unstable—electrons would quickly spiral into the nucleus—but we know that atoms in general are stable, because there is stable matter all around us. Clearly Rutherford’s model was not sufficient. Some sort of modification was needed, and Niels Bohr provided it in a model that included the quantum hypothesis. Although the Bohr model has been superceded, it did provide a crucial stepping stone to our present understanding. And some aspects of the Bohr model are still useful today, so we examine it in detail in the next Section.

37-11

The Bohr Model

Bohr had studied in Rutherford’s laboratory for several months in 1912 and was convinced that Rutherford’s planetary model of the atom had validity. But in order to make it work, he felt that the newly developing quantum theory would somehow have to be incorporated in it. The work of Planck and FEinstein had shown that in heated solids, the energy of oscillating electric charges must change discontinuously— from one discrete energy state to another, with the emission of a quantum of light. Perhaps, Bohr argued, the electrons in an atom also cannot lose energy continuously, but must do so in quantum “jumps.” In working out his model during the next year, Bohr postulated that electrons move about the nucleus in circular orbits, but that only

certain orbits are allowed. He further postulated that an electron in each orbit would have a definite energy and would move in the orbit without radiating energy (even though this violated classical ideas since accelerating electric charges are supposed to emit EM waves; see Chapter 31). He thus called the possible orbits stationary states. Light is emitted, he hypothesized, only when an electron jumps from a higher (upper) ftationary state to another of lower energy, Fig. 37-23. When such a transition occurs, single photon of light is emitted whose energy, by energy conservation, is given by hf = Ey — E., (37-9) where Ej, refers to the energy of the upper state and E, the energy of the lower state.

FIGURE 37-23 An atom emits a photon (energy = hf) whenits

energy changes from Ey; to a lower

energy E; .

Ey

E

SECTION 37-11

1003

In 1912-13, Bohr set out to determine what energies these orbits would have in the simplest atom, hydrogen; the spectrum of light emitted could then be predicted from Eq. 37-9. In the Balmer formula he had the key he was looking for. Bohr quickly found that his theory would be in accord with the Balmer formula if he assumed that the electron’s angular momentum L is quantized an: equal to an integer n times h/27. As we saw in Chapter 11 angular momentum is given by L = Jw, where [ is the moment of inertia and w is the angular velocity. For a single particle of mass m moving in a circle of radius r with speed v, I = mr?

and o = v/r; hence, L = Iw = (mr?)(v/r) = mvr. Bohr’s quantum condition is L

FIGURE 37-24

mor,

=

ni’

n=123-::",

21

(37-10)

where n is an integer and r, is the radius of the n™ possible orbit. The allowed

-

P

=

F= G

1

Ze)e) 2

Electric force

(Coulomb’s 1aw) keeps the negative electron in orbit-around the

positively charged nucleus.

orbits are numbered 1,

2,3,-:+, according to the value of n, which is called the

principal quantum number of the orbit. Equation 37-10 did not have a firm theoretical foundation. Bohr had searched for some “quantum condition,” and such tries as E = hf (where E represents the energy of the electron in an orbit) did not give results in accord with experiment. Bohr’s reason for using Eq. 37-10 was simply that it worked; and we now look at how. In particular, let us determine what the Bohr theory predicts for the measurable wavelengths of emitted light. An electron in a circular orbit of radius r, (Fig. 37-24) would have a centripetal acceleration v?/r, produced by the electrical force of attraction

between the negative electron and the positive nucleus. This force is given by Coulomb’s law,

F

=

1 (Ze)(e) )

4mey

1y

The charge on the nucleus is +Ze, where Z is the number of positive charges’ (i.e., protons). For the hydrogen atom, Z = +1. In Newton’s second law, F = ma, we substitute Coulomb’s law for F, anc. a = v*/r, for a particular allowed orbit of radius r,, and obtain 1 dmeq

F

=

ma

Zé =

_=

mv? .

ry,

r;

We solve this for r,, and v = nh/2mmr,): Iy, =

then substitute

Ze?

Ze*4m*mr?,

dmregmv?

daregn*h?

for » from

Eq. 37-10

(which

says

We solve for r, (it appears on both sides, so we cancel one of them) and find n

=

n2h2€0

mmZe?

=

nz

Z ¢

(

37-11

)

where h2€ 0 rl

=

mTme

2

.

Equation 37-11 gives the radii of all possible orbits. The smallest orbit is for n = 1, and for hydrogen (Z = 1) has the value

(1)X(6.626 x 1077 -5)%(8.85 X 1072C%/N-m?) (3.14)(9.11 x 10" kg)(1.602 x 107 C)’

1004

CHAPTER 37

*We include as the ions electron is charge +2e.

Z in our derivation so that we can treat other single-electron (“hydrogenlike) atoms suc..\ He' (Z =2) and Li**(Z = 3). Helium in the neutral state has two electrons: if one missing, the remaining He™ ion consists of one electron revolving around a nucleus of Similarly, doubly ionized lithium, Li?*, also has a single electron, and in this case Z = 3.

or

r,

=

0529 X 107 m.

37-12)

The radius of the smallest orbit in hydrogen, r;, is sometimes called the Bohr fadius. From Eq. 37-11, we see that the radii of the larger orbits’ increase as n?, so rn,

=

r3 Ty

=

4"1

=

2.12

9,

=

4.76 X 10 %m,

X

10—10m,

nzrl.

The first four orbits are shown in Fig. 37-25. Notice that, according to Bohr’s model, an electron can exist only in the orbits given by Eq. 37-11. There are no allowable orbits in between. For an atom with Z # 1, we can write the orbital radii, r,, using Eq. 37-11:

e

n*

'n = f(0~529 x 107

m),

n =123, ".

(37-13)

FIGURE 37-25

The four smallest

orbits in the Bohr model of

nhydrogen; r; = 0.529 X 107°m,

In each of its possible orbits, the electron would have a definite energy, as the following calculation shows. The total energy equals the sum of the kinetic and potential energies. The potential energy of the electron is given by U = qV = —eV, where V is the potential due to a point charge +Ze as given by Eq.23-5: V = (1/4me,)(Q/r) = (1/4me)(Ze/r). So U=

—-eV

=

1 ze

-

e,

1

The total energy E, for an electron in the n'™ orbit of radius r, is the sum of the kinetic and potential energies: Ex

=

1,2

_

2my

1

Zé

4irey

1y

hen we substitute » from Eq. 37-10 and r,, from Eq. 37-11 into this equation, we tain Z%*'m\ [ 1

‘(W)(?)

b=

n o= 123"

(37-14n)

If we evaluate the constant term in Eq. 37-14a and convert it to electron volts, as is customary in atomic physics, we obtain 2 E, = —(136eV) z = n

The lowest energy level

E,

=

(n = 1) for hydrogen

n

=

1,2,8;,

(Z =1)

(37-14b)

is

—13.6eV.

Since n? appears in the denominator of Eq. 37-14b, the energies of the larger orbits in hydrogen (Z = 1) are given by

E, For example,

—13.6eV = —— n

—13.6eV

E, = —1—346—6 = —3.40eV, E, = ——13‘9fl = —15leV. We see that not only are the orbit radii quantized, but from Egs. 37-14 so is the energy. The quantum number # that labels the orbit radii also labels the energy levels. The lowest energy level or energy state has energy E,, and is called the ound state. The higher states, E,, E;, and so on, are called excited states. The ied energy levels are also called stationary states. "Be careful not to believe that these well-defined orbits actually exist. Today electrons are better thought of as forming “clouds,” as discussed in Chapter 39.

SECTION 37-11

1005

Notice that although the energy for the larger orbits has a smaller numerical value, all the energies are less than zero. Thus, —3.4¢eV is a higher energy than

—13.6eV. Hence the orbit closest to the nucleus (r) has the lowest energy. The

reason the energies have negative values has to do with the way we defined the

zero for potential energy (U). For two point charges,

U = (1/4me,)(q; ./ ™

corresponds to zero potential energy when the two charges are infinitely far apart. Thus, an electron that can just barely be free from the atom by reaching r = (or, at least, far from the nucleus) with zero kinetic energy will have E = 0, corresponding to n = oo in Eqgs. 37-14. If an electron is free and has kinetic energy, then £ > 0. To remove an electron that is part of an atom requires an energy input (otherwise atoms would not be stable). Since E =0 for a free electron, then an electron bound to an atom needs to have E < 0. That is, energy must be added to bring its energy up, from a negative value, to at least zero in order to free it. The minimum energy required to remove an electron from an atom initially in the ground state is called the binding energy or ionization energy. The ionization energy for hydrogen has been measured to be 13.6eV, and this corresponds precisely to removing an electron from the lowest state, E; = —13.6eV, up to E = 0 where it can be free.

Spectra Lines Explained It is useful to show the various possible energy values as horizontal lines on an energy-level diagram. This is shown for hydrogen in Fig. 37-26." The electron in a hydrogen atom can be in any one of these levels according to Bohr’s theory. But it could never be in between, say at —9.0eV. At room temperature, nearly all H atoms will be in the ground state (n = 1). At higher temperatures, or during an electric discharge when there are many collisions between free electrons and atoms, many atoms can be in excited states (n > 1). Once in an excited state, an atom’s electron can jump down to a lower state, and give off a photon in the process. This is, according to the Bohr model, the origin of the emission spectra «# excited gases. "Note that above E =0, an electron is free and can have any energy (E is not quantized). Thus there is a continuum of energy states above E = 0, as indicated in the energy-level diagram of

Fig. 37-26. FIGURE 37-26 Energy-level diagram for the hydrogen atom, showing the transitions for the spectral lines of the Lyman, Balmer, and Paschen series (Fig. 37-22).

Each vertical arrow represents an atomic transition that gives rise to the

+iam b E=0 - o _085 3 _'1 5 Tn=3

photons of one spectral line (a single

-34

wavelength or frequency).

=L

-5

&

+ L T

REE

onized atom (continuous energ___y levels) = = Yy

S—~— P:::i}é‘;"

—

&g

Excited

states

Balmer

4+

series

il 4

10

4+

136 Tn=1y ~13.6 | -5

CHAPTER 37

SRR

e

el

1006

T

T

Ground state Lyman

series

Early Quantum Theory and Models of the Atom

fl

The vertical arrows in Fig. 37-26 represent the transitions or jumps that correspond to the various observed spectral lines. For example, an electron jumping from the level n = 3 to n = 2 would give rise to the 656-nm line in the Balmer series, and the jump from n = 4 to n = 2 would give rise to the 486-nm f*ne (see Fig. 37-21). We can predict wavelengths of the spectral lines emitted by combining Eq. 37-9 with Eq. 37-14a. Since hf = hc/A, we have from Eq. 37-9

1

hf

A

1

he

E(E"

= Ey)

where n refers to the upper state and n’ to the lower state. Then using Eq. 37-14a, Z%" -

T ‘

Zem

1 '

(37-15)

This theoretical formula has the same form as the experimental Balmer formula, Eq. 37-8, with n’ = 2. Thus we see that the Balmer series of lines corresponds to transitions or “jumps” that bring the electron down to the second energy level. Similarly, n' =1 corresponds to the Lyman series and n’ = 3 to the Paschen series (see Fig. 37-26). When the constant in Eq. 37-15 is evaluated with Z = 1, it is found to have the measured value of the Rydberg constant, R = 1.0974 X 10’m™! in Eq. 37-8, in accord with experiment (see Problem 58). The great success of Bohr’s model is that it gives an explanation for why atoms emit line spectra, and accurately predicts the wavelengths of emitted light for hydrogen. The Bohr model also explains absorption spectra: photons of just the right wavelength can knock an electron from one energy level to a higher one. To conserve energy, only photons that have just the right energy will be absorbed. This explains why a continuous spectrum of light entering a gas will emerge with dark (absorption) lines at frequencies that correspond to emission lines

ig. 37-20c).

The Bohr theory also ensures the stability of atoms. It establishes stability by decree: the ground state is the lowest state for an electron and there is no lower energy level to which it can go and emit more energy. Finally, as we saw above, the Bohr theory accurately predicts the ionization energy of 13.6eV for hydrogen. However, the Bohr model was not so successful for other atoms, and has been superseded as we shall discuss in the next Chapter. We discuss the Bohr model because it was an important start, and because we still use the concept of stationary states, the ground state, and transitions between states. Also, the terminology used in the Bohr model is still used by chemists and spectroscopists.

DG

EAE

Wavelength of a Lyman line. Use Fig. 37-26 to determine

the wavelength of the first Lyman line, the transition from what region of the electromagnetic spectrum does this lie?

n =2

to n = 1. In

APPROACH We use Eq. 37-9, hf = Ey — E,, with the energies obtained from Fig. 37-26 to find the energy and the wavelength of the transition. The region of the electromagnetic spectrum is found using the EM spectrum in Fig. 31-12.

SOLUTION In this case, Af = E, — E; = {-3.4eV — (—13.6eV)} = 102eV

=(10.2eV)(1.60 x 107°J/eV) = 1.63 x 1078].

Since A = c/f, we have

A=l __heh | 6.63 X 1077 X -5)(3.00 X 10°m/s/)=1.22> = 2.2.44 X Fim 10°m —. = (1097

wig&

\

So A =1/(244 X 10°m™) = 4.10 X 107m or 410 nm. This is the fourth line in the Balmer series, Fig. 37-21, and is violet in color.

EXERCISE F The energy of the photon emitted when a hydrogen atom goes from the n = 6 state to the n = 3 state is (a) 0.378 eV; (b) 0.503 eV; (c) 1.13 eV, (d) 3.06eV; (e) 13.6eV.

DOV

YESER

Absorption wavelength. Use Fig. 37-26 to determine the

maximum wavelength that hydrogen in its ground state can absorb. What would be the next smaller wavelength that would work?

APPROACH Maximum wavelength corresponds to minimum energy, and this would be the jump from the ground state up to the first excited state (Fig. 37-26). The next smaller wavelength occurs for the jump from the ground state to the second excited state. In each case, the energy difference can be used to find the wavelength. SOLUTION The energy needed to jump from the ground state to the first excited state is 13.6eV — 3.4eV = 10.2eV; the required wavelength, as we saw in Example 37-13, is 122 nm. The energy to jump from the ground state to the second excited state is 13.6eV — 1.5eV = 12.1 eV, which corresponds to a wavelength =%

=

he

fhf

_

_

hc

E-E

6.63 X 107*J+5)(3.00 X 108m/s

X

b

(12.1eV)(1.60 X 1077J/eV)

)~ 103 0m ™

He™ ionization energy. (a) Use the Bohr model to determine

the ionization energy of the He™ ion, which has a single electron. (b) Also calculate the maximum wavelength a photon can have to cause ionization. APPROACH We want to determine the minimum energy required to lift the electron from its ground state and to barely reach the free state at E = 0. The ground state energy of He" is given by Eq. 37-14b with n =1 and Z = 2. SOLUTION (a) Since all the symbols in Eq. 37-14b are the same as for the calculation for hydrogen, except that Z is 2 instead of 1, we see that E; will

be Z? = 2% = 4 times the E, for hydrogen: E,

=

4(-13.6eV)

=

—544¢V.

Thus, to ionize the He™ ion should require 54.4 eV, and this value agrees with experiment. (b) The maximum wavelength photon that can cause ionization will have energy hf = 54.4eV and wavelength

:

NOTE

Cf

c

he

hf

(6.63 x 107*J-5)(3.00 X 10°m/s)

(544eV)(1.60 X 107°J/eV)

=

22.8nm.

If the atom absorbed a photon of greater energy (wavelength shorter than

22.8 nm), the atom could still be ionized and the freed electron would have kinetic

energy of its own. If A > 22.8 nm, the photon has too little energy to cause ionization.

In this last Example, we saw that E; for the He™ ion is four times more negatiy% than that for hydrogen. Indeed, the energy-level diagram for He™ looks just like tt for hydrogen, Fig. 37-26, except that the numerical values for each energy level are 1008

CHAPTER 37

four times larger. Note, however, that we are talking here about the He™ ion. Normal

(neutral) helium has two electrons and its energy level diagram is entirely different.

CONCEPTUAL

EXAMPLE

37-17 | Hydrogen

at 20°C. Estimate the average

kinetic energy of whole hydrogen atoms (not just the electrons) at room temperature, and use the result to explain why nearly all H atoms are in the ground state at room ftemperature, and hence emit no light. RESPONSE According to kinetic theory (Chapter 18), the average kinetic energy of atoms or molecules in a gas is given by Eq. 18-4:

K = 3kT,

where

k = 1.38 X 1002 J/K

is Boltzmann’s

constant, and

(absolute) temperature. Room temperature is about

K = 3(138 X 102J/K)(300K)

7 = 300K,

7 is the kelvin so

= 6.2 X 1072],

or, in electron volts:

_

62X 1097

T 16 x 10°7/eV

=

0.04eV.

The average kinetic energy of an atom as a whole is thus very small compared to the energy between the ground state and the next higher energy state (13.6eV — 3.4eV = 10.2eV). Any atoms in excited states quickly fall to the ground state and emit light. Once in the ground state, collisions with other atoms can transfer energy of only 0.04 eV on the average. A small fraction of atoms can have much more energy (see Section 18-2 on the distribution of molecular speeds), but even a kinetic energy that is 10 times the average is not nearly enough to excite atoms into states above the ground state. Thus, at room temperature, nearly all atoms are in the ground state. Atoms can be excited to upper states by very high temperatures, or by passing a current of high energy electrons through the gas, as in a discharge tube (Fig. 37-19).

T

forrespondence Principle ‘e should note that Bohr made some radical assumptions that were at variance with classical ideas. He assumed that electrons in fixed orbits do not radiate light even though they are accelerating (moving in a circle), and he assumed that angular momentum is quantized. Furthermore, he was not able to say how an electron moved when it made a transition from one energy level to another. On the other hand, there is no real reason to expect that in the tiny world of the atom electrons would behave as ordinary-sized objects do. Nonetheless, he felt that where quantum theory overlaps with the macroscopic world, it should predict classical results. This is the correspondence principle, already mentioned in regard to relativity (Section 36-13). This principle does work for Bohr’s theory of the hydrogen atom. The orbit sizes and energies are quite different for » =1 and n = 2, say. But orbits with » = 100,000,000 and 100,000,001 would be very close in radius and energy (see Fig. 37-26). Indeed, jumps between such large orbits (which would approach macroscopic sizes), would be imperceptible. Such orbits would thus appear to be continuously spaced, which is what we expect in the everyday world. Finally, it must be emphasized that the well-defined orbits of the Bohr model do not actually exist. The Bohr model is only a model, not reality. The idea of electron orbits was rejected a few years later, and today electrons are thought of (Chapter 39) as forming “probability clouds.”

37-12

de Broglie’'s Hypothesis Applied to Atoms

Bohr’s theory was largely of an ad hoc nature. Assumptions were made so that

f-eory would agree with experiment. But Bohr could give no reason why the orbits

-ere quantized, nor why there later, a reason was proposed by 1923, de Broglie proposed that nature; and that this hypothesis

should be a stable Louis de Broglie. material particles, was confirmed by

ground state. Finally, ten years We saw in Section 37-7 that in such as electrons, have a wave experiment several years later.

SECTION 37-12

1009

o

SNe

L

One of de Broglie’s original arguments in favor of the wave nature of electrons was that it provided an explanation for Bohr’s theory of the hydrogen atom. According to de Broglie, a particle of mass m moving with a nonrelativistic speed v would have a wavelength (Eq. 37-7) of -

2

L

FIGURE 37-27 An ordinary standing wave compared to a circular standing wave. FIGURE 37-28 When a wave does not close (and hence interferes

destructively with itself), it rapidly dies out.

mv

Each electron orbit in an atom, he proposed, is actually a standing wave. As we saw in Chapter 15, when a violin or guitar string is plucked, a vast number of wavelengths are excited. But only certain ones—those that have nodes at the ends—are sustained. These are the resonant modes of the string. Waves with other wavelengths interfere with themselves upon reflection and their amplitudes quickly drop to zero. With electrons moving in circles, according to Bohr’s theory, de Broglie argued that the electron wave was a circular standing wave that closes on itself, Fig. 37-27. If the wavelength of a wave does not close on itself, as in Fig. 37-28, destructive interference takes place as the wave travels around the loop, and the wave quickly dies out. Thus, the only waves that persist are those for which the circumference of the circular orbit contains a whole number of wavelengths, Fig. 37-29. The circumference of a Bohr orbit of radius r, is 2#r,, so to have constructive interference, we need 2mwr,

=

nA,

When we substitute A = h/mv, mor,

FIGURE 37-29

Standing circular

waves for two, three, and five wavelengths on the circumference;

n, the number of wavelengths, is also the quantum number.

=

n. =

1,23,

we get 2wr, = nh/mv,

or

nh —: 27

This is just the quantum condition proposed by Bohr on an ad hoc basis, Eq. 37-10. It is from this equation that the discrete orbits and energy levels were derived. Thus we have a first explanation for the quantized orbits and energy states in the Bohr model: they are due to the wave nature of the electron, and only resonant “standing” waves can persist.” This implies that the wave—particle dualitffl is at the root of atomic structure. In viewing the circular electron waves of Fig. 37-29, the electron is not to be thought of as following the oscillating wave pattern. In the Bohr model of hydrogen, the electron moves in a circle. The circular wave, on the other hand, represents the amplitude of the electron “matter wave,” and in Fig. 37-29 the wave amplitude is shown superimposed on the circular path of the particle orbit for convenience. Bohr’s theory worked well for hydrogen and for one-electron ions. But it did not prove successful for multi-electron atoms. Bohr’s theory could not predict line spectra even for the next simplest atom, helium. It could not explain why some emission lines are brighter than others, nor why some lines are split into two or more closely spaced lines (“fine structure”). A new theory was needed and was indeed developed in the 1920s. This new and radical theory is called quantum mechanics. It finally solved the problem of atomic structure, but it gives us a very different view of the atom: the idea of electrons in well-defined orbits was replaced with the idea of electron “clouds.” This new theory of quantum mechanics has given us a wholly different view of the basic mechanisms underlying physical processes. "We note, however, that Eq. 37-10 is no longer considered valid, as discussed in Chapter 39.

| Summary Quantum theory has its origins in Planck’s quantum hypothesis that molecular oscillations are quantized: their energy E can only be integer (n) multiples of 4f, where & is Planck’s constant and f is the natural frequency of oscillation: E = nhf. 37-2) This hypothesis explained the spectrum of radiation emitted by a blackbody at high temperature.

1010

CHAPTER 37

Einstein proposed that for some experiments, light could be pictured as being emitted and absorbed as quanta (particles), which we now call photons, each with energy E = hf 37-3)

and momentum

Early Quantum Theory and Models of the Atom

E

_hf

P=_2="7=3

h

(37-5)

“

He proposed the photoelectric effect as a test for the photon

theory of light. In the photoelectric effect, the photon theory says that each incident photon can strike an electron in a material and eject it if the photon has sufficient energy. The aximum energy of ejected electrons is then linearly related to .ne frequency of the incident light. The photon theory is also supported by the Compton effect and the observation of electron—positron pair production. The wave-particle duality refers to the idea that light and matter (such as electrons) have both wave and particle properties. The wavelength of an object is given by

A= p

37-7

where p is the momentum of the object (p = mv for a particle of mass m and speed v). The principle of complementarity states that we must be aware of both the particle and wave properties of light and of matter for a complete understanding of them.

Early models of the atom include the plum-pudding model,

and Rutherford’s planetary (or nuclear) model of an atom which consists of a tiny but massive positively charged nucleus surrounded (at a relatively great distance) by electrons. To explain the line spectra emitted by atoms, as well as the stability of atoms, Bohr’s theory postulated that: (1) electrons bound in an atom can only occupy orbits for which the angular

momentum

is quantized, which results in discrete values for the

radius and energy; (2) an electron in such a stationary state emits no radiation; (3) if an electron jumps to a lower state, it emits a photon whose energy equals the difference in energy between the two states; (4) the angular momentum L of atomic electrons is quantized by the rule (37-10) where # is an integer called the quantum number. The n =1 state is the ground state, which in hydrogen has an energy E; = —13.6eV. Higher values of n correspond to excited states, and their energies are E,

=

22

—(13.6 eV)?-

(37-14b)

Atoms are excited to these higher states by collisions with other atoms or electrons, or by absorption of a photon of just the right frequency. De Broglie’s hypothesis that electrons (and other matter) have a wavelength A = h/mv gave an explanation for Bohr’s quantized orbits by bringing in the wave-particle duality: the orbits correspond to circular standing waves in which the circumference of the orbit equals a whole number of wavelengths.

I Questions r!.

of

If energy is radiated by all objects, why can we not see most of them in the dark?

3. Does a lightbulb at a temperature of 2500K white a light as the Sun at 6000 K? Explain.

produce

as

4, Darkrooms for developing black-and-white film were sometimes lit by a red bulb. Why red? Would such a bulb work in a darkroom for developing color photographs? 5. If the threshold wavelength in the photoelectric effect increases when the emitting metal is changed to a different metal, what can you say about the work functions of the two metals?

6. Explain why the existence of a cutoff frequency in the photoelectric effect more strongly favors a particle theory rather than a wave theory of light.

©

7. UV light causes sunburn, whereas visible light does not. Suggest a reason. 8. The work functions for sodium and cesium are 2.28 eV and 2.14 eV, respectively. For incident photons of a given frequency, which metal will give a higher maximum kinetic energy for the electrons?

10. r

(a) Does a beam of infrared photons always have less energy than a beam of ultraviolet photons? Explain. (b) Does a single photon of infrared light always have less energy than a single photon of ultraviolet light? Light of 450-nm wavelength strikes a metal surface, and a stream of electrons emerges from the metal. If light of the same intensity but of wavelength 400 nm strikes the surface, are more electrons emitted? Does the energy of the emitted electrons change? Explain.

1 . Explain how the photoelectric circuit of Fig. 37-4 could —

1. What can be said about the relative temperatures whitish-yellow, reddish, and bluish stars? Explain.

be used in (@) a burglar alarm, (b) a smoke detector, (c) a photographic light meter.

12. If an X-ray photon is scattered by an electron, does the photon’s wavelength change? If so, does it increase or decrease? 13. In both the photoelectric effect and in the Compton effect, a photon collides with an electron causing the electron to fly off. What then, is the difference between the two processes? 14. Consider a point source of light. How would the intensity of light vary with distance from the source according to (a) wave theory, (b) particle (photon) theory? Would this help to distinguish the two theories? 15. If an electron and a proton travel at the same speed, which has the shorter de Broglie wavelength? Explain. 16. Why do we say that light has wave properties? Why do we say that light has particle properties? 17. Why do we say that electrons have wave properties? Why do we say that electrons have particle properties? 18. What are the differences between a photon and an electron? Be specific: make a list. 19. In Rutherford’s

planetary model

of the atom, what

keeps

the electrons from flying off into space? 20. How can you tell if there is oxygen near the surface of the Sun? 21. When a wide spectrum of light passes through hydrogen gas at room temperature, absorption lines are observed that correspond only to the Lyman series. Why don’t we observe the other series? 22. Explain how the closely spaced energy levels for hydrogen near the top of Fig. 37-26 correspond to the closely spaced spectral lines at the top of Fig. 37-21. Questions

1011

23. Is it possible for the de Broglie wavelength of a “particle” to be greater than the dimensions of the particle? To be smaller? Is there any direct connection? . In a helium atom, which contains two electrons, do you think that on average the electrons are closer to the nucleus or farther away than in a hydrogen atom? Why? 25. How can the spectrum of hydrogen contain so many lines when hydrogen contains only one electron? The Lyman series is brighter than the Balmer series because this series of transitions ends up in the most common state for hydrogen, the ground state. Why then was the Balmer series discovered first?

27. Use conservation of momentum to explain why photons emitted by hydrogen atoms have slightly less energy than that predicted by Eq. 37-9. 28 Suppose we obtain an emission spectrum for hydrogen at very high temperature (when some of the atoms are ° excited states), and an absorption spectrum at room temperature, when all atoms are in the ground state. Will the two spectra contain identical lines?

| Problems 37-1 Planck’s Quantum Hypothesis |

(I) Estimate the peak wavelength for radiation from (a) ice at 273K, (b) a floodlamp at 3500 K, (c) helium at 4.2 K, (d) for the universe at 7 = 2.725 K, assuming blackbody emission. In what region of the EM spectrum is each? . (I) How hot is metal being welded if it radiates most strongly at 460 nm? . (I) An HCI molecule vibrates with a natural frequency of

8.1 X 10'* Hz. What is the difference in energy (in joules

and electron volts) between successive values of the oscillation energy? . (II) Estimate the peak wavelength of light issuing from the pupil of the human eye (which approximates a blackbody) assuming normal body temperature. . (III) Planck’s radiation law is given by:

I(AT)

2mhc?A~3

= /AT

_ |

where I(A, T) is the rate energy is radiated per unit surface area per unit wavelength interval at wavelength A and Kelvin temperature 7. (a) Show that Wien’s displacement law follows from this relationship. () Determine the value of h from the experimental value of Ap7 given in the text. [You may want to use graphing techniques.] (¢) Derive the T* dependence of the rate at which energy is radiated (as in the Stefan-Boltzmann law, Eq. 19-17), by integrating Planck’s

11. (I) What

37-2 and 37-3 Photons and the Photoelectric Effect 6. (I) What is the energy of photons (in joules) emitted by a 104.1-MHz FM radio station? 7. (I) What is the energy range (in joules and eV) of photons in the visible spectrum, of wavelength 410 nm to 750 nm? 8. (I) A typical gamma ray emitted from a nucleus during radioactive decay may have an energy of 380 keV. What is its wavelength? Would we expect significant diffraction of this type of light when it passes through an everyday opening, such as a door? . (I) About 0.1 eV is required to break a “hydrogen bond” in a protein molecule. Calculate the minimum frequency and maximum wavelength of a photon that can accomplish this, 10. (I) Calculate the momentum of a photon of yellow light of

wavelength 6.20 X 107 m,

1012

CHAPTER 37

frequency

of light is needed

to eject

12. (I) What is the longest wavelength of light that will emit electrons from a metal whose work function is 3.70 eV? 13. (IT) What wavelength photon would have the same energy as a 145-gram baseball moving 30.0 m/s? 14. (IT) The human eye can respond to as little as 10718J of light energy. For a wavelength at the peak of visual sensitivity, 550 nm, how many photons lead to an observable flash?

15. (IT) The work iron

are

2.3,

functions 2.1, 4.7,

for sodium, cesium, copper, and

and

4.5V,

respectively.

Which

of

these metals will not emit electrons when visible light shines on it? 16. (IT) In a photoelectric-effect experiment it is observed that no current flows unless the wavelength is less than 520 nm. (a) What is the work function of this material? (b) What 4 the stopping voltage required if light of wavelength 470 n1. is used? 17, (II) What is the maximum Kkinetic energy of electrons

ejected from barium

(W, = 2.48¢eV) when illuminated by

surface, the maximum

kinetic energy of emitted electrons is

white light, A = 410 to 750 nm? 18. (IT) Barium has a work function of 2.48¢eV. What is the maximum kinetic energy of electrons if the metal is illuminated by UV light of wavelength 365 nm? What is their speed? 19. (IT) When UV light of wavelength 285 nm falls on a metal

formula over all wavelengths; that is, show that

jI(A,T) d\ o< T4,

minimum

electrons from a metal whose work function is 4.8 X 107°J?

20.

21.

22.

23.

1.70 eV. What is the work function of the metal? (IT) The threshold wavelength for emission of electrons from a given surface is 320 nm. What will be the maximum kinetic energy of ejected electrons when the wavelength is changed to (a) 280 nm, (b) 360 nm? (IT) When 230-nm light falls on a metal, the current through a photoelectric circuit (Fig. 37-4) is brought to zero at a stopping voltage of 1.84 V. What is the work function of the metal? (IT) A certain type of film is sensitive only to light whose wavelength is less than 630 nm. What is the energy (eV and kcal/mol) needed for the chemical reaction to occur which causes the film to change? (IT) The range of visible light wavelengths extends from about 410 nm to 750 nm. (a) Estimate the minimum energy (eV) necessary to initiate the chemical process on the retina

that is responsible for vision. (b) Speculate as to why, at the other end of the visible range, there is a threshold photo®, energy beyond which the eye registers no sensation of sigh. Determine this threshold photon energy (eV).

Early Quantum Theory and Models of the Atom

(.

24. (IT) In a photoelectric experiment using a clean sodium surface, the maximum energy of the emitted electrons was measured for a number of different incident frequencies, with the following results.

Frequency (x 10" Hz)

Energy (eV)

11.8

2.60

8.2 6.9

1.10 0.57

10.6 9.9 9.1

2.11 1.81 1.47

25. (IT) A photomultiplier tube (a very sensitive light sensor), is based on the photoelectric effect: incident photons strike a metal surface and the resulting ejected electrons are collected. By counting the number of collected electrons, the number of incident photons (i.e., the incident light intensity) can be determined. (¢) If a photomultiplier tube is to respond properly for incident wavelengths throughout the visible range (410nm to 750 nm), what is the maximum value for the work function W, (eV) of its metal surface? (b) If W, for its metal surface is above a certain threshold value, the photomultiplier will only function for incident ultraviolet wavelengths and be unresponsive to visible light. Determine this threshold value (eV). 26. (IIT) A group of atoms is confined to a very small (point-like) volume in a laser-based atom trap. The incident laser light causes each atom to emit 1.0 X 10° photons of wavelength 780 nm every second. A sensor of area 1.0 cm? measures the light intensity emanating from the trap to be 1.6nW when placed 25 cm away from the trapped atoms. Assuming each atom emits photons with equal probability in all directions, determine the number of trapped atoms. 27. (IIT) Assume light of wavelength A is incident on a metal whose

work

function

is known

precisely

(IT) In the Compton effect, determine the ratio (AA/A) of the maximum change AA in a photon’s wavelength to the photon’s initial wavelength A, if the photon is (a) a visiblelight photon with A = 550 nm, (b) an X-ray photon with A =010nm. 32. (IT) A 1.0-MeV gamma-ray photon undergoes a sequence of Compton-scattering events. If the photon is scattered at an angle of 0.50° in each event, estimate the number of events required to convert the photon into a visible-light photon with wavelength 555 nm. You can use an expansion for small 6; see

Plot the graph of these results and find: (a) Planck’s constant; (b) the cutoff frequency of sodium; (c) the work function.

surface,

31

(i.e., its

uncertainty is better than 0.1% and can be ignored). Show that if the stopping voltage can be determined to an accuracy of AVg, the fractional uncertainty (magnitude) in wavelength is

Appendix A. [Gamma rays created near the center of the Sun are transformed to visible wavelengths as they travel to the Sun’s surface through a sequence of small-angle

Compton scattering events.]

33. (IIT) In the Compton effect, a 0.160-nm photon strikes a free electron in a head-on collision and knocks it into the forward direction. The rebounding photon recoils directly backward. Use conservation of (relativistic) energy and momentum to determine (a) the kinetic energy of the electron, and (b) the wavelength of the recoiling photon. Use Eq. 37-5, but not Eq. 37-6. 34. (III) In the Compton effect (see Fig. 37-7), use the relativistic equations for conservation of energy and of linear momentum to show that the Compton shift in wavelength is given by Eq. 37-6.

37-5 Pair Production 35. (I) How much total kinetic energy will an electron—positron pair have if produced by a 2.67-MeV photon? 36. (IT) What is the longest wavelength photon that could produce a proton-antiproton pair? (Each has a mass of

1.67 X 107" kg.)

37

(IT) What is the minimum photon energy needed to produce a u' —u” pair? The mass of each u (muon) is 207 times the mass of an electron. What is the wavelength of such a photon? 38. (I) An electron and a positron, each moving at 2.0 X 10° m/s, collide head on, disappear, and produce two photons moving in opposite directions, each with the same energy and momentum. Determine the energy and momentum of each photon. 39. (I) A gamma-ray photon produces an electron and a positron, each with a kinetic energy of 375 keV. Determine the energy and wavelength of the photon.

37-4 Compton Effect

37-7 Wave Nature of Matter 40. (I) Calculate the wavelength of a 0.23-kg ball traveling at 0.10 m/s. 41. (I) What is the wavelength of a neutron (m = 1.67 X 107?"kg)

28. (I) A high-frequency photon is scattered off of an electron and experiences a change of wavelength of 1.5 X 10~ nm. At what angle must a detector be placed to detect the scattered photon (relative to the direction of the incoming photon)? 29. (II) Determine the Compton wavelength for (4) an electron, (b) a proton. (c) Show that if a photon has wavelength equal to the Compton wavelength of a particle, the photon’s energy is equal to the rest energy of the particle. 30. (IT) X-rays of wavelength A = 0.120nm are scattered from carbon. What is the expected Compton wavelength shift for photons detected at angles (relative to the incident beam) of exactly (a) 60°, (b) 90°, (c) 180°?

42. (I) Through how many volts of potential difference must an electron be accelerated to achieve a wavelength of 0.21 nm? 43. (IT) What is the theoretical limit of resolution for an electron microscope whose electrons are accelerated through 85 kV? (Relativistic formulas should be used.) . (IT) The speed of an electron in a particle accelerator is 0.98c. Find its de Broglie wavelength. (Use relativistic momentum.) 45. (IT) Calculate the ratio of the kinetic energy of an electron to that of a proton if their wavelengths are equal. Assume that the speeds are nonrelativistic.

Determine this fractional uncertainty if A}y = 0.01V A = 550 nm.

and

traveling at 8.5 X 10*m/s?

Problems

1013

(IT) Neutrons can be used in diffraction experiments to probe the lattice structure of crystalline solids. Since the neutron’s wavelength needs to be on the order of the spacing between atoms in the lattice, about 0.3 nm, what should the speed of the neutrons be? 47. (II) An electron has a de Broglie wavelength

A =6.0x 107%m.

. 49.

50.

51

(a) What is its momentum? (b) What is

its speed? (c) What voltage was needed to accelerate it to this speed? (IT) What is the wavelength of an electron of energy (a) 20 eV, (b) 200eV, (c) 2.0keV? (IT) Show that if an electron and a proton have the same nonrelativistic kinetic energy, the proton has the shorter wavelength. (IT) Calculate the de Broglie wavelength of an electron in a TV picture tube if it is accelerated by 33,000V, Is it relativistic? How does its wavelength compare to the size of the “neck” of the tube, typically 5 cm? Do we have to worry about diffraction problems blurring our picture on the screen? (IT) After passing through two slits separated by a distance of 3.0um, a beam of electrons creates an interference pattern with its second-order maximum at an angle of 55°. Find the speed of the electrons in this beam.

*37-8 Electron Microscope *52. (IT) What voltage is needed to produce electron wavelengths of 0.28 nm? (Assume that the electrons are nonrelativistic.) * 53 (IT) Electrons are accelerated by 3450V in an electron microscope. Estimate the maximum possible resolution of the microscope.

37-10 and 37-11

Bohr Model

54. (I) For the three hydrogen transitions indicated below, with n being the initial state and n’ being the final state, is the transition an absorption or an emission? Which is higher, the initial state energy or the final state energy of the atom? Finally, which of these transitions involves the largest energy photon? (@) n =1, n' =3; (b) n =6, n'=2(c)n=4,n" =5 S5. (I) How much energy is needed to ionize a hydrogen atom inthe n = 3 state? 56. (I) (a) Determine the wavelength of the second Balmer line (n=4 to n =2 transition) using Fig. 37-26. Determine likewise (b) the wavelength of the third Lyman line and (c) the wavelength of the first Balmer line. 57. (I) Calculate the ionization energy of doubly ionized

lithium, Li**, which has Z = 3.

58. (I) Evaluate the Rydberg constant R using the Bohr model (compare Eqs. 37-8 and 37-15) and show that its value is

R =1.0974 x 10'm™,

59. (IT) What is the longest wavelength light capable of ionizing a hydrogen atom in the ground state? 60. (IT) In the Sun, an ionized helium (He™) atom makes a transition from the n = 5 state to the n = 2 state, emitting a photon. Can that photon be absorbed by hydrogen ator™) present in the Sun? If so, between what energy states wili the hydrogen atom transition occur? 61. (IT) What wavelength photon would be required to ionize a hydrogen atom in the ground state and give the ejected electron a kinetic energy of 20.0eV? 62. (IT) For what maximum kinetic energy is a collision between an electron and a hydrogen atom in its ground state definitely elastic? . (IT) Construct the energy-level diagram for the He™ ion (like Fig. 37-26). . (II) Construct the energy-level diagram (like Fig. 37-26) for

doubly ionized lithium, Li*",

. (IT) Determine the electrostatic potential energy and the kinetic energy of an electron in the ground state of the hydrogen atom. . (IT) An excited hydrogen atom could, in principle, have a diameter of 0.10 mm. What would be the value of n for a Bohr orbit of this size? What would its energy be? 67. (IT) Is the use of nonrelativistic formulas justified in the

Bohr atom? To check, calculate the electron’s velocity, v, in

terms of ¢, for the ground state of hydrogen, and then calculate \ /1 — v2/c2. 68. (IT)

A

hydrogen

atom

has

an

angular

momentum

of

5.273 x 1073*kg-m?/s. According to the Bohr model, what

is the energy (eV) associated with this state? 69. (IT) Assume hydrogen atoms in a gas are initially in the'fi ground state. If free electrons with kinetic energy 12.75 eV collide with these atoms, what photon wavelengths will be emitted by the gas? 70. (IT) Suppose an electron was bound to a proton, as in the hydrogen atom, but by the gravitational force rather than by the electric force. What would be the radius, and energy, of

the first Bohr orbit?

71. (IT) Correspondence principle: Show that for large values of n, the difference in radius Ar between two adjacent orbits (with quantum numbers n and n — 1) is given by Ar =

'n = In-1

®

2ry n’

so Ar/r, — 0 as n — oo in accordance with the correspondence principle. [Note that we can check the correspondence principle by either considering large values of n (n — o) or by letting 4 — 0. Are these equivalent?]

| General Problems 72. If a 75-W lightbulb emits 3.0% of the input energy as visible light (average wavelength 550 nm) uniformly in all directions, estimate how many photons per second of visible light will strike the pupil (4.0mm diameter) of the eye of an observer 250 m away. 73. At low temperatures, nearly all the atoms in hydrogen gas will be in the ground state. What minimum frequency photon is needed if the photoelectric effect is to be observed?

1014

CHAPTER 37

74. A beam of 125-eV electrons is scattered from a crystal, as in X-ray diffraction, and a first-order peak is observed at 6 = 38°. What is the spacing between planes in the diffracting crystal? (See Section 35-10.) 15 A microwave oven produces electromagnetic radiation at A =122cm and produces a power of 860 W. Calculate th« number of microwave photons produced by the microwave oven each second.

Early Quantum Theory and Models of the Atom

76. Sunlight reaching the Earth has an intensity of about 1350 W/m?. Estimate how many photons per square meter per second this represents. Take the average wavelength to be 550 nm. . A beam of red laser light (A = 633 nm) hits a black wall and is fully absorbed. If this light exerts a total force F = 6.5nN on the wall, how many photons per second are hitting the wall? 78. The Big Bang theory states that the beginning of the universe was accompanied by a huge burst of photons. Those photons are still present today and make up the so-called cosmic microwave background radiation. The universe radiates like a blackbody with a temperature of about 2.7 K. Calculate the peak wavelength of this radiation. 9. An electron and a positron collide head on, annihilate, and create two 0.755-MeV photons traveling in opposite directions. What were the initial kinetic energies of electron and positron? 80. By what potential difference must (¢) a proton (m =

1.67 X 107%" kg), and (b) an electron (m = 9.11 X 103! kg),

be accelerated to have a wavelength A = 6,0 X 1072 m? 81. In some of Rutherford’s experiments (Fig. 37-17) the

a particles (mass = 6.64 X 107>"kg) had a kinetic energy

82. 83. . 85.

86.

of 4.8MeV. How close could they get to the center of a silver nucleus (charge = +47¢)? Ignore the recoil motion of the nucleus. Show that the magnitude of the electrostatic potential energy of an electron in any Bohr orbit of a hydrogen atom is twice the magnitude of its kinetic energy in that orbit. Calculate the ratio of the gravitational force to the electric force for the electron in a hydrogen atom. Can the gravitational force be safely ignored? Electrons accelerated by a potential difference of 123V pass through a gas of hydrogen atoms at room temperature. What wavelengths of light will be emitted? In a particular photoelectric experiment, a stopping potential of 2.70V is measured when ultraviolet light of wavelength 380 nm is incident on the metal. If blue light of wavelength 440 nm is used, what is the new stopping potential? In an X-ray tube (see Fig. 35-26 and discussion in Section

35-10),

the

high

voltage

between

filament

89. Light of wavelength 360nm strikes a metal whose work function is 2.4 ¢V. What is the shortest de Broglie wavelength for the electrons that are produced as photoelectrons? 90. Visible light incident on a diffraction grating with slit spacing of 0.012 mm has the first maximum at an angle of 3.5° from the central peak. If electrons could be diffracted by the same grating, what electron velocity would produce the same diffraction pattern as the visible light? 91. (a) Suppose an unknown element has an absorption with lines corresponding to 2.5, 4.7, and 5.1eV ground state, and an ionization energy of 11.5eV. energy level diagram for this element. (b) If a 5.1-eV

absorbed by an atom of this substance, in which state was the

atom before absorbing the photon? What will be the energies of the photons that can subsequently be emitted by this atom? 92. Light of wavelength 424 nm falls work function of 2.28 eV. (a) How applied to bring the current to maximum speed of the emitted the de Broglie wavelength of these

(b) What is the shortest wavelength of X-ray emitted when accelerated electrons strike the face of a 33-kV television picture tube? 87. The intensity of the Sun’s light in the vicinity of Earth is about 1350 W/m?% Imagine a spacecraft with a mirrored square sail of dimension 1.0 km. Estimate how much thrust (in newtons) this craft will experience due to collisions with the Sun’s photons. [Hint: Assume the photons bounce perpendicularly off the sail with no change in the magnitude of their momentum. ] 88. Photons of energy 9.0 eV are incident on a metal. It is found that current flows from the metal until a stopping potential of 4.0V is applied. If the wavelength of the incident photons is doubled, what is the maximum kinetic energy of the ejected electrons? What would happen if the wavelength of the incident photons was tripled?

on a metal which has a much voltage should be zero? (b) What is the electrons? (c¢) What is electrons?

93. (a) Apply Bohr’s assumptions to the Earth-Moon system to calculate the allowed energies and radii of motion. (b) Given the known distance between Earth and the Moon, is the quantization of the energy and radius apparent? 94. Show that the wavelength of a particle of mass m with kinetic energy K is given by the relativistic formula A = he//K? + 2mc2K . . A small flashlight is rated at 3.0W. As the light leaves the flashlight in one direction, a reaction force is exerted on the flashlight in the opposite direction. Estimate the size of this reaction force. 96. At the atomic-scale, the electron volt and nanometer are well-suited units for energy and distance, respectively. (a) Show that the energy E in eV of a photon, whose wavelength A is in nm, is given by

_ 1240 eV -nm E

and

target is V. After being accelerated through this voltage, an electron strikes the target where it is decelerated (by positively charged nuclei) and in the process one or more X-ray photons are emitted. (a) Show that the photon of shortest wavelength will have

spectrum above its Draw an photon is

*97.

A (nm)

(b) How much energy (eV) does a 650-nm photon have? Three fundamental constants of nature—the gravitational constant G, Planck’s constant 4, and the speed of light c—

have the dimensions of [L?/MT?|,[ML*/T],

and [L/T],

respectively. (a) Find the mathematical combination of these fundamental constants that has the dimension of time. This combination is called the “Planck time” fp and is thought to be the earliest time, after the creation of the universe, at which the currently known laws of physics can be applied. (b) Determine the numerical value of fp. (c) Find the mathematical combination of these fundamental constants that has the dimension of length. This combination is called the “Planck length” Ap and is thought to be the smallest length over which the currently known laws of physics can be applied. (d) Determine the numerical value of Ap. 98. Imagine a free particle of mass m bouncing back and forth between two perfectly reflecting walls, separated by distance £. Imagine that the two oppositely directed matter waves associated with this particle interfere to create a standing wave with a node at each of the walls. Show that the ground state (first harmonic) and first excited state (second harmonic) have (non-relativistic) kinetic energies

h?/8mif? and h*/2m?, respectively.

General Problems

1015

99. (a) A rubidium atom (m = 85u) is at rest with one electron in an excited energy level. When the electron jumps to the ground state, the atom emits a photon of wavelength = 780nm. Determine the resulting (nonrelativistic) recoil speed v of the atom. (b) The recoil speed sets the lower limit on the temperature to which an ideal gas of rubidium atoms can be cooled in a laser-based atom trap. Using the kinetic theory of gases (Chapter 18), estimate this “lowest achievable” temperature. 100. A rubidium atom (atomic mass 85) is initially at room temperature and

has

a velocity

» = 290m/s

due

to its thermal

motion. Consider the absorption of photons by this atom from a laser beam of wavelength A = 780nm. Assume the rubidium atom’s initial velocity » is directed into the laser beam (the photons are moving right and the atom is moving left) and that the atom absorbs a new photon every 25 ns. How long will it take for this process to completely stop (“cool”) the rubidium atom? [Note: a more detailed analysis predicts that the atom can be slowed to about 1 cm/s by this light absorption process, but it cannot be completely stopped.]

*Numerical/ Computer

, o (III) (a) Graph Planck’s radiation formula (top of page 989) as a function of wavelength from A = 20nm to 2000 nm in 20nm steps for two lightbulb filaments, one at 2700K an the other at 3300 K. Plot both curves on the same set of axes. (b) Approximately how much more intense is the visible light from the hotter bulb? Use numerical integration. *102. (IIT) Estimate what % of emitted sunlight energy is in the visible range. Use Planck’s radiation formula (top of page 989) and numerical integration. *103. (IIT) Potassium has one of the lowest work functions of all metals and so is useful in photoelectric devices using visible light. Light from a source is incident on a potassium surface. Data for the stopping voltage V; as a function of wavelength A is shown below. (@) Explain why a graph of Vj vs. 1/A is expected to yield a straight line. What are the theoretical expectations for the slope and the y-intercept of this line? (b) Using the data below, graph V; vs. 1/A and show that a straight-line plot does indeed result. (c) Determine the slope a and y-intercept b of this line. Using your values for a and b, determine (d) potassium’s work function (eV) and (e) Planck’s constant /4 (J-s). *101.

A (pm)

0.400

0430

0460

0490

0.520

Vo (V)

0.803

0578

0402

0229

0.083

Answers to Exercises A: More 1000-nm photons (lower frequency).

B: 5.50 x 10'* Hz, 545 nm.

C: Only A.

1016

CHAPTER 37

D: Decrease. E:

(e).

F:

(¢).

Early Quantum Theory and Models of the Atom

Participants

in

the

1927

Solvay

Conference. Albert Einstein, seated at

front center, had difficulty accepting that nature could behave according to the rules of quantum mechanics. We present a brief summary of quantum mechanics in this Chapter starting with the wave function and the uncertainty principle. We examine the Schrodinger equation and its solutions for some simple cases: free particles, the square well, and tunneling through a barrier.

A. Piccard, E. Henriot,

P. Ehrenfest, Ed. Herzen, Th. De

Donder,

E. Schrodinger, E. Verschaffelt, W. Pauli, W. Heisenberg,

P. Debye, M. Knudsen, W.L. Bragg, H. A. Kramers, P. A. M. Dirac, A. H. Compton, L.de Broglie, M. Born, N. Bohr, 1. Langmuir, M. Planck, Mme Curie, H. A. Lorentz, A. Einstein, P. Langevin, Ch. E. Guye, C.T.R. Wilson, O.W. Richardson

r

R. H. Fowler,

L. Brillouin,

Quantum Mechanics CHAPTER-OPENING QUESTION—Guess now! The uncertainty principle states that (a) no measurement can be perfect because it is technologically impossible to make perfect measuring instruments. (b) it is impossible to measure exactly where a particle is, unless it is at rest. (¢) itis impossible to simultaneously know both the position and the momentum of a particle with complete certainty. (d) a particle cannot actually have a completely certain value of momentum. ohr’s model of the atom gave us a first (though rough) picture of what an atom is like. It proposed explanations for why there is emission and absorption of light by atoms at only certain wavelengths. The wavelengths of the line spectra and the ionization energy for hydrogen (and one-electron ions) are in excellent agreement with experiment. But the Bohr theory had important limitations. It was not able to predict line spectra for more complex atoms—not even for the neutral helium atom, which has only two electrons. Nor could it explain why emission lines, when viewed with great precision, consist of two or more very closely spaced lines (referred to as fine structure). The Bohr model also did not explain why some spectral lines were brighter than others. And it could not explain the bonding of atoms in molecules or in solids and liquids. From a theoretical point of view, too, the Bohr theory was not satisfactory: it was a strange mixture of classical and quantum ideas. Moreover, the wave—particle duality was not really resolved.

CONTENTS 38-1

Quantum Mechanics—

38-2

The Wave Function and Its Interpretation; the DoubleSlit Experiment

38-3

The Heisenberg Uncertainty

38-4

A New Theory

Principle

Philosophic Implications;

Probability Versus Determinism

38-5

The Schrodinger Equation in

*38-6

Time-Dependent Schrodinger

One Dimension—TimeIndependent Form Equation

38-7

Free Particles; Plane Waves and Wave Packets

38-8

Particle in an Infinitely Deep Square Well Potential (a

38-9

Rigid Box) Finite Potential Well

38-10

Tunneling through a Barrier

1017

FIGURE 38-1 Erwin Schrodinger with Lise Meitner (see Chapter 42).

FIGURE 38-2 Werner Heisenberg (center) on Lake Como with Enrico Fermi (left) and Wolfgang Pauli (right).

We mention these limitations of the Bohr theory not to disparage it—for it was a landmark in the history of science. Rather, we mention them to show why, in the early 1920s, it became increasingly evident that a new, more comprehensive theory was needed. It was not long in coming. Less than two years after de Broglie gave us his matter-wave hypothesis, Erwin Schrodinger (1887-1961; Fig. 38-1) and Werner Heisenberg (1901-1976; Fig. 38-2) independently developed a new comprehensive theory.

38-1 Quantum Mechanics—A New Theory The new theory, called quantum mechanics, has been extremely successful. It unifies the wave—particle duality into a single consistent theory and has successfully dealt with the spectra emitted by complex atoms, even the fine details. It explaing the relative brightness of spectral lines and how atoms form molecules. It is also much more general theory that covers all quantum phenomena from blackbody radiation to atoms and molecules. It has explained a wide range of natural phenomena and from its predictions many new practical devices have become possible. Indeed, it has been so successful that it is accepted today by nearly all physicists as the fundamental theory underlying physical processes. Quantum mechanics deals mainly with the microscopic world of atoms and light. But this new theory, when it is applied to macroscopic phenomena, must be able to produce the old classical laws. This, the correspondence principle (already mentioned in Section 37-11), is satisfied fully by quantum mechanics. This doesn’t mean we should throw away classical theories such as Newton’s laws. In the everyday world, the latter are far easier to apply and they give sufficiently accurate descriptions. But when we deal with high speeds, close to the speed of light, we must use the theory of relativity; and when we deal with the tiny

world of the atom, we use quantum mechanics.

Although we won’t go into the detailed mathematics of quantum mechanics, we will discuss the main ideas and how they involve the wave and particle properties of matter to explain atomic structure and other applications.

38—2 The Wave Function and Its Interpretation;

the Double-Slit Experiment

The important properties of any wave are its wavelength, frequency, and amplitude. For an electromagnetic wave, the frequency (or wavelength) determines whether

the light is in the visible spectrum or not, and if so, what color it is. We also have

1018

CHAPTER 38

seen (Eq. 37-3) that the frequency is a measure of photons (E = hf). The amplitude or displacement point is the strength of the electric (or magnetic) to the intensity of the wave (the brightness of the

the energy of the correspondir of an electromagnetic wave at any field at that point, and is related light).

For material particles such as electrons, quantum mechanics relates the wavelength to momentum according to de Broglie’s formula, A = h/p, Eq. 37-7.

But what corresponds to the amplitude or displacement of a matter wave? The

amplitude of an electromagnetic wave is represented by the electric and magnetic elds, E and B. In quantum mechanics, this role is played by the wave function, which is given the symbol ¥ (the Greek capital letter psi, pronounced “sigh™). Thus V¥ represents the wave displacement, as a function of time and position, of a new kind of field which we might call a “matter” field or a matter wave. To understand how to interpret the wave function ¥, we make an analogy with light using the wave—particle duality. We saw in Chapter 15 that the intensity / of any wave is proportional to the square of the amplitude. This holds true for light waves as well, as we saw in Chapter 31. That is,

I

x

E%

where E is the electric field strength. From the particle point of view, the intensity of a light beam (of given frequency) is proportional to the number of photons, N, ‘that pass through a given area per unit time. The more photons there are, the greater the intensity. Thus I

x

E?

«

N.

This proportion can be turned around so that we have N

«

EZ%

That is, the number of photons (striking a page of this book, say) is proportional to the square of the electric field strength. If the light beam is very weak, only a few photons will be involved. Indeed, it is possible to “build up” a photograph in a camera using very weak light so the effect of individual photons can be seen.

If we are dealing with only one photon, the relationship above (N oc E?) can

r > interpreted in a slightly different way. At any point the square of the electric reld strength, E2, is a measure of the probability that a photon will be at that location. At points where E? is large, there is a high probability the photon will be there; where E? is small, the probability is low. We can interpret matter waves in the same way, as was first suggested by Max Born (1882-1970) in 1927. The wave function W may vary in magnitude from point to point in space and time. If ¥ describes a collection of many electrons, then |W|? at any point will be proportional to the number of electrons expected to be found at that point.” When dealing with small numbers of electrons we can’t make very exact predictions, so |¥|* takes on the character of a probability. If ¥, which depends on time and position, represents a single electron (say, in an atom), then |W|? is interpreted as follows: |W|* at a certain point in space and time represents the probability of finding the electron at the given position and time. Thus |¥|? is often referred to as the probability density or probability distribution.

Consider two slits whose size and separation are on the order of the wavelength

of whatever we direct at them, either light or electrons, Fig. 38-3. We know very well what would happen in this case for light, since this is just Young’s double-slit experiment (Section 34-3): an interference pattern would be seen on the screen behind. If light were replaced by electrons with wavelength comparable to the slit size, they too would produce an interference pattern (recall Fig. 37-11). In the case of light, the pattern would be visible to the eye or could be recorded on film. For f}ectrons, a fluorescent screen could be used (it glows where an electron strikes). "The wave function ¥ is generally a complex quantity (that is, it involves i = \/—1 ) and hence is not directly observable. On the other hand, ||, the absolute value of W squared, is always a real quantity

and it is to |W|? that we can give a physical interpretation.

FIGURE 38-3

Parallel beam, of

light or electrons, falls on two slits ~ whose sizes are comparable to the wavelength.An interference pattern 15 observed. Light or electrons

I

Double-Slit Interference Experiment for Electrons To understand this better, we take as a thought experiment the familiar double-slit experiment, and consider it both for light and for electrons.

Intensity on screen

SECTION 38-2

1019

If we reduced the flow of electrons (or photons) so they passed through the

slits one at a time, we would see a flash each time one struck the screen. At first,

the flashes would seem random. Indeed, there is no way to predict just where any one electron would hit the screen. If we let the experiment run for a long time, and kept track of where each electron hit the screen, we would soon see “™ pattern emerging—the interference pattern predicted by the wave theory; see Fig. 38—4. Thus, although we could not predict where a given electron would strike the screen, we could predict probabilities. (The same can be said for photons.)

The probability, as mentioned before, is proportional to |¥|2. Where |W|? is zero,

FIGURE 38-4 Young’s double-slit experiment done with electrons— note that the pattern is not evident

we would get a minimum in the interference pattern. And where |V|* is a maximum, we would get a peak in the interference pattern. The interference pattern would thus occur even when electrons (or photons) passed through the slits one at a time. So the interference pattern could not arise from the interaction of one electron with another. It is as if an electron passed through both slits at the same time, interfering with itself. This is possible because an electron is not precisely a particle. It is as much a wave as it is a particle, and a wave could travel through both slits at once. But what would happen if we covered one of the slits so we knew that the electron passed through the other slit, and a little later we covered the second slit so the electron had to have passed through the first slit? The result would be that no interference pattern would be seen. We would see, instead, two bright areas (or diffraction patterns) on the screen behind the slits. This confirms our idea that if both slits are open, the screen shows an interference pattern as if each electron passed through both slits, like a wave. Yet each electron would make a tiny spot on the screen as if it were a particle. The main point of this discussion is this: if we treat electrons (and other

photo), but with more and more electrons (second and third photos), the familiar double-slit interference pattern (Chapter 34) is seen.

them as particles, then we must treat them on a probabilistic basis. The square of the wave function, |¥|?, gives the probability of finding a given electron at a given point. We cannot predict—or even follow—the path of a single electron preciselm, through space and time. )

with only a few electrons (top

particles) as if they were waves, then W represents the wave amplitude. If we treat

38-3 The Heisenberg Uncertainty Principle Whenever a measurement is made, some uncertainty is always involved. For example, you cannot make an absolutely exact measurement of the length of a table. Even with a measuring stick that has markings 1 mm apart, there will be an inaccuracy of perhaps mm or so. More precise instruments will produce more precise measurements. But there is always some uncertainty involved in a measurement, no matter how good the measuring device. We expect that by using more precise instruments, the uncertainty in a measurement can be made indefinitely small. But according to quantum mechanics, there is actually a limit to the precision of certain measurements. This limit is not a restriction on how well instruments can be made; rather, it is inherent in nature. It is the result of two factors: the wave—particle duality, and the unavoidable interaction between the thing observed and the observing instrument. Let us look at this in more detail. To make a measurement on an object without disturbing it, at least a little, is not possible. Consider trying to locate a Ping-Pong ball in a completely dark room. You grope about trying to find its position; and just when you touch it with your finger, you bump it and it bounces away. Whenever we measure the position of an object, whether it is a ball or an electron, we always touch it with something else that gives us the information about its position. To locate a lost Ping-Pong ball in a dark room, you could probe about with your hand or a stick; or you could shine a light and detect the light reflecting off the ball. Wheq\ you search with your hand or a stick, you find the ball’s position when you touc. it, but at the same time you unavoidably bump it and give it some momentum. 1020

CHAPTER 38

Quantum Mechanics

Thus you won’t know its future position. The same would be true if you observe the Ping-Pong ball using light. In order to “see” the ball, at least one photon must scatter from it, and the reflected photon must enter your eye or some other detector. When a photon strikes an ordinary-sized object, it does not appreciably ter the motion or position of the object. But when a photon strikes a very tiny object like an electron, it can transfer momentum to the object and thus greatly change the object’s motion and position in an unpredictable way. The mere act of measuring the position of an object at one time makes our knowledge of its future position imprecise. Now let us see where the wave—particle duality comes in. Imagine a thought experiment in which we are trying to measure the position of an object, say an electron, with photons, Fig. 38-5. (The arguments would be similar if we were using, instead, an electron microscope.) As we saw in Chapter 35, objects can be seen to a precision at best of about the wavelength of the radiation used due to diffraction. If we want a precise position measurement, we must use a short wavelength. But a short wavelength corresponds to high frequency and large momentum (p = h/A);

and

the more

momentum

the photons

have, the more

momentum

they can give the object when they strike it. If we use photons of longer wavelength, and correspondingly smaller momentum, the object’s motion when struck by the photons will not be affected as much. But the longer wavelength means lower resolution, so the object’s position will be less accurately known. Thus the act of observing produces an uncertainty in both the position and the momentum of the electron. This is the essence of the uncertainty principle first enunciated by Heisenberg in 1927. Quantitatively, we can make an approximate calculation of the magnitude of this effect. If we use light of wavelength A, the position can be measured at best to a precision of about A. That is, the uncertainty in the position measurement, Ax, is approximately

Electron

Light

source

(a)

Light

source

(b)

FIGURE 38-5 Thought experiment for observing an electron with a powerful light microscope. At least one photon must scatter from the electron (transferring some momentum to it) and enter the microscope.

Ax =~ A

r.lppose that the object can be detected by a single photon. The photon has a momentum p, = h/A (Eq.37-5). When the photon strikes our object, it will give some or all of this momentum to the object, Fig. 38-5. Therefore, the final x momentum of our object will be uncertain in the amount A Px since we

can’t

RN

o—A

tell beforehand

product of these uncertainties is

(Ax)(Apy)

how

much

momentum

will be

transferred. The

=~ h.

The uncertainties could be worse than this, depending on the apparatus and the number of photons needed for detection. A more careful mathematical calculation

shows the product of the uncertainties as, at best, about |

h

(38-1)

(Ax)(Apy) = -

UNCERTAINTY PRINCIPLE (Ax and Ap)

where Ap, is the uncertainty of the momentum in the x direction.” This is a mathematical statement of the Heisenberg uncertainty principle, or, as it is sometimes called, the indeterminancy principle. It tells us that we cannot measure both the position and momentum of an object precisely at the same time. The more accurately we try to measure the position so that Ax is small, the greater will be the uncertainty in momentum, Ap,. If we try to measure the momentum very accurately, then the uncertainty in the position becomes large. «ote, however, that quantum mechanics does allow simultaneous precise measurements of py and y:

that is, (Ay)(Apy) = 0.

SECTION 38-3

The Heisenberg Uncertainty Principle

1021

ACAUTION Uncertaimtios mot due to instrument deficiency, but inherent in nature (wave—particle)

The uncertainty principle does not forbid individual precise measurements, however. For example, in principle we could measure the position of an object exactly. But then its momentum would be completely unknown. Thus, although we might know the position of the object exactly at one instant, we could have no ideg @t all where it would be a moment later. The uncertainties expressed here ax“‘ ~ inherent in nature, and reflect the best precision theoretically attainable even with the best instruments.

| EXERCISE A Return to the Chapter-Opening Question on p. 1017, and answer it again now.

Another useful form of the uncertainty principle relates energy and time, and we examine this as follows. The object to be detected has an uncertainty in position Ax ~ A. The photon that detects it travels with speed ¢, and it takes a time At = Ax/c ~ A/c to pass through the distance of uncertainty. Hence, the measured time when our object is at a given position is uncertain by about Atzic

Since the photon can transfer some or all of its energy (=hf = hc/A) object, the uncertainty in energy of our object as a result is AE

=

to our

. A

The product of these two uncertainties is

(AE)(At)

= h.

A more careful calculation gives UNCERTAINTY

PRINCIPLE

h

(38-2)

(AE)(Af) = o—

(AE and At)

This form of the uncertainty principle tells us that the energy of an object can t uncertain (or can be interpreted as briefly nonconserved) by an amount AE for . time At ~ h/(2m AE). The quantity (h/2) appears so often in quantum mechanics that for convenience it is given the symbol A (“h-bar”). That is,

ho= o27h =

6.626 X 107347 2w

0TS -

1055 x 107-s.

By using this notation, Eqgs. 38-1 and 38-2 for the uncertainty principle can be written

(Ax)(Apy) = &

and

(AE)(At) = *.

We have been discussing the position and velocity of an electron as if it were a particle. But it isn’t simply a particle. Indeed, we have the uncertainty principle because an electron—and matter in general—has wave as well as particle properties. What the uncertainty principle really tells us is that if we insist on thinking of the electron as a particle, then there are certain limitations on this simplified view—namely, that the position and velocity cannot both be known precisely at the same time; and even that the electron does not have a precise position and momentum at the same time (because it is not simply a particle). Similarly, the energy can be uncertain in the amount AE for a time At = #/AE. Because Planck’s constant, A, is so small, the uncertainties expressed in the uncertainty principle are usually negligible on the macroscopic level. But at the level of atomic

sizes, the uncertainties

are significant. Because

we

consider

ordinary

objects to be made up of atoms containing nuclei and electrons, the uncertainty principle is relevant to our understanding of all of nature. The uncertainty principlem, expresses, perhaps most clearly, the probabilistic nature of quantum mechanics. . thus is often used as a basis for philosophic discussion. 1022

CHAPTER

38

Quantum Mechanics

a straight

H

Position uncertainty of an electron. An electron moves in

line with

a constant

speed

» = 1.10 X 10°m/s

which

has

been

measured to a precision of 0.10%. What is the maximum precision with which its

position could be simultaneously measured?

APPROACH The momentum is p = mw, and the uncertainty in p is Ap = 0.0010p. The uncertainty principle (Eq. 38-1) gives us the lowest Ax using the equals sign. SOLUTION The momentum of the electron is

p = mv

= (911 X 103 kg)(1.10 X 10°m/s)

= 1.00 X 10> kg-m/s.

The uncertainty in the momentum is 0.10% of this,or Ap = 1.0 X 107 kg-m/s. From the uncertainty principle, the best simultaneous position measurement will have an uncertainty of

or 110 nm. NOTE

1,055 X 107J - = 11 x 107 m, Ax *T m Ap-i =T 10x 107 kg m/s -

This is about 1000 times the diameter of an atom.

Position uncertainty of a baseball. What is the uncertainty

in position, imposed by the uncertainty principle, on a 150-g baseball thrown at (93 £ 2) mi/h = (42 £ 1) m/s?

APPROACH The uncertainty in the speed is Av = 1 m/s. We multiply Av by m to get Ap and then use the uncertainty principle, solving for Ax. SOLUTION The uncertainty in the momentum is Ap

=

mAv

=

(0.150kg)(1m/s)

=

0.15kg-m/s.

Hence the uncertainty in a position measurement could be as small as

A

f

Ax = — Ap

=

1.055 X 10734Js

= 7X103%m,

0.15kg-m/s

NOTE This distance is far smaller than any we could imagine observing or measuring. It is trillions of trillions of times smaller than an atom. Indeed, the uncertainty principle sets no relevant limit on measurement for macroscopic objects.

SOV RR PR

ESTIMATE | J/iy

lifetime

calculated. The

J/¢f

meson,

discovered in 1974, was measured to have an average mass of 3100 MeV/c? (note the use of energy units since E = mc?) and an intrinsic width of 63 keV/c? By this we mean that the masses of different J/¢y mesons were actually measured to be slightly different from one another. This mass “width” is related to the very short lifetime of the J/¢ before it decays into other particles. From the uncertainty principle, if the particle exists for only a time A¢, its mass (or rest energy) will be uncertain by AE ~ #/At. Estimate the J/¢ lifetime. APPROACH We use the energy-time version of the uncertainty principle, Eq. 38-2. SOLUTION The uncertainty of 63 keV/c? in the J/i’s mass is an uncertainty in its rest energy, which in joules is

AE

= (63 x 10°eV)(1.60 X 1079J/eV)

= 1.01 X 107J.

Then we expect its lifetime 7 (= At using Eq. 38-2) to be TR

A

106 X 107*7]:s

AE

1.01 X 1077

~

1%

10 %5,

Lifetimes this short are difficult to measure directly, and the assignment of very |short lifetimes depends on this use of the uncertainty principle. (See Chapter 43.)

-

The uncertainty principle applies also for angular variables:

(AL;)(Ad) = 4

/here L is the component of angular momentum along a given axis (z) and ¢ is the angular position in a plane perpendicular to that axis. SECTION 38-3

1023

38-4

Philosophic Implications; Probability Versus Determinism

The classical Newtonian view of the world is a deterministic one (see Section 6-" One of its basic ideas is that once the position and velocity of an object are known at a particular time, its future position can be predicted if the forces on it are known. For example, if a stone is thrown a number of times with the same initial velocity and angle, and the forces on it remain the same, the path of the projectile will always be the same. If the forces are known (gravity and air resistance, if any), the stone’s path can be precisely predicted. This mechanistic view implies that the future unfolding of the universe, assumed to be made up of particulate objects, is completely determined. This classical deterministic view of the physical world has been radically altered by quantum mechanics. As we saw in the analysis of the double-slit experiment (Section 38-2), electrons all prepared in the same way will not all end up in the same place. According to quantum mechanics, certain probabilities exist that an electron will arrive at different points. This is very different from the classical view, in which the path of a particle is precisely predictable from the initial position and velocity and the forces exerted on it. According to quantum mechanics, the position and velocity of an object cannot even be known accurately at the same time. This is expressed in the uncertainty principle, and arises because basic entities, such as electrons, are not considered simply as particles: they have wave properties as well. Quantum mechanics allows us to calculate only the probability” that, say, an electron (when thought of as a particle) will be observed at various places. Quantum mechanics says there is some inherent unpredictability in nature. Since matter is considered to be made up of atoms, even ordinary-sized objects are expected to be governed by probability, rather than by strict determinism. For example, quantum mechanics predicts a finite (but negligibly small) probability that when you throw a stone, its path will suddenly curve upward instead of following the downward-curved parabola of normal projectile motion. Quantum mechani predicts with extremely high probability that ordinary objects will behave just . the classical laws of physics predict. But these predictions are considered probabilities, not certainties. The reason that macroscopic objects behave in accordance with classical laws with such high probability is due to the large number of molecules involved: when large numbers of objects are present in a statistical situation, deviations from the average (or most probable) approach zero. It is the average configuration of vast numbers of molecules that follows the so-called fixed laws of classical physics with such high probability, and gives rise to an apparent “determinism.” Deviations from classical laws are observed when small numbers of molecules are dealt with. We can say, then, that although there are no precise deterministic laws in quantum mechanics, there are statistical laws based on probability. It is important to note that there is a difference between the probability imposed by quantum mechanics and that used in the nineteenth century to understand thermodynamics and the behavior of gases in terms of molecules (Chapters 18 and 20). In thermodynamics, probability is used because there are far too many particles to keep track of. But the molecules are still assumed to move and interact in a deterministic way following Newton’s laws. Probability in quantum mechanics is quite different; it is seen as inherent in nature, and not as a limitation on our abilities to calculate or to measure. The view presented here is the generally accepted one and is called the Copenhagen interpretation of quantum mechanics in honor of Niels Bohr’s home, since it was largely developed there through discussions between Bohr and other

prominent physicists.

Because electrons are not simply particles, they cannot be thought of as following particular paths in space and time. This suggests that a description of matter in space and time may not be completely correct. This deep and far-reachi=™h

1024

CHAPTER

38

"Note that these probabilities can be calculated precisely, just like exact predictions of probabilities at rolling dice or playing cards, but they are unlike predictions of probabilities at sporting events or for natural or man-made disasters, which are only estimates.

conclusion has been a lively topic of discussion among philosophers. Perhaps the most important and influential philosopher of quantum mechanics was Bohr. He argued that a space-time description of actual atoms and electrons is not possible. Yet description of experiments on atoms or electrons must be given in terms of space f id time and other concepts familiar to ordinary experience, such as waves and particles. We must not let our descriptions of experiments lead us into believing that atoms or electrons themselves actually move in space and time as classical particles.

38-5 The Schrodinger Equation in One

Dimension—Time-Independent Form

In order to describe physical systems quantitatively using quantum mechanics, we must have a means of determining the wave function ¥ mathematically. The basic equation (in the nonrelativistic realm) for determining W is the Schrodinger equation. We cannot, however, derive the Schrodinger equation from some higher principles, just as Newton’s second law, for example, cannot be derived. The relation F = ma was invented by Newton to describe how the motion of an object is related to the net applied force. As we saw early in this book, Newton’s second law works exceptionally well. In the realm of classical physics it is the starting point for analytically solving a wide range of problems, and the solutions it yields are fully consistent with experiment. The validity of any fundamental equation resides in its agreement with experiment. The Schrodinger equation forms part of a new theory, and it too had to be invented—and then checked against experiment, a test that it passed splendidly. The Schrodinger equation can be written in two forms: the time-dependent version and the time-independent version. We will mainly be interested in steadystate situations—that is, when there is no time dependence—and so we mainly deal with the time-independent version. (We briefly discuss the time-dependent rsion in the optional Section 38-6.) The time-independent version involves a wave function with only spatial dependence which we represent by lowercase psi, Y (x), for the simple one-dimensional case we deal with here. In three dimensions,

we write ¢(x, y, z) or ¢(r, 6, ¢).

In classical mechanics, we solved problems using two approaches: via Newton’s laws with the concept of force, and by using the energy concept with the conservation laws. The Schrodinger equation is based on the energy approach. Even though the Schrédinger equation cannot be derived, we can suggest what form it might take by using conservation of energy and considering a very simple case: that of a free particle on which no forces act, so that its potential energy U is constant. We assume that our particle moves along the x axis, and since no force acts on it, its momentum remains constant and its wavelength (A = h/p) is fixed. To describe a wave for a free particle such as an electron, we expect that its wave function will satisfy a differential equation that is akin to (but not identical to) the classical wave equation. Let us see what we can infer about this equation. Consider a simple traveling wave of a single wavelength A whose wave displacement, as we saw in Chapter 15 for mechanical waves and in Chapter 31 for electromagnetic waves, is given by A sin(kx — wt), or more generally as a superposition of sine and cosine: Asin(kx — wt) + Bcos(kx — wt). We are only interested in the spatial dependence, so we consider the wave at a specific moment, say ¢ = 0. Thus we write as the wave function for our free particle Y(x)

=

Asinkx

+ Bcoskx,

(38-3a)

where A and B are constants’ and k = 27/A (Eq. 15-11). For a particle of mass m and velocity », the de Broglie wavelength is A = h/p, where p = mv is the

fiarticle’s momentum. Hence k

=

2 )

—-————

2mp %

——=

p 5

—

"In quantum mechanics, constants can be complex (i.e., with a real and/or imaginary part).

— (38-3b)

SECTION 38-5

1025

One requirement for our wave equation, then, is that it have the wave function Y(x) as given by Eq. 38-3 as a solution for a free particle. A second requirement is that it be consistent with the conservation of energy, which we can express as 2

£2m +v =k

>"

where E is the total energy, U is the potential energy, and (since we are considering the nonrelativistic realm) the kinetic energy K of our particle of mass m is

K = 3mv? = p*/2m. Since p = #k (Eq.38-3b),we can write the conservation of

energy condition as

hi2k? +U = E. (38-4) 2m Thus we are seeking a differential equation that satisfies conservation of energy (Eq. 38-4) when ¢(x) is its solution. Now, note that if we take two derivatives of our expression for ¢(x), Eq. 38-3a, we get a factor —k? multiplied by #(x):

dp(x) _d i = d’

.

(Asinkx + Bcoskx)

B=

k(Acoskx

:

— Bsinkx)

d

d;zll(x) = k_ch- (Acoskx — Bsinkx) = —k*(Asinkx + Bcoskx) = —k*(x). Can this last term be related to the k* term in Eq. 38-4? Indeed, if we multiply this

last relation by —#?/2m, we obtain

The right side is just the first term on the left in Eq. 38—4 multiplied by ¢/(x). If we multiply Eq. 38-4 through by /(x), and make this substitution, we obtain

SCHRODINGER EQUATION

Lt

(time-independent form)

2 g2

2m dx

f(x) + U(x)(x) = Eg(x).

]

(38-5)

S

This is, in fact, the one-dimensional time-independent Schrodinger equation, where for generality we have written U = U(x). It is the basis for solving problems in nonrelativistic quantum mechanics. For a particle moving in three dimensions there would be additional derivatives with respect to y and z (see Chapter 39). Note that we have by no means derived the Schrodinger equation. Although we have made a good argument in its favor, other arguments could also be made which might or might not lead to the same equation. The Schrodinger equation as written (Eq. 38-5) is useful and valid only because it has given results in accord with experiment for a wide range of situations. There are some requirements we impose on any wave function that is a solution of the Schrodinger equation in order that it be physically meaningful. First, we insist that it be a continuous function; after all, if |¢/|* represents the probability of finding a particle at a certain point, we expect the probability to be continuous from point to point and not to take discontinuous jumps. Second, we want the wave function to be normalized. By this we mean that for a single particle, the probability of finding the particle at one point or another (i.e., the probabilities summed over all space) must be exactly 1 (or 100%). For a single particle, [s|* represents the probability of finding the particle in unit volume. Then

Wldv

(38-6a)

is the probability of finding the particle within a volume dV, where ¢ is the value of the wave function in this infinitesimal volume dV. For the one-dimensional case, dV = dx, so the probability of finding a particle within dx of position x is

W (x) [ dx. 1026

CHAPTER 38

Quantum

Mechanics

(38-6b)

Then the sum of the probabilities over all space—that is, the probability of finding

the particle somewhere—becomes

[

all space

(38-60)

wpav = [lpax =1

chis is called the normalization condition, and the integral is taken over whatever region of space in which the particle has a chance of being found, which is often all of space, from x = —oc0 to x = 00

*38—-6 Time-Dependent Schrodinger Equation

The more general form of the Schrodinger equation, including time dependence, for a particle of mass m moving in one dimension, is

W) 2m 9x*

U)W

= it d®l)

(38-7)

This is the time-dependent Schriédinger equation; here U(x ) is the potential energy_of the particle as a function of position, and i is the imaginary number i = V—1. For a particle moving in three dimensions, there would be additional derivatives with respect to y and z, just as for the classical wave equation discussed in Section 15-5. Indeed, it is worth noting the similarity between the Schrodinger wave equation for zero potential energy (U = 0) and the classical wave equation:

#*D/at* = v**D/ax*, where D is the wave displacement (equivalent of the wave

function). In both equations there is the second derivative with respect to x; but in the Schrodinger equation there is only the first derivative with respect to time, whereas the classical wave equation has the second derivative for time. As we pointed out in the preceding Section, we cannot derive the time-dependent Schrodinger equation. But we can show how the time-independent Schrodinger equaon (Eq. 38-5) is obtained from it. For many problems in quantum mechanics, it is possible to write the wave function as a product of separate functions of space and time:

W(x,t) = ¢(x)f(1).

Substituting this into the time-dependent Schrodinger equation (Eq. 38-7), we get: 2 d? d

-2SE s wewr) = i

0.

We divide both sides of this equation by ¢(x)f(f) and obtain an equation that involves only x on one side and only f on the other:

A2

1

di(x)

Toampmae

YW

1

df(t)

= hrma

This separation of variables is very convenient. Since the left side is a function only of x, and the right side is a function only of ¢, the equality can be valid for all values of x and all values of ¢ only if each side is equal to a constant (the same constant), which we call C:

_h

1

ma

dzt/f( )

_

(38-8a)

1 df(t) _ ih o=

(38-8b)

gl

U

=€

We multiply the first of these (Eq. 38—8a) through by (x) and obtain

2 d%y(x

"dez(

(38-8¢)

) + U(x)p(x) = C(x).

’Thls we recognize immediately as the time-independent Schrédinger equation, -q. 38-5, where the constant C equals the total energy E. Thus we have obtained the time-independent form of Schrodinger’s equation from the time-dependent form.

*SECTION 38-6

Time-Dependent Schrodinger Equation

1027

Equation 38-8b is easy to solve. Putting C = E, we rewrite Eq. 38-8b as

af(t)

. E

(note that since i> = —1, i = —1/i), and then as L(t)

f(t)

=

-

"'lgdt

ho

We integrate both sides to obtain LE Inf(t) = —i fit or

i(E

fey = )

Thus the total wave function is

V(x,t)

i(E

= ¢(x)e"(fi)’,

(38-9)

where (x) satisfies Eq. 38-5. It is, in fact, the solution of the time-independent Schrodinger equation (Eq. 38-5) that is the major task of nonrelativistic quantum mechanics. Nonetheless, we should note that in general the wave function W (x, t) is a complex function since it involves i = V/—1. It has both a real and an imaginary part.” Since W(x,t) is not purely real, it cannot itself be physically measurable. Rather it is only | W% which is real, that can be measured physically. Note also that

ol = 8 =1, i|E

so |f(t)]* = 1. Hence the probability density in space does not depend on time:

¥ (x,0)F = |p(x)P.

We thus will be interested only in the time-independent Schrodinger equation, Eq. 38-5, which we will now examine for a number of simple situations.

38-7

Free Particles; Plane Waves Wave Packets

ana

>

A free particle is one that is not subject to any force, and we can therefore take its potential energy to be zero. (Although we dealt with the free particle in Section 38-5 in arguing for Schrodinger’s equation, here we treat it directly using Schrodinger’s equation as the basis.) Schrodinger’s equation (Eq. 38-5) with U(x) = 0 becomes

2 dy(x)

2m

dx?

=

Ey(x),

which can be written

d4

Pt

2mE

b

This is a familiar equation that we encountered in Chapter 14 (Eq. 14-3) in connection with the simple harmonic oscillator. The solution to this equation, but with appropriate variable changes® for our case here, is ¢

=

Asinkx

+ Bcoskx,

[free particle]

(38-10)

where

k =

/Z’ZE :

Since U = 0, the total energy E of the particle is E = 3mv® = p?/2m is the momentum); thus B = e e i "Recall that e

1028

CHAPTER 38

= cos§ — isiné.

(38-11a) (where p (38-11b)

*In Eq. 14-3, 1 becomes x and w becomes k = \V2mE/#*. (Don’t confuse this k with the spring constant k of Chapter 14.) We could also write our solution (Eq. 38-10) as ¢ = Acos (kx + ¢)

where ¢ is a phase constant.

So a free particle of momentum p and energy E can be represented by a plane

¥

wave that varies sinusoidally. If we are not interested in the phase, we can choose

B =0

in Eq. 38-10, and we show this sine wave in Fig. 38—6a.

(‘ DGRBS Free electron. An electron with energy E = 6.3¢eV is in free space (where U = 0). Find (a) the wavelength A and (b) the wave function ¢ for the electron, assuming B = 0.

!

APPROACH The wavelength A = 27/k (Eq.38-11b) where the wave number & is given by Eq. 38-11a. The wave function is given by ¢ = Asin kx. SOLUTION @ a = 2m

_

2mh

—

)

—u%—x

2mE

k

-

k = 27

(a) "

27 (1.055 X 107%7J+s)

V2(9.11 x 107 kg)(6.3eV)(1.60 X 107 J/eV) H

10 o

¢ 1T

—

(b) FIGURE 38-6

649 mp,

Ax (a) A plane wave

describing a free particle. (b) A wave

packet of “width” Ax.

A

=

1,28 X 10"°m™?,

SO

¢ = Asinkx

= Asin[(1.28 X 10°°m")(x)].

Note in Fig. 38-6a that the sine wave will extend indefinitely” in the +x nd —x directions. Thus, since |y|* represents the probability of finding the particle, the particle could be anywhere between x = —oo and x = co. This is fully consistent with the uncertainty principle (Section 38-3): the momentum of the particle was given and hence is known precisely (p = %k), so the particle’s position must be totally unpredictable. Mathematically, if Ap =0, Ax = #/Ap = oo. To describe a particle whose position is well localized—that is, it is known to be within a small region of space—we can use the concept of a wave packet. Figure 38—6b shows an example of a wave packet whose width is about Ax as shown, meaning that the particle is most likely to be found within this region of space. A well-localized particle moving through space can thus be represented by a moving wave packet. A wave packet can be represented mathematically as the sum of many plane waves (sine waves) of slightly different wavelengths. That this will work can be seen by looking carefully at Fig. 16-17. There we combined only two nearby frequencies (to explain why there are “beats™) and found that the sum of two sine waves looked like a series of wave packets. If we add additional waves with other nearby frequencies, we can eliminate all but one of the packets and arrive at Fig. 38-6b. Thus a wave packet consists of waves of a range of wavelengths; hence it does not have a definite momentum p (= h/A), but rather, a range of momenta. This is consistent with the uncertainty principle: we have made Ax small, so the momentum cannot be precise; that is, Ap cannot be zero. Instead, our particle can be said to have a range of momenta, Ap, or to have an uncertainty in its momentum, Ap. It is not hard to show, even for this simple situation (see Problem 20), that Ap ~ h/Ax, in accordance with the uncertainty principle. 'Such an infinite wave makes problems for normalization since

[

[

dx = A*[* sin’kxdx

is

rnfinite for any nonzero value for A. For practical purposes we can usually normalize the waves (A # 0) by assuming that the particle is in a large but finite region of space. The region can be chosen large enough so that momentum is still rather precisely fixed.

SECTION 38-7

Free Particles; Plane Waves and Wave Packets

1029

38-8

Particle in an Infinitely Deep Square Well Potential (a Rigid Box)

The Schrodinger equation can be solved analytically only for a few possible forrn‘q of the potential energy U. We consider some simple cases here which at first may not seem realistic, but have simple solutions that can be used as approximations to understand a variety of phenomena. In our first case, we assume that a particle of mass m is confined to a one-dimensional box of width £ whose walls are perfectly rigid. (This can serve as an approximation for an electron in a metal, for example.) The particle is trapped in this box and collisions with the walls are perfectly elastic. The potential energy for this situation, which is commonly known as an infinitely deep square well potential or rigid box, is shown in Fig. 38-7. We can write the potential energy U(x) as =

x=

Ux) Ux)

X

=0 = =

0

nar

£

t

FIGURE 38-10

\/: sin

for the states with n = 1,2, 3, and 10.

waves on a string—see Fig. 15-26. This is not surprising since the wave function

=

solutions, Eq. 38-14, are the same as for the standing waves on a string, and the

SECTION 38-8

The probability

distribution for a particle in a rigid box

The amplitude A is the same for all the quantum numbers. Figure 38-9 shows the wave functions (Eq. 38-14) for n = 1, 2,3, and 10. They look just like standing condition k€ = nr is the same in the two cases (page 414). Figure 38-10 shows the probability distribution, [, for the same states (n=1,2,3,10) for which ¢ is shown in Fig. 38-9. We see immediately that the particle is more likely to be found in some places than in others. For example, in the ground state (n = 1), the electron is much more likely to be found near the center of the box than near the walls. This is clearly at variance with classical ideas, which predict a uniform probability density—the particle would be as likely be found at one point in the box as at any other. The quantum-mechanical orobability densities for higher states are even more complicated, with areas of low probability not only near the walls but also at regular intervals in between.

=t

Wave functions

.

|¢|2’

~

-

-{n=10

N___ N __Nn=3

/\ F=-----—----= x=0

Particle in an Infinitely Deep Square Well Potential (a Rigid Box)

n=1 x=1{

1031

three well atom from

Electron in an infinite potential well. (a) Calculate the

lowest energy levels for an potential of width £ = 1.00 in its ground state). (b) If a the n = 2 state to the n =

electron trapped in an infinitely deep square X 10 m (about the diameter of a hydrogen photon were emitted when the electron jumpsfl 1 state, what would its wavelength be? :

APPROACH The energy levels are given by Eq. 38-13.In (b), hf = hc/A = E, — E,. SOLUTION

E

(a) The ground state

K

= Po8m

(n = 1) has energy

(6.63 X 1077 -s)?

=

8(9.11 x 107 kg)(1.00 X 1070 m)?

= 6.03 X 10787,

In electron volts this is

6.03 X 10718 ] = 377eV. E LT = 160 x 10 °J/eV ¢

Then

E, = (20E, = 151eV E = (37E, = 339eV.

(b) The energy difference is E, — E; = 151 eV — 38eV =113eV or 1.81 X 1077], and this would equal the energy of the emitted photon (energy conservation). Its wavelength would be

¢

he

A1

L

(663 x1077J-5)(3.00 X 108 m/s)

s

= 110 X 10%m

or 11.0 nm, which is in the ultraviolet region of the spectrum. EXERCISE B The wavelength of a photon emitted in an

n =3

(@) 0.062 nm, (b) 620 nm, (c) 301 nm, (d) 3.2 X 107 m, (e) 4.1 nm.

DGV

S

Calculating

a

normalization

to

n =1

constant. Show

normalization constant A for all wave functions describing infinite potential well of width £ has a value of A = \/2/{. APPROACH

transition is

that

the

-

a particle in an

The wave functions for various » are Y

=

Asinn—zx-

To normalize ¢, we must have (Eq. 38-15)

1=

4

]

J WlEdx = J A sin? % g 0

0

t

SOLUTION We need integrate only from 0 to £ since ¢ = 0 for all other values of x. To evaluate this integral we let

6 = nmx/f

and use the trigonometric

identity sin?6 = 3(1 — cos26). Then, with dx = £d6/nm, we have 1 = AZJ

nw 0

2

sinzf)(i) de = —jflj nm 2na 2

= %(0

_ A

nmw 0

(1 — cos26)d

— 1sin26)

nmw

0

2 Thus A*> = 2/€ and

A 1032

CHAPTER 38

Quantum Mechanics

s2

S

SN BV A ESTIMATE | Probability near center of rigid box. An electron is in an infinitely deep square well potential of width £ = 1.00 X 107°m. If the

electron is in the ground state, what is the probability of finding it in a region of

f‘ width

Ax = 1.0 X 107> m at the center of the well (at x = 0.50 X 1071° m)?

'APPROACH The probability of finding a particle in a small region of width dx is [¢* dx (Eq. 38-6b). Using A from Example 38-6, the wave function for the ground state is 2 Y(x) = ?sin WTBC SOLUTION The n = 1 curve in Fig. 38-9 shows that ¢ is roughly constant near the center of the well. So we can avoid doing an integral over dx and just set dx ~ Ax

ly|* 2 Ax

and find

B

2

mX A PECa T [ K57 ]Ax

=

0.50 x 1071 4 sinz[w( . m) ](1.0 X 102m) (1.00 X 107'°m) (1.00 x 107" m)

= 0.02.

The probability of finding the electron in this region at the center of the well is thus 2%. NOTE Since Ax = 1.0 X 1072 m is 1% of the well width of 1.00 X 107" m, our result of 2% probability is not what would be expected classically. Classically, the electron would be equally likely to be anywhere in the box, and we would expect

\t_he probability to be 1% instead of 2%.

SOTINEET RN

Probability of e~ in j of box. Determine the probability of

findng an electron in the left quarter of a rigid box—i.e., between one wall at x = 0 and position x = £/4. Assume the electron is in the ground state. APPROACH

We

cannot

make

the assumption

we

did in Example

38-7

that

[¢|> ~ constant and Ax is small. Here we need to integrate [* dx from x = 0 to

|,

x = {/4, which is equal to the area under the curve shown colored in Fig. 38-11.

SOLUTION The wave function in the ground state is ¢, = V2/{ sin(mx/l). To find the probability of the electron in the left quarter of the box, we integrate just as in Example 38-6 but with different limits on the integral (and now we know that A = \/2/f). That is, we set § = mx/f (then x = £/4 corresponds

to

0 =m/4)

and

dx = (t/m)do,

J

use the identity

€4

9

(U4

lpfPdx == J

i 20)( £

2],

1

= —(0

1 1 = - — — 4

26

™

1

#/4

2

0

= —sin20>

2

Ground-state

we show the area under the curve

t

/ (1

FIGURE 38-11

probability distribution |¢|* for an

electron in a rigid box. Same as n = 1 graph of Fig. 38-10; but here

sinz(— x) dx

w/4

v

Thus, with

-

¢

A

sin’f = j(1 — cos26).

'

from x = 0 to x = £/4 which represents the probability of finding

the electron in that region.

= 0.091.

NOTE The electron spends only 9.1% of its time in the left quarter of the box. Eassically it would spend 25%. r

EXERCISE C What is the probability of finding the electron between x = £/4 and | x = £/2? (Do you need to integrate?) (a) 9.1%; (b) 18.2%; (c) 25%; (d) 33%; (e) 40.9%.

SECTION 38-8

Particle in an Infinitely Deep Square Well Potential (a Rigid Box)

1033

Most likely and average positions. Two interesting quantities

are the most likely position of a particle and the average position of the particle. Consider an electron in a rigid box of width 1.00 X 107°m in the first excited state n = 2. (a) What is its most likely position? (b) What is its average position? APPROACH To find (a) the most likely position (or positions), we find the maximut.. value(s) of the probability distribution |y/|? by taking its derivative and setting it equal

to zero. For (b), the average position, we integrate X = [{x[y[* dx.

2 . (2 mn(lx), { £

o)

This quantity is zero when either the sine is zero (27x/€ = 0, 7, 2m, -++), or cosine is zero (27wx/t = 7/2,3m/2, -++). The maxima and minima occur at x £/2,0,and x = £/4,32/4. The latter (£/4,3€/4) are the maxima—see n = 2 curve of Fig. 38-10; the others are minima. [To confirm, you can take

the = 0, the the

SOLUTION 2

(a@) The wave . ,(2

lr(x)? = 1 sin’

function for n =2 is . 5.1

¢(x) =/~

—%x . To find maxima and minima, we set d|y|*/dx = O d o227 dx|¢r| =7 (2) 7

. [2¢% 27 sm( 7 x) cos( 7 x).

second derivative, d2|y|*/dx?, which is >0 for minima and

h

2ml

_

In (a) where

E = h*/8 mf* so

66X10%]s

2(10“kg)(107*m)

~ 3% 10

m/s

This is a speed so small that we could not measure it and the object would seem at rest, consistent with classical physics. (b) Given v = 10 m/100s = 1075 m/s, the kinetic energy of the bacterium is

E = jmv*

= 3(10%kg)(10° m/s)* = 0.5 X 1077,

From Eq. 38-13, the quantum number of this state is o

NOTE

\

E(Smfl

W

2

)

_

\/(

0.5 X 1074

)(

J)(8)(10 ¥ )(_

g)(

(6.6 x 1077 s)?

kg)(10™*

I’l’l)

m)?

~

3

X

1010.

This number is so large that we could never distinguish between adjacent

energy states (between n = 3 X 10 and 3 X 10'° + 1). The energy states would

1034

CHAPTER 38

appear to form a continuum. Thus, even though the energies involved here ar, small ( 0?7 (¢) What is the probability of finding the particle between x = Onm and x = 0.50 nm? 59. A 7.0-gram pencil, 18cm long, is balanced on its point. Classically, this is a configuration of (unstable) equilibrium, so the pencil could remain there forever if it were perfectly placed. A quantum mechanical analysis shows that the pencil must fall. (¢) Why is this the case? (b) Estimate (within a factor of 2) how long it will take

the pencil to hit the table if it is initially positioned as well as possible? [Hinr: Use the uncertainty principle in its angular form to angle ¢g = Ad¢.]

obtain

an

expression

for

the

and R =1 — T. Consider that the particle is an electron and it is incident on a rectangular barrier of height Uy = 10eV and width £ = 1.0 X 10"'°m. (a) Calculate T and R for the electron from E/Uy = 0 to 10, in steps of 0.1. Make a single graph showing the two curves of 7 and R as a function of £/Uj.

(b) From the graph determine the energies (E/Up) at which

the electron will have

20%, 50%, and 80%.

initial

transmission

probabilities of 10%,

Ux)

*Numerical/ Computer

___________________

*60. (ITI) An electron is trapped in the ground state of an infinite

_______________ U, B__s

potential well of width

£ = 0.10nm.

estimated

38-7

The probability that

the electron will be found in the central 1% of the well was

in Example

by

|y|*Ax.

Use

Ux)=0

numerical

X x=0

methods to determine how large Ax could be to cause less than a 10% error in such an estimate.

E

FIGURE 38-21

x=4 Problem 61.

Answers to Exercises f\:

(c).

C: (e).

L H (e).

D: (a).

General Problems

1043

A

neon

tube

is a thin

glass

tube,

moldable into various shapes, filled with neon (or other) gas that glows with a particular color when a current at high voltage passes through it. Gas atoms, excited to upper energy levels, jump down to lower energy levels and emit light (photons) whose wavelengths (color) are characteristic of the type of gas. In this Chapter we study what quantum mechanics tells us about atoms, their wave functions and energy levels, including the effect of the exclusion principle. We also discuss interesting applications such as lasers and holography.

Quantum Mechanics of

Atoms

CHAPTER-OPENING

QUESTIONS—Guess

now!

1. Thousands of hydrogen atoms are all in their ground state. Which statement below is true?

CONTENTS

(a) All of the atoms have the electron in a circular orbit at the Bohr radius. (b) All of the atoms have the electron in the same orbit, but it’s not the same

orbit for the ground state in the Bohr model.

39-1

Quantum-Mechanical View of Atoms

39-2

Hydrogen Atom: Schrodinger Equation and Quantum Numbers

39-3

Hydrogen Atom Wave Functions

39-4

Complex Atoms;the Exclusion Principle

39-5

Periodic Table of Elements

39-6

X-Ray Spectra and Atomic Number

(¢) No two electrons in an atom can occupy the same quantum state.

Magnetic Dipole Moment; Total Angular Momentum

(e) Electrons must be excluded from the nucleus because only positive charges are allowed in the nucleus to give rise to stable electronic orbits according to Bohr’s model.

*39-7 39-8

Fluorescence and Phosphorescence

39-9

Lasers

#39-10

1044

Holography

(¢) The electron is not in an actual orbit, but the distance between the nucleus and the electron is the same in all of the atoms. (d) If the distance from the nucleus to the electron could be measured, it would

be

found

at different

locations

distance would be the Bohr radius.

in different

atoms. The

most

probable

2. The state of an electron in an atom can be specified by a set of quantum numbers. Which of the following statements are valid? (a) No two electrons in the universe can be identical to each other. (b) All electrons in the universe are identical.

(d) There cannot be more than one electron in a given electron orbit.

t the beginning of Chapter 38 we discussed the limitations of the Bohr theory of atomic structure and why a new theory was needed. Although the Bohr theory had great success in predicting the wavelengths of light emitted and absorbed by the hydrogen atom, it could not do so for more complex atom Nor did it explain fine structure, the splitting of emission lines into two or more close., spaced lines. And, as a theory, it was an uneasy mixture of classical and quantum ideas.

Quantum mechanics came to the rescue in 1925 and 1926, and in this Chapter we examine the quantum-mechanical theory of atomic structure, which is far more

complete than the old Bohr theory.

f3 9-1

Quantum-Mechanical View of Atoms

Although the Bohr model has been discarded as an accurate description of nature, nonetheless, quantum mechanics reaffirms certain aspects of the older theory, such as that electrons in an atom exist only in discrete states of definite energy, and that a photon of light is emitted (or absorbed) when an electron makes a transition

from one state to another. But quantum

mechanics is a much deeper theory, and

has provided us with a very different view of the atom. According to quantum mechanics, electrons

do not exist in well-defined

circular orbits as in the Bohr

theory. Rather, the electron (because of its wave nature) can be thought of as spread out in space as if it were a “cloud.” The size and shape of the electron cloud can be calculated for a given state of an atom. For the ground state in the hydrogen atom, the solution of the Schrodinger equation, as we will discuss in more detail in Section 39-3, gives

W) =

Z=eh. 1

_r

Here ¢(r) is the wave function as a function of position, and it depends only on the radial distance r from the center, and not on angular position 6 or ¢. (The constant ry happens to be equal to the first Bohr radius.) Thus the electron cloud, whose density is || for the ground state of hydrogen is spherically symmetric as shown in Fig. 39-1. The extent of the electron cloud at its higher densities

roughly istinct Not all Chapter.

= FIGURE 39-1

Electron cloud or

indicates the “size” of an atom, but just as a cloud may not have a “probability distribution” |y[* for the border, atoms do not have a precise boundary or a well-defined size. ~ ground state of the hydrogen atom, electron clouds have a spherical shape, as we shall see later in this as seen fromafar. The circle But note that ¢(r), while becoming extremely small for large r (see the ~ Tepresents the Bohr radius r,, The

equation above), does not equal zero in any finite region. So quantum mechanics . . suggests that an atom is not mostly empty space. (Indeed, since ¢y — 0 only for A : : ; . r — oo, we might question the idea that there is any truly empty space in the

gms iepifesent s hypotieted

universe.) X ) . The electron cloud can be interpreted from either the particle or the wave viewpoint. Remember that by a particle we mean something that is localized in space—it has a definite position at any given instant. By contrast, a wave is spread out in space. The electron cloud, spread out in space as in Fig. 39-1, is a result of the wave nature of electrons. Electron clouds can also be interpreted as probability distributions (or probability density) for a particle. As we saw in Section 38-3, we cannot predict the path an electron will follow. After one measurement of its position we cannot predict exactly where it will be at a later time. We can only calculate the probability that it will be found at different points. If you were to make 500 different measurements of the position of an electron, considering it as a particle, the majority of the results would show the electron at points where the probability is high (darker area in Fig. 39-1). Only occasionally would the electron be found where the probability is low.

electron (denser cloud).

etection of an electron at that point: dots closer together represent

.. probable presence of an

39-2 Hydrogen Atom: Schrodinger

Equation and Quantum Numbers

f-The hydrogen atom is the simplest of all atoms, consisting of a single electron of charge —e moving around a central nucleus (a single proton) of charge +e. It is with hydrogen that a study of atomic structure must begin.

SECTION 39-2

Hydrogen Atom: Schrédinger Equation and Quantum Numbers

1045

The Schrédinger equation (see Eq. 38-5) includes a term containing the potential energy. For the hydrogen (H) atom, the potential energy is due to the Coulomb force between electron and proton:

U

0

5

ro 2rg 3ro 4ro I

T

T

T

1

—10eV T+

R

-

2

4mrey

r

where r is the radial distance from the proton (situated at r = 0) to the electron. See Fig. 39-2. The (time-independent) Schrodinger equation, which must now be written in three dimensions, is then

—20eV + —30eV +

fiZ

FIGURE 39-2 Potential energy U(r) for the hydrogen atom. The radial distance r of the electron from the nucleus is given in terms of the Bohr radius rg.

35%3p*

3s3p°

3s%3p° 45!

22

Ti

3d%4s?

23

A%

3d*4s?

24

Cr

3d%4s!

25

Mn

3d’4s?

26

Fe

3d%s?

"Names of elements can be found in Appendix F.

Periodic Table of Elements

1053

CONCEPTUAL

EXAMPLE 39-5 | Electron configurations. Which of the following

electron configurations are possible, and (b) 15252 p®3s23p°4s% (c) 1522522 p2d"?

which

are

not:

(a)

1s%2s5%2p®3s®

RESPONSE («) This is not allowed, because too many electrons (three) are showp\ in the s subshell of the M (n = 3) shell. The s subshell has m, = 0, with tw slots only, for “spin up” and “spin down” electrons. (b) This is allowed, but it is an excited state. One of the electrons from the 3p subshell has jumped up to the 4s subshell. Since there are 19 electrons, the element is potassium. (c¢) This is not allowed, because there is no d (£ = 2) subshell in the n = 2 shell (Table 39-1). The outermost electron will have to be (at least) in the n = 3 shell. | EXERCISE F Write the complete ground-state configuration for gallium, with its 31 electrons.

The grouping of atoms in the Periodic Table is according to increasing atomic number, Z. There is also a strong regularity according to chemical properties. Although this is treated in chemistry textbooks, we discuss it here briefly because it is a result of quantum mechanics. See the Periodic Table inside the back cover. All the noble gases (in column VIII of the Periodic Table) have completely filled shells or subshells. That is, their outermost subshell is completely full, and the electron distribution is spherically symmetric. With such full spherical symmetry, other electrons are not attracted nor are electrons readily lost (ionization energy is high). This is why the noble gases are chemically inert (more on this when we discuss molecules and bonding in Chapter 40). Column VII contains the halogens, which lack one electron from a filled shell. Because of the shapes of the orbits (see Section

40-1), an additional electron can be accepted

from

another

atom,

and

hence these elements are quite reactive. They have a valence of —1, meaning that when an extra electron is acquired, the resulting ion has a net charge of —le. Column I of the Periodic Table contains the alkali metals, all of which have a single outer s electron. This electron spends most of its time outside the inner closed shells and subshells which shield it from most of the nuclear charge. Indeed, it is relatively far from the nucleus and is attracted to it by a net charge of only abo* +1e, because of the shielding effect of the other electrons. Hence this oute. electron is easily removed and can spend much of its time around another atom, forming a molecule. This is why the alkali metals are chemically active and have a valence of +1. The other columns of the Periodic Table can be treated similarly. The presence of the transition elements in the center of the Table, as well as the lanthanides (rare earths) and actinides below, is a result of incomplete inner shells. For the lowest Z elements, the subshells are filled in a simple order: first 1s, then 2s, A CAUTION followed by 2p, 3s, and 3p. You might expect that 3d (n = 3,£ = 2) would be filled Subshells are not always — next, but it isn’t. Instead, the 4s level actually has a slightly lower energy than the 3d

filled in “order”

(due to electrons interacting with each other), so it fills first (K and Ca). Only then

does the 3d shell start to fill up, beginning with Sc, as can be seen in Table 39-4. (The 4s and 3d levels are close, so some elements have only one 4s electron, such as Cr.) Most of the chemical properties of these transition elements are governed by the relatively loosely held 4s electrons, and hence they usually have valences of +1 or +2. A similar effect is responsible for the lanthanides and actinides, which are shown at the bottom of the Periodic Table for convenience. All have very similar chemical properties, which are determined by their two outer 6s or 7s electrons, whereas the different numbers of electrons in the unfilled inner shells have little effect.

39-6 X-Ray Spectra and Atomic Number The line spectra of atoms in the visible, UV, and IR regions of the EM spectrum are mainly due to transitions between states of the outer electrons. Much of the charge

of the nucleus is shielded from these electrons by the negative charge on the inner

electrons. But the innermost electrons in the n = 1 shell “see” the full charge of t}K

1054

CHAPTER 39

nucleus. Since the energy of a level is proportional to Z? (see Eq. 37-14), for ¢ atom with Z = 50, we would expect wavelengths about 50° = 2500 times shorter than those found in the Lyman series of hydrogen (around 100nm), or 107 to 10! nm. Such short wavelengths lie in the X-ray region of the spectrum.

are characteristic of the target material used. Indeed, we can explain the peaks by

imagining that the electrons accelerated by the high voltage of the tube can reach sufficient energies that when they collide with the atoms of the target, they can knock out one of the very tightly held inner electrons. Then we explain these characteristic X-rays (the peaks in Fig. 39-11) as photons emitted when an electron in an upper state drops down to fill the vacated lower state. The K lines result from transitions into the K shell (n = 1). The K, line consists of photons emitted in a transition that originates from the n =2 (L) shell and drops to the n =1 (K) shell, whereas the Ky line reflects a transition from the n = 3 (M) shell down to the K shell. An L line, on the other hand, is due to a transition into the L shell, and so on.

Measurement of the characteristic X-ray spectra has allowed a determination

of the inner energy levels of atoms. It has also allowed the determination of

Z values for many atoms, since (as we have seen) the wavelength of the shortest characteristic X-rays emitted will be inversely proportional to Z? Actually, for an electron jumping from, say, the n = 2 tothe n = 1 level (K, line), the wavelength is inversely proportional to (Z — 1)? because the nucleus is shielded by the one electron that still remains in the 1s level. In 1914, H. G. J. Moseley (1887-1915) found that a plot of V' 1/A vs. Z produced a straight line, Fig. 39-12 where A is the rflavelength of the K, line. The Z values of a number of elements were determined oy fitting them to such a Moseley plot. The work of Moseley put the concept of atomic number on a firm experimental basis. to n =1 a photon?

X-ray wavelength. Estimate the wavelength for an

transition in molybdenum

APPROACH

(Z = 42).

What

is the energy

@ X-ray intensity

X-rays are produced when electrons accelerated by a high voltage strike the metal target inside the X-ray tube (Section 35-10). If we look at the spectrum of wavelengths emitted by an X-ray tube, we see that the spectrum consists of two arts: a continuous spectrum with a cutoff at some A, which depends only on he voltage across the tube, and a series of peaks superimposed. A typical example is shown in Fig. 39-11. The smooth curve and the cutoff wavelength A, move to the left as the voltage across the tube increases. The sharp lines or peaks (labeled K, and Ky in Fig. 39-11), however, remain at the same wavelength when the voltage is changed, although they are located at different wavelengths when different target materials are used. This observation suggests that the peaks

Ag

0.1 Wavelength, A (nm)

fiGURE 39-11

0.2

Spectrum of

X.rays emitted from a molybdenum target in an X-ray tube operated at 50kV.

~ FIGURE 39-12

for K, X-ray lines.

Plot of V1/Avs.Z

: ‘/—x‘

Z

n = 2

of such

We use the Bohr formula, Eq. 37-15 for 1/A, with Z?* replaced by

(Z - 1) = (41)%

SOLUTION

Equation 37-15 gives

=A1 where

n = 2 and

e‘m

(

e)

n’ = 1. We substitute in values:

% = (1.097 x 107m™) (41)2(%— - i—) = 138 X 10°0m". So

PPREER SU 138 x 1001

_

1 . o/enm

This is close to the measured value (Fig. 39-11) photons would have energy (in eV) of:

A

E = hf = he

A

of 0.071 nm. Each

(6.63 X 107*J-5)(3.00 X 10°m/s)

(72 %10 m)(1.60 x 107°J/eV)

=

of these

17keV.

The denominator includes the conversion factor from joules to eV.

SECTION 39-6

X-Ray Spectra and Atomic Number

1055

mm

Determining

atomic

number. High-energy

electrons

are

X-ray intensity

used to bombard an unknown material. The strongest peak is found for X-rays emitted with an energy of 7.5 keV. Guess what the material is.

Ag

0.1

0.2

Wavelength, A (nm)

(Repeated.) FIGURE 39-11 Spectrum of X-rays emitted from a molybdenum target in an X-ray tube operated at SOkV.

APPROACH The highest intensity X-rays are generally for the K, line (s Fig. 39-11) which occurs when high-energy external electrons knock out K she electrons (the innermost orbit, n = 1) and their place is taken by electrons from the L shell (n = 2). We use the Bohr model, and assume the electrons “see” a nuclear charge of Z — 1 (screened by one electron). SOLUTION The hydrogen transition n = 2 to n = 1 would yield about 10.2eV (see Fig. 37-26 or Example 37-13). Energy E is proportional to Z* (Eq. 37-14), or rather (Z — 1)? because the nucleus is shielded by the one electron in a 1s state (see above), so we can use ratios:

(Z -1 2

7500eV 102eV

so Z—1=V735=27,

and

=

735,

Z = 28, which makes it cobalt.

Now we briefly analyze the continuous part of an X-ray spectrum (Fig. 39-11) based on the photon theory of light. When electrons strike the target, they collide with atoms of the material and give up most of their energy as heat (about 99%, so X-ray tubes must be cooled). Electrons can also give up energy by emitting a photon: an electron decelerated by interaction with atoms of the target (Fig. 39-13) emits radiation because of its deceleration (Chapter 31), and in this case it is called bremsstrahlung (German for “braking radiation”). Because energy is conserved, the energy of the emitted photon, 4f, must equal the loss of kinetic energy of the electron,

AK

= K — K’,

hf FIGURE 39-13 Bremsstrahlung photon produced by an electron decelerated by interaction with a target atom.

=

so

AK.

An electron may lose all or a part of its energy in such a collision. The continuous X-ray spectrum (Fig. 39-11) is explained as being due to such bremsstrahlu collisions in which varying amounts of energy are lost by the electrons. T. shortest-wavelength X-ray (the highest frequency) must be due to an electron that gives up all its kinetic energy to produce one photon in a single collision. Since the initial kinetic energy of an electron is equal to the energy given it by the accelerating voltage, V, then

K = eV.

brought to rest (K' = 0), then hfg

=

In a single collision in which the electron is

AK = eV

and

evV.

We set fy = ¢/Ay where Ay is the cutoff wavelength (Fig. 39-11) and find Ag

=

he oV

(39-10) -1

—

This prediction for A, corresponds precisely with that observed experimentally. This result is further evidence that X-rays are a form of electromagnetic radiation (light)" and that the photon theory of light is valid.

DOV

Cutoff wavelength. What is the shortest-wavelength X-ray

photon emitted in an X-ray tube subjected to S0kV?

APPROACH The electrons striking the target will have a kinetic energy of 50 keV. The shortest-wavelength photons are due to collisions in which all of the electron’s kinetic energy is given to the photon so K = eV = hfj. SOLUTION From Eq. 39-10, 63 X 1073*7-5)(3.0 x 108 07

he _ (663 ev

)

m)

(1.6 X 107°C) (5.0 x 10*V)

. s

10 m,

or 0.025 nm. NOTE This result agrees well with experiment, Fig. 39-11. If X-rays were not photons but rather neutral particles with mass m, Eq. 39-10 would not hold.

1056

CHAPTER 39

“-

*39-7

Magnetic Dipole Moment;

Total Angular Momentum

f lagnetic Dipole Moment and the Bohr Magneton An electron orbiting classically, and thus discussed in Chapter the magnetic dipole based, essentially, on p

=

the nucleus of an atom can be considered as a current loop, might be expected to have a magnetic dipole moment as 27. Indeed, in Example 27-12 we did a classical calculation of moment of the electron in the ground state of hydrogen the Bohr model, and found it to give 1A

=

Jeur.

Here v is the orbital velocity of the electron, and for a particle moving in a circle of radius r, its angular momentum is L

=

mor.

So we can write

le

}L=2m

The direction of the angular momentum L is perpendicular to the plane of the current loop. So is the direction of the magnetic dipole moment vector i, although in the opposite direction since the electron’s charge is negative. Hence we can write the vector equation

. _ -5 _lesL po=

(39-11)

This rough semiclassical derivation was based on the Bohr theory. The same sult (Eq. 39-11) is obtained using quantum mechanics. Since L is quantized in filantum mechanics, the magnetic dipole moment, too, must be quantized. The magnitude of the dipole moment is given by (see Eq. 39-3)

po=

%Vf(ffl-l).

When a magnetic dipole moment is in a magnetic field B, it experiences a torque as we saw in Section 27-5, and the potential energy U of such a system depends on B and the orientation of f relative to B (Eq. 27-12):

U=

—ji-B.

If the magnetic field B is in the z direction, then U=

—u;B;

and from Eq.39-4 (L, = my%) and Eq. 39-11, we have bz

=

eh

— %ml

(Be careful here not to confuse the electron mass m with the magnetic quantum number, m,.) It is useful to define the quantity ke

=

eh

5

(39-12)

which is called the Bohr magneton and has the value (joule/tesla). Then we can write

Rz =

—ppMy

pgz = 927 X 10°#]J/T (39-13)

r 1ere my has integer values from 0 to + £ (see Table 39-1). An atom placed in a magnetic field would have its energy split into levels that differ by AU = pg B; this is the Zeeman effect, and was shown in Fig. 39-4.

*SECTION 39-7

Magnetic Dipole Moment; Total Angular Momentum

1057

*Stern-Gerlach Experiment and the g-Factor for Electron Spin The first evidence for the splitting of atomic energy levels, known as space quantization (Section 39-2), came in 1922 in a famous experiment known as the Stern-Gerlach experiment. Silver atoms (and later others) were heated in an oven from which=, they escaped as shown in Fig. 39-14. The atoms were made to pass through &

4 5 I z

FIGURE 39-14

The Stern-Gerlach

experiment, which is done inside a vacuum chamber.

i

=

=7

N

pEpEEE

oy

Magnet

Collimator

Viewing

screen

collimator, which eliminated all but a narrow beam. The beam then passed into a nonhomogeneous magnetic field. The field was deliberately made nonhomogeneous so that it would exert a force on atomic magnetic moments: remember that the potential energy (in this case — i - B) must change in space if there is to be a force (F, = —dU/dx, etc., Section 8-2). If B has a gradient along the z axis, as in Fig. 39-14, then the force is along z: dU dB,

F.

dz

He'z

Thus the silver atoms would be deflected up or down depending on the value of p for each atom. Classically, we would expect to see a continuous distribution on the

viewing screen, since we would

expect

the atoms

to have randomly

oriente(‘fl

magnetic moments. But Stern and Gerlach saw instead two distinct lines for silve..

(and for other atoms sometimes more than two lines). These observations were the first evidence for space quantization, though not fully explained until a few years later. If the lines were due to orbital angular momentum, there should have been an odd number of them, corresponding to the possible values of m, (since py = —ugmy). For £ =0, there is only one possibility, m, = 0. For £ =1, m, can be 1,0, or —1, and we would expect three lines, and so on. Why there are only two lines was eventually explained by the concept of electron spin. With a spin of 1, the electron spin can have only two orientations in space, as we saw in Fig. 39-5. Hence a magnetic dipole moment associated with spin would have only two positions. Thus, the two states for silver seen in the Stern-Gerlach experiment must be due to the spin of its one valence electron. Silver atoms must thus have zero orbital angular momentum but a total spin of 3 due to this one valence electron. (Of silver’s 47 electrons, the spins of the first 46 cancel.) For the H atom in its ground state, again only two lines were seen on the screen of Fig. 39-14: due to the spin 3 of its electron since the orbital angular momentum is zero. The Stern-Gerlach deflection is proportional to the magnetic dipole moment,

w., and for a spin } particle we expect w, = —pugm; = —(3)(efi/2m)

as for the

case of orbital angular momentum, Eq. 39-13. Instead, ., for spin was found to be about twice as large:

KWz =

—8WBMy,

[electron spin]

(39-14)

where g, called the g-factor or gyromagnetic ratio, has been measured to be slightly larger than 2: g = 2.0023--- for a free electron. This unexpected factor of (about) 2 clearly indicates that spin cannot be viewed as a classical angular momentum. It is a purely quantum-mechanical effect. Equation 39-14 is the sam as Eq. 39-13 for orbital angular momentum with m; replacing m,. But for the orbital case, g = 1. 1058

CHAPTER 39

Quantum Mechanics of Atoms

*Total Angular Momentum J An atom can have both orbital and spin angular momenta. For example, in the 2p

state of hydrogen £ =1 and s = 3. In the 4d state, £ = 2 and s = 3. The total gular momentum is the vector sum of the orbital angular momentum L and the ! — .pin S:

According to quantum mechanics, the magnitude of the total angular momentum J is quantized:

7 = Vijj + 1)4.

(39-15)

£+s

=40+

T

Il

—.

For the single electron in the H atom, quantum mechanics gives the result that j can be

or

j

E—-s=1¢-1

but never less than zero, just as for £ and s. For the 1s state, £ = 0 and j =3 is the only possibility. For p states, say the 2p state, £ = 1 and j can be either 3 or 12 The z component for j is quantized in the usual way: m,-

=

],]

-

1,"',“j.

For a 2p state with j = },m; canbe j or —%;for j=3,m; canbe3,3, —3, =3,

_for a total of four states. Note that the state of a single electron can be specified y giving n, £, my, mg, or by giving n, j, £, m; (only one of these descriptions at a (ime). EXERCISE G What are the possibilities for j in the 3d state of hydrogen? (a) 3,3; (b) 3,3, 3;

©)3.3,3.3; @) 3.3; (0 1.3.

The interaction of magnetic fields with atoms, as in the Zeeman effect and the Stern-Gerlach experiment, involves the total angular momentum. Thus the Stern-Gerlach experiment on H atoms in the ground state shows two lines (for m; = +1 and —3), but for the first excited state it shows four lines corresponding

to the four possible m; values (3,3, —3, —3). *Spectroscopic Notation

We can specify the state of an atom, including the total angular momentum quantum number j, using the following spectroscopic notation. For a single electron state we can write nLj,

where the value of L (the orbital quantum number) is specified using the same letters as in Table 39-3, but in upper case: L=01 letter

=

S

P

2

3

4

D

F

G

ro the 2Ps, state has n = 2, £ = 1, j = 3 whereas 18, 12 specifies the ground state in hydrogen.

*SECTION 39-7

Magnetic Dipole Moment; Total Angular Momentum

*Fine Structure; Spin-Orbit Interaction A magnetic effect also produces the fine structure splitting mentioned in Section 39-2, but it occurs in the absence of any external field. Instead, it is due to a magnetic field produced by the atom itself. We can see how it occursm, by putting ourselves in the reference frame of the electron, in which case w. see the nucleus revolving about us as a moving charge or electric current that produces a magnetic field, B,. The electron has an intrinsic magnetic dipole moment p; (Eq. 39-14) and hence its energy will be altered by an amount (Eq. 27-12)

AU

= —f,-B,.

Since w, takes on quantized values according to the values of mg, the energy of a single electron state will split into two closely spaced energy levels (for m, = +3 and —1). This tiny sphttmg of energy levels produces a tiny splitting in spectral lines. For example, in the H atom, the 2P — 1S transition is split into two lines corresponding to 2P, — 1S;, and 2P;, — 1S;,,. The difference in energy between these two is only about 5 X 107 eV, which is very small compared to the 2P — 1S transition energy of 13.6eV — 3.4eV= 10.2¢eV. The magnetic field B, produced by the orbital motion is proportlonal to the orbital angular momentum L, and since fi, is proportional to the spin S, then AU = —ji, + B, can be written -

AU « L-S. This interaction, which produces the fine structure, is thus called the spin-orbit interaction. Its magnitude is related to a dimensionless constant known as the fine structure constant,

R

* 7 2eghe | 137

o

photon

———

thsorbed FIGURE 39-15

e

O ons

emied

Fluorescence.

FIGURE 39-16 When UV light (a range of wavelengths) illuminates these various “fluorescent” rocks,

they fluoresce in the visible region of the spectrum.

S

which also appears elsewhere in atomic physics.

39-8

Fluorescence and Phosphorescence

When an atom is excited from one energy state to a higher one by the absorption of a photon, it may return to the lower level in a series of two (or more) jumps if there is an energy level in between (Fig. 39-15). The photons emitted will consequently have lower energy and frequency than the absorbed photon. When the absorbed photon is in the UV and the emitted photons are in the visible region of the spectrum, this phenomenon is called fluorescence (Fig. 39-16). The wavelength for which fluorescence will occur depends on the energy levels of the particular atoms. Because the frequencies are different for different substances, and because many substances fluoresce readily, fluorescence is a powerful tool for identification of compounds. It is also used for assaying—determining how much of a substance is present—and for following substances along a natural pathway as in plants and animals. For detection of a given compound, the stimulating light must be monochromatic, and solvents or other materials present must not fluoresce in the

same region of the spectrum. Sometimes the observation of fluorescent light being emitted is sufficient to detect a compound. In other cases, spectrometers are used to measure the wavelengths and intensities of the emitted light. Fluorescent lightbulbs work in a two-step process. The applied voltage accelerates electrons that strike atoms of the gas in the tube and cause them to be excited. When the excited atoms jump down to their normal levels, they emit UV photons which strike a fluorescent coating on the inside of the tube. The light vxA see is a result of this material fluorescing in response to the UV light striking it. 1060

CHAPTER 39

Quantum Mechanics of Atoms

Materials such as those used for luminous watch dials are said to be phosphorescent. When an atom is raised to a normal excited state, it drops back

down within about 107®s, In phosphorescent substances, atoms can be excited by

hoton absorption to energy levels, called metastable, which are states that last f'p nuch longer because to jump down is a “forbidden” transition as discussed in Section 39-2. Metastable states can last even a few seconds or longer. In a collection of such atoms, many of the atoms will descend to the lower state fairly soon, but many will remain in the excited state for over an hour. Hence light will be emitted even after long periods. When you put your luminous watch dial close to a bright lamp, many atoms are excited to metastable states, and you can see the glow for a long time afterward.

39-9

Lasers

A laser is a device that can produce a very narrow intense beam of monochromatic coherent light. (By coherent, we mean that across any cross section of the beam, all parts have the same phase.) The emitted beam is a nearly perfect plane wave. An ordinary light source, on the other hand, emits light in all directions (so the intensity decreases rapidly with distance), and the emitted light is incoherent (the different parts of a beam are not in phase with each other). The excited atoms that emit the light in an ordinary lightbulb act independently, so each photon emitted can be considered as a short wave train that lasts about 107®s. Different wave trains bear no phase relation to one another. Just the opposite is true of lasers.

(a)

The action of a laser is based on quantum theory. We have seen that a photon

the emission of a photon. However, if a photon with this same energy strikes the

€ xcited atom, it can stimulate the atom to make the transition sooner to the lower

state, Fig. 39-17b. This phenomenon is called stimulated emission: not only do . . . we still have the original photon, but also a second one of the same frequency as a result of the atom’s transition. These two photons are exactly in phase, and they are moving in the same direction. This is how coherent light is produced in a laser. Hence the name “laser,” which is an acronym for Light Amplification by Stimulated Emission of Radiation. The natural population in energy states of atoms in thermal equilibrium at any temperature 7 (in K) is given by the Boltzmann distribution (or Boltzmann factor): N,

=

Ell

Ce ¥,

& hf=E, - EgEf

I

can be absorbed by an atom if (and only if) the photon energy Af corresponds to the

energy difference between an occupied energy level of the atom and an available excited state, Fig. 39-17a. If the atom is already in the excited state, it may of course jump down spontaneously (i.e., no apparent stimulus) to the lower state with

E,

hf bo

®

PO

rssn—

o Eg hf = E, - Ey

~ FIGURE 39-17

‘

(a) Absorption of a

~ Photon: (b) Stimulated emission, f:l“ and”E refer to “upper” and ower”

energy states.

(39-16a)

where N, is the number of atoms in the state with energy E,. For two states n and n', the ratio of the number of atoms in the two states is E,-E, (39-16b) N, Thus most atoms are in the ground state unless the temperature is very high. In

Ny _ o)

FIGURE 39-18

Two energy levels

the two-level system of Fig. 39-17, most atoms are normally in the lower state, SO

for a collection of atoms. Each dot

light from stimulated emission, two conditions must be satisfied. First, atoms must

atom. (a) A normal situation; (b) an

the majority of incident photons will be absorbed. In order to obtain the coherent

be excited to the higher state, so that an inverted population is produced, one in

which more atoms are in the upper state than in the lower one (Fig. 39-18). Then emission of photons will dominate over absorption. Hence the system will not be in thermal equilibrium. And second, the higher state must be a metastable state— a state in which the electrons remain longer than usual’ so that the transition to

he lower state occurs by stimulated emission rather than spontaneously. (How

mverted populations are created will be discussed shortly.)

TAn atom excited to such a state can jump to a lower state only by a so-called forbidden transition (discussed in Section 39-2), which is why the lifetime is longer than normal.

represents the energy state of one

inverted population.

S

Normal @)

SECTION 39-9

Inverted

populano (b)

Lasers

1061

Partially FIGURE 39-19 Laser diagram, showing excited atoms stimulated to emit light.

22eV

E| (metastable)

1.8eV

metastable with a lifetime of about 3 X 1073s (compared to 10®s for ordinary

FIGURE 39-21 Energy levels for He and Ne. He is excited in the electric discharge to the E state.

This energy is transferred to the E} level of the Ne by collision. E% is metastable and decays to E5 by stimulated emission. Collision

—1 B

TT

20.61 eV

20.66eV

1062

Neon

ioeevy

k& CHAPTER 39

-

L3

E;

18.70eV

Eg Quantum

S

The excitation of the atoms in a laser can be done in several ways to produce the necessary inverted population. In a ruby laser, the lasing material is a ruby rod consisting of Al,O; with a small percentage of aluminum (Al) atoms replaced by chromium (Cr) atoms. The Cr atoms are the ones involved in lasing. In a process called optical pumping, the atoms are excited by strong flashes of light of wavelength 550 nm, which corresponds to a photon energy of 2.2eV. As shown in Fig. 39-20, the atoms are excited from state E, to state E,. The atoms quickly decay either back to E; or to the intermediate state E;, which is

FIGURE 39-20 Energy levels of chromium in a ruby crystal. Photons of energy 2.2 ¢V “pump” atoms from Ey to E,, which then decay to metastable state E;. Lasing action occurs by stimulated emission of photons in transition from E; to Ej.

—

L aser outpiit beam

Creating an Inverted Population Ey

Helium

AN [enmeaws"

Figure 39-19 is a schematic diagram of a laser: the “lasing” material is placed in a long narrow tube at the ends of which are two mirrors, one of which is partially transparent (transmitting perhaps 1 or 2%). Some of the excited atoms drop down fairly soon after being excited. One of these is the blue atom shown on the far left in Fig. 39-19. If the emitted photon strikes another atom in the excited state, it stimulates this atom to emit a photon of the same frequency, moving in the same direction, and in phase with it. These two photons then move on to strike other atoms causing more stimulated emission. As the process continues, the number of photons multiplies. When the photons strike the end mirrors, most are reflected back, and as they move in the opposite direction, they continue to stimulate more atoms to emit photons. As the photons move back and forth between the mirrors, a small percentage passes through the partially transparent mirror at one end. These photons make up the narrow coherent external laser beam. Inside the tube, some spontaneously emitted photons will be emitted at an angle to the axis, and these will merely go out the side of the tube and not affect the narrowness of the main beam. In a well-designed laser, the spreading of the beam is limited only by diffraction, so the angular spread is ~A/D (see Eq. 35-1 or 35-10), where D is the diameter of the end mirror. The diffraction spreading can be incredibly small. The light energy, instead of spreading out in space as it does for an ordinary light source, can be a pencil-thin beam.

E,

0.4 eV,

transparent mirror

,

levels). With strong pumping action, more atoms can be found in the E, state than are in the Ej state. Thus we have the inverted population needed for lasing. As soon as a few atoms in the E, state jump down to Ej, they emit photons that produce stimulated emission of the other atoms, and the lasing action begins. A ruby laser thus emits a beam whose photons have energy 1.8eV and a wavelength of 694.3 nm (or “ruby-red” light). In a helium-neon (He-Ne) laser, the lasing material is a gas, a mixture of about 85% He and 15% Ne. The atoms are excited by applying a high voltage to the tube so that an electric discharge takes place within the gas. In the process, some of the He atoms are raised to the metastable state £, shown in Fig. 39-21, which corresponds to a jump of 20.61 eV, almost exactly equal to an excited state in neon, 20.66 eV. The He atoms do not quickly return to the ground state by spontaneous emission, but instead often give their excess energy to a Ne atom when they collide—see Fig. 39-21. In such a collision, the He drops to the ground state and the Ne atom is excited to the state Ej (the prime refers to neon states). The slight difference in energy (0.05eV) is supplied by the kinetic energy of the moving atoms. In thi manner, the Ej state in Ne—which is metastable—becomes more populated thai. the E3 level. This inverted population between Ej and Ej is what is needed for lasing. Mechanics of Atoms

Very common now are semiconductor diode lasers, also called pn junction lasers, which utilize an inverted population of electrons between the conduction band of the n side of the diode and the lower-energy valence band of the p side (Sections 40-7 to fi0—9). When an electron jumps down, a photon can be emitted, which in turn can imulate another electron to make the transition and emit another photon, in phase. The needed mirrors (as in Fig. 39-19) are made by the polished ends of the pn crystal. Semiconductor lasers are used in CD and DVD players (see below), and in many other applications. Other types of laser include: chemical lasers, in which the energy input comes from the chemical reaction of highly reactive gases; dye lasers, whose frequency is tunable; CO, gas lasers, capable of high-power output in the infrared; and rare-earth solid-state lasers such as the high-power Nd:YAG laser. The excitation of the atoms in a laser can be done continuously or in pulses. In a pulsed laser, the atoms are excited by periodic inputs of energy. In a continuous laser, the energy input is continuous: as atoms are stimulated to jump down to the lower level, they are soon excited back up to the upper level so the output is a continuous laser beam. No laser is a source of energy. Energy must be put in, and the laser converts a part of it into an intense narrow beam output.

A CAUTION

Laser not an energy source

«Applications The unique feature of light from a laser, that it is a coherent narrow beam, has found many applications. In everyday life, lasers are used as bar-code readers (at store checkout stands) and in compact disc (CD) and digital video disc (DVD) players. The laser beam reflects off the stripes and spaces of a bar code, and off the tiny pits of a CD or DVD as shown in Fig. 39-22a. The recorded information on a CD or DVD is a series of pits and spaces representing Os and 1s (or “off” and “on”) of a binary code that is decoded electronically before being sent to ’ihe audio or video system. The laser of a CD player starts reading at the inside /1 the disc which rotates at about 500rpm at the start. As the disc rotates, the laser follows the spiral track (Fig. 39-22b), and as it moves outward the disc must slow down because each successive circumference (C = 27r) is slightly longer as r increases; at the outer edge, the disc is rotating about 200 rpm. A 1-hour CD has a track roughly 5 km long; the track width is about 1600 nm (= 1.6 um) and the distance between pits is about 800 nm. DVDs contain much more information. Standard DVDs use a thinner track (0.7 um) and shorter pit length (400 nm). New high-definition DVDs use a “blue” laser with a short wavelength (405nm) and narrower beam, allowing a narrower track (0.3 um) that can store much more data for high definition. DVDs can also have two layers, one below the other. When the laser focuses on the second layer, the light passes through the semitransparent surface layer. The second layer may start reading at the outer edge instead of inside. DVDs can also have a single or double layer on borh surfaces of the disc.

W

e

Underside of CD

N\ Mirror N\

FIGURE 39-22

\pjt

APPLIED

(a) Readinga CD (or

DVD). The fine beam of a laser, focused

Half-silvered mirror \MS

é

N\

Laser m

Lens

(a)

PHYSICS

DVD and CD players, bar codes

Detector

even more finely with lenses, is directed at the undersurface of a rotating compact disc. The beam is reflected back from the areas between pits but reflects much less from pits. The reflected light is detected as shown, reflected by a half-reflecting mirror MS. The | strong and weak reflections correspond to the Os and 1s of the binary code representing the audio or video signal. (b) A laser follows the CD track which starts near the center and spirals outward.

SECTION 39-9

Lasers

1063

PHYSICS

APPLIED

Medical and other uses of lasers

Lasers are a useful surgical tool. The narrow intense beam can be used to

~ 9estroy tissue in a localized area, or to break up gallstones and kidney stones. Because of the heat produced, a laser beam can be used to “weld” broken tissue, such as a detached retina, Fig. 39-23, or to mold the cornea of the eye (by vaporizing‘\ tiny bits of material) to correct myopia and other eye defects (LASIK surgery). The_ laser beam can be carried by an optical fiber (Section 32-7) to the surgical point, sometimes as an additional fiber-optic path on an endoscope (again Section 32-7). An example is the removal of plaque clogging human arteries. Tiny organelles within a living cell have been destroyed using lasers by researchers studying how the absence of that organelle affects the behavior of the cell. Laser beams are used to

FIGURE 39-23

Laser being used

| @

destroy cancerous and precancerous cells; and the heat seals off capillaries and lymph vessels, thus “cauterizing” the wound to prevent spread of the disease. The intense heat produced in a small area by a laser beam is used for welding and machining metals and for drilling tiny holes in hard materials. Because a laser beam is coherent, monochromatic, narrow, and essentially parallel, lenses can be used to focus the light into even smaller areas. The precise straightness of a laser beam is also useful

to surveyors for lining up equipment accurately, especially in inaccessible places.

*39-10

PHYSICS

Holography

APppLIED One of the most interesting applications of laser light is the production of threeHolography ~ dimensional images called holograms (see Fig. 39-24). In an ordinary photograph, the film simply records the intensity of light reaching it at each point. When the photograph or transparency is viewed, light reflecting from it or passing through it gives us a two-dimensional picture. In holography, the images are formed by interference, without lenses. When a laser hologram is made on film, a broadened laser beam is split into two parts by a half-silvered mirror, Fig. 39-24a. One part goes directly to the film; the rest passes to the object to be photographed, from which it is reflected to the film. Light from every point on the object reaches each point on the film, and the interference of the two beams allows the film tc*™ record both the intensity and relative phase of the light at each point. It is crucial that the incident light be coherent — that is, in phase at all points —which is why a laser is used. After the film is developed, it is placed again in a laser beam and a three-dimensional image of the object is created. You can walk around such an image and see it from different sides as if it were the original object (Fig. 39-24b). Yet, if you try to touch it with your hand, there will be nothing material there.

FIGURE 39-24 (a) Making a hologram. Light reflected from various points on the object interferes (at the film) with light from the direct beam. (b) A boy is looking at a hologram of a woman talking on a telephone. Holograms do not photograph well—they must be seen directly. Half-silvered mirror

Film

Direct beam

Laser light

il

Reflected rays

Object

() 1064

CHAPTER 39

Quantum Mechanics of Atoms

The details of how the image is formed are quite complicated. But we can get the basic idea by considering one single point on the object. In Fig. 39-25a the rays OA and OB have reflected from one point on our object. The rays CA and DB come directly from the source and interfere with OA and OB at points A and B on the film. A set of interference fringes is produced as shown in Fig, 39-25b. The spacing between the fringes changes from top to bottom as shown. Why this happens is explained in Fig. 39-26. Thus the hologram of a single point object would have the fringe pattern shown in Fig. 39-25b. The film in this case looks like a diffraction grating with variable spacing. Hence, when coherent laser light is passed back through the developed film to reconstruct the image, the diffracted rays in the first order maxima occur at slightly different angles because the spacing changes. (Remember Eq. 35-13, sinf = A/d: . where the spacing d is greater, the angle 6 is less.) Hence, the rays diffracted upward (in first order) seem to diverge from a single point, Fig. 39-27. This is a virtual image of the original object, which can be seen with the eye. Rays diffracted in first order downward converge to make a real image, which can be seen and also photographed.

Film

C

A B

D Reflected

o (object)

rai,swe (a)

(b)

(Note that the straight-through undiffracted rays are of no interest.) Of course real ~ FIGURE 39-25 (a) Light from point O

objects consist of many points, so a hologram will be a complex interference pattern ~ On the object interferes with light of which, when laser light is incident on it, will reproduce an image of the object. Each image point will be at the correct (three-dimensional) position with respect to other points, so the image accurately represents the original object. And it can be viewed from different angles as if viewing the original object. Holograms can be made in which a viewer can walk entirely around the image (360°) and see all sides of it.

De\;_elloped ilm

. )\/ Min M| Max Min

A

r

j

.

\\

Max

AL/

|Min

Max ———

Laser -

FIGURE 39-26 Each of the rays shown leaving point O is one wavelength

\Pa A

light

Min

Max

0)

15 ubest BEdL (v G i ), {0y seriersnice fniges produced.

OBB

B

Virtual

shorter than the one above it. If the top

Real

image

ray is in phase with the direct beam (not shown), which has the same phase at all

FIGURE 39-27

points on the screen, each of the rays

image

Reconstructing the image of one point on the

object. Laser beam strikes the developed film, which is like a

shown produces a constructive interference fringe. From this diagram it can be seen that the fringe spacing increases toward the bottom.

diffraction grating of variable spacing. Rays corresponding to the first diffraction maxima are shown emerging, The angle 6, > 65 because the spacing at B is greater than at A (sinf = A/d). Hence real and virtual images of the point are reproduced as shown.

Volume or white-light holograms do not require a laser to see the image, but can be viewed with ordinary white light (preferably a nearly point source, such as the Sun or a clear bulb with a small bright filament). Such holograms must be made,

however,

with

a laser. They

are made

not on

thin

film, but

on

a thick

emulsion. The interference pattern in the film emulsion can be thought of as an array of bands or ribbons where constructive interference occurred. This array, and the reconstruction of the image, can be compared to Bragg scattering of X-rays from the atoms in a crystal (see Section 35-10). White light can reconstruct the image because the Bragg condition (mA = 2dsin#) selects out the appropriate single wavelength. If the hologram is originally produced by lasers emitting the r hree additive primary colors (red, green, and blue), the three-dimensional image can be seen in full color when viewed with white light.

*SECTION 39-10

Holography

1065

| Summary In the quantum-mechanical view of the atom, the electrons do not

have well-defined orbits, but instead exist as a “cloud.” Electron

clouds can be interpreted as an electron wave spread out in space, or as a probability distribution for electrons considered as particles. For the simplest atom, hydrogen, the Schrddinger equation contains the potential energy

1

U=

€

T 7

The solutions give the same values of energy as the old Bohr theory. According to quantum mechanics, the state of an electron in an atom is specified by four quantum numbers: n, £, m,, and my: 1. n, the principal quantum number, can take on any integer value (1,2, 3, -+ ) and corresponds to the quantum number

of the old Bohr theory;

2. ¢, the orbital quantum number, can take on integer values fromOupton 3.

— 1;

my, the magnetic quantum values from —£ to +£;

number,

can

take

on

integer

4. my, the spin quantum number, can be + 3} or — 3. The energy levels in the hydrogen atom depend on n, whereas in other atoms they depend on » and £. The orbital angular momentum of an atom has magnitude

L=V +1)A and z component L. = myh. Spin angular momentum has magnitude S = \V/s(s + 1)% and 2z component S, =m,h where s =% and m, = £+

When an external magnetic field is applied, the spectral lines are split (the Zeeman effect), indicating that the energy depends also on m, in this case. Even in the absence of a magnetic field, precise measurements of spectral lines show a tiny splitting of the lines called fine structure, whose explanation is that the energy depends very slightly on m, and m. Transitions between states that obey the selection rule Al = +1 are far more probable than other so-called forbidden transitions. The ground-state wave function in hydrogen has spherical symmetry, as do other £ = 0 states. States with £ > 0 have some directionality in space. The probability density (or probability distribution), |1/,

and the radial probability demsity,

P, = 4mr?y|?,

are both

useful to illustrate the spatial extent of the electron cloud.

The arrangement of electrons in multi-electron atoms is governed by the Pauli exclusion principle, which states that no two electrons can occupy the same quantum state—that is, th(fl cannot have the same set of quantum numbers n, £, m,, and my. As a result, electrons in multi-electron atoms are grouped into shells (according to the value of n) and subshells (according to £). Electron configurations are specified using the numerical values

of

n,

and

using

letters

for

£

s, p,

d,

f,

etc.,

for

£=0,1,2,3, and so on, plus a superscript for the number of electrons in that subshell. Thus, the ground state of hydrogen is

1s', whereas that for oxygen is 1s2’25§;p“.

In the Periodic Table, the elements are arranged in horizontal

rows according to increasing atomic number (number of electrons in the neutral atom). The shell structure gives rise to a periodicity in the properties of the elements, so that each vertical column can contain elements with similar chemical

properties.

X-rays, which are a form of electromagnetic radiation of very short wavelength, are produced when high-speed electrons strike a target. The spectrum of X-rays so produced consists of two parts, a continuous spectrum produced when the electrons are decelerated by atoms of the target, and peaks representing photons emitted by atoms of the target after being excited by collision with the high-speed electrons. Measurement of these peaks allows determination of inner energy levels of atoms and determination of atomic number Z. [*An atom has a magnetic dipole moment i related to its orbital angular momentum L, which produces a potential energy

when in a magnetic field,

U = —n-B.

Electron spin als

yields a magnetic moment, but the energy in a magnetic field is factor g = 2.0023 - times larger than expected, as determined in Stern-Gerlach experiments.]

[*Atoms have a total angular momentum

J=L+§

which is quantized as for L and S, namely J = Vj(j + 1)% where j is a half-integer equal to £ + } in hydrogen.]

Fluorescence occurs when absorbed UV photons are followed by emission of visible light, due to the special arrangement of energy levels of atoms of the material. Phosphorescent materials have metastable states (long-lived) that emit light seconds or minutes after absorption of light. Lasers produce a narrow beam of monochromatic coherent light (light waves in phase). [*Holograms are images with a 3-dimensional quality, formed by interference of laser light.]

| Questions 1. Discuss the differences between Bohr’s view and the quantum-mechanical view. 2. The probability density [|* is a maximum at the H atom (r = 0) for the ground state, radial probability density P = 4mr¥y> is point. Explain why.

of the atom the center of whereas the zero at this

3. Which model of the hydrogen atom, the Bohr model or the

quantum-mechanical model, predicts that the electron spends more time near the nucleus? 4. The size of atoms varies by only a factor of three or so, from largest to smallest, yet the number of electrons varies from one to over 100. Why?

1066

CHAPTER 39

Quantum Mechanics of Atoms

5. Excited hydrogen and excited helium atoms both radiate light as they jump down to the n =1, = 0,m; = 0 state. Yet the two elements have very different emission spectra. Why? 6. In Fig. 39-4, why do the upper and lower levels have different energy splittings in a magnetic field? 7. Why do three quantum numbers come out of the Schrodinger theory (rather than, say, two or four)? 8. The 589-nm yellow closely spaced lines. Zeeman effect. Can the reference frame

line in sodium is actually two very This splitting is due to an “internal”q you explain this? [Hint: Put yourself . of the electron.]

9. Which of the following electron configurations are not allowed: (a) 1s2s2p*3s%4p% (b)) 15s%25%2p%3s': (c) 1522522 p®3523 p°4s24d°4f1? If not allowed, explain why. 10. Give the complete electron configuration for a uranium atom (careful scrutiny across the Periodic Table on the inside back cover will provide useful hints). 11. In what column of the Periodic Table would you expect to find the atom with each of the following configurations?

(a) 152522 p%3s% (b) 1522522 p"35%3 p%; (c) 1522522 p°35%3 pO4s!;

12.

13.

14

15 16. 17. 18.

(d) 1s%252p°. On what factors does the periodicity of the Periodic Table depend? Consider the exclusion principle, quantization of angular momentum, spin, and any others you can think of. How would the Periodic Table look if there were no electron spin but otherwise quantum mechanics were valid? Consider the first 20 elements or so. The ionization energy for neon (Z = 10) is 21.6eV and that for sodium (Z = 11) is 5.1eV. Explain the large difference. Why do chlorine and iodine exhibit similar properties? Explain why potassium and sodium exhibit similar properties. Why are the chemical properties of the rare earths so similar? Why do we not expect perfect agreement between measured values of characteristic X-ray line wavelengths and those calculated using Bohr theory, as in Example 39-6?

19. Why does the Bohr theory, which does not work at all well for normal transitions involving the outer electrons for He and more complex atoms, nevertheless predict reasonably well the atomic X-ray spectra for transitions deep inside the atom? 20. Why does the cutoff wavelength in Fig. 39-11 imply a photon nature for light? 21. How would you figure out which lines in an X-ray spectrum correspond to K, , Kg, L, etc., transitions?

22, Why do the characteristic X-ray spectra vary in a systematic way with Z, whereas the visible spectra (Fig. 35-22) do not? 23. Why do we expect electron transitions deep within an atom to produce shorter wavelengths than transitions by outer electrons? *24, Why is the direction of the magnetic dipole moment of an electron opposite to that of its orbital angular momentum? *285. Why is a nonhomogeneous field used in the Stern-Gerlach experiment? 26

Compare spontaneous emission to stimulated emission.

27. Does the intensity of light from a laser fall off as the inverse square of the distance? 28. How does laser light differ from ordinary light? How is it the same? 29. Explain

how

a 0.0005-W

laser beam, photographed

at a

distance, can seem much stronger than a 1000-W street lamp

at the same distance.

l Problems f‘ 3-2 Hydrogen Atom Quantum Numbers

11. (IT) The

L (I) For n = 7, what values can £ have? 2. (I) For n =6, £ =3, what are the possible values of ni, and m;? . (I) How many different states are possible for an electron whose principal quantum number is n = 4?7 Write down the quantum numbers for each state.

magnitude

of the orbital angular momentum

in

an excited state of hydrogen is 6.84 X 107J-s and the z component is 2,11 X 1073 J-s. What are all the possible values of n, £, and m, for this state?

39-3 Hydrogen Atom Wave Functions 12. (I) Show that the ground-state wave function, Eq. 39-5, is normalized. [Hint: See Example 39-4.]

. (I) If a hydrogen atom has m, = —4, what are the possible

13. (IT) For the ground state of hydrogen, what is the value of

. (I) A hydrogen atom has values for n, my, and m;?

14. (IT) For the

values of n, £, and m,?

£ = 5.

What

are the possible

. (I) Calculate the magnitude of the angular momentum of an electron in the

n = 5,

= 3 state of hydrogen.

. (IT) A hydrogen atom is in the 7g state. Determine (a) the principal quantum number, (b) the energy of the state, (c) the orbital angular momentum and its quantum number £, and (d) the possible values for the magnetic quantum number. . (II) (a) Show that the number of different states possible for a given value of £ is equal to 2(2¢ + 1). (b) What is this number for £ =0,1,2,3,4,5, and 6? . (IT)

Show

that

the

number

of different

electron

possible for a given value of n is 21%. (See Problem 8.)

states

10. (IT) An excited H atom is in a 5d state. (a) Name all the states to which the atom is “allowed” to jump with the emission of a photon. (b) How many different wavelengths are there (ignoring fine structure)?

(a) ¥, (b) |[¢|*, and (c) P, at r = 1.5r? n =2,

£ =0

state of hydrogen, what is the

value of (a) ¢, (b) [¢|% and (c) P, at r = 4ry?

15. (II) By what factor is it more likely to find the electron in the ground state of hydrogen at the Bohr radius (ry) than at

twice the Bohr radius (2r)?

16. (II) (a) Show that the probability of finding the electron in the ground state of hydrogen at less than one Bohr radius from the nucleus is 32%. (b) What is the probability of finding a 1s electron between r = ry and r = 2ry? 17. (II) Determine the radius » of a sphere centered on the nucleus within which the probability of finding the electron for the ground state of hydrogen is (a) 50%, (b) 90%, (c) 99%. 18. (IT) (a) Estimate the probability of finding an electron, in the ground state of hydrogen, within the nucleus assuming it to be a sphere of radius » = 1.1 fm. (b) Repeat the estimate assuming the electron is replaced with a muon, which is very similar to an electron (Chapter 43) except that its mass is 207 times greater.

Problems

1067

19. (II) Show that the mean value of r for an electron in the ground state of hydrogen is 7 = 3ry, by calculating

Fe [

all space

rwedav = [ eseomra 0

20. (IT) Show that ¢r,(q as given by Eq. 39-8 is normalized. 21. (II) Determine the average radial probability distribution P, for the n = 2,£ = 1 state in hydrogen by calculating

P

= 4ar’[in1o

+ 3l

+ 301

22, (I) Use the result of Problem 21 to show that the most probable distance r from the nucleus for an electron in the 2p state of hydrogen is r = 4ry, which is just the second Bohr radius (Eq. 37-11, Fig. 37-25). 23. (IT) For the ground state of hydrogen, what is the probability of finding the electron within a spherical shell of inner radius 0.99ry and outer radius 1.017y? . (III) For the n = 2,£ = 0 state of hydrogen, what is the probability of finding the electron within a spherical shell of inner radius 4.00 ry and outer radius 5.00 ry? [Hint: You might integrate by parts.] 25. (III) Show that 190 (Eq. 39-5a) satisfies the Schrodinger equation (Eq. 39—1) with the Coulomb potential, for energy E = —me*/8e3 h 26. (IIT) Show that the probability of finding the electron within 1 Bohr radius of the nucleus in the hydrogen atom is (a) 3.4% for the n =2, £ = 0 state, and (b) 0.37% for the n =2, £ =1 state. (See Problem 21.) Z7 (III) The wave function for the n = 3, £ =0 state in hydrogen is

o = A "

2T

(12, 2 3

2713

(a) Determine the radial probability distribution P, for this state, and (b) draw the curve for it on a graph. (c) Determine the most probable distance from the nucleus for an electron in this state.

39-4 and 39-5 Complex Atoms

28. (I) List the quantum numbers

for each

electron

in the

ground state of oxygen (Z = 8). 29, (I) List the quantum numbers for each electron in the ground state of (a) carbon (Z = 6), (b) aluminum (Z = 13). . (I) How many electrons can be in the n = 6, { = 4 subshell? . (IT) An electron has m; =2 and is in its lowest possible energy state. What are the values of »n and £ for this electron? (II) If the principal quantum number »n were limited to the range

from 1 to 6, how many elements would we find in nature?

. (II) What is the full electron configuration for (a) nickel (Ni), (b) silver (Ag), (c¢) uranium (U)? [Hint: See the Periodic Table inside the back cover.] . (IT) Estimate the binding energy of the third electron in lithium using Bohr theory. [Hint: This electron has n = 2 and “sees” a net charge of approximately +1e.] The measured value is 5.36 eV. 3s. (IT) Using the Bohr formula for the radius of an electron orbit, estimate the average distance from the nucleus for an

electron in the innermost (n =1) orbit in uranium (Z = 92). Approximately how much energy would be required to remove this innermost electron? 1068

CHAPTER 39

Quantum

Mechanics of Atoms

36. (IT) Let us apply the exclusion principle to an infinitely high square well (Section 38-8). Let there be five electrons confined to this rigid box whose width is £. Find the lowest energy state of this system, by placing the electrons in the lowest available levels, consistent with the Pauli exclusion principle. fl 37

(II) Show that the total angular momentum filled subshell.

is zero for a

39-6 X-rays

38. (I) If the shortest-wavelength bremsstrahlung X-rays emitted from an X-ray tube have A = 0.027 nm, what is the voltage across the tube? 39. (I) What are the shortest-wavelength X-rays emitted by electrons striking the face of a 32.5-kV TV picture tube? What are the longest wavelengths? . (I) Show that the cutoff wavelength Ag in an X-ray spectrum is given by 1240

Ao = —;—nm, where V is the X-ray tube voltage in volts. 41. (II) Estimate the wavelength for an n =2 to n=1 transition in iron (Z = 26). 42. (II) Use the result of Example 39-6 to estimate the X-ray wavelength emitted when a cobalt atom (Z = 27) makes a transition from n =2 to n = 1. 43. (II) A mixture of iron and an unknown material are bombarded with electrons. The wavelength of the K, lines are 194 pm for iron and 229 pm for the unknown. What is the unknown material? . (II) Use Bohr theory to estimate the wavelength for an‘ n =3 to n=1 transition in molybdenum (Z = 42).Th measured value is 0.063 nm. Why do we not expect perfect agreement? 4s. (IT) Use conservation of energy and momentum to show that a moving electron cannot give off an X-ray photon unless there is a third object present, such as an atom or nucleus.

+39-7 Magnetic Dipole Moment; J %46,

(I)

Verify

that

the

Bohr

magneton

pg = 927 X 107%#J/T (see Eq.39-12).

has

the

value

*47, (I) If the quantum state of an electron is specified by (n, £, my, my), estimate the energy difference between the states (1,0,0, —5) and (1, 0,0, +3) of an electron in the 1s state of helium in an external magnetic field of 2.5 T. *48. (I1) Silver atoms (spin = 1) are placed in a 1.0-T magnetic field which splits the ground state into two close levels. (a) What is the difference in energy between these two levels, and (b) what wavelength photon could cause a transition from the lower level to the upper one? (c) How would your answer differ if the atoms were hydrogen? *49, (II) In a Stern-Gerlach experiment, Ag atoms exit the oven with an average speed of 780m/s and pass through a magnetic field gradient dB/dz = 1.8 X 10°T/m for a distance of 5.0cm. (a) What is the separation of the two beams as they emerge from the magnet? (b) What would the separation be if the g-factor were 1 for electron spin? *50. (IT) For an electron in a 5g state, what are all the possibl. values of j, mj,J, and J;?

*51. (IT) What are the possible values of j for an electron in (a) the 4p, (b) the 4f, and (c) the 3d state of hydrogen? (d) What is J in each case? *52. (IT) (a) Write down the quantum numbers for each electron r in the gallium atom. (b) Which subshells are filled? (c) The last electron is in the 4p state; what are the possible values of the total angular momentum quantum number, j, for this electron? (d) Explain why the angular momentum of this last electron also represents the total angular momentum for the entire atom (ignoring any angular momentum of the nucleus). (¢) How could you use a Stern-Gerlach experiment to determine which value of j the atom has? *53. (IIT) The difference between the 2P;;, and 2P, energy levels in hydrogen is about 5 X 107 eV, due to the spinorbit interaction. (a) Taking the electron’s (orbital) magnetic

moment

to be

1 Bohr

magneton,

estimate

the

internal magnetic field due to the electron’s orbital motion. (b) Estimate the internal magnetic field using a

simple model of the nucleus revolving in a circle about the electron.

39-9 Lasers 54. (II) A laser used to weld detached retinas puts out 23-mslong pulses of 640-nm light which average 0.63-W output during a pulse. How much energy can be deposited per pulse and how many photons does each pulse contain?

. (II) Estimate

the angular spread of a laser beam

due to

diffraction if the beam emerges through a 3.6-mm-diameter

mirror. Assume that A = 694nm. What would be the diameter of this beam if it struck (a) a satellite 380 km above the Earth, (b) the Moon? [Hint: See Chapter 35.] 56. (II) A low-power laser used in a physics lab might have a power of 050mW and a beam diameter of 3.0mm. Calculate (a) the average light intensity of the laser beam, and (b) compare it to the intensity of a very powerful lightbulb emitting 100 W of light as viewed from a distance of 2.0 m. . (II) Calculate the wavelength of a He—Ne laser. 58. (IT) Suppose that the energy level system in Fig. 39-20 is not being pumped and is in thermal equilibrium. Determine the fraction of atoms in levels E, and E; relative to E; at

T = 300K.

59. (II) To what temperature would the system in Fig. 39-20 have to be raised (see Problem 58) so that in thermal equilibrium the level E, would have half as many atoms as Ej? (Note that pumping mechanisms do not maintain thermal equilibrium.) 60. (IT) Show that a population inversion for two levels (as in a pumped laser) corresponds to a negative Kelvin temperature in the Boltzmann distribution. Explain why such a situation does not contradict the idea that negative Kelvin temperatures cannot be reached in the normal sense of temperature.

| General Problems 61. The ionization (binding) energy of the outermost electron in boron is 8.26 eV. (a) Use the Bohr model to estimate the “effective charge,” Z.¢, seen by this electron. (b) Estimate the average orbital radius.

" A

62. How many electrons can there be in an “A” subshell? 63. What is the full electron configuration in the ground state for elements with Z equal to (a) 25, (b) 34, (c) 39? [Hint: See the Periodic Table inside the back cover.] 64. What are the largest and smallest possible values for the angular momentum L of an electron in the n = 6 shell? 65. Estimate (a) the quantum number £ for the orbital angular momentum of the Earth about the Sun, and (b) the number of possible orientations for the plane of Earth’s orbit. 66. Use the Bohr theory (especially Eq. 37-15) to show that the Moseley plot (Fig. 39-12) can be written

69. In the so-called vector model of the atom, space quantization of angular momentum (Fig. 39-3) is illustrated as shown in Fig. 39-28. The angular momentum vector of magnitude L =\ +1)4 is thought of as precessing around the z axis (like a spinning top or gyroscope) in such a way that the z component of angular momentum, L, = my#, also stays constant. Calculate the possible values for the angle 6

between L and the z axis (a) for £ =1, (¢)

£

£=3.

(d) Determine

=100 and £ = 10°

Is

the minimum

Z

this consistent with the correspondence principle?

a(Z

to precess about the z axis; L and

Ly continually change. Problems 69 and 70.

—b)’

where b ~ 1, and evaluate a. 67.

Determine the most probable distance from the nucleus of an electron in the n = 2,f = 0 state of hydrogen.

68. Show that the described in expect from beam’s width aperture D, perpendicular

diffractive spread of a laser beam, ~A/D

as

Section 39-9, is precisely what you might the uncertainty principle. [Hint: Since the is constrained by the dimension of the the component of the light’'s momentum to the laser axis is uncertain. ]

value

=2, and

of 6 for

FIGURE 39-28 The vector model for orbital angular momentum. The orbital angular momentum vector L is imagined

L, remain constant, but L and

3T

(b) £

/b,

rd

x

70. The vector model (Problem 69) gives some insight into the uncertainty principle for angular momentum, which is AL;Ap =1 for the z component. Here ¢ is the angular position measured in the plane perpendicular to the z axis. Once m, for an atom is known, L; is known precisely, so AL, = 0. (a) What does this tell us about ¢? (b) What can you say about Ly and Ly, which are not quantized (only L and L; are)? (c) Show that although L, and L, are not quantized,

nonetheless (L3 + L3)"2=[£(¢ + 1) — m}]"* 4 is. General Problems

1069

71. (a) Show that the mean value for 1/r of an electron in the ground state of hydrogen equals 1/ry, and from this conclude that the mean value of the potential energy is l-]-__le2 4TT€0

(b) Using

E =U + K,

that the spectrum of an unknown element shows a lines with one out of every four matching a line Lyman series of hydrogen. Assuming that the element is an ion with Z protons and one elec-

tron, determine Z and the element in question.

o

find a relationship between the

average kinetic energy and the average potential energy in the ground state. [Hint: For (a), see Problem 19 or Example 38-9.] 72. The angular momentum in the hydrogen atom is given both by the Bohr model and by quantum mechanics. Compare the results for n = 2. 73. For each of the following atomic transitions, state whether the transition is allowed or forbidden, and why: (a) 4p — 3p; (b)3p — 1s; (c) 4d — 3d; (d) 4d — 3s; (e) 4s — 2p. 74. It is possible for atoms to be excited into states with very high values of the principal quantum number. Electrons in these socalled Rydberg states have very small ionization energies and huge orbital radii. This makes them particularly sensitive to external perturbation, as would be the case if the atom were in an electric field. Consider the n = 45 state of the hydrogen atom. Determine

75. Suppose series of from the unknown

the binding energy, the radius of the orbit,

and the effective cross-sectional area of this Rydberg state,

#76. Suppose that the splitting of energy levels shown in Fig. 39-4 was produced by a 1.6-T magnetic field. (a) What is the separation in energy between adjacent m, levels for the same £? (b) How many different wavelengths will there be for 3d to 2p transitions, if m; can change only by + 1 or 0? (c) What is the wavelength for each of these transitions? 7 Populations in the H atom. Use the Boltzmann factor (Eq. 39-16) to estimate the fraction n=2 and n =3 levels (relative to thermal equilibrium at (a) 7 = 300K [Note: Since there are eight states with

with n = 1, multiply your result for n =2 by § = 4; do similarly for n = 3.] (c) Given 1.0 g of hydrogen, estimate

the number of atoms in each state at 7 = 6000 K. (d) Estimate the numberof n =3 to n=1and n=2 to n=1 photons that will be emitted per second at 7 = 6000 K. Assume that the lifetime of each excited state is 107%s. [Hint: To evaluate a large exponent, you can use base-10

logarithms, Appendix A.]

Answers to Exercises

A: 2,1,0, -1, -2. B: (b). C: (d).

E: (b), (c). F: 15%2522p%3523p%3d1045%4p!, G: (d).

D: Add one line to Li in Table 39-2: 2,0, 0, — 3. 1070

CHAPTER 39

Quantum

Mechanics of Atoms

of H atoms in the the ground state) for and (b) 7 = 6000 K. n = 2 and only two

This computer processor chip contains over 800 million transistors, plus diodes and other semiconductor electronic

elements,

all in

a space

smaller than a penny. Before discussing semiconductors and their applications, we study the quantum theory description of bonding between atoms to form molecules,

and

how

it

explains

molecular behavior. We then examine how atoms and molecules form solids, with emphasis on metals as well as on semiconductors and their use in electronics.

~Molecules and Solids CHAPTER-OPENING

QUESTION—Guess

now!

As a metal is heated, how does the rms speed (v,y) of the electrons change? Assume a temperature change of about 30C° near room temperature. (a) vy increases linearly with temperature. (b) v decreases linearly with temperature. (¢) vy increases exponentially with temperature. (d) vy decreases exponentially with temperature. (e) vms changes very little as the temperature is increased. ince its development in the 1920s, quantum mechanics has had a profound influence on our lives, both intellectually and technologically. Even the way we view the world has changed, as we have seen in the last few Chapters. Now we discuss how quantum mechanics has given us an understanding of

the structure of molecules

and matter in bulk, as well as a number

applications including semiconductor devices.

40-1

of important

CONTENTS 40-1 Bonding in Molecules 40-2 Potential-Energy

Diagrams for Molecules

40-3

Weak (van der Waals)

40-4

Molecular Spectra

Bonds

40-5

Bonding in Solids

40-6

Free-Electron Theory of Metals; Fermi Energy

40-7

Band Theory of Solids

40-8

Semiconductors and

Doping

40-9

Semiconductor Diodes

40-10

Transistors and

Integrated Circuits

Bonding in Molecules

One of the great successes of quantum mechanics was to give scientists, at last, an understanding of the nature of chemical bonds. Since it is based in physics, and ’because this understanding is so important in many fields, we discuss it here.

1071

By a molecule, we mean together so as to function as we say that a chemical bond chemical bond: covalent and these two types.

a group of two or more atoms that are strongly held a single unit. When atoms make such an attachment, has been formed. There are two main types of strong ionic. Many bonds are actually intermediate between

Covalent Bonds Nucleus (+1e)

Nucleus (+1e)

(a)

w2

;

same atom, for then they would have identical quantum numbers in violation

%

nucleus

nucleus

(b) FIGURE 40-1 Electron probability distribution (electron cloud) for two H atoms when the spins are the same (§ =1): (a) electron cloud; (b) projection of |y[?> along the line through the centers of the two atoms.

FIGURE 40-2 Electron probability distribution for two H atoms when the spins are opposite (S = 0): (a) electron cloud; (b) projection of |* along the line through the centers of the atoms. In this case a bond is formed because the positive nuclei are attracted to the concentration of negative charge between them. This is a hydrogen molecule, H, . Nucleus

Nucleus (+1e)

(+1e)

lwl? | T

[ T

!

nucleus

1072

!

(b)

nucleus

CHAPTER 40

To understand how covalent bonds are formed, we take the simplest case, the bond that holds two hydrogen atoms together to form the hydrogen molecule, H,. The mechanism is basically the same for other covalent bonds. As two H atoms approach each other, the electron clouds begin to overlap, and the electrons from each atom can “orbit” both nuclei. (This is sometimes called “sharing” electrons.) If both electrons are in the ground state (n = 1) of their respective atoms, there are two possibilities: their spins can be parallel (both up or both down), in which case the total spin is S = 1 + 1 = 1; or their spins can be opposite (m; = +% for one, m; = —% for the other), so that the total spin § = 0. We shall now see that a bond is formed only for the S = 0 state, when the spins are opposite. First we consider the S = 1 state, for which the spins are the same. The two electrons cannot both be in the lowest energy state and be attached to the of the exclusion principle. The exclusion principle tells us that, because no two electrons can occupy the same quantum state, if two electrons have the same quantum numbers, they must be different in some other way—namely, by being in different places in space (for example, attached to different atoms). Thus, for S = 1, when the two atoms approach each other, the electrons will stay away from each other as shown by the probability distribution of Fig. 40-1. The electrons spend very little time between the two nuclei, so the positively charged nuclei repel each other and no bond is formed. For the S = 0 state, on the other hand, the spins are opposite and the £ electrons are consequently in different quantum states (m; is different, +3 for one, — 1 for the other). Hence the two electrons can come close together, and the probability distribution looks like Fig. 40-2: the electrons can spend much of their time between the two nuclei. The two positively charged nuclei are attracted to the negatively charged electron cloud between them, and this is the attraction that holds the two hydrogen atoms together to form a hydrogen molecule. This is a covalent bond. The probability distributions of Figs. 40-1 and 40-2 can perhaps be better understood on the basis of waves. What the exclusion principle requires is that when the spins are the same, there is destructive interference of the electron wave functions in the region between the two atoms. But when the spins are opposite, constructive interference occurs in the region between the two atoms, resulting in a large amount of negative charge there. Thus a covalent bond can be said to be the result of constructive interference of the electron wave functions in the space between the two atoms, and of the electrostatic attraction of the two positive nuclei for the negative charge concentration between them. Why a bond is formed can also be understood from the energy point of view. When the two H atoms approach close to one another, if the spins of their electrons are opposite, the electrons can occupy the same space, as discussed above. This means that each electron can now move about in the space of two atoms instead of in the volume of only one. Because each electron now occupies more space, it is less well localized. From the uncertainty principle with Ax larger, we see that Ap and the minimum momentum can be less. With less momentum, each electron has less energy when the two atoms combine than when they are separate. That is, the molecule has less energy than th two separate atoms, and so is more stable. An energy input is required to bre: the H, molecule into two separate H atoms, so the H, molecule is a stable entity.

Molecules and Solids

This is what we mean by a bond. The energy required to break a bond is called the bond energy, the binding energy, or the dissociation energy. For the hydrogen molecule, H,, the bond energy is 4.5eV.

r.)nic Bonds

+

An ionic bond is, in a sense, a special case of the covalent bond. Instead of the electrons being shared equally, they are shared unequally. For example, in sodium chloride (NaCl), the outer electron of the sodium spends nearly all its time around the chlorine (Fig. 40-3). The chlorine atom acquires a net negative charge as a

Na

FIGURE 40-3

result of the extra electron, whereas the sodium atom is left with a net positive charge. The electrostatic attraction between these two charged atoms holds them

Cl

Probability

distribution |* for the outermost electron of Na in NaCl.

together. The resulting bond is called an ionic bond because it is created by the attraction between the two ions (Na™ and CI7). But to understand the ionic bond, we must understand why the extra electron from the sodium spends so much of its time around the chlorine. After all, the chlorine atom is neutral; why should it attract another electron? The answer lies in the probability distributions of the two neutral atoms. Sodium contains 11 electrons, 10 of which are in spherically symmetric closed shells (Fig. 40-4). The last electron spends most of its time beyond these closed shells. Because the closed shells have a total charge of —10e and the nucleus has

= FIGURE 40-4 In a neutral sodium atom, the 10 inner electrons shield

It is not held very strongly. On the other hand, 12 of chlorine’s 17 electrons form

electronis attracted by a net charge

charge +11e, the outermost electron in sodium “feels” a net attraction due to +1e.

the nucleus, so the single outer

closed shells, or subshells (corresponding to 152s22p3s?). These 12 electrons form ~ ©f Tl

a spherically symmetric shield around the nucleus. The other five electrons are in 3p states whose probability distributions are not spherically symmetric and have a form similar to those for the 2p states in hydrogen shown in Fig. 39-9. Four of these 3p electrons can have “doughnut-shaped” distributions symmetric about the z axis, as shown in Fig. 40-5. The fifth can have a “barbell-shaped” distribution (as fpr my = 0 in Fig. 39-9), which in Fig. 40-5 is shown only in dashed outline ‘cause it is half empty. That is, the exclusion principle allows one more electron to be in this state (it will have spin opposite to that of the electron already there). If an extra electron—say from a Na atom—happens to be in the vicinity, it can be in this state, perhaps at point x in Fig. 40-5. It could experience an attraction due to as much as +5e because the +17e of the nucleus is partly shielded at this point by the 12 inner electrons. Thus, the outer electron of a sodium atom will be more strongly attracted by the +5e of the chlorine atom than by the +1e of its own atom. This, combined with the strong attraction between the two ions when the extra electron stays with the Cl”, produces the charge distribution of Fig. 40-3, and hence the ionic bond.

—10e

Last (3s) electron

FIGURE 40-5 Neutral chlorine atom. The +17e of the nucleus is shielded by the 12 electrons in the inner shells and subshells. Four of the five 3p electrons are shown in doughnut-shaped clouds (seen in cross section at left and right), and the fifth is in the dashed-line cloud concentrated about the z axis (vertical). An extra electron at x will be attracted by a net charge that can be as much as +3Se.

SECTION 40-1

Bonding in Molecules

1073

H (+)

Partial Ionic Character of Covalent Bonds A pure covalent bond in which the electrons are shared equally occurs mainly in symmetrical molecules such as H,, O,, and Cl,. When the atoms involved are different from each other, it is usual to find that the shared electrons are morc®y likely to be in the vicinity of one atom than the other. The extreme case is an ionic bond; in intermediate cases the covalent bond is said to have a partial ionic character. The molecules themselves are polar—that is, one part (or parts) of the molecule has a net positive charge and other parts a net negative charge. An example is the water molecule, H,O (Fig. 40-6). The shared electrons are more likely to be found around the oxygen atom than around the two hydrogens. The reason is similar to that discussed above in connection with ionic bonds. Oxygen has eight electrons

(1s*2s%2p*), of which four form a spherically symmetric core and the other four

could

Hit) FIGURE 40-6

H,O is polar.

The water molecule

have,

for

example,

a doughnut-shaped

distribution.

The

barbell-shaped

distribution on the z axis (like that shown dashed in Fig. 40-5) could be empty, so electrons from hydrogen atoms can be attracted by a net charge of +4e. They are also attracted by the H nuclei, so they partly orbit the H atoms as well as the O atom. The

net effect is that there is a net positive charge on each H atom (less than +1e),

because the electrons spend only part of their time there. And, there is a net

FIGURE 40-7 Potential energy U as a function of separation r for two

pist hasgesof @) ke signsnd

P

Repulsive force (two like charges)

negative charge on the O atom.

()2

Potential-Energy Diagrams for

Molecules

It is useful to analyze the interaction between two objects—say, between two atoms or molecules—with the use of a potential-energy diagram, which is a plot of the potential energy versus the separation distance. For the simple case of two point charges, g, and ¢,, the potential energy U is given by (see Chapter 23) &g 1 4mrey 1

“

)

where r is the distance between the charges, and the constant (1/47re,) is equal to

9.0 X 10°N-m?/C2 If the two charges have the same sign, the potential energy U is

U r Attractive force (unlike charges)

positive for all values of r, and a graph of U versus r in this case is shown in Fig. 40-7a. The force is repulsive (the charges have the same sign) and the curve rises as r decreases; this makes sense since work is done to bring the charges together, thereby increasing their potential energy. If, on the other hand, the two charges are of the opposite sign, the potential energy is negative because the product g, g, is negative. The force is attractive in this case, and the graph of U (x —1/r) versus r looks like Fig. 40-7b. The potential energy becomes more negative as r decreases. Now let us look at the potential-energy diagram for the formation of a covalent bond, such as for the hydrogen molecule, H,. The potential energy U of one H atom in the presence of the other is plotted in Fig. 40-8. Starting at large r, U decreases as the atoms approach, because the electrons concentrate between the two nuclei (Fig. 40-2), so attraction occurs. However, at very short distances, the electrons would be “squeezed out”—there is no room for them between the two nuclei. U

FIGURE 40-8 .

(right) Potential-energy diagram for .

H, mglec_ule, r is the separation o'f the two H atoms.

The binding energy (the energy difference between

U = 0 and the lowest energy state near the bottom

of the well) is 4.5 eV, and r, = 0.074 nm. 1074

CHAPTER 40

Molecules and Solids

"o

0

r

4 Bindi

e]rflar]gng

This part

corresponds

to repulsive force

y

This part corresponds

to attractive force

P

gy

-

Without the electrons between them, each nucleus would feel a repulsive force due to the other, so the curve rises as r decreases further. There is an optimum separation of the atoms, r, in Fig. 40-8, at which the energy is lowest. This is the point of greatest tability for the hydrogen molecule, and r, is the average separation of atoms in the I, molecule. The depth of this “well” is the binding energy,” as shown. This is how much energy must be put into the system to separate the two atoms to infinity, where U = 0. For the H, molecule, the binding energy is about 4.5¢V and r, = 0.074 nm. In molecules made of larger atoms, say, oxygen or nitrogen, repulsion also occurs at short distances, because the closed inner electron shells begin to overlap and the exclusion principle forbids their coming too close. The repulsive part of the curve rises even more steeply than 1/r. A reasonable approximation to the potential energy, at least in the vicinity of ry, is U

=

_rA”’ + r—li

(40-1)

where A and B are constants associated with the attractive and repulsive parts of the potential energy and the exponents m and n are small integers. For ionic and some covalent bonds, the attractive term can often be written with m =1

(Coulomb potential).

For many bonds, the potential-energy curve has the shape shown in Fig. 40-9.

There is still an optimum distance r, at which the molecule is stable. But when the

atoms approach from a large distance, the force is initially repulsive rather than attractive. The atoms thus do not interact spontaneously. Instead, some additional energy must be injected into the system to get it over the “hump” (or barrier) in the potential-energy diagram. This required energy is called the activation energy. The curve of Fig. 40-9 is much more common than that of Fig. 40-8. The activation energy often reflects a need to break other bonds, before the one under discussion can be made. For example, to make water from O, and H,, the H, and 0, molecules must first be broken into H and O atoms by an input of energy; this is what the activation energy represents. Then the H and O atoms can combine to rm H,O with the release of a great deal more energy than was put in initially. [he initial activation energy can be provided by applying an electric spark to a mixture of H, and O,, breaking a few of these molecules into H and O atoms. The resulting explosive release of energy when these atoms combine to form H,0O quickly provides the activation energy needed for further reactions, so additional H, and O, molecules are broken up and recombined to form H,O. The potential-energy diagrams for ionic bonds can have similar shapes. In NaCl, for example, the Na™ and Cl™ ions attract each other at distances a bit larger than some ry, but at shorter distances the overlapping of inner electron shells gives rise to repulsion. The two atoms thus are most stable at some intermediate separation, ry, and for many bonds there is an activation energy. "The binding energy corresponds not quite to the bottom of the potential energy curve, but to the lowest quantum energy state, slightly above the bottom, as shown in Fig, 40-8,

U Activation energy

p

0

T

0

r

Binding energy

r

s

N

Repulsion

=

¥

Attraction

=

FIGURE 40-9 Potential-energy diagram for a bond requiring an activation energy.

L

Repulsion

SECTION 40-2

Potential-Energy Diagrams for Molecules

1075

DOVIR

ORI

ESTIMATE | Sodium

chloride

bond. A

potential-energy

diagram for the NaCl ionic bond is shown in Fig. 40-10, where we have set U = 0 for free Na and Cl neutral atoms (which are represented on the right in Fig. 40-10). Measurements show that 5.14eV are required to remove an electron from “ neutral Na atom to produce the Na™ ion; and 3.61 eV of energy is released when a. electron is “grabbed” by a Cl atom to form the CI™ ion. Thus, forming Na* and CI~ ions from neutral Na and Cl atoms requires 5.14eV — 3.61eV = 1.53eV of energy, a form of activation energy. This is shown as the “bump” in Fig. 40-10. But note that the potential-energy diagram from here out to the right is not really a function of distance—it is drawn dashed to remind us that it only represents the energy difference between the ions and the neutral atoms (for which we have chosen U = 0). (¢) Calculate the separation distance, r;, at which the potential of the Na™ and Cl™ ions drops to zero (measured value is r; = 0.94 nm). (b) Estimate the binding energy of the NaCl bond, which occurs at a separation 7, = 0.24 nm. Ignore the repulsion of the overlapping electron shells that occurs at this distance (and causes the rise of the potential-energy curve for r < ry, Fig. 40-10). FIGURE 40-10

U(eV

Example 40-1. Potential-

5 0)_

energy diagram for the NaCl bond. Beyond about r = 1.2 nm, the diagram is schematic

only, and represents the energy difference

between ions and neutral atoms. U chosen for the two separated atoms (not for the ions). [For the two ions, CI', the zero of potential energy at

1 =0.94 nm

' 1.0 1

0

= 0 is Na and Cl Na™ and r = oo

’

—1.0 1 =2.0 1 =3.04 —4.0-

corresponds to U ~ 1.53 eV on this diagram.]

'

'

oo

-

e e ’

Tsso

o

Naand CI~

(ions)

=5.01

SOLUTION

J;/?kv

e /

Naand Cl

(atoms)

(a) The potential energy of two point charges is given by Coulomb’s law:

S

) 4irey

T

-

where we distinguish U’ from the U in Fig. 40-10. This formula works for our two ions if we set U’ = 0 at r = oo, which in the plot of Fig. 40-10 corresponds to U = +1.53eV (Fig.40-10 is drawn for U = 0 for the free atoms). The point r, in Fig. 40-10 corresponds to U’ = —1.53eV relative to the two free ions. We solve the U’ equation above for r, setting g, = +e and ¢, = —e: r,

=

'

1

g

=

dme, U

(90 x 10°N-m?*/C?)(-1.60 X 107 C)(+1.60 X 107?C)

(—1.53eV)(1.60 x 10°J/eV)

= 0.94nm, which is just the measured value. (b) At ry = 0.24 nm, the potential energy of the two ions (relative to r = co for the two ions) is ,_

e

1

4mey

qq

T

(9.0 X 10°N-m?/C?)(-1.60 X 107 C)(+1.60 x 107 C) (0.24 X 10°m)(1.60 X 10 J/eV)

=

—6.0eV.

Thus, we estimate that 6.0 eV of energy is given up when Na™ and CI™ ions form a NaCl bond. Put another way, it takes 6.0 eV to break the NaCl bond and form the Na® and CI” ions. To get the binding energy—the energy to separate the NaCl into Na and Cl atoms—we need to subtract out the 1.53 eV (the “bump” in Fig. 40-10) needed to ionize them: binding energy = 6.0eV — 1.53eV = 4.5¢eV. The measured value (shown on Fig. 40-10) is 4.2 eV. The difference can be attributed to the energy associated with the repulsion of the electron shells at this distance.

@PHYSlcs

APPLIED

ATP and energy in the cell

1076

CHAPTER 40

Sometimes the potential energy of a bond looks like case, the energy of the bonded molecule, at a separation ry, is no bond (r = co). That is, an energy input is required the binding energy is negative), and there is energy release

that of Fig. 40-11. In thi is greater than when there to make the bond (hence when the bond is broken.

Such a bond is stable only because there is the barrier of the activation energy. This type of bond is important in living cells, for it is in such bonds that energy can be stored efficiently in certain molecules, particularly ATP (adenosine triphosphate). The bond that connects the last phosphate group (designated (P) in Fig. 40-11) to the rest f‘ of the molecule (ADP, meaning adenosine diphosphate, since it contains only two phosphates) has potential energy of the shape shown in Fig. 40-11. Energy is stored in this bond. When the bond is broken (ATP — ADP + (P)), energy is released and this energy can be used to make other chemical reactions “go.” In living cells, many chemical reactions have activation energies that are often on the order of several eV. Such energy barriers are not easy to overcome in the cell. This is where enzymes come in. They act as catalysts, which means they act to lower the

Y

To

r

ATP

ADP +(®

~ FIGURE 40-11

Potential-energy

diagram for the formation of ATP

activation energy so that reactions can occur that otherwise would not. Enzymes act ~ from ADP and phosphate (®).

via the electrostatic force to distort the bonding electron clouds, so that the initial bonds are easily broken.

Bond length. Suppose a diatomic molecule has a potential energy given by U = —(1/47760)(e2/r§+ B/rS where B =1.0 X 107*J-m® and r is the distance between the centers of the two atoms. Determine the expected equilibrium separation of the two atoms (bond length of the molecule).

APPROACH The force between the two atoms is given by F = —dU/dr (analogous to Eq. 8-7). The classical equilibrium separation is found by setting F = 0. SOLUTION F = —dU/dr = (1/417&0)(—e2/r2) - B(—6/r7) = —(1/4me,)(e*/r*) + 6B/r’ = 0. Then

r’ = 6B(4mey)/e* = 2.6 X 107°m’, Sor=12x10"m = 0.12nm. NOTE

This can be shown to be a stable equilibrium point by checking the sign

of d°U/dr? or by evaluating F at positions slightly larger and smaller than the

€

equilibrium position.

40-3 Weak (van der Waals) Bonds Once a bond between two atoms or ions is made, energy must normally be supplied to break the bond and separate the atoms. As mentioned in Section 40-1, this energy is called the bond energy or binding energy. The binding energy for covalent and ionic bonds is typically 2 to 5eV. These bonds, which hold atoms together to form molecules, are often called strong bonds to distinguish them from so-called “weak bonds.” The term weak bond, as we use it here, refers to attachments between molecules due to simple electrostatic attraction—such as between polar molecules (and not within a polar molecule, which is a strong bond). The strength of the attachment is much less than for the strong bonds. Binding energies are typically in the range 0.04 to 0.3 eV—hence their name “weak bonds.” Weak bonds are generally the result of attraction between dipoles (Sections 21-11, 23-6). For example, Fig. 40-12 shows two molecules, which have permanent dipole

moments, attracting one another. Besides such dipole—dipole bonds, there can also be

dipole-induced dipole bonds, in which a polar molecule with a permanent dipole moment can induce a dipole moment in an otherwise electrically balanced (nonpolar)

molecule, just as a single charge can induce a separation of charge in a nearby object (see Fig. 21-7). There can even be an attraction between two nonpolar molecules,

because their electrons are moving about: at any instant there may be a transient separation of charge, creating a brief dipole moment and weak attraction. All these weak bonds are referred to as van der Waals bonds, and the forces involved van der Waals forces. The potential energy has the general shape shown in Fig. 40-8, with the attractive van der Waals potential energy varying as 1/r°. (A When one of the atoms in a dipole—dipole bond is hydrogen, as in Fig. 40-12, it is called a hydrogen bond. A hydrogen bond is generally the strongest of the weak bonds, because the hydrogen atom is the smallest atom and can be approached more closely. Hydrogen bonds also have a partial “covalent” character: that is, electrons between the two dipoles may be shared to a small extent, making a stronger, more lasting bond.

FIGURE 40-12 The C"— O~ and =~ H"—N~ dipoles attract each other.

(These dipoles may be part of, for

e€xample, the nucleotide bases cytosine

~ 2nd guanine in DNA molecules. See

Flg',‘?u_li‘) The: = it d_ cl;arges gggiogo?‘ée EERS

us

0.12 FW’I

' t+—0.29 nm— =

0L8

SECTION 40-3

1077

FIGURE 40-13 (a) Section of a DNA double helix. The red dots represent hydrogen bonds between the two strands. (b) “Close-up” view: cytosine (C) and guanine (G) molecules on separate strands of a DNA double helix are held together by the hydrogen bonds (red dots) involving an H on one molecule attracted to an N~ or

C"— O~ of amolecule on the adjacent

chain. See also Section 21-12 and Figs. 21-47 and 21-48.

@

PHYSICS

APPLIED DNA

(b) Weak bonds are important in liquids and solids when strong bonds are absent (see Section 40-5). They are also very important for understanding the activities of cells, such as the double helix shape of DNA (Fig. 40-13), and DNA replication. The average kinetic energy of molecules in a living cell at normal temperatures (7 ~ 300 K) is around 3k7 ~ 0.04 eV, about the magnitude of weak bonds. This means that a weak bond can readily be broken just by a molecular collision. Hence weak bonds are not very permanent—they are, instead, brief attachments. This helps them play particular roles in the cell. On the other hand, strong bonds—those that hold molecules together— are almost never broken simply by molecular collision. Thus they are relatively permanent. They can be broken by chemical action (the making of even stronger bonds), and this usually happens in the cell with the aid of an enzyme, which is a protein molecule.

Nucleotide energy. Calculate the potential energy between

a C==0 dipole of the nucleotide base cytosine and the nearby H—N dipole of guanine, assuming that the two dipoles are lined up as shown in Fig. 40-12. Dipole moment (= gf) measurements (see Table 23-2 and Fig. 40-12) give -

30 X S N = 010 de S ey T = 80 X

I and

APPROACH dipole

due

100°C-m =30 X 102C = 0.19%, x 10°m 1 e -30 = 6.77 Xx 10°2C 1002°C = 0.42¢ 0.42e. X10°°C-m o

We want to find the potential energy of the two charges in one

to the

two

charges

in the

other,

since

this will

be

equal

to

the

work needed to pull the two dipoles infinitely far apart. The potential energy U of a

charge ¢ in the presence of a charge ¢, is U = (1/4me,)(q1q2/r12) Where 1/4mey = 9.0 X 10°N-m?/C? and ry, is the distance between the two charges.

SOLUTION

The potential energy U consists of four terms:

U

= Ucy + Uex + Uon + Uon,

where Ucy means the potential energy of C in the presence of H, and similarly for the other terms. We do not have terms corresponding to C and O, or N and H, because

the two dipoles are assumed

distances shown in Fig. 40-12, we get: U

=

1

I:quH

Il

4mey |

ren

4 O ren

to be stable entities. Then, using the

| 9o

CIOQN:]

Ton

(9.0 x 10° N-mZ/CZ)(6.7)(3.O)(

ToN

11

031

041

1

019

1 \(oey

029/ (10°m)

-1.86 X 107°] = —-0.12eV. The potential energy is negative, meaning 0.12 eV of work (or energy input) is required to separate the dipoles. That is, the binding energy of this “weak” 04‘\ hydrogen bond is 0.12 eV. This is only an estimate, of course, since other charges in the vicinity would have an influence too. 1078

CHAPTER 40

_~

DNA

New protein chain of

4 amino acids (a Sth is being added) Growing end

of m-RNA

! I\

|

t-RNA

Ribosome

Anticodons

*Protein Synthesis Weak bonds, especially hydrogen bonds, are crucial to the process of protein synthesis. Proteins serve as structural parts of the cell and as enzymes to catalyze chemical reactions needed for the growth and survival of the organism. A protein molecule consists of one or more chains of small molecules known as amino acids. There are 20 different amino acids, and a single protein chain may contain hundreds of

them in a specific order. The standard model for how amino acids are connected together rin the correct order to form a protein molecule is shown schematically in Fig. 40-14.

It begins at the DNA double helix: each gene on a chromosome contains the information for producing one protein. The ordering of the four bases, A, C, G, and T, provides the “code,” the genetic code, for the order of amino acids in the protein. First, the DNA double helix unwinds and a new molecule called messenger-RNA (m-RNA) is synthesized using one strand of the DNA as a “template.” m-RNA is a chain molecule containing four different bases, like those of DNA (Section 21-12) except that thymine (T) is replaced by the similar uracil molecule (U). Near the top left in Fig. 40-14, a C has just been added to the growing m-RNA chain in much the same way that DNA replicates; and an A, attracted and held close to the T on the DNA chain by the electrostatic force, will soon be attached to the C by an enzyme. The order of the bases, and thus the genetic information, is preserved in the m-RNA because the shapes of the molecules only allow the “proper” one to get close enough so the electrostatic force can act to form weak bonds. Next, the m-RNA is buffeted about in the cell (kinetic theory) until it gets close to a tiny organelle known as a ribosome, to which it can attach by electrostatic attraction (on the right in Fig. 40-14), because their shapes allow the charged parts to get close enough to form weak bonds. Also held by the electrostatic force to the ribosome are one or two transfer-RNA (t-RNA) molecules. These t-RNA molecules “translate” the genetic code of nucleotide bases into amino acids in the following way. There is a different t-RNA molecule for each amino acid and each combination of three bases. On one end of a t-RNA molecule is an amino acid. On the other end of the t-RNA molecule is the appropriate “anticodon,” a set of three nucleotide bases that “code” for that amino acid. If all three bases of an anticodon match the three bases of the “codon” on the m-RNA (in the sense of G to C and A to U), the anticodon is attracted electrostatically to the m-RNA codon and that t-RNA molecule is held there briefly. The ribosome has two particular attachment sites which hold two t-RNA molecules while nzymes link their two amino acids together to lengthen the amino acid chain (yellow in

FIGURE 40-14

Protein synthesis.

See text for details.

FHYSICS Protein synthesis

SFPLIED

ig. 40-14). As each amino acid is connected by an enzyme (four are already connected

in Fig. 40-14, top right, and a fifth is about to be connected), the old t-RNA molecule is removed—perhaps by a random collision with some molecule in the cellular fluid. A new one soon becomes attracted as the ribosome moves along the m-RNA.

SECTION 40-3

1079

This process of protein synthesis is often presented as if it occurred in clockwork fashion—as if each molecule knew its role and went to its assigned place. But this is not the case. The forces of attraction between the electric charges of the molecules are rather weak and become significant only when the molecules can come close together and several weak bonds can be made. Indeed, if thc\ shapes are not just right, there is almost no electrostatic attraction, which is why there are few mistakes. The fact that weak bonds are weak is very important. If they were strong, collisions with other molecules would not allow a t-RNA molecule to be released from the ribosome, or the m-RNA to be released from the DNA. If they were not temporary encounters, metabolism would grind to a halt. As each amino acid is added to the next, the protein molecule grows in length until it is complete. Even as it is being made, this chain is being buffeted about in the cellular sea—we might think of a wiggling worm. But a protein molecule has electrically charged polar groups along its length. And as it takes on various

shapes, the electric forces of attraction between different parts of the molecule will

eventually lead to a particular configuration that is quite stable. Each type of protein has its own special shape, depending on the location of charged atoms. In the last analysis, the final shape depends on the order of the amino acids.

40-4 3p

N

===

2s

Afoming molectle

(a) FIGURE 40-15

(b) (a) The individual

energy levels of an isolated atom

become (b) bands of closely spaced levels in molecules, as well as in

solids and liquids.

Molecular Spectra

When atoms combine to form molecules, the probability distributions of the outer electrons overlap and this interaction alters the energy levels. Nonetheless, molecules can undergo transitions between electron energy levels just as atoms do. For example, the H, molecule can absorb a photon of just the right frequency to excite one of its ground-state electrons to an excited state. The excited electron can then return to the ground state, emitting a photon. The energy of photons emitted by molecules can be of the same order of magnitude as for atoms, typically 1 to 10eV, or less. Additional energy levels become possible for molecules (but not for atoms ™ b.ecause the molecule as a whole can rotate, and the atoms of the molecule can vibrate relative to each other. The energy levels for both rotational and vibrational levels are quantized, and are generally spaced much more closely (107 to 107" eV) than the electronic levels. Each atomic energy level thus becomes a set of closely

spaced levels corresponding to the vibrational and rotational motions, Fig. 40-15.

Transitions from one level to another appear as many very closely spaced lines. In fact, the lines are not always distinguishable, and these spectra are called band spectra.

Each type of molecule has its own characteristic spectrum, which can be used for

identification and for determination of structure. We rotational and vibrational states in molecules.

now

look in more

detail at

Rotational Energy Levels in Molecules FIGURE 40-16

Diatomic molecule

rotating about a vertical axis.

=~ We consider only diatomic molecules, although the analysis can be extended to

polyatomic molecules. When a diatomic molecule rotates about its center of mass as shown in Fig. 40-16, its kinetic energy of rotation (see Section 10-8) is 5

(Iw)?

—lo* = —/——> 2 21 where /w is the angular momentum (Section 11-1). Quantum mechanics predicts quantization of angular momentum just as in atoms (see Eq. 39-3): Eo

=

lo =

VAE+ DA,

£ =012,

where £ is an integer called the rotational angular momentum quantum number. Thus the rotational energy is quantized: Eqo

1080

CHAPTER 40

=

lw)?

o)21

72

je+ X21

=012

(40-2)

Transitions between rotational energy levels are subject to the selection rule (as in Section 39-2): Al = *1.

The energy of a photon emitted or absorbed for a transition between rotational

-

states with angular momentum quantum number

M

£ and £ — 1 will be

1

= Eo— By = 5-0(8 + 1) = 2= (€ = 1)() #?

f‘

AE = 5#2/1

[f isi for upper]

(40-3)

energy state

We see that the transition energy increases directly with £. Figure 40-17 shows some of the allowed rotational energy levels and transitions. Measured absorption lines fall in the microwave or far-infrared regions of the spectrum, and their frequencies are generally 2,3, 4, -+ times higher than the lowest one, as predicted by Eq. 40-3. EXERCISE A Determine the three lowest rotational energy states (in eV) for a nitrogen

molecule which has a moment of inertia / = 1.39 X 10 * kg -m?

The moment of inertia of the molecule in Fig. 40-16 rotating about its center of mass (Section 10-5) is

I = mri+

myr,

where ry and r, are the distances of each atom from their common center of mass. We can show (in Example 40-4 below) that / can be written I

=

ny

==

(40_4)

[.er,

=

_m_l__’nz._rz

{=4

10 h?i

AE = 4%

=3 I =

f=1 i

6 2

)

3 fi_z

1 éz ol=

AE = 3421

AE = 2121 — 2 AE=#I

FIGURE 40-17 Rotational energy levels and allowed transitions (emission and absorption) for a diatomic molecule. Upward-pointing arrows represent absorption of a photon, and downward arrows

represent emission of a photon.

ny

where r = r; + r, is the distance between the two atoms of the molecule and w = mymy/(m; + my) is called the reduced mass. If m, = m,, then p = Lmy=im,. SR IR S Reduced mass. Show that the moment of inertia diatomic molecule rotating about its center of mass can be written I

=

where

of a

p,rz, myni,

®

my

+

my

is the reduced mass, Eq. 40—4, and r is the distance between the atoms. SOLUTION The moment of inertia of a single particle of mass m a distance r from the rotation axis is 7 = mr? (Eq.10-11 or 10-13). For our diatomic molecule (Fig. 40-16) I

=

mlrf + mzr%.

Now r = r, +r, and myry the center of mass. Hence

rl

r

—

m

1+

=

—1

m,r, because the axis of rotation passes through

myr m;

my

Similarly, np,

=

!

+ m

2

myr

———— my

+ my

Then (see first equation of this Solution)

r

where

which is what we wished to show.

—

SECTION 40-4

Molecular Spectra

D¢\ | LB Rotational transition. A rotational transition £ = 1 to £ = 0 for the molecule CO has a measured absorption wavelength A; = 2.60 mm (microwave region). Use this to calculate (a) the moment of inertia of the CO molecule, and (b) the CO bond length, . (¢) Calculate the wavelengths of the next three rotational transitions, and the energies of the photon emitted for eacfl of these four transitions. APPROACH The absorption wavelength is used to find the energy of the absorbed photon. The moment of inertia 7 is found from Eq. 40-3, and the bond length r from Eq. 40-4. SOLUTION (@) The photon energy, E = hf = hc/A, equals the rotational energy level difference, AE,;. From Eq. 40-3, we can write

he

2

fiT(?:AE:hf: With

£ =1

Ay

(the upper state) in this case, we solve for I 7

=

h he

—

A

hh 4nic

=

7'

=

1.46 X 10 % kg-m? (b) The

lu

masses

=

of

C

O

and

(6.63 X 1077 -5)(2.60 X 10~ m) 4m*(3.00 X 10°m/s)

are

12.0u

and

1.66 X 10 ¥ kg. Thus the reduced mass is

~ (120)(16.0)

mym,

IJ'=

my;

+

g0

my

16.0u,

respectively,

where

(166 X 107kg) = 1.14 X 10 kg

or 6.86 u. Then, from Eq. 40-4, the bond length is

46 X 10 ®kg-m>

=

1.14 x 10 % kg

(c) From Eq. 40-3, AE «x £.

Thus,

for

£ =2

close

to

measured

£=2,

to

113 x10"m

) ==

0.113nm.

o

Hence A = ¢/f = hc/AE is proportional to 1/£. transitions, A, = 1A, = 1.30mm. For £=3 to

£=1

\3=4%A =087mm.

And for £ =4

values.

The

energies

of

to £=3, the

A, = 0.65mm.

photons,

All ar

Af = hc/A,

Ll:e_spectively 48 x 10eV, 9.5 X 10%eV, 1.4 x 103eV, and 1.9 X 103 eV.

are

Vibrational Energy Levels in Molecules The potential energy of the two atoms in a typical diatomic molecule has the shape shown in Fig. 40-8 or 40-9, and Fig. 40-18 again shows the potential energy for the H, molecule. We note that the potential energy, at least in the vicinity of the equilibrium separation r,, closely resembles the potential energy of a simple harmonic oscillator, U = $kx?, which is shown superposed in dashed

lines. Thus, for small

displacements

v harmonic oscillator (Usyo = kx?,

that Usyo

ry, each

atom

1

\

FIGURE 40-18 Potential energy for the H, molecule and for a simple 0

with |x| = |r — rg|).The 0.50-eV energy height marked is for use in Example 40-6 to estimate k. [Note

from

] —

\ ‘\

T

experiences

_1 Usho (= ikxz)

0.1 nm ,’ T } T ro = 0.074 nm )

\

a

=

\ H, molecule

= 0 is not the same as

U = 0 for the molecule.]

—45evV+

0.5 eV (for Example 40-6)

restoring force approximately proportional to the displacement, and the molecule vibrates as a simple harmonic oscillator (SHO). The classical frequency of vibration is

f 1082

CHAPTER 40

@-s

=

where k is the “stiffness constant” (as for a spring, Chapter 14) and instead of the mass

m

we

must

again

use

the reduced

mass

p = mlmz/(ml

+ mz).

(This

is

shown in Problem 19.) The Schrodinger equation for the SHO yields solutions for energy that are quantized according to

(40-6)

v o= 0,1,2,,

= (v + 3)hf

Eg

potential energy

Vibrational quantum number v

S —

fi(here f is given by Eq. 40-5 and v is an integer called the vibrational quantum - aumber. The lowest energy state (v = 0) is not zero (as for rotation) but has E = }hf. This is called the zero-point energy.’ Higher states have energy 3 hf, 3 hf, and so on, as shown in Fig. 40-19. Transitions are subject to the selection rule Av = %1, so allowed transitions occur only between adjacent states and all give off photons of energy

AEg, = hf.

(40-7)

This is very close to experimental values for small v; but for higher energies, the potential-energy curve (Fig. 40-18) begins to deviate from a perfect SHO curve, which affects the wavelengths and frequencies of the transitions. Typical transition energies are on the order of 107" eV, about 10 to 100 times larger than for rotational transitions, with wavelengths in the infrared region of the spectrum

(~ 10~ m). DOV

IR

ESTIMATE | Wavelength

for

H,. (a) Use

the

curve

of

Fig. 40-18 to estimate the value of the stiffness constant k for the H, molecule, and then (b) estimate the fundamental wavelength for vibrational transitions.

APPROACH To find k, we arbitrarily choose an energy height of 0.50 eV which is indicated in Fig. 40-18. By measuring directly on the graph, we find that this energy corresponds to a vibration on either side of 7y = 0.074 nm of about x = +0.017 nm.

SOLUTION

’L

'Z'hf

4 —1— 9if AE 3 — 2 —

Energy

7 Ehf S

Ehf

AE ] ———

3

0

ihf

FIGURE 40-19

Ehf 1

Allowed

vibrational energies of a diatomic molecule, where f is the

fundamental frequency of vibration,

given by Eq. 40-5. The energy levels are equally spaced. Transitions are allowed only between adjacent levels (Av = %1).

(a) For SHO, Ugyo = 3kx? and Ugyo = 0 at x = 0 (r = ry); then k

NOTE

Vibrational energy "

=

Wsio

2(0.50eV)(1.6 X 107°J/eV)

2

(17 X 10 my

~ 550 N/m.

This value of k would also be reasonable for a macroscopic spring.

(b) The reduced mass is w = my m,/(my +my) = my/2 = 5(1.0u)(1.66 X 10 7kg) =

0.83 X 10"% kg. Hence, using Eq. 40-5, e | B ) A= 7 2770\/; 27(3.0 X 108 m/s)

[0.83 x 107 kg _ “SONm 2300 nm,

which is in the infrared region of the spectrum. Experimentally, we do the inverse process: The wavelengths of vibrational transitions for a given molecule are measured, and from this the stiffness constant k can be calculated. The values of k calculated in this way are a measure of the strength of the molecular bond.

Vibrational energy levels in hydrogen. Hydrogen molecule

vibrations emit infrared radiation of wavelength around 2300 nm. (¢) What is the separation in energy between adjacent vibrational levels? (b) What is the lowest vibrational energy state? APPROACH The energy separation between adjacent vibrational levels is AE., = hf = hc/A. The lowest energy (Eq. 40-6) has v = 0. SOLUTION (a

AEj,

@) he (6.63 X 10737 +5)(3.00 X 108m/s) = hf = —A = (2300 x 107 s m)(1.60 x 1077J/eV) =

where the denominator includes the conversion factor A(b) The lowest vibrational energy has v = 0 in Eq. 40-6:

Eg

= (v+3)hf

from

=

0.54¢V,

joules

to eV.

= 3hf = 027eV.

"Recall this phenomenon for a square well, Fig. 38-8. *Forbidden transitions with

Av = 2

are emitted somewhat more weakly, but their observation can be

important in some cases, such as in astronomy.

SECTION 40-4

1083

Rotational plus Vibrational Levels

[}

3 2

b

————

4

L

Vibrational

R

Rotational {

When energy is imparted to a molecule, both the rotational and vibrational modes can be excited. Because rotational energies are an order of magnitude or so smaller than vibrational energies, which in turn are smaller than the electronic energy« levels, we can represent the grouping of levels as shown in Fig, 40-20. Transition. between energy levels, with emission of a photon, are subject to the selection rules:

[

S

states

states g

“3p” electronic B

§

Av

=

=*1

and

Al

Some allowed and forbidden (marked

=

1.

X) transitions are indicated in Fig. 40-20.

Not all transitions and levels are shown, and the separation between

E

41 =i

vibrational

levels, and (even more) between rotational levels, has been exaggerated. But we can clearly see the origin of the very closely spaced lines that give rise to the band spectra, as mentioned with reference to Fig. 4015 earlier in this Section. The spectra are quite complicated, so we consider briefly only transitions within the same electronic level, such as those at the top of Fig. 40-20. A transition from a state with quantum numbers v and £, to one with quantum numbers v + 1 and £ £ 1 (see the selection rules above), will absorb” a photon of energy:

; i

AE = AEg4 + AEy gt

h?

. 0>elesctt;toenic

=

§830 0

== hf -_ €5v#? where

L >

hf + (L + 1)_1"

we have

used

[(Af

0+1

Y

1)],

L > 0-1 [(M _ _1)], Egs. 40-3

and

40-7. Note

£ =012,

(40-8a)

€ == 1,23,

_ (40-8b)

that for

£ —

£ — 1 transitions,

FIGURE 40-20 Combined electronic, vibrational, and rotational

{ cannot be zero because there is then no state with £ = —1. Equations 40-8 ~ predict an absorption spectrum like that shown schematically in Fig. 40-21,

an X are not allowed by selection rules.

shows the molecular absorption spectrum of HCI, which follows this pattern very well.

energy levels. Transitions marked with

with transitions £ — £ — 1 on the left and £ — £ + 1 on the right. Figure 40-22

"Eqs. 40-8 are for absorption; for emission of a photon, the transition wouldbe

Al

Al

=—1 FIGURE 40-21

AY¢

Al

=—-1=-1

Af

=+1

Expected spectrum for

=1

f =

e =

to

Absorption

spectrum for HCI molecules. Lines on the left correspond to transitions where { — £ — 1; those on the right are for £ — £ + 1. Each line has a double peak because chlorine has two isotopes of different mass and different moment of inertia.

«—[e

—

|

hf

to

(=0 f =

-1 }———»

1

f =

AZ

=+1

P

Photon energy

to

~—

AlL=-1)

!

[=1

to

0

2

2

[2 —0+1

Al=+1)

U 1084

CHAPTER 40

£ — £ £ 3

Relative absorption

FIGURE 40-22

=+1

N

=2

— v — 1,

AL

=+1

transitions between combined rotational and vibrational states.

v

0330

.

0.340

I

0.350

“ il

0.360

Energy (eV)

.

T

0370

}

0.380

(Each line in the spectrum of Fig. 40-22 is split into two because Cl consists of two

isotopes of different mass; hence there are two kinds of HCl molecule with slightly different moments of inertia 7.)

DGR

EE N ESTIMATE | The HCl molecule. Estimate the moment of inertia

of the HCI molecule using the absorption spectrum shown in Fig. 40-22. For the purposes of a rough estimate you can ignore the difference between the two isotopes. APPROACH The locations of the peaks in Fig. Egs. 40-8. We don’t know what value of £ each peak we can estimate the energy difference between peaks SOLUTION From Egs. 40-8, the energy difference

40-22 should correspond to corresponds to in Fig. 40-22, but to be about AE’ = 0.0025eV. between two peaks is given by

fiZ AEI

AR

=

F ol

AE[+]

2

AE'"

=

-

AEI

(

=

7'

34 6.626 X 107*J-s/2

)

2

(0.0025¢V)(1.6 x 1079J/eV)

= 2.8 X 10

NOTE To get an idea of what this number means, we write I = ur? where u is the reduced mass (Example 40-4); then we calculate u:

my my

(1.0u)(35u)

“=m1+m2=

36u

kg-m?

(Eq. 40-4),

(1.66 x 10 kg/u) = 1.6 X 107 kg;

the bond length is given by (Eq. 40-4)

- (1)% ~ (2.8 X 104 kg - m? "7 \w) T \T16x107ke

1

)2 = 13X 100m,

which is the expected order of magnitude for a bond length.

40-5

Bonding in Solids

.

This U, is known as the ionic cohesive energy; it is a sort of “binding energy”—the energy (per ion) needed to take the solid apart into separated ions, one by one. A different type of bond occurs in metals. Metal atoms have relatively loosely held outer electrons. Present-day metallic bond theories propose that in a metallic solid, these outer electrons roam rather freely among all the metal atoms which, without their outer electrons, act like positive ions. The electrostatic attraction between the metal ions and this negative electron “gas” is what is believed, at least in part, to hold the solid together. The binding energy of metal bonds are typically 1 to 3 eV, somewhat weaker than ionic or covalent bonds (5 to 10eV in solids). The “free electrons,” according to this theory, are responsible for the high electrical and thermal conductivity of metals (see Sections 40-6 and 40-7). This theory also nicely accounts for the shininess of smooth metal surfaces: the electrons are free and can vibrate at any frequency, so when light of a range of frequencies falls on a metal, the electrons can vibrate in response and re-emit light of those same frequencies. Hence the reflected light will consist largely of the same frequencies as the incident light. Compare this to nonmetallic materials that have a distinct color—the atomic electrons exist only in certain energy states, and when white light falls on them, the atoms absorb at certain frequencies, and reflect other frequencies which make up the color we see. Here is a brief summary of important strong bonds: « ionic: an electron is stolen from one atom by another; covalent: electrons are shared by atoms within a single molecule; » metallic: electrons are shared by all atoms in the metal.

>

The atoms or molecules of some materials, such as the noble gases, can form only weak bonds with each other. As we saw in Section 40-3, weak bonds have very low binding energies and would not be expected to hold atoms together as a liquid or solid at room temperature. The noble gases condense only at very low temperatures, where the atomic kinetic energy is small and the weak attraction can then hold the atoms together.

40-6

Free-Electron Theory of Metals; Fermi Energy

Let us look more closely at the free-electron theory of metals mentioned in the preceding Section. Let us imagine the electrons trapped within the metal as being in a potential well: inside the metal, the potential energy is zero, but at the edges of the metal there are high potential walls. Since very few electrons leave the metal at room temperature, we can imagine the walls as being infinitely high (as in Section 38-8). At higher temperatures, electrons do leave the metal (we know that thermionic emission occurs, Section 23-9), so we must recognize that the well is of finite depth. In this simple model, the electrons are trapped within the metal, but are free to move about inside the well whose size is macroscopic—the size of the piece of metal. The energy will be quantized, but the spacing between energy levels will be very tiny (see Eq. 38-13) because the width of the potenti well £ is very large. Indeed, for a cube 1cm on a side, the number of states wit

energy between, say, 5.0 and 5.5 eV, is on the order of 10% (see Example 40-9). 1086

CHAPTER 40

Molecules and Solids

To deal with such vast numbers of states, which are so closely spaced as to seem continuous, we need to use statistical methods. We define a quantity known as the density of states, g(E), whose meaning is similar to the Maxwell distribution, Eq. 18-6 (see Section 18-2). That is, the quantity g(E)dE represents the number f states per unit volume that have energy between E and E + dE. A careful calculation (see Problem 41), which must treat the potential well as three dimensional, shows that

g(E)

=

8\/5

’7T"'l3/2

E1/2

g(E) g(E) o EV2

(40-10)

h3

where m is the mass of the electron. This function is plotted in Fig. 40-25. .

:

DOV IETN ESTIMATE | Electron states in copper. Estimate the number of states in the range 5.0 to 5.5eV available to electrons in a 1.0-cm cube of copper metal.

.

.

:

)

APPROACH Since g(E) is the number of states per unit volume per unit energy interval, the number N of states is approximately (it is approximate because AE is not small) N

~

where the volume SOLUTION

0

E

~ FIGURE 40-25

Density of states

as a function of energy E

g(E)

(Eq. 40-10).

g(E)V AE,

V = 1.0cm® = 1.0 X 10°m®

and AE = 0.50¢eV.

We evaluate g(E) at 5.25eV, and find (Eq. 40-10):

N ~ g(E)V AE

= B8V 2al01 X167 ke V/(525eV)(1.6 x 1079J/eV) (6.63 X 107347 -s)? ' ' X (1.0 x 107°m?)(0.50eV)(1.6 X 107?J/eV)

~ 8 x 10%

in 1.0 cm®. Note that the type | states Sta p of metal did not enter the calculation. r

Equation 40-10 gives us the density of states. Now we must ask: How are the states available to an electron gas actually populated? Let us first consider the

situation at absolute zero, T = 0 K.

For a classical ideal gas, all the particles would

be in the lowest state, with zero kinetic energy (= 3kT = 0)—recall Eq. 18-4. But the

situation is vastly different for an electron gas because electrons obey the exclusion principle. Electrons do not obey classical statistics but rather a quantum statistics called Fermi-Dirac statistics’ that takes into account the exclusion principle. All

particles that have spin 4 (or other half-integral spin: 3,3, etc.), such as electrons, protons, and neutrons, obey Fermi-Dirac statistics and are referred to as fermions.*

The electron gas in a metal is often called a Fermi gas. According to the exclusion principle, no two electrons in the metal can have the same set of quantum numbers.

factor of 2 has already been included in Eq. 40-10.) Thus,at 7' = 0K, the possible energy levels will be filled, two electrons each, up to a maximum level called the Fermi level. This is shown in Fig. 40-26, where the vertical axis is labeled n,(E) for “density of occupied states.” The energy of the state at the Fermi level is called the Fermi energy, Er. To determine Ep, we integrate Eq. 40-10 from E = 0

FIGURE 40-26 At 7 = 0K, all states up to energy Ef, called the Fermi energy, are filled.

n (E)

\

two

electrons: one with spin up (m, = +3%) and one with spin down (m, = —31). (This

to E = Eg (all states up to Ey are filled at T = 0K):

i N

4

f g(E) dE, 0 Ex

r=0E

[OAQ] TULIS]

Therefore, in each of the states of our potential well, there can be at most

(40-11)

where N/V is the number of conduction electrons per unit volume in the metal. "Developed independently by Enrico Fermi (Figs. 41-8, 38-2, 37-10) in early 1926 and by P. A. M. flirac a few months later. © Particles with integer spin (0, 1, 2, etc.), such as the photon, obey Bose-Einstein statistics and are called bosons, as mentioned in Section 39-4.

SECTION 40-6

0

; i ——+———+ 2 4 6

:

Free-Electron Theory of Metals; Fermi Energy

Fermi energy y

E(V)

1087

Then, solving for Ey, the result (see Example 40-10 below) is Ex

=

B\

Y

(40-12)

“

The average energy in this distribution (see Problem 35) is

E = iE. T=0K

1.0 1B

N—r-T=1200K

0

Ep

L

E

FIGURE 40-27 The Fermi-Dirac probability function for two temperatures, 7 = 0K (black line) and 7 = 1200K (blue curve). For f(E) = 1, astate with energy E is certainly occupied. For f(E) = 0.5, which occurs at £ = Ep,

the state

with Er has a 50% chance of being occupied.

(40-13)

For copper, Ex = 7.0eV (see Example 40-10) and E = 4.2 eV. This is very much greater than the energy of thermal motion at room temperature (37 ~ 0.04eV). Clearly, all motion does not stop at absolute zero. Thus, at T = 0, all states with energy below Ep are occupied, and all states above Ep are empty. What happens for 7 > 0? We expect that some (at least) of the electrons will increase in energy due to thermal motion. Classically, the distribution of occupied states would be given by the Boltzmann factor, e / E. Thus

1

EE;

T=0.

This is what is plotted in black in Fig. 40-27 and is consistent with Fig. 40-26: all states up to the Fermi level are occupied [probability f(E) = 1] and all states above are unoccupied. For T = 1200 K, the Fermi factor changes only a little, a shown in Fig. 40-27 as the blue curve. Note that at any temperature 7, wh‘% E = Eg, then Eq. 40-14 gives f(E) = 0.50, meaning the state at E = Ey has 4 50% chance of being occupied by an electron. Alternatively, at E = Eg, half the available states are filled and half are empty. To see how f(E) affects the actual distribution of electrons in energy states, we must weight the density of possible states, g(E), by the probability that those states will be occupied, f(E). The product of these two functions gives the density of occupied states,

8(E)f(E) FIGURE 40-28 The density of occupied states for the electron gas in copper. The width k7 shown above the graph represents thermal energy at 7 = 1200 K.

(40-14)

QE-E/KT 4 1

=

8\V2mwm’? X

E e(E—Ep)/kT

o

:

(40-15)

Then n,(E) dE represents the number of electrons per unit volume with energy between E and E + dE in thermal equilibrium at temperature 7. This is plotted in Fig. 40-28 for T = 1200K, a temperature at which a metal is so hot it would glow. We see immediately that the distribution differs very little from that at T = 0. We see also that the changes that do occur are concentrated about the Fermi level. A few electrons from slightly below the Fermi level move to energy states slightly above it. The average energy of the electrons increases only very slightly when the temperature is increased from 7 = 0K to 7 = 1200 K. This is very different from the behavior of an ideal gas, for which kinetic energy increases directly with 7. Nonetheless, this behavior is readily understood as follows. Energy of thermal motion at 7 = 1200K is about 347 =~ 0.1 eV. The Fermi level, on the other hand, is on the order of several eV: for copper it is Ex = 7.0eV. An electron at 7 = 1200K may have 7eV of energy, but it can acquire at most only a few times 0.1 eV of energy by a (thermal) collision with the lattice. Only electrons very near the Fermi level would find vacant states close enough to make such a transition. Essentially none of the electrons could increase in energy by, say, 3eV, Sam electrons farther down in the electron gas are unaffected. Only electrons near ti top of the energy distribution can be thermally excited to higher states. And their new energy is on the average only slightly higher than their old energy.

EXERCISE B Return to the Chapter-Opening Question, page 1071, and answer it again now. Try to explain why you may have answered differently the first time.

.

DOV IR The Fermi level. For the metal copper, determine (@) the fermi energy, (b) the average energy of electrons, and (c¢) the speed of electrons at the Fermi level (this is called the Fermi speed). APPROACH

We first derive Eq. 40-12 by combining Eqs. 40-10 and 40-11:

N

8V2 mm™?

4

K

n- £23)

E}:El/sz _ 8V2 wm*? 2E3/2 0

n

30

Solving for Eg, we obtain 2

2

8m\mwV

2

3.2

s

[note: (2V2)3 = (22)3 = 2], and this is Eq. 40-12. We calculated N/V, the

number of conduction electrons per unit volume in copper, in Example 25-14 to

be N/V = 84 X 10®m™ SOLUTION

P

(a) The Fermi energy for copper is thus

(6.63 X 1077 -5)? [3(8.4 X 1028m"3)]%

8(9.1 x 10 kg)

o

1

1.6 X 10°J/eV

=

7.0eV.

(b) From Eq. 40-13,

E=3E.=42eV. (¢) In our model, we have taken U = 0 inside the metal (assuming a 3-D infinite potential well, Section 38-8, Eq. 40-10, and Problem 41). Then E is only »kinetic energy = 3mv* Therefore, at the Fermi level, the Fermi speed is

.

/ \/ (70eV)(L6 x 10°1/eV)

91> = 0.039eV.

This result is far from correct. It is off by a factor of 100: Example 40-10 gave 4.2 eV. The ideal gas model does not work for electrons which obey the exclusion principle. flieed, we see here how important and powerful the exclusion principle is. r

EXERCISE C Determine the Fermi energy for gold (density = 19,300 kg/m?).

(b) 6.2¢V, (c) 7.2¢V, (d) 8.1¢V, (¢) 84¢V.

SECTION 40-6

(a) 5.5V,

Free-Electron Theory of Metals; Fermi Energy

1089

FIGURE 40-29 Potential energy for an electron in a metal crystal, ; ; : with deep potential wells in the vicinity of each ion in the crystal lattice.

] > &4 g 3} °

°

°

Position ——»

e~

Positions of metallic

ions in lattice

The simple model of an electron gas presented here provides good explanations for the electrical and thermal properties of conductors. But it does not explain why some materials are good conductors and others are good insulators. To provide an explanation, our model of electrons inside a metal moving in a uniform potential well needs to be refined to include the effect of the lattice. Figure 40-29 shows a “periodic” potential that takes into account the attraction of electrons for each atomic ion in the lattice. Here we have taken U = 0 for an electron free of the metal; so within the metal, electron energies are less than zero (just as for molecules, or for the H atom in which the ground state has E = —13.6eV).

The quantity eW, represents the minimum energy to remove an electron from the

metal, where W, is the work function (see Section 37-2). The crucial outcome of putting a periodic potential (more easily approximated with narrow square wells) into the Schrodinger equation is that the allowed energy states are divided into bands, with energy gaps in between. Only electrons in the highest band, close to the Fermi level, are able to move about freely within the metal crystal. In the next Section we will see physically why there are bands and how they explain th*™ properties of conductors, insulators, and semiconductors.

40-7

Band Theory of Solids

We saw in Section 40-1 that when two hydrogen the wave

functions

overlap, and

the two

1s states

atoms approach each other, (one

for each

atom)

divide

into two states of different energy. (As we saw, only one of these states, S =0, has low enough energy to give a bound H, molecule.) Figure 40-30a shows this situation for 1s and 2s states for two atoms: as the two atoms get closer (toward the left in Fig. 40-30a), the 1s and 2s states split into two levels. If six atoms come together, as in Fig. 40-30b, each of the states splits into six levels. If a large number of atoms come together to form a solid, then each of the original atomic levels becomes a band as shown in Fig. 40-30c. The energy levels are so close together in each band that they seem essentially continuous. This is why the spectrum of heated solids (Section 37-1) appears continuous. (See also Fig. 40-15 and its discussion at the start of Section 40-4.) atoms

&

fl Atomic separation ———

(a) 1090

CHAPTER 40

&-

Energy ——

2s

Allowed

Energy ——

FIGURE 40-30 The splitting of 1s and 2s atomic energy levels as (a) two atoms approach each other (the atomic separation decreases toward the left on the graph), (b) the same for six atoms, and (c) for many atoms when they come together to form a solid.

Energy ——

molecule

Molecules and Solids

Atomic separation ——

(b)

Atomic separation ——»

(c)

The crucial aspect of a good conductor is that the highest energy band containing electrons is only partially filled. Consider sodium metal, for example, whose energy bands are shown in Fig. 40-31.The 1s, 2s, and 2p bands are full (just as in a sodium atom) and don’t rgoncern us. The 3s band, however, is only half full. To see why, recall that the exclusion rinciple stipulates that in an atom, only two electrons can be in the 3s state, one with spin up and one with spin down. These two states have slightly different energy. For a solid consisting of N atoms, the 3s band will contain 2N possible energy states. A sodium atom has a single 3s electron, so in a sample of sodium metal containing N atoms, there are N electrons in the 3s band, and N unoccupied states. When a potential difference is applied across the metal, electrons can respond by accelerating and increasing their energy, since there are plenty of unoccupied states of slightly higher energy available. Hence, a current flows readily and sodium is a good conductor. The characteristic of all good conductors is that the highest energy band is only partially filled, or two bands overlap so that unoccupied states are available. An example of the latter is magnesium, which has two 3s electrons, so its 3s band is filled. But the unfilled 3p band overlaps the 3s band in energy, so there are lots of available states for the electrons to move into. Thus magnesium, too, is a good conductor. In a material that is a good insulator, on the other hand, the highest band containing electrons, called the valence band, is completely filled. The next highest energy band, called the conduction band, is separated from the valence band by a “forbidden” energy gap (or band gap), E,, of typically 5 to 10eV. So at room temperature (300 K), where thermal energies (that is, average kinetic energy—see Chapter 18) are on the order of 3kT ~ 0.04 eV, almost no electrons can acquire the 5eV needed to reach the conduction band. When a potential difference is applied across the material, no available states are accessible to the electrons, and no current flows. Hence, the material is a good insulator. Figure 40-32 compares the relevant energy bands (a) for conductors, (b) for insulators, and also (c) for the important class of materials known as semiconductors. The bands for a pure (or intrinsic) semiconductor, such as silicon or germanium,

re like those for an insulator, except that the unfilled conduction

3s 2p 25

~ FIGURE 40-31 sodium (Na).

Energy bands for

band is

separated from the filled valence band by a much smaller energy gap, E;, typically on the order of 1eV. At room temperature, a few electrons can acquire enough thermal energy to reach the conduction band, and so a very small current may flow when a voltage is applied. At higher temperatures, more electrons have enough energy to jump the gap. Often this effect can more than offset the effects of more frequent collisions due to increased disorder at higher temperature, so the resistivity of semiconductors can decrease with increasing temperature (see Table 25-1). But this is not the whole story of semiconductor conduction. When a potential difference is applied to a semiconductor, the few electrons in the conduction band move toward the positive electrode. Electrons in the valence band try to do the same thing, and a few can because there are a small number of unoccupied states which were left empty by the electrons reaching the conduction band. Such unfilled electron states are called heles. Each electron in the valence band that fills a hole in this way as it moves toward the positive electrode leaves behind its own hole, so the holes migrate toward the negative electrode. As the electrons tend to accumulate at one side of the material, the holes tend to accumulate on the opposite side. We will look at this phenomenon in more detail in the next Section.

Conduction band

Conduction band

’ 5 (a) Conductor

(b) Insulator

—

' (¢) Semiconductor

SECTION 40-7

FIGURE 40-32 Energy bands for (a) a conductor, (b) an insulator, which has a large energy gap E, and (c) a semiconductor, which has a small energy gap E . Shading represents

occupied states. Pale shading in

(c) represents electrons that can pass from the top of the valence band to the bottom of the conduction band due to thermal agitation at room temperature (exaggerated).

Band Theory of Solids

1091

Calculating the energy gap. It is found that the conduc-

tivity of a certain semiconductor increases when light of wavelength 345 nm or shorter strikes it, suggesting that electrons are being promoted from the valence band to the conduction band. What is the energy gap, E,, for this semiconductor?fl APPROACH The longest wavelength (lowest energy) photon to cause an increasc in conductivity has A = 345 nm, and its energy (= Af) equals the energy gap. SOLUTION The gap energy equals the energy of a A = 345-nm photon:

he

(663 X 10734J-5)(3.00 X 108m/s)

E, e = hf =M= = Y£ T= G55 X 10 m)(1.60 X 10-53/eV) = 36eV. ¢ Free

electrons

in

semiconductors

and

insulators. Use the Fermi-Dirac probability function, Eq. 40-14, to estimate the

Conduction band

order solid (T = Eg ~

FIGURE 40-33 Example 40-13; The Fermi energy is midway between the valence band and the conduction band.

of magnitude of the numbers of free electrons in the conduction band of a containing 10°' atoms, assuming the solid is at room temperature 300K) and is (a) a semiconductor with E; ~ 1.1eV, (b) an insulator with 5¢eV. Compare to a conductor.

APPROACH At T = 0, all states above the Fermi energy Ep are all those below. are full. So for semiconductors and insu.lators we to be about midway between the valence and conduction bands, and it does not change significantly as we go to room temperature.

empty, and can take Ep Fig. 40-33, We can thus

use Eq. 40-14 to find the fraction of electrons in the conduction band at room

temperature for the two cases.

SOLUTION (a) For the semiconductor, the gap E, ~ 1.1eV,so E — Ep ~ 0.55eV for the lowest states in the conduction band. Since at room temperature we have kT ~ 0.026 eV, then (E — Eg)/kT ~ 0.55eV/0.026eV ~ 21 and -

Thus about 1 atom in 10° can contribute an electron to the conductivity.

(b) For the insulator with E — Ep ~ 5.0eV — 3(5.0eV) = 2.5¢eV, we get

Thus

in an ordinary

sample

containing

10?! atoms,

there

would

be

no free

electrons in an insulator (10* X 107 = 107?'), about 10" (10* X 107°) free electrons in a semiconductor, and about 10*' free electrons in a good conductor.

@

PHYSICS

APPLIED Transparency

CONCEPTUAL EXAMPLE 40-14 | Which is transparent? The energy gap for silicon is 1.14 eV at room temperature, whereas that of zinc sulfide (ZnS) is 3.6 eV. Which one of these is opaque to visible light, and which is transparent? RESPONSE

Visible light photons span energies from roughly 1.8eV to 3.1eV

(E = hf = hc/A where A = 400nm to 700nm and 1eV = 1.6 X 107"°J). Light

is absorbed by the electrons in a material. Silicon’s energy gap is small enough to absorb these photons, thus bumping electrons well up into the conduction band, so silicon is opaque. On the other hand, zinc sulfide’s energy gap is so large that no visible light photons would be absorbed; they would pass right througfi the material which would thus be transparent. 1092

CHAPTER 40

Molecules and Solids

40-8

Semiconductors and Doping

Nearly all electronic devices today use semiconductors. The most common are r%ilicon (Si) and germanium (Ge). An atom of silicon or germanium has four outer electrons that act to hold the atoms in the regular lattice structure of the crystal, shown schematically in Fig. 40-34a. Germanium and silicon acquire properties useful for electronics when a tiny amount of impurity is introduced into the crystal structure (perhaps 1 part in 10° or 107). This is called doping the semiconductor. Two kinds of doped semiconductor can be made, depending on the type of impurity used. If the impurity is an element whose atoms have five outer electrons, such as arsenic, we have the situation shown in Fig. 40-34b, with the arsenic atoms holding positions in the crystal lattice where normally silicon atoms would be. Only four of arsenic’s electrons fit into the bonding structure. The fifth does not fit in and can move relatively freely, somewhat like the electrons in a conductor. Because of this small number of extra electrons, a doped semiconductor becomes slightly conducting, Silicon atom

Bl econ

Silicon atom

°

°

Q. o cQ+

¢ ° Q° . o Q-

c Q.

»

Y

c

°

Q4 . Q. o Q-

.

'

'

o

.

cQL ¢ . ¢ . Arsenic . o . atom e Qe Qo o o *‘./’El’gcrso " o . o cQ++Q+ Q-

o

.

(a)

FIGURE 40-34

Two-dimensional

representation of a silicon crystal. (a) Four (outer) electrons surround each silicon atom. (b) Silicon crystal doped with a small percentage of arsenic atoms: the extra electron doesn’t fit into the crystal lattice and soisfreetqmoveabout,Thisisan

n-type semiconductor.

(b)

The density of conduction electrons in an intrinsic (undoped) semiconductor at room rtemperature is very low, usually less than 1 per 10° atoms. With an impurity concenration of 1 in 10° or 10’ when doped, the conductivity will be much higher and it can be controlled with great precision. An arsenic-doped silicon crystal is called an n-type semiconductor because negative charges (electrons) carry the electric current. In a p-type semiconductor, a small percentage of semiconductor atoms are replaced by atoms with three outer electrons—such as gallium. As shown in Fig. 40-35a, there is a “hole” in the lattice structure near a gallium atom since it has only three outer electrons. Electrons from nearby silicon atoms can jump into this hole and fill it. But this leaves a hole where that electron had previously been, Fig. 40-35b. The vast majority of atoms are silicon, so holes are almost always next to a silicon atom. Since silicon atoms require four outer electrons to be neutral, this means that there is a net positive charge at the hole. Whenever an electron moves to fill a hole, the positive hole is then at the previous position of that electron. Another electron can then fill this hole, and the hole thus moves to a new location; and so on. This type of semiconductor is called p-type because it is the positive holes that seem to carry the electric current. Note, however, that both p-type and n-type semiconductors have no net charge on them. .

Silicon

G:ig;:nm b

e Q .

atom

*

o\

.

. . e Q-0 .

r

.

%

L

o L]

e

e

e Qe

*_Hole ./ °

. e Q-+

4

.

°

.

e L ]

L]

o L]

(@)

o

e Qo

L]

L]

o L ]

L]

o

.

o

o L]

,

L]

o L]

(b)

N Q-

structure. (b) Electrons from silicon atoms can jump into the hole and fill it. As a result, the hole

electrons, so there is an empty spot, or hole in the

moves to a new location (to the right in this

R ®

L

A p-type semiconductor, gallium-

doped silicon. (a) Gallium has only three outer

%

" L]

FIGURE 40-35

Q®

o 0O b

-

L]

o

e Qe

o

Figure), to where the electron used to be. ®

L

SECTION 40-8

Semiconductors and Doping

1093

Conduction band ®

6

o

o

o

Conduction band

Donor /—O—O—O—O——O—

level e ————_—

(a) n-type FIGURE 40-36

Acceptor level

fl

(b) p-type

Impurity energy levels in doped semiconductors.

According to the band theory (Section 40-7), in a doped semiconductor the impurity provides additional energy states between the bands as shown in Fig. 40-36. In an n-type semiconductor, the impurity energy level lies just below the conduction band, Fig. 40-36a. Electrons in this energy level need only about 0.05eV in Si (even less in Ge) to reach the conduction band; this is on the order of the thermal energy, 3k7 (= 0.04eV at 300 K), so transitions occur readily at room temperature. This energy level can thus supply electrons to the conduction band, so it is called a doner level. In p-type semiconductors, the impurity energy level is just above the valence band (Fig. 40-36b). It is called an acceptor level because electrons from the valence band can easily jump into it. Positive holes are left behind in the valence band, and as other electrons move into these holes, the holes move as discussed earlier. EXERCISE D Which of the following impurity atoms would produce a p-type semiconductor? (a) Ge; (b) Ne; (¢) Al (d) As; (e) none of the above.

CONCEPTUAL

D

(a) X

X

X

G TRl

X

:

X

i3 o0, S

X

CX

DX

X

SR,

X

X

IS, 5

X

)Sf

_l_

X

FIGURE 27-32 (Repeated.) The Hall effect. (a) Negative charges moving to the right as the current, (b) The same current, but as positive charges moving to the left.

1094

EXAMPLE

40-15

| Determining charge of conductors. Hov™

can we determine if a p-type semiconductor has a current that is really due to the motion of holes? Or, is this just a convenient model?

CHAPTER 40

RESPONSE Recall from Section 27-8 that the Hall effect can be used to distinguish the sign of the charges involved in a current. When placed in a magnetic field, the current in a particular direction can result in a voltage perpendicular to that current due to the magnetic force on the moving charges (Fig. 27-32, repeated here). The direction of this Hall voltage depends on the sign of the charges carrying the current. In this way, it has been demonstrated that it really is moving holes that are responsible for the current in a p-type semiconductor.

40-9

Semiconductor Diodes

Semiconductor diodes and transistors are essential components of modern electronic devices. The miniaturization achieved today allows many thousands of diodes, transistors, resistors, and so on, to be placed on a single chip less than a millimeter on a side. We now discuss, briefly and qualitatively, the operation of diodes and transistors. When an n-type semiconductor is joined to a p-type, a pn junction diode is formed. Separately, the two semiconductors are electrically neutral. When joined, a few electrons near the junction diffuse from the n-type into the p-type semiconductor, where they fill a few of the holes. The n-type is left with a positive charge, and the p-type acquires a net negative charge. Thus a potential difference i established, with the n side positive relative to the p side, and this prevents furthei diffusion of electrons.

Molecules and Solids

If a battery is connected to a diode with the positive terminal to the p side and the negative terminal to the » side as in Fig. 40-37a, the externally applied

voltage opposes the internal potential difference and the diode is said to be

r(orward biased. If the voltage is great enough (about 0.3 V for Ge, 0.6V for Si at oom temperature), a current will flow. The positive holes in the p-type semiconductor are repelled by the positive terminal of the battery, and the electrons in the n-type are repelled by the negative terminal of the battery. The holes and electrons meet at the junction, and the electrons cross over and fill the holes. A current is flowing. Meanwhile, the positive terminal of the battery is continually pulling electrons off the p end, forming new holes, and electrons are being supplied by the negative terminal at the n end. Consequently, a large current flows through the diode. When the diode is reverse biased, as in Fig. 40-37b, the holes in the p end are

attracted to the battery’s negative terminal and the electrons in the n end

are attracted to the positive terminal. The junction and, ideally, no current flows. A graph of current versus voltage for a can be seen, a real diode does allow a small most practical purposes, it is negligible. (At is a few uA in Ge and a few pA in Si; but

current carriers do not meet near the

typical diode is shown in Fig, 40-38. As amount of reverse current to flow. For room temperature, the reverse current it increases rapidly with temperature,

and may render a diode ineffective above 200°C.)

*+

\gfltage *°""°

P | (Conventional) current n | flow

(a) P

~

:g’dii%e

gg b n (b)

=~ FIGURE 40-37

Schematic diagram

showing how a semiconductor diode operates. Current flows when the

I(mA)

voltage is connected in forward bias,

30+ -

as in (a), but not when connected in reverse bias, as in (b).

20 +

Reverse f

L

1

-12.0

f

Frererrt

bias

}

-12-1.0 —08 -06 04

{

0T t

=02

Forward

T

0

V (volts)

+

02

bias

f

04

f

06

t

08

FIGURE40-38

Current through asilicon pn

diode as a function of applied voltage.

DENILSRITAN A diode. The diode whose current-voltage characteristics are shown in Fig. 40-38 is connected in series with a 4.0-V battery in forward bias and a resistor. If a current of 15 mA is to pass through the diode, what resistance must the resistor have? APPROACH We use Fig. 40-38, where we see that the voltage drop across the diode is about 0.7V when the current is 15mA. Then we use simple circuit analysis and Ohm’s law (Chapters 25 and 26). SOLUTION

The

voltage

drop

across

the resistor is

R=V/I=(33V)/(1.5x10%A)=220Q.

40V

— 0.7V

=33V,

so

The symbol for a diode is

—P—

[diode]

where the arrow represents the direction conventional (+) current flows readily. If the voltage across a diode connected in reverse bias is increased greatly, a point is reached where breakdown occurs. The electric field across the junction becomes so large that ionization of atoms results. The electrons thus pulled off their atoms contribute to a larger and larger current as breakdown continues. The voltage remains constant over a wide range of currents. This is shown on the far left in Fig. 40-38. This property of diodes can be used to accurately regulate a voltage supply. A diode designed for this purpose is called a zener diode. When placed across the output of an unregulated power supply, a zener diode can maintain the r'oltage at its own breakdown voltage as long as the supply voltage is always above -his point. Zener diodes can be obtained corresponding to voltages of a few volts to hundreds of volts. SECTION 40-9

Semiconductor Diodes

1095

Diode &

R o~

(a)

b

AC so}r{e Vi)

— ATANTAY /\ voltage 7 J \ L/\

A\

/\j’)‘;lgz:"

RS

(b)

FIGURE 40-39

(a) A simple

(half-wave) rectifier circuit using a

semiconductor diode. (b) AC source input voltage, and output voltage

Since a pn junction diode allows current to flow only in one direction (as long as the voltage is not too high), it can serve as a rectifier—to change ac into dc. A simple rectifier circuit is shown in Fig. 40-39a. The ac source applies a voltage across the diode alternately positive and negative. Only‘\ during half of each cycle will a current pass through the diode; only then is there a current through the resistor R. Hence, a graph of the voltage V,;, across R as a function of time looks like the output voltage shown in Fig. 40-39b. This half-wave rectification is not exactly dc, but it is unidirectional. More useful is a.full-'wave rectifier circuit,.whic!l uses two diodes (or. sometimes four) as shown in Fig. 40-40a. At any given instant, either one diode or the oth.er will cqnduct current to the? right. Therefore, the output across the' load resistor R will be as shown in Fig. 40-40b. Actually this is the voltage if the capacitor C were not in the circuit. The capacitor tends to store charge and, if the time constant RC is sufficiently long, helps to smooth out the current as shown in Fig. 40-40c. (The variation in output shown in Fig. 40-40c is called ripple voltage.) Rectifier circuits are important because most line voltage in buildings is ac,

and most electronic devices require a dc voltage for their operation. Hence, diodes

are found in nearly all electronic devices including radio and TV sets, calculators, and computers.

bt

across R, as functions of time.

L

FIGURE 40-40

(a) Full-wave rectifier

circuit (including a transformer so the magnitude of the voltage can be changed). (b) Output voltage in the absence of capacitor C. (c) Output voltage with the capacitor in the circuit.

=

@

Output

> (a)

V output

¢

I =

=

V. output

. Time

77 1

(b) Without capacitor

PHYSICs AppLIED LEDs and applications Car safety (brakes)

FIGURE 40-41

LED traffic light.

R

Sl =

7 \! 7 Vs Time

q

(c) With capacitor

Another useful device is a light-emitting diode (LED), invented in the 1960s. When a pn junction is forward biased, a current begins to flow. Electrons ~ SFO88 from the n region into the p region and combine with holes, and a photon

can

be emitted

with

an energy

approximately

equal

to the band

gap,

E, (see Figs. 40-32c and 40-36). Often the energy, and hence the wavelength, is in the red region of the visible spectrum, producing the familiar LED displays on electronic devices, car instrument panels, digital clocks, and so on. Infrared (i.e., nonvisible) LEDs are used in remote controls for TVs, DVDs, and stereos. New types of LEDs emit other colors, and LED “bulbs” are beginning to replace other types of lighting in applications such as flashlights, traffic signals, car brake lights, and outdoor signs, billboards, and theater displays. LED bulbs, sometimes called solid-state lighting, are costly, but they offer advantages: they are long-lived, efficient, and rugged. LED traffic lights, for example (Fig. 40-41), last 5 to 10 times longer than traditional incandescent bulbs, and use only 20%

of the energy for the same

light output.

As car brake lights, they light up a fraction of a second sooner, allowing a driver an extra 5 or 6 meters (15-20 ft) more stopping distance at highway speeds. Solar cells and photodiodes (Section 37-2) are pn junctions used in the reverse way. Photons are absorbed, creating electron-hole pairs if the photon energy is greater than the band gap energy, E,. The created electrons and holes produce a current that, when connected to an external circuit, becomes a source of emf and power. Particle detectors (Section 41-11) operate similarly. A diode is called a nonlinear device because the current is not proportional tc the voltage. That is, a graph of current versus voltage (Fig. 40-38) is not a straight line, as it is for a resistor (which ideally is linear). Transistors are also nonlinear devices. 1096

CHAPTER 40

Molecules and Solids

40-10

Transistors and Integrated Circuits (Chips)

r A simple junction transistor consists of a crystal of one type of doped semiconductor sandwiched between two crystals of the opposite type. Both npn and pnp transistors are made, and they are shown schematically in Fig. 40-42a. The three semiconductors

are given the names

collector, base, and emitter. The symbols for

npn and pnp transistors are shown in Fig. 40-42b. The arrow is always placed on the emitter and indicates the direction of (conventional) current flow in normal operation. Collector

,—p

Collector

Base

Base

Emitter

Emitter

Rc

8;?5:;

Input npn transistor

Base

(a)

pnp transistor

Collector

EE z

Emitter

hpn

FIGURE 40-42

Base

(b)

e

é\i ?

pnp

?;%:;1])

Collector

o

Emitter

FIGURE 40-43

.I—

8¢

An npn transistor used as an amplifier.

Iy is the current produced by €5 (in the absence of a

(a) Schematic diagram of npn and pnp

signal), iy is the ac signal current (= change in Ig).

transistors. (b) Symbols for npn and pnp transistors.

The operation of a transistor can be analyzed qualitatively—very briefly—as follows. Consider an npn transistor connected as shown in Fig. 40-43. A voltage Vg rjs maintained between the collector and emitter by the battery €é.. The voltage ipplied to the base is called the base bias wvoltage, Vgi. If Vyp is positive, conduction electrons in the emitter are attracted into the base. Since the base region is very thin (less than 1 um—much less if on a chip), most of these electrons flow right across into the collector, which is maintained at a positive voltage. A large current, I, flows between collector and emitter and a much smaller current, Iy, through the base. In the steady state, Iy and /- can be considered dc. A small variation in the base voltage due to an input signal causes a large change in the collector current and therefore a large change in the voltage drop across the output resistor R.. Hence a transistor can amplify a small signal into a larger one. Typically a small ac signal (call it ig) is to be amplified, and when added to the base bias voltage and current causes the voltage and current at the collector to vary at the same rate but magnified. Thus, what is important for amplification is the change in collector current for a given input change in base current. We label these ac signal currents (= changes in I and ) as ic and ig. The current gain is defined as the ratio _output (collector) ac current AN input (base) ac current

ic s

B, is typically on the order of 10 to 100. Similarly, the voltage gain is By

output (collector) ac voltage input (base) ac voltage

Transistors are the basic elements in modern electronic amplifiers of all sorts. A pnp transistor operates like an npn, except that holes move instead of electrons. The collector voltage is negative, and so is the base voltage in normal operation. In digital circuits, including computers, where “off” and “on” (or zero and one) make up the binary code, transistors act like a gate or switch. That is, they let current pass (“on”) or they block it (“off”).

SECTION 40-10

Transistors and Integrated Circuits (Chips)

1097

Transistors were a great advance in miniaturization of electronic circuits. Although individual transistors are very small compared to the once-used vacuum tubes, they are huge compared to integrated circuits or chips (see photo at start of this Chapter). Tiny amounts of impurities can be placed at particular locations within a single silicon crystal. These can be arranged to form diodes, transistor and resistors (undoped semiconductors). Capacitors and inductors can also be formed, although they are often connected separately. A tiny chip, a few millimeters on a side, may contain millions of transistors and other circuit elements. Integrated circuits

are

the

heart

of computers,

televisions,

calculators,

cameras,

and

the

electronic instruments that control aircraft, space vehicles, and automobiles. The “miniaturization” produced by integrated circuits not only allows extremely complicated circuits to be placed in a small space, but also has allowed a great increase in the speed of operation of, say, computers, because the distances the electronic signals travel are so tiny.

| Summary Quantum mechanics explains the bonding together of atoms to form molecules. In a covalent bond, the electron clouds of two or

more atoms overlap because of constructive interference between the electron waves. The positive nuclei are attracted to this concentration of negative charge between them, forming the bond. An ionic bond is an extreme case of a covalent bond in which one or more electrons from one atom spend much more time around the other atom than around their own. The atoms then act as oppositely charged ions that attract each other, forming the bond. These strong bonds hold molecules together, and also hold atoms and molecules together in solids. Also important are weak bonds (or van der Waals bonds), which are generally dipole attractions between molecules. When atoms combine to form molecules, the energy levels of the outer electrons are altered because they now interact with each other. Additional energy levels also become possible because the atoms can vibrate with respect to each other, and

the molecule as a whole can rotate. The energy levels for both

vibrational and

rotational motion

are quantized, and

are very

close together (typically, 107! eV to 1073 eV apart). Each atomic

energy level thus becomes a set of closely sponding to the vibrational and rotational from one level to another appear as many lines. The resulting spectra are called band The quantized rotational energy levels Eot

=

L+1)=—

spaced levels corremotions. Transitions very closely spaced spectra. are given by

£ =012,

(40-2)

where [ is the moment of inertia of the molecule. The energy levels for vibrational motion are given by

Egob = (v + Dhf,

v =0,1,2,--,

(40-6)

where f is the classical natural frequency of vibration for the molecule. Transitions between energy levels are subject to the selection rules Af = +1 and Av = +1. Some solids are bound together by covalent and ionic bonds,

just as molecules are. In metals, the electrostatic force between free electrons and positive ions helps form the metallic bond. In the free-electron theory of metals, electrons occupy the possible energy states according to the exclusion principle. At T = 0K, all possible states are filled up to a maximum energy level called the Fermi energy, Er, the magnitude of which is typically a few eV. All states above Ef are vacant at 7 = 0K.

1098

CHAPTER 40

Molecules and Solids

At normal temperatures (300K) the distribution of occupied states is only slightly altered and is given by the Fermi-Dirac probability function

f(E)

=

1

GE-EJAT 4 1

(40-14)

In a crystalline solid, the possible energy states for electrons are arranged in bands. Within each band the levels are very close together, but between the bands there may be forbidden energy gaps. Good conductors are characterized by the highest occupied band (the conduction band) being only partially full, so there are many accessible states available to electrons to move about and accelerate when a voltage is applied. In a good insulator, the highest occupied energy band (the valence band) is completely full, and there is a large energy gap (5 to 10eV) 1My the next highest band, the conduction band. At room temperature,

molecular kinetic energy (thermal energy) available due to collisions is only about 0.04 eV, so almost no electrons can jump from the valence to the conduction band. In a semiconductor, the gap between valence and conduction bands is much smaller, on the order of 1eV, so a few electrons can make the transition from the essentially full valence band to the nearly empty conduction band. In a doped semiconductor, a small percentage of impurity atoms with five or three valence electrons replace a few of the normal silicon atoms with their four valence electrons. A five-electron impurity produces an n-type semiconductor with negative electrons as carriers of current. A three-electron impurity produces a p-type semiconductor in which positive holes carry the current. The energy level of impurity atoms lies slightly below the conduction band in an n-type semiconductor, and acts as a donor from which electrons readily pass into the conduction band. The energy level of impurity atoms in a p-type semiconductor lies slightly above the valence band and acts as an acceptor level, since electrons from the valence band

easily reach it, leaving holes behind to act as charge carriers. A semiconductor diode consists of a pn junction and allows

current to flow in one direction only; it can be used as a rectifier

to change ac to dc. Common transistors consist of three semiconductor sections, either as pnp or npn. Transistors can amplify electrical signals and in computers serve as switches or gates for the 1s and Os. An integrated circuit consists of a tiny semiconductor crystal or chip on which many transistors, diodeSfl resistors, and other circuit elements have been constructed usin_

careful placement of impurities.

| Questions 1. What type of bond would you expect for (a) the N, molecule, (b) the HCI molecule, (c) Fe atoms in a solid?

rz’ Describe how the molecule CaCl, could be formed.

3. Does the H, molecule have a permanent dipole moment? Does O, ? Does H,O? Explain.

.

:

% glthlopgh t.he :}I;OIIC,CU}.C H31 is not §taplle, Hhejor RPN K TRN. eRE uswn.p.rmcql) ©

by

5 % )

da

5. The energy of a molecule can be divided into four categories. What are they?

13. Compare the resistance of a pn junction diode connected in forward bias.

bias to its resistance when

connected in reverse

14. Which aspects of Fig. 40-28 are peculiar to copper, and which are valid in general for other metals?

)

i

)

15. Explain how a transistor can be used as a switch.

6. Would you expect the molecule H," to be stable? If so, where would the single electron spend most of its time?

16. What is the main semiconductors?

7. Explain why the carbon atom

19 pray a circuit diagram showing how a pnp transistor can

bonds' with hydrogen:hke atoms.

8. Explain

chloride

on

the basis

crystal

is

a

(Z = 6) usually forms four

of energy

good

shells of Na* and Cl™ ions.]

bands

insulator.

why

[Hin[;

the

‘

sodium

Consider

the

difference

between

n-type

and p-type

operate as an amplifier.

18.

In

a

transistor,

the

base—emitter

junction

and

the

base—

collector junction are essentially diodes. Are these junctions

9. If conduction electrons are free to roam about in a metal, why don’t they leave the metal entirely?

reverse-biased or forward-biased in the application shown in Fig. 40-43?

10. Explain why the resistivity of metals increases with increasing temperature whereas the resistivity of semiconductors may decrease with increasing temperature.

19, A transistor can amplify an electronic signal, meaning it can increase the power of an input signal. Where does it get the energy to increase the power?

11. Figure 40-44 shows a “bridge-type” full-wave rectifier.

29, A silicon semiconductor is doped with phosphorus. Will

Explain how the current

is rectified and

how

current flows during each half cycle. r

12. Discuss the differences between an ideal gas and a Fermi electron gas.

these atoms be donors or acceptors? What type of semicon-

ductor will this be?

i

) ) 21. Do diodes and transistors obey Ohm’s law? Explain.

nput

FIGURF 40-44

Question 11.

.

.

.

22. Can a diode be used to amplify a signal? Explain. Output

23.

1If €c were reversed in Fig. 40-43, how would the amplifica-

tion be altered?

| Problems 40-1 to 40-3 Molecular Bonds L (I) Estimate the binding energy of a KCl molecule by calculating the electrostatic potential energy when the K* and Cl” ions are at their stable separation of 0.28 nm. Assume each has a charge of magnitude 1.0e. . (II) The measured binding energy of KCl is 4.43 ¢V. From the result of Problem 1, estimate the contribution to the binding energy of the repelling electron clouds at the equilibrium distance ry = 0.28 nm. . (II) Estimate the binding energy of the H, molecule, assuming the two H nuclei are 0.074nm apart and the two electrons spend 33% of their time midway between them.

(o

. (II) The equilibrium distance ry between two atoms in a molecule is called the bond length. Using the bond lengths of homogeneous molecules (like H,, O,, and N,), one can estimate the bond length of heterogeneous molecules (like CO, CN, and NO). This is done by summing half of each bond length of the homogenous molecules to estimate that of the heterogeneous molecule. Given the following bond lengths: H, (= 74pm), N, (= 145pm), O, (=121pm), C, (= 154pm), estimate the bond lengths for: HN, CN, and NO.

5. (II) Estimate the energy associated with the repulsion of the electron shells of a lithium fluoride

(LiF)

molecule. The

ionization energy of lithium is 5.39 eV, and it takes 3.41 eV

to remove the extra electron from an F~ ion. The bond length is 0.156 nm, and the binding energy of LiF is 5.95 eV. . (II) Binding energies are often measured experimentally in kcal per mole, and then the binding energy in eV per molecule is calculated from that result. What is the conversion factor in going from kcal per mole to eV per molecule? What is the binding energy of KCl (= 4.43 eV) in kcal per mole? . (IIT) (a) Apply reasoning similar to that in the text for the S =0 and S =1 states in the formation of the H, molecule to show why the molecule He; is not formed. (b) Explain why the He," molecular ion could form. (Experiment shows it has a binding energy of 3.1eV at ry = 0.11 nm.)

40-4 Molecular Spectra 8. (I) Show that the quantity #2/I has units of energy. 9 (I) What is the reduced mass of the molecules (a) KCI; (b) O3; (c) HCI? 10. (I1) (a) Calculate the “characteristic rotational energy,” #2/21, for the O, molecule whose bond length is 0.121 nm. (b) What are the energy and wavelength of photons emitted inan £ =2 to £ = 1 transition? Problems

1099

11. (I) The “characteristic rotational energy,” #%/21, for N, is

2.48 X 107 eV. Calculate the N, bond length.

12. (IT) Estimate the longest wavelength emitted by a lithium hydride (LiH) molecule for a change in its rotational state if its equilibrium separation is 0.16 nm. 13. (IT) The equilibrium separation of H atoms in the H, molecule is 0.074 nm (Fig. 40-8). Calculate the energies and wavelengths of photons for the rotational transitions (a)£=1to =0, () L=2tol=1 and(c) £=3 to =2 14. (II) Explain why there is no transition for AE = hf in Fig. 40-21 (and Fig. 40-22). See Eqgs. 40-8. 15 (II) The fundamental vibration frequency for the CO

molecule is 6.42 X 101*Hz. Determine

(a) the reduced

mass, and (b) the effective value of the “stiffness” constant k. Compare to k for the H, molecule. 16. (II) Li and Br form a molecule for which the lowest

vibrational frequency is 1.7 X 10'3 Hz. What is the effective

stiffness constant k? 17. (IT) Calculate the bond length for the NaCl molecule given that three successive wavelengths for rotational transitions

25. (III) (a) Starting from Eq. 40-9, show that the ionic cohesive

energy is given by Uy = —(ae’/4meory)(1—1/m). Determine U for (b) Nal (ro = 0.33nm) and (c) MgO

40-6 Free-Electron Theory of Metals 26. (IT) Estimate the number of possible electron states in a 1.00-cm® cube of silver between 0.985E and Er (= 5.48eV). 27. (I) Estimate the number of states between 7.00eV and 7.05eV that are available to electrons in a 1.0-cm® cube of copper. (II) What, roughly, is the ratio of the density of molecules in an ideal gas at 285K and 1 atm (say O,) to the density of free electrons (assume one per atom) in a metal (copper) also at 285 K? . (II) Calculate the energy which has 85.0% occupancy probability for copper at (a) T = 295K; (b) T = 750K. . (II) Calculate the energy which has 15.0% occupancy probability for copper at (@) 7 = 295K; (b) T = 950K. . (II) What is the occupancy probability electron in copper at T =295K E = 1.015Eg?

are 23.1 mm, 11.6 mm, and 7.71 mm.

18. (II) (a) Use the curve of Fig. 40-18 to estimate the stiffness

constant k for the H, molecule. (Recall that U = Jkx2)

(b) Then estimate the fundamental wavelength for vibrational transitions using the classical formula (Chapter 14), but use only } the mass of an H atom (because both H atoms move). 19. (IIT) Imagine the two atoms of a diatomic molecule as if they were connected by a spring, Fig. 40-45. Show that the classical frequency of vibration is given by Eq. 40-5. [Hint: Let xq and x, be the displacements of each mass from

initial equilibrium positions; then m;d’x,/dt*> = —kx, and

my d*x,/dt* = —kx, where x = x; + x,. Find another relationship between x; and x,, assuming that the center of mass of the system stays at rest, and then show

that wd’x/dt* = —kx.]

32.

(I1) The atoms in zinc metal (p = 7.1 X 10*kg/m?) each have

and the known

40-5 Bonding in Solids 20. (I) Estimate a =175

21.

the ionic cohesive

m=_8,

and

energy

ry = 0.28 nm.

for NaCl

taking

(IT) Common salt, NaCl, has a density of 2.165 g/cm>. The

molecular weight of NaCl is between nearest neighbor Na can be considered to have one unknown) extending out from

58.44. Estimate the distance and Cl ions. [Hint: Each ion “cube” or “cell” of side s (our it.]

22. (IT) Repeat the previous Problem for KCI whose density is

1.99 g/cm?.

23. (II) The spacing between “nearest neighbor” Na and Cl ions in a NaCl crystal is 0.24 nm. What is the spacing between two nearest neighbor Na ions? . (IlI) For a long one-dimensional chain of alternating positive and negative ions, show that the Madelung constant would be a =2In2. [Hint: Use a series expansion for In(1 + x).]

1100

CHAPTER 40

Molecules and Solids

density (2.70 X 10°kg/m’)

and atomic

mass (27.0) of aluminum. 35 (IT) Show that the average energy of conduction electrons calculating

FIGURE 40-45

for a conduction for an energy

two free electrons. Calculate (a) the density of conduction electrons, (b) their Fermi energy, and (c) their Fermi speed. 33. (II) Calculate the Fermi energy and Fermi speed for sodium, which has a density of 0.97 x 10°kg/m> and has one conduction electron per atom. . (II) Given that the Fermi energy of aluminum is 11.63 e (a) calculate the density of free electrons using Eq. 40-12, and (b) estimate the valence of aluminum using this model

in a metal at

Problem 19.

(rp = 0.21 nm).

Assume m = 10. (d) If you used m = 8 instead, how far off would your answers be? Assume o = 1.75. “

T=0K

is

E =3Fz

B

JEno(E) dE

e

Jno(E) dE

(Eq. 40-13) by

36. (II) The neutrons in a neutron star (Chapter 44) can be treated as a Fermi gas with neutrons in place of the electrons in our model of an electron gas. Determine the Fermi energy for a neutron star of radius 12 km and mass 2.5 times that of our Sun. Assume that the star is made entirely of neutrons and is of uniform density. 37. (II) For a one-dimensional potential well of width £, start with Eq. 38-13 and show that the number of states per unit energy interval for an electron gas is given by &(E) E

=

/8ml2 E

.

Remember that there can be two electrons (spin up and spin down) for each value of n. [Hint: Write the quantum number n in terms of E. Then g(E) = 2dn/dFfl where dn is the number of energy levels between E anc E + dE.]

~

38. (II) Show that the probability for the state at the Fermi energy being occupied is exactly 3, independent of temperature. 39. (IT) A very simple model of a “one-dimensional” metal consists of N electrons confined to a rigid box of width £. We neglect the Coulomb interaction between the electrons. (a) Calculate the Fermi energy for this one-dimensional metal (Er = the energy of the most energetic electron at T = 0K), taking into account the Pauli exclusion principle. You can assume for simplicity that N is even. (b) What is the smallest amount of energy, AE, that this 1-D metal can absorb? (c) Find the limit of AE/E for large N. What does this result say about how well metals can conduct? 40. (II) (a) For copper at room temperature (7 = 293K), calculate the Fermi factor, Eq. 40-14, for an electron with energy 0.12 eV above the Fermi energy. This represents the

probability that this state is occupied. Is this reasonable? (b) What is the probability that a state 0.12eV below the Fermi energy is occupied? (c) What the state in part (b) is unoccupied?

is the probability that

41. (IIT) Proceed as follows to derive the density of states, g(E), the number of states per unit volume per unit energy interval, Eq. 40-10. Let the metal be a cube of side £. Extend the discussion of Section 38-8 for an infinite well to three dimensions, giving energy levels E

=

hZ

—(n} + n}2 + nj). Smez(l 3)

(IT) Suppose that a silicon semiconductor is doped with phosphorus so that one silicon atom in 1.2 X 10° is replaced by a phosphorus atom. Assuming that the “extra” electron in every phosphorus atom is donated to the conduction band, by what factor is the density of conduction electrons increased? The density of silicon is 2330 kg/m?, and the density of conduction electrons in pure silicon is

about 10! m~3 at room temperature. 40-9 Diodes

48. (I) At what wavelength will an LED radiate if made from a material with an energy gap E; = 1.6eV? 49. (I) If an LED emits light of wavelength A = 680 nm, what is the energy gap (in eV) between valence and conduction bands? 50. (IT) A silicon diode, whose current—voltage characteristics are given in Fig. 40-38, is connected in series with a battery and an 860-() resistor. What battery voltage is needed to produce a 12-mA current? 51. (II) Suppose that the diode of Fig. 40-38 is connected in series to a 150-() resistor and a 2.0-V battery. What current flows in the circuit? [Hint: Draw a line on Fig. 40-38 representing the current in the resistor as a function of the voltage

across the diode; the intersection

of this line with

the characteristic curve will give the answer.] 52. (IT) Sketch the resistance as a function of current, V > 0, for the diode shown in Fig. 40-38.

(Explain the meaning of ny, n, n3.) Each set of values for the quantum numbers ny, n,, n3 corresponds to one state. Imagine a space where ny, ny, ny are the axes, and each state is represented by a point on a cubic lattice in this space, each separated by 1 unit along an axis. Consider the octant ny > 0, n > 0, n3 > 0. Show that the number of

states N within a radius R = (n} + n3 + n3)2 is 2(})37R?).

Then, to get Eq. 40-10, set g(E) = (1/V)(dN/dE),

V =3 is the volume of the metal.

where

40-7 Band Theory of Solids 42. (I) A semiconductor is struck by light of slowly increasing frequency and begins to conduct when the wavelength of the light is 580 nm; estimate the size of the energy gap Eg. 43. (I) Calculate the longest-wavelength photon that can cause

an electron in silicon (E, = 1.14eV)

40-8 Semiconductors and Doping 47.

to jump from the

valence band to the conduction band. (IT) The energy gap between valence and conduction in germanium is 0.72 eV. What range of wavelengths photon have to excite an electron from the top valence band into the conduction band? 45. (IT) We saw that there are 2N possible electron states

bands can a of the in the

3s band of Na, where N is the total number of atoms. How

many possible electron states are there in the (a) 2s band, (b) 2p band, and (c) 3p band? (d) State a general formula for the total number of possible states in any given electron band. 46. (II) The energy gap Eg in germanium is 0.72 eV. When used as a photon detector, roughly how many electrons can be made to jump from the valence to the conduction band by the passage of a 730-keV photon that loses all its energy in this fashion?

for

53. (IT) An ac voltage of 120V rms is to be rectified. Estimate very roughly the average current in the output resistor R (35kQ) for (a) a half-wave rectifier (Fig. 40-39), and (b) a full-wave rectifier (Fig. 40-40) without capacitor. 54. (II) A semiconductor diode laser emits 1.3-um light. Assuming that the light comes from electrons and holes recombining, what is the band gap in this laser material? 55. (IT) A silicon diode passes significant current only if the forward-bias voltage exceeds about 0.6 V. Make a rough estimate of the average current in the output resistor R of (a) a half-wave rectifier (Fig. 40-39), and (b) a full-wave rectifier (Fig. 40-40) without a capacitor. Assume that R =120 in each case and that the ac voltage is 9.0 V rms in each case. 56. (IIT) A 120-V rms 60-Hz voltage is to be rectified with a full-wave rectifier as in Fig. 40-40, where R = 28k(), and C =35uF. (a) Make a rough estimate of the average current. (b) What happens if C = 010 uF? [Hint: See Section 26-5.] 40-10 Transistors S7. (II) If the current gain of the transistor amplifier in Fig. 40-43 is B = ic/ig = 95, what value must Rc have if a 1.0-uA ac base current is to produce an ac output voltage of 0.35Vv? 58. (IT) Suppose that the current gain of the transistor in Fig. 40-43 is B = ic/ig = 85. If Rc = 4.3k(), calculate the ac output voltage for an ac input current of 2.0 uA. 59. (IT) An amplifier has a voltage gain of 65 and a 25-k() load (output) resistance. What is the peak output current through the load resistor if the input voltage is an ac signal with a peak of 0.080 V?

Problems

1101

60. (IT) A transistor, whose current gain B = ic/ig = 75, is connected as in Fig. 40-43 with Rp =38k and Rc = 7.8kQ. Calculate (a) the voltage gain, and (b) the power amplification.

61. (II) From Fig. 40-43, write an equation for the relationship

between the base current (I), the collector current (Ic), and the emitter current Assume iy = ic = 0.

(/g, not labeled

in Fig. 40-43).

“

| General Problems 62. Use the uncertainty principle to estimate the binding energy of the H, molecule by calculating the difference in kinetic energy of the electrons between when they are in separate atoms and when they are in the molecule. Take Ax for the electrons in the separated atoms to be the radius of the first Bohr orbit, 0.053 nm, and for the molecule take Ax

to be the separation

Ap ~ Apy.]

63. The

average

of the nuclei, 0.074 nm.

translational kinetic

energy

[Hint: Let

of an atom

or

molecule is about K = 3kT (see Chapter 18), where k =138 x 10°2J/K is Boltzmann’s constant. At what

temperature 7 will K be on the order of the bond energy (and hence the bond easily broken by thermal motion) for (a) a covalent bond (say H;) of binding energy 4.0 eV, and (b) a “weak” hydrogen bond of binding energy 0.12eV?

72. Do we need to consider quantum effects for everyday rotating objects? Estimate the differences between rotational energy levels for a spinning baton compared to the energy of the baton. Assume the baton consists of a uniform 32-cm-long bar with a mass of 260g and two small end masses, each of mass 380 g, and that it rotates at 1.6 rev/s about the bat’s center. 73. Consider a monatomic solid with a weakly bound cubic lattice, with each atom connected to six neighbors, each bond having a binding energy of 3.9 X 107> eV. When this solid melts, its latent heat of fusion goes directly into breaking the bonds between the atoms. Estimate the latent heat of fusion for this solid, in J/mol. [Hint: Show that in a simple cubic lattice (Fig. 40-46), there are three times as many bonds as there are atoms, when the number of atoms is large.]

. In the ionic salt KF, the separation distance between ions is

about 0.27nm. (a) Estimate the electrostatic potential energy between the ions assuming them to be point charges (magnitude le). (b) When F “grabs” an electron, it releases 3.41 eV of energy, whereas 4.34 eV is required to ionize K. Find the binding energy of KF relative to free K and F atoms, neglecting the energy of repulsion. 65. A diatomic molecule is found to have an activation energy of 1.4eV. When the molecule is disassociated, 1.6 eV of energy is released. Draw a potential energy curve for this molecule. . One possible form for the potential energy (U) of a diatomic molecule (Fig. 40-8) is called the Morse Potential:

@

U = U1 = et

(a) Show that ry represents the equilibrium distance and U, the dissociation energy. (b) Graph U from r = 0 to r = 4r,

assuming @ = 18 nm™!,

U, = 4.6eV,

is 8.66 X 10" Hz. Determine (a) the reduced mass, and

(b) the effective value of the stiffness constant k. Compare to k for the H, molecule. . For H,, estimate how many rotational states there are between vibrational states. 69. Explain, using the Boltzmann factor (Eq. 39-16), why the heights of the peaks in Fig. 40-22 are different from one another. Explain also why the lines are not equally spaced. [Hint: Does the moment of inertia necessarily remain constant?] 70. The rotational absorption spectrum of a molecule displays

peaks about 8.4 x 10! Hz apart. Determine the moment of inertia of this molecule.

71. A TV remote control emits IR light. If the detector on the

TV set is not to react to visible light, could it make use of silicon as a “window” with its energy gap E, = 1.14eV? What is the shortest-wavelength light that can strike silicon without causing electrons to jump from the valence band to the conduction band?

CHAPTER 40

@/

23

and r, = 013 nm.

67. The fundamental vibration frequency for the HCI molecule

1102

&

Molecules and Solids

FIGURE 40-46

Problem 73.

74. The energy gap between valence and conduction bands in zinc sulfide is 3.6eV. What range of wavelengths can a photon have to excite an electron from the top of the valence band into the conduction band? 75. When EM radiation is incident on diamond, it is found that light with wavelengths shorter than 226 nm will cause the diamond to conduct. What is the energy gap between the valence band and the conduction band for diamond? 76. The Fermi temperature 7y is defined as that temperature at

which the thermal energy k7 (without the ) is equal to the Fermi energy:

kTg = Eg.

(a) Determine the Fermi temper-

ature for copper. (b) Show that for 7' >> T, the Fermi factor (Eq. 40-14) approaches the Boltzmann factor. (Note: This last result is not very useful for understanding conductors. Why?)? . Estimate the number of states from 4.0eV available to electrons in a 10-cm cube of iron.

to

6.2¢

78. The band gap of silicon is 1.14 eV. (a) For what range of wave-

lengths will silicon be transparent? (See Example 40-14.) In what region of the electromagnetic spectrum does this

o

transparent range begin? (b) If window glass is transparent for all visible wavelengths, what is the minimum possible band gap value for glass (assume A = 450nm to 750nm)? [Hint: If the photon has less energy than the band gap, the photon will pass through the solid without being absorbed.] 9. For a certain semiconductor, the longest wavelength radiation that can be absorbed is 1.92 mm. What is the energy gap in this semiconductor?

80. Assume conduction electrons in a semiconductor behave as an ideal gas. (This is not true for conduction electrons in a metal.) (a) Taking mass m = 9 X 1073 kg and temperature T = 300K, determine the de Broglie wavelength of a semiconductor’s conduction electrons. (b) Given that the spacing between atoms in a semiconductor’s atomic lattice is on the order of 0.3 nm, would you expect room-temperature conduction electrons to travel in straight lines or diffract when traveling through this lattice? Explain. 81. Most of the Sun’s radiation has wavelengths shorter than 1100 nm. For a solar cell to absorb all this, what energy gap ought the material have? 82. Green and blue LEDs became available many years after red LEDs were first developed. Approximately what energy gaps would you expect to find in green (525 nm) and in blue (465 nm) LEDs? . For an arsenic donor ductor, assume

atom

in a doped

that the “extra” electron

silicon semicon-

moves

in a Bohr

orbit about the arsenic ion. For this electron in the ground state, take into account the dielectric constant

K = 12

of

the Si lattice (which represents the weakening of the Coulomb force due to all the other atoms or ions in the lattice), and estimate (a) the binding energy, and (b) the orbit radius for this extra electron. [Hint: Substitute e = Ke( in Coulomb’s law; see Section 24-5.]

8s. A zener diode voltage regulator is shown in Fig. 40-48. Suppose that R = 2.80 k() and that the diode breaks down at a reverse voltage of 130 V. (The current increases rapidly at this point, as shown on the far left of Fig. 40-38 at a voltage of —12V on that diagram.) The diode is rated at a maximum current of 120 mA. (a) If Rjoaq = 18.0k(), over what range of supply voltages will the circuit maintain the output voltage at 130 V? (b) If the supply voltage is 245V, over what range R of load resistance l MW will the voltage 4 be regulated? Vsupply

FIGURE 40-48

A

voutput

§ Rload

‘T

Problem 85.

86. A full-wave rectifier (Fig. 40-40) uses two diodes to rectify a 95-V rms 60 Hz ac voltage. If R = 7.8k{) and C = 36 uF, what will be the approximate percent variation in the output voltage? The variation in output voltage (Fig. 40-40c) is called ripple voltage. [ Hint: See Section 265 and assume the discharge of the capacitor is approximately linear.]

*Numerical/Computer *87. (II) Write a program that will determine the Fermi-Dirac probability function (Eq. 40-14). Make separate plots of this function versus E/Ep for copper at (a) T = S00K;

(b) T =1000K;

(c) T = 5000K; and (d) T = 10,000K.

For copper, Er = 7.0 eV. Interpret each plot accordingly. *88. (IIT) A simple picture of an H, molecule sharing two electrons is shown in Fig. 40-49. We assume the electrons are symmetrically located between the two protons, which are separated by rg = 0.074nm. (a) When the electrons are separated by a distance d, write the total potential energy U in terms of d and ry. (b) Make a graph of U in eV as a function of d in nm, and state where U has a minimum on your graph, and for what range of d values U is negative. (c) Determine analytically the value of d that gives minimum U (greatest stability).

84. A strip of silicon 1.8 cm wide and 1.0 mm thick is immersed in a magnetic field of strength 1.3 T perpendicular to the strip (Fig. 40-47). When a current of 0.28 mA is run through the strip, there is a resulting Hall effect voltage of 18 mV across the strip (Section 27-8). How many electrons per silicon atom are in the conduction band? The density of

+e

|

.4——~————r0

e

-

nucleus

nucleus

—e FIGURE 40-49

silicon is 2330 kg/ m’,

Problem 88.

*89. (I11) Estimate the current produced per cm? of area in a flat silicon semiconductor placed perpendicular to sunlight. Assume

the sunlight has an intensity of 1000 W/m? and that only photons

that have more energy than the band gap can create an electron-hole pair in the semiconductor. Assume the Sun is a blackbody

FIGURE 40-47

Problem 84.

emitter

at 6000 K, and find the fraction of

photons that have energy above the band gap (1.14 eV). See Section 37-1 and integrate the Planck formula numerically.

Answers to Exercises

:0,5.00 X 107*eV, 1.50 X 103 eV. €

@ (e).

(a). D: (c).

General Problems

1103

This archeologist has unearthed the remains of a sea-turtle within an ancient man-made stone circle. Carbon dating of the remains can tell her when humans inhabited the site. In this Chapter we begin our discussion of nuclear physics. We study the properties of nuclei, the

various

and how used in

determine

forms

radioactive a variety the

age

of radioactivity,

of

decay can be of fields to old

objects,

from bones and trees to rocks and other mineral substances, and obtain information on the history of the Earth.

Q>

>

v

TE

&

Nuclear Physics and Radioactivity

o

CONTENTS 41-1 41-2

Structure and Properties of the Nucleus Binding Energy and Nuclear Forces

41-3 41-4 41-5 41-6

Radioactivity Alpha Decay Beta Decay Gamma Decay

41-7

Conservation of Nucleon Number and Other Conservation Laws

41-8 41-9 41-10 41-11

Half-Life and Rate of Decay Decay Series Radioactive Dating Detection of Radiation

1104

—Guess now!

If half of an 80-ug sample of $Co decays in 5.3 years, how much $Co is left in 15.9 years?

10 pg. 20 pg.

30 ug.

40 pg.

0 pg.

n the early part of the twentieth century, Rutherford’s experiments led to the idea that at the center of an atom there is a tiny but massive nucleus. At the same time that quantum theory was being developed and scientists were attempting to understand the structure of the atom and its electron« investigations into the nucleus itself had also begun. In this Chapter and the ne. we take a brief look at nuclear physics.

4]1-1

Structure and Properties of the Nucleus

r n important question for physicists was whether the nucleus had a structure, and what that structure might be. We now understand the nucleus to be a complicated entity that is not fully understood even today. By the early 1930s, a model of the nucleus had been developed that is still useful. According to this model, a nucleus is considered to be an aggregate of two types of particles: protons and neutrons. (These “particles” also have wave properties, but for ease of visualization and language, we often refer to them simply as “particles.”) A proton is the nucleus of

the simplest atom, hydrogen. It has a positive charge (= +e = +1.60 X 107 C, the same magnitude as for the electron) and a mass

m, = 167262 X 107 kg,

The neutron, whose existence was ascertained in 1932 by the English physicist James Chadwick (1891-1974), is electrically neutral (¢ = 0), as its name implies. Its mass is very slightly larger than that of the proton:

m, = 1.67493 X 10

kg.

These two constituents of a nucleus, neutrons and protons, are referred to collectively as nucleons. Although the hydrogen nucleus consists of a single proton alone, the nuclei of all other elements consist of both neutrons and protons. The different nuclei are often referred to as nuclides. The number of protons in a nucleus (or nuclide) is called the atomic number and is designated by the symbol Z. The total number of nucleons, neutrons plus protons, is designated by the symbol A and is called the atomic mass number, or sometimes simply mass number. This name is used since the mass of a nucleus is very closely A times the mass of one nucleon. A nuclide with 7 protons and neutrons thus has Z = 7 and A = 15. The neutron number N is N = A — Z. To specify a given nuclide, we need give only A and Z. A special symbol is commonly used which takes the form A ZX’

where X is the chemical symbol for the element (see Appendix F, and the Periodic Table inside the back cover), A is the atomic mass number, and Z is the atomic number. For example, SN means a nitrogen nucleus containing 7 protons and 8 neutrons for a total of 15 nucleons. In a neutral atom, the number of electrons orbiting the nucleus is equal to the atomic number Z (since the charge on an electron has the same magnitude but opposite sign to that of a proton). The main properties of an atom, and how it interacts with other atoms, are largely determined by the number of electrons in the neutral atom. Hence Z determines what kind of atom it is: carbon, oxygen, gold, or whatever. It is redundant to specify both the symbol of a nucleus and its atomic number Z as described above. If the nucleus is nitrogen, for example, we know immediately that Z = 7. The subscript Z is thus sometimes

dropped and 3N is then written simply '°N; in words we say “nitrogen fifteen.”

For a particular type of atom (say, carbon), nuclei are found to contain different numbers of neutrons, although they all have the same number of protons. For example, carbon nuclei always have 6 protons, but they may have 5, 6, 7, 8, 9, or 10 neutrons. Nuclei that contain the same number of protons but different

numbers of neutrons are called isotopes. Thus, '}C, '2C, 3C, '¢C, 13C, and '$C are

all isotopes of carbon. The isotopes of a given element are not all equally common. For example, 98.9% of naturally occurring carbon (on Earth) is the isotope '2C, and about 1.1% is '}C. These percentages are referred to as the natural abundances.’ Even hydrogen has isotopes: 99.99% of natural hydrogen is {H, a simple proton, as

re nucleus; there are also 7H, called deuterium, and {H, tritium, which besides the proton contain 1 or 2 neutrons. "The

mass

value

for each

element

as given

in the Periodic Table

weighted according to the natural abundances of its isotopes.

(inside back

cover)

is an average

SECTION 41-1

1105

Many isotopes that do not occur naturally can be produced in the laboratory by means of nuclear reactions (more on this later). Indeed, all elements beyond uranium (Z > 92) do not occur naturally on Earth and are only produced artificially (that is, in the laboratory), as are many nuclides with Z = 92, The approximate size of nuclei was determined originally by Rutherford fronfl the scattering of charged particles by thin metal foils (Fig. 37-17). We cannot speak about a definite size for nuclei because of the wave—particle duality: their spatial extent must remain somewhat fuzzy. Nonetheless a rough “size” can be measured by scattering high-speed electrons off nuclei. It is found that nuclei have a roughly spherical shape with a radius that increases with A according to the approximate formula

1

r o~ (1.2 X 1075 m)(43).

(41-1)

Since the volume of a sphere is V = $7r°, we see that the volume of a nucleus is approximately proportional to the number of nucleons, V' oc A. This is what we would expect if nucleons were like impenetrable billiard balls: if you double the number of balls, you double the total volume. Hence, all nuclei have nearly the same density, and it is enormous (see Example 41-1).

The metric abbreviation for 107 m is the fermi (after Enrico Fermi) or the femtometer, fm (Table 1-4 or inside the front cover). Thus 1.2 X 107°m = 1.2 fm or 1.2 fermis. , Because nuclear radii vary as A3, the largest nuclei, such as uranium with A = 238, have a radius only about V238 ~ 6 times that of the smallest, hydrogen (A4 = 1).

DOV

IEA

ESTIMATE | Nuclear and atomic densities. Compare the

density of nuclear matter to the density of normal solids. APPROACH

The density of normal liquids and solids is on the order of 10° to

10*kg/m? (see Table 13-1), and because the atoms are close packed, atoms have

about this density too. We therefore compare the density (mass per volume) of “~ nucleus to that of its atom as a whole. SOLUTION The mass of a proton is greater than the mass of an electron by a factor

1.7 X 10 ¥ kg 9.1 X 103" kg

~

2 X 10%

Thus, over 99.9% of the mass of an atom is in the nucleus, and for our estimate we can say the mass of the atom equals the mass of the nucleus, m,,q/Maom = 1. Atoms have a radius of about 107°m (Chapter 37) and nuclei on the order of

107 m (Eq. 41-1). Thus the ratio of nuclear density to atomic density is about (mnucl/Vnucl)

Pnucl Patom

-

(matom/vatom)

=

(mnud)%-n-rimm Matom

%Wrgucl

(]_()—10)3

" -

(

)(10_15)3

= 10%.

The nucleus is 10" times more dense than ordinary matter. The masses of nuclei can be determined from the radius of curvature of fast-

moving nuclei (as ions) in a known magnetic field using a mass spectrometer, as discussed in Section 27-9. Indeed the existence of different isotopes of the same element (different number of neutrons) was discovered using this device. Nuclear masses can be specified in unified atomic mass units (u). On this scale, a neutral

A

CAUTION

Masses are for neutral atom

2C atom is given the precise value 12.000000 u. A neutron then has a measured mass of 1.008665 u, a proton 1.007276 u, and a neutral hydrogen atom H (proton

Plus electron) 1.007825 u. The masses of many nuclides are given in Appendix F. I}\

(nucleus plus electrons) — should be noted that the masses in this Table, as is customary, are for the neutr¢ atom (including electrons), and not for a bare nucleus.

1106

CHAPTER 41

Nuclear Physics and Radioactivity

Masses are often specified using the electron-volt energy unit. This can be done because mass and energy are related, and the precise relationship is given by Einstein’s equation E = mc* (Chapter 36). Since the mass of a proton is ’1.67262 X 107% kg, or 1.007276 u, then

1.0000u

= (1.0000 u)(

1.67262 x 102 kg 1.007276 u

> = 1.66054 x 107 kg;

this is equivalent to an energy (see Table inside front cover) in MeV (= 10°eV) of R

= me? =

(1.66054 X 107 kg)(2.9979 X 10°m/s)*

(1.6022 X 10 °/eV)

g

Bl

Thus lu

=

1.6605 X 10% kg

=

931.5MeV/c%

The masses of some of the basic particles are given in Table 41-1.

TABLE 41-1

Masses in Kilograms, Unified Atomic Mass Units, and MeV/c? Mass

Object Electron

Proton 1H atom

kg

u

9.1094 x 1073

0.00054858

167262 x 1077 1.67353 x 10777

Neutron

1.67493 x 1077

MeV/c?

1.007276 1.007825 1.008665

0.51100

938.27 938.78 939.57

Just as an electron has intrinsic spin and orbital angular momentum quantum numbers, so too do nuclei and their constituents, the proton and neutron. Both the roton and the neutron are spin-3 particles, just like the electron. A nucleus, made fi p of protons and neutrons, has a nuclear spin quantum number / that is the vector sum of the spins of all its nucleons (plus any orbital angular momentum), and can be either integer or half integer, depending on whether it is made up of an even or an odd number of nucleons. [Orbital angular momentum is integer and doesn’t affect half integer or whole for /.| The total nuclear angular momentum of a nucleus is given, as might be expected (see Section 39-2 and Eq. 39-15), by VI(I + 1)4. Nuclear magnetic moments are measured in terms of the nuclear magneton Py

=

eh

— 2m,

(41-2)

which is defined by analogy with the Bohr magneton for electrons (ug = efi/2m,, Section 39-7). Since wy contains the proton mass, m,,, instead of the electron mass, it is about 2000 times smaller. The electron spin magnetic moment is about 2 Bohr magnetons. The proton’s magnetic moment ., has been measured to be Bp

=

2.7928py .

There is no satisfactory magnetic moment

explanation

for this large factor. The

neutron

has a

we = —1.9135py, which suggests that, although the neutron carries no net charge, it may have internal structure (quarks, as we discuss later). The minus sign for p, indicates that its magnetic moment is opposite to its spin. r Important applications based on nuclear spin are nuclear magnetic resonance NMR) and magnetic resonance imaging (MRI). They are discussed in the next Chapter (Section 42-10).

SECTION 41-1

Structure and Properties of the Nucleus

1107

41-2 Binding Energy and Nuclear Forces

™

Binding Energies

The total mass of a stable nucleus is always less than the sum of the masses of its separate protons and neutrons, as the following Example shows.

ST

ZAPROBLEM

SOLVING

Keep track of electron masses

IEYE

jHe

mass

compared

to its constituents. Compare

the

mass of a $He atom to the total mass of its constituent particles. APPROACH The 4He nucleus contains 2 protons and 2 neutrons. Tables normally give the masses of neutral atoms—that is, nucleus plus its Z electrons—since this is how masses are measured. We must therefore be sure to balance out the electrons when we compare masses. Thus we use the mass of }H rather than that of a proton alone. We look up the mass of the 3He atom in Appendix F (it includes the mass of 2 electrons), as well as the mass for the 2 neutrons and 2 hydrogen atoms (= 2 protons + 2 electrons). SOLUTION The mass of a neutral $He atom, from Appendix F, is 4.002603 u. The mass of two neutrons and two H atoms (2 protons including the 2 electrons) is 2m,

=

2(1.008665u)

=

2.017330u

2m(1H)

=

2(1.007825u) sum

= =

2.015650u 4.032980u.

Thus the mass of $He is measured to be less than the masses of its constituents by an amount 4.032980u — 4.002603 u = 0.030377 u. Where has this lost mass of 0.030377 u disappeared to? It must be E = mc?. If the four nucleons suddenly came together to form a §He nucleus, the mass “loss” would appear as energy of another kind (such as 7y radiation, or kinetic energy). The mass (or energy) difference in the case of $He, given in energy uni is (0.030377u)(931.5MeV/u) = 28.30 MeV. This difference is referred to & the total binding energy of the nucleus. The total binding energy represents the amount of energy that must be put info a nucleus in order to break it apart into its constituents. If the mass of, say, a {He nucleus were exactly equal to the mass of two neutrons plus two protons, the nucleus could fall apart without any input of energy. To be stable, the mass of a nucleus must be less than that of its constituent nucleons, so that energy input is needed to break it apart. Note that the binding energy is not something a nucleus has—it is energy it “lacks” relative to the total mass of its separate constituents. We saw in Chapter 37 that the binding energy of the one electron in the hydrogen atom is 13.6 eV, so the mass of a |H atom is less than that of a single proton plus a single electron by 13.6eV/c% Compared to the total mass of the hydrogen atom

(939 MeV/c?), this is incredibly small, 1 part in 10%. The binding energies of nuclei are

on the order of MeV, so the eV binding energies of electrons can be ignored. Note that nuclear binding energies, compared to nuclear masses, are on the order of (28 MeV/4000 MeV) ~ 1 X 1072 where we used helium’s binding energy (see above) and mass ~ 4 X 940 MeV =~ 4000 MeV. EXERCISE A Determine how much less the mass of the jLi nucleus is compared to that of its constituents.

The binding energy per nucleon is defined as the total binding energy of a nucleus divided by A, the total number of nucleons. We calculated above that the binding energy of 4He is 28.3 MeV, so its binding energy per nucleon is 28.3MeV/4 = 7.1 MeV. Figure 41-1 shows the binding energy per nucleon as a function of A for stable nuclei. The curve rises as A increases and reaches a plateau at about 8.7 MeV per nucleon above A ~ 40. Beyond A ~ 80, the curve decreases slowly, indicating that larger nuclei are held together a little less tight "™ than those in the middle of the Periodic Table. We will see later that these characteristics allow the release of nuclear energy in the processes of fission and fusion.

1108

CHAPTER 41

Nuclear Physics and Radioactivity

10 ;

8

4He 2

§0

s

120

G

2l

\\

ngsU_

g

\

| Q

5

é

6

§.

FIGURE 41-1

?

nucleon for the more stable nuclides as a function of mass number A.

5

4

2 9

g&

Binding energy per

L 2

2He

H

5

30

Fm

100

750

Number of nucleons, A (mass number)

300

750

Binding energy for iron. Calculate the total binding energy and

the binding energy per nucleon for 3¢Fe, the most common stable isotope of iron.

APPROACH We subtract the mass of an 3’Fe atom from the total mass of 26 hydrogen atoms and 30 neutrons, all found in Appendix F. Then we convert mass units to energy units; finally we divide by A = 56, the total number of nucleons. SOLUTION 3:Fe has 26 protons and 30 neutrons whose separate masses are 26m(1H)

=

(26)(1.007825u)

30m, = (30)(1.008665u) =

r{

26.20345 u (includes 26 electrons)

_30.25995u

sum

=

56.46340 u.

Am

=

0.52846 u.

Subtract mass of 35Fe: = —55.93494 u (Appendix F) The total binding energy is thus (0.52846 u)(931.5MeV/u)

=

492.26 MeV

and the binding energy per nucleon is 492.26 MeV = 8.79MeV. 56 nucleons NOTE The binding energy per nucleon graph (Fig. 41-1) peaks about here, for iron, so the iron nucleus (and its neighbors) is the most stable of nuclei. | EXERCISE B Determine the binding energy per nucleon for '§O.

DGUILIRIEL N Binding energy of last neutron. What is the binding energy of the last neutron in 3C? APPROACH If '3C lost one neutron, it would be '2C. We subtract the mass of '2C from the masses of '2C and a free neutron. SOLUTION

Obtaining the masses from Appendix F, we have

Mass 2C = Mass jn = Total

=

Am

=

Subtract mass of 3C:

12.000000 u 1.008665 u

13.008665 u.

—13.003355u

0.005310 u

which in energy is (931.5MeV/u)(0.005310u) = 4.95MeV. That is, it would require 4.95MeV input of energy to remove one neutron from ';C. SECTION 41-2

1109

Nuclear Forces

T120

Neutron number (N

A—2Z)

i

160

0

;

2

40

N

60

.

Proton number (Z)

FIGURE 41-2

80

Number of neutrons

We can analyze nuclei not only from the point of view of energy, but also from the point of view of the forces that hold them together. We would not expect a collection of protons and neutrons to come together spontaneously, since protons are a positively charged and thus exert repulsive electric forces on each other. Since stablt nuclei do stay together, it is clear that another force must be acting. Because this new force is stronger than the electric force, it is called the strong nuclear force. The strong nuclear force is an attractive force that acts between all nucleons—protons and neutrons alike. Thus protons attract each other via the strong nuclear force at the same time they repel each other via the electric force. Neutrons, since they are electrically neutral, only attract other neutrons or protons via the strong nuclear force. The strong nuclear force turns out to be far more complicated than the gravitational and electromagnetic forces. One important aspect of the strong nuclear force is that it is a short-range force: it acts only over a very short distance. It is very strong between two nucleons if they are less than about 107> m apart, but it is essentially zero if they are separated by a distance greater than this. Compare this to electric and gravitational forces, which decrease as 1/r* but continue acting over any distances and are therefore called long-range forces.

The strong nuclear force has some strange features. For example, if a nuclide

contains too many or too few neutrons relative to the number of protons, the binding of the nucleons is reduced; nuclides that are too unbalanced in this regard are unstable. As shown in Fig. 41-2, stable nuclei tend to have the same number of

versus number of protons for stable

protons as neutrons

The straight line represents N = Z.

electrical repulsion increases, so a greater number of neutrons—which exert only the attractive strong nuclear force—are required to maintain stability. For very large Z, no number of neutrons can overcome the greatly increased electric repulsion. Indeed, there are no completely stable nuclides above Z = 82. What we mean by a stable nucleus is one that stays together indefinitely. Wha™ then is an unstable nucleus? It is one that comes apart; and this results in radioactive decay. Before we discuss the important subject of radioactivity (next Section), we note that there is a second type of nuclear force that is much weaker than the strong nuclear force. It is called the weak nuclear force, and we are aware of its existence only because it shows itself in certain types of radioactive decay. These two nuclear forces, the strong and the weak, together with the gravitational and electromagnetic forces, comprise the four basic types of force in nature.

nuclides, which are represented by dots.

41-3 FIGURE 41-3 Marie and Pierre Curie in their laboratory (about

1906) where radium was discovered.

(N = Z)

up to about

A = 30. Beyond this, stable nuclei

contain more neutrons than protons. This makes sense since, as Z increases, the

Radioactivity

Nuclear physics had its beginnings in 1896. In that year, Henri Becquerel (1852-1908)

made an important discovery: in his studies of phosphorescence, he found that a certain mineral (which happened to contain uranium) would darken a photographic

plate even when the plate was wrapped to exclude light. It was clear that the mineral

emitted some new kind of radiation that, unlike X-rays, occurred without any external stimulus. This new phenomenon eventually came to be called radioactivity. Soon after Becquerel’s discovery, Marie Curie (1867-1934)

and her husband,

Pierre Curie (1859-1906), isolated two previously unknown elements that were very highly radioactive (Fig. 41-3). These were named polonium and radium. Other radioactive elements were soon discovered as well. The radioactivity was found in every case to be unaffected by the strongest physical and chemical treatments, including strong heating or cooling or the action of strong chemical reagents. It was suspected that the source of radioactivity must be deep within the atom, emanating from the nucleus. It became apparent that radioactivity is the result of the disintegration or decay of an unstable nucleus. Certain isotopes are™™ not stable, and they decay with the emission of some type of radiation or “rays.”

1110

CHAPTER 41

Nuclear Physics and Radioactivity

K

X X X X

X X X X

XX

oK

X

X

X

X X

Many unstable isotopes occur in nature, and such radioactivity is called “natural radioactivity.” Other unstable isotopes can be produced in the laboratory 5 by nuclear reactions (Section 42-1); these are said to be produced “artificially” and X ¥ X ’ko have “artificial radioactivity.” Radioactive isotopes are sometimes referred to as x Magnetic field adioisotopes or radionuclides. X (into page) X X X X Rutherford and others began studying the nature of the rays emitted in X X X X radioactivity about 1898. They classified the rays into three distinct types according to their penetrating power. One type of radiation could barely penetrate a piece of paper. The second type could pass through as much as 3mm of aluminum. The third was extremely penetrating: it could pass through several centimeters of lead Lead block and still be detected on the other side. They named these three types of radiation _ alpha (a), beta (B), and gamma (7), respectively, after the first three letters of the ~ Radioactive Greek alphabet. S| (Facinm) Each type of ray was found to have a different charge and hence is bent differently =~ FIGURE 41-4 Alpha and beta rays

in a magnetic field, Fig. 41-4; « rays are positively charged, B rays are negatively are bent in opposite directions by a charged, and 7 rays are neutral. It was soon found that all three types of radiation =~ magnetic field, whereas gamma rays consisted of familiar kinds of particles. Gamma rays are very high-energy photons whose energy is even higher than that of X-rays. Beta rays are electrons, identical to those that orbit the nucleus, but they are created within the nucleus itself. Alpha rays (or « particles) are simply the nuclei of helium atoms, 3He; that is, an « ray consists of two protons and two neutrons bound together. We now discuss each of these three types of radioactivity, or decay, in more detail.

~ 3are not bent at all.

41-4 Alpha Decay Experiments show that when nuclei decay, the number of nucleons (= mass umber A) is conserved, as well as electric charge (= Ze). When a nucleus emits A n a particle (3He), the remaining nucleus will be different from the original: it has

lost two protons and two neutrons. Radium 226 (*§Ra), for example, is an

a emitter. It decays to a nucleus with Z = 88 — 2 = 86 and A = 226 — 4 = 222. The nucleus with Z = 86 is radon (Rn)—see Appendix F or the Periodic Table. Thus the radium decays to radon with the emission of an « particle. This is written

See Fig. 41-5.

« decay occurs, a different element is formed. The daughter nucleus

ZRn in this case) is different from the parent nucleus (*%§Ra in this case). This

changing of one element into another is called transmutation. Alpha decay can be written in general as

4N — 475N’ + $He

Radioactive decay

of radium to radon with emission of

an alpha particle.

Z6Ra — ZRn + 3He. When

picURE 41-5

(W

(P)

22Ra

O

(P)

()

222Rn

JHe

[« decay]

where N is the parent, N’ the daughter, and Z and A are the atomic number and atomic mass number, respectively, of the parent. EXERCISE C 34Dy decays by a emission to what element? (d) Er, (e) Yb.

(a) Pb, (b) Gd, (c) Sm,

Alpha decay occurs because the strong nuclear force is unable to hold very large nuclei together. The nuclear force is a short-range force: it acts only between neighboring nucleons. But the electric force acts all the way across a large nucleus. For very large nuclei, the large Z means the repulsive electric force becomes so rge (Coulomb’s law) that the strong nuclear force is unable to hold the nucleus together.

SECTION 41-4

Alpha Decay

1111

We can express the instability of the parent nucleus in terms of energy (or mass): the mass of the parent nucleus is greater than the mass of the daughter nucleus plus the mass of the « particle. The mass difference appears as kinetic energy, which is carried away by the « particle and the recoiling daughter nucleus. The total energy released is called the disintegration energy, O, or the Q-value of the decay. Frornq conservation of energy,

Mpc?* = Mpc® + m,c* + Q, where Q equals the kinetic energy of the daughter and « particle, and Mp, Mp, and m, are the masses of the parent, daughter, and « particle, respectively. Thus

Q

= Mpc® — (Mp + m,)c2

(41-3)

If the parent had /ess mass than the daughter plus the a particle (so @ < 0), the decay could not occur spontaneously, for the conservation of energy law would be violated.

Uranium decay energy release. Calculate the disintegration

energy when %3U (mass = 232.037156 u) decays to 235Th (228.028741 u) with the emission of an « particle. (As always, masses given are for neutral atoms.)

APPROACH We use conservation of energy as expressed in Eq. 41-3. 33U is the parent, %§Th is the daughter. SOLUTION Since the mass of the $He is 4.002603 u (Appendix F), the total mass in the final state is 228.028741u

+ 4.002603 u

=

232.031344 u.

The mass lost when the 33U decays is 232.037156 u — 232.031344u

Since

1u = 931.5MeV,

=

0.005812 u.

the energy Q released is

>

0 = (0.005812u)(931.5MeV/u) = 5.4MeV

and this energy appears as kinetic energy of the a particle and the daughter nucleus.

DOV

RIECE

Kinetic energy of the « in %2U decay. For the 33U decay

of Example 41-5, how much of the 5.4-MeV disintegration energy will be carried off by the « particle? Daughter nucleus

a aVa

mpvp

FIGURE 41-6

Momentum

conservation in Example 41-6.

APPROACH In any reaction, momentum must be conserved as well as energy. SOLUTION Before disintegration, the nucleus can be assumed to be at rest, so the total momentum was zero. After disintegration, the total vector momentum must still be zero so the magnitude of the « particle’s momentum must equal the magnitude of the daughter’s momentum (Fig. 41-6): m,v,

Thus

=

v, = mpvp/m, K,

MmMpvp.

and the o’s kinetic energy is

1, .2 _ 12ma< , (Mo%)\" = 3m,v, m, 228.028741 u

a ( 4.002603 u )KD The total Hence

disintegration K,

=

57 -5§Q

energy =

is

2

)

_ 1ZmDvD L(Mp)

_

(Mp

(ma>KD

= 37k Q = K, + Kp = 57Kp + Kp = 58Kp.

5.3MeV.

The lighter a particle carries off (57/58) or 98% of the total kinetic energy.

1112

CHAPTER 41

Nuclear Physics and Radioactivity

-

a-Decay Theory—Tunneling If the mass of the daughter nucleus plus the mass of the a particle is less than the mass of the parent nucleus (so the parent is energetically allowed to decay), why are there any parent nuclei at all? That is, why haven’t radioactive nuclei all r lecayed long ago, right after they were formed (in supernovae)? We can understand decay using a model of a nucleus that has an alpha particle trapped inside it. The potential energy “seen” by the a particle would have a shape something like that shown in Fig. 41-7. The potential energy well (approximately square) between r = 0 and r = R, represents the short-range attractive nuclear force. U Nuclear attraction Coulomb repulsion

/Energy

0

Ry

Rp

of a

r

—I .

Uin

e

FIGURE 41-7 Potential energy for alpha particle and (daughter) nucleus, showing the Coulomb barrier through which the « particle must tunnel to escape. The Q-value of the reaction is also shown. This plot assumes spherical symmetry, so the central well has

diameter 2R,

Beyond the nuclear radius, R,, the Coulomb repulsion dominates (since the nuclear force drops to zero) and we see the characteristic 1/r dependence of the Coulomb potential. The « particle, trapped within the nucleus, can be thought of as moving back and forth between the potential walls. Since the potential energy just beyond r = R, is greater than the energy of the a particle (dashed line), the a particle could not escape the nucleus if it were governed by classical physics. But according to quantum mechanics, there is a nonzero probability that the « particle can tunnel through the Coulomb barrier, from point A to point B in Fig. 41-7, as we discussed in Section 38-10. The height and width of the barrier affect the rate at which nuclei decay (Section 41-8). Because of this barrier, the lifetimes of a-unstable nuclei can be quite long, from a fraction of a microsecond to over

10" years. Note in Fig. 41-7 that the Q-value represents the total kinetic energy

when the « particle is far from the nucleus. A simple way to look at tunneling is via the uncertainty principle which tells us that energy conservation can be violated by an amount AE for a length of time At given by

(AE)(Ar) ~ o Thus quantum mechanics allows conservation of energy to be violated for brief periods that may be long enough for an « particle to “tunnel” through the barrier. AE would represent the energy difference between the average barrier height and the particle’s energy, and At the time to pass through the barrier. The higher and wider the barrier, the less time the « particle has to escape and the less likely it is f‘ .0 do so. It is therefore the height and width of this barrier that controls the rate of decay and half-life of an isotope.

SECTION 41-4

Alpha Decay

1113

Why a Particles? Why, you may wonder, do nuclei emit this combination of four nucleons called an a particle? Why not just four separate nucleons, or even one? The answer is that the a particle is very strongly bound, so that its mass is significantly less than thah of four separate nucleons. As we saw in Example 41-2, two protons and tw neutrons separately have a total mass of about 4.032980 u (electrons included). For

the decay of %3U discussed in Example 41-5, the total mass of the daughter %5Th

plus four separate

nucleons

is 228.028741 + 4.032980 = 232.061721 u, which

is

greater than the mass of the %3U parent (232.037156). Such a decay could not occur

because it would violate the conservation of energy. Indeed, we have never seen

33U — Z¥Th + 2n + 2p. Similarly, it is almost always true that the emission of a single nucleon is energetically not possible.

Smoke Detectors—An Application @

PHYSICS

APPLIED

Smoke detector

One widespread application of nuclear physics is present in nearly every home in the form of an ordinary smoke detector. The most common type of detector

contains about 0.2 mg of the radioactive americium isotope, 2Am, in the form

of AmO,. The radiation continually ionizes the nitrogen and oxygen molecules in the air space between two oppositely charged plates. The resulting conductivity allows a small steady electric current. If smoke enters, the radiation is absorbed by the smoke particles rather than by the air molecules, thus reducing the current. The current drop is detected by the device’s electronics and sets off the alarm. The radiation dose that escapes from an intact americium smoke detector is much less than the natural radioactive background, and so can be considered relatively harmless. There is no question that smoke detectors save lives and reduce property damage.

41-5

Beta Decay

fl

B~ Decay Transmutation of elements also occurs when a nucleus decays by B8 decay—that is, with the emission of an electron or 8~ particle. The nucleus '¢C, for example, emits an electron when it decays:

YC — 4N + e + neutrino,

A CAUTION B-decay e~ comes from nucleus (itis notan orbital electron) —

where e~ is the symbol for the electron. The particle known as the neutrino, whose charge ¢ = 0 and whose mass is very small or zero, was not initially detected and was only later hypothesized to exist, as we shall discuss later in this Section. No nucleons are lost when an electron is emitted, and the total number of nucleons, A, is the same in the daughter nucleus as in the parent. But because an electron has been emitted from the nucleus itself, the charge on the daughter nucleus is +1e greater than that on the parent. The parent nucleus in the decay written above had Z = +6, so from charge conservation the nucleus remaining behind must have a charge of +7e. So the daughter nucleus has Z = 7, which is nitrogen. It must be carefully noted that the electron emitted in B decay is not an orbital electron. Instead, the electron is created within the nucleus itself. What happens is that one of the neutrons changes to a proton and in the process (to conserve charge) emits an electron. Indeed, free neutrons actually do decay in this fashion: n —

p + e

+ neutrino.

To remind us of their origin in the nucleus, the electrons emitted in B decay are

often referred to as “B particles.” They are, nonetheless, indistinguishable from orbital electrons.

1114

CHAPTER 41

Nuclear Physics and Radioactivity

IETLITIESE when

Energy release in '¢C decay. How much energy is released

'¥C decays to 4N by B emission?

APPROACH We find the mass difference before and after decay, Am. The energy released is E = (Am)c®. The masses given in Appendix F are those of the ‘neutral atom, and we have to keep track of the electrons involved. Assume the parent nucleus has six orbiting electrons so it is neutral; its mass is 14.003242 u.

The daughter in this decay, 4N, is not neutral since it has the same six orbital

electrons circling it but the nucleus has a charge of +7e. However, the mass of this daughter with its six electrons, plus the mass of the emitted electron (which makes a total of seven electrons), is just the mass of a neutral nitrogen atom. SOLUTION

A CAUTION Be careful with atomic and electron masses in B decay

The total mass in the final state is

(mass of N nucleus + 6 electrons) + (mass of 1 electron), and this is equal to mass of neutral 4N (includes 7 electrons), which from Appendix F is a mass of 14.003074 u. So the mass difference is 14.003242 u — 14.003074 u = 0.000168 u, which is equivalent to an energy change Am c* = (0.000168 u)(931.5MeV/u) = 0.156 MeV or 156 keV. NOTE The neutrino doesn’t contribute to either the mass or charge balance since ithas ¢ = 0 and m = 0. According to Example 41-7, we would expect the emitted electron to have a kinetic energy of 156 keV. (The daughter nucleus, because its mass is very much larger than that of the electron, recoils with very low velocity and hence gets very little of the kinetic energy—see Example 41-6.) Indeed, very careful measurements indicate that a few emitted B particles do have kinetic energy close to this calculated alue. But the vast majority of emitted electrons have somewhat less energy. In fact, fle energy of the emitted electron can be anywhere from zero up to the maximum value as calculated above. This range of electron kinetic energy was found for any B decay. It was as if the law of conservation of energy was being violated, and indeed Bohr actually considered this possibility. Careful experiments indicated that linear momentum and angular momentum also did not seem to be conserved. Physicists were troubled at the prospect of giving up these laws, which had worked so well in USA Enrico all previous situations. 34 Fermi In 1930, Wolfgang Pauli proposed an alternate solution: perhaps a new particle that was very difficult to detect was emitted during B decay in addition to the electron. This hypothesized particle could be carrying off the energy, momentum, and angular momentum required to maintain the conservation laws. This new particle was named the neutrino—meaning “little neutral one”—by the great Italian physicist Enrico Fermi (1901-1954; Fig. 41-8), who in 1934 worked out a detailed theory of B decay. (It was Fermi who, in this theory, postuFIGURE 41-8 Enrico Fermi, as lated the existence of the fourth force in nature which we call the weak nuclear ~ portrayed on a US postage stamp.

force.) The neutrino has zero charge, spin of 1%, and was long thought to

Fermi contributed significantly to

Oth theoretical and experimental

have zero mass, although today we are quite sure that it has a very tiny mass

(< 0.14eV/c?). If its mass were zero, it would be much like a photon in that it is ~ Physics.a feat almost unique in

neutral and would travel at the speed of light. But the neutrino is very difficult to detect. In 1956, complex experiments produced further evidence for the existence of the neutrino; but by then, most physicists had already accepted its existence. The symbol for the neutrino is the Greek letter nu (v). The correct way of

HEHN B

writing the decay of '¢C is then

UC - UN + e + n r’l’he bar(7) over the neutrino symbol is to indicate that it is an “antineutrino.” ‘Why thls is called an antineutrino rather than simply a neutrino need not concern us now; it is discussed in Chapter 43.)

SECTION 41-5

Beta Decay

1115

B* Decay

T+ 120

Neutron number (N

A—2Z)

Many isotopes decay by electron emission. They are always isotopes that have too many neutrons compared to the number of protons. That is, they are isotopes that lie above the stable isotopes plotted in Fig. 41-2. But what about unstable isotopesfl that have too few neutrons compared to their number of protons—those that fak_ below the stable isotopes of Fig. 41-2? These, it turns out, decay by emitting a positron instead of an electron. A positron (sometimes called an e” or B8* particle) has the same mass as the electron, but it has a positive charge of +1e. Because it is so like an electron, except for its charge, the positron is called the antiparticle’ to

the electron. An example of a 8" decay is that of }gNe: PNe

0

0

+

20

+

t

40

t

60

Proton number (Z)

80

FIGURE 41-2 (Repeated.) Number of neutrons versus number

where e” stands for a positron. Note that the v emitted here is a neutrino, whereas that emitted in B~ decay is called an antineutrino. Thus an antielectron (= positron) is emitted with a neutrino, whereas an antineutrino is emitted with an electron; this gives a certain balance as discussed in Chapter 43. We can write 8~ and B* decay, in general, as follows:

of protons for stable nuclides, which

are represented by dots. The straight line represents N = Z.

—» BF + e + v,

IN—-> 4.N +e

+v

124N

+

—_p

é_]N’

+

e+

[B~ decay] v,

where N is the parent nucleus and N’ is the daughter.

[B* decay]

Electron Capture Besides B~ and B emission, there is a third related process. This is electron capture (abbreviated EC in Appendix F) and occurs when a nucleus absorbs one of its orbiting electrons. An example is }Be, which as a result becomes ]Li. The process is written

Be + e — ILi + v, or, in general,

“

IN +e

= 4N

+ .

[electron capture]

Usually it is an electron in the innermost (K) shell that is captured, in which case the process is called K-capture. The electron disappears in the process, and a proton in the nucleus becomes a neutron; a neutrino is emitted as a result. This process is inferred experimentally by detection of emitted X-rays (due to other electrons jumping down to fill the empty state) of just the proper energy. In B decay, it is the weak nuclear force that plays the crucial role. The neutrino is unique in that it interacts with matter only via the weak force, which is why it is so hard to detect. FIGURE 41-9

Energy-level

diagram showing how 3B can decay to the ground state of '2C by B decay

(total energy released = 13.4 MeV), or can instead B decay to an excited state of '2C (indicated by *), which subsequently decays to its ground state by emitting a 4.4-MeV Y ray. 12sB B~ (9.0 MeV) (13.4 MeV)

——

| ¢C 2O%

y (44 MeV)

1116

CHAPTER 41

41-6

Gamma Decay

Gamma rays are photons having very high energy. They have their origin in the decay of a nucleus, much like emission of photons by excited atoms. Like an atom, a nucleus itself can be in an excited state. When it jumps down to a lower energy state, or to the ground state, it emits a photon which we call a ¥ ray. The possible energy levels of a nucleus are much farther apart than those of an atom: on the order of keV or MeV, as compared to a few eV for electrons in an atom. Hence, the emitted photons have energies that can range from a few keV to several MeV. For a given decay, the Y ray always has the same energy. Since a ¥ ray carries no charge, there is no change in the element as a result of a ¥ decay. How does a nucleus get into an excited state? It may occur because of a violent collision with another particle. More commonly, the nucleus remaining after a previous radioactive decay may be in an excited state. A typical example i shown in the energy-level diagram of Fig. 41-9. '2B can decay by B decay directly "Discussed in Chapter 43. Briefly, an antiparticle has the same mass as its corresponding particle, but opposite charge. A particle and its antiparticle can quickly annihilate each other, releasing energy (7 rays).

to the ground state of '2C; or it can go by B decay to an excited state of '2C, which then decays by emission of a 4.4-MeV We can write ¥ decay as

7 ray to the ground state.

4N* = 4N + 7,

[Y decay]

~here the asterisk means “excited state” of that nucleus. What, you may wonder, is the difference between a ¥ ray and an X-ray? They both are electromagnetic radiation (photons) and, though ¥ rays usually have higher energy than X-rays, their range of energies overlap to some extent. The difference is not intrinsic. We use the term X-ray if the photon is produced by an electron-atom interaction, and 7 ray if the photon is produced in a nuclear process.

Isomers; Internal Conversion In some cases, a nucleus may remain in an excited state for some time before it

emits a ¥ ray. The nucleus is then said to be in a metastable state and is called an isomer. An excited nucleus can sometimes return to the ground state by another process known as internal conversion with no ¥ ray emitted. In this process, the excited nucleus interacts with one of the orbital electrons and ejects this electron from the atom with the same kinetic energy (minus the binding energy of the electron) that an emitted ¥ ray would have had.

and Other

In all three

Energy,

1asLe 41-2 The Three

Conservation of Nucleon Number

41-7

types

of radioactive

linear momentum,

conserved. These

.

Conservation Laws

decay, the classical conservation

angular

momentum,

quantities are the same

before

and

laws

electric charge

the decay

—

hold.

are

all

as after. But a new

nservation law is also revealed, the law of conservation of nucleon number.

.ccording to this law, the total number of nucleons (A) remains constant in any process, although one type can change into the other type (protons into neutrons or vice versa). This law holds in all three types of decay. Table 41-2 gives a summary of a, B, and ¥ decay. [In Chapter 43 we will generalize this and call it conservation of baryon number.]

41-8

es of Radioactive Decay

4N — 42N’ + 4He

B decay: 4N

>

4N

4N - 4N

+e

+ 7

+e" +v

AN + e — 4N’ ¥ decay:

+ v [EC]'

IN* — 4N + 7 " Electron capture.

*Indicates the excited state of a nucleus.

Half-Life and Rate of Decay

A macroscopic sample of any radioactive isotope consists of a vast number of radioactive nuclei. These nuclei do not all decay at one time. Rather, they decay one by one over a period of time. This is a random process: we can not predict exactly when a given nucleus will decay. But we can determine, on a probabilistic basis, approximately how many nuclei in a sample will decay over a given time period, by assuming that each nucleus has the same probability of decaying in each second that it exists. The number of decays AN that occur in a very short time interval At is then proportional to Af and to the total number N of radioactive nuclei present:

AN

=

—AN At

(41-4a)

where the minus sign means N is decreasing. We rewrite this to get the rate of decay:

AN

A

- MV

(41-4b)

In these equations, A is a constant of proportionality called the decay constant, flhiCh is different for different isotopes. The greater A is, the greater the rate of «cay and the more “radioactive” that isotope is said to be. The number of decays that occur in the short time interval At is designated AN because each decay that

SECTION 41-8

Half-Life and Rate of Decay

1117

o o Hute

3

;

;

.

s

aba

O

nuclei in a sample is continually decreasing.

Q

0?2240

g

When a '¢C nucleus emits an electron, the

Q

nucleus becomes a 4N nucleus.

O -

Q

Q

o

e ) Q

o

o

©,.9

Qo

o

Q 0

@) o)

(a)

Legend

i

O

Q o

o

(b)

o

O '4Catom

Q

(pasenty

(O 4N atom

(daughter)

()

occurs corresponds to a decrease by one in the number N of nuclei present. That is, radioactive decay is a “one-shot” process, Fig. 41-10. Once a particular parent nucleus decays into its daughter, it cannot do it again. If we take the limit At — 0in Eq. 41-4, AN will be small compared to N, and we can write the equation in infinitesimal form as dN

=

—AN dt.

(41-5)

We can determine N as a function of ¢ by rearranging this equation to dN N

-

—Adt

and then integrating from

Lw

¢t =0

to t = t:

=l

NdN

Jl

—

= —|Adt,

0

where N is the number of parent nuclei present at t = 0 and N is the number remaining at time ¢. The integration gives

or

N = Nye™.

41-6

Equation 41-6 is called the radioactive decay law. It tells us that the number c. radioactive nuclei in a given sample decreases exponentially in time. This is shown

in Fig. 41-11a for the case of '#C whose decay constant is A = 3.83 x 1072571, The rate of decay in a pure sample, or number of decays per second, is

Ndt

L}

also called the activity of the sample. We use absolute value signs to make activity a positive number (dN/dt is negative because the number of parent nuclei N is decreasing). The symbol R is also used for activity, R = |dN/dt|. FIGURE 41-11

(a) The number N of parent nuclei in a given sample of '3C decreases exponentially.

We set N, = 1.00 X 10? here, as we do in the text shortly. (b) The number of decays per second also

decreases exponentially. The half-life (Eq. 41-8) of '¢C is about 5730 yr, which means that the number of

Number of parent nuclei, N

parent nuclei, N, and the rate of decay,

1118

CHAPTER 41

|dN/dt|, decrease by half every 5730 yr.

Nk (=1.0 X 10%)

=i

104

S

3

o=

1 2N67

£2.0 X 10104 §

L

1 INE 0

0

o s —

—

o + f

5730

—r

+ }

11,460

g

P

Time, t (yr)

}

17,190

(a) Nuclear Physics and Radioactivity

5S A

L e e L 0

0

+ f

5730

-

—

+ f

11460

_

Time, t (yr)

(b)

}

17,190

-

From Egs. 41-5 and 41-6,

dN

zi—t— ’At

=

AN

=

ANe™.

(41-7a)

t = 0, the activity is

)

dN — ar |, =

ANy .

( 41-7b )

Hence, at any other time ¢ the activity is dN| i

_

|dN] ar Oe

;

(41-7¢)

so the activity decreases exponentially in time at the same rate as for N (Fig. 41-11b). Equation 41-7c is sometimes referred to as the radioactive decay law (so is Eq. 41-6), and can be written using R to represent activity, R = |dN/dt|, as

R = Rpe™

(41-7d)

The rate of decay of any isotope is often than the decay constant A. The half-life of an for half the original amount of isotope in a the half-life of '3C is about 5730years. If

specified by giving its half-life rather isotope is defined as the time it takes given sample to decay. For example, at some time a piece of petrified

wood contains, say, 1.00 X 10% ¥C nuclei, then 5730 yr later it will contain only 0.50 X 10%nuclei. After another 5730yr it will contain 0.25 X 10?nuclei, and

so on. This

is characteristic

of the

exponential

function,

and

is shown

in

Fig. 41-11a. Since the rate of decay |dN/dt| is proportional to N, it too decreases by a factor of 2 every half-life, Fig, 41-11b.

EXERCISE D The half-life of Na is 2.6 years. How much will be left of a 1.0-ug sample of #Na after 5.2 yr? (a) None. (b) §ug. (c) ug. (d) }ug. (e) 0.693 ug. EXERCISE E Return to the Chapter-Opening Question, page 1104, and answer it again now. Try to explain why you may have answered differently the first time.

r '

The half-lives of known

radioactive isotopes vary from as short as 107%s

to about 10%®s (about 10?'yr). The half-lives of many isotopes are given in

Appendix F. It should be clear that the half-life (which we designate T1) bears an inverse relationship to the decay constant. The longer the half-life of an isotope, the more slowly it decays, and hence A is smaller. The precise relation is obtained from Eq. 41-6 by setting N = N,/2 at t = fi%

No = Nye ™

or

el

= 2,

2 We take natural logs of both sides (“In” and “e” are inverse operations, meaning

In(e*) = x) and find In(e’}) = In2,

50

AT, = In2

and T.

2

=

E A

=

9@ A

(41-8)

We can then write Eq. 41-6 as N

=

Noe_°'693’/7'%.

Sometimes the mean life 7 of an isotope is quoted, which is defined as 7 = 1/A (see also Problem 80), so then Eq. 41-6 can be written

N = Nye'/ just as for RC and LR circuits (Chapters 26 and 30, where 7 is called the time constant). Since

T X T 0693

41-9 =

R = Rye ™o,

(41-9b)

the mean life and half-life differ by a factor 0.693; confusing them can cause serious error. The radioactive decay law, Eq. 41-7d, can be written simply as

SECTION 41-8

1119

Sample activity. The isotope '3C has a half-life of 5730 yr. If

a sample contains 1.00 X 10%? carbon-14 nuclei, what is the activity of the sample?

APPROACH We first use the half-life to find the decay constant (Eq. 41-8), and use that to find the activity, Eq. 41-7b or 41-5. The number of seconds in a year 1.\

(60 s/min)(60 min/h)(24 h/d)(3655d/yr) = 3.156 X 107s.

SOLUTION

9

The decay constant A from Eq. 41-8 is

i

0693

0.693

T,

(5730yr)(3.156 X 107s/yr)

From Eq. 41-7b, the activity or rate of decay is

dN ‘Za_

_

PO

B

2, g-1

= AN, = (3.83 x 102571)(1.00 X 10%) = 3.83 X 10'" decays/s.

0

Notice that the graph of Fig. 41-11b starts at this value, corresponding to the

original value of N, = 1.0 X 10% nuclei in Fig. 41-11a.

NOTE The unit “decays/s” is often written simply as s~ 1 since “decays” is not a unit but refers only to the number. This simple unit of activity is called the becquerel: 1 Bq = 1decay/s, as discussed in Chapter 42.

EXERCISE F Determine the decay constant for radium (7i = 1600 yr).

CONCEPTUAL

EXAMPLE 41-9 | Safety: Activity versus half-life. One might

think that a short half-life material is safer than a long half-life material because it will not last as long. Is that true? RESPONSE No. A shorter half-life means “radioactive” and can cause more biological half-life for the same sample size N means about it for longer and find safe storage until

A

the activity is higher damage. On the other a lower activity but we it reaches a safe (low)

and thus more hand, a longer have to worry level of activity.

sample of radioactive ';N. A laboratory has 1.49 ug of pu

3N, which has a half-life of 10.0 min (6005s). (¢) How many nuclei are present initiall, (b) What is the activity initially? (c¢) What is the activity after 1.00h? (d) After approximately how long will the activity drop to less than one per second (= 1s71)?

APPROACH We use the definition of the mole and Avogadro’s number (Sections 17-7 and 17-9) to find the number of nuclei. For (b) we get A from the given half-life and use Eq. 41-7b for the activity. For (c) and (d) we use Eq. 41-7c.

SOLUTION

(a) The atomic mass is 13.0, so 13.0 g will contain 6.02 X 10% nuclei

(Avogadro’s number). We have only 1.49 X 107 g, so the number of nuclei N, that we have initially is given by the ratio Ny 1.49 X 10°%g 6.02 x 107 13.0¢g Solving, we find N, = 6.90 X 10' nuclei. (b) From Eq. 41-8,

A = 0.693/T; = (0.693)/(600s) = 1.155 x 107*s". Then, at t = 0 (Eq. 41-7b), dN|

AN, = (1.155 x 103571)(6.90 x 10') = 7.97 X 10 decays/s.

dt |y

(c) After

£t

1.00h = 3600s, the magnitude of the activity will be (Eq. 41-7c)

AN g

= (7.97 X 1013 571)e (1159X1077)(605) = 125 % 102571,

dt dt |o (d) We want to determine the time f when —At

1120

CHAPTER 41

_

|dN/atl

"~ |dN/dt|,

|dN/dt| = 1.00s™". From Eq.41-7c, we hay\

1.00s7!

7.97 x 105!

=

125 % 107,

We take the natural log (In) of both sides

In(1.25 x 1074

& B

S

fL&

(Ine

™ = —Af)

and divide by A to find

X 10*s = 7.70h.

AEasy Alternate Solution to (c) 1.00 h = 60.0 minutes is 6 half-lives, so the activity will

decrease to (3)3)3)3)R)E) = 3)° = & ofits original value, or (7.97 x 10')/(64)

= 1.25 X 10" per second.

EXERCISE G Technicium %3Tc has a half-life of 4.2 X 10° yr. Strontium %Sr has a half-life of 28.79 yr. Which statements are true? (a) The decay constant of Sr is greater than the decay constant of Tc. (b) The activity of 100 g of Sr is less than the activity of 100 g of Tc. (¢) The long half-life of Tc means that it decays by alpha decay. (d) A Tc atom has a higher probability of decaying in 1yr than a Sr atom. (e) 28.79 g of Sr has the same activity as 4.2 X 10° g of Tc.

41-9

Decay Series

It is often the case that one radioactive isotope decays to another isotope that is also radioactive. Sometimes this daughter decays to yet a third isotope which also is radioactive. Such successive decays are said to form a decay series. An important

example is illustrated in Fig. 41-12. As can be seen, 35U decays by emission to 24Th, which in turn decays by 8 decay to %{Pa. The series continues as shown, with several possible branches near the bottom, ending at the stable lead isotope, 26Pb. The two last decays can be

or

26T1 — %6Pb + ¢~ + ¥,

(T, = 4.2min)

%0

(T, =

— 2%Pb + a.

138days)

Other radioactive series also exist.

"

A 238

T »

LESEND

-

\Q?%

LW

296 22

222

218

&

)

214,

518 210

210

27 miry, #20 mig,

Pid

G

.V 1.3 min[22

'2% 202l 8142ming>

Tl

= yg#

S

’

206

\b@*

L

4

Pb

83

Bi

o

FIGURE 41-12

Rn 4

series are specified by a dot

representing A and Z values. Half-lives are given in seconds (s),

minutes (min), hours (h), days (d),

or years (yr). Note that a horizontal arrow represents 8 decay (A does not change), whereas a diagonal line represents a decay (A changes by 4, Z changes by 2). For the four nuclides shown that can decay by both « and $ decay,

o 218, 218, o

the more prominent decay (in these four cases, >99.9%) is shown as a

x

Po

Decay series

beginning with Z§U. Nuclei in the

£

’

/ 3-\\94 &

Bi

4

2305,

‘

218, |

+\°fl

[6.7h/

24d

B Ziecay

adecay

230

-

238, 2

solid arrow and the less common

decay (< 0.1%) as a dashed arrow.

b+\°’

Y

[5d

N 84

Po

85

At

86

Rn

87

Fr

88

Ra

8

Ac

90

Th

91

Pa

92

U

Z

SECTION 41-9

1121

Because of such decay series, certain radioactive elements are found in nature that otherwise would not be. When the solar system (including Earth) was formed about 5 billion years ago, it is believed that nearly all nuclides were present, having been formed (by fusion and neutron capture, Sections 42-4 and 44-2) in a nearby\ supernova explosion (Section 44-2). Many isotopes with short half-lives decayec quickly and no longer are detected in nature today. But long-lived isotopes, such as

%8U with a half-life of 4.5 X 10° yr, still do exist in nature today. Indeed, about half of the original %3U still remains. We might expect, however, that radium (%3$Ra), with a half-life of 1600yr, would long since have disappeared from the Earth.

Indeed, the original %Ra nuclei must by now have all decayed. However, because 23U decays (in several steps) to %¢Ra, the supply of %2¢Ra is continually replenished, which is why it is still found on Earth today. The same can be said for many other radioactive nuclides.

CONCEPTUAL EXAMPLE 41-11 | Decay chain. In the decay chain of Fig. 41-12, if we start looking below 33U, we see four successive nuclides with half-lives of

250,000 yr, 75,000 yr, 1600 yr, and a little under 4 days. Each decay in the chain has an alpha particle of a characteristic energy, and so we can monitor the radioactive decay rate of each nuclide. Given a sample that was pure %33U a million years ago, which alpha decay would you expect to have the highest activity in the sample?

RESPONSE The first instinct is to say that the process with the shortest half-life would show the highest activity. Surprisingly, perhaps, the activities of the four nuclides in this sample are all the same. The reason is that in each case the decay of the parent acts as a bottleneck to the decay of the daughter. Compared to the

1600-yr half-life of %3§Ra, for example, its daughter 32Rn decays almost immediately,

but it cannot decay until it is made. (This is like an automobile assembly line: if worker A takes 20 minutes to do a task and then worker B takes only 1 minute to do the next task, worker B still does only one car every 20 minutes.)

41-10 @

PHYSICS

Radioactive Dating

R

Radioactive decay has many interesting applications. One is the technique of radioactive dating by which the age of ancient materials can be determined. APPLIED The age of any object made from once-living matter, such as wood, can be Carbon-14 dating ~ determined using the natural radioactivity of '¢C. All living plants absorb carbon dioxide (CO,) from the air and use it to synthesize organic molecules. The vast

majority of these carbon atoms are '2C, but a small fraction, about 1.3 X 107", is the radioactive isotope '¢C. The ratio of '{C to '2C in the atmosphere has remained roughly constant over many thousands of years, in spite of the fact that '¢C decays with a half-life of about 5730 yr. This is because energetic nuclei in the cosmic radi-

ation, which impinges on the Earth from outer space, strike nuclei of atoms in the

atmosphere and break those nuclei into pieces, releasing free neutrons. Those neutrons can collide with nitrogen nuclei in the atmosphere to produce the

nuclear transformation absorbed

by a “N

n + “N —

%C + p. That is, a neutron strikes and is

nucleus, and a proton

is knocked

out in the process. The

remaining nucleus is '¢C. This continual production of '{C in the atmosphere roughly balances the loss of '¢C by radioactive decay.

As long as a plant or tree is alive, it continually uses the carbon from carbon dioxide in the air to build new tissue and to replace old. Animals eat plants, so they too are continually receiving a fresh supply of carbon for their tissues.

Organisms cannot distinguish’ C

from '2C, and since the ratio of "}C to C

in

the atmosphere remains nearly constant, the ratio of the two isotopes within the living organism remains nearly constant as well. When an organism dies, carbon dioxide is no longer absorbed and utilized. Because the '$C decays radioactively, the ratio of #C to '2C in a dead organism decreases over time. Since the half-life o

1C is about 5730 yr, the '¢C/'2C ratio decreases by half every 5730 yr. If, for examplec 1122

CHAPTER 41

*Organisms operate almost exclusively via chemical reactions—which involve only the outer orbital electrons of the atom; extra neutrons in the nucleus have essentially no effect.

the '¥C/'2C ratio of an ancient wooden tool is half of what it is in living trees, then the object must have been made from a tree that was felled about 5730 years ago.

Actually, corrections must be made for the fact that the '3C/'2C ratio in

he atmosphere has not remained precisely constant over time. The determination f what this ratio has been over the centuries has required techniques such as comparing the expected ratio to the actual ratio for objects whose age is known, such as very old trees whose annual rings can be counted reasonably accurately. DGR IERPE An ancient animal. The mass of carbon in an animal bone fragment found in an archeological site is 200 g. If the bone registers an activity of 16 decays/s, what is its age?

APPROACH

200-g sample

First we when

determine

the

how

animal

was

many

'{C atoms

alive, given

there

the known

were

in our

fraction

PHYSICS APPLIED Archeological dating

of ¢C,

1.3 X 107", Then we use Eq. 41-7b to find the activity back then, and Eq. 41-7c

to find out how long ago that was by solving for the time ¢.

SOLUTION The 200g of carbon is nearly 6.02 X 10% atoms, so 200 g contains 6.02 X 10* atoms/mol

all '?C; 12.0g

)(200 g)

12 g/mol

=

of }C

contains

1.00 X 10* atoms.

When the animal was alive, the ratio of '¢C to '2C in the bone was 1.3 X 1072, The number of 'C nuclei at that time was

N, = (1.00 x 10¥ atoms)(1.3 X 107?) = 1.3 X 10" atoms. From Eq. 41-7b, with A = 3.83 X 107?s7!

for '{C (Example 41-8), the magni-

tude of the activity when the animal was still alive (f = 0) was

(L

Z—tN

0

= AN, = (3.83 X 1072s7!)(1.3 X 10¥) = 50s7%

From Eq. 41-7c,

dN| _ |dN] dt

dat |

where |dN/dt| is given as 16s7". Then _|aN/dtly 50s7! Al |dN/dt|

165

We take the natural log (In) of both sides (and divide by A) to get

1. | |dN/dt|

t = —In A

|dN/dt|

=

1

505

g 3.83 X 107%*s 2.98 X 10''s

16 -

= 9400yr,

which is the time elapsed since the death of the animal.

e

Geological Time Scale Dating Carbon dating is useful only for determining the age of objects less than about

60,000 years old. The amount of '¢C remaining in objects older than that is usually

too small to measure accurately, although new techniques are allowing detection of

even smaller amounts of '¢C, pushing the time frame further back. On the other

hand, radioactive isotopes with longer half-lives can be used in certain circumstances to obtain the age of older objects. For example, the decay of 35U, because

PHYSICS

Geological dating

APPLIED

of its long half-life of 4.5 X 10° years, is useful in determining the ages of rocks on a geologic time scale. When molten material on Earth long ago solidified into r)ck as the temperature dropped, different compounds solidified according to the melting points, and thus different compounds separated to some extent. Uranium

SECTION 41-10

Radioactive Dating

1123

PHYSICS

APPLIED

Oldest Earth rocks ~and

carliestlife

iy

present in a material became fixed in position and the daughter nuclei that result from the decay of uranium were also fixed in that position. Thus, by measuring the amount of 23U remaining in the material relative to the amount of daughter nuclei, the time when the rock solidified can be determined.

Radioactive dating methods using 35U and other isotopes have shown the agfl

of the oldest Earth rocks to be about 4 X 10° yr. The age of rocks in which the oldest fossilized organisms are embedded indicates that life appeared more than

31 billion years ago. The earliest fossilized remains of mammals are found in rocks 200 million years old, and humanlike creatures seem to have appeared about 2 million years ago. Radioactive dating has been indispensable for the reconstruction of Earth’s history.

41-11

Gas in tube

Detection of Radiation

Individual particles such as electrons, protons, « particles, neutrons, and ¥ rays are

Wire electrode

ments have detected directly by our senses. Consequently, a variety of instru not been developed to detect them.

Metal tube (cathode = —)

Counters One of the most common detectors is the Geiger counter. As shown in Fig. 41-13, it

(anode = +)

consists of a cylindrical metal tube filled with a certain type of gas. A long wire

runs down the center and is kept at a high positive voltage (~ 10* V) with respect l 1 To counter .

FIGURE 41-13

Geiger counter.

Diagram of a

FIGURE 41-14 Scintillation counter with a photomultiplier tube. Incoming

particle

Photon ov

Scintillation

crystal

Photocathode

E-

L +200V

+400V —

1600V

+800V —

—+ 1000V

+1200V — +1600 V —iof i T——

Exitgt

pulse (to counter)

1124

— + 1400 V

Phomg}géflpher

CHAPTER 41

to the outer cylinder. The voltage is just slightly less than that required to ionize the gas atoms. When a charged particle enters through the thin “window” at one end of the tube, it ionizes a few atoms of the gas. The freed electrons are attracted toward the positive wire, and as they are accelerated they strike and ionize additional atoms. An “avalanche” of electrons is quickly produced, and when it reaches the wire anode, it produces a voltage pulse. The pulse, after being amplified, can be sent to an electronic counter, which counts how many particles have bee

detected. Or the pulses can be sent to a loudspeaker and each detection of'

particle is heard as a “click.” Only a fraction of the radiation emitted by a sample is detected by any detector. A scintillation counter makes use of a solid, liquid, or gas known as a scintillator or phosphor. The atoms of a scintillator are easily excited when struck by an incoming particle and emit visible light when they return to their ground states. Typical scintillators are crystals of Nal and certain plastics. One face of a solid scintillator is cemented to ao photomultiplier tube, and the . whole is wrapped i . . : e with opaque material to keep it light-tight (in the dark) or is placed within a light-tight container. The photomultiplier (PM) tube converts the energy of the

geintillator-emitted photon(s) into an electric signal.

A PM

tube is a vacuum

tube containing several electrodes (typically 8 to 14), called dynodes, which are maintained at successively higher voltages as shown in Fig. 41-14. At its top surface is a photoelectric surface, called the photocathode, whose work function (Section 37-2) is low enough that an electron is easily released when struck by a photon from the scintillator. Such an electron is accelerated toward the positive voltage of the first dynode. When it strikes the first dynode, the electron has acquired sufficient kinetic energy so that it can eject two to five more electrons. These, in turn, are accelerated toward the higher voltage second dynode, and a multiplication process begins. The number of electrons striking the last dynode may be 10° or more. Thus the passage of a particle through the scintillator results in an electric signal at the output of the PM tube that can be sent to an electronic counter just as for a Geiger tube. Solid scintillators are much more dense than the gas of a Geiger counter, and so are much more efficient detectors—

especially for ¥ rays, which interact less with matter than do a or B particles.

Scintillators that can measure the total energy deposited are much used today am‘ are called calorimeters.

Nuclear Physics and Radioactivity

In tracer work (Section 42-8), liquid scintillators are often used. Radioactive samples taken at different times or from different parts of an organism are placed directly in small bottles containing the liquid scintillator. This is particularly conve-

nient for detection of B rays from {H and '¢C, which have very low energies and

rlave difficulty passing through the outer covering of a crystal scintillator or Geiger tube. A PM tube is still used to produce the electric signal from the liquid scintillator. A semiconductor detector consists of a reverse-biased pn junction diode (Sections 40-8 and 40-9). A particle passing through the junction can excite electrons into the conduction band, leaving holes in the valence band. The freed charges produce a short electrical pulse that can be counted just as for Geiger and scintillation counters. Hospital workers and others who work around radiation may carry film badges which detect the accumulation of radiation exposure. The film inside is periodically replaced and developed, the darkness being related to total exposure (see Section 42-6).

Visualization

The devices discussed so far are used for counting the number of particles (or

decays of a radioactive isotope). There are also devices that allow the track of charged particles to be seen. Very important are semiconductor detectors. Silicon wafer semiconductors have their surface etched into separate tiny pixels, each providing particle position information. They are much used in elementary particle physics (Chapter 43) to track the positions of particles produced and to determine their point of origin and/or their momentum (with the help of a magnetic field). The pixel arrangement can be CCD or CMOS (Section 33-5), the latter able to incorporate electronics inside, allowing fast readout. One of the oldest tracking devices is the photographic emulsion, which can be small and portable, used now particularly for cosmic-ray studies from balloons. A charged particle passing through an emulsion ionizes the atoms along its path. r “hese points undergo a chemical change, and when the emulsion is developed (like film) the particle’s path is revealed. In a cloud chamber, used in the early days of nuclear physics, a gas is cooled to a temperature slightly below its usual condensation point (“supercooled”), and gas molecules condense on any ionized molecules present. Tiny droplets form around ions produced when a charged particle passes through (Fig. 41-15). Light

scattering from these droplets reveals the track of the particle.

The bubble chamber, invented in 1952 by D. A. Glaser (1926- ), makes use of a superheated liquid kept close to its normal boiling point. Bubbles characteristic of boiling form around ions produced by the passage of a charged particle, revealing paths of particles that recently passed through. Because a bubble chamber uses a liquid, often liquid hydrogen, many more interactions can occur than in a cloud chamber. A magnetic field is usually applied across the chamber so the momentum of the moving particles can be determined from the radius of curvature of their paths. A wire drift chamber consists of a set of closely spaced fine wires immersed in a gas (Fig. 41-16). Many wires are grounded, and the others between are kept at very high voltage. A charged particle passing through produces ions in the gas. Freed electrons drift toward the nearest high voltage wire, creating an “avalanche” of many more ions, and producing an electric pulse or signal at that wire. The positions of the particles are determined electronically by the position of the wire and by the time it takes the pulses to reach “readout” electronics at the ends of the wires. The paths of the particles are reconstructed electronically by computers which can “draw” a picture of the tracks, as shown in the photo at the start of Chapter 43. An external magnetic field curves the paths, allowing the rmomentum of the particles to be measured. In many detectors, the energy of the particles can be measured by the strength of the electronic signal; such detectors are referred to as calorimeters, as already mentioned. SECTION 41-11

Path of particle

FIGURE 41-15 In a cloud chamber or bubble chamber, droplets or bubbles are formed around ions produced by the passage of a charged particle. FIGURE 41-16

Wire-drift chamber

inside the Collider Detector at

Fermilab (CDF). The photo at the start of Chapter 43 (page 1164) was done with this detector.

Detection of Radiation

1125

| Summary Nuclear physics is the study of atomic nuclei. Nuclei contain protons and neutrons, which are collectively known as nucleons.

The total number of nucleons, A, is the nucleus’s atomic mass number. The number of protons, Z, is the atomic number. The number of neutrons equals A — Z. Isotopes are nuclei with the same Z, but with different numbers of neutrons. For an element X, an isotope of given Z and A is represented by

2X.

i The nuclear radius is approximately proportional to A3, indicating that all nuclei have about the same density. Nuclear masses are specified in unified atomic mass units (u), where the mass of '3C (including its 6 electrons) is defined as exactly 12.000000u. In terms of the energy equivalent (because

E = mc?),

lu = 931.5MeV/c?

Nuclei are held together by the strong nuclear force. The weak nuclear force makes itself apparent in 8 decay. These tw forces, plus the gravitational and electromagnetic forces, are thy four known types of force. Electric

linear

and

angular

momentum,

mass—

the minus sign means N decreases in time.

The proportionality constant A is called the decay constant

and is characteristic of the given nucleus. The number nuclei remaining after a time ¢ decreases exponentially

= 1.66 X 107 kg.

The mass of a stable nucleus is less than the sum of the masses of its constituent nucleons. The difference in mass (times c?) is the total binding energy. It represents the energy needed to break the nucleus into its constituent nucleons. The binding energy per nucleon averages about 8 MeV per nucleon, and is lowest for low mass and high mass nuclei. Unstable nuclei undergo radioactive decay; they change into other nuclei with the emission of an a, B3, or ¥ particle. An a particle is a 3He nucleus; a B particle is an electron or positron; and a Y ray is a high-energy photon. In B decay, a neutrino is also emitted. The transformation of the parent into the daughter nucleus is called transmutation of the elements. Radioactive decay occurs spontaneously only when the mass of the products is less than the mass of the parent nucleus. The loss in mass appears as kinetic energy of the products. Alpha decay occurs via the purely quantum mechanical process of tunneling through a barrier.

charge,

energy, and nucleon number are conserved in all decays. Radioactive decay is a statistical process. For a given type of radioactive nucleus, the number of nuclei that decay (AN) in a time A7 is proportional to the number N of parent nuclei present: AN = —AN Az (41-4a)

N = Npe™,

N of

(41-6)

as does the activity, R = |dN/dt|:

aN| ’ _= |aN||u T

Oe

.

(41-7¢)

The half-life, 7}, is the time required for half the nuclei of a radioactive sample to decay. It is related to the decay constant by

5 2 - 0653 A

Radioactive

decay can be used to determine

(41-8) the age of

certain objects, such as once-living biological material ('¢C) or geological formations (%5U).

Particle detectors include Geiger counters, scintillators wit.. attached photomultiplier tubes, and semiconductor detectors. Detectors that can image particle tracks include semiconductors, photographic emulsions, bubble chambers, and wire drift chambers.

I Questions 1. What do different isotopes of a given element have in common? How are they different? 2. What are the elements represented by the X in the

11. Describe, in as many ways as you can, the difference between a, B, and 7 rays. 12. What element is formed by the radioactive decay of

3. How many protons and how many neutrons do each of the

13. What element is formed by the decay of (a) 2P (B8 )

4. Identify the element that has 88 nucleons and 50 neutrons. 5. Why are the atomic masses of many elements (see the Periodic Table) not close to whole numbers? 6. How do we know there is such a thing as the strong nuclear force? 7. What are the similarities and the differences between the strong nuclear force and the electric force? 8. What is the experimental evidence in favor of radioactivity being a nuclear process? 9. The isotope $4Cu is unusual in that it can decay by ¥, 87, and B emission. What is the resulting nuclide for each case?

14. Fill in the missing particle or nucleus:

following: (a) Z5X; (b) §X; (¢) 1X; (d) %X; (e) 27X?

isotopes in Question 2 have?

10. A 28U nucleus decays via a decay to a nucleus containing how many neutrons?

1126

CHAPTER 41

Nuclear Physics and Radioactivity

(a) 1{Na (8"): (b) iNa (B"): (c) 3P0 (a)? (b) 138 (B7); (¢) *}Bi (a)? (@) $Ca - 2+e+ ¥ (b) BCu*

—»

2+

v

(c) $8Cr —» %KV +? (d) BiPu > ? + a

(e) %3Np — %iPu + ?

15. Immediately after a 233U nucleus decays to 2¢Th + 3He,

the daughter thorium nucleus may still have 92 electrons circling it. Since thorium normally holds only 90 electrons, what do you suppose happens to the two extra ones? 16. When a nucleus undergoes either B~ or B decay, wha happens to the energy levels of the atomic electrons? Wha is likely to happen to these electrons following the decay?

17, The alpha particles from a given alpha-emitting nuclide are generally monoenergetic; that is, they all have the same kinetic energy. But the beta particles from a beta-emitting nuclide have a spectrum of energies. Explain the difference between these two cases. Do isotopes that undergo electron capture generally lie above or below the stable nuclides in Fig, 41-27 Can hydrogen or deuterium emit an « particle? Explain. Why are many rare in nature?

artificially produced

radioactive

isotopes

21. An isotope has a half-life of one month. After two months, will a given sample of this isotope have completely decayed? If not, how much remains? 22, Why are none of the elements with Z > 92 stable?

23, A proton strikes a §Li nucleus. As a result, an « particle and another particle are released. What is the other particle?

24, Can '4C dating be used to measure the age of stone walls and tablets of ancient civilizations? Explain.

25. In both internal conversion and B decay, an electron is emitted. How could you determine which decay process occurred? 26. Describe how the potential energy curve for an « particle in an a-emitting nucleus differs from that for a stable nucleus. 27. Explain the absence of B emitters in the radioactive decay series of Fig. 41-12. 28. As %22Rn decays into 20682 2%Pb, how many alpha and beta particles are emitted? Does it matter which path in the decay series is chosen? Why or why not?

| Problems 41-1

Nuclear Properties

L (I) A pi meson has a mass of 139 MeV/c?. What is this in atomic mass units? 2.

(I) What is the approximate radius of an alpha particle (3He)?

B (I) By what % is the radius of 233U greater than the radius

of 3U?

4. (IT) (a) What is the approximate radius of a '}3Cd nucleus? (b) Approximately what is the value of A for a nucleus whose radius is 3.7 X 107 m? . (II) What is the mass of a bare « particle (without electrons)

in MeV/c*?

r.fi.

(IT) Suppose two alpha particles were held together so they were just touching. Estimate the electrostatic repulsive force each would exert on the other. What would be the acceleration of an alpha particle subjected to this force? . (IT) (a) Show that the density of nuclear matter is essentially the same for all nuclei. (b) What would be the radius of the Earth if it had its actual mass but had the density of nuclei? (¢) What would be the radius of a 25U nucleus if it had the density of the Earth? (IT) What stable nucleus has approximately half the radius of a uranium nucleus? [Hint: Find A and use Appendix F to get Z.] (IT) If an alpha particle were released from rest near the

surface of a 134Fm nucleus, what would its kinetic energy be when far away?

10. (IT) (a) What is the fraction of the hydrogen atom’s mass that is in the nucleus? (b) What is the fraction of the hydrogen atom’s volume that is occupied by the nucleus? 11. (IT) Approximately how many nucleons are there in a 1.0-kg object? Does it matter what the object is made of? Why or why not? 12. (II) How much kinetic energy must an « particle have to just

“touch” the surface of a 35U nucleus?

41-2 Binding Energy 13. (I) Estimate the total binding energy for $3Cu, using Fig. 41-1.

@

14. (IT) Use Appendix F to calculate the binding energy of $H (deuterium). (IT) Determine the binding energy of the last neutron in a

2P nucleus.

16. (IT) Calculate

the total binding

energy, and the binding

energy per nucleon, for (a) JLi, (b) '%Au. Use Appendix F.

17. (IT) Compare

the average binding energy of a nucleon in

#INa to that in 7{Na.

18. (III) How much energy is required to remove (a) a proton,

(b) a neutron, from N?

Explain the difference in your

answers. 19. (II1) (a) Show that the nucleus

§Be (mass = 8.005305 u) is

unstable and will decay into two « particles. (b) Is ‘2C stable against decay into three a particles? Show why or why not.

41-3 to 41-7

Radioactive Decay

20. (I) How much energy is released when tritium, $H, decays by B~ emission? 21. (I) What is the maximum kinetic energy of an electron emitted in the B decay of a free neutron? 22. (I) Show that the decay '$C — 'YB + p is not possible because energy would not be conserved. . (I) The Li nucleus has an excited state 0.48 MeV above the ground state. What wavelength gamma photon is emitted when the nucleus decays from the excited state to the ground state? . (IT) Give the result of a calculation that shows whether or not the following decays are possible:

() U - %BU + n () "N (c) K

— -

BN + n; ¥K + n.

25. (I) #Na is radioactive. (¢) Is it a B~ or B* emitter? (b) Write down the decay reaction, and estimate the maximum kinetic energy of the emitted g. 26. (II) When #Ne (mass = 22.9945u) decays to 3Na (mass = 22.9898 u), what is the maximum kinetic energy of the emitted electron? What is its minimum energy? What is the energy of the neutrino in each case? Ignore recoil of the daughter nucleus. 27. (I1) A Z8U nucleus emits an « particle with kinetic energy = 420MeV. (a) What is the daughter nucleus, and (b) what is the approximate atomic mass (in u) of the daughter atom? Ignore recoil of the daughter nucleus. 28. (IT) What is the maximum Kkinetic energy of the emitted

B particle during the decay of $9Co?

Problems

1127

29. (I1) A nucleus of mass 256u, initially at rest, emits an a particle with a kinetic energy of 5.0 MeV. What is the kinetic energy of the recoiling daughter nucleus? 30. (IT) The isotope 2§§Po can decay by either @ or 8~ emission. What is the energy release in each case? The mass of

218Po is 218.008965 u. ] (IT) The nuclide 2P decays by emitting an electron whose

maximum Kkinetic energy can be 1.71 MeV. (a) What is the daughter nucleus? (b) Calculate the daughter’s atomic mass (in u).

32.

(I) A photon with a wavelength of 1.00 X 10 ¥ m is

ejected from an atom. Calculate its energy and explain why it is a ¥ ray from the nucleus or a photon from the atom. . (II) How much energy is released in electron capture by

beryllium: }Be + e~ — ILi + »?

. (IT) How much recoil energy does a {JK nucleus get when it emits a 1.46-MeV gamma ray? . (II) Determine the maximum Kkinetic energy of B* particles

released when '}C decays to '{B. What is the maximum energy the neutrino can have? What is the minimum energy of each?

. (IIT) The a particle emitted when 233U decays has 4.20 MeV

of kinetic energy. Calculate the recoil kinetic energy of the daughter nucleus and the Q-value of the decay. 37. (III) What is the energy of the a particle emitted in the

decay %JPo — 2%Pb + a? Take into account the recoil of

the daughter nucleus. . (IIT) Show that when a nucleus decays by 8" decay, the total energy released is equal to (MP

-

MD

il

-8 to 41-10

Dating

39,

Half-Life, Decay Rates, Decay Series,

(I) (a) What is the decay constant of 235U whose half-life is

4.5 X 10°yr? (b) The decay constant of a given nucleus is

3.2 X 107571, What is its half-life?

40. (I) A radioactive material produces 1280 decays per minute at one time, and 3.6 h later produces 320 decays per minute. What is its half-life?

. (I) What fraction of a sample of $§Ge, whose half-life is about 9 months, will remain after 2.0 yr?

. (I) What is the activity of a sample of '¢C that contains

(I1) '2Cs has a half-life of 30.8s. (a) If we have 7.8 ug

initially, how many Cs nuclei are present? () How many are present 2.6 min later? (c) What is the activity at this time? (d) After how much time will the activity drop to less than about 1 per second?

49. (I1) Calculate the mass of a sample

1.265 X 10%yr.

50

(IT)

What

6 half-lives?

fraction

of

a

sample

is

left

after

exactly

. (II) A sample of $9Co and a sample of '3iI both have N

atoms at ¢ = 0. How long will it take until both have the same activity? (Use Appendix F for half-life data.)

4s. (IT) How many nuclei of 2§U remain in a rock if the activity registers 340 decays per second?

. (II) In a series of decays, the nuclide 23U becomes %3Pb. How many a and B8~ particles are emitted in this series?

47.

(IT) The iodine isotope '31I is used in hospitals for diagnosis

of thyroid function. If 782 ug are ingested by a patient, determine the activity () immediately, (b) 1.00 h later when the thyroid is being tested, and (c¢) 4.0months later. Use Appendix F.

1128

CHAPTER 41

Nuclear Physics and Radioactivity

Calculate

the

activity

2P (1 = 1.23 x 10%s).

of a pure

8.7-ug

sample

of

2

)B

52.

(IT) The activity of a sample of 338 (7, = 8732 days) is

3.65 X 10* decays sample?

(I)

A

sample

550 X 10"® nuclei.

per second. What

of %3U (a)

is the mass

(7, = 1.59 X 105yr)

What

is

the

decay

of the

contains

constant?

(b) Approximately how many disintegrations will occur per minute? 53. (IT) The activity of a sample drops by a factor of 4.0 in 8.6 minutes. What is its half-life? 54.

(IT) A 385-g sample of pure carbon contains 1.3 parts in 10'2

(atoms) second?

of

%C.

How

many

disintegrations

occur

per

55. (I1) A sample of /U is decaying at a rate 3.70 X 107 decays/s. What is the mass of the sample?

of

56. (IT) Rubidium-strontium dating. The rubidium isotope §Rb,

a B emitter with a half-life of 4.75 x 100yr, is used to determine the age of rocks and fossils. Rocks

containing

fossils of ancient animals contain a ratio of §Sr to §/Rb of

0.0260. Assuming that there was no §}Sr present when themy rocks were formed, estimate the age of these fossils. 57

(IT) The activity of a radioactive source decreases by 2.5% in 31.0 hours. What is the half-life of this source?

58. (IT) 7Be decays with a half-life of about 53 d. It is produced in the upper atmosphere, and filters down onto the Earth’s surface. If a plant leaf is detected to have 350 decays/s of IBe, (a) how long do we have to wait for the decay rate to drop to 15per second? (b) Estimate the initial mass of

1Be on the leaf.

59. (I)

Two

of

the

naturally

occurring

radioactive

decay

sequences start with 233Th and with 233U. The first five decays of these two sequences are: a,B,B,aa

8.1 X 10%° nuclei?

. (I)

of pure {0K with an

initial decay rate of 2.0 X 10°s™!. The half-life of {JK is

zme)czi

where Mp and Mp are the masses of the parent and daughter atoms (neutral), and m, is the mass of an electron or positron. 41

48.

and a, B, a, B, a.

Determine the resulting intermediate daughter nuclei in each case. 60. (IT) An ancient wooden club is found that contains 85 g of carbon and has an activity of 7.0 decays per second. Determine its age assuming that in living trees the ratio of 14C/12C atoms is about 1.3 X 10712, 61. (IIT) At ¢ = 0, a pure sample of radioactive nuclei contains N, nuclei whose decay constant is A. Determine a formula for the number of daughter nuclei, Np, as a function o time; assume the daughter is stable and that Np =0 a t=0.

| General Problems «

62. Which radioactive isotope of lead is being produced if the measured activity of a sample drops to 1.050% of its original activity in 4.00 h? 63. An old wooden tool is found to contain only 6.0%

74. (a) Calculate the kinetic energy of the a particle emitted

when 28U decays. (b) Use Eq. 41-1 to estimate the radius of an a particle and a %3Th nucleus. Use this to estimate (c) the maximum height of the Coulomb barrier, and (d) its width

AB in Fig. 41-7.

of the '3C that an equal mass of fresh wood would. How old

is the tool? . A neutron star consists of neutrons at approximately nuclear density. Estimate, for a 10-km-diameter neutron star, (a) its mass number, (b) its mass (kg), and (c) the acceleration of gravity at its surface. 65. Tritium dating. The 7H isotope of hydrogen, which is called tritium (because it contains three nucleons), has a half-life of 123 yr. It can be used to measure the age of objects up to about 100 yr. It is produced in the upper atmosphere by cosmic rays and brought to Earth by rain. As an application, determine approximately the age of a bottle of wine whose

1H radiation is about 1 that present in new wine.

66. Some elementary particle theories (Section 43-11) suggest

that the proton may be unstable, with a half-life = 10* yr. How long would you expect to wait for one proton in your body to decay (approximate your body as all water)?

67. Show,

using

the decays

given

neutrino has either spin § or 3-

in Section

41-5, that the

68. The original experiments which established that an atom has a heavy, positive nucleus were done by shooting alpha particles through gold foil. The alpha particles used had a kinetic energy of 7.7 MeV. What is the closest they could get to a gold nucleus? How does this compare with the size of the nucleus? 69. How long must you wait (in half-lives) for a radioactive sample to drop to 1.00% of its original activity? . If the potassium isotope {9K gives 45 decays/s in a liter of

75.

The nuclide '%Os decays with B~ energy of 0.14 MeV

accompanied by ¥ rays of energy 0.042 MeV and 0.129 MeV. (a) What is the daughter nucleus? (b) Draw an energy-level diagram showing the ground states of the parent and daughter and excited states of the daughter. (¢) To which of

the daughter states does B~ decay of 19Os occur?

76.

Determine the activities of (a) 1.0g of '3 (71 = 8.02 days) and (b) 1.0g of %3U (7 = 447 x 10%yr). ~

7

Use Fig. 41-1 to estimate the total binding energy for copper and then estimate the energy, in joules, needed to break a 3.0-g copper penny into its constituent nucleons.

78. Instead of giving atomic masses for nuclides as in Appendix F, some Tables give the mass excess, A, defined as

A = M — A, where A is the atomic mass number and M is the mass in 1. Determine the mass excess, in u and in MeV/c2,

for: (a) $He; (b) 2C; (c) £Sr; (d) 33U. (e) From a glance at Appendix F, can you make a generalization about the sign of A as a function of Z or A?

79. When water is placed near an intense neutron source, the neutrons can be slowed down by collisions with the water molecules and eventually captured by a hydrogen nucleus to form the stable isotope called deuterium, 7H, giving off a gamma ray. What is the energy of the gamma ray? 80. (a) Show that defined as

milk, estimate how much {JK and regular {9K are in a liter

s

71. (a) In « decay of, say, a 2§Ra nucleus, show that the nucleus

carries away a fraction 1/(1 + jAp) of the total energy

a particle when 38Ra decays?

72. Strontium-90 is produced as a nuclear fission product of uranium in both reactors and atomic bombs. Look at its location in the Periodic Table to see what other elements it might be similar to chemically, and tell why you think it might be dangerous to ingest. It has too many neutrons, and it decays with a half-life of about 29 yr. How long will we have to wait for the amount of 33Sr on the Earth’s surface to reach 1% of its current level, assuming no new material is scattered about? Write down the decay reaction, including the daughter nucleus. The daughter is radioactive: write down its decay. 73. Using the uncertainty principle and the radius of a nucleus, estimate the minimum possible kinetic energy of a nucleon in, say, iron. Ignore relativistic corrections. [Hint: A particle can have a momentum at least as large as its momentum uncertainty. ]

mean

J

of milk. Use Appendix F.

available, where Ap is the mass number of the daughter nucleus. [Hint: Use conservation of momentum as well as conservation of energy] (b) Approximately what percentage of the energy available is thus carried off by the

the

life

of a radioactive

nuclide,

t N(t) dt 0

[

N(t) dt 0

is

7 =1/A.

(b) What

fraction of the original number

nuclei remains after one mean life?

of

81. (a) A 72-gram sample of natural carbon contains the usual fraction of '¢C. Estimate how long it will take before there

is only one '¢C nucleus left. (b) How does the answer in (a) change if the sample is 270 grams? What does this tell you about the limits of carbon dating?

82. If the mass of the proton were just a little closer to the mass of the neutron, the following reaction would be possible even at low collision energies: e

+p—>n+u

(a) Why would this situation be catastrophic? (b) By what percentage would the proton’s mass have to be increased to make this reaction possible? 83. What is the ratio of the kinetic energies for an alpha particle and a beta particle if both make tracks with the same radius of curvature in a magnetic field, oriented perpendicular to the paths of the particles?

General Problems

1129

84. A 1.00-g sample of natural samarium emits a particles at a

rate of 120s™! due to the presence of '¢}Sm. The natural abundance of '¢JSm is 15%. Calculate the half-life for this decay process.

85. Almost all of naturally occurring uranium is %3§U with a half-life of 4.468 X 10°yr.

Most

of the rest of natural

uranium is 23U with a half-life of 7.04 X 10%yr. Today a sample contains 0.720% 23U. (a) What was this percentage

1.0 billion years ago? (b) What percentage of 23U will

remain 100 million years from now? 86. A typical banana contains 400 mg of potassium, of which a small fraction is the radioactive isotope fJK (see Appendix F). Estimate the activity of an average banana

due to {IK.

87. Some radioactive isotopes have half-lives that are larger than the age of the universe (like gadolinium or samarium). The only way to determine these half-lives is to monitor the decay rate of a sample that contains these isotopes. For example, suppose we find an asteroid that currently contains

about 15,000kg of 33Gd (gadolinium) and we detect an

*Numerical/Computer *89. (I) A laboratory has a 1.80-ug sample of radioactive 3N

whose decay constant A = 1.16 X 1073s!,

of nuclei, N,, present in the sample. U

the radioactive

decay

the

form

4n,

4n + 1,4n

+ 2,

or

to Exercises

A: 0.042130 u.

E: (a).

B: 7.98 MeV/nucleon.

F: 137 X 10711571, G: (a).

D: (¢).

1130

CHAPTER 41

Nuclear Physics and Radioactivity

tl

than }Be).] To reduce the amount of data, for A = 20 plot only points for even values of A, and plot to a maximum of A =142,

4n + 3,

C: (b).

to determine

stable than {H), He, §Li, and Li (since it is more stable

where n is an integer. Justify this statement and show that for a nuclide in any family, all its daughters will be in the same family.

Answers

N = Nye™,

*90. (IT) Construct a spreadsheet (or other numerical tool) that will reproduce Fig. 41-1, the graph of binding energy per nucleon (in MeV) vs. the mass number A. Using Appendix F, calculate the binding energy per nucleon for the most stable isotope of each possible mass number A =2, [The first few values will be for H, 3He (it is more

88. Decay series, such as that shown in Fig. 41-12, can be classified into four families, depending on whether the mass have

law,

number of nuclei N present at time ¢ for =0 to 30 minutes (1800s) in steps of 0.5 min (30s). Make a graph of N versus ¢ and from the graph determine the half-life of the sample.

activity of 1 decay/s. What is the half-life of gadolinium (in years)?

numbers

Calculate the

initial number

Diagram of ITER (International Thermonuclear Experimental Reactor), which will hopefully begin operation in the 2020s. Inside its cavity, over 12m in diameter, a plasma of electrons and light nuclei will be heated to high temperatures that rival the Sun. Confining a plasma by magnetic fields has proved

difficult, and intense research is needed

if

the fusion of smaller nuclei is to fulfill its promise as a source of abundant and relatively clean power. This Chapter covers the basic physics topics of nuclear reactions, nuclear fission, nuclear fusion, and

how we obtain nuclear energy. We also examine the health aspects of radiation dosimetry, therapy, and imaging by CAT, PET, SPET, and MRIL

:

e

4~}

“Nuclear Energy; Effects

y b B

®

4

and Uses of Radiation

1. The Sun is powered by

42-1

(a) nuclear alpha decay.

2

Nuclear Reactions and the

Transmutation of Elements

(b) nuclear beta decay.

42-2

Cross Section

(¢) nuclear gamma decay. (d) nuclear fission.

42-3

Nuclear Fission; Nuclear Reactors

(e) nuclear fusion.

2. Which radiation induces the most biological damage for a given amount energy deposited in tissue?

42-4 of

(a) Alpha particles.

(b) Gamma radiation. (¢) Beta radiation.

(d) They all do the same damage for the same deposited energy. (e) It depends on the type of tissue. e continue our study of nuclear physics in this Chapter. We begin with a discussion of nuclear reactions, and then we examine the important large energy-releasing processes of fission and fusion. This

Chapter also deals with the effects of nuclear radiation passing

rlrough matter, particularly biological matter, and how radiation is used medically for therapy, diagnosis, and imaging techniques.

42-5 42-6

Nuclear Fusion

Passage of Radiation Through Matter; Radiation Damage Measurement of

Radiation—Dosimetry

#42-7

Radiation Therapy

#42-8

Tracers in Research and Medicine

#42-9

Imaging by Tomography: CAT Scans and Emission

Tomography #42-10 Nuclear Magnetic Resonance (NMR);

Magnetic Resonance Imaging (MRI)

1131

47—1

Nuclear Reactions and the

Transmutation

of Elements

When a nucleus undergoes a or B decay, the daughter nucleus is a differenfl element from the parent. The transformation of one element into another, called transmutation, also occurs by means of nuclear reactions. A nuclear reaction is said to occur when a given nucleus is struck by another nucleus, or by a simpler particle such as a ¥ ray or neutron, and an interaction takes place. Ernest Rutherford was the first to report seeing a nuclear reaction. In 1919 he observed that some of the a particles passing through nitrogen gas were absorbed and protons emitted. He concluded that nitrogen nuclei had been transformed into oxygen nuclei via the reaction

‘He + N > 0 + IH, where 3He is an « particle, and {H is a proton.

Since then, a great many nuclear reactions have been observed. Indeed, many

of the radioactive isotopes used in the laboratory are made by means of nuclear reactions. Nuclear reactions can be made to occur in the laboratory, but they also occur regularly in nature. In Chapter 41 we saw an example: '¥{C is continually

being made in the atmosphere via the reaction n + YN — %¥C + p.

Nuclear reactions are sometimes written in a shortened form: for example, the reaction

n+ YN > %C +p can be written

YN (n,p) '§C. The symbols outside the parentheses on the left and right represent the initial and final nuclei, respectively. The symbols inside the parentheses represent th=my bombarding particle (first) and the emitted small particle (second). In any nuclear reaction, both electric charge and nucleon number are conserved. These conservation laws are often useful, as the following Example shows. CONCEPTUAL

EXAMPLE 42-1

| Deuterium reaction.

A neutron is observed to

strike an 'O nucleus, and a deuteron is given off. (A deuteron, or deuterium, is the isotope of hydrogen containing one proton and one neutron, 7H; it is sometimes given the symbol d or D.) What is the nucleus that results?

RESPONSE We have the reaction

n + 'O —

? + ?H. The total number of

nucleons initially is 1 + 16 = 17, and the total charge is 0 + 8 = 8. The same totals apply after the reaction. Hence the product nucleus must have Z = 7 and A = 15.

From the Periodic Table, we find that it is nitrogen that has

the nucleus produced is 3N,

| EXERCISE A Determine the resulting nucleus in the reaction n + '¥Ba —

Z = 7, so

? + 7.

Energy and momentum are also conserved in nuclear reactions, and can be used to determine whether or not a given reaction can occur. For example, if the

total mass of the final products is less than the total mass of the initial particles,

this decrease in mass (recall AE = Am c?) is converted to kinetic energy (K) of the outgoing particles. But if the total mass of the products is greater than the total mass of the initial reactants, the reaction requires energy. The reaction will then not occur unless the bombarding particle has sufficient kinetic energy. Consider a nuclear reaction of the general form a+X—>Y+0b,

(42-1",

where particle “a” is a moving projectile (or small nucleus) that strikes nucleus X, producing

1132

CHAPTER 42

Nuclear Energy; Effects and Uses of Radiation

nucleus Y and particle b (typically, p, n, @, V). We or Q-value, in terms of the masses involved, as Q

=

(M, + Mx

define the reaction energy,

— My, — My)c%

(42-2a)

r)ra“/ray, M = 0. Because energy is conserved, Q has to be equal to the change in kinetic energy (final minus initial):

Q0 = Ky + Ky — K, — Kx.

(42-2b)

If X is a target nucleus at rest (or nearly so) struck by incoming particle a, then Kx = 0. For O > 0, the reaction is said to be exothermic or exoergic; energy is released in the reaction, so the total kinetic energy is greater after the reaction than before. If Q is negative (Q < 0), the reaction is said to be endothermic or endoergic: the final total kinetic energy is less than the initial kinetic energy, and an energy input is required to make the reaction happen. The energy input comes from the kinetic energy of the initial colliding particles (a and X).

m

A slow-neutron reaction. The nuclear reaction n + 9B — JLi + jHe

is observed to occur even when very slow-moving neutrons (mass M, = 1.0087 u) strike boron atoms at rest. For a particular reaction in which K; ~ 0, the outgoing helium (M, = 4.0026 u) is observed to have a speed of 9.30 X 10°m/s. Determine (a) the kinetic energy of the lithium (M;; = 7.0160u), and (b) the Q-value of the reaction. APPROACH

Since the neutron and boron are both essentially at rest, the total

momentum before the reaction is zero; momentum zero afterward as well. Thus,

A'

Myv

=

is conserved and so must be

Myevye.

We solve this for v;; and substitute it into the equation for kinetic energy. In (b) we use Eq. 42-2b. SOLUTION (a) We can use classical kinetic energy with little error, rather than relativistic formulas, because vy, = 9.30 X 10°m/s is not close to the speed of light ¢, and v; will be even less since M;; > My, . Thus we can write: 1

1

Kii = 5 Myl

MHchc

.

= “Mu Me Viie 2M;;

We put in numbers, changing 1.60 X 107*J = 1 MeV: Li =

=

; fince,

=

mass

in

u

to

kg

and

recall

that

(4.0026 u)*(1.66 x 1072 kg/u)*(9.30 X 10° m/s)’ 2(7.0160 u)(1.66 x 1072 kg/u)

1.64

X 10°%]

(b) We are given the data Kye

the

=

1.02MeV.

K, = Kx = 0 in Eq.42-2b,s0

Q = K;; + Ky,

where

%Mfle U%—le

= 1(4.0026 u)(1.66 X 1072 kg/u)(9.30 X 10°m/s)’ = 2.87 x 10¥] = 1.80 MeV. O = 1.02MeV

+ 1.80 MeV

= 2.82 MeV.

SECTION 42-1

Nuclear Reactions and the Transmutation of Elements

1133

Will the reaction “go"?

Can the reaction

p+R3C > 5N+n occur when 3C is bombarded by 2.0-MeV protons?

-

APPROACH The reaction will “go” if the reaction is exothermic (Q > 0) and even if @ < 0 if the input momentum and kinetic energy are sufficient. First we calculate Q from the difference between final and initial masses using Eq. 42-2a, and look up the masses in Appendix F. SOLUTION

The total masses before and after the reaction are: Before

M(3C) = 13.003355 M(IH) = 1.007825 14.011180

After

M(N) = 13.005739 M(n) = 1.008665 14.014404

(We must use the mass of the |H atom rather than that of the bare proton

because the masses of 12C and 3N include the electrons, and we must include an

equal number of electron masses on each side of the equation since none are created or destroyed.) The products have an excess mass of (14.014404 — 14.011180)u

=

0.003224u X 931.5MeV/u

=

3.00 MeV.

Thus Q = —3.00MeV, and the reaction is endothermic. This reaction requires energy, and the 2.0-MeV protons do not have enough to make it go. NOTE The incoming proton in this Example would have to have somewhat more than 3.00 MeV of kinetic energy to make this reaction go; 3.00 MeV would be enough to conserve energy, but a proton of this energy would produce the 3N and n with no kinetic energy and hence no momentum. Since an incident 3.0-MeV proton has momentum, conservation of momentum would be violated. A calculati using conservation of energy and of momentum, as we did in Examples 41-6 an. 42-2, shows that the minimum proton energy, called the threshold energy, is 3.23MeV in this case (= Problem 16).

Neutron Physics Neutron captured by 23§ U.

(b)

@+@+Q+Q 239U decays by B decay to neptunium-239.

(c)

@—’@‘*e‘*@

233Np itself decays by

B decay to produce plutonium-239. FIGURE 42-1 Neptunium and

plutonium are produced in this

series of reactions, after bombardment of Z§U by neutrons.

The artificial transmutation of elements took a great leap forward in the 1930s when Enrico Fermi realized that neutrons would be the most effective projectiles for causing nuclear reactions and in particular for producing new elements. Because neutrons have no net electric charge, they are not repelled by positively charged nuclei as are protons or alpha particles. Hence the probability of a neutron reaching the nucleus and causing a reaction is much greater than for charged projectiles,” particularly at low energies. Between 1934 and 1936, Fermi and his co-workers in Rome produced many previously unknown isotopes by bombarding different elements with neutrons. Fermi realized that if the heaviest known element, uranium, is bombarded with neutrons, it might be possible to produce new elements with atomic numbers greater than that of uranium. After

several years of hard work, it was suspected that two new elements had been

produced, neptunium (Z = 93) and plutonium (Z = 94). The full confirmation that such “transuranic” elements could be produced came several years later at the University of California, Berkeley. The reactions are shown in Fig. 42-1. It was soon shown that what Fermi had actually observed when he bombarded

uranium was an even stranger process—one that was destined to play an extraordinary role in the world at large. We discuss it in Section 42-3. "That is, positively charged particles. Electrons rarely cause nuclear reactions because they do interact via the strong nuclear force.

1134

CHAPTER 42

Nuclear Energy; Effects and Uses of Radiation

“>

49~

Cross Section

Some reactions have a higher probability of occurring than others. The reaction f)robability is specified by a quantity called the cross section. Although the size sf a nucleus, like that of an atom, is not a clearly defined quantity since the

5

edges are not distinct like those of a tennis ball or baseball, we can nonetheless

—

define a cross section for nuclei undergoing collisions by using an analogy. Suppose that projectile particles strike a stationary target of total area A and thickness £, as shown in Fig. 42-2. Assume also that the target is made up of identical objects (such as marbles or nuclei), each of which has a cross-sectional area o, and we assume the incoming projectiles are small by comparison. We assume that the target objects are fairly far apart and the thickness £ is so small that we don’t have to worry about overlapping, This is often a reasonable assumption because nuclei have diameters

Projectiles —

are n nuclei per unit volume, the total cross-sectional area of all these tiny targetsis

=~ FIGURE 42-2

since

=

=

is the total number of targets and o is the cross-

RO.»_:_

_

o

A

o)

Y

ke

Projectile particles

stltike a target of area A and . thickness £ made up of n nuclei per

sectional area of each. If A’ H+e

(0.42 MeV)

(42-7a)

- 3He + 7

(5.49MeV)

(42-7b)

3He + 3He — 3He + |H + IH

(12.86 MeV)

(42-7c¢)

H+?2H

+v

where the energy released (Q-value) for each reaction is given in parentheses, keepiny, track of atomic electrons as for B-decay (Section 41-5). The net effect of this sequence, 1142

CHAPTER 42

Nuclear Energy; Effects and Uses of Radiation

which is called the proton—proton chain, is for four protons to combine to form one 4He nucleus plus two positrons, two neutrinos, and two gamma rays:

4'H — $He + 2" + 2v + 27.

(42-8)

r Tote that it takes two of each of the first two reactions

(Eqgs. 42-7a

and b) to

produce the two 3He for the third reaction. Also, each of the two e* formed (Eq. 42-7a) quickly annihilates with an electron to produce 27 rays (Section 37-5) with total energy 2m.c? = 1.02MeV. So the total energy release for the net

reaction, Eq. 42-8, is 2(0.42MeV)

+ 2(1.02MeV)

+ 2(5.49MeV)

+ 12.86 MeV

= 26.7 MeV.

The first reaction, the formation of deuterium from two protons (Eq. 42-7a), has a very low probability and so limits the rate at which the Sun produces energy. (Thank goodness; this is why the Sun and other stars have long lifetimes and are still shining brightly.) EXERCISE C Return to the first Chapter-Opening Question, page 1131, and answer it again now. Try to explain why you may have answered it differently the first time. EXERCISE D If the Sun is generating a constant amount of energy via fusion, the mass of the Sun must be (a) increasing, (b) decreasing, (c) constant, (d) irregular.

GV IZR YRS M ESTIMATE | Estimating fusion energy. Estimate the energy

released if the following reaction occurred:

‘H + ?H — jHe. APPROACH We use Fig. 42-13 for a quick estimate. SOLUTION We see in Fig. 42-13 that each 7H has a binding energy of about

14MeV/nucleon, which for 2 nuclei of mass 2is 4 X (13) ~ 5MeV. The {He has

a binding energy per nucleon of about 7MeV for a total of 4 X 7MeV Hence the energy release is about 28 MeV — 5MeV = 23 MeV.

= 28 MeV.

In stars hotter than the Sun, it is more likely that the energy output comes principally from the carbon (or CNO) cycle, which comprises the following sequence of reactions:

C+H->HN+Y BN

> BC+e"+v

B0

> BN +e"

RC+1H > YN +7 UN +H - 50 + 7

BN + IH - %C + {He.

+v

No carbon is consumed in this cycle and the net effect is the same as the proton—proton chain, Eq. 42-8 (plus one extra V). The theory of the proton—proton chain and of the carbon cycle as the source of energy for the Sun and stars was first worked out by Hans Bethe (1906-2005) in 1939.

CONCEPTUAL EXAMPLE 42-9 | Stellar fusion. What is the heaviest element likely to be produced in fusion processes in stars?

RESPONSE Fusion is possible if the final products have more binding energy

(less mass) than the reactants, because then there is a net release of energy. Since the binding energy curve in Fig. 42-13 (or Fig. 41-1) peaks near A ~ 56 to 58 which corresponds to iron or nickel, it would not be energetically favorable to Aproduce elements heavier than that. Nevertheless, in the center of massive stars or in supernova explosions, there is enough initial kinetic energy available to Iirive endothermic reactions that produce heavier elements, as well. SECTION 42-4

Nuclear Fusion

1143

Possible Fusion Reactors @

PHYSICS APPLIED Fusion energy reactors

The possibility of utilizing the energy released in fusion to make a power reactor is very attractive. The fusion reactions most likely to succeed in a reactor involve the isotopes of hydrogen, H (deuterium) and $H (tritium), and are as follows, with th&\ energy released given in parentheses:

(4.03MeV) (327MeV)

(42-9a) (42-9b)

(17.59 MeV)

(42-9¢)

iH+iH —> {H + {H iH + {H — 3He + n ’H + H

— 3$He + n.

Comparing these energy yields with that for the fission of 33U, we can see that the

energy released in fusion reactions can be greater for a given mass of fuel than in fission. Furthermore, as fuel, a fusion reactor could use deuterium, which is very plentiful in the water of the oceans (the natural abundance of H is 0.0115% on average, or about 1 g of deuterium per 80 L of water). The simple proton—proton reaction of Eq. 42-7a, which could use a much more plentiful source of fuel, {H, has such a small probability of occurring that it cannot be considered a possibility on Earth. Although a useful fusion reactor has not yet been achieved, considerable progress has been made in overcoming the inherent difficulties. The problems are associated with the fact that all nuclei have a positive charge and repel each other. However, if they can be brought close enough together so that the short-range attractive strong nuclear force can come into play, it can pull the nuclei together and fusion will occur. For the nuclei to get close enough together, they must have large kinetic energy to overcome the electric repulsion. High kinetic energies are easily attainable with particle accelerators (Chapter 43), but the number of particles involved is too small. To produce realistic amounts of energy, we must deal with matter in bulk, for which high kinetic energy means higher temperatures. Indeed, very high temperatures are required for fusion to occur, and fusion devices are often referred to as thermonuclear devices. The interiors of the Sun and other stars are very hot, many millions of degrees, so the nuclei are moving fast enoug}‘. for fusion to take place, and the energy released keeps the temperature high so that further fusion reactions can occur. The Sun and the stars represent huge self-sustaining thermonuclear reactors that stay together because of their great gravitational mass; but on Earth, containment of the fast-moving nuclei at the high temperatures and densities required has proven difficult. It was realized after World War II that the temperature produced within a fission (or “atomic”) bomb was close to 10® K. This suggested that a fission bomb could be used to ignite a fusion bomb (popularly known as a thermonuclear or hydrogen bomb) to release the vast energy of fusion. The uncontrollable release of fusion energy in an H-bomb (in 1952) was relatively easy to obtain. But to realize usable energy from fusion at a slow and controlled rate has turned out to be a serious challenge.

DOV

YRR

ESTIMATE | Temperature needed for d-t fusion. Estimate

the temperature required for deuterium-tritium fusion (d-t) to occur.

APPROACH We assume the nuclei approach head-on, each with kinetic energy K, and that the nuclear force comes into play when the distance between their centers equals the sum of their nuclear radii. The electrostatic potential energy (Chapter 23) of the two particles at this distance equals the minimum total kinetic energy of the two particles when far apart. The average kinetic energy is related to Kelvin temperature by Eq. 18-4. SOLUTION The radii of the two nuclei (A4 =2 and A, = 3) are given by

Eq. 41-1:

rg = 1.5fm, r, ~ 1.7fm, so rq + r, = 3.2 X 107"m. We equate the

kinetic energy of the two initial particles to the potential energy when at this distance: 2K

=

~ 1144

CHAPTER 42

1 47T€0

(

e? (rd

+

9.0 x 10°

rt)

N-m’

(1.6 X 10°C)?

Cct ) (32 X 107°¥ m)(1.6 X 107°J/eV)

~

0.45MeV.

Thus, K ~ 0.22MeV, and if we ask that the average kinetic energy be this high,

“

then from Eq. 18-4, 3kT = K, we have a temperature of

" r

28

C 3k

2(022MeV)(1.6 X 10 3J/MeV)

3(1.38 X 102 J/K)

~

2 X 10°K.

NOTE More careful calculations show that the temperature required for fusion is actually about an order of magnitude less than this rough estimate, partly because it is not necessary that the average kinetic energy be 0.22 MeV—a small percentage of nuclei with this much energy (in the high-energy tail of the Maxwell distribution, Fig. 18-3) would be sufficient. Reasonable estimates for a usable fusion reactor are in the range 7 = 1 to 4 x 10K. It is not only a high temperature that is required for a fusion reactor. But there must also be a high density of nuclei to ensure a sufficiently high collision rate. A real difficulty with controlled fusion is to contain nuclei long enough and at a high enough density for sufficient reactions to occur that a usable amount of energy is obtained. At the temperatures needed for fusion, the atoms are ionized, and the resulting collection of nuclei and electrons is referred to as a plasma. Ordinary materials vaporize at a few thousand degrees at most, and hence cannot be used to contain a high-temperature plasma. Two major containment techniques are magnetic confinement and inertial confinement. In magnetic confinement, magnetic fields are used to try to contain the hot plasma. A simple approach is the “magnetic bottle” shown in Fig. 42-14. The paths of the charged particles in the plasma are bent by the magnetic field; where magnetic field lines are close together, the force on the particles reflects them back toward the center. Unfortunately, magnetic bottles develop “leaks” and the charged particles slip out before sufficient fusion takes place. The most promising design today is the tokamak, first developed in Russia. A tokamak (Fig. 42-15)

is toroid-shaped

and

involves

complicated

magnetic

fields: current-carrying

conductors produce a magnetic field directed along the r(“toroidal” field); an additional field is produced by currents tself (“poloidal” field). The combination produces a helical Fig. 42-15, confining the plasma, at least briefly, so it doesn’t chamber’s metal walls.

axis of the toroid within the plasma field as shown in touch the vacuum

Current-carrving

wires & \

Magnetic field lines riGURE 42-14

Plasma

“Magnetic bottle” used to confine a plasma.

Toroidal vacuum chamber

FIGURE 42-15 Tokamak configuration, showing the total B field due to external Btoroidal

current plus current in the plasma itself.

Bpoloidal

External current

Plasma

Plasma

current

In 1957, J. D. Lawson showed that the product of ion density n and confinement time 7 must exceed a minimum value of approximately nr = 3 X 10%s/m’ This Lawson criterion must be reached to produce ignition, meaning fusion that continues after all external heating is turned off. Practically, it is expected to be

achieved with

n ~ 1 to 3 X 10°m™

and 7 ~ 1-3s. To reach break-even, the

point at which the energy output due to fusion is equal to the energy input to heat the plasma, requires an nr about an order of magnitude less. The break-even point r vas very closely approached in the 1990s at the Tokamak Fusion Test Reactor (TFTR) at Princeton, and the very high temperature needed for ignition (4 X 10°K) was exceeded—although not both of these at the same time. SECTION 42-4

Nuclear Fusion

1145

Magnetic confinement fusion research continues throughout the world. This research will help us in developing the huge multinational test device (European Union, India, Japan, South Korea, Russia, China, and the U.S.), called ITER (International Thermonuclear Experimental Reactor), situated in France. It is expected that ITER

will produce temperatures above 108K for extended periods (minutes or hours) an“™

to begin running by the 2020s, with an expected power output of about 500 MW, 10 times the input energy. ITER (see Chapter-Opening Photograph on page 1131) is planned to be the final research step before building a commercial reactor. The second method for containing the fuel for fusion is inertial confinement fusion (ICF): a small pellet or capsule of deuterium and tritium is struck simultaneously from hundreds of directions by very intense laser beams. The intense influx of energy heats and ionizes the pellet into a plasma, compressing it and heating it to temperatures at which fusion can occur (> 10%K). The confinement

time is on the order of 107" to 10~%s, during which time the ions do not move appreciably because of their own inertia, and fusion can take place.

42-5

Passage of Radiation Through

Matter; Radiation Damage

When we speak of radiation, we include «, B, 7, and X-rays, as well as protons, neutrons, and other particles such as pions (see Chapter 43). Because charged particles can ionize the atoms or molecules of any material they pass through, they are referred to as ionizing radiation. And because radiation produces ionization, it can cause considerable damage to materials, particularly to biological tissue. Charged particles, such as @ and B rays and protons, cause ionization because of electric forces. That is, when they pass through a material, they can attract or repel electrons strongly enough to remove them from the atoms of the material. Since the @ and B rays emitted by radioactive substances have energies on t order of 1MeV (10* to 107 eV), whereas ionization of atoms and moleculés requires on the order of 10eV, it is clear that a single @ or B particle can cause thousands of ionizations. Neutral particles also give rise to ionization when they pass through materials. For example, X-ray and Y-ray photons can ionize atoms by knocking out electrons by means of the photoelectric and Compton effects (Chapter 37). Furthermore, if a Y ray has sufficient energy (greater than 1.02MeV), it can undergo pair production: an electron and a positron are produced (Section 37-5). The charged particles produced in all of these processes can themselves go on to produce further ionization. Neutrons, on the other hand, interact with matter mainly by collisions with nuclei, with which they interact strongly. Often the nucleus is broken apart by such a collision, altering the molecule of which it was a part. The fragments produced can in turn cause ionization. Radiation passing through matter can do considerable damage. Metals and other structural materials become brittle and their strength can be weakened if the radiation is very intense, as in nuclear reactor power plants and for space vehicles that must pass through areas of intense cosmic radiation.

*Biological Damage PHYSICS

APPLIED

The

radiation

damage

produced

in biological

organisms

is due

primarily

to

Biological radiation damage — jonization produced in cells. Several related processes can occur. Ions or radicals

are produced that are highly reactive and take part in chemical reactions that interfere with the normal operation of the cell. All forms of radiation can ionize atoms by knocking out electrons. If these are bonding electrons, the molecule may break apart, or its structure may be altered so it does not perform its normal function or may perform a harmful function. In the case of proteins, the loss of one molecule

not serious if there are other copies of it in the cell and additional copies can be

1146

CHAPTER 42

Nuclear Energy; Effects and Uses of Radiation

made from the gene that codes for it. However, large doses of radiation may damage so many molecules that new copies cannot be made quickly enough, and the cell dies. Damage to the DNA is more serious, since a cell may have only one copy. Each alteration in the DNA can affect a gene and alter the molecule it codes for ection 40-3), so that needed proteins or other materials may not be made at all. Again the cell may die. The death of a single cell is not normally a problem, since the fi)dy can replace it with a new one. (There are exceptions, such as neurons, which are “hostly not replaceable, so their loss is serious.) But if many cells die, the organism may not be able to recover. On the other hand, a cell may survive but be defective. It may go on dividing and produce many more defective cells, to the detriment of the whole organism. Thus radiation can cause cancer—the rapid uncontrolled production of cells. The possible damage done by the medical use of X-rays and other radiation must be balanced against the medical benefits and prolongation of life as a result of their use.

42—-6

Measurement of Radiation— Dosimetry

Although the passage of ionizing radiation through the human body can cause considerable damage, radiation can also be used to treat certain diseases, particularly cancer, often by using very narrow beams directed at a cancerous tumor in order to destroy it (Section 42-7). It is therefore important to be able to quantify the amount, or dose, of radiation. This is the subject of dosimetry. The strength of a source can be specified at a given time by stating the source activity: how many nuclear decays (or disintegrations) occur per second. The traditional unit is the curie (Ci), defined as

1Ci

@ PHYSICS Dosimetry

APPLIED

= 3.70 x 10" decays per second.

rkThis number comes from the original definition as the activity of exactly one ram of radium.) Although the curie is still in common use, the SI unit for source activity is the becquerel (Bq), defined as 1Bq

=

1decay/s.

Commercial suppliers of radionuclides (radioactive nuclides) specify the activity at a given time. Since the activity decreases over time, more so for short-lived

isotopes, it is important to take this into account.

The magnitude of the source activity |dN/dt| is related to the number radioactive nuclei present, N, and to the half-life, T3, by (see Section 41-8): dN dN dt

-t

of

0.693 5,

Radioactivity taken up by cells. In a certain experiment, 0.016 uCi of 73P is injected into a medium containing a culture of bacteria. After

1.0h the cells are washed and a 70% efficient detector (counts 70% of emitted B rays) records 720 counts per minute from the cells. What percentage of the orig-

inal 3P was taken up by the cells?

APPROACH The half-life of {2P is about 14 days (Appendix F), so we can ignore any loss of activity over 1hour. From the given activity, we find how many g rays are emitted. We can compare 70% of this to the (720/min)/(60s/min) = 12 per second detected.

SOLUTION

The

total

number

of

decays

per

second

originally

was

of this, or 410 per second. Since it counted 720/60 = 12 per second, 1L2/410 = 0.029 or 2.9% was incorporated into the cells.

then

A(0.0w X 107°)(3.7 x 10'°) = 590. The counter could be expected to count 70%

SECTION 42-6

Measurement of Radiation—Dosimetry

1147

Another type of measurement is the exposure or absorbed dose—that is, the effect the radiation has on the absorbing material. The earliest unit of dosage was the roentgen (R), defined in terms of the amount of ionization produced by the radiation (1R = 1.6 X 10" ion pairs per gram of dry air at standard conditions). Today, 1R is defined as the amount of X or ¥ radiation that deposit™™ 0.878 X 107%J of energy per kilogram of air. The roentgen was largely superseded- ., by another unit of absorbed dose applicable to any type of radiation, the rad: 7 rad is that amount of radiation which deposits energy per unit mass of 1.00 X 107 J/kg in any absorbing material. (This is quite close to the roentgen for X- and ¥ rays.) The proper SI unit for absorbed dose is the gray (Gy): 1Gy

TABLE 42-1 Quality Factor &Qa?ioé Different Kinds of i QF X and vz 1 B (electrons) S

o

1J/kg

=

100rad.

effective dose (inrem) 1

Fast neutrons

Up to 10

ah;;?:;cilgi:n ;

VRt

(42-10)

The absorbed dose depends not only on the strength of a given radiation beam (number of particles per second) and the energy per particle, but also on the type of material that is absorbing the radiation. Bone, for example, absorbs more of the radiation normally used than does flesh, so the same beam passing through a human body deposits a greater dose (in rads or grays) in bone than in flesh. The gray and the rad are physical units of dose—the energy deposited per unit mass of material. They are, however, not the most meaningful units for measuring the biological damage produced by radiation because equal doses of different types of radiation cause differing amounts of damage. For example, 1rad of a radiation does 10 to 20 times the amount of damage as 1rad of B8 or ¥ rays. This difference arises largely because « rays (and other heavy particles such as protons and neutrons) move much more slowly than B and ¥ rays of equal energy due to their greater mass. Hence, ionizing collisions occur closer together, so more irreparable damage can be done. The relative biological effectiveness (RBE) or quality factor (QF) of a given type of radiation is defined as the number of rads of X or Y radiation that produces the same biological damage as 1rad of the given radiation. Table 42-1 gives the QF for several types of radiation. The numbers are approximate since they depend somewhat on the energy of the particles and on the type of damage that is used as the criterion. fl The effective dose can be given as the product of the dose in rads and the QF, and this unit is known as the rem (which stands for rad equivalent man):

~1

Fast protons

=

:

5 5

;

=

dose (inrad) X QF. g

This unit is being replaced by the SI unit for effective dose (Sv)

=

o

;

“effective dose,”

dose (Gy) X QF.

(42-11a) 5

.

the sievert (Sv):

.

(42-11b)

By these definitions, 1 rem (or 1 Sv) of any type of radiation does approximately the same amount of biological damage. For example, 50 rem of fast neutrons does the same damage as 50 rem of ¥ rays. But note that 50 rem of fast neutrons is only Srads, whereas 50 rem of ¥ rays is 50 rads.

EXERCISE E Return to the second Chapter-Opening Question, page 1131, and answer it again now. Try to explain why you may have answered it differently the first time.

Human Exposure to Radiation @

@

PHYSICS

APPLIED Radon

PHYSICS APPLIED Human radiation exposure

1148

CHAPTER 42

We are constantly exposed to low-level radiation from natural sources: cosmic rays, natural radioactivity in rocks and soil, and naturally occurring radioactive

isotopes in our food, such as {0K. Radon, %3;Rn, is of considerable concern today. It

is the product of radium decay and is an intermediate in the decay series from uranium (see Fig. 41-12). Most intermediates remain in the rocks where formed, but radon is a gas that can escape from rock (and from building material like concrete) to enter the air we breathe, and attack the interior of the lung. The natural radioactive background averages about 0.30 rem (300 mrem) per year per person in the US., although there are large variations. From medical X-rays and scans, the average person receives about 50 to 60 mrem per year, givin“™h an average total dose of about 360 mrem (3.6 mSv) per person. Governmem regulators suggest an upper limit of allowed radiation for an individual in the

general populace at about 100 mrem (1 mSv) per year in addition to natural back-

ground. It is believed that even low doses of radiation increase the chances of

cancer or genetic defects; there is no safe level or threshold of radiation exposure. The upper limit for people who work around radiation—in hospitals, in power lants, in research—has been set higher, a maximum of 5 rem (50 mSv) whole-body dose in any one year, and significantly less averaged over more years (below 2rem/yr averaged over 5years). To monitor exposure, those people who work around radiation generally carry some type of dosimeter, one common type being a radiation film badge which is a piece of film wrapped in light-tight material. The passage of ionizing radiation through the film changes it so that the film is darkened upon development, and thus indicates the received dose. Newer types include the thermoluminescent dosimeter (TLD). Dosimeters and badges do not protect the worker, but high levels detected suggest reassignment or modified work practices to reduce radiation exposure to acceptable levels. Large doses of radiation can cause unpleasant symptoms such as nausea, fatigue, and loss of body hair. Such effects are sometimes referred to as radiation sickness. Large doses can be fatal, although the time span of the dose is important. A short dose of 1000 rem (10 Sv) is nearly always fatal. A 400-rem (4-Sv) dose in a short period of time is fatal in 50% of the cases. However, the body possesses remarkable repair processes, so that a 400-rem dose spread over several weeks is

@ PHYSICS APPLIED Radiation worker exposure Film badge

@ PHYSICS Radiation sickness

APPLIED

usually not fatal. It will, nonetheless, cause considerable damage to the body.

The effects of low doses over a long time are difficult to determine and are not well known as yet.

Whole-body dose. What whole-body dose is received by a

70-kg laboratory worker exposed to a 40-mCi $Co source, assuming the person’s body has cross-sectional area 1.5 m? and is normally about 4.0 m from the source for 4.0h per day? $9Co emits ¥ rays of energy 1.33MeV and 1.17 MeV in quick succession. Approximately 50% of the ¥ rays interact in the Abody and deposit all their energy. (The rest pass through.) APPROACH Of the given worker, equal to her area 4.0m (Fig. 42-16). SOLUTION The total Y-ray so the total energy emitted "

energy emitted, only a fraction passes through the divided by the total area over a full sphere of radius energy per decay is (1.33 + 1.17) MeV = 2.50 MeV, by the source per second is 10

.

9

FIGURE 42-16

o

Radiation spreads

(0.040 Ci)(3.7 x 10" decays/Ci-s)(2.50MeV) = 3.7 X 10°MeV/s. utin all directions. A person 4.0m The proportion of this energy intercepted by the body is its 1.5-m? area divided by the =~ away intercepts only a fraction: her cross-sectional area divided by the area of a sphere of radius 4.0 m (Fig. 42-16): 1.5

oL drr?

m?

S

1.5

m2

> M 47(4.0m)?

=

area of a sphere of radius 4.0 m.

Example 42-12.

7.5 %1073,

So the rate energy is deposited in the body (remembering that only 50% of the Y rays interact in the body) is

E = (575 x 1073)(3.7 x 10°MeV/s)(1.6 X 1073J/MeV) Since

1Gy = 1J/kg,

the

whole-body

dose

rate

for

this

= 2.2 X 107°/s. 70-kg

person

(22 x 107°%J/s)/(70kg) = 3.1 X 10 Gy/s. In 4.0h, this amounts to a dose of (4.0h)(3600s/h)(3.1 X 108Gy/s) = 4.5 X 10*Gy.

is

Since QF ~ 1 for gammas, the effective dose is 450 uSv (Egs. 42-11b and 42-10) or:

NOTE

(100 rad/Gy)(4.5 X 10*Gy)(1rem/rad)

= 45mrem

= 0.45mSv.

This 45-mrem effective dose is almost 50% of the normal allowed dose for

a whole year (100 mrem/yr), or 1% of the maximum one-year allowance for radition workers. This worker should not receive such a large dose every day and should seek ways to reduce it (shield the source, vary the work, work farther

| from the source, work less time this close to source, etc.).

SECTION 42-6

Measurement of Radiation—Dosimetry

1149

@rHYSics

APPLIED Radon exposure

Radon exposure. In the US, yearly deaths from radon

exposure (the second leading cause of lung cancer) are estimated to be on the order of 20,000 and maybe much more. The Environmental Protection Agency recommends taking action to reduce the radon concentration in living areas if 1% exceeds 4 pCi/L of air. In some areas 50% of houses exceed this level fro1 naturally occurring radon in the soil. Estimate (a) the number of decays/s in

1.0m? of air and (b) the mass of radon that emits 4.0 pCi of 222Rn radiation.

APPROACH We can use the definition of the curie to determine how many decays per second correspond to 4 pCi, then Eq. 41-7b to determine how many nuclei of radon it takes to have this activity |dN/dt|.

SOLUTION We saw at the start of this Section that 1 Ci = 3.70 X 10'° decays/s.

Thus

o

dt

4.0pCi (4.0 x 1072 Ci)(3.70 X 10'° decays/s/Ci) = 0.148s7!

per liter of air. In 1.0m?> of air (1m® = 10°cm = 10°L)

there

(0.148571)(1000) ~ 150 decays/5. (b) From Eqs. 41-7a and 41-8 dN

‘dT

=

AN

=

0.693

—"--T%

would

be

N.

Appendix F tells us T = 3.8232 days, so

-

‘if\_"_T_ dt | 0.693

(01485 = (014857)

(3.8232 days)(8.64 x 10*s/day)

-

0.693

7.05 X 10* atoms of radon-222. The molar mass (222 u) and Avogadro’s number are used to find the mass:

_ (705 x 10* atoms)(222 g/mol) 6.02 x 10% atoms/mol

= 2,6 x 107V bl

or 26 attograms in 1L of air at the limit of 4pCi/L. This 2.6 X 107 grams of radon per m® of air.

2.6 X 1077 g/L is

NOTE Each radon atom emits 4 « particles and 4 B particles, each one capable of causing many harmful ionizations, before the sequence of decays reaches a stable element.

*4)—7 @rHYSsics

APPLIED

Radiation therapy

Radiation Therapy

The medical application-of radioactivity and radiation to human beings involves two basic aspects: (1) radiation therapy—the treatment of disease (mainly cancer)—which we discuss in this Section; and (2) the diagnosis of disease, which we discuss in the following Sections of this Chapter. Radiation can cause cancer. It can also be used to treat it. Rapidly growing cancer cells are especially susceptible to destruction by radiation. Nonetheless, large doses are needed to kill the cancer cells, and some of the surrounding normal

cells are inevitably killed as well. It is for this reason that cancer patients receivir™ radiation therapy often suffer side effects characteristic of radiation sickness. To

1150

CHAPTER 42

Nuclear Energy; Effects and Uses of Radiation

minimize the destruction of normal cells, a narrow beam of ¥ or X-rays is often used when a cancerous tumor is well localized. The beam is directed at the tumor, and the source (or body) is rotated so that the beam passes through various parts of the body to keep the dose at any one place as low as possible—except at the rumor and its immediate surroundings, where the beam passes at all times

(Fig. 42-17). The radiation may be from a radioactive source such as $Co, or it

may be from an X-ray machine that produces photons in the range 200keV to 5MeV. Protons, neutrons, electrons, and pions, which are produced in particle accelerators (Section 43-1), are also being used in cancer therapy. Protons used to Kkill tumors have a special property that makes them particularly useful. As shown in Fig. 42-18, when protons enter tissue, most of their energy is deposited at the end of their path. The protons’ initial kinetic energy can be chosen so that most of the energy is deposited at the depth of the tumor itself, to destroy it. The incoming protons deposit only a small : ; ; ; amount of energy in the tissue in front of the tumor, and none at all behind the tumor, thus having less negative effect on healthy tissue than X- or 7 rays. Because tumors have physical size, even several centimeters in diameter, a range of proton energies is often used. Heavier ions, such as a particles or carbon ions, are similarly useful. This proton therapy technique is more than a half century old, but the necessity of having a large accelerator has meant that few hospitals have used the technique until now. Many such “proton centers” are now being built.

FIGURE 42-17 Radiation source rotates so that the beam always

. " gasses' uitaugh the diseassik tssue, ut minimizes the dose in the rest of dy. @ PHYSICS Proton therapy

APPLIED

100 +

2

S

r

)

E 3 &

751

FIGURE 42-18

507 251

-

Energy deposited in tissue

as a function of depth for 170-MeV protons (red curve) and 190-MeV protons (green). The peak of each curve is often called the Bragg peak.

5

10

15

Depth in tissue (cm)

20

Another form of treatment is to insert a tiny radioactive source directly inside a tumor, which will eventually kill the majority of the cells. A similar technique is

used to treat cancer of the thyroid with the radioactive isotope '3I. The thyroid

gland concentrates iodine present in the bloodstream, particularly in any area where abnormal growth is taking place. Its intense radioactivity can destroy the defective cells. Another application of radiation is for sterilizing bandages, surgical equipment, and even packaged foods, since bacteria and viruses can be killed or deactivated by large doses of radiation.

*42—-8

Tracers in Research and Medicine

Radioactive isotopes are commonly used in biological and medical research as tracers. A given compound is artificially synthesized using a radioactive isotope

such as '§C or {H. Such “tagged” molecules can then be traced as they move

through an organism or as they undergo chemical reactions. The presence of these tagged molecules (or parts of them, if they undergo chemical change) can be detected by a Geiger or scintillation counter, which detects emitted radiation see Section 41-11). How food molecules are digested, and to what parts of the vody they are diverted, can be traced in this way. *SECTION 42-8

@

PHYSICS

APPLIED

Tracers in medicine and biology

Tracers in Research and Medicine

1151

g‘

Radioactive tracers have been used to determine how amino

-

(41

‘ » : FIGURE 42-19

]

e 3:

acids and other

essential compounds are synthesized by organisms. The permeability of cell walls to various molecules and ions can be determined using radioactive isotopes: the tagged molecule or ion is injected into the extracellular fluid, and the radioactivity present inside and outside the cells is measured as a function of time. In a technique known as autoradiography, the position of the radioactive isotopes is detected on film. For example, the distribution of carbohydrates produced in the leaves of plants from absorbed CO, can be observed by keeping the plant in an atmosphere where the carbon atom in the CO, is '¢C. After a time, a leaf is placed firmly on a photographic plate and the emitted radiation darkens the film most strongly where the isotope is most strongly concentrated (Fig. 42-19a). Autoradiography using labeled nucleotides (components of DNA) has revealed much about the details of DNA replication (Fig. 42-19b).

"

29

-

For medical diagnosis, the radionuclide commonly used today is *37'Tc, a

long-lived excited state of technetium-99

'

(the “m” in the symbol

“metastable” state). It is formed when Mo

stands for

decays. The great usefulness of

inTc derives from its convenient half-life of 6 h (short, but not too short) and the fact that it can combine with a large variety of compounds. The compound F

(b)

to be labeled with the radionuclide is so chosen because it concentrates in the

‘

(a) Autoradiograph

organ or region of the anatomy to be studied. Detectors outside the body then record, or image, the distribution of the radioactively labeled compound. The detection could be done by a single detector (Fig. 42-20a) which is moved across the body, measuring

the intensity of radioactivity at a large number

of points.

of a leaf exposed for 30s to 1“CO,. The photosynthetic (green) tissue has become radioactive; the

The image represents the relative intensity of radioactivity at each point. The relative radioactivity is a diagnostic tool. For example, high or low radioactivity may represent overactivity or underactivity of an organ or part of an organ, or

is free of '¢C and therefore does not

cameras make use of many detectors which simultaneously record the radioac-

nonphotosynthetic tissue of the veins blacken: the X-ray sheet. This technique is useful in following

in another tivity

y

case may

at many

yp

represent

points. The

a lesion or tumor.

measured

intensities

can

More

complex

be displayed e

on

gamma

a TV

or

atiersof ittt iransnort i glants (b) Autoradio graplljl of chromosomal DNA. The dashed arrays of film grains show the Y-shaped

computer monitor. The image is sometimes called a scintigram (after scintillator« Fig. 42-20b. Gamma cameras are relatively inexpensive, but their resolution !imited—by non-pgrfe.ct cc?llimation*.; but they allow “dynamic” studies: images that change in time, like a movie.

growing point of replicating DNA.

"To “collimate” means to “make parallel,” usually by blocking non-parallel rays with a narrow tube inside lead, as in Fig. 42-20a.

_____+ —Photomultiplier tube

Scintillator crystal —

| Lead collimator

71—

Collimating hole

/_—_—‘-\\

Patient

(a)

(b)

FIGURE 42-20 (a) Collimated gamma-ray detector for scanning (moving) over a patient. The collimator selects ¥ rays that come in a (nearly) straight line from the patient. Without the collimator, ¥ rays from all parts of the body could strike the scintillator, producing a poor image. Detectors today usually have many collimator tubes and are called gamma cameras. (b) Gamma camera image (scintigram), of both legs of a patient with shin splints, detecting ¥s from *§Tc. 1152

CHAPTER 42

-

“42-9 Imaging by Tomography: CAT

&

Scans and Emission Tomography

.Jormal X-ray Image For a conventional medical or dental X-ray photograph, the X-rays emerging from the tube (Section 35-10) pass through the body and are detected on photographic film or a fluorescent screen, Fig. 42-21. The rays travel in very nearly straight lines

@

diffraction or refraction. There is absorption (and scattering), however; and the

(120 lenses involved)

through the body with minimal deviation since at X-ray wavelengths there is little

difference in absorption by different structures in the the image produced by the transmitted rays. The less the transmission and the darker the film. The image is, in a the rays have passed through. The X-ray image is not with lenses as for the instruments discussed in Chapter

body is what gives rise to absorption, the greater the sense, a “shadow” of what produced by focusing rays 33.

PHYSICS

APPLIED

Normal X-ray image is a sort of shadow

FIGURE 42-21 Conventional X-ray imaging, which is essentially shadowing.

*Tomography Images (CT) In conventional X-ray images, the entire thickness of the body is projected onto the film; structures overlap and in many cases are difficult to distinguish. In the f9705, a revolutionary new X-ray technique was developed called computed .mography (CT), which produces an image of a slice through the body. (The word tomography comes from the Greek: tomos = slice, graph = picture.) Structures and lesions previously impossible to visualize can now be seen with remarkable clarity. The principle behind CT is shown in Fig. 42-22: a thin collimated beam of X-rays (to “collimate” means to “make parallel”) passes through the body to a detector that measures the transmitted intensity. Measurements are made at a large number of points as the source and detector are moved past the body together. The apparatus is then rotated slightly about the body axis and again scanned; this is repeated at (perhaps) 1° intervals for 180°. The intensity of the transmitted beam for the many points of each scan, and for each angle, are sent to a computer that reconstructs the image of the slice. Note that the imaged slice is perpendicular to the long axis of the body. For this reason, CT is sometimes called computerized axial tomography (CAT), although the abbreviation CAT, as in CAT scan, can also be read as computer-assisted tomography. The use of a single detector as in Fig. 42-22 would require a few minutes for the many scans needed to form a complete image. Much faster scanners use

FrpHYSics

APPLIED

Computed tomography images

FIGURE 42-22 Tomographic imaging: the X-ray source and detector

Video monitor

move together across the body, the

Computer

Collimator

Collimator

*SECTION 42-9

X-ray source

(

transmitted intensity being measured at a large number of points. Then the

Source—detector assembly is rotated

slightly (say, 1°) and another scan is made. This process is repeated for perhaps 180°. The computer reconstructs the image of the slice and it is presented on a TV or computer monitor.

Imaging by Tomography: CAT Scans and Emission Tomography

1153

Magnetic Electron source

X-ray

Detector ring

deflection ../ coil ////// ), 7/ AN

Tungsten target rings (X-rays created)

Electron

source

(a)

(b)

FIGURE 42-23 (a) Fan-beam scanner. Rays transmitted through the entire body are measured simultaneously at each angle. The source and detector rotate to take measurements at different angles. In another type of fan-beam scanner, there are detectors around the entire 360° of the circle which remain fixed as the source moves. (b) In still another type, a beam of electrons from a source is directed by magnetic fields at tungsten targets surrounding the patient.

a fan beam, Fig. 42-23a, in which beams passing through the entire cross section of the body are detected simultaneously by many detectors. The source and detectors are then rotated about the patient, and an image requires only a few seconds. Even faster, and therefore useful for heart scans, are fixed source machines wherein an electron beam is directed (by magnetic fields) to tungsten targets surrounding the patient, creating the X-rays. See Fig. 42-23b. FIGURE 42-24 Two CT images, with different resolutions, each showing a cross section of a brain.

Photo Photo shows color

(a) is of low resolution. (b), of higher resolution, a brain tumor, and uses false to highlight it.

*Image

Formation

But how is the image formed? We can think of the slice to be imaged as being divided into many tiny picture elements (or pixels), which could be squares. (Se*™ Fig. 35-42.) For CT, the width of each pixel is chosen according to the width of the detectors and/or the width of the X-ray beams, and this determines the resolution of the image, which might be 1 mm. An X-ray detector measures the intensity of the transmitted beam. When we subtract this value from the intensity of the beam at the source, we obtain the total absorption (called a “projection™) along that beam line. Complicated mathematical techniques are used to analyze all the absorption projections for the huge number of beam scans measured (see the next Subsection), obtaining the absorption at each pixel and assigning each a “grayness value” according to how much radiation was absorbed. The image is made up of tiny spots (pixels) of varying shades of gray. Often the amount of absorption is color-coded. The colors in the resulting false-color image have nothing to do, however, with the actual color of the object. The real medical images are monochromatic (various shades of gray). Only visible light has color; X-rays and ¥ rays don’t. Figure 42-24 illustrates what actual CT images look like. It is generally agreed that CT scanning has revolutionized some areas of medicine by providing much less invasive, and/or more accurate, diagnosis. Computed tomography can also be applied to ultrasound imaging (Section 16-9) and to emissions from radioisotopes and nuclear magnetic resonance, which we discuss in Section 42-10.

*Tomographic Image Reconstruction

1154

(b) CHAPTER 42

How can the “grayness” of each pixel be determined even though all we can measure is the total absorption along each beam line in the slice? It can be done only by using the many beam scans made at a great many different angles. Suppose the image is to be an array of 100 X 100 elements for a total of 10* pixels. If we have 100 detectors and measure the absorption projections at 100 different angle«* then we get 10* pieces of information. From this information, an image can be reconstructed, but not precisely. If more angles are measured, the reconstruction of the image can be done more accurately.

To suggest how mathematical reconstruction is done, we consider a very simple case using the “iterative” technique (“to iterate” is from the Latin “to repeat”).

Suppose our sample slice is divided into the simple 2 X 2 pixels as shown in

ig. 42-25. The number inside each pixel represents the amount of absorption by the H 1aterial in that area (say, in tenths of a percent): that is, 4 represents twice as much absorption as 2. But we cannot directly measure these values—they are the unknowns we want to solve for. All we can measure are the projections—the total absorption along each beam line—and these are shown in the diagram as the sum of the absorptions for the pixels along each line at four different angles. These projections (given at the tip of each arrow) are what we can measure, and we now want to work back from them to see how close we can get to the true absorption value for each pixel. We start our analysis with each pixel being assigned a zero value, Fig. 42-26a. In the iterative technique, we use the projections to estimate the absorption value in each square, and repeat for each angle. The angle 1 projections are 7 and 13. We divide each of these equally between their two squares: each square in the left column gets 33 (half of 7), and each square in the right column gets 63 (half of 13); see Fig. 42-26b. Next we use the projections at angle 2. We FIGURE 42-26

FIGURE 42-25

A simple2 X 2

image showing true absorption values and measured projections.

Reconstructing the image using projections

in an iterative procedure.

6 Angle 2

(measured)

(d)

(b)

r

calculate the difference between the measured projections at angle 2 (6 and 14) and the projections based on the previous estimate (top row: 33 + 6% = 10; same for bottom row). Then we distribute this difference equally to the squares in that row. For the top row, we have 6 — 10

33+———

=11

and

63+

——

=43

and for the bottom row,

3%+————14;10=5%

and

6%+¥=8%-

These values are inserted as shown in Fig, 42-26c¢. Next, the projection at angle 3 gives

(upper left)

15 + 1—1-—_2——19 = 2

and

(lowerright)

83 + ll._;fl

=9

and that for angle 4 gives

(lower left)

53 + & _210

= 5

and

(upperright)

43 + 2 ;10

= 4,

The result, shown in Fig. 42-26d, corresponds exactly to the true values. (In real situations, the true values are not known, which is why these computer techniques are required.) To obtain these numbers exactly, we used six pieces of information (two each at angles 1 and 2, one each at angles 3 and 4). For the much larger number of pixels used for actual images, exact values are generally not attained. Many iterations may be needed, and the calculation is considered sufficiently recise when the difference between calculated and measured projections is sufficiently small. The above example illustrates the “convergence” of the process: the first iteration (b to c in Fig. 42-26) changed the values by 2, the last iteration

(cto d) by only &-

*SECTION 42-9

1155

*Emission Tomography It is possible to image the emissions of a radioactive tracer (see Section 42-8) in a single plane or slice through a body using computed tomography techniques. A basic gamma detector (Fig. 42-20a) can be moved around the patient to measurim, the radioactive intensity from the tracer at many points and angles; the data ar. processed in much the same way as for X-ray CT scans. This technique is referred to as single photon emission tomography (SPET), or SPECT (single photon emission computed tomography); see Fig. 42-27. Another important technique is positron emission tomography (PET), which

makes use of positron emitters such as '}C, 3N, 130, and '§F. These isotopes are

FIGURE 42-27

SPECT scan of

brain (false color) with epilepsy,

labeled with *ATc. @

PHYSICS APPLIED Emission tomography — (SPECT.PET) FIGURE 42-28 Positron emission tomography (PET) system showing a ring of detectors to detect the two annihilation ¥ rays (e + e~ — 27)

emitted at 180° to each other.

incorporated into molecules that, when inhaled or injected, accumulate in the organ or region of the body to be studied. When such a nuclide undergoes B" decay, the emitted positron travels at most a few millimeters before it collides with a normal electron. In this collision, the positron and electron are annihilated, producing two ¥ rays (e + e~ — 27). The two ¥ rays fly off in opposite directions (180° £ 0.25°) since they must have almost exactly equal and opposite momenta to conserve momentum (the momenta of the initial e* and e~ are essentially zero compared to the momenta of the ¥ rays). Because the photons travel along the same line in opposite directions, their detection in coincidence by rings of detectors surrounding the patient (Fig. 42-28) readily establishes the line along which the emission took place. If the difference in time of

arrival of the two photons could be determined accurately, the actual position

of the emitting nuclide along that line could be calculated. Present-day electronics can measure times to at best +300ps, so at the ¥ ray’s speed (¢ =3 x10°m/s), the actual position could be determined to an accuracy

on the order of about d = of ~ (3 X 10°m/s)(300 X 1072s) ~ 10cm,

which

is not very useful. Although there may be future potential for time-of-flight measurements to determine position, today computed tomography techniques are used instead, similar to those for X-ray CT, which can reconstruct PET image with a resolution on the order of 3-5mm. One big advantage of PET is th no collimators are needed (as for detection of a single photon—see Fig. 42-20a). Thus, fewer photons are “wasted” and lower doses can be administered to the patient with PET. Both PET and SPET systems can give images related to biochemistry, metabolism, and function. This is to be compared to X-ray CT scans, whose images reflect shape and structure—that is, the anatomy of the imaged region.

*42-10 Nuclear Magnetic Resonance (NMR);

Magnetic Resonance Imaging (MRI)

FIGURE 42-29

Schematic picture

of a proton in a magnetic field B

ointing upward) with the tw gz;ssiblegstalzes of ;roton spin ¢

up and down.

s é Up

1156

CHAPTER 42

Ran

Nuclear magnetic resonance (NMR) in 1946 became a powerful research istry and biochemistry. It is also an briefly discuss the phenomenon, and

is a phenomenon which soon after its discovery tool in a variety of fields from physics to chemimportant medical imaging technique. We first then look at its applications.

*Nuclear Magnetic Resonance (NMR)

B . . We saw in Chapter 39 (Section 39-2) that when atoms are placed in a magnetic field B,

atomic energy levels are split into several closely spaced levels (the Zeeman effect) according to the angular momentum or spin of the state. The splitting is proportional to B and to the magnetic moment, . Nuclei too have magnetic moments (Section 41-1), and we examine mainly the simplest, the hydrogen (}H) nucleus because it is the one most used, even for medical imagining. The H nucleus consists of a single proton. Its spin angular momentum (and its magnetic moment), like that of the electron, can take on only two values when placed in a magnetic field: We call these “spin up” (parall ™ to the field) and “spin down” (antiparallel to the field) as suggested in Fig. 42-29.

Nuclear Energy; Effects and Uses of Radiation

r

EO-(-B—=—0)—-(

,/

F

el

FIGURE 42-30

hf = AE

N

= 2upB

Energy K, in the

absence of a magnetic field splits into two levels in the presence of a

magnetic field.

N Spin_ | up When a magnetic field is present, an energy state of a nucleus splits into two levels

as shown in Fig. 42-30 with the spin-up state (parallel to field) having the lower energy. The spin-down state acquires an additional energy w,Br and the spin-up state has its energy changed by —p,Br (see Eq. 27-12, Example 27-12, and Section 39-7), where By is the total magnetic field at the nucleus. The difference in energy between the two states (Fig. 42-30) is thus

AE = 2pu,Br, where u,, is the magnetic moment of the proton.

In a standard nuclear magnetic resonance (NMR) setup, the sample to be examined is placed in a static magnetic field. A radiofrequency (RF) pulse of elec-

tromagnetic radiation (that is, photons) is applied to the sample. If the frequency, f, of this pulse corresponds precisely to the energy difference between the two energy levels (Fig. 42-30), so that

hf = AE = 2p,Br,

(42-12)

then the photons of the RF beam will be absorbed, exciting many of the nuclei from the lower state to the upper state. This is a resonance phenomenon since there is significant absorption only if fis very near f = 2u,Br/h. Hence the name “nuclear magnetic resonance.” For free {H nuclei, the frequency is 42.58 MHz for a f ield By = 1.0T (Example 42-14). If the H atoms are bound in a molecule, the total magnetic field By at the H nuclei will be the sum of the external applied field

(Bex) plus the local magnetic field (By,.) due to electrons and nuclei of neighboring

atoms. Since f is proportional to By, the value of f for a given external field will be slightly different for bound H atoms than for free atoms: hf

=

2/“"p(cht

+

B]oc)-

This small change in frequency can be measured, and is called the “chemical shift.” A great deal has been learned about the structure of molecules and bonds using this NMR technique.

SOV

ERLN

NMR for free protons. Calculate the resonant frequency

for free protons in a 1.000-T magnetic field. APPROACH We use (Section 41-1) is Hp SOLUTION

_AE

fis o

=

Eq. 42-12, where

2.7928

N

=

2.7928|

the magnetic et —

| =

moment

2.7928

¢

eh

(477mp>

of the proton .

We solve for fin Eq. 42-12 and find

2B

a—

= (2.7928)( o

2mm,

) — 27

[(1-6022X10_19C)(1.000T):| 2m(1.6726 X 10 kg)

-

42.58 MHz.

*SECTION 42-10

Nuclear Magnetic Resonance (NMR); Magnetic Resonance Imaging (MRI)

1157

* Magnetic Resonance Imaging (MRI) @rHYsics

APPLIED

NMR imaging (MRI)

For producing medically useful NMR

images—now

commonly

called MRI, or

magnetic resonance imaging—the element most used is hydrogen since it is the

commonest element in the human body and gives the strongest NMR signals. Th experimental apparatus is shown in Fig. 42-31. The large coils set up the static magnetic field, and the RF coils produce the RF pulse of electromagnetic waves (photons) that cause the nuclei to jump from the lower state to the upper one (Fig. 42-30). These same coils (or another coil) can detect the absorption of energy or the emitted radiation (also of frequency f = AE/h) when the nuclei jump back down to the lower state. Magnetic field coils

RF coils

__ Patient

-~

-~

P

pr

~

—

B high

| J low

G

N

Jf high

"

~— FIGURE 42-32 A static ficld that is stronger at the bottom than at the top. The frequency of absorbed or

emitted radiation is proportional to Bin NMR.

FIGURE 42-33 False-color NMR image (MRI) of a vertical section

through the head showing structures . . in the normal brain.

1158

CHAPTER 42

FIGURE 42-31 :

Typical MRI imaging setup: (a) diagram; (b) photograph. ¢

:

:

y

:

The formation of a two-dimensional or three-dimensional image can be don using techniques similar to those for computed tomography (Section 42-9, The simplest thing to measure for creating an image is the intensity of absorbed and/or reemitted radiation from many different points of the body, and this would be a measure of the density of H atoms at each point. But how do we determine from what part of the body a given photon comes? One technique is to give the static magnetic field a gradient; that is, instead of applying a uniform magnetic field, By, the field is made to vary with position across the width of the sample (or patient). Since the frequency absorbed by the H nuclei is proportional to By (Eq. 42-12), only one plane within the body will have the proper value of By to absorb photons of a particular frequency f. By varying f, absorption by different planes can be measured. Alternately, if the field gradient is applied after the RF pulse, the frequency of the emitted photons will be a measure of where they were emitted.

See Fig. 42-32. If a magnetic field gradient in one direction is applied during excitation (absorption of photons) and photons of a single frequency are transmitted, only H nuclei in one thin slice will be excited. By applying a gradient

during reemission in a direction perpendicular to the first, the frequency f of the reemitted radiation will represent depth in that slice. Other ways of varying the

magnetic field throughout the volume of the body can be used in order to correlate NMR frequency with position. A reconstructed image based on the density of H atoms (that is, the intensity of absorbed or emitted radiation) is not very interesting. More useful are images based on the rate at which the nuclei decay back to the ground state, and such images can produce resolution of 1 mm or better. This NMR technique (sometimes called spin-echo) produces images of great diagnostic value, both in the delineation of structure (anatomy) and in the study of metabolic processes. An NMR image is shown in Fig. 42-33, color enhanced—no medical imaging uses\ visible light, so the colors shown here are added. The original images, those lookec at by your doctor, are various shades of gray, representing intensity (or counts).

Nuclear Energy; Effects and Uses of Radiation

NMR imaging is considered to be noninvasive. We can calculate the energy of the photons involved: as determined in Example 42-14, in a 1.0-T magnetic field, f = 42.58 MHz for {H. This corresponds to an energy of

gt

= (6.6 x 104J+5)(43 X 10°Hz) ~ 3 X 107J or about 1077 eV. Since molec-

r bonds are on the order of 1 eV, it is clear that the RF photons can cause little cellular disruption. This should be compared to X- or ¥ rays whose energies are 10* to 10°eV and thus can cause significant damage. The static magnetic fields, though often large (typically around 2.0 T), are believed to be harmless (except for people wearing heart pacemakers). Table 42-2 lists the major techniques we have discussed for imaging the interior of the human body, along with the optimum resolution attainable today. Resolution is just one factor that must be considered, as the different imaging techniques provide different types of information, useful for different types of diagnosis.

TABLE 42-2

Medical

Imaging Techniques

Technique

Resolution

Conventional X-ray CT scan, X-ray Nuclear medicine (tracers) SPET (single photon emission)

51 mm 0.2 mm lcm

PET (positron emission)

2-5mm

lem

NMR (MRI)

I-1mm

Ultrasound (Section 16-9)

0.3-2 mm

| Summary A nuclear reaction occurs when two nuclei collide and two or more other nuclei (or particles) are produced. In this process, as in radioactivity, transmutation (change) of elements occurs. The reaction energy or Q-value of a reaction a+X > Y+bis

Q = (My + Mx — My — My)c? =

Ky + Ky

(42-2a)

- K, — Kx.

(42-2b)

The effective cross section o for a reaction is a measure of the reaction probability per target nucleus. In fission, a heavy nucleus such as uranium splits into two

intermediate-sized nuclei after being struck by a neutron. 23U is

"ssionable

by slow

neutrons, whereas

some

fissionable

nuclei

-equire fast neutrons. Much energy is released in fission (= 200MeV per fission) because the binding energy per nucleon is lower for heavy nuclei than it is for intermediatesized nuclei, so the mass of a heavy nucleus is greater than the total mass of its fission products. The fission process releases neutrons, so that a chain reaction is possible. The critical mass is the minimum mass of fuel needed to sustain a chain reaction. In a nuclear reactor or nuclear bomb, a moderator is used to slow down the released neutrons. The fusion process, in which small nuclei combine to form larger ones, also releases energy. The energy from our Sun originates in the fusion reactions known as the proton—proton chain in which four protons fuse to form a He nucleus producing 25MeV of energy. A useful fusion reactor for power generation has not yet proved possible because of the difficulty in containing the fuel (e.g., deuterium) long enough at the extremely

high temperature required (~ 10°K). Nonetheless, great progress has been made in confining the collection of charged ions known as a plasma. The two main methods are magnetic confinement, using a magnetic field in a device such as the toroidal-shaped tokamak, and inertial confinement in which intense laser beams compress a fuel pellet of deuterium and tritium. Radiation can cause damage to materials, including biological tissue. Quantifying amounts of radiation is the subject of dosimetry. The curie (Ci) and the becquerel (Bq) are units that measure the source activity or rate of decay of a sample:

1Ci = 3.70 x 10

decays

per

second,

whereas

1Bq = 1decay/s. The absorbed dose, often specified in rads, measures the amount of energy deposited per unit mass of absorbing material: 1rad is the amount of radiation that deposits energy at the rate of 1072 J/kg of material. The SI unit

of absorbed dose is the gray:

1Gy = 1J/kg = 100rad.

The

effective dose is often specified by the rem = rad X QF, where QF is the “quality factor” of a given type of radiation; 1rem of any type of radiation does approximately the same amount of biological damage. The average dose received per person per year in the United States is about 360 mrem. The SI unit for effective dose is the sievert: 1Sv = 10°rem. [*Nuclear radiation is used in medicine for cancer therapy, and for imaging of biological structure and processes. Tomographic imaging of the human body, which can provide 3-dimensional detail, includes several types: CT scans, PET, SPET (= SPECT), and MRI; the latter makes use of nuclear magnetic resonance (NMR).]

| Questions (NOTE: Masses are found in Appendix F.) 1. Fill in the missing particles or nuclei:

(@) (b) (c) @

3. When #Na is bombarded by deuterons (}H), an « particle is emitted. What is the resulting nuclide? 4. Why are neutrons such good projectiles nuclear reactions?

n + 3Ba — 2 + 7, n + 3Ba — 3ICs + 7 d+3H - $He + 2 a+ YAu —» 2 +d

5 A proton

where d stands for deuterium.

2, The

n+?

isotope —

{3P

is

produced

by

the

reaction:

?5P + p. What must be the target nucleus?

for producing

strikes a 7gNe nucleus, and an a particle is

observed to emerge. What is the residual nucleus? Write down the reaction equation. 6. Are fission fragments 8 or 8~ emitters? Explain. 7. The energy from nuclear fission appears in the form of thermal energy—but the thermal energy of what?

Questions

1159

8. 28U

releases

an

average

of 2.5

neutrons

per

fission

compared to 2.9 for 233Pu. Which of these two nuclei do you

.o

think would have the smaller critical mass? Explain.

. 11. . .

. 15. .

If 23U released only 1.5 neutrons per fission on the average,

would a chain reaction be possible? If so, how would the chain reaction be different than if 3 neutrons were released per fission? Why can’t uranium be enriched by chemical means? How can a neutron, with practically no kinetic energy, excite a nucleus to the extent shown in Fig. 42-4? Why would a porous block of uranium be more likely to explode if kept under water rather than in air? A reactor that uses highly enriched uranium can use ordinary water (instead of heavy water) as a moderator and still have a self-sustaining chain reaction. Explain. Why must the fission process release neutrons if it is to be useful? Why are neutrons released in a fission reaction? What is the reason for the “secondary system” in a nuclear reactor, Fig. 42-9? That is, why is the water heated by the fuel in a nuclear reactor not used directly to drive the turbines?

17. What is the basic difference between fission and fusion?

18. Discuss the relative merits and disadvantages, including pollution and safety, of power generation by fossil fuels, nuclear fission, and nuclear fusion.

19. A higher temperature is required for deuterium-deuterium q ignition than for deuterium-tritium. Explain. 20. Light energy emitted by the Sun and stars comes from the fusion process. What conditions in the interior of stars make this possible? 21. How do stars, and our Sun, maintain confinement of the plasma for fusion? Why is the recommended maximum radiation dose higher for women beyond the child-bearing age than for younger women? . People who work around metals that emit alpha particles are trained that there is little danger from proximity or touching the material, but they must take extreme precautions against ingesting it. Why? (Eating and drinking while working are forbidden.) . What is the difference between absorbed dose and effective

dose? What are the SI units for each? 25, Radiation is sometimes used to sterilize medical supplies and even food. Explain how it works. *26. How might radioactive tracers be used to find a leak in a pipe?

| Problems (NOTE: Masses are found in Appendix F.)

11.

42-1 Nuclear Reactions, Transmutation 1 (I) Natural aluminum is all Al If it absorbs a neutron, what does it become? Does it decay by 87 or 87? What will be the product nucleus?

(II) Radioactive '¢C is produced in the atmosphere when a

neutron is absorbed by '4N. Write the reaction and find its

Q-value. 12. (II) An

d+§Li

example

of

a

stripping

nuclear

reaction

iq

- X + p. (a) What is X, the resulting nucleus?

. (I) Is the reaction n + 35U — 23U + ¥ possible with slow

(b) Why is it called a “stripping” reaction? (c) What is the Q-value of this reaction? Is the reaction endothermic or exothermic? 13. (I) An example of a pick-up nuclear reaction is

(IT) Does the reaction p + jLi — $He + a require energy,

reaction? (b) What is the resulting nucleus? (c¢) What is the

. (I) Determine whether the reaction 7H + {H — 3He + n requires a threshold energy. neutrons? Explain.

or does it release energy? How much energy? . (II) Calculate the energy released required) for the reaction @ + jBe —

. (II) (a) Can the reaction n + Mg

(or energy '2C + n.

input

— #Na + d occur if

the bombarding particles have 16.00 MeV of kinetic energy? (d stands for deuterium, 7H.) (b) If so, how much energy is released? If not, what kinetic energy is needed?

. (II) (a) Can the reaction p + JLi — $He + a occur if the

incident proton has kinetic energy = 3500keV? (b) If so, what is the total kinetic energy of the products? If not, what kinetic energy is needed?

. (II) In the reaction

& + 4N — 'O + p, the incident

a particles have 9.68 MeV of kinetic energy. The mass of {0 is 16.999132 u. (a) Can this reaction occur? (b) If so, what is the total kinetic energy of the products? If not, what kinetic energy is needed?

. (IT)

a+

Calculate

0

the

Q-value

— #Ne + 7.

for

the

“capture”

reaction

10. (IT) Calculate the total kinetic energy of the products of the

reaction

d + 3C — 4N + n

has kinetic energy

1160

CHAPTER 42

if the incoming deuteron

K = 44.4 MeV.

3He + 2C —» X + a. (a) Why Q-value

of this reaction?

exothermic? (a) Complete 14. (I)

the

is it called a “pick-up”

Is the reaction

following

endothermic

nuclear

or

reaction,

p+? — 128 + 7. (b) What is the Q-value? 15. (IT) The reaction p + '§O0 — '§F + n requires an input

of energy equal to 2.438 MeV. What is the mass of §F? 16. (IIT) Use conservation of energy and momentum to show that a bombarding proton must have an energy of 3.23 MeV in order to make the reaction 2C(p,n)3N occur. (See Example 42-3.) 17. (III) How much kinetic energy (if any) would the proton

require for the reaction '¢C(p,n)"JN

to proceed?

42-2 Cross Section 18. (I) The cross section for the reaction n + 'YB — JLi + §He is about 40bn for an incident neutron of low energy (kinetic energy ~ 0). The boron is contained in a gas with

n =17 X 10’ nuclei/m®

and

the

target

has

thickness

£ =12.0cm. What fraction of incident neutrons will be fl scattered? 19. (I) What is the effective cross section for the collision of two hard spheres of radius Ry and R;?

Nuclear Energy; Effects and Uses of Radiation

f‘

20. (IT) When the target is thick, the rate at which projectile particles collide with nuclei in the rear of the target is less than in the front of the target, since some scattering (i.e., collisions) takes place in the front layers. Let R be the rate at which incident particles strike the front of the target, and R, be the rate at a distance x into the target (Ry = Ry at x = 0). Then show that the rate at which particles are scattered (and therefore lost from the incident beam) in a thickness dx is —dRy = Rynodx, where the minus sign means that Ry is decreasing and » is the number of nuclei per unit volume. Then show that Ry = Rye "?*, where o is the total cross section. If the thickness

what does Ry = Rye ! represent? 21. (IT) rays will 22. (II)

of the target is £,

A 1.0-cm-thick lead target reduces a beam of gamma to 25% of its original intensity. What thickness of lead allow only one 7 in 10 to penetrate (see Problem 20)?

Use Fig. 42-3 to estimate what thickness of '}$Cd

p = 8650kg/m’)

will

cause

a

2.0%

reaction

rate

R/Rqy = 0.020) for (a) 0.10-eV neutrons (b) 5.0-eV neutrons.

42-3 Nuclear Fission 23. (I) What is the energy released in the fission reaction of

Eq. 42-5? (The masses of '¢{Ba and $Kr are 140.914411 u and 91.926156 u, respectively.)

24. (I) Calculate

n+ 23U

assume 25. (I) How reactor? (I) The

the energy

released

in the fission reaction

- $Sr + ¥Xe + 12n. Use Appendix

F, and

the initial kinetic energy of the neutron is very small. many fissions take place per second in a 200-MW Assume 200 MeV is released per fission. energy produced by a fission reactor is about

34. (II) Suppose that the neutron multiplication factor is 1.0004. If the average time between successive fissions in a chain of reactions is 1.0ms, by what factor will the reaction rate increase in 1.0s?

42-4

iH + H — 3He + n is 17.57 MeV.

37. (II) Show that the energy released when two deuterium nuclei fuse to form 3He with the release of a neutron is 3.23 MeV. 38. (IT) Verify the Q-value stated for each of the reactions of

Eqs. 42-7. [Hint: Be careful with electrons.]

39. (II) (a) Calculate the energy release per gram of fuel for the reactions of Egs. 42-9a, b, and c. (b) Calculate the energy

release per gram of uranium 23U in fission, and give its

ratio to each reaction in (a). 40. (I1) How much energy is released when 35U absorbs a slow

neutron (kinetic energy ~ 0) and becomes %33U?

41. (IT) If a typical house requires 850W of electric power on average, what minimum amount of deuterium fuel would have to be used in a year to supply these electrical needs? Assume the reaction of Eq. 42-9b. 42. (II) 1f §Li is struck by a slow neutron, it can form He and another isotope. (a) What is the second isotope? (This is a method of generating this isotope.) () How much energy is released in the process? 43. (II) Suppose a fusion reactor ran on “d-d” reactions, Egs. 42-9a and b in equal amounts. Estimate how much natural water, for fuel, would be needed per hour to run a

200MeV per fission. What fraction of the mass of a 23U nucleus is this?

. (II) Suppose that the average electric power consumption, day and night, in a typical house is 880 W. What initial mass

of 23U would have to undergo fission to supply the elec-

trical needs of such a house for a year? (Assume 200 MeV is released per fission, as well as 100% efficiency.) 28, (IT) Consider the fission reaction

1250-MW reactor, assuming 33% efficiency.

. (II) Show that the energies carried off by the $He nucleus and the neutron for the reaction of Eq. 42-9c are about 3.5MeV

and

14 MeV,

respectively. Are

these fixed values,

independent of the plasma temperature? 45. (II) How much energy (J) is contained in 1.00 kg of water if its natural deuterium is used in the fusion reaction of Eq. 42-9a? Compare to the energy obtained from the burning of 1.0 kg of gasoline, about 5 x 1077J. 46 (1IIT) (a) Give the ratio of the energy needed for the first reaction of the carbon cycle to the energy needed for a deuterium-tritium reaction (Example 42-10). (b) If a deuterium-tritium reaction requires 7 ~ 3 X 108K, estimate the temperature needed for the first carbon-cycle reaction. 47. (III) The energy output of massive stars is believed to be due to the carbon cycle (see text). (a) Show that no carbon is consumed in this cycle and that the net effect is the same as for the proton-proton cycle. (b) What is the total energy release? (c) Determine the energy output for each reaction and decay. (d) Why might the carbon cycle require a higher by

ZBU +n — 3Sb + $Nb + .

Nuclear Fusion

35. (I) What is the average kinetic energy of protons at the center of a star where the temperature is 2 X 107 K? [Hint: See Eq. 184.] 36. (II) Show that the energy released in the fusion reaction

(a) How many neutrons are produced in this reaction? (b) Calculate the energy release. The atomic masses for Sb and Nb isotopes are 132.915250 u and 97.910328 u, respectively. 29. (I1) How much mass of %3§U is required to produce the same amount of energy as burning 1.0 kg of coal (about 3 X 107 J)? 30. (II) What initial mass of %3U is required to operate a 950-MW reactor for 1yr? Assume 38% efficiency. 31 (IT) If a 1.0-MeV neutron emitted in a fission reaction loses one-half of its kinetic energy in each collision with moderator

nuclei,

how

many

collisions

thermal energy (34T = 0.040eV)?

must

it make

to reach

32. (II) Assuming a fission of 20U into two roughly equal fragments, estimate the electric potential energy just as the fragments separate from each other. Assume that the fragments are spherical (see Eq. 41-1) and compare your calculation to the nuclear fission energy released, about 200 MeV. r‘fl. (II) Estimate the ratio of the height of the Coulomb barrier for @ decay to that for fission of %SU. (Both are described by a potential energy diagram of the shape shown in Fig. 41-7.)

temperature (~2 X 10’K) than the proton-proton cycle (~ 1.5 x 10'K)?

42-6

Dosimetry

48. (I) 250rads of a-particle radiation is equivalent to how many rads of X-rays in terms of biological damage? 49. (I) A dose of 4.0Sv of ¥ rays in a short period would be lethal to about half the people subjected to it. How many grays is this? 50. (I) How much energy is deposited in the body of a 65-kg adult exposed to a 3.0-Gy dose?

Problems

1161

S1. (I) How many rads of slow neutrons will do as much biological damage as 65 rads of fast neutrons? s2. (IT) A cancer patient is undergoing radiation therapy in which protons with an energy of 1.2 MeV are incident on a 0.25-kg tumor. (a) If the patient receives an effective dose of 1.0 rem, what is the absorbed dose? (b) How many protons are absorbed by the tumor? Assume QF ~ 1. 53. (IT) A 0.035-uCi sample of 32P is injected into an animal for tracer studies. If a Geiger counter intercepts 25% of the emitted B particles, what will be the counting rate, assumed 85% efficient? . (IT) About 35V is required to produce one ion pair in air. Show that this is consistent with the two definitions of the roentgen given in the text. §5. (1) A 1.6-mCi source of 2P (in NaHPOy), a 8 emitter, is implanted in a tumor where it is to administer 36 Gy. The

half-life of 2P is 14.3days, and 1.0mCi delivers about 10 mGy/min. Approximately remain implanted?

56. S7.

how long should the source

(IT) What is the mass of a 2.00-uCi '¢C source? (IT) Huge amounts of radioactive '3 were released in the

accident at Chernobyl in 1986. Chemically, iodine goes to the human thyroid. (Doctors can use it for diagnosis and

treatment of thyroid problems.) In a normal thyroid, '3}I absorption can cause damage to the thyroid. (z) Write down

the reaction for the decay of '3I. (b) Its half-life is 8.0d;

how long would it take for ingested '2}I to become 7.0% of

the initial value? (c) Absorbing 1mCi of '3}I can be harmful; what mass of iodine is this? 58. (II) Assume a liter of milk typically has an activity of 2000 pCi due to {JK. If a person drinks two glasses (0.5L) per day, estimate the total effective dose (in Sv and in rem) received in a year. As a crude model, assume the milk stays in the stomach 12 hr and is then released. Assume also that very roughly 10% of the 1.5MeV released per decay is absorbed by the body. Compare your result to the normal allowed dose of 100 mrem per year. Make your estimate for (a) a 60-kg adult, and (b) a 6-kg baby.

59. (I1) 3}Co emits 122-keV ¥ rays. If a 58-kg person swallowed

1.55 uCi of 3]Co, what would be the dose rate (Gy/day)

averaged over the whole body? Assume that 50% of the Y-ray energy is deposited in the body. [Hinr: Determine the rate of energy deposited in the body and use the definitio‘ of the gray.] 60. (IT) Ionizing radiation can be used on meat products the levels of microbial pathogens. Refrigerated meat to 4.5kGy. If 1.2-MeV electrons irradiate 5Skg of many electrons would it take to reach the allowable 61.

to reduce is limited beef, how limit?

(IT) Radon gas, %2Rn, is considered a serious health hazard

(see discussion in text). It decays by a-emission. (a) What is the daughter nucleus? (b) Is the daughter nucleus stable or

radioactive? If the latter, how does it decay, and what is its

half-life? (See Fig. 41-12.) (c) Is the daughter nucleus also a noble gas, or is it chemically reactive? (d) Suppose 1.6 ng of

Z2Rn seeps into a basement. What will be its activity? If the basement is then sealed, what will be the activity 1 month later?

*42-9

Imaging by Tomography

*62. (IT) (a) Suppose for X-ray beam consists magnification of the X-rays come from a 15cm

a conventional X-ray image that the of parallel rays. What would be the image? (b) Suppose, instead, that the point source (as in Fig. 42-21) that is

in front of a human

body which is 25 cm thick, and

the film is pressed against the person’s back. Determine and discuss the range of magnifications that result.

*42-10

NMR

*63. (I) Calculate the wavelength of photons needed to produce NMR transitions in free protons in a 1.000-T field. In whg‘ region of the spectrum is this wavelength? *64. (IT) Carbon-13 has a magnetic moment

w = 0.7023uy

(see

Eq. 41-2). What magnetic field would be necessary if '7C were

to be detected in a proton NMR spectrometer operating at 42,58 MHz? (This large field necessitates that a '3C spectrometer operate at a lower frequency.)

| General Problems 65. J. Chadwick discovered the neutron by bombarding jBe with the popular projectile of the day, alpha particles. (a) If one of the reaction products was the then unknown neutron, what was the other product? (b) What is the Q-value of this reaction?

. Fusion temperatures are often given in keV. Determine the conversion factor from kelvins to keV

in this field,

using, as is common

K = kT without the factor 3-

67. One means of enriching uranium is by diffusion of the gas UFg. Calculate the ratio of the speeds of molecules of this

gas containing %3U and %5U, on which this process depends.

. (a) What mass of 23U was actually fissioned in the first

atomic bomb, whose energy was the equivalent of about 20 kilotons of TNT (1 kiloton of TNT releases 5 X 10'2J)? (b) What was the actual mass transformed to energy? 69. The average yearly background radiation in a certain town consists of 29 mrad of X-rays and ¥ rays plus 3.6 mrad of particles having a QF of 10. How many rem will a person receive per year on the average?

1162

CHAPTER 42

70. Deuterium makes up 0.0115% of natural hydrogen on average. Make a rough estimate of the total deuterium in the Earth’s oceans and estimate the total energy released if all of it were used in fusion reactors. 71. A shielded 7Y-ray source yields a dose rate of 0.052 rad/h at a distance of 1.0m for an average-sized person. If workers are allowed a maximum dose of 5.0 rem in 1 year, how close to the source may they operate, assuming a 35-h work week? Assume that the intensity of radiation falls off as the square of the distance. (It actually falls off more rapidly than 1/r? because of absorption in the air, so your answer will give a better-than-permissible value.) 72. Radon gas, 2%Rn, is formed by a decay. (a) Write the decay equation. (b) Ignoring the kinetic energy of the daughter nucleus (it’s so massive), estimate the kinetic energy of the a particle produced. (¢) Estimate the momentum of the alpha and of the daughter nucleus. (d) Estimate the kinetic energy of the daughter, and show that your approximatic*™ in (b) was valid.

Nuclear Energy; Effects and Uses of Radiation

75 Consider a system of nuclear power

plants that produce

83. A large amount of 33Sr was released during the Chernobyl

required to operate these plants for 1yr, assuming that 200MeV is released per fission? (b) Typically 6% of the

through the food chain. How long will it take for 85% of the 38Sr released during the accident to decay? See Appendix F.

2400 MW.

(a) What total mass of 23U fuel would be

nuclear reactor accident in 1986. The 33Sr enters the body

28U nuclei that fission produce 3Sr, a B~ emitter with a

74

half-life of 29 yr. What is the total radioactivity of the $3Sr, in curies, produced in 1 yr? (Neglect the fact that some of it decays during the 1-yr period.) In the net reaction, Eq. 42-8, for the proton—proton cycle in the Sun, the neutrinos escape from the Sun with energy of about 0.5 MeV. The remaining energy, 26.2 MeV, is available within the Sun. Use this value to calculate the “heat of combustion” per kilogram of hydrogen fuel and compare it

. Three

reaches

Earth

from

the Sun

ingestion of

at a rate of about

and (b) the number of protons consumed per second in the

reaction of Eq. 42-8, assuming that this is the source of all the

Sun’s energy. (c) Assuming that the Sun’s mass of 2.0 x 10’ kg

be released in such a reaction? (b) What kinetic energy must two carbon nuclei each have when far apart, if they can then approach each other to within 6.0fm, center-to-center? (c) Approximately what temperature would this require? . An average adult body contains about 0.10uCi of {K, which comes from food. (¢) How many decays occur per second? (b) The potassium decay produces beta particles with energies of around 1.4 MeV. Estimate the dose per year in sieverts for a 55-kg adult. Is this a significant fraction of the 3.6-mSv/yr background rate? 80. When the nuclear reactor accident occurred at Chernobyl in 1986, 2.0 X 10’ Ci were released into the atmosphere. Assuming that this radiation was distributed uniformly over the surface of the Earth, what was the activity per square meter? (The actual activity was not uniform; even within Europe wet areas received more radioactivity from rainfall.) 81. A star with a large helium abundance can burn helium in

the

82.

reaction

jHe + 3He + jHe —

Q-value for this reaction?

!2C.

What

is

i Tc (Section 42-8) which decays by emitting

suspected of being a new forgery, and an examiner has decided to use neutron activation analysis to test this hypothesis. After placing the painting in a neutron flux

of 5.0 X 102 neutrons/cm?/s for 5.0 minutes, an activity of 55decays/s of $Co (7i =527yr) is observed. Assuming 39Co has a cross section of 19 bn, how much cobalt (in grams) does the paint contain?

isotope 23U.

12C nuclei into one Mg nucleus. (2) How much energy would

activity, 35 mCi.

of cobalt (37Co) than paint today. A certain “old” painting is

midnight on midsummer night. [Hint: See Problems 74 and 75.] Estimate how much total energy would be released via fission if 2.0kg of uranium were enriched to 5% of the

78. Some stars, in a later stage of evolution, may begin to fuse two

the same

86. Centuries ago, paint generally contained a different amount

180-m? ceiling of a room, at latitude 38°, for an hour around

71

have

a 140-keV gamma. The half-life for this decay is 6 hours. Assuming that about half the gamma photons exit the body without interacting with anything, what must be the initial activity of the Tc sample if the whole-body dose cannot exceed 50mrem? Make the rough approximation that biological elimination of Tc can be ignored.

1300 W/m?, Calculate (a) the total power output of the Sun,

was originally all protons and that all could be involved in nuclear reactions in the Sun’s core, how long would you expect the Sun to “glow” at its present rate? See Problem 74. 76. Estimate how many solar neutrinos pass through a

sources

8s. A 60-kg patient is to be given a medical test involving the

to the heat of combustion of coal, about 3 x 107 J/kg.

8. Energy

radioactive

Source A emits 1.0-MeV 7 rays, source B emits 2.0-MeV Y rays, and source C emits 2.0-MeV alphas. What is the relative danger of these sources?

87. Show, using the laws of conservation of energy and momentum, that for a nuclear reaction requiring energy, the minimum kinetic energy of the bombarding particle (the

threshold energy) is equal to [~Qmp/(mp; — my)], where

—Q is the energy required (difference in total mass between products and reactants), my, is the mass of the bombarding particle, and m, is the total mass of the products. Assume the target nucleus is at rest before an interaction takes place, and

that all speeds are nonrelativistic.

88. The early scattering experiments performed around 1910 in Ernest Rutherford’s laboratory in England produced the first evidence that an atom consists of a heavy nucleus surrounded by electrons. In one of these experiments, a particles struck a gold-foil target 4.0 X 107> cm thick in

which there were 5.9 X 10 gold atoms per cubic meter.

Although most a particles either passed straight through the foil or were scattered at small angles, approximately 1.6 X 1073 percent were scattered at angles greater than 90°—that is, in the backward direction. (a) Calculate the cross section, in barns, for backward scattering. (b) Rutherford concluded that such backward scattering could occur only if an atom consisted of a very tiny, massive, and positively charged nucleus with electrons orbiting some distance away. Assuming that backward scattering occurs for nearly direct collisions (i.e., o ~ area of nucleus), estimate the diameter of a gold nucleus.

the

A 1.2-uCi 3ICs source is used for 1.6hours by a 65-kg worker. Radioactive 21Cs decays by B~ decay with a half-

life of 30yr. The average energy of the emitted betas is about 190 keV per decay. The B decay is quickly followed by a Y with an energy of 660 keV. Assuming the person absorbs all emitted energy, what effective dose (in rem) is received?

Answers to Exercises

A: '3¥Ba. r!: 2 x 10V,

» '

T

Fa

o

El CONTENTS 43-1 High-Energy Particles and Accelerators 43-2 Beginnings of Elementary 43-3 43-4

43-5

43-6 43-7

pions.

Particles and Antiparticles Particle In_teractions and Conservation Laws

quarks. bosons.

Exchange

leptons.

Neutrinos—Recent Results

photons.

Particle Classification Particle Stability and Resonances

,Sr“angg Pagide?\; %hfl”m?

4310

The *Standard Model

Qou\:?;:a

Thus the elementary particles as we see them today are atoms and electrons. protons, neutrons, and electrons.

protons, neutrons, electrons, and photons.

ewode

quarks, leptons, and gauge bosons.

hadrons, leptons, and gauge bosons.

%’é’[’)‘;?nghmm‘)dy“amics Electroweak Theory 43-11 Grand Unified Theories 43-12 Strings and Supersymmetry

1164

Particl

_ . —Guess now! Electrons are still considered fundamental partxcl_es (in the group called leptons). But protons and neutrons are no longer considered fundamental; they have substructure and are made up of

Particle Physics—Particle

43-8 .

tary

|

n the final two Chapters of this book we discuss two of the most exciting areas of contemporary physics: elementary particles in this Chapter, and cosmology and astrophysics in Chapter 44. These are subjects at the forefron™™ of knowledge—elementary particles treats the smallest objects in the universe; cosmology treats the largest (and oldest) aspects of the universe.

In this penultimate Chapter we discuss elementary particle physics, which represents the human endeavor to understand the basic building blocks of all matter, and the fundamental forces that govern their interactions. By the mid1930s, it was recognized that all atoms can be considered to be made up of ’“tutrons, protons, and electrons. The basic constituents of the universe were no ionger considered to be atoms but rather the proton, neutron, and electron. Besides these three “elementary particles,” several others were also known: the positron (a positive electron), the neutrino, and the ¥ particle (or photon), for a total of six elementary particles. By the 1950s and 1960s many new types of particles similar to the neutron and proton were discovered, as well as many “midsized” particles called mesons whose masses were mostly less than nucleon masses but more than the electron mass. (Other mesons, found later, have masses greater than nucleons.) Physicists felt that all of these particles could not be fundamental, and must be made up of even smaller constituents (later confirmed by experiment), which were given the name quarks. Today, the basic constituents of matter are considered to be quarks (they make up protons and neutrons as well as mesons) and leptons (a class that includes electrons, positrons, and neutrinos). In addition, there are the “carriers of force” including gluons, the photon, and other “gauge bosons.” The theory that describes our present view is called the Standard Model. How we came to our present understanding of elementary particles is the subject of this Chapter. One of the exciting recent developments of the last few years is an emerging synthesis between the study of elementary particles and astrophysics (Chapter 44). In fact, recent observations in astrophysics have led to the conclusion that the greater part of the mass—energy content of the universe is not ordinary matter but two mysterious and invisible forms known as “dark matter” and “dark energy” which cannot be explained by the Standard Model in its present form. Indeed, we are now aware that the Standard Model is not sufficient. There are

problems and important questions still unanswered, and we will mention some of fngzm in this Chapter and how we hope to answer them.

43—-1

High-Energy Particles and Accelerators

In the years after World War II, it was found that if the incoming particle in a nuclear reaction has sufficient energy, new types of particles can be produced. The earliest experiments used cosmic rays—oparticles that impinge on the Earth from space. In the laboratory, various types of particle accelerators have been constructed to accelerate protons or electrons to high energies, although heavy ions can also be accelerated. These high-energy accelerators have been used to probe more deeply into matter, to produce and study new particles, and to give us information about the basic forces and constituents of nature. Because the projectile particles are at high energy, this field is sometimes called high-energy physics.

Wavelength and Resolution Particles accelerated to high energy can probe the interior of nuclei and nucleons or other particles they strike. An important factor is that faster-moving projectiles can reveal more detail. The wavelength of projectile particles is given by de Broglie’s wavelength formula (Eq. 37-7), h A > (43-1) showing that the greater the momentum p of the bombarding particle, the shorter its wavelength. As discussed in Chapter 35 on diffraction, resolution of details in images is limited by the wavelength: the shorter the wavelength, the finer the r\tail that can be obtained. This is one reason why particle accelerators of higher ~and higher energy have been built in recent years: to probe ever deeper into the structure of matter, to smaller and smaller size.

SECTION 43-1

High-Energy Particles and Accelerators

m

High resolution with electrons. What is the wavelength,

and hence the expected resolution, for 1.3-GeV electrons?

APPROACH

Because

1.3 GeV

is much

larger than

the electron mass, we must

be dealing with relativistic speeds. The momentum of the electrons is found fropfl

Eq. 36-13, and the wavelength is A = A/p. SOLUTION

Each

electron

has

K = 1.3GeV = 1300 MeV,

2500 times the rest energy of the electron

which

(mc? = 0.51 MeV).

is

abou

~

Thus we can

ignore the term (mc?) in Eq. 36-13, E* = p*c* + m’c*, and we solve for p: p

_

[E-mt c?

|E? _E c?

c

Therefore the de Broglie wavelength is

where

E = 1.3 GeV.

Hence

_ (6.63 X 107*7-5)(3.0 X 10°m/s)

© (13 x 10%eV)(1.6 X 1079 /eV)

e

cyclotron (we see the vacuum chamber enclosing it).

0.96 X 10 5m,

or 0.96 fm. This resolution of about 1 fm is on the order of the size of nuclei (see Eq. 41-1). NOTE The maximum possible resolution of this beam of electrons is far greater than for a light beam in a light microscope (A =~ 500 nm). —

| EXERCISE FIGURE 43-1 Ernest O. Lawrence, around 1930, holding the first

=

A What is the wavelength of a proton with K = 1.00 TeV?

Another major reason for building high-energy accelerators is that new part.. cles of greater mass can be produced at higher energies, transforming the kinetic energy of the colliding particles into massive particles by E = mc?, as we will discuss shortly. Now we look at particle accelerators.

Cyclotron FIGURE 43-2

Diagram of a

cyclotron. The magnetic field,

applied by a large electromagnet, points into the page. The protons start at A, the ion source. The red electric field lines shown are for the alternating electric field in the gap at a certain moment. Gap ]

g

e

4~

Path

2 |03

~_

/
ptp+tptp

B=+1+1=+1+4+1-1+1. As indicated, B = +2 on both sides of this equation. From these and other reactions, the conservation of baryon number has been established as a basic principle of physics. Also useful are conservation laws for the three lepton numbers, associated with weak interactions including decays. In ordinary B decay, an electron or positron is emitted long with a neutrino or antineutrino. In another type of decay, a particle known as a W or mu meson, or muon, can be emitted instead of an electron. The muon (discov-

ered in 1937) seems to be much like an electron, except its mass is 207 times larger

(106 MeV/c?). The neutrino (».) that accompanies an emitted electron is found to be different from the neutrino (v,) that accompanies an emitted muon.

A

CAUTION

The different types of neutrinos

are nof identical SECTION 43-4

1175

Each of these neutrinos has an antiparticle: ». and for example, n—->p+e + v but not n s

»,. In ordinary B decay we have,

p + e + v,. To explain why these do not occur, the concept o

electron lepton number, L., was invented. If the electron (e”) and the electron

neutrino (v,) are assigned L. = +1, and e and 7, are assigned L. = —1, whereas all other particles have L, = 0, then all observed decays conserve L. . For example, inn — p+e + ¥, initially L, = 0, and afterward L, = 0 + (+1) + (—=1) = 0. Decays that do not conserve L., even though they would obey the other conservation laws, are not observed to occur. In a decay involving muons, such as 7 > ut + oy, a second quantum number, muon lepton number (L,,), is conserved. The .~ and v, are

assigned L, = +1, and their antiparticles u " and v, have L, = —1, whereas all other particles have L, = 0. L, too is conserved in interactions and decays. Similar assign-

ments can be made for the tau lepton number, L., associated with the 7 lepton (discovered in 1976 with mass more than 3000 times the electron mass) and its neutrino, v, . Keep in mind that antiparticles have not only opposite electric charge from their particles, but also opposite B, L., L,, and L.. For example, a neutron has B = +1, an antineutron has B = —1 (and all the L’s are zero).

CONCEPTUAL EXAMPLE 43-5 | Lepton number in muon decay. Which of the

following By =

decay schemes is possible for muon decay: (a) u~ — e + v.; e + v +v,(c)p — e + v.?Allof these particles have L, = 0.

RESPONSE A y

has L, = +1 and L. = 0. This is the initial state for all decays

given, and the final state must also have

L, = +1,

indeed

to

has L,=0+0=0, this

decay

L,=0+0+1=+1

L. = 0. In (a), the final state

and L. = +1 — 1= 0; L, would not be conserved and

is not

and

observed

occur.

The

L.=+1-1+0=0,

final

so

both

state

L,

of

and

(b)

L.

has

arqq

conserved. This is in fact the most common decay mode of the u™. Lastly, (c) doe. not occur because L. (= +2 in the final state) is not conserved, nor is L,,.

Energy and momentum are conserved. In addition to the

“number” conservation laws which help explain the decay schemes of particles, we can also apply the laws of conservation of energy and momentum. The decay of a 2™ particle at rest with a mass of 1189 MeV/c?* (Table 43-2 in Section 43-6) commonly

yields a proton (mass = 938 MeV/c?) and a neutral pion, 7° (mass = 135 MeV/c?): St > p+7 What are the kinetic energies of the decay products, assuming the = particle was at rest?

parent

APPROACH We find the energy release from the change in mass (E = mc?) as

we did for nuclear processes (Eq. 41-3 or 42-2a), and apply conservation of energy and momentum. SOLUTION The energy released, or Q-value, is the change in mass times c*

Q = [my — (my + myp)|c* = [1189 — (938 + 135)]| MeV = 116 MeV.

This energy Q becomes the kinetic energy of the resulting decay particles, p and % Q0

=

K,

+

K

with each particle’s kinetic energy related to its momentum by (Egs. 36-11 and 13):

K, = E, — myc® = V(ppel + (mpc?)’ — myc?,

and similarly for the pion. From momentum conservation, the proton and pion have the same magnitude of momentum since the original =" was at rest: Pp = P»* = P-

Then, @ = Kp + K+ gives 116 MeV = [\/(pc)? + (938 MeV)? — 938 MeV]| +

[\/ (pc)? + (135MeV)? — 135 MeV]. We solve this for pc, which givesem, pc = 189 MeV. Substituting into the expression above for the kinetic energy, firsu for the proton, then for the pion, we obtain K, = 19MeV and K, = 97 MeV.

1176

CHAPTER 43

Elementary Particles

43-5

Neutrinos—Recent Results

We first met neutrinos with regard to B~ decay in Section 41-5. The study of neutrinos is a “hot” subject today. Experiments are being carried out in deep r inderground laboratories, sometimes in deep mine shafts. The thick layer of earth above is meant to filter out all other “background” particles, leaving mainly the very weakly interacting neutrinos to arrive at the detectors. Some very important results have come to the fore in recent years. First there was the solar neutrino problem. The energy output of the Sun is believed to be due to the nuclear fusion reactions discussed in Chapter 42, Eqs. 42-7 and 42-8. The neutrinos emitted in these reactions are all v, (accompanied by e*). But the rate at which v, arrive at Earth is measured to be much less than expected based on the power output of the Sun. It was then proposed that perhaps, on the long trip between Sun and Earth, v, might turn into », or v,. Subsequent experiments confirmed this hypothesis. Thus the three neutrinos, v, v,, v,, can change into one another in certain circumstances, a phenomenon called neutrino flavor oscillation (each of the three neutrino types being called, whimsically, a different “flavor™). This result suggests that the lepton numbers L., L, , and L, are not perfectly conserved. But the sum, L. + L, + L,, is believed to be always conserved. The second exceptional result has long been speculated on: are neutrinos massless as originally thought, or do they have a nonzero mass? Rough upper limits on the masses have been made. Today astrophysical experiments show that the sum of all three neutrino masses is less than about 0.14eV/c? But can the masses be zero? Not if there are the flavor oscillations discussed above. It seems likely that at least one neutrino type has a mass of at least 0.04 eV/c% As a result of neutrino oscillations, the three types of neutrino may not be exactly what we thought they were (e, u, 7). If not, the three basic neutrinos, called 1, 2, and 3, are linear combinations of v., v, , and v,. Another outstanding question is whether or not neutrinos are in the category alled Majorana particles,” meaning they would be their own antiparticles. If so, a (ot of other questions (and answers) would appear.

*Neutrino Mass Estimate from a Supernova The supernova of 1987 offered an opportunity to estimate electron neutrino mass. If neutrinos

do

have

mass, then

v < ¢ and

neutrinos

of different

energy

would

take different times to travel the 170,000 light-years from the supernova to Earth. To get an idea of how such a measurement could be done, suppose two neutrinos from “SN1987a” were emitted at the same time and detected on Earth (via the reaction .+ p — n+e’) 10seconds apart, with measured kinetic energies of about 20 MeV and 10 MeV. Since we expect the neutrino mass to be surely less than 100 eV (from other laboratory measurements), and since our neutrinos have kinetic energy of 20 MeV and 10 MeV, we can make the approximation m,c* s, then its mass (i.e., its rest energy) will be uncertain by an amount

AE ~ h/(2m At) ~ (6.6 X 107*7:5)/(6)(10%s)

~ 1071J ~ 100MeV,

which is what is observed. Actually, the lifetimes of ~ 107®s for such resonances are inferred by the reverse process: from the measured width being ~ 100 MeV.

43-8

Strange Particles? Charm?

Towards a

New Model

In the early 1950s, the newly found particles K, A, and 2 were found to behave rather strangely in two ways. First, they were always produced in pairs. For example, the reaction

r

™

+p—>

K

+ A°

occurred with high probability, but the similar reaction 7~ + p > K° + n, was never observed to occur even though it did not violate any known conservation law. The second feature of these strange particles, as they came to be called, was that they were produced via the strong interaction (that is, at a high interaction rate), but did not decay at a fast rate characteristic of the strong interaction (even though they decayed into strongly interacting particles). To explain these observations, a new quantum number, strangeness, and a new conservation law, conservation of strangeness, were introduced. By assigning the strangeness numbers () indicated in Table 43-2, the production of strange particles in pairs was explained. Antiparticles were assigned opposite strangeness from their particles. For example, in the reaction 7~ + p — K® + A°, the initial state has

r

strangeness

S = 0 + 0 = 0,

and

the

final state

has

S = +1

-1

=0,

so

strangeness is conserved. But for 7~ + p > K’ + n, the initial state has S = 0 and the final state has S = +1+ 0= +1, so strangeness would not be conserved; and this reaction is not observed. To explain the decay of strange particles, it is assumed that strangeness is conserved in the strong interaction but is not conserved in the weak interaction. Thus, strange particles were forbidden by strangeness conservation to decay to nonstrange particles of lower mass via the strong interaction, but could dec?y by means of the weak interaction at the observed longer lifetimes of 1071° to 10™"s. The conservation of strangeness was the first example of a partially conserved quantity. In this case, the quantity strangeness is conserved by strong interactions but not by weak.

SECTION 43-8

A CAUTION

Partially conserved quantities

Strange Particles? Charm? Towards a New Model

CONCEPTUAL EXAMPLE 43-8 | Guess the missing particle. Using the conser-

vation laws for particle interactions, determine the possibilities for the missing particle in the reaction ™

+p—>

K

+2?

-

in addition to K® + A% mentioned above.

'

RESPONSE We write equations for the conserved numbers in this reaction, with B, L., S, and Q as unknowns whose determination will reveal what the possible particle might be: Baryon number: 0+1=0+8B Lepton number: 0+0 =0+ L, Charge: -1+1=0+0 Strangeness: 0+0=1+S§S. The unknown product particle would have to have these characteristics: B

=

+1

L.

=0

0

=0

S

=

-1

In addition to A° a neutral sigma particle, =°, is also consistent with these numbers. In the next Section we will discuss another partially conserved quantity which was given the name charm. The discovery in 1974 of a particle with charm helped solidify a new theory involving quarks, which we now discuss.

43-9

Quarks

All particles, except the gauge bosons (Section 43-6), are either leptons or hadrons. One difference between these two groups is that the hadrons interact via the strong interaction, whereas the leptons do not.

There is another major difference. The six leptons (e, u™, 77, v, v,, v,) are

considered to be truly fundamental particles because they do not show any internal structure, and have no measurable size. (Attempts to determine the size of leptor have put an upper limit of about 107'® m.) On the other hand, there are hundreds or hadrons, and experiments indicate they do have an internal structure. In 1963, M. Gell-Mann and G. Zweig proposed that none of the hadrons, not even the proton and neutron, are truly fundamental, but instead are made up of combinations of three more fundamental pointlike entities called (somewhat whimsically) quarks.” Today, the quark theory is well-accepted, and quarks are considered truly fundamental particles, like leptons. The three quarks

originally proposed

down, and strange. The based on a presumed charmed, bottom, and (quantum numbers c, b, Table 43-3), and which interactions.

were labeled u, d, s, and have the names

up,

theory today has six quarks, just as there are six leptons— symmetry in nature. The other three quarks are called top. The names apply also to new properties of each f) that distinguish the new quarks from the old quarks (see (like strangeness) are conserved in strong, but not weak,

"Gell-Mann chose the word from a phrase in James Joyce’s Finnegans Wake.

TABLE 43-3

Properties of Quarks (Antiquarks have opposite sign @ B, S, ¢, t, b) Quarks

Name

Up Down Strange Charmed Bottom Top 1182

CHAPTER 43

Symbol

(MeV/c?)

Mass

Charge

Baryon Number

Strangeness

Charm

Bottomness

Topness

u d s ¢ b t

2 5 95 1250 4200 173,000

+2e —1e —le +2e —ie +2e

i ! ! i ! !

0 0 -1 0 0 0

0 0 0 +1 0 0

0 0 0 0 -1 0

0 0 0 0 0 +1

Elementary Particles

0

B

c

b

t

&

TABLE 43-4 Category

Partial List of Heavy Hadrons, with Charm and Bottomness (L, = L,=L,=0)

Particle

(k Aesons

Spin

D~ D’ Dy

0 0 0

1869.4 1864.5 1968

0 0 0

D' D’ D}

Baryons

Mass (MeV/c?

Baryon Number B

Antiparticle

Strangeness S

+1

Charm c

Bottomness b

Mean life (s)

0 0

+1 +1 +1

0 0 0

106 X 1078 41 x 107 50 x 1078

Principal Decay Modes

K + others, e + others K + others, wor e + others K + others

I/ (3097)

Self

1

3096.9

0

0

0

0

~ 107

Hadrons, e*e”,

Y (9460)

Self

1

9460

5279 5279

0

0 0

0

0

0

~ 107

Hadrons, u*u~, e'e”, 7'~

2286 2454 2453 2454 5620

+1 +1 +1 +1 +1

B~ B’

B* B’

0 0

AJ

AF ) by b Ap

1 : i i i

> ig 33 =0 AY

0 0

0 0

-1 -1

0 0 0 0 0

+1 +1 +1 +1 0

0 0 0 0 -1

p u”

1.6 X 1072 1.5x 1072

DY + others DY + others

2.0 X 107* =102 ~ 1071 ~ 107 12X 1072

Hadrons (e.g., A + others) Afpt Afn® Aln” J/gAS, pD%~, Afwtnm”

All quarks have spin } and an electric charge of either +3e or —le

(that is, a fraction of the previously thought smallest charge e). Antiquarks have opposite sign of electric charge @, baryon number B, strangeness S, charm c, bottomness b, and topness t. Other properties of quarks are shown in Table 43-3.

All hadrons are considered to be made up of combinations of quarks (plus the

gluons that hold them together), and their properties are described by looking at

their quark content. Mesons consist of a quark-antiquark pair. For example,

a 7

meson

is a ud

Q=%e+ie=+le, anda

K" = us, with

combination:

B=}i-1=0,

Q= +1,

note

that

for the

S=0+0=0,

B=10, S = +1.

ud

pair

(Table

43-3),

as they must for a ="

Baryons, on the other hand, consist of three quarks. For example, a neutron is

f‘ = ddu, whereas an antiproton is p = uud. See Fig. 43-15. Strange particles all contain an s or s quark, whereas charmed particles contain a c or ¢ quark. A few of these hadrons are listed in Table 43-4. After the quark theory was proposed, physicists began looking for these fractionally charged particles, but direct detection has not been successful. Current models suggest that quarks may be so tightly bound together that they may not ever exist singly in the free state. But observations of very high energy electrons scattered off protons suggest that protons are indeed made up of constituents. Today, the truly fundamental particles are considered to be the six quarks, the six leptons, and the gauge bosons that carry the fundamental forces. See Table 43-5, where the quarks and leptons are arranged in three “families” or “generations.” Ordinary matter—atoms made of protons, neutrons, and electrons— is contained in the “first generation.” The others are thought to have existed in : the. very egrly universe, but are seen by us today only at powerful acceleratmjs or in cosmic rays. All of the hundreds of hadrons can be accounted for by combinations of the six quarks and six antiquarks.

FIGURE 43-15

Quark

=~ compositions for several particles. o B

e B

3',?/. ‘\ 3¢ \

~

=3F

+ % e -0

%&

(@ .\) \\\. S _

42,7 ‘ Ry

¢ )

vt

Neutron

pa

N ‘39/@ @“ 3€ = . ) i +2e/7 3? ‘ N\+i ., 3e \ ) N’

EXERCISE D Return to the Chapter-Opening Questions, page 1164, and answer them again now. Try to explain why you may have answered differently the first time.

TABLE 43-5 The Fundamental Particlest as Seen Today Gauge bosons

Force

Gluons

Strong

™ 7 (photon)

EM

W7,

Weak

First generation Quarks

Leptons

u,d

€, Ve

Second generation s,C

By vy

Third generation b, t

TV,

"The quarks and leptons are arranged into three generations each.

SECTION 43-9

Quarks

1183

CONCEPTUAL EXAMPLE 43-9 | Quark

combinations.

Find

the

baryon

number, charge, and strangeness for the following quark combinations, and identify the hadron particle that is made up of these quark combinations: (@) udd, (b) u, (c) uss, (d) sdd, and (e) bu. RESPONSE We use Table 43-3 to get the properties of the Table 43-2 or 43-4 to find the particle that has these properties. (a) udd has

quarks,

the.q

QB=1+ = +le—to -}= S=0+0+0

=0,

as well as ¢ = 0, bottomness = 0, topness = 0. The only baryon that has Q = 0, § = 0, etc,, is the neutron (Table 43-2).

(B = +1)

(b) ui has Q=3%e—3e=0, B=31-1=0, and all other numbers = 0. Sounds like a 7° (dd also gives a 7°). (c)usshas Q =0, B= +1, S = -2, others = 0. This is a .

(d)ysddhas Q = =1, B= +1, S = -1, somustbea X", () bu has Q= -1, B=0, §=0, ¢ =0, bottomness = —1, This must be a B~ meson (Table 43-4).

quantum

topness = 0.

| EXERCISE E What is the quark composition of a K~ meson?

43—10 The “Standard Model”: Quantum Chromodynamics (QCD) and Electroweak Theory Not long after the quark theory was proposed, it was suggested that quarks have another property (or quality) called color, or “color charge” (analogous to electric charge). The distinction between the six types of quark (u, d, s, c, b, t) was referr to as flavor. According to theory, each of the flavors of quark can have three colors, usually designated red, green, and blue. (These are the three primary colors which, when added together in appropriate amounts, as on a TV screen, produce white.) Note that the names “color” and “flavor” have nothing to do with our senses, but are purely whimsical—as are other names, such as charm, in this new field. (We did, however, “color” the quarks in Fig. 43-15.) The antiquarks are colored antired, antigreen, and antiblue. Baryons are made up of three quarks, one of each color. Mesons consist of a quark-antiquark pair of a particular color and its anticolor. Both baryons and mesons are thus colorless or white. Originally, the idea of quark color was proposed to preserve the Pauli exclusion principle (Section 39-4). Not all particles obey the exclusion principle. Those that do, such as electrons, protons, and neutrons, are called fermions. Those that don’t are called bosons. These two categories are distinguished also in their spin (Section 39-2): bosons have integer spin (0, 1, 2, etc.) whereas fermions have halfinteger spin, usually } as for electrons and nucleons, but other fermions have spin 3,3, etc. Matter is made up mainly of fermions, but the carriers of the forces (Y, W, Z, and gluons) are all bosons. Quarks are fermions (they have spin 1) and therefore should obey the exclusion principle. Yet for three particular baryons (uuu, ddd, and sss), all three quarks would have the same quantum numbers, and

at least two quarks have their spin in the same direction (since there are only two

choices, spin up [m; = +3| or spin down [m, = —}]). This would seem to violate

1184

CHAPTER 43

the exclusion principle; but if quarks have an additional quantum number (color), which is different for each quark, it would serve to distinguish them and allow the exclusion principle to hold. Although quark color, and the resulting threefold increase in the number of quarks, was originally an ad hoc idea, it also served ‘t% bring the theory into better agreement with experiment, such as predicting t correct lifetime of the #° meson, and the measured rate of hadron production in observed e*e™ collisions at accelerators. The idea of color soon became a central feature of the theory as determining the force binding quarks together in a hadron.

u (blue)

d (red)

p

n

d (blue)

n

p

Gluon u (red)

(a)

(b)

FIGURE 43-16 (a) The force between two quarks holding them together as part of a proton, for example, is carried by a gluon, which in this case involves a change in color. (b) Strong interaction n + p — n + p with the exchange of a charged 7 meson (+ or —, depending on whether it is considered moving to the left or to the right). (c) Quark representation of the same interaction n + p — n + p. The blue coiled lines between quarks represent gluon exchanges holding the hadrons together. (The exchanged meson may be regarded as ud emitted by the n and absorbed by the p, or as ud emitted by p and absorbed by n, because a u (or d) quark going to the left in the diagram is equivalent to a U (or d) going to the right.)

Each quark is assumed to carry a color charge, analogous to electric charge, and the strong force between quarks is referred to as the color force. This theory of the strong force is called quantum chromodynamics (chroma = color in Greek),

or QCD, to indicate that the force acts between color charges (and not between,

say, electric charges). The strong force between two hadrons is considered to be a force between the quarks that make them up, as suggested in Fig. 43-16. The particles that transmit the color force (analogous to photons for the EM force) are called gluons (a play on “glue”). They are included in Tables 43-2 and 43-5. There are eight gluons, according to the theory, all massless and all have color charge.” You might ask what would happen if we try to see a single quark with color by .caching deep inside a hadron and extracting a single quark. Quarks are so tightly bound to other quarks that extracting one would require a tremendous amount of energy, so much that it would be sufficient to create more quarks (E = mc?). Indeed, such experiments are done at modern particle colliders and all we get is more hadrons (quark—antiquark pairs, or triplets, which we observe as mesons or baryons), never an isolated quark. This property of quarks, that they are always bound in groups that are colorless, is called confinement. The color force has the interesting property that, as two quarks approach each other very closely (equivalently, have high energy), the force between them

becomes small. This aspect is referred to as asymptotic freedom.

The weak force, as we have seen, is thought to be mediated by the W*, W, and Z° particles. It acts between the “weak charges” that each particle has. Each elementary particle can thus have electric charge, weak charge, color charge, and gravitational mass, although one or more of these could be zero. For example, all leptons have color charge of zero, so they do not interact via the strong force. CONCEPTUAL EXAMPLE 43-10 | Beta decay. Draw showing what happens in beta decay using quarks.

a

Feynman

diagram,

RESPONSE Beta decay is a result of the weak interaction, and the mediator is either a W* or Z° particle. What happens, in part, is that a neutron (udd quarks) decays into a proton (uud). Apparently a d quark (charge — 1e) has turned into a u quark (charge +%e). Charge conservation means that a negatively charged particle, namely a W™, was emitted by the d quark. Since an electron and an antineutrino appear in the final state, they must have come from the decay of the virtual W, as shown in Fig. 43-17. Compare to the EM interaction, where the photon has no electric charge. Because gluons have color charge, they could attract each other and form composite particles (photons cannot). Such “glueballs” are being searched for.

SECTION 43-10

FIGURE 43-17 Quark representation of the Feynman diagram for B decay of a neutron into a proton. P /’_JH

udu

e~ L

Fe

'llw_

udd

iyl

The “Standard Model": Quantum Chromodynamics (QCD) and Electroweak Theory

1185

To summarize, the Standard Model says that the truly fundamental particles (Table 43-5) are the leptons, the quarks, and the gauge bosons (photon, W and Z, and the gluons). The photon, leptons, W*, W™, and Z° have all been observed in experiments. But so far only combinations of quarks (baryons and mesons) havefl been observed, and it seems likely that free quarks and gluons are unobservable. One important aspect of theoretical work is the attempt to find a unified basis for the different forces in nature. This was a long-held hope of Einstein, which he was never able to fulfill. A so-called gauge theory that unifies the weak and electromagnetic interactions was put forward in the 1960s by S. Weinberg, S. Glashow, and A. Salam. In this electroweak theory, the weak and electromagnetic forces are seen

as two different manifestations of a single, more

interaction.

prediction

The

electroweak

of the W*

theory

has

had

particles as carriers

many

fundamental, electroweak

successes,

of the weak

including

the

force, with masses

of 80.38 + 0.02GeV/c? in excellent agreement with the measured values of 80.403 + 0.029 GeV/c? (and similar accuracy for the Z°). The combination of electroweak theory plus QCD for the strong interaction is often referred to today as the Standard Model.

EXAMPLE 43-11

Range of weak force. The weak nuclear force

is of very short range, meaning it acts over only a very short distance. Estimate its range using the masses (Table 43-2) of the W* and Z: m ~ 80 or

90 GeV/c* ~ 10> GeV/c%

APPROACH We assume the W* or Z° exchange particles can exist for a time At given by the uncertainty principle, At ~ #/AE, where AE ~ mc? is the energy needed to create the virtual particle (W

N(4yr),

where N is the number of protons in 3300 tons of water. To determine N, we note that each molecule of H,O contains 2 + 8 = 10 protons. So one mole of water

(18 g, 6 X 10 molecules) contains 10 X 6 X 10* protons in 18 g of water, or

about 3 X 10 protons per kilogram. One ton is 10° kg, so the chamber contair;\

FIGURE 43-19

(3.3 X 10°kg)(3 X 10% protons/kg) ~ 1 x 10* protons. Then our very rough est mate for a lower limit on the proton mean life is 7 > (10%)(4yr) = 4 X 10¥yr.

Time and energy

plot of the four fundamental forces, unified at the Planck time, and how each condensed out. The symbol

GUT

and Cosmology

t,ps = time after the birth of the universe. Note that the typical

Al'l interesting pred%ction of. unified theories relates. to cqsmology (Chapter 44). It is thought that during the first 107> s after the theorized Big Bang that created the

temperature of the universe) decreases to the right, as time after

corresponding to the unification scale. Baryon number would not have been conserved then, perhaps allowing an imbalance that might account for the observed

particle energy (and average

universe,

the Big Bang increases. We discuss

Relative strength

TK)

: .

Tabu (8)

109

.

014

.

1010

1025

107435

106

was

so

extremely

high

that

particles

had

energies

that we are surrounded by matter, with no significant antimatter in sight, is considered

Grand Unification

(GeV)EI'O“?

temperature

predominance of matter (B > 0) over antimatter (B < 0) in the universe. The fact

the Big Bang in the next Chapter.

T—

the

102 1

1 013

a problem in search of an explanation (not given by the Standard Model). See also Chapter 44. We call this the matter-antimatter problem. To understand it may require still undiscovered phenomena—perhaps related to quarks or neutrinos, or the Higgs boson or supersymmetry (next Section). This last example is interesting, for it illustrates a deep connection between investigations at either end of the size scale: theories about the tiniest objects (elementary particles) have a strong bearing on the understanding of the universe on a large scale. We will look at this more in the next Chapter. Figure 43-19 is a rough diagram indicating how the four fundamental forces in nature “condensed out” (a symmetry was broken) as time went on after the Big Bang (Chapter 44), and as the mean temperature of the universe and the typical

particle energy decreased.

10-1's This is much larger than the age of the universe (= 14 X 10°yr). But we don’t have to wait 10° yr

see. Instead we can wait for one decay among 10°' protons over a year (see Eqs. 41-4 and 41-9...

AN = AN At = N At/7).

1188

CHAPTER 43

Elementary Particles

43-12

Strings and Supersymmetry

We have seen that the Standard Model is unable to address important experimental f‘ssues, and that theoreticians are attacking the problem as experimenters search for new data, new particles, new concepts. Even more ambitious than grand unified theories are attempts to also incorporate gravity, and thus unify all four forces in nature into a single theory. (Such theories are sometimes referred to misleadingly as theories of everything.) There are consistent theories that attempt to unify all four forces called string theories, in which each fundamental particle (Table 43-5) is imagined not as a point but as a one-dimensional string, perhaps 1075 m long, which vibrates in a particular standing wave pattern. (You might say each particle is a different note on a tiny stretched string.) More sophisticated theories propose the fundamental entities as being multidimensional branes (after 2-D membranes). A related idea that also goes way beyond the Standard Model is supersymmetry, which applied to strings is known as superstring theory. Supersymmetry, developed by Bruno Zumino (1923- ) and Julius Wess (1934-2007), predicts that interactions exist that would change fermions into bosons and vice versa, and that each known fermion would have a supersymmetric boson partner of the same mass. Thus, for each quark (a fermion), there would be a squark (a boson) or “supersymmetric” quark. For every lepton there would be a slepton. Likewise, for every known boson (photons and gluons, for example), there would be a supersymmetric fermion (photinos and gluinos). Supersymmetry predicts also that a graviton, which transmits the gravity force, has a partner, the gravitino. Supersymmetric particles are sometimes called “SUSYs” for short, and are a candidate for the “dark matter” of the universe (discussed in Chapter 44). But why hasn’t this “missing part” of the universe ever been detected? The best guess is that supersymmetric particles might be heavier than their conventional counterparts, perhaps too rheavy to have been produced in today’s accelerators. A search for supersymmetric particles is already in the works for CERN’s new Large Hadron Collider. Versions of supersymmetry predict other interesting properties, such as that space has 11 dimensions, but 7 of them are “coiled up” so we normally only notice the 4-D of space-time. We would like to know if and how many extra dimensions there are, and how and why they are hidden. We hope to have some answers from the new LHC and the future ILC (Section 43-1). The world of elementary particles is opening new vistas. What happens in the future is bound to be exciting.

| Summary Particle accelerators are used to accelerate charged particles, such as electrons and protons, to very high energy. High-energy particles have short wavelength and so can be used to probe the structure of matter in great detail (very small distances). High kinetic energy also allows the creation of new particles through collisions (via E = mc?). Cyclotrons and synchrotrons use a magnetic field to keep the particles in a circular path and accelerate them at intervals by high voltage. Linear accelerators accelerate particles along a line. Colliding beams allow higher interaction energy. An antiparticle has the same mass as a particle but opposite charge. Certain other properties may also be opposite: for example, the antiproton has baryon number (nucleon number) opposite (B = —1) to that for the proton (B = +1). In all nuclear and particle reactions, the following conservation laws hold: momentum, angular momentum, mass—energy, electric charge, baryon number, and lepton numbers. Certain particles have a property called strangeness, which 1s conserved by the strong force but not by the weak force. The properties charm, bottomness, and topness also are conserved by the strong force but not by the weak force.

Just as the electromagnetic force can be said to be due to an exchange of photons, the strong nuclear force is carried by massless gluons. The W and Z particles carry the weak force. These fundamental force carriers (photon, W and Z, gluons) are called gauge bosons. Other particles can be classified as either leptons or hadrons. Leptons participate only in gravity, the weak, and the electromagnetic interactions. Hadrons, which today are considered to be made up of quarks, participate in all four interactions, including the strong interaction. The hadrons can be classified as mesons, with baryon number zero, and baryons, with nonzero baryon number.

All particles, except for the photon, electron, neutrinos, and proton, decay with measurable mean lives varying from 10> s to 10® 5. The mean life depends on which force is predominant. Weak

decays usually have mean lives greater than about 101* s, Electro-

magnetic decays typically have mean lives on the order of 107 to 10715, The shortest lived particles, called resonances, decay via

the strong interaction and live typically for only about 107,

Today’s Standard Model of elementary particles considers quarks as the basic building blocks of the hadrons. The six quark

Summary

1189

“flavors”

are

called

up, down,

strange, charmed,

bottom,

and

top. It is expected that there are the same number of quarks as leptons (six of each), and that quarks and leptons are the truly fundamental particles along with the gauge bosons (¥, W, Z, gluons). Quarks are said to have color, and, according to quantum chromodynamics (QCD), the strong color force acts between their color charges and is transmitted by gluons. Electroweak theory views the weak and electromagnetic forces as two aspects of a single underlying interaction. QCD plus the electroweak theory are referred to as the Standard Model.

Grand unified theories of forces suggest that at very short distance (10‘31 m) and very high energy, the weak, electromagnetic, and strong forces appear as a single force, and the fundamental difference between quarks and leptons disappears.

According to string theory, the fundamental particles ma~*™

be tiny strings, 107 m long, distinguished by their standing

wave pattern. Supersymmetry hypothesizes that each fermion (or boson) has a corresponding boson (or fermion) partner.

I Questions

1. Give a reaction between two nucleons, similar to Eq. 43-4, that could produce a 7. 2. If a proton is moving at very high speed, so that its kinetic

energy is much greater than its rest energy (mc?), can it then decayvia p = n + 77?

3. What would an “antiatom,” made up of the antiparticles to the constituents of normal atoms, consist of? What might happen if antimatter, made of such antiatoms, came in contact with our normal world of matter? 4. What particle in a decay signals the electromagnetic interaction? 5. (a) Does the presence of a neutrino among the decay products of a particle necessarily mean that the decay occurs via the weak interaction? (b) Do all decays via the weak interaction produce a neutrino? Explain. 6. Why is it that a neutron decays via the weak interaction even though the neutron and one of its decay products (proton) are strongly interacting? 7. Which of the four interactions (strong, electromagnetic, weak, gravitational) does an electron take part in? A neutrino? A proton? 8. Check that charge and baryon number are conserved in each of the decays in Table 43-2. 9. Which of the particle decays listed in Table 43-2 occur via the electromagnetic interaction? 10. Which of the particle decays listed in Table 43-2 occur by the weak interaction?

11. The A baryon has spin 3» baryon number 1, and charge Q = +2, +1, 0, or —1.Why s there no charge state Q = —2? 12. Which of the particle decays in Table 43-4 occur via the electromagnetic interaction? 13. Which of the particle decays in Table 43-4 occur by the weak interaction? 14. 15.

Quarks

have spin 3 How

do you

account for the fact

that baryons have spin 3 or 3, and mesons have spin 0 or 1? Suppose there were a kind of “neutrinolet” that was massless, had no color charge or electrical charge, and did not feel the weak force. Could you say that this particle even exists?

16. Is it possible for a particle to be both (a) a lepton and a baryon? (b) a baryon and a hadron? (c) a meson and a quark? (d) a hadron and a lepton? Explain. 17. Using the ideas of quantum chromodynamics, would it be possible to find particles made up of two quarks and no antiquarks? What about two quarks and two antiquarks?

18, Why can neutrons decay when they are free, but not whe *™ they are inside a stable nucleus?

19

Is the reaction

20

Occasionally, the A will decay by the following reaction: A% - p* + e + V.. Which of the four forces in nature is responsible for this decay? How do you know?

e + p —

n + ¥, possible? Explain.

| Problems 43-1

Particles and Accelerators

1. (I) What is the total energy of a proton whose energy is 4.65 GeV? 2. (I) Calculate the wavelength of 28-GeV electrons.

kinetic

9. (II) What magnetic field is required for the 7.0-TeV protons in the 4.25-km-radius Large Hadron Collider (LHC)? 10. (II) A cyclotron with a radius of 1.0m is to accelerate

deuterons ({H) to an energy of 12MeV. (a) What is the

required magnetic field? (b) What frequency is needed for the voltage between the dees? (c) If the potential difference between the dees averages 22 kV, how many revolutions will the particles make before exiting? (d) How much time does it take for one deuteron to go from start to exit? (e) Estimate how far it travels during this time.

3. (I) What strength of magnetic field is used in a cyclotron in which protons make 3.1 X 107 revolutions per second? 4. (I) What is the time for one complete revolution for a very high-energy proton in the 1.0-km-radius Fermilab accelerator? 5. (I) If a particles are accelerated by the cyclotron of Example 43-2, what must be the frequency of the voltage applied to the dees? 6. (II) (a) If the cyclotron of Example 43-2 accelerated a particles, what maximum energy could they attain?

What would their speed be? (b) Repeat for deuterons ({H). (c) In each case, what frequency of voltage is required?

7. (II) Which is better for resolving details of the nucleus: 25-MeV alpha particles or 25-MeV protons? Compare each of their wavelengths with the size of a nucleon in a nucleus. 8. (II) What magnetic field intensity is needed at the 1.0-kmradius Fermilab synchrotron for 1.0-TeV protons?

1190

CHAPTER 43

Elementary Particles

11.

(I)

What

is

the

wavelength

(= minimum

resolvable

size) of 7.0-TeV protons? 12. (II) The 1.0-km radius Fermilab Tevatron takes about 20 seconds to bring the energies of the stored protons from 150 GeV to 1.0 TeV. The acceleration is done once per turn. Estimate the energy given to the protons on each turn. (You can assume that the speed of the protons is essentially ¢ the whole time.) 13. (ITI) Show that the energy of a particle (charge e) in afl synchrotron, in the relativistic limit (v = c), is given by E (ineV) = Brc, where B is the magnetic field and r is the radius of the orbit (SI units).

43-2 to 43-6

Particle Interactions, Particle Exchange

14. (I) About how much energy is released when a A° decays to

n + 7% (See Table 43-2.)

(I) How much energy is released in the decay

7t = ut + w2

16. 17. 18. 19.

20.

2L

See Table 43-2. (I) Estimate the range of the strong force if the mediating particle were the kaon in place of a pion. (I) How much energy is required to produce a neutron— antineutron pair? (IT) Determine the energy released when = decays to A° and then to a proton. (IT) Two protons are heading toward each other with equal speeds. What minimum kinetic energy must each have if a 7" meson is to be created in the process? (See Table 43-2.) (IT) What minimum Kinetic energy must two neutrons each have if they are traveling at the same speed toward each other, collide, and produce a K"K~ pair in addition to themselves? (See Table 43-2.)

(I1) For the decay

K® — 7~ + e’ + v., determine

the

maximum kinetic energy of (a) the positron, and (b) the 7.

Assume the KU is at rest.

26.

27. 28.

ptp-op+tp+tp+p

p+tp—opt+te +e +p

. (IIT) Calculate when a muon [Hint: In what to the electron kinetic energy?

the maximum kinetic energy of the electron decays from rest via p~ — e + v + v,. direction do the two neutrinos move relative in order to give the electron the maximum Both energy and momentum are conserved;

use relativistic formulas.]

32. (III) Could a 7" meson be produced if a 110-MeV proton struck a proton at rest? What minimum kinetic energy must the incoming proton have?

43-7 to 43-11 Resonances, Standard Model, Quarks, QCD, GUT 33

(I) The mean life of the = particle is 7 X 1072°s, What is the

uncertainty in its rest energy? Express your answer in MeV. 34. (I) The measured width of the i (3686) meson is about 300 keV. Estimate its mean life. 35. (I) The measured width of the J/¢ meson is 88 keV. Estimate its mean life. 36. (I) The B™ meson is a bu quark combination. (a) Show that this is consistent for all quantum numbers. (b) What are the

(IT) For the decay A° — p + 7, calculate (a) the Q-value (energy released), and (b) the kinetic energy of the p and 7, assuming the A° decays from rest. (Use relativistic formulas.) (IT) (a) Show, by conserving momentum and energy, that it is impossible for an isolated electron to radiate only a single photon. (b) With this result in mind, how can you defend the photon exchange diagram in Fig. 43-87 (IT) What would be the wavelengths of the two photons produced when an electron and a positron, each with 420keV of kinetic energy, annihilate in a head-on collision? (II) In the rare decay #* — e* + v, what is the kinetic energy of the positron? Assume the 7~ decays from rest. (IT) Which of the following reactions and decays are possible? For those forbidden, explain what laws are violated. (@7 +p—->n+x° b)yn"+p—o>n+a°

37. (I) What is the energy width (or uncertainty) of (a) n°, and (b) p™? See Table 43-2. 38. (IT) Which of the following decays are possible? For those that are forbidden, explain which laws are violated.

(@) A°> n + 7~ (b) A p + K~

25.

ptp—=>p+p ptp—opt+tp+p

the two photons produced rest annihilate? following reactions. What each of the reactions?

22. (IT) What are the wavelengths when a proton and antiproton 23. (IT) The A° cannot decay by conservation laws are violated

€.

29. (IT) Calculate the kinetic energy of each of the two products in the decay 2~ — A® + 7. Assume the E~ decays from rest. 30. (IT) Antiprotons can be produced when a proton with sufficient energy hits a stationary proton. Even if there is enough energy, which of the following reactions will not happen?

()

A’

of at the in

7" + 7~

quark combinations for B¥, B%, B%?

Hp—o>n+e

()

A"+ v+

-

22>

20+

7+

. (II) What quark combinations produce (a) a E° baryon and (b) a E” baryon? . (II) What are the quark combinations that can form (a) a

neutron, (b) an antineutron, (c) a A%, (d) a =%

. (II) What particles do the following quark combinations produce: (a) uud, (b) uus, (c) us, (d) du, (e) cs? . (II) What is the quark combination needed to produce a

Dmeson(Q=B=S8=0,

¢c = +1)?

. (II) The D§ meson has S = ¢ = +1, B = 0. What quark

combination would produce it? . (IT) Draw a possible Feynman diagram using quarks (as in

Fig. 43-16c¢) for the reaction 7~ + p — #° + n.

+r

() Wt > e+,

3T + 7~

bB)YQ

)7 +p—>p+e’ dp—>e

(@) B -

. (I)

+u

n+v,

Draw

a

>p+tu.

Feynman

diagram

for

the

reaction

| General Problems r

46. The mean lifetimes listed in Table 43-2 are in terms of proper time, measured in a reference frame where the particle is at rest. If a tau lepton is created with a kinetic energy of 950 MeV, how long would its track be as measured in the lab, on average, ignoring any collisions?

47.

Assume there are 5.0 X 10" protons at 1.0 TeV stored in

the 1.0-km-radius ring of the Tevatron. (¢) How much current (amperes) is carried by this beam? (b) How fast would a 1500-kg car have to move to carry the same kinetic energy as this beam? General Problems

1191

. (a) How much energy is released when an electron and a positron annihilate each other? (b) How much energy is released when a proton and an antiproton annihilate each other? (All particles have K = 0.) 49. Protons are injected into the 1.0-km-radius tron with an energy of 150 GeV. If they by 2.5 MV each revolution, how far do approximately how long does it take for 1.0 TeV?

Fermilab Tevaare accelerated they travel and them to reach

50. Which of the following reactions are possible, and by what interaction could they occur? For those forbidden, explain why. @mnm +p->K+p+a°

(bB)K

+p

@K

»

)

K'+n—>

> A2+ 7°

7+

" +a%+7 7%+

7"

59. A proton and an antiproton annihilate each other at rest and produce two pions, 7~ and 7. What is the kinetic energy of each pion? 60. For the reaction p + p — 3p + p, where one of the initial protons is at rest, use relativistic formulas to show that thcq threshold energy is 6my, ¢?, equal to three times the magnitude of the Q-value of the reaction, where my, is the proton mass. [Hint: Assume all final particles have the same velocity.] 61. What is the total energy of a proton whose kinetic energy is 15 GeV? What is its wavelength? 62. At about what kinetic energy (in eV) can the rest energy of a proton be ignored when calculating its wavelength, if the wavelength is to be within 1.0% of its true value? What are the corresponding wavelength and speed of the proton? 63. Use the quark model to describe the reaction

(e) #" = et + v,

p+n—o 7 + 7%

SL Which of the following reactions are possible, and by what interaction could they occur? For those forbidden, explain why. (@ 7 +p > K+ =7

. Identify the missing particle in the following reactions.

@p+tp—o>pt+tn+a+2?

B)7n"+p > K+ =

Bb)p+?

@7 +p—>A"+K + 70 dr +p—> 20+ 70 () m+tp—o>pte + 1

. Calculate the Q-value for each of the reactions, Eq. 43-4, for producing a pion.

56. The mass of a 7° can be 7~ +p — 7%+ n at (assume it is zero). The with a kinetic energy energy and momentum 57.

fermions

are

there

in a water

two fragments with rest energies of m; c? and m;,c?. Show that the kinetic energy of fragment 1 is K

unified

is equivalent

to an

energy

10" GeV.

58.

Students have sampled the individual lifetimes of muons decaying within a time interval between 1 us and 10 us after being stopped in the scintillator. It is assumed that the muons obey the radioactive decay law R = Rye /7 where Ry is the unknown activity at ¢ = 0 and R is the activity (counts/us) at time ¢. Here is their data:

of

Time (us) R()

Calculate the Q-value for the reaction 7~ + p — A° + KO,

Answers to Exercises

1192

CHAPTER 43

Elementary Particles

1.5 55

25 35

35 23

45 18

5.5 12

6.5 5

Make a graph of In (R/R;) versus time ¢ (us), and from the best fit of the graph to a straight line find the mean life 7. The accepted value of the mean life of the muon is 7 = 2.19703 us = 0.00004 us. What is the percentage error of their result from the accepted value?

when negative pions strike stationary protons. Estimate the minimum pion kinetic energy needed to produce this reaction. [Hint: Assume A° and K° move off with the same velocity.]

B: =2 X 10°m/0.1m ~ 10% €t (a).

N

lifetime of muons, the muons enter a scintillator and decay.

explain how they corresponding to

A: 124 X 1078m = 1.24 am.

[(mc® = mic®f? = (myc?P).

*67. (IT) In a particle physics experiment to determine the mean

about 10'GeV. Use the uncertainty principle, and also de Broglie’s wavelength formula, and apply. (b) Calculate the temperature

1 2mc?

*Numerical/ Computer

measured by observing the reaction very low incident 7~ kinetic energy neutron is observed to be emitted of 0.60 MeV. Use conservation of to determine the 7° mass.

theory

b—

!

(a) Show that the so-called unification distance of 103! m

in grand

of a

66. A particle at rest, with a rest energy of mc?, decays into

about 107 m. Show that this corresponds to an energy that is on the order of the mass of the W=, fundamental

n+u’

65. What fraction of the speed of light ¢ is the speed 7.0-TeV proton?

52, One decay mode for a 7" is 77 — u* + »,. What would be the equivalent decay for a 7~ ? Check conservation laws. 53. Symmetry breaking occurs in the electroweak theory at

58. How many molecule?

>

D: (¢); (). E:

su.

This map of the entire sky (WMAP) is color-coded to represent slight temperature variations in the almost perfectly uniform 2.7-kelvin microwave background radiation that reaches us from all directions in the sky. This latest version (2006) is providing detailed information on the origins of our universe and its structures. The tiny temperature variations, red slightly hotter, blue slightly cooler (on the order of 1 part in 10%) are “quantum fluctuations” that are the seeds on which galaxies and clusters of galaxies eventually grew. To discuss the nature of the universe as we understand it today, we examine the latest theories on how

stars and galaxies form and evolve, including the role of nucleosynthesis. We briefly discuss Einstein’s general theory of relativity, which deals with gravity and curvature of space. We take a thorough look at the evidence for the expansion of the universe, and the Standard Model of the universe evolving from an initial Big Bang. Finally we point out some unsolved problems, including the nature of dark matter and dark energy that make up most of our universe.

~Astrophysics and Cosmology CHAPTER-OPENING QUESTION—Guess now!

Until recently, astronomers expected the expansion rate of the universe would be decreasing. Why? (a) Friction.

(b) The second law of thermodynamics. (¢) Gravity. (d) The electromagnetic force.

n the previous Chapter, we studied the tiniest objects in the universe—the elementary particles. Now we leap to the grandest objects in the universe— stars, galaxies, and clusters of galaxies. These two extreme realms, elementary particles and the cosmos, are among the most intriguing and exciting subjects in science. And, surprisingly, these two extreme realms are related in a fundamental way, as already hinted in Chapter 43. Use of the techniques and ideas of physics to study the heavens is often referred to as astrophysics. Central to our present theoretical understanding of the universe (or cosmos) is Einstein’s general theory of relativity which represents our most complete understanding of gravitation. Many other aspects of physics are involved, from electromagnetism and thermodynamics to atomic and nuclear physics as well as elementary particles. General Relativity serves also as the foundation for modern cosmology, which is the study of the universe as a whole. Cosmology deals especially with the search for 1 theoretical framework to understand the observed universe, its origin, and its future.

CONTENTS

44-1 44-2

Stars and Galaxies

44-3 44-4

Distance Measurements

44-5

The Expanding Universe:

44-6

The Big Bang and the Cosmic Microwave Background The Standard Cosmological Model: Early History of the Universe Inflation: Explaining Flatness, Uniformity, and

44-7

44-8

Stellar Evolution: Nucleosynthesis

and the Birth and Death of Stars

General Relativity: Gravity

and the Curvature of Space Redshift and Hubble’s Law

Structure

44-9

Dark Matter and Dark Energy

44-10

Large-Scale Structure of the Universe

44-11

Finally ...

1193

The

questions posed by cosmology

are profound

and difficult; the possible

answers stretch the imagination. They are questions like “Has the universe always existed, or did it have a beginning in time?” Either alternative is difficult to imagine: time going back indefinitely into the past, or an actual moment when the universe began (but, then, what was there before?). And whe*™ about the size of the universe? Is it infinite in size? It is hard to imagine infinity. Or is it finite in size? This is also hard to imagine, for if the universe is finite, it does not make sense to ask what is beyond it, because the universe is all there is. In the last few years, so much progress has occurred in astrophysics and cosmology that many scientists are calling recent work a “Golden Age” for cosmology. Our survey will be qualitative, but we will nonetheless touch on the major ideas. We begin with a look at what can be seen beyond the Earth.

44-1

Stars and Galaxies

According to the ancients, the stars, except for the few that seemed to move

relative to the others (the planets), were fixed on a sphere beyond the last planet. The universe was neatly self-contained, and we on Earth were at or near its center. But in the centuries following Galileo’s first telescopic observations of the night sky in 1610, our view of the universe has changed dramatically. We no longer place ourselves at the center, and we view the universe as vastly larger. The distances involved are so great that we specify them in terms of the time it takes light to travel the given distance: for

example, 1 light-second = (3.0 X 10°m/s)(1.0s) = 3.0 X 10®m = 300,000 km; 1 light-minute = 18 X 10°km; and 1 light-year (ly) is 1ly = (2.998 x 10®m/s)(3.156 x 107 s/yr) 9.46 X 10°m

=~ 10" km.

For specifying distances to the Sun and Moon, we usually use meters or kilometers but we could specify them in terms of light. The Earth-Moon distance > 384,000 km, which is 1.28 light-seconds. The Earth—Sun distance is 1.50 X 10" m, or 150,000,000 km; this is equal to 8.3 light-minutes. Far out in our solar system,

Pluto is about 6 X 10°km from the Sun, or 6 X 107*ly. The nearest star to us,

other than the Sun, is Proxima Centauri, about 4.3 ly away. On a clear moonless night, thousands of stars of varying degrees of brightness can be seen, as well as the long cloudy stripe known as the Milky Way (Fig. 44-1). Galileo first observed, with his telescope, that the Milky Way is comprised of countless individual stars. A century and a half later (about 1750), Thomas Wright suggested that the Milky Way was a flat disk of stars extending to great distances in a plane, which we call the Galaxy (Greek for “milky way”).

FIGURE 44-1 Sections of the Milky Way. In (a), the thin line is the trail of an artificial Earth satellite in this long time exposure. The dark diagonal area is due to dust absorption of visible light, blocking the view. In (b) the view is toward the center of the Galaxy (taken in summer from Arizona).

(a) 1194

CHAPTER 44

Astrophysics and Cosmology

(b)

Our Galaxy has a diameter of almost 100,000 light-years and a thickness of roughly 20001y. It has a bulging central “nucleus” and spiral arms (Fig. 44-2). Our Sun, which is a star like many others, is located about halfway from the galactic center to the edge, some 26,0001y from the center. Our Galaxy contains roughly 100 billion

#™10") stars. The Sun orbits the galactic center approximately once every 250 million

years, so its speed is about 200 km/s relative to the center of the Galaxy. The total

mass of all the stars in our Galaxy is estimated to be about 3 X 10* kg, which is

ordinary matter. In addition, there is strong evidence that our Galaxy is surrounded by an invisible “halo” of “dark matter,” which we discuss in Section 44-9. FIGURE 44-2

Our Galaxy, as it would appear from the

100,000

outside: (a) “edge view,” in the plane of the disk; (b) “top view,” looking down on the disk. (If only we could see it like this— from the outside!) (c) Infrared photograph of the inner reaches of the Milky Way, showing the central bulge and disk of our Galaxy. This very wide angle photo taken from the COBE satellite (Section 44-6) extends over 180° of sky, and to be viewed properly it should be wrapped in a semicircle with your eyes at the center. The white dots are nearby stars.

e

ly———

EXAMPLE 44-1 Our Galaxy’s mass. Estimate the total mass of our Galaxy using the orbital data above for the Sun about the center of the Galaxy. Assume that most of the mass of the Galaxy is concentrated near the center of the Galaxy. APPROACH We assume that the Sun (including our solar system) has total mass m and moves in a circular orbit about the center of the Galaxy (total mass M), and that the mass M can be considered as being located at the center of the Galaxy. We then apply Newton’s second law, F = ma, with a being the centripetal acceleration, @ = v*/r, and F being the universal law of gravitation (Chapter 6). SOLUTION Our Sun and solar system orbit the center of the Galaxy, according to the best measurements as mentioned above, with a speed of about v = 200km/s at a distance from the Galaxy center of about r = 26,000 ly. We use Newton’s second law: F

=

ma

Mm

v?

G-r

sr

where M is the mass of the Galaxy and m is the mass of our Sun and solar system. Solving this, we find M=

LNOTE

2

rv-

G

s

(

26,0001y)(10'6 m/ly)(2 .x 10°m/s)’ y)( ‘l/IY)( g /) 6.67 X 107"

N-m*/kg

~

2 X

1041kg

In terms of numbers of stars, if they are like our Sun (m = 2.0 X 10*kg),

€ there would be about (2 X 10" kg)/(2 X 10*kg) ‘1_00 billion stars.

~ 10" or on the order of SECTION 44-1

Stars and Galaxies

1195

: Hevanles

s

glotular star

cluster is located in the constellation

'

This gaseous nebula, found in the constellation Carina,

is about 9000 light-years from us.

In addition to stars both within and outside the Milky Way, we can see by telescope many faint cloudy patches in the sky which were all referred to once as “nebulae” (Latin for “clouds”). A few of these, such as those in the constellations Andromeda and Orion, can actually be discerned with the naked eye on a clear night. Some are star clusters (Fig. 44-3), groups of stars that are so numerou ™ they appear to be a cloud. Others are glowing clouds of gas or dust (Fig. 44-4), and it is for these that we now mainly reserve the word nebula. Most fascinating are those that belong to a third category: they often have fairly regular elliptical shapes and seem to be a great distance beyond our Galaxy. Immanuel Kant (about 1755) seems to have been the first to suggest that these latter might be circular disks, but appear elliptical because we see them at an angle, and are faint because they are so distant. At first it was not universally accepted that these objects were extragalactic—that is, outside our Galaxy. The very large telescopes constructed in the twentieth century revealed that individual stars could be o . . . . resolved within these extragalactic objects and that many contain spiral arms. Edwin Hubble (1889-1953) did much of this observational work in the 1920s using the 2.5-m (100-inch) telescope’ on Mt. Wilson near Los Angeles, California, then the world’s largest. Hubble demonstrated that these objects were indeed extragalactic because of their great distances. The distance to our nearest large

galaxy,” Andromeda, is over 2 million light-years, a distance 20 times greater than

the diameter of our Galaxy. It seemed logical that these nebulae must be galaxies similar to ours. (Note that it is usual to capitalize the word “galaxy” only when it refers to our own.) Today it is thought there are roughly 10" galaxies in the observable universe—that is, roughly as many galaxies as there are stars in a galaxy. See Fig. 44-5. Many galaxies tend to be grouped in galaxy clusters held together by their mutual gravitational attraction. There may be anywhere from a few to many thousands of galaxies in each cluster. Furthermore, clusters themselves seem to be 2.5m (= 100inches) refers to the diameter of the curved objective mirror. The bigger the mirror the more light it collects (greater intensity) and the less diffraction there is (better resolution), so m01A and fainter stars can be seen. See Chapters 33 and 35. Until recently, photographic films or plates were used to take long time exposures. Now large solid-state CCD or CMOS sensors (Section 33-5) are available containing hundreds of millions of pixels (compared to 10 million pixels in a good-quality digital camera). *The Magellanic clouds are much closer than Andromeda, but are small and are usually considered small satellite galaxies of our own Galaxy.

Photographs of galaxies. (a) Spiral galaxy in the constellation Hydra. (b) Two galaxies: the larger and more dramatic one is known as the Whirlpool galaxy. (c) An infrared image (given “false” colors) of the same galaxies as in (b), here showing the arms of the spiral as having more substance than in the visible light photo (b); the different colors correspond to different light intensities. Visible light is scattered and absorbed by interstellar dust much more than infrared is, so infrared gives us a clearer image.

(b) 1196

CHAPTER 44

Astrophysics and Cosmology

organized into even larger aggregates: clusters of clusters of galaxies, or superclusters. The farthest detectable galaxies are more than 10ly distant. See Table 44-1.

fCONCEPTUAL EXAMPLE 44-2 | Looking back in time. Astronomers often think of their telescopes as time machines, looking back toward the origin of the universe.

How far back do they look? RESPONSE The

distance

in

light-years

measures

light has been traveling to reach us, so Table time we are looking. For example, if we saw supernova today, then the event would have most distant galaxies emitted the light we

how

long

in

years

the

44-1 tells us also how far back in Proxima Centauri explode into a really occurred 4.3 years ago. The see now roughly 10'°years ago.

What we see was how they were then, 10'°yr ago, or about 10°years after the

Mverse was born in the Big Bang.

TABLE 44-1 Astronomical Distances )

Olject Moon

Sun

Approx. Distance

trom Hankdy) 4 x 108

1.6 X 107

Size of solar system (distance to Pluto) Nearest star

(Proxima Centauri)

6 x 107 4.3

Center of our Galaxy

2.6 X 10*

Nearest large galaxy

2.4 X 10°

Farthest galaxies

1)

EXERCISE A Suppose we could place a huge mirror 1 light-year away from us. What would we see in this mirror if it is facing us on Earth? When did it take place? (This might be called a “time machine.”)

Besides the usual stars, clusters of stars, galaxies, and clusters and superclusters of galaxies, the universe contains many other interesting objects. Among these are stars known as red giants, white dwarfs, neutron stars, exploding stars called novae and supernovae, and black holes whose gravity is so strong even light can not escape them. In addition, there is electromagnetic radiation that reaches the Earth but does not emanate from the bright pointlike objects we call stars: particularly important is the microwave background radiation that arrives nearly uniformly from all directions in the universe. We will discuss all these phenomena. Finally, there are active galactic nuclei (AGN), which are very luminous pointlike sources of light in the centers of distant galaxies. The most dramatic examples of AGN are quasars (“quasistellar objects” or QSOs), which are so lumirous that the surrounding starlight of the galaxy is drowned out. Their luminosity thought to come from matter falling into a giant black hole at a galaxy’s center.

44-7)

Stellar Evolution: Nucleosynthesis, and the Birth and Death of Stars

The stars appear unchanging. Night after night the night sky reveals no significant variations. Indeed, on a human time scale, the vast majority of stars change very little (except for novae, supernovae, and certain variable stars). Although stars seem fixed in relation to each other, many move sufficiently for the motion to be detected. Speeds of stars relative to neighboring stars can be hundreds of km/s, but at their great distance from us, this motion is detectable only by careful measurement. Furthermore, there is a great range of brightness among stars. The differences in brightness are due both to differences in the rate at which stars emit energy and to their different distances from us.

Luminosity and Brightness of Stars A useful parameter for a star or galaxy is its intrinsic luminosity, L (or simply luminosity), by which we mean the total power radiated in watts. Also important is the apparent brightness, b, defined as the power crossing unit area at the Earth perpendicular to the path of the light. Given that energy is conserved, and ignoring any absorption in space, the total emitted power L when it reaches a distance d from the star will be spread over a sphere of surface area 4md>. If d is the distance from the star to the Earth, then L must be equal to 47d? times b (power per unit area at Earth). That is,

«

b = —fl{‘— SECTION 44-2

(44-1) Stellar Evolution: Nucleosynthesis, and the Birth and Death of Stars

1197

Apparent brightness. Suppose a particular star has intrinsic

luminosity equal to that of our Sun, but is 101y away from Earth. By what factor will it appear dimmer than the Sun?

APPROACH The luminosity L is the same for both stars, so the apparent brightfi ness depends only on their relative distances. We use the inverse square law a. stated in Eq. 44-1 to determine the relative brightness. SOLUTION Using the inverse square law, the star appears dimmer by a factor

byar _ dhm

bsun

diar

(L5 X 10°km)*

(101y)%(10" km/ly)’

> x 10-12

'

Careful study of nearby stars has shown that the luminosity for most stars depends on the mass: the more massive the star, the greater its luminosity’. Indeed, we might expect that more massive stars would have higher core temperature and pressure to counterbalance the greater gravitational attraction, and thus be more luminous. Another important parameter of a star is its surface temperature, which can be determined from the spectrum of electromagnetic frequencies it emits (stars are “good” blackbodies—see Section 37-1). As we saw in Chapter 37, as the temperature of a body increases, the spectrum shifts from predominantly lower frequencies (and longer wavelengths, such as red) to higher frequencies (and shorter wavelengths such as blue). Quantitatively, the relation is given by Wien’s law (Eq. 37-1): the peak wavelength Ap in the spectrum of light emitted by a blackbody (we often approximate stars as blackbodies) is inversely proportional to its Kelvin temperature

T; that is, Ap7

= 2,90 X 10 m-K.

The surface tempera-

tures of stars typically range from about 3000 K (reddish) to about 50,000 K (UV).

Determining star temperature and star size. Suppose that

the distances from Earth to two nearby stars can be reasonably estimated, and that their measured apparent brightnesses suggest the two stars have about the same luminosity, L. The spectrum of one of the stars peaks at about 700nm (so it reddish). The spectrum of the other peaks at about 350 nm (bluish). Use Wien’s la. (Eq.37-1) and the Stefan-Boltzmann equation (Section 19-10) to determine (a) the surface temperature of each star, and (b) how much larger one star is than the other. APPROACH We determine the surface temperature T for each star using Wien’s law and each star’s peak wavelength. Then, using the Stefan-Boltzmann equation (power output or luminosity oc AT* where A = surface area of emitter), we can find the surface area ratio and relative sizes of the two stars. SOLUTION (a) Wien’s law (Eq. 37-1) states that Ap7 = 2.90 X 10°m-K. So the temperature of the reddish star is

T, =

'

290 X 10°m-K

700 X 10 m

= 4140 K.

The temperature of the bluish star will be double this since its peak wavelength is half (350 nm vs. 700 nm):

T, = 8280K. (b) The Stefan-Boltzmann equation, Eq. 19-17, states that the power radiated per unit area of surface from a blackbody is proportional to the fourth power of the Kelvin temperature, 7% The temperature of the bluish star is double that of the

reddish star, so the bluish one must radiate (2*) = 16 times as much energy per

unit area. But we are given that they have the same luminosity (the same total power output); so the surface area of the blue star must be i that of the red one. The

ufi

surface

area

of a sphere

is 4712,

so

the

radius

of the

reddish

star

is

= 4 times larger than the radius of the bluish star (or 4°> = 64 times the volume).

*Applies to “main-sequence” stars (see next page). The mass of a star can be determined by observir,, its gravitational effects. Many stars are part of a cluster, the simplest being a binary star in which two stars orbit around each other, allowing their masses to be determined using rotational mechanics.

1198

CHAPTER 44

1029

28 L

.

10

2

ot

g

~

e « o*%..

OurSun

Wias

i

>‘

.'.

g

2| E S

10%

z

!

é

025

A"?I,)

-

%y,

T

. o

"‘

'y

e

", Red

.

.

‘ e+

'

'.

"

o0

el

i

e

FIGURE 44-6 Hertzsprung—Russell (H-R) diagram is a logarithmic graph of luminosity vs. surface temperature 7 of stars

_,_:'-3" ‘

BTV 1023

giants

..

(note that 7 increases to the left).

R

e White dwarfs | |

|

10,000

.

|

7000 5000 Surface temperature 7 (K)

|

3500

o

H-R Diagram An important astronomical discovery, made around 1900, was that for most stars, the color is related to the intrinsic luminosity and therefore to the mass. A useful way to present this relationship is by the so-called Hertzsprung—Russell (H-R) diagram. On the H-R diagram, the horizontal axis shows the surface temperature 7 whereas the vertical axis is the luminosity L; each star is represented by a point on the diagram, Fig. 44-6. Most stars fall along the diagonal band termed the main sequence. Starting at the lower right we find the coolest stars, reddish in color; they are the least luminous and therefore of low mass. Farther up toward the left we find hotter and more luminous stars that are whitish, like our Sun. Still farther up e find even more massive and more luminous stars, bluish in color. Stars that fall n this diagonal band are called main-sequence stars. There are also stars that fall outside the main sequence. Above and to the right we find extremely large stars, with high luminosities but with low (reddish) color temperature: these are called red giants. At the lower left, there are a few stars of low luminosity but with high temperature: these are the white dwarfs.

EXAMPLE 44-5

Distance to a star using the H-R diagram and

color. Suppose that detailed study of a certain star suggests that it most likely fits on the main sequence of an H-R diagram. Its measured apparent brightness is

b =1.0 X 1072 W/m?

and the peak wavelength of its spectrum is Ap ~ 600 nm.

Estimate its distance from us. APPROACH is estimated the distance SOLUTION

We find the temperature using Wien’s law, Eq. 37-1. The luminosity for a main sequence star on the H-R diagram of Fig. 44-6, and then is found using the relation between brightness and luminosity, Eq. 44-1. The star’s temperature, from Wien’s law (Eq. 37-1), is 290 x107°m:K ~ 4800 K.

600 X 107m

A star on the main sequence of an H-R diagram at this temperature has intrinsic luminosity of about L ~ 1 X 10 W, read off of Fig. 44—6. Then, from Eq. 44-1, d

=

[L\|— 4mrb

=~ \/

1 x 10°W 4(314)(1.0 x 10 2W/m?)

~

3 X 10%m.

"

Its distance from us in light-years is

A —

3 X 10%m d =T 10" m/ly 5 a0y,¥

EXERCISE B Estimate

the distance

brightness of 2.0 X 1072 W/m?

to a 6000-K

main-sequence

star with an apparent

SECTION 44-2

1199

Stellar Evolution; Nucleosynthesis Why are there different types of stars, such as red giants and white dwarfs, as well as main-sequence stars? Were they all born this way, in the beginning? Or might each different type represent a different age in the life cycle of a star& Astronomers and astrophysicists today believe the latter is the case. Not. however, that we cannot actually follow any but the tiniest part of the life cycle of any given star since they live for ages vastly greater than ours, on the order of millions or billions of years. Nonetheless, let us follow the process of stellar evolution from the birth to the death of a star, as astrophysicists have theoretically reconstructed it today. Stars are born, it is believed, when gaseous clouds (mostly hydrogen) contract due to the pull of gravity. A huge gas cloud might fragment into numerous contracting masses, each mass centered in an area where the density was only slightly greater than that at nearby points. Once such “globules” formed, gravity would cause each to contract in toward its center of mass. As the particles of such a protostar accelerate inward, their kinetic energy increases. When the kinetic energy is sufficiently high, the Coulomb repulsion between the positive charges is not strong enough to keep the hydrogen nuclei apart, and nuclear fusion can take place. In a star like our Sun, the fusion of hydrogen (sometimes referred to as “burning”)’ occurs via the proton—proton cycle (Section 42-4, Egs. 42-7), in which four protons fuse to form a jHe nucleus with the release of ¥ rays, posi-

trons, and neutrinos: 4 {H — 3He + 2e* + 2v, + 27. These reactions require

a temperature of about 10’ K, corresponding to an average kinetic energy (~ k7) of about 1keV (Eq. 18-4). In more massive stars, the carbon cycle produces the same net effect: four {H produce a He—see Section 42-4. The fusion reactions take place primarily in the core of a star, where T may be on the order of 107 to 108 K. (The surface temperature is much lower—on the order of a few thousand kelvins.) The tremendous release of energy in these fusion reactions produces a outward pressure sufficient to halt the inward gravitational contraction. O protostar, now really a young star, stabilizes on the main sequence. Exactly where the star falls along the main sequence depends on its mass. The more massive the star, the farther up (and to the left) it falls on the H-R diagram of Fig. 44-6. Our Sun required perhaps 30 million years to reach the main sequence,

and is expected to remain there about 10 billion years (10'°yr). Although most

stars are billions of years old, evidence is strong that stars are actually being born at this moment.

More

massive stars have shorter lives, because they are hotter

and the Coulomb repulsion is more easily overcome, so they use up their fuel faster. If our Sun remains on the main sequence for 10'°years, a star ten times more massive may reside there for only 107 years. FIGURE 44-7 A shell of “burning” As hydrogen fuses to form helium, the helium that is formed is denser and hydrogen (fusing to become helium) tends to accumulate in the central core where it was formed. As the core of helium surrounds the core where the newly grows, hydrogen continues to fuse in a shell around it: see Fig. 44—7. When much of formed helium gravitates. the hydrogen within the core has been consumed, the production of energy Hydrogen Nonburning decreases at the center and is no longer sufficient to prevent the huge gravitational fusion en?/létlzr forces from once again causing the core to contract and heat up. The hydrogen in P€ " the shell around the core then fuses even more fiercely because of this rise in temperature, allowing the outer envelope of the star to expand and to cool. The surface temperature, thus reduced, produces a spectrum of light that peaks at longer wavelength (reddish).

"The word “burn” is put in quotation marks because these high-temperature fusion reactions occur via a nuclear process, and must not be confused with ordinary burning (of, say, paper, wood, or coal) in q'a which is a chemical reaction, occurring at the atomic level (and at a much lower temperature). Helium

1200

CHAPTER 44

Astrophysics and Cosmology

redder, and

fHe + jHe — §Be {He + §Be — '2C

(44-2)

with the emission of two ¥ rays. These two reactions must occur in quick succession (because §Be is very unstable), and the net effect is

34He — -C.

(Q = 73MeV)

-

-

/

] | 1 1 1

Luminosity

By this time the star has left the main sequence. It has become

as it has grown in size, it has become more luminous. So it will have moved to the right and upward on the H-R diagram, as shown in Fig. 44-8. As it moves ward, it enters the red giant stage. Thus, theory explains the origin of red giants . a natural step in a star’s evolution. Our Sun, for example, has been on the main sequence for about 43 billion years. It will probably remain there another 4 or Sbillion years. When our Sun leaves the main sequence, it is expected to grow in diameter (as it becomes a red giant) by a factor of 100 or more, possibly swallowing up inner planets such as Mercury. If the star is like our Sun, or larger, further fusion can occur. As the star’s outer envelope expands, its core continues to shrink and heat up. When the temperature reaches about 10 K, even helium nuclei, in spite of their greater charge and hence greater electrical repulsion, can come close enough to each other to undergo fusion. The reactions are

\

ed giant

\

dwarf

Surface temperature FIGURE 44-8 Evolutionary “track” of a star like our Sun represented on an H-R diagram.

This fusion of helium causes a change in the star which moves rapidly to the “horizontal branch” on the H-R diagram (Fig. 44-8). Further fusion reactions

are possible, with $He fusing with '2C to form '§O. In more massive stars, higher Z elements like 3)Ne or Mg can be made. This process of creating heavier nuclei

from lighter ones (or by absorption of neutrons which tends to occur at higher Z) is called nucleosynthesis. The final fate of a star depends on its mass. Stars can lose mass as parts of their outer envelope move off into space. Stars born with a mass less than about 8 #r perhaps 10) solar masses eventually end up with a residual mass less than .oout 1.4 solar masses, which is known as the Chandrasekhar limit. For them, no further fusion energy can be obtained. The core of such a “low mass” star (original mass < 8-10 solar masses) contracts under gravity; the outer envelope expands again and the star becomes an even larger red giant. Eventually the outer layers escape into space, the core shrinks, the star cools, and typically follows the dashed route shown in Fig. 44-8, descending downward, becoming a white dwarf. A white dwarf with a residual mass equal to that of the Sun would be about the size of the Earth. A white dwarf contracts to the point at which the electron clouds start to overlap, but no further because, by the Pauli exclusion principle, no two electrons can be in the same quantum state. At this point the star is supported against further collapse by this electron degeneracy pressure. A white dwarf continues to lose internal energy by radiation, decreasing in temperature and becoming dimmer until it glows no more. It has then become a cold dark chunk of extremely dense material. Stars whose residual mass is greater than the Chandrasekhar limit of 1.4 solar masses (original mass greater than about 8 or 10 solar masses) are thought to follow a quite different scenario. A star with this great a mass can contract under gravity and heat up even further. In the range 7 = (2.5-5) X 10°K, nuclei as heavy as 35Fe and 3¢Ni can be made. But here the formation of heavy nuclei from lighter ones, by fusion, ends. As we saw in Fig. 41-1, the average binding energy per nucleon begins to decrease for A greater than about 60. Further fusions would require energy, rather than release it. Elements heavier than Ni are thought to form mainly by neutron capture, particularly in exploding stars called supernovae (singular is supernova). Large numbers of free neutrons, resulting from nuclear reactions, are present inside these highly evolved stars and they can readily combine with, say, a 3¢Fe nucleus to form (if three

e captured) %Fe, which decays to 33Co. The 3Co can capture neutrons, also

~ecoming neutron rich and decaying by B~ to the next higher Z element, and so on to the highest Z elements, SECTION 44-2

Stellar Evolution: Nucleosynthesis, and the Birth and Death of Stars

1201

Yet at these extremely high temperatures, well above 10° K, the kinetic energy of the nuclei is so high that fusion of elements heavier than iron is still possible even though the reactions require energy input. But the high-energy collisions can also cause the breaking apart of iron and nickel nuclei into He nuclei, and‘ eventually into protons and neutrons: .

¥Fe —

134He + 4n

iHe — 2p + 2n. These are energy-requiring (endothermic) reactions, but at such extremely high temperature and pressure there is plenty of energy available, enough even to force electrons and protons together to form neutrons in inverse B decay: e

+tp—>n+uw

As a result of these reactions, the pressure in the core drops precipitously. As the core collapses under the huge gravitational forces, the tremendous mass becomes essentially an enormous nucleus made up almost exclusively of neutrons. The size of the star is no longer limited by the exclusion principle applied to electrons, but rather by neutron degeneracy pressure, and the star contracts rapidly to form an enormously dense neutron star. The core of a neutron star contracts to the point at which all neutrons are as close together as they are in an atomic nucleus. That is, the density of a neutron star is on the order of

10" times greater than normal solids and liquids on Earth. A cupful of such

FIGURE 44-9 The star indicated by the arrow in (a) exploded as a supernova, as shown in (b). It was detected in 1987, and named

SN1987A. The large bright spot in (b) indicates a huge release of energy but does not represent the physical size. Part (c) is a photo taken a few years later, showing shock waves moving outward from where SN1987A was (blow-up in corner). Part (c) is magnified

dense matter would weigh billions of tons. A neutron star that has a mass 1.5 times that of our Sun would have a diameter of only about 20 km. (Compare this to a white dwarf with 1 solar mass whose diameter would be ~10* km, as already mentioned.) The contraction of the core of a massive star would mean a great reduction in gravitational potential energy. Somehow this energy would have to be released. Indeed, it was suggested in the 1930s that the final core collapse to a neutron star may be accompanied by a catastrophic explosion (a supernova—see previous page)\ whose tremendous energy could form virtually all elements of the Periodic Table an. blow away the entire outer envelope of the star (Fig. 44-9), spreading its contents into interstellar space. The presence of heavy elements on Earth and in our solar system suggests that our solar system formed from the debris of such a supernova explosion.

relative to (a) and (b).

(b)

1202

CHAPTER 44

(c)

If the final mass of a neutron star is less than about two or three solar masses, its subsequent evolution is thought to resemble that of a white dwarf. If the mass is greater than this, the star collapses under gravity, overcoming even the neutron exclusion principle. Gravity would then be so strong that even light emitted fromem, the star could not escape—it would be pulled back in by the force of gravity. Sincc no radiation could escape from such a star, we could not see it—it would be black. An object may pass by it and be deflected by its gravitational field, but if it came too close it would be swallowed up, never to escape. This is a black hole.

Novae and Supernovae Novae (singular is nova, meaning “new” in Latin) are faint stars that have suddenly increased in brightness by as much as a factor of 10* and last for a month or two fbefore fading. Novae are thought to be faint white dwarfs that have pulled mass from . nearby companion (they make up a binary system), as illustrated in Fig. 44-10. The captured mass of hydrogen suddenly fuses into helium at a high rate for a few weeks. Many novae (maybe all) are recurrent—they repeat their bright glow years later. Supernovae are also brief explosive events, but release millions of times more energy than novae, up to 10'° times more luminous than our Sun. The peak of brightness may exceed that of the entire galaxy in which they are located, but lasts only a few days or weeks. They slowly fade over a few months. Many supernovae form by core collapse to a neutron star as described above. See Fig. 44-9. Type Ia supernovae are different. They all seem to have very nearly the same luminosity. They are believed to be binary stars, one of which is a white dwarf that pulls mass from its companion, much like for a nova, Fig. 44-10. The mass is higher, and as mass is captured and the total mass reaches the Chandrasekhar limit of 14 solar masses, it explodes as a supernova by undergoing a “thermonuclear runaway”—an uncontrolled chain of nuclear reactions. What is left is a neutron star or (if the mass is great enough) a black hole.

44-3

Main-sequence compdion

White

N Mass transfer

FIGURE 44-10 Hypothetical model for novae and type Ia

supernovae, showing how a white

dwarf could pull mass from its normal companion.

Distance Measurements

We have talked about the vast distances of objects in the universe. But how do we measure these distances? One basic technique employs simple geometry to measure the parallax of a star. By parallax we mean the apparent motion of a star, against the background of much more distant stars, due to the Earth’s motion about the Sun. As shown in Fig. 44-11, the sighting angle of a star relative to the plane of Earth’s orbit (angle 6) can be determined at different times of the year. Since we Anow the distance d from Earth to Sun, we can reconstruct the right triangles 1own in Fig. 44-11 and can determine’ the distance D to the star. "This is essentially the way the heights of mountains are determined, by “triangulation.” See Example 1-7.

K N3

3 *

Distant stars

*

»

|

N

Nearby

7\ star

* sk

FIGURE 44-11 (a) Simple example of determining the distance D to a relatively nearby star using parallax. Horizontal distances are greatly exaggerated: in reality ¢ is a very small angle. (b) Diagram of the sky showing the apparent position of the “nearby” star relative to more distant stars, at two different times (January and July). The viewing angle in January puts the star more to the right relative to distant stars, whereas in July it is more to the left (dashed circle shows January location).

Sky as seen

from

Earth in January

As seen from Earth in July

(a)

Earth’s orbit

(b) SECTION 44-3

Distance Measurements

1203

DGR

TR

ESTIMATE | Distance to a star using parallax. Estimate

the distance D to a star if the angle 6 in Fig. 44—11 is measured to be 89.99994°, APPROACH

From

trigonometry,

distance is d = 1.5 X 108km.

tan¢ = d/D

in Fig. 44-11. The

Sun-Earth

-

SOLUTION The angle ¢ = 90° — 89.99994° = 0.00006°, or about 1.0 X 107 radians.

We can use tan¢ ~ ¢ since ¢ is very small. We The distance D to the star is

D

d

=

N

tané

or about 151y.

d —

solve for D in

tan¢

= d/D.

1.5 X 10°km = ———————— = 1.5 X 10"*km,

¢ _ 1.0 x 10-5rad

m

Distances to stars are often specified in terms of parallax angle (¢ in Fig. 44-11a) given in seconds of arc: 1second (1”) is z; of one minute (1') of arc, which is 4 of a degree, so 1” = 545 of a degree. The distance is then specified in parsecs (pc) (meaning parallax angle in seconds of arc): D = 1/¢ with ¢ in

seconds of arc. In Example 44-6, ¢ = (6 x 107°)°(3600) = 0.22" of arc, so we would say the star is at a distance of 1/0.22” = 4.5pc.

One parsec is given by (recall

D = d/¢, and we set the Sun-Earth distance (Fig. 44-11a) as d = 1.496 X 10" m): d

—

1.2

=

(1)

1.496 X 10" m

4 1° \ [ 2w rad 60" /\ 60’ 360°

11 (3.086 X 1016m) 9.46 X 10" m

]

Il

1pc

=

Il

Lpe

!

Ao

X

x |

16m

3.26ly.

Parallax can be used to determine the distance to stars as far away as about 100 light-years (= 30 parsecs) from Earth, and from an orbiting spacecraft perhaps 5 to 10 times farther. Beyond that distance, parallax angles are too small to measure. For greater distances, more subtle techniques must be employed. We might compare the apparent brightnesses of two stars, or two galaxies, and use thfl inverse square law (apparent brightness drops off as the square of the distance) . roughly estimate their relative distances. We can’t expect this technique to be very precise because we don’t expect any two stars, or two galaxies, to have the same intrinsic luminosity. When comparing galaxies, a perhaps better estimate assumes the brightest stars in all galaxies (or the brightest galaxies in galaxy clusters) are similar and have about the same intrinsic luminosity. Consequently, their apparent brightness would be a measure of how far away they were. Another technique makes use of the H-R diagram. Measurement of a star’s surface temperature (from its spectrum) places it at a certain point (within 20%) on the H-R diagram, assuming it is a main-sequence star, and then its luminosity can be estimated off the vertical axis (Fig. 44—6). Its apparent brightness and Eq. 44-1 give its approximate distance; see Example 44-5. A better estimate comes from comparing wvariable stars, especially Cepheid variables whose luminosity varies over time with a period that is found to be related to their average luminosity. Thus, from their period and apparent brightness we get their distance. The largest distances are estimated by comparing the apparent brightnesses of type Ia supernovae (SNIa). Type Ia supernovae all have a similar origin (as described on the previous page, Fig. 44-10), and their brief explosive burst of light is expected to be of nearly the same luminosity. They are thus sometimes referred to as “standard candles.” Another important technique for estimating the distance of very distant stars is from the “redshift” in the line spectra of elements and compounds. The amount of redshift is related to the expansion of the universe, as we shall discuss in

Section 44-5. It is useful for objects farther than 107 to 10° ly away.

As we look farther and farther away, the measurement techniques are less a less reliable, so there is more and more uncertainty in the measurements of larg. distances.

1204

CHAPTER 44

Astrophysics and Cosmology

44-4

General Relativity: Gravity and the Curvature of Space

r‘ Ne have seen that the force of gravity plays an important role in the processes that occur in stars. Gravity too is important for the evolution of the universe as a whole. The reasons gravity plays a dominant role in the universe, and not one of the other of the four forces in nature, are (1) it is long-range and (2) it is always attractive. The strong and weak nuclear forces act over very short distances only, on the order of the size of a nucleus; hence they do not act over astronomical distances (they do act between nuclei and nucleons in stars to produce nuclear reactions). The electromagnetic force, like gravity, acts over great distances. But it can be either attractive or repulsive. And since the universe does not seem to contain large areas of net electric charge, a large net force does not occur. But gravity acts as an attractive force between all masses, and there are large accumulations in the universe of only the one “sign” of mass (not + and — as with electric charge). The force of gravity as Newton described it in his law of universal gravitation was modified by Einstein. In his general theory of relativity, Einstein developed a theory of gravity that now forms the basis of cosmological dynamics. In the special theory of relativity (Chapter 36), Einstein concluded that there is

no way for an observer to determine whether a given frame of reference is at rest or is moving at constant velocity in a straight line. Thus the laws of physics must be

the same in different inertial reference frames. But what about the more general

case of motion where reference frames can be accelerating? Einstein tackled the problem of accelerating reference frames in his general theory of relativity and in it also developed a theory of gravity. The mathematics of General Relativity is complex, so our discussion will be mainly qualitative. We begin with Einstein’s principle of equivalence, which states that r.

/'

~ FIGURE 44-12

In an elevator

falling freely under gravity,

(&) @ person releaseskahboc’k;

512(:}:2 ifilee?)svsge?zohangfl?;; 8 fow ————xy ’

no experiment can be performed that could distinguish between a uniform gravitational field and an equivalent uniform acceleration.

If observers sensed that they were accelerating (as in a vehicle speeding around a sharp curve), they could not prove by any experiment that in fact they weren’t simply experiencing the pull of a gravitational field. Conversely, we might think we are being pulled by gravity when in fact we are undergoing an acceleration having nothing to do with gravity. As a thought experiment, consider a person in a freely falling elevator near the Earth’s surface. If our observer held out a book and let go of it, what would happen? Gravity would pull it downward toward the Earth, but at the same rate (¢ = 9.8m/s?) at which the person and elevator were falling. So the book would hover right next to the person’s hand (Fig. 44-12). The effect is exactly the same as if this reference frame was at rest and no forces were acting. On the other hand, if the elevator was out in space where the gravitational field is essentially zero, the released book would float, just as it does in Fig. 44-12. Next, if the elevator (out in space) is accelerating upward at an acceleration of 9.8 m/s?, the book as seen by our observer would fall to the floor with an acceleration of 9.8m/s? just as if it were falling due to gravity at the surface of the Earth. According to the principle of equivalence, the observer could not determine whether the book fell because the elevator was accelerating upward, or because a gravitational field was acting downward and the elevator was at rest. The two descriptions are equivalent. The principle of equivalence is related to the concept that there are two types of mass. Newton’s second law, F = ma, uses inertial mass. We might say that inertial mass represents “resistance” to any type of force. The second type of mass is gravitational mass. When one object attracts another by the gravitational force Newton’s law of universal gravitation, F = Gm,m,/r?, Chapter 6), the strength of the force is proportional to the product of the gravitational masses of the two objects.

SECTION 44-4

(b)

g

General Relativity: Gravity and the Curvature of Space

1205

This is much like Coulomb’s law for the electric force between two objects which is proportional to the product of their electric charges. The electric charge on an object is not related to its inertial mass; so why should we expect that an object’s gravitational mass (call it gravitational charge if you like) be related to its inertial mass? All along we have assumed they were the same. Why? Becaus no experiment—not even of high precision—has been able to discern any measurable difference between inertial mass and gravitational mass. (For example, in the absence of air resistance, all objects fall at the same acceleration, g, on Earth.) This is another way to state the equivalence principle: gravitational mass is equivalent to inertial mass.

FIGURE 44-13 (a) Light beam goes straight across an elevator which is not accelerating. (b) The light beam bends (exaggerated) in an accelerating elevator whose speed increases in the upward direction. Both views are as seen by an outside observer in an inertial reference frame.

FIGURE 44-14 (a) Three stars in the sky observed from Earth. (b) If the light from one of these stars passes very near the Sun, whose gravity

bends the rays, the star will appear

higher than it actually is (follow the ray backwards). 1

Stars 2 >

Observer on Earth

A

3

(a)

Apparent

position of star

Observer

on Earth

1206

(b)

CHAPTER 44

The principle of equivalence can be used to show that light ought to b" deflected due to the gravitational force of a massive object. Consider another thought experiment, in which an elevator is in free space where virtually no gravity acts. If a light beam is emitted by a flashlight attached to the side of the elevator, the beam travels straight across the elevator and makes a spot on the opposite side if the elevator is at rest or moving at constant velocity (Fig. 44-13a). If instead the elevator is accelerating upward, as in Fig. 44-13b, the light beam still travels straight across in a reference frame at rest. In the upwardly accelerating elevator, however, the beam is observed to curve downward. Why? Because during the time the light travels from one side of the elevator to the other, the elevator is moving upward at a vertical speed that is increasing relative to the light. Next we note that according to the equivalence principle, an upwardly accelerating reference frame is equivalent to a downward gravitational field. Hence, we can picture the curved light path in Fig. 44-13b as being due to the effect of a gravitational field. Thus, from the principle of equivalence, we expect gravity to exert a force on a beam of light and to bend it out of a straightline path! That light is affected by gravity is an important prediction of Einstein’s general theory of relativity. And it can be tested. The amount a light beam would be deflected from a straight-line path must be small even when passing a massive object. (For example, light near the Earth’s surface after traveling 1 km is predicted

to drop only about 107'°m, which is equal to the diameter of a small atom and

not detectable.) The most massive object near us is the Sun, and it was calculated that light from a distant star would be deflected by 1.75" of arc (tiny but detectable) as it passed by the edge of the Sun (Fig. 44-14). However, such a measurement could be made only during a total eclipse of the Sun, so that the Sun’s tremendous\ brightness would not obscure the starlight passing near its edge. An opportunt eclipse occurred in 1919, and scientists journeyed to the South Atlantic to observe it.

Astrophysics and Cosmology

f‘

image s Quasar

Observer

Galaxy

fight from Quasar False o _—--""~ image

(a)

(b)

FIGURE 44-15 (a) Hubble Space Telescope photograph of the so-called “Einstein cross”, thought to represent “gravitational lensing”: the central spot is a relatively nearby galaxy, whereas the four other spots are thought to be images of a single quasar behind the galaxy. (b) Diagram showing how the galaxy could bend the light coming from the quasar behind it to produce the four images. See also Fig. 44-14. [If the shape of the nearby galaxy and distant quasar were perfect spheres, we would expect the “image” of the distant quasar to be a circular ring or halo instead of the four separate images seen here. Such a ring is called an “Einstein ring.”]

Their photos of stars around the Sun revealed shifts in accordance with Einstein’s prediction. Another example is gravitational lensing, as shown in Fig. 44-15. Fermat showed that optical phenomena, including reflection, refraction, and effects of lenses, can be derived from a simple principle: that light traveling between two points follows the shortest path in space. Thus if gravity curves the path of light, then gravity must be able to curve space itself. That is, space itself can be curved, and it is gravitational mass that causes the curvature. Indeed, the

f.;urvature of space—or rather, of four-dimensional space-time—is a basic aspect [ Einstein’s General Relativity (GR). What is meant by curved space? To understand, recall that our normal method of viewing the world is via Euclidean plane geometry. In Euclidean geometry, there are many axioms and theorems we take for granted, such as that the sum of the angles of any triangle is 180°. Non-Euclidean geometries, which involve curved space, have also been imagined by mathematicians. It is hard enough to imagine three-dimensional curved space, much less curved four-dimensional space-time. So let us try to understand the idea of curved space by using two-dimensional surfaces. Consider, for example, the two-dimensional surface of a sphere. It is clearly curved, Fig. 44-16, at least to us who view it from the outside—from our threedimensional world. But how would hypothetical two-dimensional creatures determine whether their two-dimensional space was flat (a plane) or curved? One way would be to measure the sum of the angles of a triangle. If the surface is a plane, the sum of the angles is 180°, as we learn in plane geometry. But if the space is curved, and a sufficiently large triangle is constructed, the sum of the angles will not be 180°. To construct a triangle on a curved surface, say the sphere of Fig. 44-16, we must use the equivalent of a straight line: that is, the shortest distance between two points, which is called a geodesic. On a sphere, a geodesic is an arc of a great circle (an arc in a plane passing through the center of the sphere) such as the Earth’s equator and the Earth’s longitude lines. Consider, for example, the large triangle of Fig. 44-16: its sides are two longitude lines passing from the north pole to the equator, and the third side is a section of the equator as shown. The two longitude lines make 90° angles with the equator (look at a world globe to see this more clearly). They make an angle with each other at the north pole, which could be, say, 90° as shown; the sum

of these angles is

Ona

the sum of the angles of a triangle may not be 180°.

“North pole”

90° + 90° + 90° = 270°,

r’['his is clearly not a Euclidean space. Note, however, that if the triangle is small in omparison to the radius of the sphere, the angles will add up to nearly 180°, and the triangle (and space) will seem flat.

SECTION 44-4

FIGURE 44-16

two-dimensional curved surface,

Earth

General Relativity: Gravity and the Curvature of Space

1207

FIGURE 44-17 On a spherical surface (a two-dimensional world) a circle of circumference C'is drawn (red) about point O as the center. The radius of the circle (not the sphere) is the distance r along the surface. (Note that in our three-dimensional view, we can tell that C = 27a. Since r > a, then C < 27r.)

FIGURE 44-18 Exampleofa two-dimensional surface with negative curvature.

Another way to test the curvature of space is to measure the radius r and circumference C of a large circle. On a plane surface, C = 2zrr. But on a twodimensional spherical surface, C is less than 27rr, as can be seen in Fig. 44-17. The proportionality between C and r is less than 27r. Such a surface is said to have positive curvature. On the saddlelike surface of Fig. 44-18, the circumference of a circle is greater than 27r, and the sum of the angles of a triangle is less than 180°. Such a surface is said to have a negative curvature.

Curvature of the Universe What about our universe? On a large scale (not just near a large mass), what is the overall curvature of the universe? Does it have positive curvature, negative curvature, or is it flat (zero curvature)? We perceive our world as Euclidean (flat), but we can not exclude the possibility that space could have a curvature so slight thaq we don’t normally notice it. This is a crucial question in cosmology, and it can be answered only by precise experimentation. If the universe had a positive curvature, the universe would be closed, or finite in volume. This would not mean that the stars and galaxies extended out to a certain boundary, beyond which there is empty space. There is no boundary or edge in such a universe. The universe is all there is. If a particle were to move in a straight line in a particular direction, it would eventually return to the starting point—perhaps eons of time later. On the other hand, if the curvature of space was zero or negative, the universe would be open. It could just go on forever. An open universe could be infinite; but according to recent research, even that may not necessarily be so. Today the evidence is very strong that the universe on a large scale is very close to being flat. Indeed, it is so close to being flat that we can’t tell if it might have very slightly positive or very slightly negative curvature.

Black Holes FIGURE 44-19 Rubber-sheet analogy for space-time curved by matter.

According to Einstein’s theory, space-time is curved near massive objects. We might think of space as being like a thin rubber sheet: if a heavy weight is hung from it, it curves as shown in Fig. 44-19. The weight corresponds to a huge mass that causes space (space itself!) to curve. Thus, in Einstein’s theory” we do not speak of the “force” of gravity acting on objects. Instead we say that objects and light rays move as they do because space-time is curved. An object starting at rest or moving slowly near the great mass of Fig. 44—-19 would follow a geodesic (the

equivalent of a straight line in plane geometry) toward that great mass.

Weight

1208

CHAPTER 44

"Alexander Pope (1688-1744) wrote an epitaph for Newton: “Nature, and Nature’s laws lay hid in night: God said, Let Newton be! and all was light.” Sir John Squire (1884-1958), perhaps uncomfortable with Einstein’s profound thoughts, added: “It did not last: the Devil howling ‘Ho!/ Let Einstein be! restored the status quo.”

q

The extreme curvature of space-time shown by a black hole. A black hole, as we mentioned even light cannot escape from it. To become a must undergo gravitational collapse, contracting vithin a radius called the Schwarzschild radius: R

=

2GM C2

in Fig. 44-19 could be produced in Section 44-2, is so dense that black hole, an object of mass M by gravitational self-attraction to

’

where G is the gravitational constant and ¢ the speed of light. If an object collapses

to within this radius, it is predicted by general relativity to rapidly (= 107%s)

collapse to a point at r = 0, forming an infinitely dense singularity. This prediction is uncertain, however, because

in this realm we need

to combine

quantum

mechanics with gravity, a unification of theories not yet achieved (Section 43-12). | EXERCISE € What is the Schwarzschild radius for an object with 2 solar masses?

The Schwarzschild radius also represents the event horizon of a black hole. By event horizon we mean the surface beyond which no emitted signals can ever reach us, and thus inform us of events that happen beyond that surface. As a star collapses toward a black hole, the light it emits is pulled harder and harder by gravity, but we can still see it. Once the matter passes within the event horizon, the emitted light cannot escape but is pulled back in by gravity. All we can know about a black hole is its mass, its angular momentum (there could be rotating black holes), and its electric charge. No other information, no details of its structure or the kind of matter it was formed of, can be known because no information can escape. How might we observe black holes? We cannot see them because no light can escape from them. They would be black objects against a black sky. But they do exert a gravitational force on nearby objects. The black hole believed to be at the enter of our Galaxy (M =~ 4 X 10° Mg,,) was discovered by examining the motion of matter in its vicinity. Another technique is to examine stars which appear to move as if they were one member of a binary system (two stars rotating about their common

center of mass), but without a visible companion. If the unseen star is a black hole,

it might be expected to pull off gaseous material from its visible companion (as in Fig. 44-10). As this matter approached the black hole, it would be highly accelerated and should emit X-rays of a characteristic type before plunging inside the event horizon. Such X-rays, plus a sufficiently high mass estimate from the rotational

motion,

can

provide

evidence

for

a black

hole.

One

of the

many

candidates for a black hole is in the binary-star system Cygnus X-1. It is widely believed that the center of most galaxies is occupied by a black hole with a mass

10° to 10° times the mass of a typical star like our Sun.

EXERCISE D A black hole has radius R. Its mass is proportional to (a) R, (b) R?, (c) R®. Justify your answer.

44-5

The Expanding Universe:

Redshift and Hubble’s Law

We discussed in Section 44-2 how individual stars evolve from their birth to their death as white dwarfs, neutron stars, and black holes. But what about the universe as a whole: is it static, or does it change? One of the most important scientific discoveries of the twentieth century was that distant galaxies are racing away from us, and that the farther they are from us, the faster they are moving away. How flstronomers arrived at this astonishing idea, and what it means for the past history ©of the universe as well as its future, will occupy us for the remainder of the book.

SECTION 44-5

The Expanding Universe: Redshift and Hubble's Law

1209

Intensity

Low redshift galaxy spectrum z = 0.004

FIGURE 44-20 Atoms and molecules emit and absorb light of particular frequencies depending on the spacing of their energy levels, as we saw in Chapters 37 to 40. (a) The

spectrum of light received from a

(a)

relatively slow-moving galaxy. (b) Spectrum of a galaxy moving away from us at a much higher speed. Note how the peaks (or lines) in the

500

600

Wavelength (i)

700

Higher redshift galaxy spectrum z=0.104

spectrum have moved to longer wavelengths. The redshift is

2 g

=

E

()‘obs - )‘rest)/Arest .

(b)

500

600

Wavelength (nm)

700

That the universe is expanding was first put forth by Edwin Hubble in 1929. This idea was based on distance measurements of galaxies (Section 44-3), and determination of their velocities by the Doppler shift of spectral lines in the light received from them (Fig. 44-20). In Chapter 16 we saw how the frequency of sound is higher and the wavelength shorter if the source and observer move toward each other. If the source moves away from the observer, the frequency is lower and the wavelength longer. The Doppler effect occurs also for light, and we saw in Section 36-12 (Eq. 36-15) that according to special relativity, the Doppler shift is given by Aobs

=

Arest

1+ v/c 1 - v/c

’

source and observer moving away f from each h other

-

(44-5,

where A is the emitted wavelength as seen in a reference frame at rest with respect to the source, and A, is the wavelength observed in a frame moving with velocity v away from the source along the line of sight. (For relative motion toward each other, v < 0 in this formula.) When a distant source emits light of a particular wavelength, and the source is moving away from us, the wavelength appears longer to us: the color of the light (if it is visible) is shifted toward the red end of the visible spectrum, an effect known

HUBBLE'S LAW |

as a redshift. (If the source moves

toward us, the color shifts toward the blue or shorter wavelength.) In the spectra of stars in other galaxies, lines are observed that correspond to lines in the known spectra of particular atoms (see Section 37-10 and Figs. 35-22 and 37-20). What Hubble found was that the lines seen in the spectra from distant galaxies were generally redshifted, and that the amount of shift seemed to be approximately proportional to the distance of the galaxy from us. That is, the velocity v of a galaxy moving away from us is proportional to its distance d from us:

v = Hd.

This is constant The between precisely

(44-49)

Hubble’s law, one of the most fundamental astronomical ideas. The H is called the Hubble parameter. value of H until recently was uncertain by over 20%, and thought to be 50 and 80 km/s/Mpc. But recent measurements now put its value more at H = 71km/s/Mpc

(that is, 71 km/s per megaparsec of distance). The current uncertainty is about 5%, or +4 km/s/Mpc. If we use light-years for distance, then H = 22km/s per millio=™ light-years of distance: H = 22km/s/Mly 1210

CHAPTER 44

with an estimated uncertainty of £1 km/s/Mly.

Redshift Origins Galaxies very near us seem to be moving randomly relative to us: some move towards us (blueshifted), others away from us (redshifted); their speeds are on the rprder of 0.001c. But for more distant galaxies, the velocity of recession is much _reater than the velocity of local random motion, and so is dominant and Hubble’s law (Eq. 44-4) holds very well. More distant galaxies have higher recession velocity and a larger redshift, and we call their redshift a cosmological redshift. We interpret this redshift today as due to the expansion of space itself. We can think of the originally emitted wavelength A as being stretched out (becoming longer) along with the expanding space around it, as suggested in Fig. 44-21. Although Hubble thought of the redshift as a Doppler shift, now we understand it in this sense of expanding space. Contrast the cosmological redshift, due to the expansion of space itself, with an ordinary Doppler redshift which is due to the relative motion of emitter and observer in a space that can be considered fixed over the time interval of observation. There is a third way to produce a redshift, which we mention for completeness: a gravitational redshift. Light leaving a massive star is gaining in gravitational potential energy (just like a stone thrown upward from Earth). So the kinetic energy of each photon, Af, must be getting smaller (to conserve energy). A smaller frequency f means a larger (longer) wavelength A (= ¢/f), which is a redshift. The amount of a redshift is specified by the redshift parameter, z, defined as Aobs

7z = where A

=

— A

Arest

L

AA

)‘rcst

)

(44-5a)

is @ wavelength as seen by an observer at rest relative to the source, and

Aobs 18 the wavelength measured by a moving observer. Equation 44-5a can also be written as 7z _=

and r

S

Aobs -1 —

(44-5b)

Arest

giGURE 24-21 Simplified model of a 2-dimensional universe, ~ imagined as a balloon. As you blow up the balloon (= expanding

universe), the wavelength of a wave

onits surface gets longer (redshifted).

~ FIGURE 44-22 Hubble Ultra Deep Field photograph showing

what :may be among the most distant galaxies from us (small red dots,

highlighted by squares), with

Aobs

Arest'

{44=30)

z =~ 5or 6,existing when the

universe was only about 800 million

For low speeds not close to the speed of light (v < 0.1¢), the Doppler formula ~ ears old. The two distant galaxies in (Eq. 44-3) can be used to show (Problem 29) that z is proportional to the speed of

the source toward or away from us:

g = Aot T A

D

)\rest

~this photo are shown enlarged below.

[v sme (?z?)' [Hint: See Eqs. 40-12

and 40-13; we

assume

electrons fill

energy levels from 0 up to the Fermi energy.] (b) Thh nucleons contribute to the total energy mainly via the gray itational force (note that the Fermi energy for nucleons is negligible compared to that for electrons—why?). Use a gravitational form of Gauss’s law to show that the total gravitational potential energy of a uniform sphere of radius R is

5

R

by considering the potential energy of a spherical shell of radius r due only to the mass inside it (why?) and integrate from r =0 to r = R. (See also Appendix D.) (c) Write the total energy as a sum of these two terms, and set dE/dR = 0 to find the equilibrium radius, and evaluate it

and answer the same questions as in Problem 49. Calculate the Schwarzschild radius using a semi-classical (Newtonian) gravitational theory, by calculating the minimum radius R for a sphere of mass M such that a photon can escape from the surface. (General Relativity gives R = 2GM/c?.) 52. How large would the Sun be if its density equaled the

57. Determine the radius of a neutron star using the same argument as in Problem 56 but for N neutrons only. Show that the

your answer in light-years and compare with the Earth—Sun distance and the diameter of our Galaxy.

It is thought that physics as we know it can say nothing about the universe before this time.

)

critical density of the universe, p. ~ 10726 kg/m3? Express

for a mass equal to the Sun’s (2.0 x 10¥kg).

radius of a neutron star, of 1.5 solar masses, is about 11 km.

58. Use dimensional analysis with the fundamental constants c,

G, and # to estimate the value of the so-called Planck time.

Answers to Exercises A: Ourselves, 2 years ago.

C: 6km.

B: 600ly (estimating L from Fig, 44-6 as . ~ 8 X 10%°W;

D: (a); not the usual R?, but R: see formula for the

note that on a log scale, 6000 K is closer to 7000K than it is

to 5000 K).

1228

CHAPTER 44

Astrophysics and Cosmology

Schwarzschild radius.

E: (o).

-

e

A

A-1

Mathematical Formulas

Quadratic Formula

If

ax?+bx

then

x

A-2

+c¢c

=0

—b+\/b*

=

— dac

2a

Binomial Expansion (1£x)"

=

(x+y)”=x”1+X

A-3

nn—-1)

1 £ nx +

T

"

3!

=x”1+nX+(

x

n(n— 1)

x

Other Expansions

y* ’

)y—+

E L E e x _=1+x+7+57 r.

xz

x3

x4

ln(1+x)=x—7+?—:+ 03

sinf

=

cosh

=1

= In general:

A-4

:

f(x)

=

—

;

—

5

0

;

+

E

6?

-

o

315

f(0)

df ) =) (dx Ox

(@)a™) = (@)(e") = (ab) m

—

n+m

a™)(b")

=

(ab)"

(a")m

65

+

==

2

0 =0 +—+—60+

Exponents n

A-5

6

a"m

x+

T

|0 0 cos 8 =x/r>0 tan 6 = y/x>0

sin 6>0 cos 6 a]

[ [

|

aI\)

~

=

x* L a®

-1

al

dy

xdx

" 3 fsm ax dx

;sin ax 1

Jtanax dx

N/

(#2£ az)%

——cosax a 5

=

1

J

1

=

Jcos ax dx

[m # —1]

*

f————g)fi——g = sin“(%) - —cos'1 gl — ¥

[if x* = d?

~B-5 A Few Definite Integrals L

*

L

x"e

I

e

) jo

B-6

xe

—ax

—a

dx

2

dx

—ax”

5 dx

fl!

=

a"“

=

A /ZI‘

—

L

Tr

*©

L o

1 2a

L

*

xze

x3e

—ax?

—ll.\’:

2n,—ax x—e

dx

=

dx

=

s dx

-

m™

Tal‘s'

1

fi

1'35(2,1_1) 2n+1an

P — a

Integration by Parts

Sometimes a difficult integral can be simplified by carefully choosing the functions « and v in the identity: Ju dv

=

uv — Jv du.

[Integration by parts]

This identity follows from the property of derivatives fl—(uv) dx

=

ufl dx

+ vfl dx

or as differentials: d(uv) = udv + vdu. For example [xe™* dx can be integrated by choosing u = x and dv = e”*dx in the “integration by parts” equation above: Jxe”‘dx f

=

(x)(—e™¥) + fe""dx

=—-xe*—eF

=

—(x+

1) " SECTION B-6

A-7

-

More on Dimensional Analysis™ An important use of dimensional analysis (Section 1-7) is to obtain the form of an equation: how one quantity depends on others. To take a concrete example, let us try to find an expression for the period 7 of a simple pendulum. First, we try to figure out what 7 could depend on, and make a list of these variables. It might depend on its length £, on the mass m of the bob, on the angle of swing 6, and on the acceleration due to gravity, g. It might also depend on air resistance (we would use the viscosity of air), the gravitational pull of the Moon, and so on; but everyday experience suggests that the Earth’s gravity is the major force involved, so we ignore the other possible forces. So let us assume that 7 is a function of £, m, 6, and g, and that each of these factors is present to some power:

T = Cl¥m*6’g*. C is a dimensionless constant, and w, x, y, and z are exponents we want to solve for. We now write down the dimensional equation (Section 1-7) for this relationship:

[T] = [LI"[M*[L/T*.

Because 6 has no dimensions (a radian is a length divided by a length—see Eq. 10-1a), it does not appear. We simplify and obtain

[T] = [L]*“[MP[T]™*

To have dimensional consistency, we must have 1 0 0

= -2z =w+z = x

“

We solve these equations and find that desired equation must be

z = —3,w

=3,

T = CVi/gf(6),

and

x = 0. Thus our

(C-1

where f(6) is some function of 6 that we cannot determine using this technique. Nor can we determine in this way the dimensionless constant C. (To obtain C and f, we would have to do an analysis such as that in Chapter 14 using Newton’s laws, which reveals that C = 27 and f ~ 1 for small 6). But look what we have found, using only dimensional consistency. We obtained the form of the expression that relates the period of a simple pendulum to the major variables of the situation, ¢ and g (see Eq. 14-12c), and saw that it does not depend on the mass m. How did we do it? And how useful is this technique? Basically, we had to use our intuition as to which variables were important and which were not. This is not always easy, and often requires a lot of insight. As to usefulness, the final result in our example could have been obtained from Newton’s laws, as in Chapter 14. But in many physical situations, such a derivation from other laws cannot be done. In those situations, dimensional analysis can be a powerful tool.

In the end, any expression derived by the other means, for that matter) must be checked derivation of Eq. C-1, we can compare the lengths, £, and ¢, , whose amplitudes () are the I

L

Because C and experimentally square roots of at least in part;

A-8

APPENDIXC

_

CVh/gf(6)

cVb/gfe)

use of dimensional analysis (or by any against experiment. For example, in our periods of two pendulums of different same. For, using Eq. C-1, we would have

\/E

Vb

f(6) are the same for both pendula, they cancel out, so we cak determine if the ratio of the periods varies as the ratio of tt the lengths. This comparison to experiment checks our derivation, C and f(6) could be determined by further experiments.

More on Dimensional Analysis

Gravitational Force due to a

Spherical Mass Distribution In Chapter 6 we stated that the gravitational force exerted by or on a uniform sphere acts as if all the mass of the sphere were concentrated at its center, if the other object (exerting or feeling the force) is outside the sphere. In other words,

the gravitational force that a uniform sphere exerts on a particle outside it is F

=

=5 [m outside sphere of mass M] r where m is the mass of the particle, M the mass of the sphere, and r the distance of m from the center of the sphere. Now we will derive this result. We will use the concepts of infinitesimally small quantities and integration. First we consider a very thin, uniform spherical shell (like a thin-walled basketball) of mass M whose thickness ¢ is small compared to its radius R (Fig. D-1). The force on a particle of mass m at a distance r from the center of the shell can be calculated as the vector sum of the forces due to all the particles of the shell. We imagine the shell divided up into thin (infinitesimal) circular strips so that all points on a strip are equidistant from our particle m. One of these circular strips, labeled AB, is shown in Fig. D-1. It is R df wide, ¢ thick, and has a radius Rsin 6. The force on our particle m due to a tiny piece of the strip at point A is represented by the vector F, shown. The force due to a tiny piece of the strip at point B, which is diametrically opposite A, is the force Fy. We take the two pieces it A and B to be of equal mass, so

and Fy are each equal to

F, = Fy.

The horizontal components

of F,

Fycos¢ and point toward the center of the shell. The vertical components of F, and Fy are of equal magnitude and point in opposite directions, and so cancel. Since for every point on the strip there is a corresponding point diametrically opposite (as with A and B), we see that the net force due to the entire strip points toward the center of the shell. Its magnitude will be m dM dF = G cos ¢, £2 where points density where

dM on we dV

is the mass of the entire circular strip and £ is the distance from all the strip to m, as shown. We write dM in terms of the density p; by mean the mass per unit volume (Section 13-2). Hence, dM = p dV, is the volume of the strip and equals (2R sin 6)(¢)(R d6). Then the

force dF due to the circular strip shown is mp2m Rt sin 6 do dF

=

G

[2

cos ¢.

(D-1)

FIGURE D-1 Calculating the gravitational force on a particle of mass m due to a uniform spherical shell of radius R and mass M.

A-9

FIGURE D-1 (repeated) Calculating the gravitational force on a particle of mass m due to a uniform spherical shell of radius R and mass M.

To get the total force F that the entire shell exerts on the particle m, we must

integrate over all the circular strips: that is, we integrate dF = G

mp2mR%t sin 6 do

cos ¢

[2

(D-1)

from 6 = 0° to 6 = 180°. But our expression for dF contains £ and ¢, which are functions of 6. From Fig. D-1 we can see that fcos¢p = r — Rcosé. Furthermore, we can write the law of cosines for triangle CmA: r + R* - P cosf = 2rR

D-2)

With these two expressions we can reduce our three variables (¥, 6, ¢) to only one, which we take to be £. We do two things with Eq. D-2: (1) We put it into the equation for £ cos ¢ above: cos¢p

=

1

rr+ £ - R?

z(r — Rcosf)

= B

and (2) we take the differential of both sides of Eq. D-2 (because sin 6 df appears in the expression for dF, Eq. D-1), considering r and R to be constants when summing over the strips: 20 df 'R

=

. sinfdf

or

_bde = R

We insert these into Eq. D-1 for dF and find 2

_

2

dF = Gmpmfiz(1 # = PR )dl?. r

Now we integrate to get the net force on our thin shell of radius R. To integrate over all the strips (6 = 0° to 180°), we must go from { =r — R to £ =r + R (see Fig. D-1). Thus,

=

=

RZ

{=r+R

l

rZ

2

Q 3 S

R

Il

F

r’ |:2

R

-

£

:I£=r—R

Gmpmt = (4R).

The volume V of the spherical shell is its area (4 R?) times the thickness 7. Hence

the mass M = pV = p4wR*, and finally

Foomt

mM 7

|

particle of mass m outside a ] thin uniform spherical shell of mass M

This result gives us the force a thin shell exerts on

a particle of mass

m

a

distance r from the center of the shell, and outside the shell. We see that the force

is the same as that between m and a particle of mass M at the center of the shell. In other words, for purposes of calculating the gravitational force exerted on or by a uniform spherical shell, we can consider all its mass concentrated at its center. What we have derived for a shell holds also for a solid sphere, since a solic™ sphere can be considered as made up of many concentric shells, from R =0 to R = R, where Ry is the radius of the solid sphere. Why? Because if each shell has A-10

APPENDIXD

mass dM, we write for each shell, dF = Gm dM/r?, where r is the distance from the center C to mass m and is the same for all shells. Then the total force equals the sum or integral over dM, which gives the total mass M. Thus the result r

mM

F=G

[particle of mass m outside

2

solid sphere of mass M

] (D-3)

is valid for a solid sphere of mass M even if the density varies with distance from the center. (It is not valid if the density varies within each shell—that is, depends not only on R.) Thus the gravitational force exerted on or by spherical objects, including nearly spherical objects like the Earth, Sun, and Moon, can be considered to act as if the objects were point particles. This result, Eq. D-3, is true only if the mass m is outside the sphere. Let us next consider a point mass m that is located inside the spherical shell of Fig. D-1. Here, r would be less than R, and the integration over £ would be from £ = R — r tof=R+r, so

[

IS

r2

-

RZ]R+r

f

R-r

=

0.

Thus the force on any mass inside the shell would be zero. This result has particular importance for the electrostatic force, which is also an inverse square law. For the gravitational situation, we see that at points within a solid sphere, say 1000 km below the Earth’s surface, only the mass up to that radius contributes to the net force. The outer shells beyond the point in question contribute zero net gravitational effect. The results we have obtained here can also be reached using the gravitational analog of Gauss’s law for electrostatics (Chapter 22).

r

APPENDIX D

Gravitational Force due to a Spherical Mass Distribution

A-11

Differential Form of

5

Maxwell’s Equations

Maxwell’s equations can be written in another form that is often more convenient than Egs. 31-5. This material is usually covered in more advanced courses, and is included here simply for completeness. We quote here two theorems, without proof, that are derived in vector analysis textbooks. The first is called Gauss’s theorem or the divergence theorem. It relates the integral over a surface of any vector function F to a volume integral over the volume enclosed by the surface:

35

Area A

rv.dA':J

VolumeV

V-Fav.

The operator V is the del operator, defined in Cartesian coordinates as

The quantity ax

+ — ay

+

0z

“

is called the divergence of F. The second theorem is Stokes’s theorem, and relates a line integral around a closed path to a surface integral over any surface enclosed by that path:

SE

Line

F-di=f

Area A

¥ x F-dA.

The quantity V X F is called the curl of F. (See Section 11-2 on the vector product.) We now use these two theorems to obtain the differential form of Maxwell’s equations in free space. We apply Gauss’s theorem to Eq. 31-5a (Gauss’s law):

3€ £.4k = [v.i:dv < 2.€9 A

Now the charge Q can be written as a volume integral over the charge density p:

Q = [pdV. Then

JV-F:dV

A

€0

fpdv.

Both sides contain volume integrals over the same volume, and for this to be true

over any volume, whatever its size or shape, the integrands must be equal:

VE

= 2.

This is the differential form of Gauss’s law. The second

¢§B-dA = 0, is treated in the same way, and we obtain

V-B A-12

(E-1)

€9

= 0.

of Maxwell’s

equations,

(E-2)

“

Next, we apply Stokes’s theorem to the third of Maxwell’s equations,

3€17:.di - JVXFI-dK o dd,, dt riince the magnetic flux &4 = [B-dA,

we have

JVXE-d[& . -—j where we use the partial derivative, 9B/dt, since B may also depend on position. These are surface integrals over the same area, and to be true over any area, even

a very small one, we must have

VxE = -2, This is the third of Maxwell’s equations in differential form. Finally, to the last of

Maxwell’s equations, I

%B’d?

=

[.Lol

+

dd

Moeodt—Ea

we apply Stokes’s theorem and write =5

va

e

B’dA

=

Mol

&, = fE-dK: +

i)

=5

e

flofiO-JE'dA.

ot

The conduction current / can be written in terms of the current density j, using

Eq.25-12:

] =

Jj-d}l.

r[hen Maxwell’s fourth equation becomes: s

J'V

e

X

e

B-dA

=

-

-)

[.Lon‘dA

J

+

.

[L()E()EJE'dA.

For this to be true over any area A, whatever its size or shape, the integrands on each side of the equation must be equal:

VXB

.

4 IE = uoj + poeoot -

Equations E-1, 2, 3, and 4 are Maxwell’s equations in differential form for free space. They are summarized in Table E-1.

TABLE E-1

Maxwell’s Equations in Free Space’

Integral form

Differential form

{)E-d‘& a2

Vil

§f;.d; &g

$B=0

. dd 3€E-¢fi=——”

ol B Fxi=-2

()]

dt

ot

a5 do B-dl = puol + poeg E dt 1V stands for the del operator V = i

0

+

ax

28

—

'lay

PR . oE VXB=puj+ poeo-—ot 9 . . + k 5% in Cartesian coordinates.

APPENDIX E

Differential Form of Maxwell’s Equations

A-13

‘

¢ ¥

]

2y

+

Selected

1

8))

2

3

V4

Element

Symbol

0 1

2

(Neutron) Hydrogen Deuterium Tritium Helium

n H dorD torT He

3

Lithium

Li

4

Beryllium

Be

5

Boron

B

6

Carbon

C

7

Nitrogen

N

8

Oxygen

(@)

9 10

Fluorine Neon

F Ne

11

Sodium

Na

12 13 14

Magnesium Aluminum Silicon

Mg Al Si

15

Phosphorus

P

Atomic Number

@

O]

A-14

)

Decay* Mode)

(if radioactive)

B~ 99.9885% 0.0115% /2 0.000137% 99.999863% 7.59% 92.41% EC.Y. 100% 19.9% 80.1% B, EC 98.93% 1.07% B~ B*,EC 99.632% 0.368% B*, EC 99.757% 0.205% 100% 90.48% 9.25% B, EC,Y 100% B .Y 78.99% 100% 92.2297% B 100% B~

10.23 min

Atomic

% Abundance (or Radioactive

1 1 9 3 3 4 6 7 7 9 10 11 1 12 13 14 13 14 15 15 16 18 19 20 22 22 23 24 24 27 28 31 31 32

1.008665 1.007825 2.014102 3.016049 3.016029 4.002603 6.015123 7.016005 7.016930 9.012182 10.012937 11.009305 11.011434 12.000000 13.003355 14.003242 13.005739 14.003074 15.000109 15.003066 15.994915 17.999161 18.998403 19.992440 21.991385 21.994436 22.989769 23.990963 23.985042 26.981539 27.976927 30.975363 30.973762 31.973907

A

Mass'

"The masses given in column (5) are those for the neutral atom, including the Z electrons. Chapter 41; EC = electron capture.

(6)

Mass Number

Half-life

12.312yr

53.22 days

20.370 min

5730 yr 9.9670 min

122.5 min

2.6027 yr 14.9574 h

157.3 min 14.284 days

(§))

Atomic Number VA

2

3)

Element

Symbol

C))

%)

(6)

Mass Number A

Atomic Mass

32

31.972071

94.9%

35 35 7 40 39 40

34.969032 34.968853 36.965903 39.962383 38.963707 39.963998

B~ 75.78% 24.22% 99.600% 93.258% 0.0117%

40 45 48 51 52 55 56 59 60 58 60 63 65 64 66 69 72

39.962591 44955912 47.947946 50.943960 51.940508 54.938045 55.934938 58.933195 59.933817 57.935343 59.930786 62.929598 64.927790 63.929142 65.926033 68.925574 71.922076

96.94% 100% 73.72% 99.750% 83.789% 100% 91.75% 100% B.Y 68.077% 26.223% 69.17% 30.83% 48.6% 27.9% 60.108% 27.5%

73.921178 74.921596 79.916521 78.918337 83.911507 84.911790 85.909260 87.905612 89.907738 88.905848 89.904704 92.906378 97.905408 97.907216 101.904349 102.905504 105.903486 106.905097

36.3% 100% 49.6% 50.69% 57.00% 72.17% 9.86% 82.58% B~ 100% 51.4% 100% 24.1% B .Y 31.55% 100% 27.33% 51.839%

108.904752 113.903359

48.161% 28.7% 32.58% 57.21%

16

Sulfur

S

17

Chlorine

Cl

18 19

Argon Potassium

Ar K

20 21 22 23 24 25 26 27,

Calcium Scandium Titanium Vanadium Chromium Manganese Iron Cobalt

Ca Sc Ti \Y% Cr Mn Fe Co

28

Nickel

Ni

29

Copper

Cu

30

Zinc

Zn

31 32

Gallium Germanium

Ga Ge

33 34 35 36 37 38

Arsenic Selenium Bromine Krypton Rubidium Strontium

As Se Br Kr Rb Sr

39 40 41 42 43 44 45 46 47

Yttrium Zirconium Niobium Molybdenum Technetium Ruthenium Rhodium Palladium Silver

Y Zr Nb Mo Te Ru Rh Pd Ag

74 75 80 79 84 85 86 88 90 89 90 93 98 98 102 103 106 107

48

Cadmium

Cd

109 114

49

Indium

In

115

114.903878

50 51

Tin Antimony

Sn Sb

120 121

119.902195 120.903816

% Abundance (or Radioactive Decay Mode)

B~ ,EC,7,8"

95.71%; B~

APPENDIX F

Y Half-life (if radioactive) 87.32 days

1.265 x 10° yr

52710 yr

28.80 yr

42 X 105 yr

441 x 10%yr

Selected Isotopes

A-15

@

Atomic Number Z

3)

Element

Symbol

2 53

Tellurium Todine

1e I

54

Xenon

Xe

55 56

Cesium Barium

Cs Ba

57 58 59 60 61 62 63 64 65 66 67 68 69 70 q1 72 73 74

Lanthanum Cerium Praseodymium Neodymium Promethium Samarium Europium Gadolinium Terbium Dysprosium Holmium Erbium Thulium Ytterbium Lutetium Hafnium Tantalum Tungsten (wolfram)

La (Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Hf Ta W

76

Osmium

Os

a1

Iridium

Ir

78 79 80

Platinum Gold Mercury

Pt Au Hg

81 82

Thallium Lead

Tl Pb

75

A-16

(b))

Rhenium

Re

83

Bismuth

Bi

84

Polonium

Po

85

Astatine

At

APPENDIXF

Selected Isotopes

@

)

(©)

Mass Number A

Atomic Mass

% Abundance (or Radioactive Decay Mode)

130 127 131 132 136 133 137 138 139 140 141 142 145 152 153 158 159 164 165 166 169 174 175 180 181 184

129.906224 126.904473 130.906125 131.904154 135.907219 132.905452 136.905827 137.905247 138.906353 139.905439 140.907653 141.907723 144.912749 151.919732 152.921230 157.924104 158.925347 163.929175 164.930322 165.930293 168.934213 173.938862 174.940772 179.946550 180.947996 183.950931

34.1%; 8B~ 100% B 26.89% 8.87%; BB 100% 11.232% 71.70% 99.910% 88.45% 100% 27.2% EC,a 26.75% 5219%% 24.84% 100% 28.2% 100% 33.6% 100% 31.8% 97.41% 35.08% 99.988% 30.64%; a

191 192 191 193 195 197 199 202 205 206 207 208 210 211 212 214 209 211 210 214 218

190.960930 191.961481 190.960594 192.962926 194.964791 196.966569 198.968280 201.970643 204.974428 205.974465 206.975897 207.976652 209.984188 210.988737 211.991898 213.999805 208.980399 210.987269 209.982874 213.995201 218.008694

BeY 40.78% 37.3% 62.7% 33.832% 100% 16.87% 29.9% 70.476% 24.1% 22.1% 52.4% B, «a BT B B&Y 100% a;YiB: a, 5V, [EC a,Y a, B~

187

186.955753

62.60%; B~

0 Half-life (if radioactive) >9.7 X 102 yr 8.0233 days >85

% 107y

17.7yr

>8.9 X 10% yr

435 x 10V yr

15.4 days

22.23yr 36.1 min 10.64h 26.8 min 2.14 min 138.376 days 162.3 us 14s

@

r

Atomic Number VA

2

3

Element

Symbol

O}

(6)

Atomic Mass

% Abundance (or Radioactive Decay Mode) a,Y B, «a a;? B .Y, «a a,Y

86 87 88 89 90

Radon Francium Radium Actinium Thorium

Rn Fr Ra Ac Th

222 223 226 227 228

222.017578 223.019736 226.025410 227.027752 228.028741

91

Protactinium

Pa

231

231.035884

a,Y

2]

Uranium

U

232

232.037156

a,Y

235

235.043930

0.720%; a, ¥

238.050788 239.054293 237.048173 239.052939 239.052163

99.274%; a, ¥ B Y a,Y B,Y a,Y

232

233

236

r

)

Mass Number A

232.038055 233.039635

236.045568

100%; a, ¥ a,Y oY

93

Neptunium

Np

94

Plutonium

Pu

238 239 237 239 239

244

244064204

95 96 97 98

Americium Curium Berkelium Californium

Am Cm Bk Cf

243 247 247 251

243.061381 247.070354 247.070307 251.079587

Y a,Y a,Y a,Y

99

@

Y] Half-life (if radioactive) 3.8232 days 22.00 min 1600 yr 21.772 yr 698.60 days

1.405 x 100 yr

3.276 x 10* yr

68.9 yr

1.592 X 10° yr

7.04 x 10% yr

2342 X 107 yr 4.468 x 10° yr 23.46 min 2.144 X 10 yr 2.356 days 24,100 yr

8.00 X 107 yr

7370 yr 1.56 x 107 yr 1380 yr 898 yr

Einsteinium

Es

252

252.082980

a, EC,7

471.7 days

100 101 102

Fermium Mendelevium Nobelium

Fm Md No

257 258 259

257.095105 258.098431 259.10103

a,y a,Y a, EC

100.5 days 51.5 days 58 min

103

Lawrencium

ILx

262

262.10963

a, EC, fission

104 105 106 107 108 109 110 111

Rutherfordium Dubnium Seaborgium Bohrium Hassium Meitnerium Darmstadtium Roentgenium

Rf Db Sg Bh Hs Mt Ds Rg

263 262 266 264 269 268 271 272

263.11255 262.11408 266.12210 264.12460 269.13406 268.13870 271.14606 272.15360

fission a, fission, EC a, fission a a a a a

10 min 35s ~21s ~0.44s ~10s 21 ms ~70 ms 3.8 ms

Uub

277

277.16394

a

~0.7 ms

112

~4h

Preliminary evidence (unconfirmed) has been reported for elements 113, 114, 115,116 and 118.

APPENDIX F

Selected Isotopes

A-17

Answers to Odd-Numbered Problems CHAPTER 1

1. (a) 1.4 X 10Vy; (b) 4.4 x 10", 3. (a) 1.156 x 100

(b) 3.16 x 10' ns; (c) 3.17 x 107 8y.

25. (a) 21.2m/s;

(b) 2.00 m/s?.

27. 17.0m/s%,

47. 9cm/y.

(d) 3.2865 x 10%

49. 2 x 10%kg/y. 51. 75 min.

29. (a) m/s, m/s%; (b) 2Bm/s%; (c) (A + 10B) m/s, 2B m/s%;

(e) 2.19 x 1074

53. 4 X 10° metric tons, 1 X 108 gal.

(d) A — 3Bt™ 31. 1.5m/s% 99m. 33, 240m/s?,

59. (a) 0.10 nm;

37. 45.0m. 39. (a) 560 m;

(b) 2.18 x 10%;

(c) 6.8 X 1073;

(f)4.44 x 10% 5. 4.6%. 7. 1.00 X 10°s. 9. 0.24 rad.

11. (a) 0.2866 m; (b) 0.000085 V; (c) 0.00076 kg;

(d) 0.0000000000600 s; (e) 0.0000000000000225 m; (f)2,500,000,000V. 13. 5'10” = 1.8m, 1651bs = 75.2kg. 15. (a) 1 ft> = 0.111 yd?; (b) 1 m? = 10.8 ft2. 17. (a) 3.9 X 10 %in.; (b) 1.0 x 108 atoms.

19. (a) 1 km/h = 0.621 mi/h; (b)

1m/s = 3.28 ft/s;

(c) 1km/h = 0278 m/s. 21. (a) 9.46 x 10 m;

(b) (c) 23. (a) (b)

6.31 X 10*AU; 720 AU/h. 3.80 x 10 m? 13.4.

25. 6 X 10° books.

27.5 X 10*L.

29. (a) 1800. 31. 5 X 10°m.

33. 6.5 X 10°m. 35, [M/L3). 37. (a) Cannot; (b) can; (c) can.

39. (1 x 107°)%, 8 significant figures.

A-18

41. (a) 3.16 x 107s;

43.2 x 10™*m. 45. 1 x 10" gal/y.

55. 1 x 10° days 57. 210yd, 190 m. (b) (c) (d) 61. (a) (b) 63. 8 X

1.0 X 10° fm; 1.0 X 10'° A; 9.5 x 10¥ A, 3%, 3%; 0.7%, 0.2%. 1072 m°,

65. L/m,L/y, L.

67. (a) 13.4; (b) 49.3. 69. 4 x 105 kg. CHAPTER 2 1. 61 m.

3. 0.65cm/s, no.

5. 300 m/s, 1 km every 3 sec. 7. (a) 9.26 m/s; (b) 3.1 m/s. 9. (a) 0.3 m/s; (b) 1.2 m/s; (c) 0.30m/s; (d) 1.4 m/s; (e) —0.95m/s.

11. 2.0 X 10's.

13. (a) 54 X 10°m; 15. 17. 19. 21 23.

(b) 72 min. (a) 61 km/h; (b) 0. (a) 16 m/s; (b) +5m/s. 6.73 m/s. 5. (a) 48s; (b) 90s to 108 s; (c)

0to 425,655 to 83s,90s to 108s;

(d) 655 to 83s.

-

35, 441 m/s?, 2.61s.

(b) 47 s;

(c) 23m, 21 m. 41. (a) 96 m; (b) 76 m. 43, 27 m/s. 45, 117 km/h.

47, 0.49 m/s2. 49, 1.6s. 51. (a) 20m; (b) 4s. 53. 1.16s. 55. 5.18s. 57. (a) 25 m/s; (b) 33m; (c) 1.25; (d)5.2s. 59. (a) 14 m/s; (b) fifth floor. 61. 1.3 m. 63. 18.8m/s, 18.1 m. 65. 52m. 67. 106 m.

69. (a) £ (1 = &™), g () e 71. 6. 73. 1.3 m. 75. (b) 10 m; (c) 40m. 77. 52 X 102 m/s% 79. 4.6 6.7 81. (a) (b) (c)

m/s to 5.4m/s, 5.8 m/s to m/s, smaller range of velocities. 5.39s; -\ 40.3m/s; 90.9m.

(b) —22.8,9.85; (c) 24.8,23.4° above the —x axis.

625 km/h, 553 km/h; 1560 km, 1380 km. 4.2 at 315°; 1.0i — 5.0j or 5.1 at 280°. —53.7i + 1.31j or 53.7 at 1.4°

(b) 53.71 — 1.31j or 53.7 at 1.4°

(d) 38.3m/s at 25.7°.

13. (a) —92.5i — 19.4j or 94.5 at 11.8° . below —x axis;

5

15. (—2450m)i + (3870 m)j + (2450 m)k, 5190 m. 17. (9.60i — 2.00tk) m/s, (—2.00k) m/s?.

65.

3

4

5

~ 86 5 84

£ 82

\\

280 5,78

O

£76

Fa2

AN

F = (8.0i + 6.0j) m.

~

2 4 6 8 Distance along pool (m)

10

29, 44 m, 6.9 m.

a1, 18°,72%

75.

79. 1.8 m/s%. 81. 1.9m/s, 2.7s. Dv

83. (a) (vz—_uz-), D

() —\/fi 85. 54°.

87. [(1.5m)i — (2.0¢ m)i] + [(=3.1m)j + (1.75¢> m)j,

D

parabolic.

89. Row at an angle of 24.9° upstream and run 104 m along the bank in a total time of 862 seconds.

1. 286 km, 11° south of west.

3. 10.1, —39.4°.

131 km/h, 43.1° north of east.

77. 7.0m/s.

(3.5m/s?)j,

CHAPTER 3

5. (a)

horizontal.

(b) 421 cm/s at 41.3°, 25. (a) (6.0 — 18.0¢%)) m/s, (6.0i — 36.0¢)) m/s%; (b) (191 — 94j)m, (15i — 110j) m/s. 27. 414m at —65.0°.

7 \\

0

76 m, 28° south of east, 9° below the

v = 10.0m/s,

23. (a) (3.161 + 2.78j) cm/s;

97. (b) 6.8 m.

I

73. (66 m)i — (35m)j — (12m)k,

(d) vy = 8.0m/s, vy = 6.0m/s, Time (s)

N

71. 43.6° north of east.

(c) (2.0% + 1.5¢%) m; 2


1.5kg.

mg

53. ¢

11. 89.8N. 13. 1.8 m/s? up. 15. Descend with @ = 2.2 m/s2.

17. —2800 m/s?, 280 g’s, 1.9 X 10°N. 19. (a) 7.55,135,7.5ss;

(b) 12%, 0%, —12%; (c) 55%. 21. (a) 3.1 m/s%

mamsg

+

mB’g

+

VS

)

mA+mB'

ma + mg + me

57. 1.52m/s% 183N, 19.8N.

59.

(mA

+ mpg

8-

g =

4 2

g6 [3}

+ mc)mB

(mA — mb)

23.33 X 10°N. 25. (a) 150N;

b

3 220

8 210

£

0

2

50

(b) masin® > mpg

i

~

L —

(b) 35N, 71N.

35. 63 X 10°N,84 X 10°N.

37. (a) 19.0N at 237.5°,1.03 m/s? at 237.5%

(b) 14.0N at 51.0°,0.758 m/s? at 51.0°

(c) 0.74. 71, 9.9°.

N

73. (a) 41 m_/s;

41. 40 x 10°m.

77. 87 X 10°N,

45. (a) 9.9N; (b) 260 N. 47. (a) mgg — Fr = mga;

Fr = mcg = mca;

(b) 0.68 m/s?, 10,500 N. 49. (a) 2.8 m; (b) 2.5s.

A-20

72° above the horizontal.

79. (a) 0.6 m/s% (b) 1.5 X 10°N. 81. 1.76 X 10*N. 83, 3.8 X 10°N, 7.6 X 10>N. 85. 3.4 m/s.

87. (a) 23N; (b) 3.8N.

Answers to Odd-Numbered Problems

75

90

%

-

15

T

—

I

|

30 45 60 Angle (degrees)

75

|| I

90

.

28 9000

b

< 6000

E 3000

0

0

15

30

o= 1

45

.

BN

60

Angle (degrees)

N

N

75

| |

L

90

The graphs are all consistent with the results of the limiting cases. CHAPTER 5 3. 0.20.

39. E;t%

43, 12°.

|;

1. 65N, 0.

(b) 1.4 x 102N, 75. (a) Mg/2; (b) Mg/2, Mg/2,3Mg/2, Mg.

5K

o

‘

— —

15000

ma + mp

(b) 6.8kg, 26 N;

60

Angle (degrees)

Z 12000 Z

90

[

0

69. (a) mpsin g — mp sin O "

33. (a) 31N,63N;

45

20l 0

(ma up the plane).

?mg

(b) 11.5kN, no.

30

[

(©) 2 Val.

(ma down the plane),

31. (a) 1.5m;

1

15

LT / £ 0

(ma + mp)

)

75

I:

\

mpsin@ < mp mg

30 45 60 Angle (degrees)

T ]

(®) 2gyo

\

(€) pa < ppia =

[(mA+ mg)sin 6 — (mama+ upmg) cos 6

=g

(ma+ mp)

ma mg oaB ] (uB = ma)cosb, g(mA + mg

! pa > ppias = g(sin@ — pacos),

ag = g(sin@ — ugcosh), Fr = 0. (b) 6.7 kg.

22} 2dgcos 6 B (b) ps = tané.

tan 6,

33. (M + m)g

(sin® + ucosf)

(cos® — wsing)’ 35, (a) 1.41 m/s%;

(b) 317N.

(c) 9.42 m/s® approximate,

9.43 m/s? exact. 21. 9.78 m/s?, 0.099° south of radially inward.

81. Yes, 14 m/s.

23, 7.52 X 10° m/s. 25. 1.7 m/s? upward.

28.3m/s, 0.410rev/s. 3500 N, 1900 N. 35°. 132 m. (a) 55s; (b) centripetal component of the normal force.

93, (a) 6 = cos™!

(b) 4.5m/s2,

Earth.

19. (b) g decreases as r increases;

75. 10 m. 77. 0.46. 79. 102N, 0.725.

27. (a) 0.22s;

(b) 0.16 m. 29, 0.51. 31 (a) 82N;

15. 3.46 x 108 m from the center of the

1. —Eg [t + %(e‘%‘ - 1)],ge“rli,1".

83. 85. 87. 89. 91.

23. (a) 5.0kg;

v

|

13. 2" ~ 1.26 times larger.

69. (a) 14 kg/m; (b) STON.

(b) (¢) 95. 82°. 97. (a) (b)

73.6% no.

27. 720 % 1Ps, 29. (a) 520N; (b) 520N; (c) 690N; (d) 350N; (e) 0.

B,

4arf?’

31, (a) S9N, toward the Moon;

(b) 110N, away from the Moon. 33. (a) They are executing centripetal motion;

16 m/s; 13m/s.

(b) 9.6 X 10¥kg.

99, (a) 0.88 m/s%; (b) 0.98 m/s?.

37. Vrg. 39, 30 m.

20 e

(b) 5.4 X 13N

39. 160y.

41. 2 x 10%y. 43. Europa: 671 X 10° km;

I \ T Varying friction

=

7. (a) 1.9 X 10°m


e"7fi'.

g

62

’

slope = 1.50 as predicted,

51. 2.64 X 10°m

11. sz{[% + —3—"9——3/-2]

( )-n-;)g- + ij

(x5 + yd)

+[i+_§_y9_]j} Yo

(< +

)

53, (a) 438 X 10’ m/s% (b) 2.8 X 10°N; (c) 9.4 x 10°m/s. 55, Tipner = 2.0 X 10%s,

Tonter = T X 104s,

Answers to Odd-Numbered Problems

A-21

57. 5.4 X 10" m, it is still in the solar system, nearest to Pluto’s orbit. 59. 23 g’s.

37. 3.0 X 10°7J. 400 300

61. 7.4 X 10%kg, 3.7 X 10° Mgy, .

[

/

65. 1.21 X 105m. 67. Vaeposit = 5 X 107 m?, Tdeposit = 200 m;

Meposic = 4 X 101k,

I

~ 20,000 &~

&

39. 41. 43. 45.

10,000

0

20,000 3 (AU3)

30.000

(b) 39.44 AU. CHAPTER 7

L 7.7 % 10P7. 3. 1.47 X 10%]. 5. 60007. 7.45 X 10°7J. 9. 590J.

53, —4.5 X 10°J. 55. 3.0 X 10°N, 57. (a)

%;

()

am

(d) 43m/s.

67. (a) %mv%(l 4 2fl>;

27. 52.5°,48.0°, 115°. 29, 113.4° or 301.4°. 31. (a) 130°;

(b) negative sign says that the angle is obtuse.

|

Force

kx),

{

! !

| | | | 1

|

i

4+—————[

X

Stretch distance

A-22

73. (a) 32.217; (b) 5547, (c) —3337;

() 0;

|

X3

(e) 2531. 75. 12.31.

Answers to Odd-Numbered Problems

\

0.4

k

k

13. 6.5m/s. 15, (a) 93N/m;

(b) 22 m/s>. (b) 3.01 m. 21. No, D = 2d.

(c) 2.37m.

1

0.3

11. 499 m/s.

(d) the ball moves different distances during the throwing process in the two frames of reference.

71. 17101.

0.2 Stretch (m)

—Ex—z-f-s—mz.

19. (a) 7.47m/s;

(b) 21.0 m/s; F = kx

9.U(x)=

Earth, mv} relative to train;

69. (a) 2.04 X 10°J;

35. 0.111. kxy4————

)

(¢) %mv%(l + 2%21—) relative to

25. —1.4i + 2.0j.

0.1

Tkx? —tax* - 1bx’ + C.

65. 27 m/s.

(b) 3mv3;

0

3. 54 cm. 5. (a) 42.07; (b)1117; (c) same as part (a), unrelated to part (b). 7. (a) Yes, the expression for the work depends only on the endpoints; (b)U (1x) =

3Fx

(c) O;

(d) 0. (a) 1.1 X 1077; (b) 5.0 X 10717. —4907, 0,490 1. 1.5i — 3.0j. (a) 7.1; (b) —250; (c) 2.0 x 10,

(b) 10.0N/m; (c) 2.00N.

1. 0.924 m.

(b) —47017;

(c) 66007;

(b) 18 m/s. 93, (a) F = 10.0x;

CHAPTER 8

63. (a) 6407;

(b) —6600 J;

15. 21. 23.

86 kJ, 42°. 1.5N. 2 X 10’ N/m. 6.7°,10°. (a) 130N, yes (=29 1bs);

0.0

(b)3.

59. 8.3 X 10*N/m. 61. 14001,

11. (a) 1700N;

13.

28007. 670J. 5k X? + tax* + 1bX3, 401, \/37RF

4y, L= 2 49. 721 51. (a) V/3;

I

10,000 0

79. 81, 83. 85. 87.

(b) 470N, perhaps not (~1101bs).

71. 0.44r. 73. (a) 53N; (b) 3.1 X 10%kg. 77. 1 X 107%m3/kg 0: unbounded;

© resa < (2

GMEms

48, (a) —— S

@ —.

19. (4.0i + 3.3j — 3.3k) m/s. 21. (a) (116i + 58.0j) m/s; (b) 5.02 X 10°J.

25. 2.1 kg m/s, to the left.

27. 0.11N.

49. (a) 6.2 X 10° m/s;

(b) 42 X 10*m/s, 2VEarth orbit -

Vesc at Earth orbit —

53, (a) 1.07 X 10*m/s;

79. 2.52 X 10*W, 81. (a) 42m/s; (b)2.6 X 10°W.

(b) 1.16 X 10*m/s; (c) 1.12 X 10*m/s.

83. (a) 28.2m/s; (b) 116 m.

GMg 55. (a) — 28 (b) 1.09 X 10*m/s.

85. (a) \/2—gz; (b) V1.2g¢.

89. (a) 89 X 10°I; (b) 5.0 X 10' W, 6.6 X 102 hp;

GMm ’

59. 1.12 X 10*m/s.

(c) 330 W, 0.44 hp.

63. 510N. 65. 2.9 X 10*W or 38 hp. 67. 42 x 10°N, opposing the velocity.

91. (a) 29°%

69. 510 W.

93. 5800 W or 7.8 hp.

712 X 105W.

(b) 480N; (c) 690N. 95, (a) 2.8 m;

73. (a) —2.0 X 1°W (b) 3800 W; (c) —120W; (d) 1200 W. 75. The mass oscillates between +x, and —xg, with a maximum speed at x = 0.

(b) 1.5m;

(¢) 1.5m.

97. 1.7 x 10° m® 99. (a) 5220 m/s; (b) 3190 m/s. 101. (@) 1500 m; (b) 170 m/s.

U(x)

103. 60 m. 105. (a) 79 m/s;

E

iy

K

(@) rumin

=

(b) 2.4 X 10'W 107. (a) 2.2 X 10°J;

I | |

i

0 77.

17. %voi - voj.

(b) 5.8 X 10*>N, forward.

47. 1.

12"5

15. 1:2.

23. (a) 2.0kg-m/s, forward;

_ GMEms 5

57.

7. 1.40 X 10%kg. 9, 2.0 X 10*kg.

_\a

11. 4.9 X 10°m/s.

43. (a) 0.13m;

(b)

5. 435kg-m/s (j — i).

_ (2_},)1/6

Unmin

X0

2b\1

7

s Tymax

* =

0;

(b) 22m/s; (¢) —1.4m.

109. x = JZ'a

29. 1.5kg-m/s.

3L (a) zfl, 2mv

(b)——

33. (a) 0.98N + (1L4N/s)t;

(b) 13.3N;

() [(0.62N/m2) x

V2.5m - (0.070 m/s)t] + (14N/s)t, 132N. 35. 1.60m/s (west), 3.20 m/s (east). 37. (a) 3.7m/s;

(b) 0.67 kg. 39. (a) 1.00; (b) 0.890; (c) 0.286;

(d) 0.0192.

41. (a) 0.37 m; (b) —1.6m/s, 6.4m/s; (c) yes.

-M

43. (a) m+ M’ (b) —0.96.

45.3.0 X 10°J,4.5 X 10°J. 47. 0.11 kg-m/s, upward.

49. (b)e = \/% 51. (a) 890 m/s; (b) 0.999 of initial kinetic energy lost.

Answers to Odd-Numbered Problems

A-23

53. (a) 7.1 X 102 m/s; (b) (¢) (d) (e)

—54m/s,41m/s; 0,0.13 m/s, reasonable; 0.17m/s, 0, not reasonable; in this case, —4.0m/s, 3.1 m/s, reasonable.

113. 0.2 km/s, in the original direction of my. CHAPTER 10

1. (a) %rad, 0.785 rad;

55. 1.14 X 1072 kg-m/s, 147° from the

w

(b) 3 rad, 1.05 rad;

electron’s momentum, 123° from the neutrino’s momentum.

(c) % rad, 1.57 rad;

57. (a) 30°;

(d) 27 rad, 6.283 rad;

(b) vp = vg = %: ’

U

v

(0) 3.

59. 39.9u.

63. 6.5 X 1071 m, 65. (1.2m)i — (1.2m)j. 67.oi+gfj. m™

69. 0i + 0j + 3 hk. ® 0

3.53 X 10°m. 5. (a) 260 rad/s;

(b) 46m/s, 1.2 X 10*m/s%.

7. (a) 1.05 X 107 rad/s; (b) 1.75 x 103 rad/s;

(c) 145 X 107*rad/s;

4R . + —j.

7 . 0i + 3—j

73. (a) 4.66 X 10°m from the center of the Earth.

75. (a) 5.7 m;

(b) 4.2 m; (c) 43m. 77. 0.41 m toward the initial position of the 85-kg person. 79. v

897 (e) 6 rad, 7.77 rad.

it

m

M upward, balloon also

(d) 0. 9. (a) 464 m/s; (b) 185 m/s; (c) 328 m/s. 11. 36,000 rev/min.

13. (@) 1.5 X 104 rad/s%; () 1.6 X 1072 m/s?,

6.2 X 1074 m/s2.

15. (a) —i, k;

(b) 56.2rad/s, 38.5° from —x axis

stops. 81. 0.93 hp.

towards +z axis;

(c) 1540 rad/s?, —j.

83. —76 m/s.

85. Good possibility of a “scratch” shot. 87. 11 bounces. 89. 1.4m.

93. (a) v = _.MOL;

Mo+ 2=t dt

(b) 8.2m/s, yes. 95. 112 km/h or 70 mi/h. 97. 21m. 99. (a) 1.9 m/s; (b) —=03m/s,1.5m/s; (c) 0.6 cm, 12 cm.

101. m < {M or m < 2.33kg. 103. (a) 8.6 m; (b) 40 m. 105. 29.6 km/s.

107. 0.38 m, 1.5 m.

109. (a) 1.3 X 10°N;

(b) —83 m/s%.

A-24

19. (a) —0.47 rad/s% (b) 190s.

21. (a) 0.69 rad/s%;

91. 50%.

111. 12kg.

17. 28,000 rev.

(b) 9.9s.

23. (a) w = 15.08° — 18.5¢%

(b) 6 = 55.00* — 1853

(c) w(2.0s) = —4rad/s, 6(2.0s) = —5rad. 25. 1.4m-N, clockwise.

27. mg(&, — 4,), clockwise. 29, 270N, 1700 N,

31 1.81kg-m?, 33. (@) 9.0 X 102m"N; (b) 12, 35, 56 m-N.

37. (a) 0.94 kg -m?; (b)2.4 X 102m-N. 39, (a) 78 rad/s%

(b) 670N.

41. 22 x 10°m-N.

Answers to Odd-Numbered Problems

43. 17.5m/s.

45. (a) 14M1%

(b) & Ml

(c) perpendicular to the rod and the axis.

q

47. (a) 1.90 X 103kg-m?%; (b) 7.5 X 10°m"N.

49. (a) Ry

(b) V3RS + w?; () \/%Ro;

(d) V3(R} + R}); (e) Vin;

N VL ® Vig

(h) V(2

+ w?).

(mB

- mA)

compared to a

_ (’"B - mA)

=0

(ma + mp)

53. (a) 9.70rad/s?% (b) 11.6 m/s?;

(c) 585m/s?; (d) 427 x 10°N;

(e) 1.14°.

57. (a) 5.3Mr};,

(b)) —15%.

59. (a) 3.9 cm from center along line “™ connecting the small weight anu the center;

(b) 0.42kg-m?,

61. (b) 5 M, 5 Mu?. 63. 22,200J. 65. 14,200, 67. 1.4m/s. 69. 822 m/s.

71. 7.0 X 10'J. 73. (a) 8.37m/s, 32.9rad/sec.

(b) 3;

(c) the translational speed and the energy relationship are independent of both mass and radius, but the rotational speed depends on the radius.

75. V7 e(Ry - o). 77. (a) 4.06 m/s;

(b) 8.997; (c) 0.158.

79. (a) 41 X 10°J;

(b) 18%;

(c) 1.3m/s% (d) 6%.

81. (a) 1.6 m/s;

(b) 048 m.

“

83

r

g

e 2,

11. (a) 0.55rad/s;

2 .

85. (a) 0.84 m/s; (b) 96%. 87. 2.0 m-N, from the arm swinging the sling. 89. (a) ar|_ or _ Nr, Ng’

15. f .

17. (a) 3.7 X 10'¢7J; (b) 1.9 X 10¥kg-m?/s.

9. (a) 1.7 X 1087J;

(b) 2.2 x 10°rad/s; (c) 25 min.

Mg\/2Rh — K R-h

AP

T.

41.

97. 5.0 X 10°m"N.

(a)

[(MA

MB)R(]

M BE

(b)

99, (a) 1.6 m;

+

F

(b) mR%ak.

skater’s center of mass;

(b) fsingle axel = 2.51ad/s, ftrip]e axel = 6.51ad/s.

81. (a) 17,000 rev/s; (b) 4300 rev/s. rad/s 83 (a) w=|12——|x; m

()

%O]v;

-6

o

Mp A + Mg B + —R%

(b) 1.1m. X

101. (a) — g; ()yg

(b) x should be as small as possible,

45. F, = i

y should be as large as possible,

and the rider should move upward and toward the rear of the bicycle;

(c) 3.6m/s%

e

47.

(d

+

=

rAcos¢)mArszsin¢o

-

(d = racos ¢p)marsw’sin ¢

2d

m*? — glm + M)(m +iM)

,

250

from above. 30

45

60

Angle (degrees)

75

CHAPTER 11

1. 3.98 kg -m?/s. 3. (a) L is conserved: If I increases, must decrease; (b) increased by a factor of 1.3. 5, 0.38 rev/s.

7. (a) 7.1 X 10¥3 kg-m?/s; (b) 2.7 x 10¥ kg - m?/s. I

a@—fw; ()

Iy 2 wy;

(c) ww -Ilfl:

(d) 0.

3

90

59. (a) 9.80 m/s?, along a radial line;

(b) 9.78 m/s?, 0.0988° south from a radial line;

() 9.77 m/s?, along a radial line. 61, Due north or due south. 63. (mrcu2 - F[,)f + (F spoke

mev)j

+ (Fy — mg)k. 65. (a) (—24i + 28] — 14k) kg-m?/s; (b) (16§ — 8.0k) m-N. 67. (b) 0.750. 69. v[—sin(wt)i + cos(wt)j],

o~ (3

71. (a) The wheel will turn to the right; (b) AL/Ly = 0.19.

=

77, 0.69m.

79. (a) t = Os;

(b) slope = lz, (2

o]

:

4

g s

y-intercept = — D% v

41. (a)

[in\ PR 1fs

-10

()

—

TN

=5

0 x (m)

(b) D =

5

CHAPTER 16 1. 340 m. 3. (a) 1.7cm to 17 m; 10

4.0m?

(b) 11 m; (c) 0.5%. 7. 41m.

(c) all kinetic energy.

859

9. (a) 8%;

g

yn =123,

fn = n(0.67Hz),

15

.

n = 1,23,

41 foso/fioo = V2. 49. 80 Hz.

-10

(b) 8.45sin(0.84x + 2.1) cos(47¢). 59. (a)

Q

0.1

-0.2 =03

A-28

i

s

%

SN

25

3

0

x (m)

5

\

13. (a) 53 m; 10

(b) 102

19.29 X 107°7. 21. 124 dB.

[\ / -10

t(s)

Answers to Odd-Numbered Problems

-5

(¢) 3600 m/s; 15. 63 dB. 17. (a) 10%

ib

A

(b) 675 Hz;

(d)1.0 X 10 B m.

i oay + 20m

Q

Y'A/

1

\

4.0m’

B

0.3 O

11. (a) 44 X 107 Pa; (b) 4.4 X 1073 Pa.

t = 1.0s, moving left.

57. 315Hz.

4

(b) 4%.

/\

0

=5

DD =

55. (a) D, = 4.25in(0.84x + 47t + 2.1);

E 01

VA

0,(

53. 11.

02

(b)23 X105 m. 5. (a) 0.17 m;

(x — 2.4t)* + 2.0m? (c) t = 1.0, moving right;

a7, =

-

I~ -~

e ~

V25 = 16.

(c) 0.59m.

15

87. D(x,t) = (3.5cm) cos(0.10x — 1.57t), with xincmand f in s.

dense rod, by a factor of

(b) 026 N;

43. 662 Hz.

83. 6.3 m from the end where the first pulse originated. 4 85. A = n=123,-2n—1°

93. speed = 0.50m/s; direction of motion = +x, period = 27 s,

71. (a) 0.042 m; 73. The speed is greater in the less

37. (a) 0.84 m;

(d) 0.630.

89. 12 min.

69. 44°. (b) 0.55 radians.

35. Yes, it is a solution.

(c) 1.26;

23. (a) 9.4 X 10°°W;

(b) 8.0 x 10° people.

0.5 nO x (m)

5

10

25. (a) 122dB, 115dB; (b) no. 27. 7dB.

-

L=t

3

t(s)

“ ——!.zh{

\?\

i

——

—0.5

2

/

-—_—

7NN

\)’{1/1’5

Dy, D,

T

el

I\\\]

7

T

fm=] L I

/"

15

— -

4

05

l\

=

‘I\‘

BINT

Q_OZS

—-0.2 -03

"

5—~ )fi)?\\\ e

\

T

(b) 1.59;

—

e

0

0

/[

-

E

.

0191

E

D, + D,

™ f—

025

M.

~\

-

0.5

£

o

-~

(@)D= (0.45m) cos[2.6(x + 2.0r) + 1.2].

1480 Hz;

03 0.2 01

=N pea ]

(0.45 m)cos[2.6(x — 2.0r) + 1.2];

81. (a) G: 784 Hz, 1180 Hz, B: 988 Hz,

_.m:-—i:

27. (b) D =

29. (a) The higher frequency wave, 2.6; (b) 6.8.

85. 51dB.

31 (a) 32 X 109 m;

(b) copper.

89. (a) 280m/s, 5STN;

(b) 3.0 X 10' Pa.

21. (a) 2.7 cm;

(6) 0.19 m;

33. 1.24m. 35. (a) 69.2 Hz, 207 Hz, 346 Hz, 484 Hz; (b) 138 Hz, 277 Hz, 415 Hz, 553 Hz. 37. 8.6 mm to 8.6 m.

(b) 0.3 cm.

(c) 880Hz, 1320 Hz. 91.

23. 55 min.

3Hz.

25. 3.0 X 10’ N/m?.

93. 141 Hz, 422 Hz, 703 Hz, 984 Hz.

27. (a) 27°C;

95. 22 m/s.

39. (a) 0.18 m;

(b) 5500 N.

97. (a) No beats;

(b) 1.1m;

29. —459.67°F.

(b) 20 Hz;

(c) 440Hz, 0.78 m.

31. 1.35m’.

(c) no beats.

41. -3.0%.

33, 1.25kg/m>.

99. 55.2kHz.

43. (a) 1.31 m;

35. 181°C.

101, 11.5m.

(b) 3,4,5,6. 45, 3.65cm, 7.09 cm, 10.3 cm, 13.4 cm,

37. (a) 2.8 m%;

103. 2.3 Hz.

(b) 1.88 atm.

105. 17 km/h.

16.3cm, 19.0 cm. 47. 4.3 m, open.

39. 1660 atm.

107. (a) 3400 Hz;

49. 21.4 Hz, 42.8 Hz. 51. 3430 Hz, 10,300 Hz, 17,200 Hz, relatively sensitive frequencies. 53. +0.50 Hz.

59. (a) 221.5 Hz or 218.5 Hz; (b) 1.4% increase, 1.3% decrease. (a) 1470 Hz;

45, —130°C.

109. (a)

47. 7.0 min.

2s

e 04

g5

I

/

N\

Lok

0.0

(b) 1230 Hz.

43, 3.49 atm.

(c) 0.10 m.

1.0

57. 10 beats/s.

41. 313°C.

(b) 1.50 m;

1.2

55. 346 Hz.

r61.

17. (a) 5.0 X 1073/C>;

87. 1.07.

49, Ideal = 0.588 m’,

actual = 0.598 m® (nonideal

N

Y

behavior).

51. 2.69 X 10% molecules/m>.

53. 4 x 1077 Pa.

T

55, 300 molecules/cm?.

63. (a) 2430 Hz, 2420 Hz, difference of

57. 19 molecules/breath.

10 Hz;

(b) 4310 Hz, 3370 Hz, difference of 940 Hz;

59. (a) 71.2 torr;

(c) 34,300 Hz, 4450 Hz, difference of 29,900 Hz;

61. 223 K.

(d) fsource moving

_ f(l

(b) 180°C. 63. (a) Low;

~ féabscrver moving

(b) 0.025%.

© Uobject)

65. 20%.

VUsound

67. 9.9 L, not advisable.

65. (a) 1420 Hz, 1170 Hz; (b) 1520 Hz, 1080 Hz; (c) 1330 Hz, 1240 Hz. 67. 3Hz. 69. (a) Every 1.3s; (b) every 15s. 71. 8.9 cm/s. 73. (a) 93; (b) 0.62°. 77. 19 km.

69. (a) 1100 kg;

CHAPTER 17

(b) 100 kg.

1. Nau = 0.548Njpg.

71. (a) Lower;

3. (a) 20°C;

(b) 0.36%.

(b) 3500°F.

73. 1.1 X 10“ molecules.

5. 102.9°F.

75. 3.34 nm.

7. 0.08 m.

9. 1.6 X 10 °m for Super Invar™, 9.6 X 10 m for steel, steel is

60X

as much.

77. 13 h.

79. (a) 0.66 X 10> kg/m>; (b) —3%.

79. (a) 57 Hz, 69 Hz, 86 Hz, 110 Hz, 170 Hz.

11. 981 kg/m’,

81. +0.11C". 83. 3.6 m.

83. 11'W.

13. —69°C. 15. 3.9 e’

™ 81. 904dB.

85. 3% increase.

Answers to Odd-Numbered Problems

A-29

41. (a) 3.1 x 10°Pa; 2

||

—

(b) 3.2 x 10°Pa.

g 1015

g

2

100

0

10

43. (b) a = 0.365 N-m*/mol?, b = 428 x 1075 m?/mol.

; 20

45, (a) 0.10 Pa;

30

Temperature (°C)

40

50

Slope of the line: 4.92 X 1072 ml/°C, relative B: 492 x 107%/°C, B for the liquid: 501 X 107%/°C, which liquid: glycerin.

(b) 3 x 107 Pa. 47. 2.1 X 107" m, stationary targets, effective radius of ry, + ry;.

21. (a) 51 X 10°J;

(b) 1.5 X 10°J. 23. 4700 kcal. 25. 360 m/s. 27.

49, (b) 4.7 x 107s71, 51. %. 53. 3.5 h, convection is much more

CHAPTER 18

important than diffusion.

1. (a) 5.65 X 1072'J;

(c) 0.6s.

3. 1.29.

57. 260m/s, 3.7 X 107 atm.

5.3.5 x 10°m/s.

59. (a) 290 m/s;

7. (a) 4.5;

(b) 9.5m/s.

(b)5.2.

61. 50 cm.

9. V3.

63. Kinetic energy = 6.07 x 107217, potential energy = 5.21 x 10737,

13. (b) 5.6%. 15. 1.004.

yes, potential energy can be

17. (a) 493 m/s; (b) 28 round trips/s.

neglected.

19. Double the temperature.

65. 0.07%.

21. (a) 710 m/s;

67. 1.5 X 10°K. 69. (a) 2800 Pa;

(b) 240K; (c) 650m/s, 240K, yes. 23. Vapor.

(b) 650 Pa.

71. 2 X 108 m,

25. (a) Vapor;

(b) —365kJ. 31. (a) 4801J;

(b) 0 (c) 48017 into gas.

33. (a) 43507; (b) 4350J; (c) 0. 35. —4.0 X 10°K. 37. 2361.

39. (a) 3.0 X 10'J;

(b) 6817; (c) —8417; (d) —1147J; (e) —151.

75. (b) 4.6 X 10° Hz,

27. 3600 Pa.

29. 355 torr or 4.73 X 10*Pa or 0.466 atm. 31. 92°C. 33. 1.99 x 10° Pa or 1.97 atm. 35. 70 g.

2.3 x 10 times larger.

77. 0.21. 41 CHAPTER

19

1. 10.7°C. 3. (a) 1.0 X 10"7;

37. 16.6°C.

(b) 2.9 kWh;

39. (a) Slope = —5.00 X 10°K, y intercept = 24.9. Let By = 1 Pa in this graph:

{y = —5000x + 24.91

0.0022 0.0024 0.0026 0.0028 0.0030 0.0032 UT (K1)

A-30

29. (a) 0;

73. 0.36 kg.

(b) solid.

. fi\

0.5

55. (b) 4 X 107" mol/s;

(b) 3.7 X 10°1.

14 13

1.5

P (atm)

I

E 1025 17—y = .92 x 1079x + 100]

(c) $0.29 per day, no.

5.42 x 10°7,1.0 X 10?kcal. 7. 6.0 x 1097. 9. (a) 3.3 X 10°J; (b) 56 min. 11. 6.9 min. 13. 39.9°C.

Answers to Odd-Numbered Problems

= N~

< g

15. 2.3 x 103 J /kg-C°. 17. 54 C°. 19. 0.31kg.

43. 43C°. 45, 83.7 g/mol, krypton.

47. 48 C°. 49. (a) 62307; (b) 24907; (c) 87201. 51. 0.457 atm, —39°C. 53. (a) 404K, 195K; (b) —1.59 X 10*J; (c) 0 (d) —1.59 x 1047,

08

(c)

AN ¥

%06

I

0.5

03004

005

006

\\\

Z V (m3)

~

007

= = =

AEz_,,:;

_2480J,

(c) these are many orders of

=

5. 0.21.

_1490J,

=

7. (b) 0.55. 9. 0.74.

4720.],

=0

13. 1.4 x 108 7J/h. 15. 1400 m. 17. 660°C.

Weyele = 9901.

19. (a) 41 x 10°Pa, 2.1 X 10° Pa; (b)34L,17L;

57. (a) 5.0 X 10' W; (b) 17 W.

65. (a) 110kg/s;

(c) 2100J;

59. 21 h.

(b) 9.3 x 107 gal/h. 67. (a) 18 km3/days; (b) 120 km?. 69. (a) 0.19; (b) 0.23. 7. (a) 5.0C°; (b) 72.81/kg K. 73. 1700 J /K. 75. 57 W or 0.076 hp.

(d) —15007; (e) 60017,

61. (a) Ceramic: 14 W, shiny: 2.0 W; (b) ceramic: 11 C°, shiny: 1.6 C°.

(f)03.

’-63. (a) 1.73 x 1017 W;

21. 8.55.

(b) 278K or 5°C.

23. 54.

65. 28%.

25. (a) —4°C;

67. (b) 4.8 C°/s;

(b) 29%. 27. (a) 2307;

(c) 0.60 C°/cm. 69. 6.4 Cal.

(b) 3901J.

7. 4 X 1097,

29. (a) 3.1 X 10*7;

73. 1C°.

71. Csterling =

(b) 2.7 min.

75. 3.6 kg.

(27"

31. 91 L.

71. 0.14 C°.

33. 0.20J/K.

79. (a) 800 W;

T

35. 5 x 10*J/K.

(b)53¢g.

37. 549 x 10*25/—K.

81. 1.1 days.

83. (a) 4.79 cm;

2.0

\

515 1.0

43

20

30 40 V (mL)

e 50

Ty

+ 3w 2

— h Tu

’

Ty

- a

b

d

c

(a) 1010J/K; (b) 1020J/K; (c) —9.0 X 10*J/K.

0.5 0.0 10

Vi Vi

=]

T

(b) =93 mJ/K, no; m in kg (SI).


2

49, E

(b) 44 x 10°J;

iy

107

(C)

T

89. (a) 1.9 X 10°J;

55. (a) 11 1.0 0.9

60

70

(c) Q=4997, AE = 0, W = 4.997. ™ss. 110°C. 87. 305 1.

T, 4=

45. (a) Adiabatic; (b) ASadiabatic ASisothermal (¢)

L

= 0, =

S

—nRIn2;

(b) Whet.

ASenvironment adiabatic = 0,

ASenvironment isothermal = nRIn2.

47. (a) All processes are reversible,

81. 16 kg.

83. 3.61 x 1072J/K.

Answers to Odd-Numbered Problems

A-31

CHAPTER 21

1.08 x 107

39.

" [3.00 = cos(13.90)]?

1.27 X 1073 N.

81. 5 x 107°C.

3. 7200 N.

83. 8.0 x 107 C.

5 (49 x 107%)%.

85, 18°.

7. 4.88 cm.

87. E5 Eg Ec Ep

9. -58 x 10°C, 0.

1L (a) g1 = g = 3Or: (b)q1 = 0,9, = Q7. 13. F; = 0.53 N at 265°,

F, = 033N at 112°, F; = 026N at 53°. 15. F = 2.96 x 107 N, away from

41, 1

17. 1.0 x 10" electrons.

(@) ——————=

@ 2meg(y? + )2

45. 1.8 X 10°N/C, away from the wire. 4

8z

" el

+ 422 V4 + 28

3.4 2.3 56 3.4

X X x X

10°N/C, 10* N/C, 10°N/C, 10°N/C,

to to to to

the the the the

right; left; right; left.

» vertical.

91. (a) 9.18 X 10°N/C, down; (b) 1.63 X 1074 C/m?. a

93, (a) —=

V2

= 7.07 cmy;

(b) yes;

3 2Asin 6y , 4meg R t

49.

51. (a) —

| md® @,

= = = =

“

89, —7.66 X 107 C, unstable.

Qy

43

center of square.

A

drregx(x? + )12

(b) 0.2 ps.

b4 (fi + [x - (x2 + fz)l/zlj).

21. 3.08 X 10716 N west. 23. 1.10 X 10" N/C up.

25. (172j)N/C.

. *dmregx(x s

55

27. 1.01 x 10" m/s?, opposite to the field.

+ ) 2an

Ot —7J) ’ 47730(x2 + a2)3/2

57. (a) (3.5 x 105 m/s?) i

— (141 x 10" m/s?) §;

29,

(b) 166° counterclockwise from the

(c) and (d) Electric field (10% N/C) o - w & o > n o in O

19. (a) =

N/C (upwards).

LY

e

Ring

===« Point

initial direction.

59, —23°, 61.

.

(b)2

(e) 37 cm. dmregmR> | ——

qQ 63. (a) 34 X 10°2°C;

(b) no; (c) 85 X 107 m-N;

A

N 31 (-4.7 x 10" )N/C

- (1.6 x 10" j)N/C; or

(d)2.5 x 107 7. 65. (a) 0 very small;

®) 55 1

5.0 X 10" N/C at 199°. 33. E = 2.60 X 10*N/C, away from the center.

E

67. (a) In the direction of the dipole. 69. 3.5 X 10° C.

71. 6.8 X 10° C, negative. 73. 1.0 x 107 electrons. 75. 5.71 x 101 C. 77. 1.6 m from Q,, 3.6 m from Q; .

A-32

Answers to Odd-Numbered Problems

CHAPTER 22

1. (a) 31 N-m?/C; (b) 22N-m?/C;

(c) 0. 3. (a) O;

(b)0,0,0,0, Eof?, —Eof%.

5. 1.63 x 1078¢C, 7. (a) —1.1 X 10 N-m?/C; (b) 0. 9, -83 x 1077 C. 11. 43 X 107 C/m. 13. -852 x 107! C, 15. (a) —2.6 X 10*N/C (toward wire); ™ (b) —8.6 X 10*N/C (toward wire).

17. (a) —(1.9 x 10" N/C-m)r; (b) —(1.1 X 108 N-m?/C)/r?

R 33. (a) E—g, radially outward; BoR

+ (3.0 x 10" N/C-m)r;

35. (a) O;

(d) yes.

g % S

20 1.0

2

00

g

_10

S

ol

10

N

20

|

'

30

40

I A |

50

}

|

1

e

CHAPTER 23 1. -0.71 V. 3. 3280V, plate B has a higher potential. 5. 30 m. 7. 1.4 uC. 9. 1.2 cm, 46 nC. 11. (a) 0; (b) =294V, (c) =294 V.

(@

37. (a) 1.9 X 10" m/s;

39. (a) 2= /A

1/

:

/

0

5

13. (a) —9.6 X 108V

~

‘

10

15

r(cm)

20

25

30

® )Q Ry if A = 2w Ryo.

(c) (41 x 108N -m%/C)/r%;

I

¢

65. (a) On inside surface of shell. (b) r < 0.10m,

15 r/ry

20

~ 25

30

7 ket

6

207‘0

Ro

23. (a)-—;l

+ Vo

(b) Vo3

(c) no, from part (a) V — to length of wire. 25. (a) 29 V; (b) —4.6 X 107187, 27. 0.34 1.

Answers to Odd-Numbered Problems

—oo due

A-33

29, 42 MV.

(b) Ex =0,

looks

Ey Y = L. e, __R__ (2 + R2)3/2

like a dipole.

o

ol

\/Ri I+ +

\/

x%)x7).

3. o~ " 1n

55. —62.5kV. 57. 1.3 eV.

x? AC

(b)~

(b) m;

() 1'81 g‘vv (€) =1. s

(a

6

19. (a) 022 um = x = 220 pm;

g

g\2

AL é (R? = 20 0

) -2],

V3 )

79. (a) 1.2 MV;

il S,

53.

(\/i

81. (a) pe(r3 = r)

+ x*x =

2+

Q

21780f megl

(

33. (a) 0;

35 230(\/ — (VR

\/§Q

", -

31. 9.64 X 10° m/s.

24 -8

55.1.1 x 107*1.

“

57. (a) 0.32 um?;

97. (a) 32 nF; (¢) 7.0 mm;

A

59. e;—d(K1 + Ky). ) AK1

Kz

" (d Ky + dy Ky) 63. (a)s‘;—gz[1 + (K - 1)%]; ) ©

[1 4K - 1)%];

V(Z)Eoez

—

V%SOE

(K — 1), left.

g

80A

67.

d-{+

E

Paper

=

|

Paper

]

[

Paper

]

[

Paper

|

\

Paper

|

[

Paper

47. 0.15kg/s = 150 mL/s.

= 0.345 uC, Qing

(b) 6.9 A/m?% (c) 12 X 1077 V/m.

1. 8.13 x 10'® electrons/s.

Eglass = 4.64 X 10° V/m, e

59, 2.5 A/m?, north.

3.55x 101 A,

= 0.286 uC. e e

61. 35 m/s, delay time from stimulus to

5. (a) 28 A; )

action.

(b) 84 x 10*C.

63. 11 hr.

7. 1.1 X 10% electrons/min. 9. (a) 2.0 x 10'

65. 1.8 m, it would generate 540

67. 0.16 S.

(b) 430 1.

69. (a) $35/month;

13. 0.64.

s

uF.

73. 15 V. 75. 840 V. 77. 3.76 X 10~°F, 0.221 m?, 1 79. K

. work done by the electric

1 field, ield X—-

83. (a) 257; (b) 940 kW. 85. (a) Parallel; (b) 7.7 pF to 35 pF. 87. 5.15 pE. = 11 uC, Q, = 13 uC,

Q;=13uC, V=11V,

V, =63V, V5 =52V. 91

Q’x

‘289 A

93. 9 x 107 m, no.

~

95. (a) 0.27 uC, 15kV/m, 5.9 nF, 6.0 uJ; (b) 0.85 uC, 15kV/m, 19 nF, 19 uJ.

71. (a) —19% change;

(b) yes, R = 1.39 Q;

031y

P

0.1

0.0

e

(b) % change would be slightly less.

>

73. (a) 190 () (b) 15 Q. 75. (a) 1500 W;

Pl

0

01

02

03

V(V)

04

05

06

17. At 1/5.0 of its length, 2.0 £2, 8.0 Q. 19. 2400°C.

21 V2.

71, 2:1.

() 2.0 X 10's. (c) 0.17 cents. 81. 36.0 m, 0.248 mm. 83. (a) 1200 W;

23. 44.1°C.

(b) 100 W.

25. One-quarter of the original.

85. 1.4 X 10'? protons.

27, 1 (l _ l)_ dmo

(b) 12 A. 79. (a) 21 ()

(c) 1.0 X 10°® Q-m, nichrome.

81. 1.2.

89. 0

(b) 1300 kg/year.

15. (a) Slope = 1/R, y-intercept = 0; 04

W of

heat and could start a fire.

11. 0.47 mm. 3

49. 0.12 A. 51. (a) oo; (b) 96 Q. 53. (a) 930 V: (b) 3.9 A. 55. (a) 1.3 kW; (b) max = 2.6 kW, min = 0. 57. (a) 5.1 X 10710 m/s;

CHAPTER 25

69. Eny = 2.69 X 10* V/m, Ofree

|

(d) 450 V.

.

{

s o

41. 0.90kWh = 32 X 1097, 43. 24 lightbulbs. 45, 11 kW.

(b) 14 uC;

(b) 59 megabytes.

¢

39. 0.055 kWh, 7.9 cents/month.

n

87. (a) 3.1 kW;

r

29. (a) 0.14 O,

(b) 24 W;

(b) 0.60 A;

(c) 15W;

(c) Va1 = 52mV, V¢ = 33 mV. 31. 0.81 W. 33. 29 V. 35. (b) As large as possible. 37. (a) 0.83 A; (b) 140 Q.

(d) 38 cents/month. 89. (a) $55/kWh; (b) $280/kWh, D-cells and AA-cells are 550X

and 2800 X,

respectively, more expensive.

91. 1.34 X 1074 Q. Answers to Odd-Numbered Problems

A-35

41.

@

RGR' +3R).

S&E TR

89. (a) 6.8V, 15 uC; (b) 48 s. 91. 200 MQ. 93. 4.5 ms. 2.5

S

o

(b) 14V;

output voltage circuit across four

(c) 80V;

consecutive resistors.

11. 0.3 Q.

(a) 80V,

(d) 4.8 uC. 53. 29 nA.

13. 450 Q, 0.024. 15. Solder a 1.6-k{) resistor in parallel

55

with 480-(Q resistor.

(b) place in series with 45-kQ) resistor.

17. 120 Q.

19. B R.

57

R=r.

23. (a) Vst decreases,

(a) Place in parallel with 0.22-m() shunt resistor;

100 kQ.

59. Vig =24V, V5, =15V, -15%, —15%.

Viniddle increases,

61

Viight = 0;

63. 12V.

(b) Legt decreases,

65

Imiddle increases,

Connect a 9.0-k(} resistor in series with human body and battery.

67 25V,117 V.

Liight = 0; (c) terminal voltage increases; (d) 85V, (e) 86 V. 25, (a) V; and V; increase, V3 and V decrease;

69. 92 kQ. Ry R, 71. (a) R (b) 121 Q.

73. Terminal voltage of mercury

cell (3.99 V) is closer to 4.0 V than terminal voltage of dry cell (3.84 V).

(b) I and I increase, I and I decrease; (c) increases;

(d) before: I = 117mA, I, = 0, I = Iy = 59mA; after: [; = 132 mA, =1y

0.960 mA, 4.8 V.

= 44mA, yes.

27. 0.38 A. 29. 0. 31. (a) 29V;

75. 150 cells, 0.54 m2, connect in series; connect four such sets in parallel to total 600 cells and deliver 120 V. 77

Counterclockwise current: —24 V, clockwise current: +48 V.

79. 10.7 V. 83. 9.0 Q. 85. (b)1.39V; (c) 0.42 mV;

(b) 43V, 73 V. 37. 0.70 A.

(d) no current from “working” battery is needed to “power” galvanometer.

39. 0.17 A.

87. 1.0 mV, 2.0 mV, 4.0 mV, 10.0 mV.

33. I} = 0.68A left, I, = 0.33 A left.

A-36

Answers to Odd-Numbered

Problems

Current (mA)

51

.

series with battery; then connect

L =1

I

0

9. Connect nine 1.0-) resistors in

2.

W

e

1. (a) 5.93 V: (b) 5.99 V. 3. 0.060 Q. 5.93V. 7. (a) 2.60kQ; (b) 270 Q.

A\

D o o

Charge (uC)

CHAPTER 26

060.5

0.4!

2

AN

0.3

4 6 Time (ms)

I -

D

02

8

[

P

0: 1

\'\L

|i

0.0 0

2

4

6

8

Time (ms)

| 1 10

CHAPTER 27 1. (a) 8.5N/m; (b) 49N/m. 3.26 X 107*N. 5. (a) South pole; (b) 341 A; (c) 7.39 X 1072N. 7. 2.13 N, 41.8° below negative y axis. ~

9. (—2IrBysin 6p)j.s

13. 6.3 X 107N, north. 15. 18 T. 17. (a) Downward; (b) into page; (c) right. 19. (a) 0.031 m; (b)3.8 X 1077s. 23. 1.8 m.

25. (0.78i — 1.0j + 0.1k) x 107N, 27. Liinal = 3 Linitial 29, (a) Negative;

(b) qBo(

2+ d? 2d )

31. 1.3 x 108m/s, yes.

33. (a) 45° (b)23 X 107 m. 35. (a) 2NIAB; (b) 0.

37. (a) 485 X 107 m*N; (b) north.

39. (a) (—4.3k) A-m?% (b) (2.6 — 2.4)) m"N;

(c) —2.81.

41, 12%.

43. 39 pA.

S

55. (a) 2.7 X 10°°T; (b)53 X 10°°T;

45, 6 electrons.

47. (b) 0.05 nm, about ; the size of a typical metal atom;

r

() no, no Newton’s third-law-type

(c) 10 mV.

of relationship;

(d) both 1.1 X 107 N/m, yes, Newton’s third law holds.

49, 0.820 T. 51. 70 u, 72 u, 73 u, and 74 u.

Holj 2

53. 1.5 mm, 1.5 mm, 0.77 mm, 0.77 mm.

§7. ——,

55, ’H, {He.

current coming toward you).

57. 2.4 T, upwards.

NugIR?

61. (a) “OT

59. (a) I_’id_ t;

g

o1/

(c) east.

61. 1.1 X 10" °m/s, west.

mb(3a + b) 12 65 7| ——————— 3NIBa(a + b) | . 67. They do not enter second tube, 12°. 69. 1.1 A, down. 71. 7.3 X 1073 T.

(b) 39. (a)

(b) 28 mT;

() 0.12T.

(b)

1. 0.37 mT, 7.4 times larger. 3. 0.15 N, toward other wire. 7. 0.12 mT, 82° above directly right. 9, 3.8 X 107°T, 17° below the horizontal to north.

11 (a) (2.0 X 107%)(25 = I) T;

(b) (2.0 X 107%)(25 + I)T.

15. Closer wire: 0.050 N/m, attractive,

farther wire: 0.025 N/m, repulsive. 2

B o (x(d %) )J‘ 21. 46.6 uT. 23, ()() 3P2 vesv Teoks ks B trots long straight wire. 25. 0.160 A. 27, (a) 5.3 mT; (b) 3.2 mT;

(c) 1.8 mT. ’59.

(a) 0.554 m;

(b) 10.5 mT.

20

R (cm)

wI(R? + R3

25

30

QCURZ

4l

R2

2R

(D)=

( )47ry

43. (a)

+

2X2

=

(

’

d

| =———=

, into the page.

69. 0.10 N, south.

25 20 15 1.0

os

F

0.0

/[

)A

k.

Vd + y?

27R

N

xy

Ma-y)

T—

10 O X (cm)

20

30

40

CHAPTER 29

, into the page.

VPGP (b - x)y

6=

(a—y)b - x) (@a-yP++

out of page.

1. —460 V. 3. Counterclockwise. 5. 1.2 mm/s. 7. (a) 0.010 Wb (b) 55° (c) 5.8 mWhb. 9. Counterclockwise. 11. (a) Clockwise;

|

(b) 43 mV; (c) 17 mA. 13. (a) 8.1 mlJ; (b) 4.2 x 1073 Ce. 15. (a) 0.15 A;

47. (a) 16 A-m?,

(b) 13m-N. 49.24T. 51. (F/f)y = 6.3 X 107* N/m at 90°, (F/f)y = 3.7 x 10™* N/m at 300°, (F/0)p = 3.7 X 107* N/m at 240°,

53. 170 A.

2ma

—40 -30 =20 -10

npo I tan(m/n)

+

pol V5

73. (¢) 1.5 A. 75.

x)*];.

\\/R? +

. g_figl\/xz Y7 4

67.

b

/.LQQ&)(

H«ol

65. (a) 46 turns;

. 3.

'i\.

3

63.3 X 10°A.,

(b) 0.83 mT; (c) no.

, into the page.

2

17. 17 A, downward. d—2x

15

(c) yes.

CHAPTER 28

I-LLI

10

I 37. (a) HZ_ (R% + ;};) into the page;

75. 0.083 N, northerly and 68° above the horizontal.

P

05

(R2+(x—R)2)3/2’

(b) 4.5 mT.

33.36 X 10°T. 35. 0.075 o I/R.

73. —6.9 X 10727,

77. (a) Downward;

(R2+x2)3/2

\ I

00

63. 3.8 X 10*m-N.

e

B(1073T)

(22

to the left above sheet (with

(b) 1.4 mW. 17. 8.81 C. 19. 21 pJ. 21. 23 mV, 26 mV. 23. (a) 0O; (b) 0.99 A, counterclockwise.

Answers to Odd-Numbered Problems

A-37

28, (a) & ;

a

111(1 %+ ’5), ’.L()Iaz'v

Bt 2wb(a + b)

73. 3 Bwl’. 77. BwR, radially in toward axis. 7. (a) 7Td 22 B fv

(c) clockwise; 2724 ugl 600 nm overlap with A3 < 467 nm. 41. Ay = 614 nm, A, = 899 nm. 43. 7 cm, 35 cm, second order. 45. (c) —32°,0.9°. 47. (a) 16,000 and 32,000; (b) 26 pm, 13 pm. 49, 14.0°. 51. No. 53. 45°. 55. 61.2°, 57. (a) 35.3% (b) 63.4°.

51. I,cos® (%Tx )

59. 36.9°, smaller than both angles.

53. (a) 81.5 nm; (b) 0.130 um.

61. [ = Zsm

10

.2

(20), 45°.

63. 28.8 um.

A

55. 0 = sin‘1 km/h;

19. 2.66 eV.

(b) 67 Hz. 69. 75 us.

23, (a) 1.66¢V,;

11. (a) 0.141c;

75. (@) tan~'

(b) 0.140c¢. 13. (a) 3.4 yr;

(b) 3.03eV. 25. (a) 1.66 eV, (b) 3.03eV.

(b) 0.049.

(b) 0.21m.

33. (a) 229¢V;

79. 1.022 MeV.

21. (a) (820 m, 20 m, 0);

83. (a) 4 X 10°kg/s; (b) 4 x 107 yr:

(b) (2280 m, 20 m, 0).

(b) 0.165 nm. 35, 1.65MeV. 37. 212 MeV, 5.86 fm.

(c) 1 X 108yr.

23. (a) 0.88c; (b) —0.88c.

39. 1.772 MeV, 702 fm.

85, 28.32 MeV. 87. (a) 2.86 X 10 ¥ kg-m/s;

25. (a) 0.97c; (b) 0.55c¢.

41. 4.7 pm.

43. 4.0pm.

(b) 0;

27. 0.93¢ at 35°.

B /1 — —5 cos®6; c

2

v2 o2

1

(c) 42 V.

91. 0.987c.

93, 5.3 X 10%'J, 53 times as great.

51. 590 m/s.

95. (a) 6.5yr;

53. 20.9 pm.

(b) 2.31y.

3Lth— 1ty = ————”—? C

47. (a) 1.1 X 10 kg-m/s; (b) 1.2 X 10°m/s;

89. 3 x 107 kg.

tan @

| ==

45. 1840.

(c) 3.31 X 107" kg-m/s.

l

-

'U_

B is turned on first.

33. Not possible in boy’s frame of reference.

(b) —20%. 37. 0.95c. 39. 820 x 10717, 0.511 MeV. 41. 900 kg.

12

I

L]

1.0

=

5 0.8

Classical

- == Relativistic

206

< 04

3

”,

¥

i

57. 122 eV.

]

59, 91.4 nm. 61. 37.0nm.

63.

B

0so ¢ # 0 exactly;

(b) 4s.

Answers to Odd-Numbered Problems

A-43

1.0

20873 = 06

&

..

29. (a) (1,0,0, -%),(1,0,0, +3), (200 ),(2oo+)

Joose

"‘°

2 04

(2.1,

e Transmission T

& 02

-1, —5).(2.1, -1, +3);

() (1, 00 —7),(100 +) (200 ;),(200+)

o Reflection R

20%:

E/U,

(210 ),(210+) (2,1,1,-%),(2,1,1, +3), (300 -3,(3,0,0, +3),

S

EIU, (b) 10%: E/U, = 0.146;

)

»

o

(S

(21,-1,-3),(2,1, -1, +3),

[e)}

61. (a)

= 0.294;

50%: E/U, = 0.787, 80%: E/U, = 1.56.

31.

(3,1, -1,-3)

n=3,

f =2,

33. (a) 15%25%2p®3523p®3d34s?;

CHAPTER 39

(b) 15s%252p®35%3p°3d 4524 p®4d'5s";

1. 0,1,2,3,4,5,6. 3. 32 states; the 18 given in Example 391,

(c) 1525223573 p03d'%4s5%4 po44d'°-

4145525 p55d'%65%6p°5 f36d" 75,

with n = 3 now replaced by n = 4,

35, 5.75 X 107 m, 115 keV.

(4,3,-3,-13),(4,3,-3, +1),

39,

0.0383 nm, 1 nm.

43.

Chromium.

plus these additional 14 states:

41. 0.194nm.

(4,3, 2 -1), (4,3, -2, +1), (4,3,-1,-1),(4,3, -1, +1), (4,3,0 —%),(4 3,0, +3), (4,3,1,-3),(4,3,1, +3),

47, 2.9 x 10™*eV.

49. (a) 038mm, (b) 019mm 5L

(4,3,2,-1).(4,3,2, +3), (4,3,3,- 1), (4, 3 3, +3).

S.n=6;

m=-5-4,-3,-2,

(b) —0.278 eV; (c) 472 X 1073#7Js, 4:

53. (a) 04T (b)0.5T. 55, 4.7 X 10 *rad; (a) 180 m; (b) 1.8 X 10°m

(d) —4,-3,-2,-1,0,1,2,3,4.

57.

7. (a) 7;

13. (a)

™o

= e15;

(b) 1.36 X 1070 m,

63. (a) 15252p%3s%3p%3d°4s?;

(c) 0.9801 u. 11. 1.10 X 107 %m 13. (a) 1.5 X 1072 eV, 0.082 mm; (b) 3.0 X 102 ¢V, 0.041 mm; (c) 46 X 102 eV, 0.027 mm. 15. (a) 6.86 u;

(b) 1850N/m, kco/ky, = 3.4. 17. 236 X 107 m. 19. myx; = myx,. 21. 0.2826 nm. 23. 0.34 nm.

25. (b) —6.9eV; (c) —11eV;

(d) —2.8%. 27. 9.0 x 10%,

29. (a) 6.96¢V;

(b) 6.89¢V. 31. 1.6%. 33.32¢eV,1.1 X 10°m/s. N2 39. (a) 32me?’

(b)

65. (a) 2.5 x 107 (b) 5.1 x 107, 67. 5.24r,. 69. (a) 45°,90°, 135° (b) 35.3°,65.9°,90° 114.1°, 144.7°; (c) 30°,54.7°,73.2°,90°, 106.8°, 125.3°,150°; (d) 5.71°,0.0573°, yes.

15. 1.85. 17. (a) 1.3ry;

(b) 2.7ry;

(c) 4.2r,.

21 —4—e_’/’° 24)‘0

23. 1.1%.

7. @227r3 (1- 23ro 4 227r§ o

71. (b) K = —1U. 73. (a) Forbidden; (b) allowed; (c) forbidden; (d) forbidden; (e) allowed. 75. 4, beryllium.

() 20 =S

£ 10 05

0

s

(c) 13.1rg.

10

rlry

15

20

25

77. (a) 3 ()1 (c) 7 (d) 4 7

Answers to Odd-Numbered Problems

X x X x X

1071711 x 10720% 1078,6 x 1071% 105,4 x 10 10?2 photons/s, 10% photons/s.

8mf*>

() N 43. 1.09 um. 45. (a) 2N,

(b) 6N; (c) 6N;

(c) 15225%2p®35?3p®3d1%4s°4p%4d'5s°.

,

K (N +1) 4

(b) 15225%2p%35%3p®3d 4524 p*,;

k)

-3

9. (a) 18.59 u; (b) 8.00 u;

634nm.

61. (a) 1.56;

(b) =™ 4 (c)r—oe

1. 51eV. 3. 47¢eV. 5.1.28eV.

59. 3.7 X 10*K.

1. n=7,0=6, m=2.

A-44

(d) 4p: 2, B s df: VB 5, VB 1 3d: Vg, V54,

-1,

0,1,2,3,4,5; ms——%,+%

(a) g0 2, (b)2? 2; (C)22 2,

CHAPTER 40

47. 49, 51. 53. 55. 57. 59. 61 63. 65.

(d) 2N (2L + 1). 4 x 10°. 1.8¢eV. 8.6 mA. (a) 1.7mA; (b) 3.4 mA. (a) 35mA; (b) 70mA. 3700 Q. 021 mA. Iy + Ic = Ig. (a) 3.1 X 10*K; (b) 930K.

17. }iNa: 8.113 MeV/nucleon; ##Na: 8.063 MeV/nucleon.

71. Yes, 1.09 um.

73. 1100 J/mol.

19. (b) Yes, binding energy is positive.

23.2.6 X 102 m.

79. 81. 83. 85.

6.47 X 1074 eV. 1.1eV. (a) 0.094 ¢V; (b) 0.63 nm. (a) 150V =V =486 V; (b) 3.16 kKQ = Rjppq < 0. 87. (a)

25. (a) B; (b)HiNa — Mg + B~ + 7, 5.52 MeV. 27. (a) 3{Th; (b) 234.04368 u. 29. 0.079 MeV.

FE)

T

0.50 +

=500K

0.25 1

3

f(E)

o

T

0.50

T

T

1.00

T

1.50

1

49. 0.76 g. 51.230 x 107 'g, 53. 4.3 min. 55.2.98 x 107%g. 57. 35.4d. 59. 2Z8Ra, Z8Ac, 3Th, %Z¢Ra, ZIRn; Z{Th, Z1Pa, &l Ac, %{Th, %Ra.

69. 6.64T ;. 71. (b) 98.2%. 73. 1 MeV.

2.10 x 10*MeV/g; (b) 5.13 X 102 MeV/g; Eq. 42-9a

(b)

gives about 17% more energy per gram, 42-9b gives about 6% less,

B~ (0.14 MeV)

Y (0.042 MeV)

—

89. 32 mA.

Y (0.129 MeV)

191 [ 3§ s

1

5. 3727 MeV/c% 7. (b) 180m; (c) 2.58 X 107 10m.

9. 30 MeV.

11. 6 X 10% nucleons, no, mass of all

nucleons is approximately the same.

i

(c) The higher excited state.

77. 550 MeV, 2.5 x 10'27]. 79. 2.243 MeV.

81. (a) 2.4 X 107 yr; (b) no significant change, maximum

age is on the order of 10° yr.

83. 548 x 1074,

and 42-9c gives about 4X more. 41. 0.35g. 43. 6100 kg/h.

45. 2.46 x 10°7J, 50 times more than

CHAPTER 41

1. 0.149 u. 3. 0.85%.

0.34 . 5 X 107 kg. 25 collisions. 0.11. 3000 eV.

4.83 x 102 MeV/g,

75. (a) 'Pr;

2.00

27. 29. 31. 33. 35.

39. (a) 5.98 x 10*MeV/g,

191260s

1.50

15. 17. 19. 21. 23.

(c) 1.856 MeV, exothermic. 18.000938 u. 0.671 MeV. 7(R; + R, 10 cm. 173.3 MeV.

25. 6 X 10'® fissions/s.

63. 2.3 X 10*yr. 65. 41 yr.

=10,000K

1800

(b) '§C;

61 Np = No(1 — ™).

T

T

from the C;

(b) 3.58 x 10'? decays/s; (c) 9.72 x 107 decays/s.

2.00

T

1200

13. (a) The He has picked up a neutron

47. (a) 3.60 X 10" decays/s;

1) @ e B &

0

L T

o

600

3. Yes, because Q = 4.807 MeV. 5. 5.701 MeV released. 7. (a) Yes; (b) 20.8 MeV. 9. 4730 MeV. 11. n + YN — %C + p, 0.626 MeV.

45. 6.9 X 10" nuclei.

n T

2.0

1. BAl g7, Iisi.

(b) 6.0h. 41. 0.16. 43. 0.015625.

| T

6.0 4.0

CHAPTER 42

39. (a) 1.5 X 107 0yr 1

T = 1000K

8.0

600 s.

(b) 31.972071 u. 33. 0.862 MeV. 35. 0.9612 MeV, 0.9612 MeV, 0, 0. 37. 531 MeV.

0.75 1

10.0

0

31 (a) 13S:

1.00

89. 8.33 x 10' nuclei,

N (106 nuclei)

21. 0.782 MeV.

=] (58] W | T

’

85. (a) 1.63%; (b) 0.663%. 87. 1.3 X 10% yr.

13. 550 MeV. 15. 7.94 MeV.

67. (a) 0.9801 u; (b) 482 N/m, kyci/kn, = 0.88.

gasoline. 47. (b) 26.73 MeV; (c) 1.943 MeV, 2.218 7.551 MeV, 7.296 2.752 MeV, 4.966 (d) larger Coulomb overcome. 49. 4.0 Gy. 51. 220 rad.

MeV, MeV, MeV; repulsion to

Answers to Odd-Numbered Problems

A-45

65. v/c =1 — (9.0 x 10°).

53. 280 counts/s.

27. 69.3 MeV.

55. 1.6 days.

29. Ko = 8.6 MeV,

57. (a) %I -

YiXe + B~ + v;

(b) 31 d;

(c) 8 X 10 kg.

5.0

40

33. 9keV.

37. (a) 700eV; (b) 70 MeV. 39. (a) uss;

(c) chemically reactive;

71. 44 m.

1.0 0.0

b)) T

—

3

»W


5d'6s!

|Th

90|Pa

| 232.03806

6d%75>

91

|231.03588

|U

92

[238.0289

|Np (237)

93(Pu

94|Am (244)

(243)

95|Cm (247)

96

|Bk

97

(247)

|Cf (251)

98

|Es (252)

99

|Fm 100| Md 101| (257)

(258)

No 102 | Lr (259)

103 (262)

5£26d'7s> | 53641752 | 5£46d'7s> | 5£°6d°7s% | 5£764°752 | 5£76d7s? | 5£°6d°7s* |5£196d7s2| 5£116d"752| 5£126d°752| 5£136d°752| 5£146d°752| 571464752

¥ Atomic mass values averaged over isotopes in percentages they occur on Earth’s surface. For many unstable elements, mass of the longest-lived known isotope is given in parentheses. 2006 revisions. (See also Appendix F.) Preliminary evidence (unconfirmed) has been reported for elements 113, 114, 115, 116 and 118.