B.Sc. Physics-IV | Edition-1 | Pages-460 | Code- 1428 |Concept+ Theorems/Derivation + Solved Numericals + Practice Exercise | Text Book

Table of contents :
B.Sc. Physics-IV (Kumaun)
Dedication
Preface
Syllabus 1 (Heat Transfer Mechanism)
Syllabus 2 (Physical Optics)
Syllabus 3 (Statistical Physics)
Brief Contents
Book 1: Heat Transfer Mechanism
UNIT-1: Conduction and Convection
UNIT-2: Kinetic Theory of Gases
UNIT-3: Thermal Radiation
UNIT-4: Low Temperature Physics
Book 2: Physical Optics
UNIT-1: Interference
UNIT-2: Diffraction
UNIT-3: Polarization
UNIT-4: Associated Optical Instruments
Book 3: Statistical Physics
UNIT-1: Basic Concepts
UNIT-2: Ensembles and Thermodynamic Connections
UNIT-3: Classical Statistics
UNIT-4: Quantum Statistics (BES & FDS)

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Krishna's TEXT BOOK

B.Sc.

Physics-IV (For B.Sc. IVth Semester Students) As per Kumaun University Latest Semester Syllabus (w.e.f. 2017-18)

By

Avinash Sharma M.Sc., Ph.D.

Ashok Kumar M.Sc., N.E.T

Head, Department of Physics

Assistant Professor

J.L.P. G. College, Hasanpur

J.L.P. G. College, Hasanpur

Amroha (U.P.)

Amroha (U.P.)

KRISHNA Prakashan Media (P) Ltd. KRISHNA HOUSE, 11, Shivaji Road, Meerut-250 001 (U.P.), India

Jai Shri Radhey Shyam

Dedicated to

Lord

Krishna Authors & Publishers

P reface

W

e are happy to present this book entitled “B.Sc. Physics-IV”. It has been written according to the latest Kumuan University Syllabus to fulfil the requirement of B.Sc. IVth semester students of all Colleges & Universities in Kumuan.

The book is written with the following special features: 1.

It is written in a simple language so that all the students may understand it easily.

2.

It has an extensive and intensive coverage of all topics.

3.

In each Unit, Solved Examples are given based on different Topics.

4.

The book has been divided into three parts.

5.

Sufficient Numerical Problems, Subjective Questions and Objective type questions with Hints & Solutions given at the end of each unit will enable students to understand the concept . We are extremely grateful to my respected and beloved Parents whose incessant inspiration guided us to

accomplish this work. we also express gratitude to our Family for its moral support. We are immensely thankful to Mr. S.K. Rastogi (Managing Director), Mr. Sugam Rastogi (Executive Director), Mrs. Kanupriya Rastogi (Director) and entire team of Krishna Prakashan Media (P) Ltd., for taking keen interest in this project and outstanding Management in getting the book published. The originality of the ideas is not claimed and criticism and suggestions are invited from the Students, Teaching community and other Readers.

–Authors

(iv)

Syllabus Heat Transfer Mechanism B.Sc. IVth Semester; Ist Paper Kumaun University (w.e.f. 2017-2018) MM-60

Unit-I- Conduction and Convection Modes of heat transfer via Conduction, Convection and Radiation, Conduction: Fourier’s law, One dimensional steady state conduction, Heat conduction through plane and composite walls, Cylinders and spheres, Electrical analogy, Thermal conductivity and its experimental detection, Convection: Newton’s law of cooling, Dimensional analysis applied to forced and free convection, Dimensionless numbers and their physical significance. Unit-II- Kinetic Theory of Gases Kinetic theory of gases, Microscopic View of an Ideal gas, Degrees of freedom, Law of Equipartition of Energy, Distribution law of velocities, Most probable speed, Average speed and root mean square velocity of molecules, Pressure exerted by a perfect gas, Kinetic Interpretation of Temperature. Unit-III- Thermal Radiation Physical quantities associated with Radiation, Black body, Radiation from non-black-bodies, Thermodynamics of radiations inside a hollow enclosure, Kirchoff’s Laws, Derivation of Stefan Boltzmann Law, Wein’s displacement law, Black body spectrum formula-early attempts, Raleigh Jeas’s Law, Quantum theory of Radiation, Planck ’s formula for black body spectrum, Wien’s law, Radiation as a photon gas. Unit-IV-Low Temperature Physics Methods of producing low temperatures via Joule- Kelvin Expansion and Adiabatic demagnetization, Joule Thomson Coefficient, Inversion temperature, Introduction to cryogenics and refrigeration, Air conditioning Mechanism, Air compression Machine, Hampson’s and Linde’s regenerative cooling machine, Liquification of air, Hydrogen and Helium, Solidification of Helium.

(v)

Syllabus Physical Optics B.Sc. IVth Semester; IInd Paper Kumaun University (w.e.f. 2017-2018) MM-60

Unit-I- Interference The principle of superposition, Two slit interference, coherence, Division of wave front and amplitude, Optical path retardations lateral shift of fringes, Fresnel biprism, Interference with multiple reflection, Thin films, Application for precision measurements, Haidinger fringes, Fringes of equal thickness and equal inclination. Unit-II- Diffraction Fresnel’s and Fraunhofer diffraction: Diffraction of single slit, Zone plates, intensity distribution, Resolution of image, Rayleigh criterion, Resolving power of telescopes and microscopes, Diffraction due to 2-slits and N-slits, Diffraction grating, Resolving power of grating and comparison with resolving powers of prisms. Unit-III- Polarization Plane polarized, Circular polarized and elliptically polarized light, Malus law, Brewster’s law, Double reflection and uniaxial crystals, Application of bi-refringence, Dichroism, Optical rotation, Rotation of plane of polarization, Optical rotation in liquids and crystals, Polarimeter. Unit-IV-Associated Optical Instruments Michelson intereferometer and its application for precise measurement of wavelength, Wavelength difference and width of spectral lines, Twyman-Green interferometer, Tolansky fringes, Fabry-Perot interferometer and Etalon.

(vi)

Syllabus Statistical Physics B.Sc. IVth Semester; IIIrd Paper Kumaun University (w.e.f. 2017-2018)

MM-60

Unit-I- Basic Concepts Basic postulates of Statistical Physics, Specification of states, Macro state, Micro State, Phase Space, Density distribution in phase space, µ space representation and its division, Statistical average values, Condition of equilibrium, Stirling’s Approximation, Entropy and Thermodynamic probability (S = K ln W ), Boltzmann entropy relation. Unit-II- Ensembles and Thermodynamic Connection Definition, Micro -canonical, Canonical and Grand Canonical ensembles, Their thermodynamic connections, Statistical definition of temperature and interpretation of second law of thermodynamic, Pressure, Entropy and Chemical potential. Entropy of mixing and Gibb’s paradox, Partition function and Physical significance of various statistical quantities. Unit-III-Classical Statistics Maxwell-Boltzmann statistics and Distribution law, Energy distribution function, Maxwell-Boltzmann law of velocity distribution (most probable velocity, average velocity, RMS velocity), Limitations of M-B statistics. Unit-IV-Quantum Statistics (BES & FDS) Bridging Microscopic and Macroscopic behaviour, indistinguishability of particles and its consequences, Transition to quantum statistics and its implications, , Bose –Einstein and Fermi-Dirac gases, Gas Degeneracy, Applications to liquid helium, Free electrons in a metal and Photon gas, Fermi level and Fermi energy.

(vii)

Brief Contents (Dedication).................................................................................... (iii) (Preface). .......................................................................................(iv) (Syllabus).... .....................................................................................(v-vii) (Brief Contents)....................................................................................(viii)

Book 1:

Heat Transfer Mechanism

UNIT-I: Conduction and Convection..............................................................................H–01-54 UNIT-II: Kinetic Theory of Gases....................................................................................H–55-90 UNIT-III: Thermal Radiation........................................................................................H–91-120 UNIT-IV: Low Temperature Physics............................................................................H–121-156

Book 2:

Physical Optics

UNIT-I: Interference....................................................................................................P–01-46 UNIT-II: Diffraction......................................................................................................P–47-94 UNIT-III: Polarization.................................................................................................P–95-126 UNIT-IV: Associated Optical Instruments....................................................................P–127-148

Book 3:

Statistical Physics

UNIT-I: Basic Concepts................................................................................................S–01-32 UNIT-II: Ensembles and Thermodynamic Connections....................................................S–33-62 UNIT-III: Classical Statistics.........................................................................................S–63-96 UNIT-IV: Quantum Statistics (BES & FDS).....................................................................S–97-148

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Book-1 Heat Transfer Mechanism Unit-1 : Conduction and Convection Unit-2 : Kinetic Theory of Gases Unit-3 : Thermal Radiation Unit-4 : Low Temperature Physics

Unit

1

Conduction and Convection

1. Introduction Heat transfer is a science that studies the energy transfer between two bodies due to temperature difference. Heat is capable of being transmitted through solids and fluids by conduction, through fluids by convection and through empty pace by radiation. Now, we will study in this lesson about the heat transfer between two bodies and role of heat transfer in thermal conductivity.

1.1 Heat Transfer Heat can be transferred from one place to another place by three methods. These are : 1. Conduction

2. Convection

3. Radiation

2. Modes of Heat Transfer via Conduction, Convection and Radiation 2.1 Conduction Conduction is the process in which heat is transferred from one place of a body of higher temperature to another place of body of lower temperature. Here molecules of body plays important role to transferred the heat but they do not migrate from their place. Hence heat flows from high temperature to low temperature due to conduction process. Example : If we heated up one end of a metal rod the molecules at the hot end vibrates from their mean position of equilibrium with high amplitude and transmit the heat energy to their neighbours and so on. Free electrons are responsible for conduction.

2.2 Convection In this process the heat is transmitted from a place to another place (High temperature to low temperature) by means of particles with their movements from one place to other place. Example for this process is gas and liquid. When a liquid in a vessel is heated, the molecules at the bottom are heated first causing a decrease in their density. These lighter molecules rise up to the surface and heavier molecules at the surface come down. Thus the movement of molecules in the liquid gives rise to convection current.

H-4 Example : Ventilation, gas filled electric lamps and heating of building by hot water circulations.

2.3 Radiation It is the process in which heat is transferred from one place to another place directly without help of any medium. Example : Heat from the sun reaches earth due to the radiation without affecting the intervening medium. Hence, it is clear from the above discussion that we need medium for conduction and convection and no need of medium for radiation. Here, it is important to mention that heat flows in conduction and convection may be zig-zag while for radiation path is only the straight line.

2.3.1 Temperature Gradient When one point of body is placed at a higher temperature than the other, the flow of heat occurs from the point at higher temperature to lower temperature. The rate of change of temperature with the variation of distance in the direction of heat flow is called temperature gradient. Let a rod AB of length x is heated up from point A. At steady state the temperature of each section of the rod comes at constant but not same. Let two sections are P and Q on the rod Q

P

A

dx

Q

B

Q

x T1

T2 Fig. 1

Where, Temperature at P is T1 °C or θ1 °C Temperature at Q is T2 °C or θ 2 °C Section P is near to A i. e. T1 > T2 Let P and Q have distance dx between them T − T2 dT Temperature gradient 1 = dx dx Thus the temperature gradient is the ratio of a temperature difference to variation of distance between two sections. dT Temperature gradient = − dx ‘−’ ve sign shows that the temperature decrease as distance increases. °C k Its units (SI) is = m m Dimensional formula is [M 0L− T0θ] = [L−1θ]

H-5

3. Fourier’s Law The mathematical theory of heat conduction was developed by Joseph Fourier in 1880. Conduction heat transfer exists in materials due to the presence of temperature difference within the material. It represents heat transfer due to random molecular motion or diffusion in the absence of velocity gradients. Thus conduction is possible in solids, liquids and gases, but it is maximum for solids. The relationship between the rate of heat transfer and temperature for conduction in unit time is known as Fourier’s law of heat transfer. For one dimensional plane heat flow is given as Q= −k

dT dx

or in general form Q=k

(T1 − T2) t x

where k is coefficient of thermal conductivity.

3.1 Derivation for Fourier’s Law The theory of heat conduction is based on the results of experiments similar to that illustrated in fig. 2. In which one side of a rectangular solid is held at temperature T1 while other side is at T2 and rest sides of this rectangular solid is insulated. Hence heat flows only in x-direction. For a given material, it is found that the rate Q at which heat is transferred from the hot side to the cold side is proportional to area A. Insulated

T2

T1

Q

Q

x

Insulated Fig. 2 : One dimensional heat conduction in solids

Where, A — Cross-sectional area (T1 − T2) — The temperature difference t — Time for heat flow 1 — x is the thickness of rectangular piece x

H-6 Thus, Heat flow A(T1 − T2) t x A(T1 − T2) Q=k t x

Q∝ or

…(1)

Where k is coefficient of thermal conductivity We can write equation (1) in this form also Q = kA

(θ1 − θ 2) t d

where θ1 → T1, θ 2 → T2 and x → d Equation (1) is known as fundamental equation of heat flow and known as Fourier’s Law in heat conduction. Here if we take, A = 1 m 2,

T1 − T2 1° = x m

t = 1 sec Then,

Q=k

3.2 Thermal Resistance (Electrical Analogy) The thermal resistance of a conductor is inversely propotional to the thermal conductivity of the material the resistance offered by a material of flow of heat is called thermal resistance. Let Q be the heat is flowing in a conductor of length x, area of cross section of conductor is A and thermal conductivity for time t is k. If temperature difference between its two ends of the conductor is (T1 − T2), then heat flow equation is T1 − T2 t x dT Q = kA x t

Q = kA or

…(1)

Rate of heat flow in a conductor dT Q = kA L t Q dT = H = L t kA H = or

…(2)

Now, according to ‘Ohm’s law i= Comparing equation (2) and eqnation (3) We have some conclusions,

V R

…(3)

H-7 1.

Since,

q Q  is rate of flow of heat and i is the rate of flow of charge i =   t t

Thus, H or 2. 3.

Q will be thermal current as i is the electric current. t

A thermal current will flow due to the temperature difference dT , same as the voltage difference V. L is the thermal resistance as R is electrical resistance. kA R = L / kA K °C The SI unit of thermal resistance = = cal K cal Thus,

Dimension is [M −1 L−2 T3 θ]. Table 1 : Coefficient of Thermal Conductivity of Some Materials S.No.

Material

Thermal Conductivity J/m.s°C

1.

Aluminium

209

2.

Copper

385

3.

Iron

74.4

4.

Mercury

29.1

5.

Glass

.78

6.

Silver

418.7

7.

Air

0.024

8.

Water

0.556

9.

Paper

.14

10.

Gold

312

11.

Steel

5.4

12.

Asbestos

0.57

13.

Brass

85.5

14.

Ice

2.21

15.

Helium

1558 × 10 −4

16.

Hydrogen

1754 × 10 −4

17.

Oxygen

262 × 10 −4

18.

Ethanol (C2 H 5 OH)

.188

19.

Methanol (CH 3 OH)

.214

20.

Card board

.14

H-8 Hence, coefficient of thermal conductivity of a material is equal to the rate of flow of heat across it per unit area per unit temperature gradient. Unit of k in CGS is cm −1°C sec −1 in MKS. The unit is kilo Cal/m sec °C or Joule/m-sec °C. Dimensions of k is [MLT−3 θ −1].

4. One Dimensional Steady State Conduction Let a long bar of conducting material having uniform cross-sectional area A, ρ be the density and c is specific heat of rod is taken along X-axis and treating hot body as origin. T+ Hot body

dt δx dx

M1 M2 x δx x=0 Direction of flow of heat Fig. 3

Let two plane M 1 and M 2 are parallel to each other at distance x and x + δx from hot body. Hence, δx is the distance between M 1 and M 2. dT If T is the excess of temperature at M 1 and T + at M 2 dx dT where is temperature gradient dx dT Now, temperature at δx M2 = T + dx Now, temperature gradient at

M2 =

d dx

 dT  T dx δx 

Let Q1 is the heat entering at M 1 and Q2 is the heat leaving from M 2 in per second. dT Now, Q1 = − kA dx Here k is the thermal conductivity of the material Again Q2 = − kA

d dx

dT   T + dx δx 

Now, per second heat gain (change in heat) Q = Q1 − Q2 = − kA Q = kA

d dT  −  −kA dx dx 

d 2T dx 2

δx

dT   T + dx δx   …(1)

H-9 Now Case I : Before reaching steady state condition. Heat Q is distributed into two types : 1.

It will increases the temperature of section.

2.

Small amount of this heat is lost to the surroundings due to radiations.

Now, we will calculate both of above. dT If is the rate of rise of temperature per sec. dt Hence, the heat used to rise the temperature of planes = mass of section M 1M 2 × specific heat × rise of temperature per sec dT =m× C× dt Now, mass of section m = ( Aδxρ) [m = volume × density] dT = AδxρC dt

…(2)

Let the heat loss per second per unit area of the surface per unit temperature excess over the surroundings is E or it may say emissivity of the surface. If p be the surface area of the section (M 1M 2) thus the value of heat loss due to radiation process = emissivity × surface area of surface × excess of temperature. = E × p δx × T

…(3)

Now, the total heat dT + E × pδxT dt

Q = AδxρC ×

…(4)

Using equation (1) and equation (4), we get kA

d 2T dx 2

δx = AδxρC

dT + EpδxT dt

Dividing this equation by kAδx d 2T dx or

d 2T dx

2

−

2

=

ρC dT E p + T k dt kA

ρC dT E p T =0 − k dt kA

This is known as Fourier’s differential equation. If heat loss is due to radiation is negligible hence emissivity ‘E ’ will becomes to zero. i. e. E = 0 So, equation (6) becomes d 2T dx 2 or

−

ρC dT ⋅ =0 k dt dT k d 2T = ⋅ dt ρC dx 2

…(5) …(6)

H-10 d 2T dT =h dt dx 2

or Where h =

k is known as thermal diffusivity. ρC

It gives the rate of change at temperature of bar. Case II. After Steady State At stationary temperature dT =0 dt Using equation (6) d 2T dx or or

2

−

Ep T=0 kA dT 2 dx 2 d 2T dx 2

d 2T dx 2

Ep T kA

= µ 2T

µ2 =

Where or

=

Ep kT

= µ 2T

…(7)

This is second order differential equation. We will solve this equation. Let the solution of equation (7) is T = al αx where a and α are the constants, Now, Again

dT = aαl αx dx d 2T dx 2

= aα 2l αx

Using equation (7) aαl αx = µ 2al αx α2 = µ 2 α= ±µ Hence, solution of equation number (7) will be T = al ± µx The general solution will be T = a1l µx + a 2− µx

…(8)

H-11 Let the bar is of infinite length its cold end will be nearly temperature of surroundings. i. e., at x = ∞, T = 0 and if T0 is maximum temperature of hot end above. Surroundings, thus at x = 0 , T = T0 Using x = ∞, T = 0 0 = a1, l µα + a 2l − µ∞ a1 = 0 Now using equation (8) T = a 2l − µx Using II boundary At x = 0, T = T0 T0 = a 2l − µ 0

or

l −µ0 = 1

Also ∴

a 2 = T0

So, exact solution of differential equation T = T0 l − µx This equation gives the temperature distribution in the slab. and equation (9) shows the plot between temperature and distance is negative exponential.

Temp. T

T = T0

Exposed bar

Distance x from hot end Fig. 4 : Temperature vs distance graph

If radiation loss is negligible i. e.

E= 0

So,

µ2 = 0

or

d 2T dx 2

=0

On integrating, dT = C1 dx Here C1 is constant.

…(9)

H-12 Again integrating, T = C1x + C 2

…(10)

x = l at T = 0

…(A)

x = 0 at T = T0

…(B)

Boundary conditions are

Applying first boundary, 0 = C1l + C 2

…(11)

T0 = 0 + C 2

…(12)

C 2 = T0 Using equation (11) 0 = C1l + T0 T C1 = − 0 l T0 x + T0 T=− l

or So,

this equation gives excess temperature at any point at distance x from where the heating started.

5. Heat Conduction Through Plane and Composite Walls [Slab] Let us consider a composite slab (wall) made of two different materials are attached together both are having different thickness x 1 and x 2. Let the temperature of first wall of slab I is T1 and outer wall of slab II to is T2. Let the temperature at interface is T Hence, there are arising two cases which is discussed below :

5.1 If slabs are connected in series If both slabs are connected in series then we can calculate different parameters of heat flow. T1

T

T2

T1

T2

x1

x2

Fig. 5 : Slabs in series

5.1.1 Temperature of Interface Rate of flow of heat for I slab Q=

k1(T1 − T ) x1

…(1)

H-13 For II slab Q=

k 2 A(T − T2) x2

…(2)

Here, k1 and k 2 are the thermal conductivities of I and II slab respectively. Using equation (1) and equation (2) k1 A(T1 − T ) k 2 A(T − T2) = x2 x1 or

k k1 (T1 − T ) = 2 (T − T2) x2 x1

or

k1T1 k1T k 2T k 2T2 − = − x1 x1 x2 x2

or

k1T1 k 2T2 k 2T k1T + = + x1 x2 x2 x1

or

k k  kT k T T  1 + 2 = 1 1 + 2 2 x x x x2  1 2 1

or

k x + k 2x 1  k1T1x 2 + k 2T2x 1 T 1 2 = x 1x 2 x 1x 2   T=

k1T1x 2 + k 2T2x 1 k 1x 2 + k 2x 1

…(3)

Thus, we can calculate interface temperature from equation (3).

5.1.2 Equivalent Thermal Resistance of Both Slab Putting the value of T into equation number (1) Q=

or

k1 A  k T x + k 2T2x 1  T1 − 1 1 2  x1  k1x 2 + k 2x 1 

=

k1 A T1(k1x 2 + k 2x 1) − (k1T1x 2 + k 2T2x 1)  x 1  k 1x 2 + k 2x 1 

=

k1 A k1T1x 2 + k 2T1x 2 − k1T1x 2 − k 2T2x 1   x 1  k 1x 2 + k 2x 1 

=

k1 A k 2T1x 1 − k 2T2x 1  x 1  k1x 2 + k 2x 1 

=

k1 A k 2x 1 (T1 − T2) x 1 k 1x 2 + k 2x 1

Q=

k1k 2 A(T1 − T2) k 1x 2 + k 2x 1

…(4)

H-14 or

Q=

or

Q=

Q=

Where

(T1 − T2) k 1x 2 + k 2x 1 k1k 2 A (T1 − T2) k 1x 2 k x + 2 1 k1k 2 A k1k 2 A T1 − T2 T − T2 = 1 x2 x1 R2 + R1 + k 2 A k1 A

x2 = R2 k2A x1 = R1 k1 A

or

Q=

T1 − T2 R2 + R1

…(5)

T1 − T2 R

…(6)

If equivalent resistance at combined slab is ‘R’ then,

H=

So, comparing equation (5) and equation (6) We get

R1 = R1 + R2

5.1.3 Equivalent Conductance (k) Using equation (4) Q= =

or

Q=

k1k 2 A(T1 − T2) k1k 2 A(T1 − T2) = k 1x 2 + k 2x 1 k 2x 1 + k 1x 2 A(T1 − T2) k 2x 1 k 1x 2 + k1k 2 k1k 2 A(T1 − T2) x1 x 2 + k1 k 2

…(7)

If k is equivalent conductance of both slab Then,

Q= Q=

k ⋅ A(T1 − T2) x1 + x 2 A(T1 − T2) (x 1 + x 2) k

Comparing equation (7) and equation (8) x1 + x 2 x1 x 2 = + k k1 k 2

…(8)

…(9)

H-15 k=

or

x1 + x 2 x1 x 2 + k1 k 2

If x 1 = x 2 = x Using equation (9) So,

x 2x x = + k1 k 2 k

or

2 1 1 = + k k1 k 2

or

2 k1 + k 2 = k1k 2 k k=

2k1k 2 k1 + k 2

…(10)

5.2 If Slabs are Connected in Parallel When both slabs are connected in parallel the rate of flow of heat will be different in both slabs Q1 and Q2 respectively.

T1

T2

Q1

A1

5.2.1 Thermal Resistance of Composite Slab For first slab Q1 =

k1 A1(T1 − T2) x

For second slab Q2 =

k 2 A2(T1 − T2) x

Q2

A2

x Fig. 6 : Slabs in parallel

So, rate of flow of heat Q = Q1 + Q2 k A k A = 1 1 (T1 − T2) + 2 2 (T1 − T2) x x or

k A  k A Q = (T1 − T2) 1 1 + 2 2  x   x

or

   1 1  Q = (T1 − T2) +  x   x  k1x 1 k 2x 2 

or

1 1 Q = (T1 − T2) +   R1 R2 

or

x =R k 1x 1

…(1)

…(2)

H-16 If equivalent resistance is R Q=

Then,

T1 − T2 R

…(3)

Comparing equation (2) and equation (3) We get,

1 1 1 = + R R1 R2

5.2.2 Equivalent Thermal Conductivity of Composite Slabs Q=

k ( A1 + A2) (T1 − T2) x

…(4)

Comparing equation (1) and equation (4) k ( A1 + A2) k1 A1 k 2 A2 = + x x x k A + k 2 A2 or k= 1 1 A1 + A2 If A1 = A2 = A k=

Then,

k1 + k 2 2

6. Uses of Heat Conductivity in Our Daily Life There are several uses of heat conductivity in our daily life. Some of them are given as follows : 1.

At same temperature in summer iron rod has high temperature than wooden block We know that iron is the good conductor of heat and wood is the bad conductor. In summers our body temperature is less than room temperature. Hence if we touch an iron piece then heat is easily transferred and we feels that iron piece becomes heated; but wooden piece or block give heat very slow. Hence it feels little bit cold in compare to iron block and vice-versa in winters iron block seen cold in compare to wooden block.

2.

Water in wells feels cold in summer while warm in winter We know that earth is the conductor of heat. In summer the heat of atmosphere cannot enter inside the earth. So earth becomes cold from inside and hence the water in well seems to be cold. Again due to this reason the vegetables that are growing from inside the earth being safe from heat. In winter the heat of earth inside does not come outside from earth inside and hence temperature of inside earth is maintained as warm, that’s why ground water becomes warm in winter.

3.

The thick glass pot breaks into pieces if pot is filled by hot tea If we filled hot tea in a pot having thick walls of glass then the inner part of glass spread after getting heat but as glass is the insulator of heat then heat of glass cannot comes outside so outer layer of glass does not spread. Hence, just because of anamolus expansion of walls glass will break.

H-17 4.

Cooking pots have copper coating on the bottom Copper is a good conductor of heat and promotes the distribution of heat over the bottom of a pot for uniform cooking.

5.

Ice is covered with saw dust of wood To prevent the melting of ice, it is covered with saw dust of wood. The reason is wood is the bad conductor of heat, so saw dust does not allow heat to reach at the ice.

7. Cylindrical Flow of Heat Let a cylindrical tube of length x and radius r1 and r2 where r1 is inner radius and r2 is outer radius. Let the tube be heated along its length by means of electrically heated wires. The direction of flow of heat will be from inner side to outer side across the walls of the tube. Let T1 and T2 are the temperature of inner and outer surfaces respectively. T1

r

A

dr T

r1

T1

B

r2 Fig. 7 : Experimental cylinder

If dr is the thickness of a shell of cylinder at distance from its axis AB. T is the temperature at distance r and (T − dT ) at distance (r + dr ) Now heat flowing through the shell of radius r in per second → Q = − kA × temperature gradient or dT …(1) Q = − k 2πrx dr dT where is temperature gradient in the direction of flow of heat and 2πrx is the surface area of cylinder; k is dr thermal conductivity and ‘−’ ve sign show that T will decreases as r increases. Q Again using equation (1) r dr Q = − 2πkxdT r Integrating this equation, Q∫

r2 dr

r1

r

= − 2πkx ∫

T2

T1

or

Q log r|rr2 = − 2πkxT|TT2

or

Q log r

1

dT

1

r2 = − 2πkx (T2 − T1) r1

H-18 or

Q log

r2 = 2πkx (T1 − T2) r1

…(2)

r2 r1 k= 2πx (T1 − T2) Q log e

or

…(3)

We can determine thermal conductivity of the tube with the help of equation (2). Using equation (2) we get, Q=

2πkx (T1 − T2) r log e 2 r1

…(4)

Now, using equation (4) and equation (1) −k 2πrx

We get,

dT =

or

As r = r1 and T = T1 T1 = −

T =−

=

=

or

C1 = T1 +

T1 − T2 r log e 2 r1

T − T2 T1 − T2 log e r + T1 + 1 log e r1 r r2 log e log e 2 r1 r1

  r −(T − T2) log e r + T1 log e 2 + (T1 − T2) log e r1 r2  1 r1  log e r1 1

1 log e

T =

T1 − T2 log e r + C1 r log e 2 r1

T1 − T2 log e r1 + C1 r log e 2 r1

Using C1 we get,

∴

−(T1 − T2) dr ⋅ r r log e 2 r1

T=−

On integrating,

So,

dT 2πkx (T1 − T2) = r dr log e 2 r1

r2 r1

1 log e

r2 r1

[−(T1 − T2) log e r + T1 log e r2 − T1 log e r1 + T1 log e r1 − T2 log e r1]

[−(T1 − T2) log e r + T1 log e r2 − T2 log e r1]

This equation gives the temperature distribution in the given cylinder.

H-19

8. Temperature Distribution in a Spherical Shell If we heated up a spherical shell from the centre i. e. heat will flow in radial direction. Let shell is divided into many shells of different radius. Let r1 and r2 are the radius of two shells. dr

The heat source is at centre O the flow of hot is in radial direction. Let the inner temperature of shell is T1 and outer temperature of shell is T2 and the temperature of shell of radius r and thickness dr is T and (T − dT )

T2

r2 O r1 T1 rT

Now, the heat of flow in per sec, Q = − kA

dT dr

Q = − k 4 πr 2

or

dr

or

r

2

r2 dr

∫r

On integrating,

r

1

−

or

1 r

2

r2

Fig. 8 : Spherical shell

dT dr

=−

4πk ⋅ dT Q

=−

4 πk Q

=−

r1

T2

∫T

4 πk T Q

(where A = 4 πr 2) …(1)

dT

1

T2 T1

1 1 4 πk − = (T − T2) r1 r2 Q 1

or or

k=

Q(r2 − r1) 4πr1r2(T1 − T2)

…(2)

or

Q=

4 πkr1r2(T1 − T2) k (r2 − r1)

…(3)

Using equation (1) and equation (3) −k 4 πr 2 or On integrating,

dT 4 πkr1r2(T1 − T2) = k (r2 − r1) dr dT = − T=−

(T1 − T2) r1r2 dr ⋅ 2 (r2 − r1) r (T1 − T2) r1r2  1  ×  − + C1   r (r2 − r1)

At r = r1 and T = T1 (T1 − T2) r1 r2 1 ⋅ + C1 r2 − r1 r1

or

T1 =

∴

C1 = T1 − (T1 − T2)

r2 r2 − r1

H-20 So,

T =

(T1 − T2)r1r2 r + T1 − (T1 − T2) 2 r2 − r1 r (r2 − r1)

or

T =

1 r2 − r1

∴

T =

1 r1r2 (T1 − T2)  + (r2T2 − r1T1)  r2 − r1  r 

r1r2(T1 − T2)  + r2T1 − r1T1 + r2T2 − r2T1  r  

9. Some Important Experimental Methods to Determine Thermal Conductivity There are several methods to determine the thermal conductivity for different kind of materials. 1.

Searle’s method

2.

Lee’s method

9.1 Searle’s Method With the help of this method we can determine thermal condutivity for good conductors e. g . metals like copper. Experimental arrangment of the apparatus is shown in fig. (9). A cylindrical bar of copper metal whose thermal conductivity is to be determined. Let MN is a steam tube and its X and is covered by steam chamber and Y end is wounded by helical tube through which cold water is allowed to pass. A non-conducting material is used to covered the copper tube this is because that non conducting material can avoid loss of heat coming from inside. Thermometers

X

T3

Y

T1

T4

T2

Water inlet Water outlet Steam in

O

M

Steam chamber

x

P

Non-conducting materials Steam out Fig. 9 : Searle’s apparatus

N

Helical tube

H-21 Four thermometers are connected with the copper tube when heat is started to flow then at steady state the thermometer records temperature T1T2T3 and T4 . Heat will pass from O to P and when it becomes constant then same amount of mass enters the tube and same amount will comes out. Let mass of the water is m in time t which is passing through the tube. Thus amount of heat gained by the water in this time = m[T4 − T3]

…(1)

Thus amount of heat of flow of water =

kA(T1 − T2)t x

…(2)

Where A is the cross-sectional area of tube. Both equation (1) and (2) are giving the result for heat amount, by equating them kA(T1 − T2)t m[T4 − T3] We get, = x x mx (T4 − T3) or k= A t (T1 − T2) If we have x , T1, T2, T3 and T4 Area of tube A and time of flow of heat t are known then we can determine the coefficient of thermal conductivity ‘k’.

9.2 Lee’s Disc Method for the Thermal Conductivity of Bad Conductors The thermal conductivity of poor or bad conductors can be determined by using Lee’s Disc method. The examples of bad conductors are ebonite, cardboard rubber, galss, wood etc. The apparatus for the determination of thermal conductivity is shown in below figure.

Steam in

Steam (S) chamber

Steam out

T1 Bad conductor T2

Disc (D)

Stand Fig. 10 : Lee’s disc apparatus

The given bad conductor (B) is shaped with the diameter as that of the circular slab or disc ‘D’. The bad conductor is placed in between the steam chamber (S) and the disc (D), provided the bad conductor, steam chamber and the slab should be of same diameter. Holes are provided in the steam chamber (S) and

H-22 the disc (D) in which thermometer are inserted to measure the temperatures. The total apparatus is hanged over the stand as shown in fig.(10). When steam is passed through the steam chamber till the steady state is reached. Let the temperature of the steam chamber (hot end) and the disc (cold end) be T1 and T2 respectively. Let x be the thickness of the bad conductor B, m is the mass of the slab, S be quite specific heat capacity of the slab, r is the radius of slab and h be the height of the slab, then Amount of heat conducted by the bad conductor per second kA (T1 − T2) = x

…(1)

If area of cross section is A = πr 2 Thus amount of heat =

kπr 2(T1 − T2) x

…(2)

Amount of heat lost by slab per second = mSRc

…(3)

where RC is rate of cooling. Under steady state, equation (2) and (3) are same Thus, or

kπr 2(T1 − T2) = mSRc x mSxRc k= 2 πr (T1 − T2)

…(4)

Now the slab is allowed to cool, simultaneously a stop watch is switched ON. We plot a curve between time and temperature, the slope of the curve gives  dT  rate of cooling  .  dt  The rate of cooling is directly proportional to surface area exposed. Case (i) Steam chamber and bad conductor are placed over slab, in which radiation takes place from the bottom surface of area (πr 2) of the slab and the side of the area (2πrh) So,

Temperature

To measure the rate of cooling for the disc alone, bad conductor is removed and the steam chamber is directly placed over the disc and heated. When the temperature of slab attains 5°C higher than T2, the steam chamber is removed.

dT

dt Time Fig. 11

RC = 2πr 2 + 2πrh RC = πr (r + 2h)

…(5)

Case (ii) The heat is radiated by the slab alone i. e. from the bottom of area (πr 2), top surface of the slab of area (πr 2) and also through the sides of slab of area (2πrh)

H-23  dT  2 2   = πr + πr + 2πrh  dt 

So,

= 2πr 2 + 2πrh  dT    = 2π(r + h)  dt 

or

…(6)

Using equation (5) and equation (6), we get, Rc πr (r + 2h) = dT 2 πr (r + h)      dt  Rc =

or

(r + 2h)  dT    2(r + h)  dt 

Hence thermal conductivity of bad conductor will be k=

 dT  mSx   (r + 2h)  dt  πr 2(T1 − T2)2(r + h)

Wm −1k −1

10. Newton’s Law of Cooling This law states that how much time an object takes to cool or heated up. According to Newton’s law of cooling, the rate of cooling of a body is directly proportional to the difference in temperature of the body (T ) and the surrounding (T0) provided difference in temperature should not exceed by 30°C or 303 k. Hence, the rate of cooling

i. e.

dQ ∝ (T − T0) dt dQ = k (T − T0) dt

…(1)

Let the mass of the body is m, specific heat is C and temperature T in surrounding of temperature T0 So, heat will be Q = mCT Now rate of cooling dT dQ = mC dt dt dT Using equation (1) and equation (2) mC = (T − T0) dt dT or ∝ (T − T0) dt

…(2)

…(3)

Where mass and specific heat are taken as constant. Equation (3) explains that as the time increases the difference in temperature of the body and surroundings decreases. Hence the rate of fall of temperature also decreases.

H-24

Temperature°C

100 80 60 40 100

50

150

200

Time in minutes Fig. 12

A graph between time and temperature is shown in figure, where time is taken in minute and temperature is in °C.

11. Dimensional Analysis Applied to Forced and Free Convection We know that convection takes place in fluids, (liquid, vapours and gases). It is due to the bulk motion of the fluid. It is at macro level i. e. visible to the naked eye and driving force is temperature difference. It takes place from higher temperature to lower temperature. It is governed by the Newton’s law of cooling. Convection is of two types : 1.

Free convection

2.

Forced convection

11.1 Free or Natural Convection In this, bulk motion of the fluid is caused by density difference or by the bouyancy force. Thermal boundary layer coincide with the hydrodynamic boundary layer. Velocity of fluid is zero at the solid surface. Also, at the boundary layer free convection is governed by Grashoft’s number and Prandtl number. If it is further of two types : 1.

Laminar free convection

2.

Turbulent free convection

Example : Thunderstorms, glider planes, sea breeze, land breeze, cooling of electric motor, cooling of pumps, cooling of transformers, cooling of compressors, motion of hot baloons etc.

11.2 Forced Convection Heat Transfer When bulk motion of liquid caused by a pump or bulk motion of vapours and gases is caused by a fan/blower then forced convection produced . It is visible due to bulk motion, thus it is at macro level. Driving forces for convection are also temperature differences. It is governed by the value of the Reynolds number.

H-25 It has same two types as free convection. Examples : Flow in boilers, flow in refrigeration, air conditioning plant, cooling in nuclear reactions, cooling heat exchangers etc.

12. Forced Convection Correlations Since, the heat transfer coefficient is a direct function of temperature gradient next to the wall, the physical variables on which it depends can be expressed as follows : h = f (fluid, properties, velocity field, geometry, temperature etc.) As the function is dependent on several parameters the heat transfer coefficient is usually expressed in the terms of correlations involving pertinent non-dimensional numbers.

13. Types of Non-Dimensional Numbers 13.1 Nusselt Number The Nusselt number belongs to conduction/convection heat transfer strength, hx Convective heat transfer Nu = = k Conductive heat transfer Where,

h → convective heat transfer k → thermal conductivity

13.2 Prandtl Number This number is related with momentum/diffusivity/thermal diffusivity. V Viscous diffusion rate Pr = = α Thermal diffusion rate

13.3 Reynolds Number This number is related with inertia force/viscous force, Ux Re = v Viscous force provides the dampening effect for disturbances in the fluid. If dampening is strong then flow is Laminar otherwise flow is turbulent. In the force convection heat transfer is related as Nu = f [Re , Pr ]

14. Physical Significance of Dimensionless Numbers There are many reasons to use dimensionless numbers for a system. It is important to understand the physical significance of dimensionless groups.

H-26

14.1 Physical Significance of Nusselt Number In heat transfer at a boundary with in a fluid, the Nusselt number (Nu) is the ratio of convective to conductive heat transfer across the boundary. In this context, convection includes both advection and diffusion, named after Wihlem Nusselt. A Nusselt number close to one, namely convection and conduction of similar magnitude, is characteristic of slug flow or laminar flow. A larger Nusselt number corresponds to more active convection with trubulent flow typically in (100-1000) range.

14.2 Physical Singnificance of Prandtl Number It is a dimensionless number named after Ludwig Prandtl; it is defined as the ratio of momentum diffusivity to thermal diffusivity. For most gases over a wide range of temperature and pressure (Pr ) is constant. Thus it is used to determine the thermal conductivity of gases at high temperature, which is difficult to measure experimentally due to the formation of convection currents. Typical values for Pr is 0.003 for molten potassium at 975 k around 0.015 for mercury, 1.38 for ammonia. Small value of Prandtl number Pr < 1 means thermal diffusivity dominates. For large value i. e. Pr > 1 the momentum diffusivity dominates. For example the listed value for mercury indicates that the heat conduction is more significant compared to convection. So, thermal diffusivity is dominant. In heat transfer problems, the Prandtl number controls the relative thickness of the momentum and thermal boundary layers. When Pr is small, it means that the heat diffuses quickly compared to the velocity. This means that for liquid metals the thickness of the thermal boundary layer is much bigger than the velocity boundary layer.

14.3 Physical Significance of Reynolds Number Reynolds number of a flowing fluid could be defined as the ratio of inertia force and viscous or friction force. We will be able to determine the type of flow i. e. laminar flow, transient flow or turbulent flow on the basis of Reynolds number. 1.

Fluid flow be laminar, if Re < 2300.

2.

Fluid flow will be transient, if Reynolds number is in between 2300 and 4000.

3.

Fluid flow will be turbulent, if Re > 4000.

Example 1. A slab of stone of area 0.36 m 2 thickness 0.10 m is exposed on the lower surface of steam at 100°C. A block of ice at 0°C restes on the upper surface of the slab. If in one hour 4.8 kg of ice is melted, calculate the thermal conductivity of the stone. Solution : The amount of heat flowing through the slab,

Q= Also according to problem

kA(θ1 − θ 2)t d

Q = mL

H-27 ∴

mL =

kA(θ1 − θ 2) t d

Here, m = 4.8 kg, L = 80 kilo cal/kg A = 0.36 m 2, θ1 − θ 2 = 100 − 0 = 100 °C t = 1 hour = 3600 sec, d = 01 . m, k = ? k × 0.36 × 100 × 3600 010 . 4.8 × 80 × 010 . k= 0.36 × 100 × 3600

∴

4.8 × 80 =

or

= 3 × 10−4 kilo cal m −1(°C) −1S −1 Example 2. A closed cubical box made of perfectly insulating material has walls of thickness 8 cm and the only way for the heat to enter or leave the box is through two solid cylindrical, metallic plugs each of cross-sectional area 12 cm 2 and length 8 cm fixed in the opposite walls of the box as shown in fig. Outer surface A is kept at 100°C while the outer surface B of the other plug is maintained at 4°C.

Source of energy

A

36 cal s–1

θ1 = 4°C

θ

8 cm

B θ2 = 4°C 12 cm2

Fig. 13

The thermal conductivity of the material of the plugs is 0.5 cal s −1 cm −1 (°C) −1 . A source of energy generating 36 cal s −1 is enclosed inside the box. Find the euilibrium temperature of inner surface of the box assuming that it is same at all points on the inner surface. Q Q Solution : Let H 1 = 1 and H 2 = 2 be the rates of heat conduction through the surfaces A and B. The total t t heat available per second inside the box is (H 1 + 36). This heat is conducted through B. Therefore in steady state, H 1 + 36 = H 2 or

kA(θ − 4) kA(100 − θ) + 36 = d d

Here, k = 0.5 cal s −1cm −1(°C) −1, A = 12 cm 2, d = 8 cm ∴

0.5 × 12(θ − 4) 0.5 × 12(100 − θ) + 36 = 8 8

or

075 . (100 − θ) + 36 = 075 . (θ − 4)

H-28 75 − 075 . θ + 36 = 075 . θ−3 1.5 θ = 114 114 θ= 1.5

or or ∴

θ = 76ºC

Example 3. Water is being boiled in a flat bottom kettle placed on a stone. The area of the bottom is 300 cm 2 and the thickness is 2 mm. If the amount of steam produced is 1 g/min. Calculate the difference of temperature between the inner and outer surfaces of the botttom. The thermal conductivity of material of kettle = 0.5 cal (°C) −1 sec −1 cm −1 and latent heat of steam = 540 cal/g Solution : Amount of heat flowing through the bottom of kettle per second,

Q mL 1 × 540 cal/sec = 9 cal/sec = = t t 60 Q kA(θ1 − θ 2) H= = t d kA(θ1 − θ 2) =9 d H=

∴

k = 0.5 cal(°C) −1sec −1cm −1, A = 300 cm 2 d = 2 mm = 0.2 cm, (θ1 − θ 2) = ? 0.5 × 300(θ1 − θ 2) =9 0.2 θ1 − θ 2 =

9 × 0.2 0.5 × 300

∴

θ1 − θ 2 =

1.8 = 0.012° C 150

Example 4. An electric heater is used in a room of total wall area 137 m 2 to maintain a temperature of 20°C inside it when the outside temperature is −10 °C. The walls have three layers of different materials. The innermost layer is of wood of thickness 2.5 cm the middle layer is of cement of thickness 25 cm. Find the power of the electric heater. Assume that there is no heat loss through the floor and ceiling. The thermal conductivities of wood, cement and brick are 0.125, 1.5 and 1 watt m −1 (°C) −1 respectively.

θ1 = 20°C

or

k1

k2

k3

Brick

l1

θ3 l2 θ4

l3

2.5 cm

1 cm

Solution : Let l 1, l 2 and l 3 be the thickness of wood, cement

and brick layers respectively. Also let θ 3 and θ 4 be the temperature of wood-cement interface and cement-brick interface respectively.

25 cm Fig. 14

θ2 = –10°C

Here,

Cement

i. e.

Wood

We have,

H-29 Let P be the power of electric heater. In steady state, the rate of heat flowing is the same in all layers. Therefore we have, Q k1 A(θ1 − θ 3) k 2 A (θ 3 − θ 4 ) = = t l1 l2 =

k 3 A(θ 4 − θ 2) l3

i. e.

k1 A(θ1 − θ 3) k 2 A(θ 3 − θ 4 ) = l2 l1

or

k1 l (θ1 − θ 3) = 1 (θ 3 − θ 4 ) k2 l2

or or Also or or or

…(1)

2.5 × 10−2 m 0.125 (θ 3 − θ 4 ) (20 − θ 3) = 1.5 1.0 × 10−2 m 31θ 3 − 30 θ 4 = 20 k1 A(θ1 − θ 3) k 3 A(θ 4 − θ 2) = l3 l1

…(2)

l k1 (θ1 − θ 3) = 1 (θ 4 − θ 2) l3 k3 2.5 × 10−2 m 0.125 (θ 4 + 10) (20 − θ 3) = 1.0 25 × 10−2 m 5 θ 3 + 4θ 4 = 60

…(3)

On solving, we get,

∴

θ 3 = 6.86°C, θ 4 = 6.4 °C Q Power of heater P = t k1 A(θ1 − θ 3) = l1 =

0.125 × 137(20 − 6.86) 2.5 × 10−2

watt

= 9000 watt = 9 kW Example 5. A room is maintained at 20°C by a heater of resistance 20 ohms connected to 200 volt-mains. The temperature is uniform throughout the room and the heat is transmitted through a glass window of area 1 m 2 and thickness 0.2 cm. Calculate the temperature outside (Thermal conductivity of glass is 0.2 cal m −1 sec −1 and mechanical equivalent of heat J = 4.2 J/cal).  V 2  (200)2 2000 cal/sec = 2000 J/S = = 20 4.2  R

Solution : Heat generated by heater per sec, H = 

…(1)

H-30 But heat passing through glass window per sec is, Q kA(θ1 − θ 2) = H= t d

…(2)

From equation (1) and equation (2), we have, kA(θ1 − θ 2) 2000 = d 4.2 Here, k = 0.2 cal m −1(°C) −1sec −1 A = 1 m 2, θ1 = 20°C, d = 0.2 cm = 0.2 × 10−2 m 0 .2 × (20 − θ 2)

∴

0 .2 × 10

−2

=

(20 − θ 2) =

or ∴

2000 4.2 2000 × 10−2 = 4.76 4.2

θ 2 = 20 − 4.76 = 15.24 °C.

Example 6. A bar of copper of length 75 cm and a bar of steel of length 125 cm are joined together end to end. Both are of circular cross-section with diameter 2 cm. The free ends of copper and steel are maintained at 100°C and 0°C respectively. The surface of the bars are thermally insulated. θ1 = 100°C

θ2 = 0°C

θ

k1

k2

Copper

A l1

2 cm

Steel

B

l2

C

Fig. 15

What is the temperature of copper junction ? What is the heat transmitted per unit time across the junction ? Thermal conductivity of copper is 9.2 × 10 −2 kilo cal/m°C sec and that of steel is 1.1 × 10 −3 kilo cal/m °C sec. Solution : Let l 1 and l 2 be lengths of copper and steel bars AB, BC joined together end to end.

Let θ1 and θ 2 be temperature of face ends A and C of the composite bar and θ be the temperature of junction B. In steady state the heat flowing per second through the two bars is the same. i. e. ∴ or

H1 = H 2 k1 A(θ1 − θ 2) k 2 A(θ − θ 2) = l2 l1 k k k1 k θ − 1 θ = 2 θ − 2 θ2 l2 l2 l1 l2

H-31  k1 k 2  k k  +  θ = 1 θ1 + 2 θ 2 l l2 l l  1 2 2

or

∴ Temperature of junction

k1 k θ1 + 2 θ 2 l1 l2 θ= k1 k 2 + l1 l2

Here, k1 = 9.2 × 10−2, k 2 = 11 . × 10−2 kilo cal/m°C sec l 1 = 75 cm = 075 . m, l 2 = 125 cm = 1.25 m θ1 = 100°C, θ 2 = 0°C.

∴

θ=

 9.2 × 10−2   1.1 × 10−2    × 100 +   ×0 .  075   1.25   9.2 × 10−2   1.1 × 10−2    +  .   1.25   075

θ = 93.31°C. Example 7. The thickness of the ice in a lake is 5 cm and the temperature of air is −20°C. Calculate how long will it take for the thickness of ice to be doubled. Thermal conductivity of ice = 5 × 10 −4 kilo cal/(°C) sec, density of ice = 0.92 × 10 3 kg/m 3 and latent heat of ice = 80 kilo cal/kg. Solution : Time taken for the thickness of ice layer to increase by dx is given by,

dt =

ρL × dx kθ

∴ Time taken for the thickness of ice to increase from X1 to X 2 is, ρL X 2 X ⋅ dX t= kθ ∫X 1 ρL ( X 22 − X12) = 2kθ Substituting given values,

t= =

0.92 × 103 × 80 2 × 5 × 10

−4

× 20

3

0.92 × 10 × 80 2 × 10

−2

[(10 × 10−2)2 − (5 × 10−2)2]

× 10−4 (100 − 25)

= 27600 sec t = 7 hours 40 min Example 8. Show that the rate of flow of heat through a spherical shell, the inner and outer walls of which have radii r1 and r2 and are maintained at different uniform temperatures θ1 and dQ 4kr1 r2 (θ1 − θ 2 ) . Where k is thermal conductivity of material of θ 2 respectively is given by = dt r2 − r1 shell. Hence obtain the expression for temperature difference.

H-32 B

r2 A

r1 O (r +

r

dr)

Fig. 16 Solution : Let O be the centre of spherical shell, having inner and outer radii r1 and r2 respectively i. e.,

OA = r1, OB = r2. Let θ1 and θ 2 be the steady temperatures of inner and outer walls (θ1, θ 2) respectively. The shell may be supposed to be formed of large number of thin concentric spherical shells. Consider one such shell of radius r and thickness dr. Let θ and (θ + dθ) be respective temperatures at distances r and (r + dr ) from centre O. dθ The temperature gradient = − dr ∴

Rate of flow of heat across the surface of the shell, dθ dQ = kA dr dt

or

dθ  dQ  = − k  4 πr 2   dr  dt

k being thermal conductivity of the material of shell. Due to symmetric nature of shell

…(1) dQ is same in all dt

directions in steady state. Therefore from equation (1), −(dQ / dt) dr ⋅ 2 4 πk r (dQ / dt) 1 θ= ⋅ +C 4 πk r

dθ = Integrating, we get,

…(2)

where C is a constant of integration. At r = r1, θ = θ1 ∴

θ1 =

(dQ / dt) 1 ⋅ +C 4 πk r1

…(3)

θ2 =

(dQ / dt) 1 ⋅ +C 4 πk r2

…(4)

At r = r2, θ = θ 2 ∴

H-33 Subtracting equation (4) from equation (3), we get, θ1 − θ 2 =

(dQ / dt)  1 1   −  4 πk  r1 r2 

This gives rate of flow of heat, dQ 4 πk (θ1 − θ 2) 4 πkr1r2(θ1 − θ 2) = = (r2 − r1) dt  1 1  −   r1 r2 

…(5)

Temperature distribution : Substituting dQ / dt from equation (5) in equation (3), we get, 4 πk (θ1 − θ 2) 1 θ1 = ⋅ +C  1 1  r1 4 πk  −   r1 r2  i. e.,

Substituting this value of C and value of

or

C = θ1 −

(θ1 − θ 2)  1 1  −   r1 r2 

 1 ⋅   r1 

dQ from equation (5) in equation (2), we get, dt 4 πk (θ1 − θ 2) 1 (θ − θ 2) 1 ⋅ + θ1 − 1 ⋅ θ =  1 1 r  1 1  r1 − 4 πk  −     r1 r2   r1 r2 

θ = θ1 + (θ1 − θ 2) ⋅

1 1   −   r r2   1 1  −   r1 r2 

…(6)

This is the required expression. Example 9. show that the radial rate of flow of heat through a uniform cylindrical shell, the inner and outer walls of which have radii r1 and r2 are are maintained at different uniform dQ 2πkl(θ1 − θ 2 ) temperatures θ1 and θ 2 respectively is given by where k is thermal = r dt loge 2 r1 conductivity of material of shell. Hence, obtain the expression for temperature distribution. Solution : Consider a cylindrical shell of length l, inner and outer radii r1 and r2 maintained at steady

temperatures θ1 and θ 2 respectively (θ1 > θ 2). Heat will flow from the inner side towards the outer side along the radii of cylinder. Consider an elementary cylindrical shell of radius r and thickness dr. Let θ and (θ + dθ) be the temperatures at r dθ (negative sign shows that θ decreases with increases in r). and (r + dr ). The temperature gradiant is dr

H-34 θ2 θ+dθ

x

θ θ1

r

r1

(r+dr)

r2

x

l Fig. 17

The radial rate of flow of heat through this elementary shell in steady state is, dQ dθ …(1) = − k (2πrl ) dt dt dQ is same for all values of r. where k = thermal conductivity and 2πrl is surface area of shell. In steady state, dt Re-arranging equation (1), we get, dθ =

−(dQ / dt) dr ⋅ 2πkl r

Integrating, we get, θ=

−(dQ / dt) log e r + C 2πkl

…(2)

θ1 =

−(dQ / dt) log e r1 + C 2πkl

…(3)

θ2 =

−(dQ / dt) log e r2 + C 2πkl

…(4)

where C is a constant of integration. At r = r1, θ = θ1 ∴ At r = r2, θ = θ 2 ∴

Subtracting equation (4) from equation (3), we get, r dQ / dt θ1 − θ 2 = log e 2 2πkl r1 This gives for radial rate of heat flow, dQ 2πkl (θ1 − θ 2) = dt r  log e  2   r1  Temperature distribution dQ from equation (3) in equation (2), we get, Substituting dt

…(5)

H-35 θ1 =

− (θ1 − θ 2) log e r1 + C  r2  log e    r1 

C = θ1 +

∴

Substituting

(θ1 − θ 2) log e r1 r  log e  2   r1 

…(6)

dQ from equation (5) and C from equation (6) in equation (2), we get, dt (θ − θ 2) log e r1 (θ − θ 2) log e r + θ1 + 1 θ= − 1 r  r  log e  2  log e  2  r  1  r1  r log e    r1  θ = θ1 − (θ1 − θ 2)  r2  log e    r1 

∴

This is required expression. Example 10. A body which has a surface area 5 cm 2 and a temperature of 727°C radiates 300 joules of energy each minute. What is its emmissivity? (Stefan-Boltzmann constant = 5.67 × 10 −8 watt/m 2 (°k) 4 ). Solution : The total energy radiated by the body of emmissivity e, surface area A, at temperature T in

surroundings of temperature T0 in time t is given by, Q = EAt = eσ(T 4 − T04 ) At = eσT 4 AT (if T0 copper > aluminium > iron (b) silver < aluminium < copper < iron (c) silver < aluminium < copper > iron (d) copper > silver > iron > aluminium

H-47 20.

21.

Which of the following sets of qualities is most useful 4. for cooking kettles ? 5. (a) high specific heat and low conductivity 6. (b) high specific heat and high conductivity 7. (c) low specific heat and high conductivity (d) low specific heat and low conductivity 8. A wall is made of two equally thick layers A and B of different materials the thermal conductivity of A is 9. twice of B. In the steady state the temperature difference across the wall is 36°C. The temperature 10. 11. difference across layer A will be : (a) 6°C (c) 18°C

(b) 12°C (d) 24°C

Q = ……… is the rate of heat flow. Thermal conduction in metal is taken by ……… ……… method is used for bad conductors. The destruction offered by a material in the flow of heat is called……… ………is the dimension formula for thermal resistance. Ice is covered with………to prevent of melting. Copper is a ………of heat. Thermal conductivity of copper is……… J/ms °C.

12.

In Lee’s method the rate of cooling is proportional to……… .

22.

Thermal conductivity of bad conductors is measured 13. by : (a) Lee’s method (b) Searle’s method (c) both (d) none of these 14.

According to Newton's law of cooling the rate of cooling of a body is directly proportional to the difference in ………of body.

23.

Two rods of length x1 and x 2 and coefficients of 15. ……… is the example of forced convection. thermal conductivity are k1 and k 2 are kept touching 16. Flow in boilers is the example of ………convection. each other. The area of both rods is same thus equivalent thermal conductivity is : True/False (a) k1 x1 + k 2 x 2 (b) k1 + k 2 1. It is sometimes possible for a block of wood and a k 2 x1 + k1 x 2 x1 + x 2 (d) (c) block of metal to feel equally cold or hot when x1 x x1 + x 2 + 2 touched. k1 k2 2. Two layers of cloth of equal thickness provides a In Searle’s method for finding conductivity of metal warm covering as a single layer of double the the temperature gradient along the rod : thickness. (a) is greater near the hot end 3. The thermal conductivity of air being less than that of (b) is greater near the cold end felt we prefer felt to air for thermal insulation. (c) constant along the bar 4. On a winter night we fell warmer when clouds cover (d) none of these the sky than when the sky is clear. Transportation of heat in solids by : 5. Two thermometers are contructed in such a way that (a) radiation (b) conduction one has a spherical bulb and the other an elongated (c) convection (d) none of these cylinderical bulb. The bulb are made of the same material and their walls have equal thickness. The In steady state condition : spherical bulb will respond more quickly to (a) temperature is constant temperature changes. (b) temperature increases 6. A sphere, a cube and a thin circular plate of same (c) temperature decreases mass are made of the same material. If all of them (d) none of the above are heated upto the same high temperature, the rate of cooling is maximum for which shape.

24.

25.

26.

Fill in the Blank 1.

Thermal conductivity is ………… to the distance between planes.

2.

In steady state the temperature is …………

3.

……… is the SI unit of k.

7.

………is the example of free convection.

Before the steady state is reached, the rate of heat flow is proportional to the thermal diffusivity (or thermometric conductivity = k/Cρ the ratio between the thermal conductivity specific heat and CP the ratio between the thermal conductivity specific heat and ρ the density.

H-48 8.

9.

Boilers are occasionally scrubbed by rapidly and artificially circulating water inside them to remove any thin water film that may have formed on their inside. 10. Newton’s law of cooling is an approximate form of Stefan’s law is valid for both for forced and natural

convection provided the difference of temperature between the body and its surroundings is smaller than about 30°C. During clear nights, the temperature rises steadily upwards near the ground level, i. e. the lapse rate there is negative.

H-49

Objective Type Questions Multiple Choice Questions 1.

(c)

2.

(a)

3.

(b)

4.

(c)

5.

(a)

6.

(d)

7.

(a)

8.

(c)

9.

(c)

10.

(d)

11.

(b)

12.

(a)

13.

(b)

14.

(a)

15.

(d)

16.

(d)

17.

(b)

18.

(d)

19.

(a)

20.

(c)

21.

(b)

22.

(a)

23.

(d)

24.

(c)

25.

(b)

26.

(a)

Fill in the Blank 1.

inversely proportional

2.

constant

3.

kilo cal/metre-sec °C

4.

dT Q = − kA dx

5.

free electrons

6.

Searle’s

7.

thermal resistance

8.

[M −1 L −1 T 3 θ]

9.

saw dust

10.

good conductor

11.

385

12.

surface area

13.

temperature

14.

Thunderstorms

15.

Cooling innuclear reaction

16.

forced

True / False 1.

True

2.

False

3.

False

4.

True

6.

True

7.

True

8.

True

9.

False

5.

False

10. True

H-50

H ints and Solutions =

Numerical Solutions 1.

The quantity of heat flowing through the stone in one hour = 4800 × 80 = 384 × 103 cal Now, we know that, Q = kA

=

384 × 103 10 1 × × 3600 100 3600

The temperature T of the junction Q of the bars PQ and QR can be determined by remembering that, in the steady state, the heat flowing per second through PQ must equal the heat flowing per second through QR.

T

T1 = 100°C

9.2 × 10− 2 × π × (10− 2 )2 × 67 . 075 .

= 258.2 × 10−6 k cal s −1 = 0.2582 cal s −1 Let α the the temperature of the interface in the steady state. Since the same amount of heat must flow through wood and cork, we have, k1 A (12 − α) k 2 A (α − 0) = l1 l2 where l1 and l 2 are the thickness of wood and cork respectively. Their respective conductivities are k1 and k 2 .

= 2.96 × 10−3 calcm −1 s −1 (°C) −1 2.

Thus the temperature of the junction is 93.3 °C. The heat flowing through the junction per second is now given by, k A (100 − 93.3) Q= 1 075 .

∆T t ∆x

where k is the thermal conductivity of the stone, A its surface area and ∆x its thickness and ∆T is the temperature difference across the two faces of the 3. slab and t is the time for which heat flows. Substituting the values we get Q ∆x 1 k= t ∆T A =

1150 = 93.3°C 0.825 + 11.50

Substituting the values we get, 6 × 10− 4 12 × 10−5 × (12 − α) = ×α 175 2 . . α × 10−1 12 − α = 175

or

α=

T2 = 0°C

12 = 10.2 C 1175 .

Copper P

R

Q 75 cm

125 cm

Since the area of cross-section of the two bars are the same, we have, K1 A (T1 − T) K 2 A (T − T2 ) = 75 125 K2 K1  K1 T1 K T  4. + + 2 2 T  =  125 75  75 125 Hence, T = =

K1 T1 × 125 + K 2 T2 × 75 K 2 × 75 + K1 × 125 −2

9.2 × 10

. × 10−2 × 0 × 75 × 100 × 125 + 11

11 . × 10−2 × 75 + 9.2 × 10−2 × 125

0°C 12°C

Let θ1 and θ2 be the temperature at the two ends of the bar. Since the same amount of the heat must be flowing per second throughout the bar in the steady state, we have k1 A (100 − θ1 ) k 2 A (θ1 − θ2 ) k1 A (θ2 − 0) = = 0.01 50 0.01

H-51

Copper θ1°C

y dy =

Water 0°C

Water

Hence, if the thickness increases form y1 to y2 in time t, we have, y2 kθ t ∫y1 y dy = ρL ∫0dt

θ2°C

Here, k1 = 0.0014 cal cm −1 s −1 (°C) −1 = thermal conductivity of water, k 2 = 1.04 cal cm −1 s −1 (°C) −1

t=

= thermal conductivity of copper. Hence, we get, 100 − θ1 = θ2 k1 A (100 − θ1 ) k 2 A [(θ1 − (100 − θ1 )] and = 0.01 50 k1 or 5000 × × (100 − θ1 ) = (2θ1 − 100) k2 or

5000 ×

or 5000 ×

k1 k2

ρ = 0.91 g cm −1 , k = 0.005 cal cm −1 s −1 (°C) −1 , θ = 20° C, y2 = 3.1 cm and y1 = 3.0 cm We get, t=

6.

The temperature gradient along the bar is θ1 − θ2 88.54 − 11.46 = 50 l = 1542 ° C per cm Let a layer of ice of surface area A in crease in thickness from y to (y + dy) in time dt. The amount of heat flowing ice from the water at the bottom (at θ1 ° C) to the surroundings (at θ2 ° C) in time dt is given by, kAθdt ∆Q = y where θ = (θ1 − θ2 ) = 20° C in this case. The mass of ice formed when –20°C the layer of ice increases in dy thickness by dy is A dy ρ, where ρ is the y ICE density of ice. Hence, the heat released to the 0°C surroundings also equals (A dy ρ) X L, where L is the latent 7. heat of ice. Thus, Water we have, 8. kAθ dt = AρLdy y

. )2 − (3)2 0.91 × 80 (31 × 0.05 × 20 2

= 222.04 s = 3 min 42 s

Which gives θ1 = 88.54°C, θ2 = 100 − θ1 = 11.46 °C

5.

2 2 ρ L(y2 − y1 ) 2 kθ

Substituting the numerical values of L = 80 cal g −1 ,

× (100 − θ1 ) = (2 θ1 − 100)

0.0014 × (100 − θ1 ) = (2 θ1 − 100) 1.4

kθ dt ρL

According to Newton’s law of cooling, we have, dθ = − k(θ − θ0 ) dt where θ is the temperature of the body and θ0 that of the surroundings. Hence, dθ = − kdt θ − θ0 Integrating we have, θ 2 dθ t ∫θ1 θ − θ = − k ∫0dt 0  θ − θ0  log e  2  = − kt  θ1 − θ0   60 − 25  log e   = − k × 10 × 60  50 − 25  If α is the temperature at the end of the next 10 minutes : We have,  50 − 25  log e   = − k × 10 × 60  α − 25  Hence,

35 25 = 25 α − 25 α = 42.85°C

See example 1 Ans. 3 × 10−4 kilo cal m −1 °C −1 s −1 We have Q =

kA (θ1 − θ2 )t d

Here θ represents temperature T But

Q = mL

H-52 mL =

kA (θ1 − θ2 )t

14.

d

H=

Here m = 5 kg, L = 80 kilo cal/kg for ice A = 0.46 m 2 (θ1 − θ2 ) = 200 ° C − (− .3 ° C)

V 2 (220)2 = 1936 Joule/sec = 25 R 1936 = = 460.9 cal sec 4.2

Heat passing through glass window per sec Q KA (θ1 − θ2 ) = H= t d KA (θ1 − θ2 ) = 460.9 d 0.2 × 1 (25 − θ2 ) = 460.9 01 . × 10−2

= 203° C t = 1 hour = 60 × 60 = 3600 sec d = 0.2 m k=? k × 0.46 × 203 × 3600 0.20 5 × 80 × 0.20 80 = k= 0.46 × 203 × 3600 336168

5 × 80 = or

Heat generated by heater per sec

(25 − θ2 ) =

. × 10−2 460.9 × 01 0.2

=

460.9 × 01 . × 10−2 0.2

= 0 .000237 kilo cal m −1 °C −1 s −1 9.

See example-2 Ans. θ = 76°C

10.

(25 − θ2 ) = 230.45 × 10−2

See example-3

θ2 = 25 − 2.30

Ans. θ1 − θ2 = 0.012°C 11.

Amount of heat flowing through the bottom of kettle Q mL = H= t t 2 × 540 cal/sec = 60 = 18 cal/sec Q kA(θ1 − θ2 ) We have = H= t d kA (θ1 − θ2 ) i. e. = 18 d

θ2 = 22.7 °C 15.

Ans. 93.31°C 16. 17.

=

7.2 300

18.

See example-4

400 − 4θ = θ 400 = 5θ θ = 80° C 19.

20. 21.

See example-5 Ans. 15.24°C

See example-12 Ans. 200 °C/m T1 − T2 1000 − 500 = 500° C/m = 1 dx See example-13 Ans. − 4°C

22.

Ans. 9 kW 13.

Let the temperature of interface is θ 4 KA [100 − θ] KA (θ − 0) = d d 4 × (100 − θ) = (θ − 0)

= 0.024°C 12.

See example-11 Ans. θ2 = 16 °C and θ3 = 43 °C

A = 600 cm 2 , d = 0.4 cm

0.4 × 18 (θ1 − θ2 ) = 0.5 × 600

See example −7 Ans. 7 hours 40 min

k = 0.5 cal °C −1 sec −1 cm −1

θ1 − θ2 = ? 0.5 × 600 × (θ1 − θ2 ) So, = 18 0.4

See example −6

See example-13 Ans. 0 °C

23.

See example-14 Ans. 5/3

H-53 24.

5.

See example-14 Ans. 9/5

25.

See example-15 Ans. 64 °C

26.

Growth of ice on ponds : When the atmospheric temperature falls below 0°C (say θ°C), the cold air above water extracts heat from the water. As a result the water begins to freeze into the ice layers.

AIR At θ°C

See example-15 θ1 − θ2 θ1′ − θ2′ Ans. = d1 d2

x

80 (θ1′ − θ′2 ) = 1 2

dx

(θ1′ − θ2′ ) = 80 × 2 (θ1′ − θ′2 ) = 160 °C 27.

See example-13 Ans. −10 °C

28.

Consider at any time the thickness of ice is X and further layer of ice of thickness dx is formed in time dt.

See example-12 Ans. 160 °C/m

29.

See example-17

If ρ is density of ice and L, the latent heat of fusion, dθ then from the relation Q = mL = kA t gives dx θ ( Adxρ)L = kA dt X ρL × dx dt = ∴ kθ

Ans. 116 °C and 74°C 30.

See example-18 Ans. 34.8 °C

Short Answer Type Questions 1.

Read 2.1, 2.2 and 2.3

2.

Read 2.1 and 3

3.

Read 2.1

4.

When the temperature at each point attains a constant value this condition is called steady state. 6. When one point of a body is maintained at high 7. temperature than the other, the flow of heat occurs 8. from the point at higher to lower temperature. The temperature at the midway points decreases in the 9. 10. direction of flow of heat.

The time for the thickness of ice growing from X1 to X2 is given by, ρL X 2 ρL 2 X dx = t= ( X2 − X12 ) 2kθ ∫X1 2kθ See example-8 See example-8 See example-9 See 3 See 5.1.1

The rate of change of temperature with distance in 11. T − T2 direction of two points distance x apart, then 1 12. x 13. is known as temperature gradient. If T and (T + dT) are the temperature at points x and 14.

See 5.2

(x + dx) this temperature gradient T − (T + dT) = x + dx − x

15.

See 9.2

16.

See 9.2

17.

See 10

18.

See 11

=−

dT dx

See 6 See 9.1 See 9.1

19. Negative sign indicates that the temperature 20. decreases with increates in x.

See 13.1 and 14.1

21.

See 13.3 and 14.3

See 13.2 and 14.2

H-54 (iii) Reynolds number

Very Short Answer Type Questions 1.

(i) Conduction (ii) Convection (iii) Radiation

13.

2.

Radiation

14.

3.

When the temperature at each point attains a 15. constant value this condition is called steady state. 16. When the temperature of each cross-section varies with time then its called variable state

4. 5.

The rate of variation of temperature with respect to dT distance is known as temperature gradient i. e. − . 17. dx

6.

k cal m −1 C −1 s −1 −3 −1

Fluid flow is Laminar Re > 4000 Diffusivity dominates (a) Air — 0.024 w/mk (b) Glass — 0.78 w/mk (c)

Water — 0.556 w/mk

(d) Copper — 385 w/mk Radiation

18.

(i) Copper (ii) Silver

7.

[MLT θ ]

19.

The resistance offered by the body in the flow of heat

8.

k=∞

20.

[M−1 L−2 T 3θ]

9.

Silver

21.

10.

Lee’s Disc method

To prevent the melting of ice because the wood is the insulator of heat.

11.

Searle’s method

22.

Because copper is a good conductor of heat.

12.

(i) Nusselt number (ii) Prandtl number

mmm

H-55

Unit

2

Kinetic Theory of Gases

1. Ideal Gas Gases that follows Boyle’s and Charles law at all temperature and pressure are known as ideal gases. The molecules of ideal gas does not impose any attractional of force on each other. Hence they have only kinetic energy which depends upon temperature. The ideal gas equation is given as, PV = nRT PV = RT

or where

(for n = 1 mole)

P → Pressure T → Temperature R → Gas constant

The behaviour of real gas is too different from ideal gas. Some real gases have very much deviation while some gases are nearly ideal gases but not perfect. Generally at low pressure and high temperature all gases are closer to ideal gases but as the pressure rise and temperature becomes slow down then gases deviates excessively from ideal gas. At sufficient temperature and pressure we can liquify gases.

2. Kinetic Theory of Gases According to this theory : 1.

Each gas is made up of small particles called molecules.

2.

Molecules are far from each other hence maximum part of gas is blank.

3.

Molecules can move anywhere in blank space.

4.

During the motion the molecules collides with the walls of vessel; after collision, speed and direction of molecules is altered.

5.

The pressure of a gas arise when the molecules strikes continuously on the walls of vessel.

H-56

2.1 Basic Assumptions There are several basic assumptions for ideal gas : 1.

Gas molecules are rigid, smooth and spherical, hence their rotation is not possible.

2.

Gas molecules are very small in size.

3.

Attraction force between gas molecules is zero.

4.

Each and every collision is elastic.

5.

Large number of molecules and large average separation (molecular volume is negligible).

6.

All of the gas molecules are identical.

3. Microscopic View of an Ideal Gas An ideal gas is a theoretical gas composed of many randomly moving point particles whose only interactions are perfectly elastic collisions. The ideal gas concept is very useful because it obeys the ideal gas law, a simplified equation of state and is amenable to analysis under statistical mechanics. One mole of ideal gas has a volume of 22.710947 litres at STP of 273.15 K and an absolute pressure of exactly 105 Pa. Some important microscopic properties are given below :

3.1 Pressure An ideal gas exerts pressure on a container is given by P=

2 1 Nm nv rms 3 V

where, N → Total number of molecules of the gas m → Mass of each molecules V → Volume of container Vrms → The rms speed of gas molecules

3.2 Temperature of an Ideal Gas We have pressure 2 1 N mv rms 3 V 1 2 PV = N mv rms 3

P=

or

=

2 1 2  N nv rms  3  2 

The average kinetic energy of the molecules is 1 3 PV 2 K = mv rms = 2 2 N

H-57 But according to equation of state  N PV = nRT =   RT = NkT  NA

We have, Where N A → Avogadro number k → Boltzmann’s constant

k=

R NA

Thus

K=

1 3 2 mv rms = kT 2 2

or

K ∝T

Kinetic energy is proportional to temperature of gas.

4. Degree of Freedom The total number of coordinates or independent quantities required of completely specify the configuration of molecules of a system are called its degrees of freedom. 1.

Monotomic gas : Such gases as helium, argon, etc. consists of a single atom which has only three translational degrees of freedom along x, y and z axis. Thus a monotomic gas has only three degrees of freedom.

2.

Diatomic gas : Such as hydrogen, oxygen, hydrogen etc. consists of two atoms and has three translational and two rotational motions i. e. 3 + 2 = 5 degrees of freedom.

3.

Triatomic gas : It has three translational motions and three rotational motions = 3 + 3 = 6 total degrees of freedom.

5. Law of Equipartition of Energy This law has been stated by Maxwell. According to this law when two gases at different temperature mixed together then after sometimes both gases comes in steady state condition. Hence the temperature of both gases becomes equal. Maxwell states, that each collision between molecules will decrease in energy and when both gases comes at some temperature then average kinetic energy becomes equal. Hence two gases of equal kinetic energy has same temperature. Hence only kinetic energy is the best measurement of temperature. This is called law of equipartition of energy. Boltzmann states this law in the form of degree of freedom. According to this law, the average energy of molecules of a dynamical system in thermal equilibrium is

kT 2

where k is Boltzmann’s constant and T is absolute temperature. Limitation of this law is that it is applicable only for normal and high temperature. It is not valid for very low temperature when gas comes at meting point.

H-58

5.1 Proving of Law of Equipartition of Energy Let a system has many molecules, their position coordinate are q1, q2, q3, …q f and momentum coordinates p1, p2, p3, …p f . The average energy of molecules − βE i

∫ E ie E1 = − βE ∫e − Ei =

i

dpi

dpi

∂ [ e −βE i dpi] ∂β ∫ − βE ∫ e i dpi

=−

∂ log e [∫ e −βE i dpi] ∂β

Now the momentum space limits are from −∞ to + ∝ +∞ ∂ log e [∫ e −βE i dpi] Ei = − −∞ ∂β

…(1)

pi 2 2m 1 b= 2m

Ei =

Now we have energy Let

E i = bpi 2

So, +∞

∫−∞ e

or

− βE i

dpi =

+∞

∫−∞ e

−βbpi2

dpi

y = β pi

Let

dy = B dpi +∞

∫−∞ e

So,

So, using equation (1)

− βE i

1

Ei = − =−

The integral So,

+∞

∫−∞ e

− by 2 dy

So,

− by 2 dy

 1 ∂ log e  ∂β  β ∂ ∂β

+∞

∫−∞ e

Ei = −

∂ ∂β

1 2β

1 kT kT Ei = 2 β=

This is law of equipartition of energy.

− by 2 dy

  

+∞ − by 2 dy   1 − 2 log e β + log e ∫−∞ e 

= 0 with respect to β.

Ei = But

+∞

∫ e β −∞

dpi =

  1 − 2 log e β

H-59

6. Relation between Atomicity and γ of a Gas We know that the degree of freedom changes as the number of atom changes in gas and γ also depends upon atomicity. Cp Where γ= Cv

6.1 Monotomic Gas For monotomic gas there are three degrees of freedom (only 3 translational). kT energy. Then kinetic energy of one mole According to law of equipartition of energy, each molecule has 2 gas kT E=N× ×3 2 3 or E = NkT 2 3 or E = RT 2 Where

Nk = R

Now specific heat at constant volume, Cv =

dE 3 = R dT 2

But according to Mayer's formula, C p − Cv = R

For monotomic gas, ∴

Cp = R + CV 3 Cp = R + R 2 5 Cp = R 2 Cp 5 / 2 R 5 = = γ= Cv 3 /2 R 3 γ = 1.66

6.2 Diatomic Gas For diatomic gas number of degrees of freedom are five (3 translational + 2 rotational). Energy of 1 mole gas E=N or

E=

kT 5 × 5 = NkT 2 2

5 RT 2

[‡ Nk = R]

H-60 Now specific heat, Now,

Cv =

dE 5 = R dT 2

C p − Cv = R

or

Cp = R +

or

Cp =

7 R 2 Cp

So,

γ=

∴

γ = 1.4

Cv

5 R 2

=

7 /2 R 7 = 5 /2 R 5

6.3 Triatomic Gas For triatomic gas there are six degrees of freedom (3 translational + 3 rotational). kT Energy, E = N×6 2 = 3NkT E = 3RT

or Now specific heat,

Cv = or Now, ∴

dE = 3R dT

C P = R + Cv = R + 3R = 4R Cp 4 R 4 γ= = = 3R 3 Cv γ = 1.33

7. Relation between γ and Degree of Freedom ‘f’ It we have one mole gas at temperature ‘T ’ and the degree of freedom of the molecules are ‘f ’ then total energy of one mole gas 1 E = kT × N × f 2 1 or [‡ Nk = R] E = f RT 2 Now specific heat, Cv =

dE 1 = fR dT 2

Now we have, C p − Cv = R or

C p = R + Cv =

1 fR+R 2

H-61 1   C p = 1 + f  R  2 

or

Cp

Now,

γ=

or

γ=

∴

γ = 1 + 2 /f

Cv (1 + f / 2 )R fR / 2

8. Distribution Law of Velocity Let us consider a gas in a container of volume V, then according to Maxwell's–Boltzmann's distribution law, the number of molecules in i th cell of energy E i n i = g i e − α e − βE i Where e

−α

…(1)

and β are the constants.

Let the molecules of gas are identical and distinguishable because they have finite separation between them. Now writing equation (1) for energy range E and (E + dE) n(E) dE = g (E)e − α e −βE dE

Thus,

…(2)

Now energy of molecules, p2 2m 2 p dp pdp dE = = m 2m E=

Now,

Now writing equation (2) for momentum range n( p) dp = g ( p)e − α e −βp

2

/ 2m

…(3)

dp

Here n( p) dp are the number of molecules between momentum range p to dp. Here g ( p) dp are number of phase cells, Volume of phase space in momentum range g ( p) dp = Volume of a phase cell g ( p) dp =

py

∫ ∫ ∫ ∫ ∫ ∫ dx dy dz dpx dpy dpz h3

p+

where each phase cell has volume h 3. But ∴

B

∫ ∫ ∫ dx dy dz = V g ( p) dp =

V h3

dp

p

∫∫∫

dpx dpy dpz

A

Now we will find the value of momentum space i. e. ∫ ∫ ∫ dpx dpy dpz . Momentum space is the volume in between spare of radius ( p + dp) and p.

Fig. 1 : Momentum space

px

H-62 Now volume between two concentric circle, = Volume of circle of radius ( p + dp) − Volume of circle of radius p 4 4 3 3 ∫ ∫ ∫ dpx dpy dpz = 3 π ( p + dp) − 3 πp 4 = π ( p + dp)3 − p3 3 4 = π[ p3 + dp3 + 3 pdp ( p + dp) − p3] 3 4 = π[3 p2dp + 3 p(dp)2 + (dp)3] 3 4 [ ‡ dp2 & dp3 are negligible because dp is very small] = π 3 p2 dp 3 or

∫ ∫ ∫ dpx dpy dpz = 4 πp dp 2

g ( p)dp =

So, now

V h3

4πp2dp

Using y( p) dp into equation (3), we get, n( p) dp =

4 π p2 h

Ve − α e −βp

3

2

/ 2m

…(4)

dp

This equation states number of molecules in momentum range p and ( p + dp). This is known as M.B. distribution law in momentum range. Now total number of molecules in whole system, N=

∞

∫0 n( p) dp ∞

∫0

4 π p2

⋅ Ve − α e −βp

2

/ 2m dp h 2 ∞ 4 πV = 3 e − α ∫ p2e −βp / 2m dp 0 h

N=

=

∴

N=

or

e− α =

3

4 πV h

3

e− α ×

π

1 4

 2π m  e− α   3  β  h V

Nh 3  β    V  2π m 

 β     2m 

3

…(5)

1 π  ∞ 2 − ax 2 dx = ∫0 x e 4 a3

3/ 2

…(6)

3/ 2

…(7)

Putting e − α into equation (4), we get, n( p) dp =

4 π p2 h

3

V

Nh 3  β    V  2π m 

3/ 2

e − βp

2

/ 2m

dp

H-63  β  n( p)dp = 4 πN    2π m 

3/ 2

p2e −βp

2

/ 2m

…(8)

dp

p2 = 2mE

Now using

2 pdp = 2mdE mdE dp = p

or or

dp =

or

mdE 2mE

So, equation (8) will change into energy range  β  n(E)dE = 4 πN    2π m  or

n(E)dE =

2N π

3/ 2

2mE e −βE

md E 2 mE

β 3/ 2E1/ 2e −βE dE

…(9)

Now total energy of gas of N molecules ∞

E=

∫0

or

E=

∫0

or

E=

2N

or

E=

3N 2β

…(11)

3 NkT 2

…(12)

∞

En(E) dE 2N π

π

β 3/ 2 E 3/ 2 e −βE dE

β 3/ 2 ×

3 4β

2

π β

3 π  ∞ …(10) ‡ ∫ x 3/ 2e − ax dx =  0 4 a2 a  

Now total energy of molecules of an ideal gas E=

Comparing equation (11) and equation (12) we get, 1 β= kT Putting this value into equation (10)  1  n( p)dp = 4 πN    2πmkT   1  n(E)dE = 2πN    πkT  Now to convert the above formula into velocity range, 1 Putting, E = mv 2 2 dE = mvdv

3/ 2

3/ 2

p2 e − β p

2

/ 2m

E1/ 2e − E / kT dE

dp

…(13)

…(14)

H-64 3/ 2

 m  n(v)dv = 4 πN    2πkT 

Now,

v

2

− mv 2 e 2kT

…(15)

dv

This formula gives the number of molecules in velocity range v to (v + dv). Equation (15) known as Maxwell-Boltzman distribution law.

9. Velocity Distribution Graph We have Maxwell Boltzmann's velocity distribution law  m  n(v)dv = 4 πN    2πkT 

3/ 2

v 2e − mv

n(v)  m  = F (v) = 4 π    2πkT  N

Now

3/ 2

2

/ 2kT

v 2e − mv

dv 2

/ 2kT

where F (v) is known as number factor between velocity range v and (v + dv). The graph between n(v) and v at two temperature T1 and T2 given as follows.

n(v)

T 2 > T1

vavg

vmp T1

vrms

A

T2

B vrms

v

v+dv v

vmp1 vavg

vmp2

Area = n(v)dv Fig. 2 : Velocity distribution graph

It is clear from figure that first of all graph increases a parabolic and comes at peak after it becomes down exponentially. We can get some information from both curves : 1.

Shaded area gives number of molecules between velocity range v and (v + dv).

2.

From graph v mpi < v avg .

3.

The area of both curves are same and both areas are proportional to number of molecules.

4.

If v = 0, so, x v = 0 Hence no molecules have zero velocity

5.

For small values of v mv 2 > 1 2kT e So,

7.

−

mv 2 2kT

n(v) ∝ e

−

mv 2 2kT

Speed of molecules from 0 to ∞.

10. Expression for Most Probable, rms and Mean Speed 10.1 Average Speed or Mean Speed If n1, n 2, n 3 … n n are the molecules in a gas and v1, v 2, v 3… v n are the velocities Thus average speed v=

n1v1 + n 2 v 2 + ... + n i v i n1 + n 2 + n 3 + ... + n i

v = v avg = So,

Where

v avg =

∞

∫0 vn(v) dv

 m  n(v)dv = 4 πN    2πkT  v avg

Let

1 N

1 Σ n i vi N i

3/ 2 2

v e

1  m  4 πN  =   2πkT  N

−

3/ 2

mv 2 2kT

dv

1 mv 2 ∞ 2 − v e 2 2kT dv 0

∫

m =a 2kT  a v avg = 4 π    π

3/ 2

 a = 4π   π

3/ 2

=

∞ 2 − av 2

∫0 v ×

e

1 2a

2

1   ∞ 2 − av 2 dv = ∫0 v e 2a 2 

4 πa

m  4 × 2 kT  = = a + kT  2 πm  v avg = 1.59

kT m

dv

8kT πm

H-66

10.2 rms Speed ∞

v

2

∫ n (v) dv v = 0

v2 =

2

N

1  m  4 πN    2πkT  N

 a v 2 = 4π    π

3/ 2

 a v 2 = 4π    π

3/ 2

∞ 4

∫0 v ×

3/ 2

∞ 4

∫0 v

e − mv

2

/ 2kT dv

2

e − av dv

3 π  ∞ 4 − av 2 dv 3 π  =  v e  8 a 5 ∫0 8 a5 

3 2a 3 2kT v2 = 2m 3kT v2 = m v2 =

v rms = v 2 v rms =

3kT m

. v rms = 173

kT m

10.3 Most Probable Speed To measure most probable speed we need to use differential calculus d n(v) = 0 dv mv 2 3/ 2   − d   m  2kT v 2 = 0 e 4 πN    2πkT  dv    

or

 m  4 πN    2πkT 

3/ 2

 mv 2   d  − 2kT 2 e v =0    dv  

 mv  2 v =0 2v + −  kT  or

2=

or

v=

mv 2 kT 2kT m

H-67 Most probable speed vp =

2kT kT = 1.41 m m

v p = 1.41

kT m

Hence it is clear v p < v < v rms

11. Pressure Exerted by a Perfect Gas According to kinetic theory of gases, molecules of gas moves randomly in gas and due to this random motion of molecules, they collides not only with each other but also with the walls of vessel. After collision the momentum of the molecules is altered. According to Newton's second law the rate of change of linear momentum is equal to the force.

E

C

D Y W P v C

Hence a force is imposed on wall by collision of molecules. ‘‘A force acted on the wall by colliding the molecules. This force on per unit area of wall is known as pressure of gas.’’

F

v

X G

H

Y v

A

Hence, according to kinetic theory of gases the pressure of gases is equal to the change of momentum given to unit area of walls of vessel. Now we will drive on expression for this pressure.

W C

B O

u

X

Fig. 3

Let a cube shape vessel ABCDEFGH. Each side of cube is ‘l’ and each wall is perfectly elastic. Let the gas is made of small molecules of rigid and elastic molecules. All molecules are identical (same mass, size and properties). These molecules moves in gas with all possible speed in all possible directions. These molecules collides with each other and also with walls of vessel. The volume of cube is V = l 3 and each face of cube has area A = l 2 Let a molecule P is moving with velocity C in any direction. This velocity is distributed into three axis X, Y and Z and the velocities components in these axis are u1 v1and w1. From fig. C 2 = u12 + v12 + w12

…(1)

the particle with velocity u moving in X-direction collides with BCFG and its momentum is mu, where m is the mass of molecules.

H-68 Now the collision of molecule and wall is perfectly elastic, thus molecule comes back with same speed in opposite direction and collides with wall ADEH after travelling ‘l’ distance. Now momentum at BCFG = + mu1 After collision, momentum = − mu1 Change in momentum = − mu1 − (+ mu1) = − 2 mu1 The molecule after colliding ADEH it collides BCFG. Now the velocity in X-direction is u1, thus time interval at colliding again at BCFG is ∆t = Hence molecules will collides at BCFG at every Now number of collision at plane BCFG =

u1 2l

2 l  Distance u1  Time  2l time. Thus a force will act on plane. u1

1   Frequency = Time peried   

Now change in momentum in each collision in per second or rate of change in linear momentum = change in momentum in each collision X number of collision in one oscillation u = − 2 mu1 × 1 2l =−

mu12 l

Now force acted on molecule = rate of change of linear momentum of molecule =−

mu 2 l

Now, according to Newton’s third law the force acted on plane BCFG (the rate of change of momentum) mu12 F1( x ) = l Similarly for other molecule which is moving with velocity C2 and components are u 2, v 2, w 2. Thus force imposing on BCFG F2( x ) =

mu 22 l

Hence the force imposing on BCFG by all n molecules F( x ) = F1( x ) + F2( x ) + F3( x ) + ... + Fn( x ) mu12 mu 22 mu n2 + + ... l l l m 2 2 2 = (u1 + u 2 + ... + u n ) l

F( x ) = F( x )

This force in per unit area will be same as pressure

H-69 i. e.

Px =

or

Px =

Fx l2 m l3

(u12 + u 22 + ... + u n2)

Similarly pressure in Y-direction i. e., on plane CDEF or ABGH m Py = 3 (v12 + v 22 + ... + v n2) l Now pressure in Z-direction i. e. on plane ABCD or EFGH m Pz = 3 (w12 + w 22 + ... + w n2) l

…(2)

…(3)

…(4)

But ideal gas exerts same pressure in each direction i. e.

Px = Py = Pz = P (Let)

Now total pressure,

or

3P = Px + Py + Pz m = 3 [(u12 + u 22 + ... + u n2) + (v12 + v 22 + ... + v n2) + (w12 + w 22 + ... + w n2)] l m 2 = 3 (u1 + v12 + w12) + ... + (u n2 + v n2 = w n2) l m 2 [‡ u12 + v12 + w12 = C12] 3P = (C1 + C 22 + ... + C n2) V

and

l3 = V

So,

P=

m (C12 + C 22 + ... + C n2) 3V

C12 + C 22 + ... + C n2 n 1m nC 2 P= 3V M But density, ρ = V 1 P = ρC 2 3 2 1 P = × × ρC 2 3 2 C2 =

or

1 ρC 2 kinetic energy of per unit volume gas 2 2 P = × kinetic energy of per unit volume gas 3 2 Hence the total pressure of gas is of its per unit volume kinetic energy. 3 But

H-70

12. Kinetic Interpretation of Temperature According to kinetic theory of gas, every gas is a collection of molecules and each molecules is identical. Due to this concept the average per molecules is proportional to the kinetic energy temperature of gas 1 i. e. kinetic energy of gas molecules M C 2 ∝ T 2 Proving : We have the formula of pressure of gas, 1 P = ρC 2 3 P=

1 mNC 2 3 V

or

PV =

1 mNC 2 3

Now

mN = M

or where

P → Pressure of gas m → Mass of molecules N → Avogadro number V → Volume of gas C 2 → Average speed of molecule

But for ideal gas PV = RT 1 RT = M C 2 3 3RT C2 = M or

C2 ∝ T

Hence, the rms speed of moelcules is directly proportional to T . If T = 0, then

C2 = 0

or C = 0

Hence it is clear that absolute zero is the temperature at which the movement of molecule will becomes zero. 1 3 But M C 2 = RT 2 2 Dividing by ‘N’, we get,

But

1M 2 3R C = T 2 N 2N M =m N 1 3 m C 2 = kT 2 2

[‡

R = k] N

H-71 1 m C2 ∝ T 2

Hence,

This is kinetic interpretation of temperature.

Example 1. Calculate the number of molecules of an ideal gas in 1 m 3 at 27°C temperature and 10 mm pressure of mercury. The kinetic energy of molecules is 6.2 × 10 −21 J at temperature 27° C, density of mercury is 13.6 × 10 3 kg/m 3 and g = 9.8 N/kg. Solution : We have pressure P = P=

So,

1 ρC 2 3 2 1 2  ρC    3 2

2 × kinetic energy of per unit volume 3 3 3 K. E. = × PV = pressure × volume 2 2 =

P = 10−2 mm of mercury

Here,

= 10−2 × 13.6 × 103 × 9.8 = 1.33 × 103 N/m 2 Volume, V = 1 m 3 Kinetic energy of gas =

3 × 1.33 × 103 × 1 = 1.995 × 103 J 2

Now, average kinetic energy = 6.2 × 10−21 J So,

number of molecules =

1.995 × 103 6.2 × 10−21

≈ 3.2 × 1023 Example 2. Calculate the Avogadro’s numbers if kinetic energy of hydrogen molecules is 5.64 × 10 −21 J at 0°C and gas constant R = 8.32 J/mole-K. Solution : We know that, kinetic energy of molecules, 1 3 mC 2 = kT 2 2 =

3R T 2N

At 0°C i. e. T = 273 K The translational kinetic energy = 5.64 × 10−21 J R = 8.32 J/mole-K

R  where = k = N 

H-72 3 × 2 3 N= × 2

5.64 × 10−21 =

8.32 × 273 N 8.32 × 273 5.64 × 10−21

N = 6.02 × 1023 Example 3. Calculate the kinetic energy of each molecules of an ideal gas at NTP given k = 1.38 × 10 −23 J/K. Solution : For a monoatomic gas, 3×

1 3 kT = kT 2 2 k = 1.38 × 10−23 J/k T = 273 K 3 = × 1.38 × 10−23 × 273 J 2 = 5.65 × 10−21 J

Example 4. Calculate the number of oxygen gas molecules in 1 m 3 volume at NTP. Avogadro’s number, N = 6.02 × 10 23 / mole. Solution : At NTP 1 mole gas has 22.4 litre and N = 6.02 × 1023 molecules Thus

volume in 1 m 3 =

6.02 × 1023 22.4 × 10−3

= 2.69 × 1025 molecules

Example 5. An ideal monoatomic gas at temperature 27°C, pressure 10 6 N/m 2 and volume of gas is 10 litre. If we added 10,000 cal without changing the volume. Calculate final temperature of gas if gas constant R = 8.31 J/mole-K and J = 4.18 J/cal. Solution : We have, PV = nRT Where n → number of moles P = 106 N/m 2 V = 10 litre = 10−2 m 3 T = 27 °C + 273 = 300 K n= For monoatomic gas, or

PV 106 × 10−2 = 4 mole = RT 8.31 × 300

3 R 2 3 Cv = × 8.31 J/mole-K 2 3 8.31 = × = 3 cal/m-K 2 4.18 Cv =

H-73 Q = nCv dT Q 10, 000 = dT = nC V 4×3

Now,

= 833 K Example 6. Calculate rms speed at nitrogen molecule at 20°C, the molecular weight of nitrogen molecules is 28 and R = 8.31 × 10 3 J/K-mole °C. Solution : We have rms speed, Here,

v rms =

3RT M

T = 20° C + 273 = 293 K, R = 8.31 × 103 J/kilo m-K and M = 28 kg/kilo-mole v rms =

∴

3 × 8.31 × 103 × 293 28

v rms = 510 .75 m/sec.

Example 7. Discuss the Avogadro’s law on the basic of kinetic theory of gases. Solution : According to this law, equal volumes of all gases under the same temperature and pressure has same number of molecules. Proving : Let we have two same volumes V of two different gases and both contains n1 and n 2 molecules respectively. The masses of the molecules are m1 and m2. Thus, we have the formula of pressure exserted by a gas, 1 mn 2 C P= 3 V Thus, for first gas, PV =

1 m1n1 C12 3

…(1)

PV =

1 m n C2 3 2 2 2

…(2)

For second gas,

Where C12 and C 22 are the rms speeds of molecules of first and second gas respectively. Now from equation (1) and equation (2) 1 1 m1n1 C 2 = m2n 2 C 22 3 3 or or or ∵

m1n1 C11 = m2n 2 C 22 1 m1n1 C12 = 2 1 m1 C12n1 = 2 1 m C2 = 2 1 1

1 m2n 2 C 22 2 1 m2 C 22n 2 2 1 3 m C 2 = kT 2 2 2 2

H-74 or

3 3 kTn1 = kTn 2 2 2

or

n1 = n 2

This is Avogadro’s law. Example 8. Find the rms velocity of nitrogen molecules at 25°C if molecular weight of nitrogen is 28 and R = 8.31 × 10 3 J/kilo mole-degree. Solution : We have v rms =

3RT M

Here, R = 8.31 × 103 J/kilo mole degree, M = 28 gram/mole = 28 kg/kilo mole T = 25 ° C + 273 = 298 K 3 × 8.31 × 103 × 298 28

v rms =

v rms = 515 m/sec Example 9. Find out the rms speed of colloidial particles of molecular weight 3.2 × 10 5 , if speed of an hydrogen molecule is 1000 m/sec. Solution :

or

or or

M2 M1

v rms = v rms v1, rms v 2, rms v1 rms v 2 rms

= =

2 3.2 × 10 1 400

5

or

=

1 400

v1 rms =

[Weight of hydrogen is 2] 1 1 × 100 , v 2 rms = 400 400

v1 rms = 2.5 m/sec

Example 10. Particle of an inorganic substance each having a mass of 1.55 × 10 −14 gm are suspended in a liquid at 300 K and have a rms speed of 2.8 cm/sec. Calculate the Avogadro’s number. 3 Solution : We know that translational energy of gas molecules is kT . 2 If v rms is rms velocity of gas molecules 1 3 2 So, mv rms = kT 2 2 m k= (v rms )2 3T =

1.55 × 10−17 kg × 2.8 × 10−2 m / sec 3 × 300

k = 1.35 × 10− 23 J/K But we have,

Nk = R

H-75 =

or

R 8.31 J / mole-K = N 1.35 × 10−23 J / K

= 6.05 × 1023 molecules/mole Example 11. Using Maxwell’s velocity distribution law find v x for which the probability falls to 1 times its maximum value. l Solution : We have Maxwell’s law of distribution,  m  n(v)dv = 4 πN    2πkT  or

n(v)dv  m  = 4π   2πkT  N

3/ 2

−

2

v e

3/ 2 2

v e

−

mv 2 2kT dv

mv 2 2kT dv

If probability of molecules between the velocity v to (v + dv) is P(v)dv So,

 m  P(v) = 4 π    2πkt 

Thus for velocity v,

For x component of velocity,

3/ 2

 m  P(v)dv = 4 π    2πkT 

2

v e

1/ 2

 m  P(v x ) = 4 π    2πkT 

e

−

1/ 2

e

−

mv 2 2kT dv

mv 2 2kT

−

mv x2 2kT

Exponent term has maximum value 1 So,

 m  P(vmax ) =    2πkT  mv x2

So,

− P(v x ) =e P(v x )max

But

P(v x ) 1 = = e −1 P(v x )max e

So,

1/ 2

2kT

mv x2 = 1 or v x = 2kT

At this value of speed velocity becomes

2kT m

1 of its maximum value. e

Example 12. Calculate the rms speed of molecules of following gases of NTP (i) air (ii) hydrogen, given that densities of air and hydrogen are 0.001293 and 0.00009 gm/cm 3 . Solution : Gas pressure, P=

1 3P ρC 2 or C 2 = 3 ρ

H-76 C2 = (i)

3P ρ

For air, P = 76 cm of mercury = 76 × 13.6 × 981 dyne/cm 2 ρ = 0.001293 gm/cm 3 C rms =

So,

3P = ρ

3 × 76 × 13.6 × 981 0.001293

= 4.856 × 104 cm/sec (ii)

For hydrogen, P = 76 × 13.6 × 981 dyne/cm 2 ρ = 0.00009 gm/cm 3 C rms =

3P = ρ

3 × 76 × 13.6 × 981 0.00009

= 1.84 × 105 cm/sec Example 13. A data is given for number of molecules and velocity of molecules Ni

vi (m/sec)

4

1

2

2

8

3

6

4

5

5

Calculate average, rms and probable speed. Solution : Average speed v=

4(1) + 2(2) + 8(3) + 6(4) + 5(5) 4 + 2+ 8+ 6+ 5

4 + 4 + 24 + 24 + 25 25 81 v= = 3.24 m/sec 25 =

Now,

Now,

v2 =

4(1)2 + 2(2)2 + 8(3)2 + 6(4)2 + 5(5)2 4 + 2+ 8+ 6+ 5

v2 =

305 = 12.2 m/sec 25

v rms = v 2 = 12.2 v rms = 3.49 m/sec

Maximum number of molecules travels with 3 m/sec so its most probable speed.

H-77 Example 14. We know that molecules at a gas travels with three different velocities v avg, vrms and v p , which velocity is related with kinetic energy? Solution : We know that average kinetic energy, 1 (v12 + v 22 + ... + v n2) m 2 N 1 K. E. = mv 2 2 1 2 K. E. = mv rms 2 K. E. =

or

[‡v rms = v 2 ]

Thus v rms is the velocity which is related with kinetic energy. Example 15. Calculate vrms and v p for hydrogen gas molecules at NTP if Boltzmann’s constant k = 1.38 × 10 −23 J/mole-K and Avogadro’s number N = 6 × 10 23 /mole. Solution : v rms =

3kT m

Given, k = 1.38 × 10−23 J/mole-K T = 0° C + 273 = 273 K For hydrogen moelcules,

So,

m=

2 gm / mol 1 M = = × 10−23 gm N 6 × 1023 mol 3

m=

1 × 10−26 kg 3 3 × (1.38 × 10−23) × 273 1 × 10−26 3

v rms =

= 9 × 1.38 × 273 × 103 = 1841 m/sec Most probable speed v p =

2 1.414 v rms = × 1841 . 3 1732

= 1503 m/sec Example 16. Calculate the temperature at which H 2 and N 2 gas molecules moves with same average speed where N 2 and H 2 has molecular weights 28 and 2 respectively. Solution : For same average speed, 8kTH = πmH or

TH =

8kTN πmN mH × TN mN

H-78 mH 2 = mN 28 TN = 40° C + 273 = 313 K 2 TH = × 313 = 22.4 K 28 = − 250.6 °C Example 17. At what temperature oxygen molecule and hydrogen molecule have same rms velocity if temperature of hydrogen molecule is 100°C. Solution : We have energy gas molecule, 1 3 mc 2 = kT 2 2 For hydrogen molecule, 1 3 mH C H2 = kTH 2 2

…(1)

1 3 M C 2 = kT 2 O O 2 O

…(2)

For oxygen molecule,

Dividing equation (1) and equation (2) We get, Here

mH C H2 mOC O2

=

TH TO

CH = CO TH = − 100 °C + 273 = 173 K mO = 16 mH TO =

mOC O2 TH mH C H2

= 16 × 173

= 2768 K T2 = 2768 − 273 T2 = 2495 °C Example 18. Calculate the number of molecules in 1 m 3 for an ideal gas at NTP. Solution : We have ideal gas, PV = RT Where,

Nk = R

or

PV = NkT N P = kT V

or or

P = nkT

H-79 Where n are number of molecules in per c.c. P kT R k= N PN n= RT =

or But

Number of molecules is 1 m 3. x = n × 106 x= x=

PN × 106 RT 76 × 13.6 × 980 × 6.023 × 1023 × 106 8.31 × 107 × 273

x = 2.688 × 1025 Example 19. Calculate the total kinetic energy of 1 gm nitrogen at 300 K. Solution : We have kinetic energy of a gas, 3 E = RT 2 3 RT or E= 2 M M = 28 gm =

3 × 8.3 × 107 × 300 = 133.4 × 107 erg 2 × 28

E = 133.4 J

1.

An electric bulb of value 250 cm 3 was sealed off during manufacture at a pressure of 10− 3 mm of Hg at 27°C. Find the number of molecules in the bulb.

2.

You are given the following group of particles, ni representing the number of molecules with speed Vi ni

Vi (m/sec)

2

1

4

2

8

3

6

4

3

5

Compute (i) the average speed Vavg, (ii) root mean square speed, (iii) the most probable speed.

H-80 3.

If the root mean square value of the molecules of hydrogen at NTP is 1.84 km/s, calculate the root mean square velocity of oxygen molecular at NTP, molecular weights of hydrogen and oxygen are 2 and 32 respectively.

4.

The first excited state of hydrogen atom is 10.2 eV above its ground state. What temperature is needed to excite hydrogen atoms to first excited level ?

5.

At what temperature the root mean square velocity is equal to escape velocity from the surface of earth for hydrogen and for oxygen. Given radius of earth = 6.4 × 106 m and g = 9.8 m/s 2

6.

Assuming hydrogen to be an ideal gas, calculate the root mean square speed of the gas molecules at 0.0°C and a pressure of 1 atm. The density of hydrogen at 0.0° C = 9 × 10−2 kgm −3 .

7.

At what temperature would the root mean square speed of a gas molecule have twice its value at 100°C.

8.

Calculate the average kinetic energy of a gas molecule at a temperature of 300 K. The Boltzmann’s constant k = 1.38 × 10−23 JK −1 .

9.

Argon is a monoatomic gas. Assuming that the gas is an ideal one, calculate CP and CV for argon. Given R = 8.32 JK −1 mol −1 .

10.

Given that densities of air and hydrogen are 0.001293 and 0.00009 gm/cm 3 then calculate the rms speed of molecules at NTP for (i) air and (ii) hydrogen.

11.

Calculate the number of moelcules of an ideal gas in 1 m 3 at 27°C temperature and 10 mm pressure of mercury. Kinetic energy of molecules is 6.2 × 10−21 J at 27°C. The density of mercury is 13.6 × 103 kg/m 3 and g = 9.8 N/kg.

12.

Calculate the Avogadro’s number if K.E. of hydrogen molecules is 5.64 × 10−21 J at 0°C and gas constant is R = 8.31 J/mole-K.

13.

Calculate the kinetic energy of each molecules of an ideal gas at NTP given k = 1.38 × 10−23 Joule/K.

14.

Calculate the number of molecules of oxygen 1 m 3 volume at NTP given Avogadro’s number N = 6.02 × 1023 mole.

15.

An ideal monoatomic gas at 27°C temperature and pressure 106 N/m 2 . Volume of gas is 10 litre and if we add 10,000 cal without changing the volume then calculate final temperature of gas given gas constant R = 8.31 J/mole-K.

16.

Calculate rms speed of nitrogen molecule at 20°C, the molecular weight of nitrogen molecules is 28. Given R = 8.31 × 103 J/K-mole°C.

17.

Find the rms speed of collidal particles of molecular weight 3.2 × 105 if speed of hydrogen molecules is 1000 m/sec.

18.

Using the formula Ω =

n! prove that thermodynamical probability for (4, 1) state in 5 while for (3, 2) state, it n1 ! n2 ! n3 !

is 10. 19.

Calculate the rms and most probable speed of hydrogen molecules if Boltzmann’s constant k = 1.38 × 10−23 J/mole-K and Avogadro’s number N = 6 × 1023 mole.

20.

Calculate the temperature at which average speed of H 2 molecules equals with average speed of molecules of N 2 at 40°C. Molecular weight of N 2 and H 2 are 28 and 2 respectively.

21.

At what temperature oxygen molecules and hydrogen molecule have same rms velocity if temperature of hydrogen molecule is 100° C.

22.

Calculate the number of molecules in m 3 of an ideal gas at NTP.

23.

Calculate the total kinetic energy of 1 g nitrogen gas at 300 K.

24.

Calculate the number of molecules in litre of an ideal gas at 13.6°C and 3 atmospheric pressure.

25.

Calculate the total kinetic energy of 1 g molecules of O2 gas at 300 K.

26.

Calculate total kinetic energy of 2 g He gas at 200 K.

27.

Calculate total kinetic energy of 10 g He gas at 100 K.

H-81

Long Answer Type Questions 1.

2.

3.

4.

5.

12.

What is kinetic theory of gases ? Discuss its assumptions and give microscopic properties of ideal 13. gas. What is an ideal gas ? Discuss kinetic theory of gases. Explain : (i) Pressure of ideal gas (ii) Temperature of 14. ideal gas. What do you mean by degree of freedom of a gas ? State the law of equipartion of energy and give its 15. one example. kT Derive law of equipartition of energy E = and 2 show that for monoatomic gas γ = 1.6, for diatomic gas γ = 1.4 and for triatomic gas γ = 1.33.

6.

What is Maxwell’s distribution law of velocity? Explain in detail.

7.

Show that number of molecules in velocity range v to (v + dv) are m  n(v)dv = 4 πN    2 πkT 

3/ 2

v2 e

−

mv2 2 kT

Discuss the kinetic theory of gases? Give its assumptions and hence gives interpretation of temperature on the basis of kinetic theory. Write a short note on the following : (i) Kinetic theory of gases (ii) Interpretation of temperature on the basis of kinetic theory (iii) Law of equipartition of energy

Derive Maxwell’s distribution law of velocity and hence plot the graph between number of molecules and velocity of molecules.

9.

Distinguish between mean speed, rms speed, and most probable speed and show that v = vavg = 1.59 vrms = 173 .

kT m

vp = 1.4i

kT m

2.

Given samples of 1 cm 3 of hydrogen and 1 cm 3 of oxygen both at NTP, which sample has a larger number of molecules ?

3.

If a container with porous walls filled with a mixture of the two gases and is placed in an evacuated space, then lighter of the two gases will escape out sooner. Write the relevant equations which you use for explanation.

4.

The kinetic temperature of the gas in the upper atomosphere is of the order of 1000 K, but it is quite cooled up there. Resolve this paradox.

5.

Will the temperature of gas in a container increase when we put the container on a moving train ? Explain.

6.

The air that leaks out into vacum system contains a larger fraction of H2 and He than that found in the outside atmosphere. Explain this using kinetic theory.

7.

During the inelastic collisions between the macroscopic bodies, mechanical energy is converted to heat through internal friction.

dv

8.

kT m

Calculate the average rms and most probable speed and show that vrms > ravg > vp

11.

Show that the pressure exerted by an ideal gas is 1M 2 C P= 3V

What do you mean by law of equipartition of Short Answer Type Questions C  2 1. Explain on the basis of kinetic theory, that the energy ? Show that γ = p =  1 +  Cv  f pressure of gas increases with increase of temperature. where ‘f’ → degree of freedom. Hence calculate molar specific heat and values of γ for various gases.

10.

Show that the pressure exerted by an ideal gas is 2/ 3 of its kinetic energy per unit volume.

Derive the expression for the pressure of on ideal gas.

Does this apply in the case of inelastic collisions between molecules ? 8.

It does not matter whether the walls of the containing vessel of the gas are elastic or inelastic so long as the walls are at the same temperature as the gas. Explain.

H-82

13.

Absolute zero degree temperature is not the zero 13. energy temperature. Explain. 14. Equal masses of monoatomic and diatomic gases at 15. the same temperature are given equal quantities of 16. heat. Which gas will undergo a larger temperature rise and why ? 17. What do you mean by an ideal gas ? 18. Discuss the kinetic theory of gases. 19. Write the basic assumptions of kinetic theory of gas.

14.

Discuss the microscopic properties of an ideal gas.

15.

Discuss (i) Pressure (ii) Temperature of gas

16.

Study degree of freedom for different kind of gases.

17.

State the law of equipartition of energy.

18.

Calculate the value of γ for different kind of gases.

19.

Give one application of law of equipartition of energy.

20.

Derive an expression Cp 2 = 1 + where ‘f’ is degree of freedom. Cv f

9. 10.

11. 12.

21.

Draw the velocity distribution graph and explain it.

22.

Derive expressions for average speed, rms speed and most probable speed.

23.

Discuss Avogadro’s law.

25.

Which velocity among these v, rrms and vp is related with average kinetic energy.

What is the value of rms speed ? What is the value of most probable speed ? What is the order of average speed, most probable speed and rms speed. What is the value of pressure exert by the gas ? What is the value of Boltzmann’s constant ? Total kinetic energy associated with a molecules of a diatomic gas is given as

20.

How many numbers of degree of freedom are th ere in nitrogen gas ?

21.

A mosquito is flying in a room what will be the degree of freedom ?

Objective Type Questions Multiple Choice Questions 1.

The root mean square speed of gas molecules : (a) is the same for all gases at the same temperature (b) depends on the mass of the gas moelcules and its temperature (c) is independent of the density and pressure of the gas

Discuss the interpretation of temperature on the basis of kinetic theory.

24.

What is the value of average speed ?

(d) depends only on the temperature and volume of the gas 2.

The relation between volume V, pressure P and absolute temperature T of an ideal gas is PV = xT, where x is a constant. The value of x depends upon : (a) the mass of the gas molecule

Very Short Answer Type Questions

(b) the average kinetic energy of the gas molecules

1.

Write the gas equation.

2.

What is the formula of pressure exert by a gas ?

(d) the number of gas molecules in volume V

3.

What kind of of collision takes place between 3. molecules of an ideal gas ?

The equation of state for n moles of an ideal gas is PV = nRT where R is a constant. The SI unit of R is :

4.

Give two properties of molecules of an ideal gas.

(a) JK −1 per molecule

5.

Write two microscopic properties of gas.

(b) J kg −1 K −1

6.

What is the formula between γ and degree at freedom f ?

(c) JK −1 mol −1

7.

What is the degree of freedom of a monoatomic gas ?

8.

What is the degree of freedom of a diatomic gas ?

9.

What is the degree of freedom of a triatomic gas ?

10.

State the law of equipartition of energy.

11.

What is the value of γ for monotomic, diatomic and triatomic gas.

12.

What is the velocity distribution law ?

(c) P, V and T

(d) 4.

JK −1 g −1

Four molecules of a gas have speeds 1, 2, 3 and 4 kms −1 . The value of the root mean square speed of the gas molecules is : 1 1 (b) (a) 15 kms −1 10 kms −1 2 2 15 (c) 2.5 kms −1 (d) kms −1 2

H-83 5.

The average kinetic energy of a molecule of a gas at 12. absolute temperature T is proportional to : 1 (b) (a) T T (c) T

6.

The rootmean square speed of the molecules of a gas 13. at absolute temperature T is proportional to : 1 (b) (a) T T (c) T

7.

8.

9.

(d) T 2

(d) T 2

(b) v/ 4

(c) 4v

(d) 16v

The average kinetic energy of hydrogen molecules at 300 K is E. At same temperature, the average kinetic energy of oxygen molecules will be : E E (a) (b) (c) E (d) 4E 16 4 The average kinetic energy of a gas molecules at 27°C is 6.21 × 10−21 J. The average kinetic energy at 227°C

(a) Hydrogen

(b) Oxygen

will be :

(c) Nitrogen

(d) Carbon dioxide

(a) 9.35 × 10−21 J

(b) 10.35 × 10−21 J

(c) 11.35 × 10−21 J

(d) 12.35 × 10−21 J

If k is the Boltzmann’s constant, the average kinetic energy of a gas molecule at absolute temperature T is : 15. 3 kT kT (a) (b) 4 2 3 kT (c) kT (d) 2

By what percentage should the pressure of a given mass of a gas be increased so as to decrease its volume by 10% at a constant temperature ?

E o and E h respectively the average kinetic energy of 16. oxygen and hydrogen. If the two gases are at the same temperature, which of the following statements is true ?

A gas at a temperature 250 K is contained in a closed vessel. If the gas is heated through 1°C, the percentage increase in its pressure is :

(b) E o = E h

17.

(c) E o < E h (d) Nothing can be said about the magnitude of E o and E h as the information is insufficient Choose the correct statement from the following :

(a) the average kinetic energy of molecule of any 18. gas is the same at the same temperature (b) the average kinetic energy of a molecule of a gas is independent of its temperature (c) the average kinetic energy of 1 g of any gas is the same at the same temperature

11.

(a) v/16

The following four gases are at the temperature. In which gas do the molecules have the maximum root 14. mean square speed?

(a) E o > E h

10.

The mass of an oxygen molecule is about 16 times that of hydrogen molecule. At room temperature the rms speed of oxygen molecules is v. the rms speed of the hydrogen molecule at the same temperature will be :

(a) 81 . %

(b) 91 . %

(c) 101 . %

(d) 111 . %

(a) 0.4%

(b) 0.6%

(c) 0.8%

(d) 1%

A vessel containing 0.1 m 3 of air at 76 cm of Hg pressure is connected to an evacuated vessel of capacity of 0.09 m 3 . The resultant air pressure will be : (a) 20 cm of Hg

(b) 30 cm of Hg

(c) 40 cm of Hg

(d) 50 cm of Hg

A vessel is filled with a gas at a pressure of 10 atm and a temperature of 27°C. One half of the mass of the gas is removed from the vessel and the temperature of the remaining gas is increased to 87°C. At this temperature the pressure of the gas in the vessel will be :

(d) the average kinetic energy of 1 g of a gas is independent of its temperature

(a) 5 atm

(b) 6 atm

(c) 7 atm

(d) 8 atm

The root mean square speed of the molecules of a 19. enclosed gas is v. What will be the root mean square speed if the pressure is doubled, the temperature remaining the same ? v (a) (b) v 2

Figure shows a horizontal tube sealed at both ends and containing a pellet of mercury which occupies negligible volume. The region marked I contains a mass m of a gas and the region marked II contains a mass 2m of the same gas. If both region have the same temperature, what fractions of the volume of the tube will be occipied by the gas of mass 2m when the pellet is in equilibnium.

(c) 2v

(d) 4v

H-84 (a)

II

I

1 PV 2

(b) P V

(c) 2P V

Pellet of mercury

20.

23.

(a) 1 / 4

(b) 1 / 2

(c) 2 / 3

(d) 3 / 4

(a) brownian movement of colloidal particles

Figure shows the P-V curves for a certain mass of an ideal gas at two constant temperature T1 and T2 . Which one of the following inferences is correct?

P T1

(b) gas equation (c) tracks of particles in cloud chamber (d) none of the above 24.

The average energy of the molecules of monoatomic gas at temperature T is : 5 3 (a) kT (b) kT (c) kT (d) zero 2 2

25.

A honeybee is flying in a room, the number of degrees of freedom of motion of honeybee will be :

T2

V

(a) 1

(a) T1 = T2 (b) T1 > T2

1. (d) No inference can be drawn due to insufficient 2. information. 3. Figure shows a uniform tube open at one end and 4. closed at the other. A pellet of mercury is introduced 5. in it so as to enclose a column of air as shown. The length of air column is 18 cm at 27°C. Its length at 6. 7. 87°C will be : 8.

Pellet of mercury

22.

(a) 21.5 cm

(b) 21.6 cm

(c) 21.7 cm

(d) 21.8 cm

(c) 6

(d) 5

2PV B

2P.2V C

A PV

D

Molecules can move………in blank space. Shape of molecule is……… . The molecules collision is……… . Monoatomic gas has………degree of freedom. Diatomic gas has………degree of freedom. Triatomic gas has………degree of freedom. According to law of equipartition average energy is…...... . Specific heat for monatomic gas……… .

10.

Specific heat for diatomic gas……… .

11.

Specific heat for triatomic gas……… .

12.

The value of for monatomic gas is……… .

13.

For diatomic gas γ = ……… . For triatomic gas γ = ……… . The relation between v, vrms and vp ……… . Relation between γ and f is ……… .

17.

……… is the avg speed of molecules.

18.

……… is the vp speed of molecules.

19.

………is the vp of molecules.

20.

Kinetic energy is proportional to……… .

True/False P.2V

V

An ideal gas equation is ……… .

9.

An ideal monoatomic gas is taken around the cycle 14. ABCD as shown in the P-V diagram. The work done 15. during the cycle is given by : 16.

P

(b) 3

Fill in the Blank

(c) T1 < T2

21.

(d) 4P V

The first evidence in favour of the molecular structure of gas came from the experimental observation of :

1.

The pressure exerted by an enclosed ideal gas depends on the shape of the container.

H-85 2.

3.

The root mean square speeds of the molecules of 6. different ideal gases, maintained at the same temperature are the same. 7. The average kinetic energy of the molecules in one mole if all ideal gases, the same temperature is the same.

4.

The average kinetic energy of 1f of all ideal gases, at 8. the same temperature is same.

5.

The volume V versus temperature T graphs for a certain amount of a perfect gas at two pressures P1 and P2 are shown in figure. It follows from the graph that P1 > P2 .

Electrons in a conductor have no motion in the absence of potential difference across it. One mole of a monatomic ideal gas is mixed with one mole of a diatomic ideal gas. The molar specific heat of the mixture at constant volume is 2R, where R is the molar gas constant. The curves A and B in figure shows the PV graphs for an isothermal process and an adiabatic process for an ideal gas. The isthermal process is represented by the curve A

P

V P2 A

P1 B

V O

T

H-86

Objective Type Questions Multiple Choice Questions 1.

(b)

2.

(d)

3.

(c)

4.

(d)

5.

(c)

6.

(b)

7.

(a)

8.

(d)

9.

(b)

10.

(a)

11.

(b)

12.

(c)

13.

(c)

14.

(b)

15.

(d)

16.

(a)

17.

(c)

18.

(b)

19.

(c)

20.

(c)

21.

(b)

22.

(b)

23.

(a)

24.

(c)

25.

(b)

Fill in the Blank 1.

pv = nRT

2.

anywhere

3.

sphere

4.

elastic

6.

5

7.

6

8.

kT/2

9.

3/ 2 R

10. 5/2 R

5.

3

11.

3R

12. 1.66

13.

1.44

14.

1.33

15. v p > v avg > v p

16.

v = 1 + 2/ f

17. 1.59 kT /m

18.

1.73 kT / m

19.

1.44 kT /m

20. temperature

True/False 1.

False

2.

False

3.

True

6.

False

7.

True

8.

True

4.

False

5.

True

H-87

H ints and Solutions =

Numerical Questions 1.

Vrms = (V 2 ) mean

Let n be the number of molecules. From perfect gas equation, we have

=

PV = nkT V = 250 cm 3

Given,

= 250 × 10 m −23

K = 1.38 × 10

3

(iii) The most probable speed is the speed possessed by maximum number of particles, accordingly most probable speed = 3 m/s

J/K

T = 27° C = 300 K 3.

P = 10−3 mm of Hg

∴ Coxygen =

=

10−3 × 1.013 × 105 N/m 2 760

(Since at NTP T is same)

= n × 1.38 × 10− n=

∴

23

∴

× 300

10−3 × 1.013 × 105 × 250 × 10−6 1.38 × 10−23 × 300 × 760

4.

The average speed is given by Σni vi Vav = Σni

= =

Coxygen Chydrogen

or

2×1+ 4× 2+ 8× 3+ 6× 4+ 3× 5 2 + 4+ 8 + 6 3 2 + 8 + 24 + 24 + 15 73 = = 317 . m/s 23 23

(V 2 ) mean =

i

Σ ni i

2

=

2 × 1 + 4 × 22 + 8 × 32 + 6 × 42 + 3 × 52 2+ 4+ 8+ 6+ 3

=

2 + 16 + 72 + 96 + 75 23

 3RT  ,   M  hydrogen 

 Mhydrogen    = 2 =1  M   oxygen  32 4 Chydrogen 4

=

1.84 = 0.47 km/s. 4

According to kinetic interpretation of temperature,

∴

n1 + n2 + n3 + n4 + n5

Σ niVi2

=

Coxygen =

Given,

n1v1 + n2v2 + n3v3 + n4v4 + n5v5

(ii) The mean square speed is

 3RT    , Chydrogen = M   oxygen 

average kinetic energy per atom, 3 1 E K  = mV 2  = kT.  2  2

n = 8.05 × 1015

∴

=

 3RT     M 

10−3 atmospheres 760

10−3 × 1.013 × 105 × 250 × 10−6 760

(i)

We have C = Vrms =

=

Substituting these values, we get

2.

281 23

= 3.36 m/s −6

or

281 23

E i = 10.2 eV = 10.2 × 1.6 × 10−19 J 3 kT = 10.2 × 1.6 × 10−19 J 2 T=

2 10.2 × 1.6 × 10−19 × k 3

=

2 10.2 × 1.6 × 10−19 × 3 1.38 × 10−23

= 7.88 × 104 K 5.

We have from kinetic interpretation of temperature 1 3 2 …(1) MVrms = kT 2 2 Escape Velocity V0 = (2gR), R being radius of earth. Given,

Vrms = V0 (2gR)

∴ From equation (1), we have

H-88 1 m 2gR = 3kT 2 2 mgR This gives T= 3 k

9.

The heat energy is completely used up in increasing the kinetic energy of the molecules of the gas if it is not allowed to expand. Thus, if the volume is kept constant the gain in kinetic energy for 1 K rise in temperature is the same as the amount of heat energy gained by the gas for 1 K rise in temperature. Thus, from the definition of Cv, we have, 3 3 3R Cv = R(T + 1) − RT = 2 2 2

For hydrogen 2 mhydrogen gR Thydrogen = k 3   2  × 10−3  × 9.8 × 6.4 × 106 23  2  6.02 × 10 . = × 3 1.38 × 10− 23 oxygen = 1.0 × 104 K

For

2 moxygen gR 3 k 32  × 1023 × 10−3  × 9.8 × 6.4 × 106   2  = × 6.02 3 1.38 × 10−23

Now

oxygen =

∴

P = 1 atm = 076 . m of Hg = 076 . × 13600 × 9.8 = 1.01 × 105 Nm −2 P = 9 × 10−2 Kg m −1

Crms = 7.

3 × 1.01 × 105 9 × 10−2

10.

11.

= 1840 m/s

R T k C =3 =3 T Nm M 3kT1 C12 = m kT2 3 C22 = m

Here,

C22

=

T1 T2 C22 C12

= 373 × 4

= 1492 K = 1219°C 8.

13.

T = 300 K −23

k = 1.38 × 10

JK

−1

The average kinetic energy of gas molecule is given 3 by E = kT 2 3 = × 1.38 × 10−23 × 300 = 6.21 × 10− 21 J 2

See example-1

See example-2

See example-3 Ans. 5.65 × 10−21 Joule

14.

See example-4 Ans. 2.69 × 1025 molecules

15.

See example-5 Ans. 833 K

16.

See example-6 Ans. 570.75 m/s

17.

See example-9 Ans. 2.5 m/s n! Ω= n1 ! n2 !

C2 = 2C1 , T1 = 273 + 100 = 373 K T2 = T1 ×

See example-12,

Ans. 6.02 × 1023

2

C12

3 × 8.32 = 12.48 J K −1 mol −1 2

Ans. 3.2 × 1023 12.

We know that

∵

Cv =

Ans. 4.856 × 104 cm/sec, 1.84 × 105 cm/sec

The root mean square speed is given by 3p = ρ

R = 8.32 J K −1 mol −1 ,

Since Cp = Cv + R we have 3 5 Cp = R + R = R 2 2 5 = × 8.32 = 20.80 J K −1 mol −1 2

= 16 × 104 K 6.

The kinetic energy of 1 mole of a gas is given by 3 U = RT 2

18.

For (4, 1) state n = 5, n1 = 4, n2 = 1 5× 4× 3× 2×1 5! =5 Ω= = 4! 1! 4 × 3 × 2 × 1 × 1 For (3, 2) state n = 5 n1 = 3, n2 = 2 5× 4× 3× 2×1 5! = 10 Ω= = 3! 2! 3 × 2 × 1 × 2 × 1

H-89 19.

of upper atmosphere is much lower, therefore the number of molecules per unit volume in the upper atmosphere is much smaller than on the earth. Consequently the total kinetic energy of translation or total heat content per unit volume in upper atmosphere is much smaller than that on earth. Hence we feel quite cold there.

See example-15 Ans. 1841 m/sec, 1503 m/sec

20.

See example-16 Ans. − 250.6 ° C

21.

See example-17 Ans. TQ = T2 = 2495 °C

22.

5.

According to kinetic interpretation temperature, the average kinetic energy per molecule is proportional to absolute temperature. But the temperature of a gas is determined by the total translational kinetic energy measured with respect to the centre of mass of the gas. Therefore the motion of centre of mass of the gas does not affect the temperature. Hence in this case there will be no increase or decrease of gas temperature.

6.

H 2 and He are much lighter than other gases like N 2 , O 2 etc. found in air and so have larger root mean square speeds. Hence they leak faster.

7.

The inelastic collisions of molecules of a substance do not involve the loss of mechanical energy to heat. The reason is that the mechanical energy of molecules is itself heat. Therefore any conversion of mechanical energy into heat is simply the conversion of mechanical energy into mechanical energy.

8. The pressure of a gas is the rate of change of momentum of molecules striking the walls of container per unit area, when the temperature increases, the root mean square speed of gas molecules also increases. This causes an increase in change in momentum of the molecules as well as the frequency of collisions. Consequently the increase in 9. temperature results in an increase of rate of change of momentum of striking molecules per unit area ans hence and increase in pressure.

If the gas molecules make inelastic collisions with the walls of the containing vessel, some kinetic energy may be lost as heat. But if the walls are in thermal equilibrium with the gas, this heat will be gained by molecules in the form of kinetic energy. Hence inelastic collisions do not matter so long as the walls are at the same temperature as the gas.

See example-18 Ans. 2.688 × 1025

23.

See example-19 Ans. 133.4 J

24.

Similar to example-18 Ans. 5.376 × 1022

25.

Similar to example-19 Ans. 3739 . × 103 J

26.

27.

Similar to example-19 Ans. 1245 J 2 × 3RT 3RT 3 × 8.3 × 107 × 200 = = 5M 10 M 5× 4 = 3984 J

Short Answers Type Questions 1.

2.

According to Avogadro hypothesis equal volumes of different gases contain equal number of molecules at 10. given temperature and pressure. Hence both samples have equal number of molecules.

3.

As root mean square speed of molecules is inversely proportional to square root of molecular weight Vmbs ∝ 1/ M . Accordingly lighter gas has greater rms speed and hence escapes sooner.

4.

The gas kinetic temperature in the upper atmosphere 11. is of the order of 1000 K and as such that the average 12. 3 kinetic energy of translation per molecule  kT  is 13.  2  higher as compared to that on earth. But the density 14.

According to kinetic theory of gases, the average translational kinetic energy per molecules is zero at absolute temperature; however the molecules, may passess potential energy; hence the absolute zero temperature is not zero energy temperature. As average translational kinetic energy per molecule α absolute temperature; and the monoatomic gas molecules posses any translational kinetic energy while diatomic gas molecules may posses rotational and vibrational energies in addition to translational kinetic energy; therefore the monotomic gas will undergo a larger temperature rise. See 1 See 2 See 2.1 See 3.2

H-90 15.

See 3.1 and 3.2

8.

f =5

16.

See 4

9.

17.

See 5

10.

18.

See 6

f =6 kT E= 2

19.

See 6

20.

See 7

21.

See 9

22.

See 10

23.

See 12

24.

See Example-7

25.

See Example- 14

11.

γ = 1.66, γ = 1.4, γ = 1.33

12.

m  n(v) dv = 4 π N    2 πkT 

13.

Vavg = 1.59

kT m

14.

Qrms = 173 .

kT m

15.

Vp = 1.59

PV = nRT

16.

where n → number of molecules 1 mn 2 C P= 3 v

17.

Vp < V < Vrms 2 P = × kinetic energy of per unit volume 3 R k= N

Very Short Answer Type Questions 1.

2. 3.

Elastic collision

4.

(i) molecules are rigid smooth and spherical (ii) attraction force between gas molecules is zero

5. 6. 7.

(i) Temperature (ii) Pressure C 2 V = p =1+ Cv f f =3

18.

3 /2

v2 e

−

mv2 2 kT

dv

kT m

where k = 1.38 × 10−23 J/mole-K 19.

5 kT 2

20.

5

21.

3 mmm

H-91

Unit

3

Thermal Radiation

1. Thermal Radiation The heat comes to the earth from the sun is found by the radiation, hence radiation is the method of heat transfer in which there is no need of any medium for transition of heat in which every substance emits the heat continuously from its thermal property and it is known as thermal radiation or radiation heat. The radiation heat from any substance in per second is depend upon the nature of the surface of substance and also depend on its temprature. At low temprature the rate of emission of radiation heat is low, at the increase of temperature the rate of emission increases rapidly. In emitted radiation heat there are mixture of different wavelength in which at normal temperature high wavelength and at high temperature short wavelength is emit. Thermal radiation is moving in the form of electromagnetic wave with the speed of light and they transmit one place to another place without any change in medium. It can be also transmitted into vacuum or medium. When thermal radiation is incident at such substance which can not transmitted it then some part of radiation is absorbed by substance which are of in the form of heat and rest part of the radiation is reflected by the radiation.

1.1 Properties of Thermal Radiation Main properties of thermal radiation is given as follows : 1.

Thermal radiation is moving in the straight line same as light travels from sun to earth.

2.

They can be transmitted in vacuum.

3.

The wavelength of thermal radiation is greater than the wavelength of light and rest properties are similar to light.

4.

Thermal radiation follows the inverse square law.

5.

When thermal radiation is incident on a surface then it produce pressure on the surface.

6.

Reflection and refraction occurs in thermal radiation and thermal radiation can be polarised same as light.

7.

The spectrum of radiation can be found with the help of prism.

H-92

2. Physical Quantities Associated with Radiation 1.

Energy density (u): Energy density is equal to the heat on any point of its per unit volume. It is expressed by u and it has unit joule/meter.

2.

Spectrumic energy density (u λ ) : The quantity of any fixed wavelength present in per unit volume of any substance is known as spectrumic energy density, it is denoted by u λ . Relation between spectrum energy density and energy density is given by ∞

u = ∫ u λ ⋅ dλ 0

Where d λ is wavelength range 3.

Emissive power or emissivity : Emitting radiation by a unit area of a surface in per second is known as emissive power of that surface. It is expressed by E and having unit joule/meter.

4.

Spectrumic emissive power ( E λ ) : For a particular wavelength spectrumic emissiue power of a surface is equal to such radiation energy which is emitted by per unit area of surface for unit energy range. It is denoted by E λ . The relation between emissive power and spectrum emissive power is given as follows : ∞

E=

∫ Eλ ⋅ dλ 0

5.

Absorption power : The ratio between absorbed radiation energy by any surface in per second and incident radiation energy from the same substance in per second is known as absorptive power. It is denoted by ‘a’ and it has no unit.

6.

Spectrumic absorptive power ( a λ ) : In unit wavelength range for any wavelength, the ratio between absorbed energy by any substance in per second and incident energy from same substance is called spectrum absorptive power. It is denoted by a λ .

3. Black Body When radiation energy is incident on the surface of any body then some part of energy is absorbed by substance and some is reflected by substance and rest part of energy is transmitted by substance. Besides it the substance which absorb all wavelength of incident radiation energy is called black body. Black body is not effected by the angle and frequency of incident radiation.

3.1 Ideal Black Body Substances which absorb all radiation incident on its surface is called ideal black body. It is not effected by the volume angle and wavelength of incident radiation. For a ideal black body the value of spectrumic absorption power a λ is equal to one for every value of wavelength. Its perfect emitter as well as perfect absorber. Lamp black is nearly equal to an ideal black body. Black body is only a imagination. In behaviour any substance is not ideal black body. It is developed only far practicals. Ferry and Wien has developed ideal black body and its structure is given as :

H-93

4. Practical Black Bodies (Non-perfect Black Bodies) 4.1 Ferry’s Black Body A body is developed by Ferry which has the properties same as an ideal black body. It is a hollow copper ball with double wall, in which the inside wall is polished while the outside wall is polished by nikel. Vacuum is present between both the walls.

Incident Radiation

P

Emitted Radiation

H

Fig. 1 : Ferry’s black body

When a radiating heat is incident on hole H then it becomes absorbed after many successive reflectins. In the presence of cone shape P bump some radiation can come outside by reflection but by the presence of it this condition is not possible. So, hole H is behave like an ideal black body. For using this sphere as a radiation source we put it in fixed temperature in a heater. By this process, heat radiation come outside from the hole which is nearly equal to a black body radiation.

4.2 Wein’s Black Body In Wein’s black body there is a cylindrical hollow chamber platinum which is black inside and it is totally covered by thermal loop of platinum wire in which the temperature of chamber is increased by flowing the electric current in the wire. It is measured by thermoelectric couple pipe of porcelain are placed outside the chamber. When the temperature rises of the chamber, emission of radiation take place from it which is passes through diaphragms and come outside from the hole H. This radiation is called black body radiation.

4.2.1 Thermodynamics of Radiation Inside a Hollow Enclosure The nature of radiation is that it depends upon temperature as well as on the nature of surface. If we consider a non conducting enclosure maintained at a constant temperature i. e. its called isothermal enclosure, the properties (Quality and quantity) of radiation inside such kind of enclosure depends upon temperature only does not depends upon nature of substance which may be inside it. We can prove this in following way : Let two hollow enclosures A and B of different materials. If we maintain both A and B at the same temperature T. Let each one is filled with radiation energy. Let the radiation energy density (Radiation per unit volume) depends W on temperature and on nature of walls of enclosure too.

H-94 Let the density of radiation energy of A is greater than B. Let both enclosures A and B are joined together with a tube fitted with a screen S transparent to all radiations. Now, the radiation energy of A is greater than B. So more radiation will fall on screen from A in compare to B. Thus heat will flow from A to B. After sometime make both enclosures seprated and allowed them to attain equilibrium. Now in such case energy density of A will be less than before and hence its walls will loose the heat into A enclosures and cool down till equilibrium is again attained. Now, the energy density in B is greater than before, therefore its walls will absorb energy from within and get heated. Hence, there will be a temperature difference between A and B.

S

B Fig.2

Now, we can assume B as source (hot body of a heat engine) and A as sink (cold body of heat engine). If we obtain the work without expenditure of any energy i. e. without giving to sink (cold body). This is violent of thermodynamics second law. Hence, our assumption that energy density of A is greater than in B i. e. u A > u B is incorrect. Thus the energy densities in A and B must be identical. Hence, the energy density of the radiation in a uniform temperature enclosure depends on its temperature and is entirely independent of the nature of the walls. Let we have three bodies A, B and C are placed inside on isothermal enclosure. The body will attain the temperature of the enclosure irrespective of its nature of original temperature. As the bodies A, B and C are at the same temperature as that of enclosure, each body must radiate energy at the same rate and exactly of same kind as it absorbs. Hence, the bodies contained inside the enclosure have no effect on qantity and quality of the radiant energy filled in the enclosure.

5. Kirchoff’s Law For a particular wavelength the ratio of emissive power to absorption power is same for all object at same temperature and this is equal to the emissive power to black body at that temperature eλ = Eλ i. e. aλ Where e λ → Spectrumic emissive power of body a λ → Spectrumic absorption power of body E λ → Spectrumic emissive power of black body Let a body whose spectrumic absorption power is a λ . The radiation incident on the body from wavelength λ to dλ whose total heat is dQ. Hence, the total quantity of the radiation falling on the body = a λ dQ Let the spectrumic emissive power of body = e λ . Hence, the quantity of radiation comes out through the body = e λ d λ

H-95 The radiation falls down on the body is equal to the same amount of radiation comes out from body. i. e.

…(1)

a λ dQ = e λ d λ

If perfect black body is kept in place of that body whose spectrumic emissive power is E λ . whose spectrumic emissive power is E λ . So, equation number (1) will be (2)

1dQ = E λ d λ Where, a λ = 1 (absorption power of a perfect black body is 1) E λ = Spectrumic emissive power of black body Dividing equation (1) by (2) We get,

aλ =

eλ Eλ

or

Eλ =

eλ = constant aλ

This is Kirchoff’s law.

6. Stefan-Boltzmann Law In 1879 Stefan gave a law for rate of emitted heat from a hot substance is called Stefan’s law. According to this law, emitted radiation energy by a unit surface area from an ideal black body in per second is proportional to the fourth power of absolute temprature. That is

Piston

E ∝T4 E = σT 4

Where, E is emitted radiation energy, T is the absolute temperature of black body and σ is Stefan constant. Stefan law is confirmed from the Boltzmann law, so it is also called Stefan-Boltzmann law. If the substance is not an ideal black body E = σeT 4

u, v T Cylinder Fig. 3

Where, E is emission power of substance. It has value from 0 to 1 or according to the nature of surface and for ideal black body it is 1. This law is not applicable for the nature of cooling of substance because the rate of coolness also depend upon the surrounding temperature of the substance. If at temperature T an ideal black body is placed in a chamber of temperature T0 then σT 4 radiation energy at per unit area in per second absorb σT04 radiation energy. So, total emission radiation energy for per unit area of a black body will be, or

E = σT 4 − σT04

or

E = σ (T 4 − T04 )

H-96 Thermodynamic proof of Stefan’s law : Let a cylinder is filled with u energy density at diffused radiation at absolute temperature T and u is depend only on T . The walls of cylinder and piston is totally reflected by which there is no transition of heat because radiation acts as a gas. So, we can say that this cylinder is a thermodynamic. If v is the volume of radiation in cylinder then total internal energy (U) of radiation, …(1)

U =u ×v

Let, a small quantity of heat dQ flows in the cylinder from outside, by this expansion of radiation will occur then the volume of radiation is increase from v to v + dv and piston is lifted upside. Because of this, if change in internal energy is dv, change in entropy dS and work done dW for volume expansion then from law of thermodynamics, dQ = dU + dW = dU + PdV But

dQ = TdS TdS = dU + PdV 1 dS = (dU + PdV) T

or

…(2)

By the Maxwell’s second law of thermodynamics  ∂S   ∂P    =   ∂V  T  ∂T  v

…(3)

Putting the value of dS in equation (2) 1  ∂U + P∂V   ∂P   =    ∂T  v   T ∂V T or

 ∂P   ∂U   + P =T    ∂T  v  ∂V  T

…(4)

By the Maxwell’s electromagnetic principle of radiation, the pressure of radiation P is one third of energy density u. u So, …(5) P= 3 u T  ∂U   + =   ∂v  T 3 3

 ∂u     ∂T  T

Putting the value of v in equation (1) u T  ∂u   ∂ (u × V)  + =     ∂v  T 3 3  ∂T  v Because T is the function of u only and it is not depend on v So, or

u+

u T  ∂u  =   3 3  ∂T 

4u T  ∂ u  =   3 3  ∂T 

[‡ U = u ⋅ V]

H-97  ∂u  4u = T    ∂T  ∂u ∂T =4 u T On integration, log e u = 4 log e T + log e a Where a is integral constant log e 4 = log e aT 4

or,

u = aT 4

…(6)

But emissive power for diffused radiation is given as

E=

1 uc 4

E=

1 ac aT 4 c T 4 4 4

(where c is speed of light)…(7)

From equation (6) and equation (7)

E = σT 4 Where σ =

…(8)

ac Stefan’s constant and equation (5) represent the Boltzmann law. 4

7. Spectrum Formula–Early Attempts The emissive radiation energy distribution in different wavelength by a black body at different temperature is studied by Lummer and Pringsheim is 1819. They represent an experiment for it. S1 Flurospar prism

M1

M2

P B G

S2 Bolometer

Fig. 4 : Lummer and Pringsheim’s experiment

In this a chamber which is heated up by electric energy, consist of a small gate used as a black body. The emissive radiation of black body passes through slit S1, and it is incident on a reflected mirror M 1, is incident on fluorspar prism or rock slit. This arrangement is placed on a rotating table of spectrometer. After refraction from prism P the parallel rays beam is focused on bolometer B by the second mirror M 2. A comfortable current meter is connected with bolometer M which the intensity E λ of produced radiation is measured. Now rotate slowly prism table

H-98 and radiation is made to full on it (on different part of bolometer). Hence, read the deflection in current meter to measure spectral intensity. Hence, we can measure different intensities and emission power of back body E λ . If we plott the graph between E λ and λ then we find a curve, which is called spectral distribution curve. The experiment is repeated for very high temperature. Many points is found by the analysis of this curve : 1.

2.

3.

The distribution of energy is not equal in the spectrum and the distribution of radiation energy is continuous in any wavelength range. Initially the value of E λ increases with the λ for a given temperature and the value of E λ is maximum for a particular value of λ and after it the value of E λ decreases while the value of λ increases. When we increase the value of temperature the value of wavelength λ m correspond to the maximum energy. It is called Wein’s displacement law and according to this relation between λ m and T is given by, λ mT = constant

13 12 11 10 9 8 7 Eλ 6 5 4 3 2 1

1646 K 1449 K 1259 K 1095 K 904 K 1

2

5 3 4 λ (in micron) Fig. 5

6

7

4.

At constant temperature, horizontal axis X and middle area of curve shows the total radiation energy and this area is proportional to the fourth power of absolute temperature that is E ∝ T 4 . This is the Stefan’s law.

5.

When increasement of temprature the value of E λ increases rapidly for all values of wavelengths and it is given by relation E λT −5 = constant

8. Wein’s Distribution Law The law of physics which explained the spectrum of thermal radiation is known as Wein’s distribution law. It is also called Wein’s law or black body function. This law is derived by Willem Wein in 1986. Its equation is able to explained correctly the thermal emission of any substance of short wavelength (high frequency) but unable to explained correctly the emission of high wavelength (low frequency). This law is derived by Wein before the Planck’s radiation law. According to this law E(u, T ) =

2hν 3 c2

e − hν / kT

…(1)

In terms of wavelength I (λ , T ) =

2hc 2 λ5

e − hc / λkT

Where, E is the emission energy of per unit area in per unit time, T is the absolute temperature of black body and λ and k are Planck and Boltzmann constant, respectively.

H-99

9. Rayleigh–Jeans Law Rayleigh and Jeans gives a formula in 1900 for explanation of distribution of energy in spectrum of black body, which is known as Rayleigh-Jeans law. According to this law radiation present in the form of electromagnetic wave for zero to infinite wavelength in a black body chamber, reflected from the wall of black body again and again. Rayleigh and Jeans gives a formula for distribution of energy in the spectrum of radiation of a black body which is given as, 8πkT …(1) E λ dλ = dλ λ4 Where k is Boltzmann constant and T is absolute temprature. It is able to explained the energy distribution only in high wavelength region of black body spectrum and it is unable to explained the energy distribution in short wavelength region Now, written the equation (1) in terms of frequency c −c and dλ = 2 dν λ= ν ν So,

E ν dν =

−8πkTν 2 c3

dν

…(2)

10. Wein’s Displacement Law Wein studied for the distribution of energy for black body radiation in different wavelength and stated that ‘‘The wavelength correspond to maximum energy is inversely proportional to the absolute temprature of black body.’’ 1 …(1) That is λ max ∝ T b …(2) or λ max = T Where b is Wein’s constant and its values is 2.897 × 10−3 meter Kelvin. From equation (2) λ maxT = b It is called the Wein’s displacement law. According to Wein spectrum emissive power E λ is proportional to the fifth power of absolute temprature T . That is, or

Eλ ∝ T 5 E λT −5 = constant

From equation (1) and equation (4) E λ λ 5 = constant or Where f (λT ) is the function of λT .

E λ λ 5 = Af (λT )

…(3) …(4)

H-100 or

E λ dλ =

A λ5

…(5)

F (λT )

Equation (5) shows the Wein’s radiation law. On the basis of experimental result Wein’s distribution law correct only for short wavelength.

10.1 Quantum Theory of Radiation In 1900 Sir Max Planck introdue a new idea to explain the distribution of energy among the various wave lengths of the cavity radiation. He assumed that the walls of cavity radiator behaves as an oscillator and each oscillator oscillates with a frequency. These oscillator emits electromagnetic radiation energy into the cavity and also absorbs the same from it and maintain an equilibrium state. Sir Max Planck gave two assumptions : 1.

An oscillator can have only discrete energy Quantum energy, E = nhν where n → An integer h → Planck’s constant ν → Frequency of vibrations Hence, energy can have only hν, 2hν .... . Hence, the energy of oscillator is quantised.

2.

The oscillator do not emits or absorbs energy continuously but only in jumps. According to this, oscillator can emits or absorbs packets of energy not in continuous form.

11. Average Energy of Planck’s Oscillator and Formula If the number of Planck’s oscillator is N and the total energy of oscillation E in a black body chamber then average energy of Planck oscillations, E …(1) E= N According to Boltzmann, the number of oscillation of energy E is proportional to the e − hν/ kT , where T is absolute temprature and k is Boltzmann constant. Now, if the number of oscillations in a chamber is N 0, N1, N 2, ..., N n and 0, hν, 2hν, ..., nhν is the energy of oscillations respectively then the number of oscillation of nhν N n = N 0e − nhν/ kT

…(2)

Now, total energy of oscillations E = N 0(0 ) + N1(hν) + N 2(2hν) + N 3(3hν) + ... = [( N 0e − hν / kT ⋅ hν) + ( N 0e −2hν / kT ⋅ 2hν) + ( N 0e −3hν / kT ⋅ 3hν) + ... By using equation (2) = N 0e − hν / kT ⋅ hν [1 + 2e − hν / kT + 3e −2hν / kT + ...] or,

E=

N 0e − hν / kThν (1 − e − hν / kT )

 1  2 …(3) ‡ 1 + 2x + 3x + ... = 1 ( x )2  − 

H-101 Total number of oscillations N = N 0 + N1 + N 2 + ... = N 0 + N 0e − hν / kT + N 0e − hν / kT + ... = N 0(1 + e − hν / kT + 2e −2hν / kT + ...) or

N=

N0

…(4)

(1 − e − hν / kT )

Using equation (3) and equation (4) in equation (1), average energy of oscillations E=

= or

E=

N e − hν / kT ⋅ hν /(1 − e − hν / kT )2 E = 0 N0 N (1 − e − hν / kT ) e − hν / kT ⋅ hν 1 − e − hν / kT hν

…(5)

(e hν / kT − 1)

This is the derivation for average energy of Planck’s oscillations. In terms of wavelength, ν= E=

c λ hc λ (e hν / kT − 1)

Special Case : 1.

If the frequency is low and temprature is high then the value of hν/ kT is very small. e hν / kT = 1 +

So, Neglecting higher order of

2

1  hν  hν +   + ... kT 2 !  kT 

hν kT hν kT hν = kT E= hν 1+ −1 kT

E hν / kT = 1 + Now,

E = kT 2.

It the frequency is high and the temperature is low then the value of term So,

e hν / kT = 1 = e hν / kT E=

hν e

hν / kT

hv is very high. kT

H-102

12. Planck’s Radiation Formula Unit volume energy for frequency range ν and ν + dν or energy density volume is equal to the product of number of oscillations and average energy of oscillations. No. of oscillations

Nν d ν =

8πν 2 c3

…(1)

⋅ dν

The formula of average energy per oscillation E=

hν

…(2)

e hν/ kT − 1

Per unit volume energy or energy density of frequency range ν and ν + dν is equal to the product of number of oscillations and average energy of oscillations i. e., E ν dν = N ν dν × E = or,

E ν dν =

8πhc 3

×

c3

8πν 2 c3

⋅

hν e hν / kT − 1

⋅ dν

dν e hν / kT − 1

…(3)

Equation (3) is called Planck’s formula for radiation is the term of frequency. Now, in the term of wavelength it is given as, c −c ν = and dν = 2 λ and because wavelength is decreases as frequency increases. So, λ λ E ν dν = − E λ dν Putting all these values in equation (3) 3

We get,

E λ dλ = E λ dλ =

− 8π h  c   − c  dλ  ⋅  2  hν / kT 3      λ −1 λ e c 8π h ⋅ c λ5

⋅

dλ e hν / kT − 1

…(4)

Equation (4) is known as Planck radiation is the term of wavelength.

12.1 Deduction of Wein’s law and Rayleigh-Jeans law from Planck Law Wein’s law and Rayleigh-Jeans law both law are the special state of Planck’s law. Planck’s radiation formula in the term of wavelength is given as 8πhc dλ E λ dλ = 5 hc / kT λ e −1 Special case : 1.

For short wavelength, the value of e hc / kT is greater than 1. So taking 1 as negligible e hc / kT − 1 = e hc / kT From equation (1) Let

E λ dλ =

8πhc λ5

⋅

dλ

e hc / kT hc 8πhc = A and =B k

…(1)

H-103 then,

2.

E λ dλ =

A λ5

e − B / λT d λ

…(2)

This is the law of Wein. So, at short wavelength Planck’s law is equivalent to Wein’s law. hc For high wavelength the value of is high λkT 2

hc 1  hc  +  + ...  λkT 2 !  λkT 

e hc / λkT = 1 +

So,

hc hc (neglecting higher order term of ) λkT λkT hc e hc / λkT − 1 = λkT 8πhc dλ E λ dλ = λ 5 hc / λkT =1+

or From equation (1) or

8πkT

E λ dλ =

λ4

…(3)

dλ

This is the Rayleigh- Jeans law. So for high wavelength Planck’s law is equivalent to Rayleigh-Jeans law.

13. Deduction of Stefan’s Law From Planck’s Law Emission law of Stefan-Boltzman can be derived by Planck’s radiation formula. According to the Planck radiation formula E λ dλ =

8πhc λ

5

1 (e

hc / kT

…(1)

− 1)

On integrating equation (1) from 0 to ∞ value of λ, we get total energy density ∞

E=

∫

0

Let

hc = x then, λkT

λ=

8πhc λ

⋅

5

hc xkT

dλ (e

hc / kT

or dλ =

…(2)

− 1) −hc x 2kT

⋅ dx

From equation (2) ∞

1 hc ⋅ 2 ⋅ dx 5 x ( / ) ( ) ( 1 hc xkT ⋅ e − x kT ) 0

E = − 8πhc ∫ =

8πk 4T 4 h 3c 3 5 4

or

E=

This is total energy density.  ∞ x5 x 5 dx =  ‡ ∫ x 15   0 e − 1

8π k T 15h 3c 3

∞

x5

∫ e x − 1 ⋅ dx 0

4

…(3)

H-104 For total emissive power of black body ET = or

ET =

Ec 8π 5k 4T 4 ⋅ c = 4 4 × 15 × h 3c 3 2π 5k 4 3 2

15 × h c

T4

or

ET ∝ T 4

This is the Stefan-Boltzmann law.

14. Radiation of Photon Gas Let a hollow chamber (black painted from inside) is in the form of a black body. Let the temperature of this black body is T . Let it has a hole the radiation which comes out from this hole is called black body radiation. The temperature of radiation inside the chamber depends upon the nature of walls. These thermal radiations are electromagnetic and depends upon temperature. These radiations are made of photons every photon has energy hν. Where h is Planck’s constant, ν is the frequency of radiation. Photons are also known as Bosons, hence the radiation which is coming out from hole is also known as ‘‘photon gas’’.

hole Incident radiation

Fig.6 : Black body chamber

Number of photons in a chamber are not finite because in each emission new photons will be formed. If a photon of frequency ν is absorbs then photons of frequency ν1, ν 2, ν 3 ..... will be formed. Hence,

hν = hν1 + hν 2 + .......

Example 1. A black body whose surface area is 5 × 10 −5 m 2 and temperature is 727° C. Find the value of emissive heat per minute from it. Stefan constt. σ = 5.67 × 10 −8 J/m 2 s K 4 . Solution : At absolute temperature T , according to Stefan emissive heat per sec from unit area of a black body is given as, E = σT 4

H-105 Given,

T = 727 °C = 1000 K

and emissive heat per minute from black body of area A, E = σT 4 × A × 60 J/m = 5.67 × 10−8 × 5 × 10−5 × (1000)4 × 60 J/m = 170 .1 J/m or

E = 40.5 cal/m

Example 2. It two body x and y are placed in a vacuum vessel whose temperature are 427°C and 227°C respectively and the temprature of vessel is 27°C, then compare the heat loss from both bodies. Solution : Let the rate per second energy loss from unit area of bodies x and y is E1 and E 2 respectively, then from the Stefan’s law, E = σ(T 4 − T04 ) Given,

T0 = 27 °C = 300 K

For body x,

T = 427 °C = 700 K

For body y,

T = 227 °C = 500 K

Heat loss comparison from both bodies E1 σ(T 4 − T04 )x = E 2 σ(T 4 − T04 )y or

E1 (700)4 − (300)4 = E 2 (500)4 − (300)4 E1 = 4 .27 E2

Example 3. If the luminosity of Rizel star is 17000 times of Sun and if temperature of sun’s surface is 6000 K then calculate the temperature of star. Solution : If the luminosity of sun and star E1 and E 2 respectively and absolute temprature is T1 and T2. Consider it as a black body then according to Stefan’s law, E1 T14 = E 2 T24 E2 4 T1 E1

or

T24 =

Given,

E2 = 17000, T1 = 6000 K E1 T24 = 17000 × (6000)4 T2 = (17000)1/ 4 × 6000 T2 = 68512 K

H-106 Example 4. Find the temprature at which a black body will emit heat at the rate of 1 watt/m 2 , given σ = 5 .6 × 10 −5 square/cm 2 sec−1 × K −4 . Solution : According to Stefan’s law rate of emission of heat at unit area in per sec E = σT 4  E T =  σ

1/ 4

Given, E = 1 watt/cm 2 = 1 J/cm × cm 2 = 102 erg/cm × cm 2 and σ = 5 .6 × 10−5 erg/cm 2 × cm −1 × K −4   107 T =  −5  5 .6 × 10  T = 650 K Example 5. If two wavelengths of energy spectrum radiated by moon are 14 × 10 −6 m and 5 × 10 −7 m, then calculate the temperature of each correspondance wavelength. Given that b = 0.3 × 10 −2 mK. Solution : From Wein’s displacement law λ mT = b or

λm =

b 0 .3 × 10−2 = T 3000

λ m = 10−6 m Example 6. According to Wein’s displacement law, λ mT =

hc = b, if h = 6 .63 × 10 −34 J 5k

sec, c = 3 × 10 8 m/sec, k = 1 .38 × 10 −23 J/K then, (a) Calculate the value of constant b. (b) Solar radiation λ m = 4753 A°, calculate the temperature corresponded to it. Solution : (i) b =

hc 6.63 × 10−34 × 3 × 108 = = 0.29 × 10−2 mK −23 5k 5 × 1.38 × 10

b = 0.29 × 10−2 mK

or (ii)

λ mT = b or

T =

b 0.29 × 10−2 = 6101 K = λ m 4753 × 10−10

Example 7. Calculate the energy radiated per minute from the filament of an incandescent lamp at 2000 K if the surface area is 5 × 10 −5 meter 2 and its relative emittance is 0.85. Solution : According to Stefan-Boltzmann’s law, The energy radiated by a body per unit area per second placed in an enclosure is given by E = σ[T 4 − T04 ] = σT 4 (if T >> T0) where T is the temperature of the body and T0 is the temperature of the enclosure.

H-107 If the body has on emissivity, then E = eT 4 The total energy radiated by the body per second = σe(T 4 − T04 ) × surface area = 57 . × 10−8 × 0.85 × (2000)4 × (5 × 10−5) The energy radiated by the body per minute = 57 . × 10−8 × 0.85 × (2000)4 × (5 × 10−5) × 60 = 2325 J Example 8. A body of mass 10 gram is kept in an enclosure of temperature 27°C. If the temperature of the body is 127°C, its specific heat is 0.1 kilo calorie per kg per degree C and area of emitting surface of the body is 10 −3 m 2 , find out the rate of cooling of the body. (σ = 5.72 × 10 −8 Jule m −2 s −1 (°C) −4 ) Solution : We have, Q = Aσ(T 4 − T04 )t dQ = Aσ(T 4 − T04 ) dt But,

Q = msT d(msT ) = Aσ(T 4 − T04 ) dt

So,

Rate of cooling of the body, dT Aσ = ⋅ (T 4 − T04 ) dt ms

= =

 5.72 × 10−8  (10−2 m 2)  kilo cal. m −2s−1(° C)−4  (4004 − 3004 )(0° C)4 3  4.2 × 10  (10 × 10−3 kg){0.1 kilo cal. kg −1(0° C)−1} 10−3 × 572 . × 10−8 × (4004 − 3004 ) 10 × 10−3 × 01 . × 4.2 × 10−3

° C/s

= 0.23°Cs −1 = 0.23 Ks −1 Example 9. A black body with an initial temperature of 300°C is allowed to cool inside an evacuated enclosure surrounded by melting ice at the rate of 0.35°C per second. If the mass, specific heat and surface area of the body are 32 g, 0.10 cal/g/°C and 8 cm 2 respectively, calculate the Stefan’s constant. Solution : We have,

E = σ(T 4 − T04 )

If A is the surface area of the body, the loss of energy radiated by the whole body per second, Q = σ(T 4 − T04 ) A J/sec

H-108 =

σ(T 4 − T04 ) kilo cal/sec J

…(1)

J = 4.2 × 10−3 J/kilo cal

where We know that,

Q = mass × specific heat × fall in temperature per sec = 32 × 10−3 kg × 010 . kilo cal/kg°C × (0.35°C) = 112 . × 10−3 kilo cal/sec

…(2)

From equation (1) and equation (2), we have σ(T 4 − T04 ) A = 112 . × 10−3 J σ=

112 . × 10−3 × J (T 4 − T04 )A

=

1.12 × 10−3 × 4.2 × 103 [(573)4 − (273)4 ] × 8 × 10−4

= 57 . × 10−8 J m −2sec −1(K) −4 Example 10. Two ideal black bodies A and B temperature 227°C and 327°C respectively are placed in an evacuated enclosure whose walls are blackened and kept at 27°C. Compare their rates of loss of heat. Solution : According to Stefan’s Law, E = σ(T 4 − T04 ) For black body A,

T = 227 ° C = 500 K T0 = 27 ° C = 300 K E1 = σ[(500)4 − (300)4 ] = 544 × 108σ

For black body B,

T = 327 ° C = 600 K T0 = 27 ° C = 300 K E 2 = σ[(600)4 − (300)4 ] = 1215 × 108σ E1 544 × 108 σ 544 = = E 2 1215 × 108 σ 1215

∴ Ratio of rates of loss

A : B = 544 : 1215

Example 11. If each square cm of sun’s surface radiates energy at the rate of 1.5 × 10 3 cal s −1 cm −2 and Stefan’s constant is 5.7 × 10 −8 Js −1 m −2 (K) −4 , calculate the temperature of sun’s surface. Solution : We have,

E = σT 4

The energy radiated by sun, E = 1.5 × 103 cal cm −2s −1 = 1.5 × 103 × 4.2 J (10 −2m) −2 s −1

…(1)

H-109 = 1.5 × 4.2 × 107 J m −2 s −1 σ = 57 . × 10−8 J s −1m −2 (K) −4 Substituting these values in equation (1), We get,

1.5 × 4.2 × 107 = 57 . × 10−8T 4 T4 =

1.5 × 4.2 × 107 57 . × 10−8

 1.5 × 4.2 T = × 1015 . 57  

1/ 4

= 5764 K

T = 5492°C Example 12. Luminosity of Rigel star in Orion constellation is 17000 times that of our sun. If the surface temperature of the sun is 6000 K, calculate the temperature of the star. Solution : We have,

E = σT 4

Therefore, if E1 and E 2 represent the luminosities and T1 and T2 the absolute temperatures of the star and the sun respectively, then we have, E1 = σT14 E 2 = σT24 E1 T4 = − 14 E2 T2 Given,

E1 = 17000 E2 T2 = 6000 K 17000 =

T14

(6000)4

T14 = (6000)4 × 17000 T1 = 6000 × (17000)1/ 4 = 68512 K Example 13. Calculate the maximum amount of heat which may be lost per second by radiation from a sphere of 10 cm in diameter at a temperature of 227°C when placed in an enclosure at a temperature of 27°C [ σ = 5.7 × 10 −12 watts × cm −2 (°C) −4 ]. Solution : We know that Stefan’s Law, E = σ(T 4 − T04 ) Where T and T0 are the absolute temperature of sphere and the surroundings respectively and σ is Stefan’s constant. Total surface area of sphere = 4 πr 2 The net rate of loss of heat from the total surface of the sphere Q = E × 4 πr 2

H-110 Q = σ(T 4 − T04 ) × 4 πr 2 =

σ(T 4 − T04 ) × 4 πr 2 cal/s J

Here, r = 5 cm, = 5 × 10−2 m, T = 227 °C = (227 + 273) K = 500 K T0 = 27 ° C = (27 + 273) K = 300 K σ = 5.7 × 10−12 watt cm 2 (K) −4 = 57 . × 10−8 watt m −2 (K) −4 J = 4.2 J/cal Q=

−8

57 . × 10 [(500)4 − (300)4 ] × 4 × 314 . × (5 × 10−2)2 = 23.3 cal/s 4.2

Example 14. Calculate the surface temperature of the sun. Given solar radiations λ m = 4753 Å and Wien’s constant b = 2 .89 × 10 −3 mK. Solution : We know that Wien’s displacement law, λ mT = constant = 0.2898 × 10−2 mK For the sun given, λ m = 4753 Å = 4753 × 10−10 m Temperature of sun Ts =

0.2898 × 10−2 4753 × 10−10

= 6097 K Example 15. From Wein’s displacement law we have, λ mT =

hc =b 5k

(i) Calculate the constant b what are the units of b ? (ii) Radiation from moon gives λ m = 4700 Å and 14 × 10 −6 m [Given, h = 6.6 × 10 −34 J-s, k = 1.37 × 10 −23 J/K. C = 3 × 10 8 m/s] Calculate temperature for each radiation. Solution : (i) As b = λ mT , The unit of b is meter-kelvin (mK), we have, b= =

hC 5k 6.6 × 10−34 × 3 × 108 5 × 1.37 × 10−23

= 0.002890 mK (ii)

When

λ m = 4700 Å = 4700 × 10−10 m

We have,

T1 =

0.002890 4700 × 10−10

= 6148 K

…(1)

H-111 When

λ m = 14 × 10−6 m

We have,

T2 =

0.002890 14 × 10−6

= 207 K

Example 16. The earth receives 2 cal cm −2 min −1 from the sun. If angular diameter of the sun is 32′ and it is treated as a black body, deduce its surface temperature. [σ = 5.7 × 10 −12 watt cm −2 ( K) −1 ] . Solution : The temperature of the sun is given by,  R  2 s 1 ×  T =   × 60 σ   r   where

1/ 4

radius of the sun r = R distance of sun from the earth = angular radius

Given, Angular diameter = 32′ Angular radius =

r  16  = 16′ =    60 R

°

π  16  radians =  ×  60 180 Hence,

R 180 × 60 = 16π r

Given,

s = 2 cal cm −2min −1 = 2 × 4.2 J × (10−2m) × min −1 = 2 × 4.2 × 104 J m −2min −1 σ = 57 . × 10−12 watt cm −2(K) −4 = 57 . × 10−8 watt m −2 min −1

Now,

 180 × 60 2 2 × 4.2 × 104  1 × T =    × − 8 60 57 . × 10   16π 

1/ 4

= 5803 K Example 17. Find the pressure exerted by sunlight on earth’s surface if the value of solar constant is 2 × 10 4 cal/min m 2 . Solution : We have radiation pressure, P = Here

E C

E = 2 × 104 cal/min m 2 E=

2 × 104 cal/s m 2 60

H-112 =

2 × 104 × 4.2 J/s m 2 60

= 1.4 × 103 J/s m 2 So

Pressure, P = =

E C 1.4 × 103 J / s m 2 3 × 198 m / s

= 4.66 × 10−6 N/m 2

1.

The black wall of a vacuum is of temperature 27°C. A black body of temperature 127°C is placed in it. Calculate the energy dissipiation at per unit area in per second by the body? [σ = 5 ⋅ 62 × 10−5 erg/cm 2 × sec × K 4 ]

2.

When at 27°C temperature a sphere of diameter 10 cm of temperature 127°C is placed in a closed vessel then calculate the per second energy dissipiation. [Given σ = 5 ⋅ 7 × 10−12 watt/cm 2 °C 4 ]

3.

Calculate the temperature at which anybody look like red and blue. Correspond to it the wavelength of maximum energy are λ m = 7500 Å respectively, given b = 0 ⋅ 3 cm K.

4.

If the rate of emissive energy is 16 watt at 1800 K temperature then find the emissive energy of a bulb which have temperature 3600 K.

5.

The temperature of a black is 1000 K. Find the frequency corresponding to maximum energy density in the emissive radiation of that body, given k = 1 ⋅ 38 × 10−23 J/K, h = 6 ⋅ 6 × 10−34 Js.

6.

Find the average energy of plank oscillator it’s frequency is 0 ⋅ 6 × 1014 cm −1 at 1800 K temperature. K = 1.38 × 10−23 J/K

7.

A black body whose surface area is 5 × 10−5 m 2 and temperature is 727°C. Find the value of emissive heat per minute from it. Given that Stefan’s constant σ = 5.67 × 10−8 J/m 2 sec K 4

8.

If two bodies X and Y are placed in a vaccumed vessel whose temperature are 427° and 227°C respectively and temperature of vessel is 27°C then compare heat loss from both the bodies.

9.

If the luminosity of Rigel star is 17000 times of sun and if temperature of sun’s surface is 6000 K then calculate the temperature of Rigel star.

10.

Find the temperature at which a black body will emit heat at the rate of 1 watt/m 2 . Given Stefan’s constant σ = 5.67 × 10−5 erg/cm 2 sec −1 K −4 .

11.

If two wavelengths of energy spectrum radiated by moon are 14 × 10−6 m and 5 × 10−7 m then calculate the temperature of each correspondance wavelength. Given b = 0.3 × 10−2 mK.

12.

According to Wein’s law λ m T =

hc =b 5k

If h = 6.63 × 10−34 J-s, c = 3 × 108 m/s and k = 1.38 × 10−23 J/K then calculate : (a) The value of b (b) If solar radiation wavelength λ m = 4753 Å, calculate the temperature correspondance to it. 13.

Calculate the energy radiated per minute from the filament of an incandescent temperature at 2000 K if the surface area is 5 × 10−5 meter 2 and its relative emittance is 0.85.

H-113 14.

A body of mass 10 gram is kept in an enclosure of temperature 27°C. If the temperature of the body is 127°C, its specific heat is 0.1 kcal/kg/°C and area of emitting surface of the body is 10−3 m 2 , find out the rate of cooling of the body. [σ = 572 . × 10−8 Jm −2 s −1 (°C) −4 ].

15.

A black body with an initial temperature of 300°C is allowed to cool inside an evacuted enclosure surrounded by melting ice at the rate of 0.35°C per second. If the mass, specific heat and surface area of the body are 32 g, 0.10 cal/g/°C and 8 cm 2 respectively, calculate the Stefan’s constant.

16.

Two ideal black bodies A and B at temperature 227°C and 327ºC respectivley are placed in an evacuted enclosure whose walls are blackened and kept at 27°C. Compare their rates of loss of heat.

17.

If each square cm. of sun’s surface radiates energy at the rate of 1.5 × 103 cal s −1 cm −2 and Stefan’s constant is . × 10−8 Js −1 m −2 (K) −4 , calculate the temperature of the Sun’s surface. 57

18.

Luminosity of Rigel star in Orion constellation is 17000 times that of our sun. If the surface temperature of the sun is 6000 K, calculate the temperature of the star.

19.

Calculate the maximum amount of heat which may be lost per second by radiation from sphere of 10 cm in diameter at a temperature of 227°C when placed in an enclosure at a temperature of 27°C [σ = 47 . × 10−12 watt cm −2 , (°C) −4 ].

20.

Calculate the surface temperature of the sun. Given solar radiations wavelength λ m = 4753 Å and Wien’s constant b = 2.898 × 10−3 mK.

21.

From Wein’s displacement law we have λ m ⋅ T = (i)

hc =b 5k

Calculate the constant b. What are the units of b ?

(ii) Radiation from moon gives λ m = 4700Å and 14 × 106 m. 22.

Calculate the temperature of each radiation. The earth receives 2.0 cal cm −2 min −1 , from the sun. If angular diameter of the sun is 32′ and is treated as a black body, deduce its surface temperature [σ = 57 . × 10−12 watt cm −2 (k) −2 , h = 6.6 × 10−34 Js, k = 1.37 × 10−23 J/K, C = 3 × 108 m/s].

23.

Find the pressure exerted by sunlight on earth’s surface if the value of solar constant is 2 × 104 cal/min m 2 .

H-114

Long Answer Type Questions

12.

What is Kirchoff’s law? Discuss it.

1.

13.

What is Stefan’s law? Give its thermodynamical proof.

14.

What is spectral formula ? Discuss early attempts.

15.

Derive the energy of Planck’s oscillator.

16.

What is hollow enclosure ? Give the thermodynamics of radiation inside a hollow enclosure.

2.

3. 4.

5.

6. 7. 8. 9.

What do you mean by thermal radiation? Discuss the property of thermal energy. Given the Stefan-Boltzmann law for black body radiation and prove it with the help of thermodynamics law. Explained the energy distribution in spectrum of black body radiation. Derive the Planck law for black body radiation and shows that Wein’s law and Rayleigh-Jeans law are the special state of Planck’s law. Explain the concept for emission and absorption of radiation. Derive the formula for average energy of Planck’s oscillator and explained it. Write the Rayleigh-Jeans law for black body radiation and derive it. Derive the Stefan’s law with the help Planck’s radiation formula. Derive the Rayleigh-Jeans and Wein’s displacement law with the help of Planck’s is radiation formula. Write the Wein’s displacement law for black body radiation and also derive it.

Short Answer Type Questions 1. 2. 3. 4.

5. 6. 7. 8. 9. 10. 11.

Write the definition and properties of thermal radiation. What is ideal black body? Explained the Ferry’s black body construction. What is the emissive power of any black body? Explained it. What will be colour seen of red and blue glass if we see them from a hole when they are inside red hot chamber? Write the Stefan-Boltzmann radiation law. Derive the Rayleigh-Jeans law weigh the help of Planck’s radiation formula. Derive Wein’s displacement law with the help of Planck’s radiation formula. What is ultraviolet Catstraff? What is a black body ? Define ideal black body and hence discuss about practical black bodies.

17.

What is Planck’s concept ?

18.

Discuss quantum theory of radiation.

19.

What is a photon gas ?

20.

What do you mean by photon gas ? Which kind of particle is a photon ?

Very Short Answer Type Questions 1.

What is radiation?

2.

What is absorptive power of a surface?

3.

Write the Wein’s displacement law.

4.

Write the Planck’s radiation formula.

5.

What is energy density?

6.

What is black body?

7.

Write the formula of Rayleigh-Jeans law.

8.

What is the value of Wein’s constant?

9.

Write two properties of radiation.

10.

Define quanta.

11.

Which method of transformation of heat has no need of any medium?

12.

By which method radiation from sun to earth is coming?

13.

What is Planck’s radiation formula in terms of wavelength?

14.

What is the value of average energy of Planck’s oscillator?

15.

Planck’s radiation formula converts into which formula at high wavelength side?

16.

What is the formula of Stefan’s law?

17.

What is Wein’s law formula ?

18. What are practical black bodies ? Discuss Wein’s 19. black body and Ferry’s black body. 20. Show that Kirchoff’s law is : 21. l λ / aλ = G 22.

What is the unit of Wein’s constant ? Who gave quantum theory of radiation ? What is the energy of photon or quantum particle? What is photon gas ? What is Rayleigh-Jean’s law ?

H-115 23.

Which meter is used for measuring the temperature 11. of sun?

Objective type questions

12.

Which surface will emit high radiation energy? (a) rough black surface (b) polished black surface (c) rough white surface (d) polished white surface

13.

Which one is not true ? (a) All bodies emit thermal radiations at all temperatures (b) Thermal radiations are electromagnetic waves (c) Thermal radiations are not reflected from a mirror (d) Thermal radiations travel in free space with a velocity of 3 × 108 ms −1

Multiple Choice Questions 1.

Absorption coefficient of an ideal black body : (a) 1 (b) 0 (c) ∞ (d) none of these

2.

Bodies having unit absorption coefficient are called : (a) black body (b) red body (c) white body (d) green body

3.

Emissive radiation energy per sec from anybody depends upon : (a) only in the nature of Surface (b) only in the area of Surface 14. (c) one in the temprate of Surface (d) all of them

4.

Thermal radiation are electromagnetic wave which is found in : (a) ultraviolet region (b) infrared region (c) visible region (d) x-Ray region

5.

Which colour has lowest temperature? (a) blue (b) red (c) green (d) yellow

6.

According to Stefan emissive radiation energy by a substance is proportional to : (a) (c) T 2 (d) T (b) T 4 T

7.

The spectrum of black body : (a) continuous (b) band (c) sharp (d) none of there

8.

According to Wein’s displacement law : λm (a) = constant (b) λ m T = constant T I (c) = constant (d) none of these λm

9.

Wein’s law is a special state of Planck’s formula : (a) for high wavelength (b) for all wavelength (c) for short wavelength (d) none of the above

10.

The value of Wein’s constant is : 5k hc hc hc (a) (d) (b) (c) hc k 5 5k

Which of the following devices will you use to detect thermal radiations? (a) Liquid thermometer (b) Six’s maximum and minimum thermometer (c) Constant volume air thermometer (d) Thermopile

15.

The absorptive power of a perfectly black body is : (a) zero (b) 0.5 (c) 1 (d) infinite

16.

The amount of energy radiated by a body depends upon : (a) the nature of its surface (b) the area of its surface (c) the temperature of its surface (d) all of the above factors

17.

The wavelength of the radiation emitted by a body depends upon : (a) the nature of its surface (b) the area of its surface (c) the temperature of its surface (d) all the above factors

18.

The SI units of Stefan’s constant is : (b) Jm −2 K −4 (a) Nm −2 K −4 −1 −1 −4 (c) Jm s K (d) Wm −2 K −4

19.

Which one of the following devices is used for measuring very high temperatures ? (a) Mercury thermometer (b) Gas thermometer (c) Platinum-resistance thermometer (d) Pyrometer

Rayleigh-Jeans law is equivalent to Planck law at which wavelength is : (a) for high wavelength (b) for short wavelength (c) for all wavelength 20. (d) none of the above

Nature of spectrum of a black body is : (a) linear

H-116 (b) bond (c) continuous (d) linear and continuous 21.

14.

The high wavelength side of Planck’s formula is ……… law.

15.

The formula E ∝ T 4 represents ……… law.

Planck’s formula reduces to Rayleigh-Jeans formula for : (a) high wavelength (b) low wavelength (c) at both (d) none of the above

16.

High temperature can be measured by ……… .

17.

……… is the formula of Planck’s oscillator average energy.

18.

Thermal radiations are ……… wave.

19.

The speed of radiations is equal to ……… .

According to Stefan’s law energy E is proportional to : (a) T 2 (b) T 4 −1 (c) T (d) T 0

20.

Good absorbers are good emitters is given by ………

The Wein’s displacement law is : (b) λ m T 2 = b (a) λ m T = b

1.

Black body is a good emission of heat.

2.

The rate of heat dissipiation by a radiation of a body is not depend upon the temperature of body.

The instrument used to measure solar constants : (a) pyrheliometers (b) thermometers (c) differentromate (d) none of these

3.

The unit of Wein constant is meter kelvin.

4.

For a red hot body all the radiated wavelengths are not same.

Planck’s formula reduce to Wein’s law at : (a) high wavelength (b) low wavelength (c) at all wavelength (d) none of the above

5.

For all wavelength the absorptivity of a body is not unit.

6.

The pressure of diffused radiation is equal to the two third of its energy density.

7.

Electromagnetic radiations are electromagnetic waves.

Fill in the Blank

8.

Speed of radiation is 3 × 104 m/sec.

1.

Colour of a star is defined it with ........... .

9.

2.

If we increases temperature of a black body then frequency related with emissive bower .......... . 10. Energy radiation depends ........... . hc 11. is ............ 5k 12.

Wein’s law is the short wavelength side of Planck’s law.

22.

23.

(c) λ2m T = A 24.

25.

3. 4.

True / False

(d) E ∝ T 4

Rayleigh-Jeans law is high wavelength side of Planck’s law. A perfect black body radiates all of its energy. Gas thermometer can measure temperature of sun.

5.

Absorption power is ............ for all wavelengths.

13.

Wein’s law is λ m T = b.

6.

The spectrum of black radiation is ..............

14.

7.

Stefan’s law is .............

A body that absorbs all the radiation falling on it is known as perfect black body.

8.

The unit of solar constant is ........... .

15.

9.

Good absorbers are good ………… of radiation.

Energy emitted by a hot body is distributed equally among different wavelengths.

10.

Wein’s constant λ m T = b is nearly ……… mK.

16.

A black body is poor radiator of heat.

11.

Three stars A, B and C appear green, red and blue, 17. 18. the star having minimum temperature ………

12.

Spectral distribution of energy in a black body 19. radiation is successfully explained by ……… formula. 20. The short wavelength side of Planck’s formula is ……… law.

13.

Thermal radiations cannot travel in vaccum. Thermal radiation can exert pressure. eλ = E λ is known as Stefan’s law. aλ To measure the temperature of stars we need Wein’s law.

H-117

Objective Type Questions Multiple Choice Questions 1.

(a)

2.

(a)

3.

(d)

4.

(b)

5.

(b)

6.

(b)

7.

(a)

8.

(b)

9.

(c)

10.

(a)

11.

(d)

12.

(c)

13.

(c)

14.

(d)

15.

(c)

16.

(d)

17.

(c)

18.

(d)

19.

(d)

20.

(c)

21.

(d)

22.

(b)

23.

(a)

24.

(a)

25.

(b)

Fill in the Blank 1.

temperature

2.

increase

3.

absolute temperature

4.

Wein’s constant

5.

unit

6.

continuous

7.

σT 4

8.

σT 4

9.

emitters

10.

3 × 10 −3

11.

Red

12.

Planck’s

13.

Weins

14.

Rayleigh-Jeans

15.

Stefan’s

16.

pyrometer

17.

E=

18.

electromagnetic

19.

speed of light

20.

hν hν (e kT

− 1)

Kirchoff’s

True / False 1.

True

2.

False

3.

True

4.

True

6.

False

7.

True

8.

False

9.

True

10. True

5.

False

11. False

12. False

13. True

14. True

15. False

16. False

17. False

18. True

19. False

20. True

H-118

H ints and Solutions λ m T = constant =

Numerical Questions 1.

Given T = 127°C = 4000 K, T0 = 27 ° C = 300 K −5

2

σ = 5 ⋅ 62 × 10

erg/ cm × sec × K

=

According to Stefan’s law energy dissipation per cm 2 area in per sec

= 9 ⋅ 8 × 105 erg/cm 2 sec

6.

Given, ν = 0.6 × 1014 cm −1 , T = 1800 K k = 1.38 × 10−23 J/K

Given T = 227 °C = 500 K, T0 = 27 ° C = 300 K σ = 5 ⋅ 7 × 10

2

watt / cm °C

2

4

2

and area A = 4 πr = 4 × 3.14 × 5 = 314 cm

and

Average energy of Planck oscillator hν E = hν e kT − 1

E = σA (T 4 − T04 )

hν 6.62 × 10−34 × 0.6 × 1014 = kT 1.38 × 10−23 × 1800

= 5.7 × 10−12 × 314 [(500)4 − (300)4 ] = 79 ⋅ 4 J/sec

= 1.599

According to Wein’s law, λ m T = b or T =

E=

b

Temperature for body to look blue 0.3 T= = 600 K 5000 × 10−8 Given, E1 = 16 watt, T1 = 1800 K E 2 = ?, T2 = 3600 K T  Stefan’s law, =  1 E 2  T2  E1

E2 =

T24 T14

4

= or,

3600  E1 =   × 16  1800 

E = 1.03 × 10−20 J

See Example-1 See Example-2

9.

See Example-3

10.

See Example-4

11.

See Example-5

12.

See Example-6

13.

See Example-7 Ans. 2315 J

14.

See Example-8 Ans. 0.23 K s −1

Given, T = 1000 K, k = 1.38 × 10−23 J/K h = 6.6 × 10

3.972 × 10−20 4.853 − 1

8.

4

−34

e1 . 599 − 1

7.

E 2 = 256 watt 5.

6.62 × 10−34 × 0.6 × 1014

λmax

Temperature for body to look red 0.3 cm K T= = 4000 K 37500 × 10−8 cm

4.

h = 6.62 × 10−34 sec

2

According to Stefan’s law, energy dissipation from a surface

3.

6.6 × 10−34

So, νm = 1.04 × 1014 cm −1

E = 5 ⋅ 62 × 10−5 [(400)4 − (300)4 ]

−12

5 × 1.38 × 10−23 × 1000

= 1.04 × 1014 cm −1

E = σ (T 4 − T04 )

2.

5k c 5 kT = hc h

νm =

4

hc 5k

Js

15.

See Example-9 Ans. 57 . × 10−8 J m −2 s −1 K −4

H-119 16.

See Example-10 Ans. 544 : 1215

17.

See Example-11 emission radiation

Ans. 5492°C 18.

See Example-12 Ans. 68520 K

19.

See Example-13

20.

See Example-14

Ans. 23.3 cal/sec

Ferry’s black body is a hollow sphere of copper with double wal whose inside walls are polished while outside walls are polished by nikel. There is a hole in it and in front of this hole there is a cone bump P. When a thermal radiation is incident on it then it becomes absorbes after successive reflections. Hence, hole H is behaves like a black body.

Ans. 6097 K 21.

See Example-15 Ans. 0.002890 mK and 207 K

22.

See Example-16 Ans. 5803 K

23.

3.

Emissive power of a surface is the emissive radition energy in per sec from unit area of that surface, it is denoted by E and its unit is J/m 2 cm.

4.

When an object is put into a black body then both black body and object has in thermal and body behaves like a perfect black body. Hence red and blue glass pieces look like black.

See Example-17 Ans. 4.66 × 10−6 N/m 2

Short Answer Type Questions 1.

Every substance emits the energy continuously because of its thermal properties which is called 5. thermal radiation. 6. Properties : 7. (i) It moves in a straight line like light. 8. (ii) It can also transmitted in vacuum. (iii) Its wavelength is greater than wavelength of light. (iv) We can found its spectrum with the help of prism.

2.

Ideal black body is a substance which absorbed incident radiation of all wavelength on its surface. For an ideal black body the value of absorptivity is unit for all values of wavelength. In behaviour any black body is not ideal. For experiments an ideal black body is constructed by Ferry’s whose construction is given as.

P

9.

For this solution see part 3. For this solution see special case (II) of part 10. For this solution see special case (I) of part 10. According to Rayleigh and Jeans’ law, in black body emission electromagnetic waves are present for all wavelength from 0 to ∞. It is reflected again and again from the wall of black body and produce non-progressive waves for which all black bodies has degree of freedom. Rayleigh-Jeans law is only correct for high wavelength for shorter wavelengths, it is not valid. See 3, 3.1 and 4

10.

See 4.1 and 4.2

11.

See 5

12.

See 5

13.

See 6

14.

See 7

Incident radiation

15.

See 11

16.

See 4.2.1

H

17.

See 12

18.

See 10.1

19.

See 14

20.

See 14

H-120

Very Short Answer Type Questions

10.

1.

Radiation is the process of energy transmission by which energy is transfered from one place to another place without any changing in medium.

2.

Ratio between per sec absorbed radition energy by a surface and per sec incident radition energy, is called the absorptive power. 11. According to Wein’ displacement law wavelength 12. corresponding to maximum energy is inversely proportional to the absolute temprature of black body. 13. 1 λmax ∝ T 14. b λmax = i. e. T

3.

4.

According to Planck radiation formula E ν dν =

8 πhν3 c

3

⋅

dν e

hν / kT

The substance which absorb incident radiation 18. energy of all wavelength is called black body. 19. The formula of energy distribution in spectrum given 20. by Rayleigh and Jeans is as follows : 8 πkT 21. E λ dλ = dλ λ4 The value of Wein’s constant is (i)

Radiation is electromagnetic energy.

(ii)

Radiation moves with speed of light.

Radiation E λ dλ = = E=

8 πhc ⋅ λs

dλ hc e kT

22. 23.

−1

hν hν

e kT E ∝ T 4 or E = σT 4

6.

9.

Radiation

16. −1

b = 0 ⋅ 3 × 10−2 mK

ν → Frequency of light

Rayleigh-Jean’s

Energy density is equal to the energy in unit volume 17. along a point

8.

h → Planck’s constant

15.

5.

7.

Sir Max Planck introduced the idea that radiation of black body is emitted not in continuous, it emits in the form of small packets of energy called quanta. The energy of each quanta is hν, where

where σ is Stefan’s constant hc λ mT = =b 5k mk Sir Max Planck in 1900 E = hν The Boson particles comes in the form of radiation comes out from a hole of a chamber is called photon gas. 8 πkT E λ dλ = dλ λ4 Pyrometer mmm

H-121

Unit

4

Low Temperature Physics

1. Methods of Producing Low Temperature Long time ago ice was only known coldest matter on earth, at that time only 0°C was known as lowest temperature. Fahrenheit was the first person observed that temperature comes at −18 °C, when they put a thermometer inside the mixture of ice and salt. Hence, they developed a scale on the basis of this called Fahrenheit scale of temperature. Now a days minimum possible temperature is − 273°C which is called 0 K. Some important methods for cooling are : 1.

Freezing mixture

2.

Evaporation at low pressure

3.

Adiabatic expansion

4.

Joule-Thomson effect

5.

Absorption

6.

Peltier effect

7.

Adiabatic demagnetisation

8.

Nuclear demagnetisation

2. Joule-Kelvin Expansion (Joule-Thomson Expansion) When a gas passess through a porous plug from high pressure to low pressure, then after passing gas experiences a change in temperature. This phenomenon is called Joule-Thomson or Joule-Kelvin effect and the process is called adiabatic throttling. This change in temperature in gases is based upon the basic nature of gas and initial temperature. All the gases except hydrogen and helium shows the cooling effect while hydrogen and helium shows, heating effect i. e. after passing through the porous plug both hydrogen and helium becomes warm. Rest gases becomes cool after passing through the porous plug. The fall in temperature is directly proportional to the difference of pressure on the either sides of porous plug. i. e.,

dT ∝ (P2 − P1)

H-122 The fall in temperature becomes zero at a particular temperature called the inversion temperature i. e. at a temperature where gas shows neither cooling nor heating effect. If temperature of gas is above the inversion temperature this temperature is different for different gases. Any gas having temperature lower than inversion temperature will becomes cool after passing porous plug. Let a gas is suffering with Joule-Thomson expansion and this is repeated for many times. The initial pressure and temperature is P1 and T1 are chosen arbitary P1 and P2 is also fixed but P2 < P1 then the temperature T2 is measured, keeping P1 and T1 the same. P2 is varied to another value and measured the value of T2. Now we will plot the curve between P2 and T2 graph as shown in figure. One point on curve is (P1, T1) and others are indicated by x , y, z , m, n ....... represented various P2 ′ and T2 ′ s. Since in the J−K expansion the initial and final enthalpies are equal i. e. H1 = H 2

Temperature T

This experiment is now repeated up to a number of times and we will find many values of T2. n

m

z

y x (P1T1)

The enthalpy at all points x , y, z , ..... is the same and curve through these points represents a constant enthalphy and is called isenthalpic curve. Pressure P

Now the temperature T1 on the high pressure side of the plug is changed to Fig. 1 another value keeping P1 the same, P2 is again varied and corresponding T2’s measured. By repeating this experiment with initial temperature T1 same for previous but different for other experiment. A group of curves corresponding to different values of H are shown in fig. 2

Cooling

T He atin g Inv e

rsio n

cur v

e

P Fig. 2

If the temperature is not too high, such curves passing through a maximum called inversion curve. Since towards the left of maximum side gas shows cooling while on other hand gas shows heating.

2.1 Expression for Joule-Kelvin Coefficient Let a porous plug is shown in the figure. On left side the pressure is P1 at volume V1 while other side pressure

H-123 porous plug P2

P1

P1

1 Fig. 3

is P2 at gas volume V2. If gas has 1 mole thus net work done after passing through porous plug P1 is P2V2 − P1V1. Internal energy of gas at first point is U 1 and at second point is U 2. But according to First Law Thermodynamics dQ = dU + dW For adiabatic process dQ = 0 dW = − dU or

dW = − (U 2 − U 1)

or

dW = (U 1 − U 2)

But we have work,

dW = P2V2 − P1V1 U 1 − U 2 = P2V2 − P1V1

So,

U 1 + P1V1 = U 2 + P2V2 = U + PV = constant U + PV = H (Enthalpy)

Hence,

Hence under the Joule-Thomson cooling enthalpy becomes constant. H = U + PV = Constant

Hence

…(1)

On differentiating dH = dU = PdV + VdP = 0 or

dU + PdV + VdP = 0

…(2)

But according to first thermodynamics law dQ = dU + dW or

dQ = dU + PdV

[‡ dW = PdV]

Second thermodynamics law dQ = Tds or

dU = Tds − PdV

Using equation (2) Tds − PdV + PdV + VdP = 0 or

Tds + VdP = 0

Let entropy S is the function of temperature and pressure both i. e.

S = S (T , P)

…(3)

H-124 On differentiating  dS   dS  dS =   dT +   dP  dT  P  dP  T Using dS in equation (3)  dS   dS  T  dT + T   dP + VdP = 0  dP  T  dT  P We have second law of thermodynamics dQ = TdS  dS   dQ  dT + T   dP + VdP = 0   dP  T  dT  P But specific heat at constant pressure  ∂Q  C P =   for m = 1 mole  ∂T  P So, C P using in equation (4)  dS  C P dT + T   dP + VdP = 0  dP  T We have Maxwell’s fourth relation  dV   dS     =−  dT   dP  T  dV  C p dT − T   dP + VdP = 0  dT  P

So, or

  dV   C P dT − dP T   − V = 0   dT   P

or

dT 1 = dP CP

Here

  dV   T  dT  − V P  

dT is known as Joule-Kelvin coefficient. It is denoted by µ. dP

So,

µ=

1 CP

  dV   T  dT  − V   P

2.2 Joule-Kelvin Coefficient for Ideal Gas We have an ideal gas equation

So,

PV = RT RT on differentrating V= P R  dV   =   dT  P P

…(4)

H-125 So,

µ=

1 CP

 R  T P − V

µ=

1 CP

 PV   P − V or µ = 0

Thus Joule-Kelvin coefficient for ideal gas is zero.

2.3 Joule-Kelvin Coefficient for van der Waal’s Gas We have van der Waal’s gas equation a   P + 2  (V − b) = RT  V 

…(5)

Differentiate with respect to T a   dV  2a  dV   P + 2   − (V − b) 3   = R     dT P V V  dT  P a  2a  dV     P + 2  − 3 (V − b) = R  dT  P  V  V R  dV    =  dT  P  a  2a  P + 2  − 3 (V − b)  V  V

or

Using equation (5), we get, R  dV   =   dT  P  RT  2a (V − b)  −   V − b V 3 Multiplying and dividing by (V − b) So,

 dV   =   dT  P

R(V − b) 2a RT − 3 (V − b)2 V

 dV  T  =  dT  P

RT (V − b) 2a RT − 3 (V − b)2 V

Multiplying by T both sides

Subtracting both sides with V  dV  T  −V=  dT  P

RT (V − b) −V 2a RT − 3 (V − b)2 V

Taking LCM, we get,  dV  T  −V=  dT  P

RTV − RTb − RTV + RT −

2a V3

2a V2

(V − b)2

(V − b)2

H-126 2a (V − b)2 − RTb 2 V = 2a RT − 3 (V − b)2 V

or

2a  2 3 − RTb + V 2 (V − b)  V  dV  T  −V=  dT  P 2a  2 3 RT − V 3 (V − b)  V

or

− RTbV 3 + 2aV(V − b)2  dV  T  −V=  dT  P RTV 3 − 2a(V − b)2

On comparing with RTV 3, 2a(V − b)2 can neglected and on comparing with V, b can be neglected. So, we get − RTbV 3 + 2aV 3  dV  T  −V=  dT  P RTV 3 2a − RTbV 3 2aV 3  dV  T = − b+ +  −V=  dT  P RT RTV 3 RTV 3 2a  dV  T −b  −V=  dT  P RT Now Joule-Kelvin coefficient for van der Waal’s gas µ=

1  2a  − b C P  RT 

2.4 Importance of Joule-Thomson Effect 2a =b RT

If

T = 2a / Rb

So,

This term is known as inversion temperature it is denoted by Ti. 2a i. e. Ti = Rb 2a 2a 2a , then gas will show the cooling effect and if T > then If gas temperature T is less than i. e., T < Rb Rb Rb gas will show heating effect.

3. Adiabatic Demagnetisation 3.1 Approach to Absolute Zero Temperature very near to the absolute zero can be produced by the evaporation of liquid helium under greatly reduced pressure. Prof. K. Onnes attained a temperature of 80 K by evaporating liquid helium under a pressure of 0.01 mm of mercury. Kessom reduced the pressure on the surface of liquid helium down to 0.0036 mm of mercury by powerful keeps and attained a temperature of 0.726 K.

H-127

3.2 Adiabatic Demagnetisation Debye and Criauque independently pointed out in 1926 that still lower temperature can be achieved salts (magnetic cooling). It is based on ‘magneto-caloric effect; when a paramagnatic salt is magnetised. Its molecules are set in the converted into heat. The temperature of the salt, therefore, rises. If the salt is now allowed to cool to its intial temperature and then demagnetised under adiabatic condition, the molcules return to their original random distribution. Hence there is a corresponding fall in temperature. The fall in temperature is of the order of fraction of a degree if the substance is already at a low temperature of about 1 K.

3.3 Method Fig. 4 illustrates the experimental procedure. This is as follows : 1.

2.

3.

The paramagnetic salt procedure D is put in an aluminium container and suspended in a vessel. A surrounded by a dewar flask B containing liquid helium at about 1 K. The arrangement is kept between the poles N, S of an electromagnet giving a magnetising field of about 10,000 oersteds.

To vacuum pump Solenoid

Paramagnatic salt

N

S A

The vessel A is first filled with helium gas. Since the gas is conducting at low temperature the salt A comes in thermal contact of the liquid helium field in B. Hence its temperature falls to 1 K. Now, the magnetising field is switched on. The salt D is magnetised and becomes warmed. The heat of magnetisation flows out through the helium gas into the liquid helium. So that the temperature of the salt D again falls to 1 K.

A B

C Liquid helium

Liquid hydrogen Fig. 4

4.

The helium gas is now pumped out from the vessel A so that the salt D becomes thermally insulated.

5.

The magnetising field is then switched off. The salt D is now demagnetised and its temperature sharply falls.

3.4 Magnetic Cooling at Ordinary Temperature The drop in temperature by adiabatic demagnetisation is inversly proportional to the initial temperature. It is of the order of fraction of a degree when the initial temperature of the paramagnetic salt is about 1 K. Clearly, it would be negligible at ordinary temperature 300 K. Hence this method cannot be used for cooling at ordinary temperature.

3.5 Measurment of Temperature near Absolute Zero The temperature obtained by adiabatic demagnetisation are measured magnetically by using Curie’s law. According to this law, the magnetic susceptibility x of a paramagnetic salt is inversly proportional to its absolute temperature T

H-128 x=

i. e.

C T

where C is Curie’s constant. Hence, if the susceptibility be measured, the corresponding temperature can be obtained. For this purpose, the vessel A which contains the paramagnatic salt D, is surrounded by two coaxial coils. An alternating current is passed through one of them. The current induced in the other, which depends upon the susceptibility of the salt, is amplified and measured. This current gives a measure of the susceptibility from which the temperature can be calculated by the above relation. Usually, the galvanometer which reads the induced current is calibrated to read the temperature directly. Temperature so obtained is called the magnetic temperature; It somewhat differs from the kelvin temperature. Recently, acoustic thermometers have been developed in which the speed of ultrasonic waves in gaseous hydrogen and helium is measured. Such thermometers are expected to be useful between 5-10 K.

4. Introduction to Cryogenics Cryogenics is the branch of physics which deals at very low temperature i. e. below 123 K (Kryo-very cold, Genics-to produce). This is also known as low temperature physics. In general, cryogenic is the study of the production and behaviour of materials at very low temperature.

5. Refrigerator and Air Conditioning Mechanism A refrigerator is a machine used for the aritificial production of cold. It is designed to maintain a chamber at a low temperature or to keep a brine-bath cell below 0°C for freezing water into ice. Its action is base on the cooling produced by rapid evaporation of a liquid. The liquid undergoing evaporation is called ‘refrigerant’, and is a readily liquefiable substance like ammonia, carbon dioxide, sulphur dioxide from CCl 2F2. A vapour-compression type of refrigerator, named ‘frigidaire’, is shown in fig. 5(a) and the corresponding refrigeration cycle (P - V diagram) in Fig. 5(b). Compressor Water

Q

P

Brine-bath

Condenser Evaporator

Water

Fig. 5(a) Frigidaire

The frigidaire consists of an electrically driven compressor, a condenser, an evaporator (refrigerating chamber) and an expansion value V.

H-129

5.1 Refrigeration Cycle As the piston of the compressor moves downwords, the value P is closed and Q is opened. The refrigerant say, ammonia vapour, is compressed adiabatically. The P C process is shown by the curve AB in P − V diagram. The compressed vapour is delivered to the copper coils in the condenser where it liquefies, the heat of compression being carried off by a stream of circulating water. (The coils may also be cooled by tan-driven air currents). The process of liquifaction is shown by the straight line BC in the P − V diagram. The liquefied ammonia passes through the expansion value into the lower pressure side of the refrigerator. The process is shown by the line CD in the P − V diagram. Finally, the liquid ammonia evaporates in the evaporator coils, extracting its latent heat of evaporation from the brine-bath surrounding the coils. The evaporation is indicated by the line DA in the P − V diagram.

B

A

D V Fig. 5(b)

The cooled vapour is drawn into the compressor as the piston moves up. The cycle is then repeated. The brine solution which is thus cooled to a temperature below 0°C is circulated through pipes to maintain a low temperature in a cold storage clamber, or circulated past tanks containing water which is freezed into ice.

5.2 Meaning of Efficiency In principle, the refrigerant extracts heat Q from a cold reservoir (evaporator), has work W done on it by the compressor, and rejects the heat Q + W to a hotter reservoir (condenser). The refrigerator is said to be most efficient if it extracts as much heat as possible from the cold reservoir with the expenditure of little work as possible. A measure of efficiency of a refrigerator is expressed by the coefficient of performance k, where Heat extracted Q = k= Work expanded W

5.3 Air Compression Machine It is a mechanical machine to compress the air with raise its pressure. The air compressor sucks air from the atmosphere and compresses it then further delivers with a high pressure to a storage vessel. From the storage vessel, it may be transmit by the channel (pipeline) to a place where the supply of compressed air is required. Afterward the compression of air requires some work to be done on it, therefore a compressor must be driven by some prime mover.

5.3.1 Classification of Air Compressors 1.

According to working (i) Reciprocating compressors (ii) Rotary compressors

2.

According to action (i) Single acting compressors (ii) Double acting compressors

H-130 3.

According to number of stages (i) Single-stage compressors (ii) Multi-stage compressors

5.3.2 Important Definitions 1.

Inlet pressure : It is the absolute pressure of air at the inlet of a compressor.

2.

Discharge : It is the absolute pressure of air at the outlet of a compressor.

3.

Compression ratio (or pressure ratio) : It is the ratio of discharge pressure to the inlet pressure. Since, there discharge pressure is always more than the inlet pressure, therefore the value of compression ratio is more than unity.

4.

Compressor capacity : It is the volume of air delivered by the compressor and is expressed in m 3 / min or m 3/s.

5.

Free air delivery : It is the actual volume delivered by a compressor when reduced to the normal temperature and pressure condition. The capacity of a compressor is generally given in terms of free air delivery.

6.

Swept volume : It is the volume of air sucked by the compressor during its suction stroke. Mathematically, the swept volume or displacement of a single acting air compressor is given by  π Vs =   × D 2 × L, where D = Diameter of cylinder bore and L = Length of piston stroke.  4

A. Single Cylinder Compressor Piston compressors are available as single or double acting, oil lubricated or oil free with different number of cylinders in different configurations. With the exception of really small compressors with vertical cylinders, the V configuration is the most common for small compressors. On double acting, large compressors the L type with vertical low pressure cylinder and horizontal high pressure cylinder, offer immense benefits and that is why it is the most common design. The construction and working of a piston type reciprocating compressor is very much similar to that of an internal combustion engine. (i)

Construction : Piston type compressor consists of cylinder, cylinder head and piston with piston rings, inlet and outlet spring loaded valves, connecting rod, crank, crankshaft and bearings.

(ii)

Working : Compression is accomplished by the reciprocating movement of a piston within a cylinder. This motion alternately fills the cylinder and then compresses the air. A connecting rod transforms the rotary motion of the crankshaft into the reciprocating motion of piston in the cylinder. Depending on the application, the rotating crank (or eccentric) is driven at constant speed by a suitable prime mover (usually electric motor). Schematic diagram of single cylinder compressor is shown in figure 6.

(iii)

Inlet Stroke : Suction or inlet stroke begins with piston at top dead centre (a position providing a minimum or clearance volume). During the downward stroke, piston motion reduces the pressure inside the cylinder below the atomospheric pressure. The inlet valve then opens against the pressures of its spring and allows air to flow into the cylinder. The air is drawn into the cylinder unitil the piston reaches to a maximum volume position (bottom deod centre). The discharge valve remains closed during this stroke.

H-131 (iv)

Outlet stroke : During compression stroke piston moves in the opposite direction (bottom dead centre to top dead centre), decreasing the volume of the air. As the piston starts moving upwards, the inlet valve is closed and pressure starts to increase continuously until the pressure inside the cylinder is above the pressure of the delivery side which is connected to the receiver. Then the outlet valve opens and air is delivered during the remaining upward motion of the piston to the receiver.

Inlet air

Outlet air

Cooling fins

Fig. 6 : Single cylinder compressors

B. Analysis of Single Cylinder Single Stage Air Compressor A typical indicator diagram for reciprocating compressor with three different types of compression is shown in the Figure 7. Clearance volume is neglected.

3

2' 2 2'' Adiabatic pV γ = c

Constant pressure line 4-1 represents the suction stroke. Polytrophic pV n = c p The air is then compressed adiabatically (process line 12’’) and is then forced out of the cylinder at constant pressure Isothermal pV = c (process 2’’3). Area 12’’34 represents the work. If the 1 4 compression is carried out isothermally, then it follows the curve 12’ which has less slope than both isotropic and polytrophic processes. This work done that is area 12’34 in V isothermal process is considerably less than that due to Fig. 7 : Types of compression adiabatic compression. Thus compressor will have higher efficiency if compression follows isothermal process. It is not possible in practice as to achieve isothermal process, as the compressor must run very slowly. In practice compressors run at high speeds which results in polytrophic process. The cold water spray and multi stage compression are used for approximating to isothermal compression while still running the compressor at high speed.

C. Work Done in a Single Stage Compressor Neglecting Clearance Figure 8 shows that PV diagram of the air in the cylinder of an air compressor. Constant pressure line ab represents the suction stroke. The air is then compressed adiabatically (process line bc) and is then forced out of the cylinder at constant pressure (process cd). Area abcd represents the work.

H-132

p2

c

d

p

p1

b

a v

0

f v1

e v2

Fig. 8 : P-V diagram

There are three types of compression processes possible in compressor they are : (i)

Isothermal compression : Compression of air takes place at constant temperature p  Work done during compression = p1V1 ln  2   p1 

…(1)

Where p1 = inlet pressure, p2 = outlet pressure (ii)

Adiabatic compression : There is no flow of heat energy into or out of the gas during expansion or compression. γ Work done during compression = …(2) ( p2V2 − p1V1) γ −1 γ p1V1 Work done during compression = γ −1

n−1     p2  n −1     p    1

γ is the ratio of specific heat = 1.4 for air. C p = 1.005 (iii)

kJ kJ k, Cv = 0.718 k kg kg

Polytropic compression : This process lies between isothermal and adiabatic. In pneumatics, most compression/expansions are neither adiabatic (very fast) nor isothermal (very slow). It is polytropic n workdone during compression = ( p2V2 − p1V1) n −1 np V work done during compression = 1 1 n −1

n−1     p2  n −1   p    1

n is the polytrophic = 1.4 for air Efficiency of compressor is a reciprocating compressor the work is minimum when compression follows the isothermal process. The ratio of isothermal work done to the actual work done is called isothermal efficiency. Isothermal work ηisothermal = Actual work

H-133

D. Work Done in a Single Stage Compressor Considering Clearance In practical design of compressors, some clearance is required between the cylinder and piston to prevent hitting of piston to crown of the cylinder. Figure. 9 shows a PV diagram of single stage compressor with clearance. 3

p2

2 Compression

Expansion pVn = constant

p1

4 Vc

pVn = constant

(V1 – V4) Free air volume

1 Intake pressure

Swept volume V s = V 1 – V3

Fig. 9 : P-V diagram with clearance

Thus when the compressed air is delivered during the delivery stroke, some amount of air corresponding to clearance volume V3 at a pressure p2 will be left over in the cylinder. During the next suction stroke this air expands back to initial pressure p1 and volume V4 . Thus before the fresh air enters the cylinder some air corresponding to volume V4 will be already there in the cylinder. Thus the volume inhaled during the suction stroke will be V1 − V4 which is less the swept volume Vs . The work done on the air delivered is not affected by the clearance volume as the work required to compress the clearance volume is theoretically regained during its expansion from V3 to V4 . Thus the work done is given by n−1   np1(V1 − V4 )  p2  n  work done during compression = −1   p  n −1 1  

5.4 Derivation for Efficiency If Q1 is the heat rejected from refrigerator and Q2 is the amount of heat is accepting by the refrigerator The coefficient of performace, k=

Q2 Q2 = W Q1 − Q2

Work done = (Q1 − Q2) But in carnot’s cycle Q1 T1 = Q2 T2 Where T1 is the temperature of hot pot and T2 is the temperature of cold pot. T2 So, β= T1 − T2

H-134 If efficiency of a carnot’s engine operating between same temperature T T − T2 η=1− 2 = 1 T1 T1 or

T1 1 = η T1 − T2

or

T1 1 −1= −1 η T1 − T2

or

1 − η T1 − T2 + T2 = η T1 − T2 T2 =k T1 − T2

⇒ Thus efficiency of refrigerator

k=

1− η η

6. Liquefaction of Gases The work on the liquefaction of gases started in 1823 when Faraday could performed an experiment to liquefy, chlorine, carbon dioxide, hydrogen sulphide etc. They applied, some pressure at the temperature below room temperature. But same gases as H 2, O2, N 2 and He etc. could not be liquefied even by applying high pressure. These gases are called permanent gases. An experiment in 1862 has been carried out by Andrew for liquefaction of CO2. According to this experiment a gas must be cooled below critical temperature before being liquefied the gas under pressure. It resulted in the liquefaction of oxygen in 1877 by picket. But the real process has been developed by Joule-Kelvin effect and regenerative cooling. Linde and Hampson developed their liquefier on commercial basis. Hydrogen and Helium were also liquefied on same principle.

7. Hampson’s and Linde’s Regenerative Cooling Machine 7.1 Principle of Regenerative Cooling This principle is based upon Joule Thomson effect i. e. when a gas having temperature less than inversion temperature passess through porous plug from high pressure side to low pressure side then gas shows the cooling effect. Hence, first of all we cooled the gas by using Joule-Thomson effect and then we pass again in porous plug by compressing it. Hence, if we repeate this up to many times then gas will becomes so much cool and becomes liquified after passing porous plug this is called regenerative cooling. This process of regenerative cooling is first developed in 1857 by Simens but it is used by Linde and Hampson in 1895.

H-135

7.2 Apparatus and Method First of all air is made to passes through KOH and KCl because of making it free from CO2 and water vapour. This is necessary because if this impurity is mixed with air are not removed than these impurities becames solid before liquified the air and these solid particle will oppose the flow of air flow. After this dried air is made to flow in compression pump P1. Here, the pressure of gas becomes 40 atmospheric pressure.

CO2 Free dried air 1 Atomospheric P1 40 Atomospheric P2

150 Atomospheric

A

B

– 183°C

Ammonia flux

– 188°C

C

Liquified air

D

Fig. 10

Now, air moves towards second compressor pump P2 where its pressure increases up to 150 atmospheric pressure. Due to compression there is some increasing in temperature occurs now passing through this air in a

H-136 spiral tube which is immersed in ammonia then gas becomes cooled. Now, this compressed gas (150 atm) passes through spiral tube ‘B’ and this will reach to N. But either side of N is at 40 atmospheric pressure then Thomson expansion takes place and hence air temperature becomes more low. Now, this air will enter in pump P2 at 40 atmospheric pressure now compressed upto 150 atmosphere and comes again in cycle now due to again and again. This process when air temperature becomes −188°C then gas becomes liquified at 40 atmospheric. Now, cooled this liquid air into chamber C and then in dewar flaks D.

8. Hampson’s Liquefier This is also based upon regenerative cooling. The basic difference between Linde’s liquified is that it has many fold copper spiral tube B, It has high efficiency than Linde’s liquefier. It can liquefy 5 litre air in one hour while Linde’s liquefier only 1 litre in one hour but Linde’s liquefier is low in cost. So, highly used in industries.

9. Method of Producing Liquefying Hydrogen Dewar liquefying hydrogen in 1898 by the process of regenerative cooling. This apparatus is developed by Travers, Olszewski, Nernst and others.

9.1 Basic Principle Hydrogen could not be liquefied for a long time. Its critical temperature is −240°C. No independant cooling agent can cool it to below this temperature. Hence, the cascade process cannot be employed here. The discovary at Joule-Thomson effect opened a new field for liquefication of gases. If we passes a gas through a porous plug then gas will show cooling effect. If the temperature of gas before entering is lower to inversion temperature while if gas temperature B is higher than inversion temperature then gas will show heating effect. The inversion temperature for hydrogen is −80°C.

9.2 Hydrogen Liquefier An experimental set up for producing liquid hydrogen is shown in figure. Pure hydrogen free from dust, CO2 and water vapour is compressed to 150 atmospheric pressure by using compressor, the heat of compression being removed by external cooled bath. Now, this cooled and compressed gas is sent into a spiral A immersed in liquid air where it is cooled to −170°C. It is then pass in spiral B the temperature here now reduce to −208°C. Now, gas will pass into C from which it expands through the nozzle N to a pressure of 1 atmosphere and suffers J-K cooling. This expanded and cooled gas flows back over the spiral C and round the outside of the liquid air vessels. Thus cools the incoming gas and return to compressure. And finally hydrogen becomes liquified at −253°C at 1 atomsphere pressure. Now, hydrogen is connected in dewar flasks. F, H 2 should be free from O2, N 2 and other impurities.

H-137 Compressor 1 atm

150 atm

A

Liquid air

– 170°C

To suction

B

Liquid air at – 208°C

C

1 atm – 253°C

Liquid hydrogen (otherwise this will solidify before liquefication) Fig. 11

H-138

10. Liquefaction of Helium Helium was the last gas to be liquefied. All attempts made before 1908 to liquefy helium failed. The reason is obvious. The critical temperature of helium is as low as −268°C. Hence, it was necessary to cool it below −268°C before it could be liquefied. In the system of cascade the lowest attainable temperature was −218°C. Hence, this method was not applicable to liquefy helium. Claude’s method could also not be used to liquefy helium because there was no lubricant for the expansion engine which could resist solidification at such low temperatures. Prof. K. Onnes, in 1908, pointed out that the temperature of inversion for helium was −240°C and it was possible to cool it below this temperature by means of liquid hydrogen evaporating under reduced pressure. He actually developed a helium liquefier in which he pre-cooled helium to −258°C and subjected it to Joule-Kelvin expansion and obtained liquid helium.

10.1 Kapitza’s Helium Liquefier Kapitza, in 1934, developed another helium liquefier using Claude’s method in which both the adiabatic expansion and the Joule-Kelvin effect are utilised. He used a special expansion engine in which nolubricant was used. Kapitza’s arrangement is shown in fig. 12 which the spirals carrying the helium gas are indicated by lines. The pure gas is first compressed to 40 atm and then passed through the heat-exchanger A. It is then made a flow round a ring shaped vessel NN filled with liquid nitrogen under reduced pressure, where it is cooled to −208°C, from there it passes through the heat exchanger B. At a point K, the gas is divided into two parts : 8% of the gas passes through an expansion value V1, and the remaining 92% to an expansion engine E which cools it to −263°C. This cooled part follows back through the heat exchanges C, B, A and cools the incoming gas before returning to the compressor at 1 atm. The gas which has passed through V1 undergoes Joule-Kelvin expansion after which it follows through the heat exchanges C and D and undergoes a second Joule-Kelvin expansion through the value V2 when its temperature falls to −268.8°C. It is now partly liquefied and collected in the flask F. The gas which has not yet liquified flows back through the heat exchanges D, C, B, A and returns to the compressor.

10.2 Solidification of Helium

Compressor Helium 1 atm

F – 268°C Fig. 12 : Kapitza’s Apparatus

Critical temperature of helium is very low in comparison to all gases, it is −268°C. This was only reason that helium could be liquefied in last. Liquid helium is colourless, transparent and easily vaporized liquid. Its melting point is very low 4.2 K at Atmospheric pressure. It has no freezing point. Helium has such kind of categories that it has different nature with other materials. It is important to mention here for helium that it has an extra ordinary feature. It could not becomes solid at any lowest liquid helium up to 0.8 K but they could not comes into solid state. But Keesom now applied some high pressure on the helium and they found the helium becomes solid. At 4.2 K and 23 atmospheric pressure and temperature and at 1.1 K temperature and at 23 atmospheric pressure the helium becomes solid. Hence, for making helium solid we not only decrease the temperature but also need to increase pressure.

H-139

11. Relation between Boyle Temperature, Inversion Temperature and Critical Temperature We have inversion temperature Ti =

2a Rb

…(1)

Boyle’s temperature a Rb

…(2)

TC =

8a 27 Rb

…(3)

Ti = TC

2a Rb = 27 8a 4 27 Rb

TB = Critical temperature

Using equation (1) and equation (3)

Ti = 675 . TC

…(4)

Ti = 2TB

…(5)

Using equation (1) and equation (2) We get

Ti = 6.75 TC = 2TB Ti > TB > TC

So,

Example 1. A refrigerator works between temperature 300 K and 400 K. Calculate (i) thermal efficiency (ii) coefficient of performance. T 300 Solution : (i) η=1− 2 =1− = 0.25 or 25% T1 400 (ii)

Coefficient performance Q T2 300 k= 2 = = =3 W T1 − T2 400 − 300

Example 2. If coefficient performance k = 3 calculate the power for refrigerating effect of 100 watt if the refrigerator works (i) 100% coefficent (ii) 60% coefficient. Solution : Given k = 3 (i)

for first case kmax = 3

H-140 k= (ii)

Q2 100 = = 33.3 watt 3 W

for second case 60 60 × kmax = × 3 = 1.8 100 100 Q 100 W= 2 = = 55.5 watt 1.8 k k=

Example 3. Show that the expression for cooling in Joule-Thomson effect is, dT =

1  2a  − b( P1 − P2 ) JCP  RT 

Solution : Let a cylinder AB has a porous plug and this is divided into two parts by this porous plug. Let two insulator pistons 1 and 2. Piston 1 sents 1 mole gas at pressure P1 and this gas pushes the Piston 2 with pressure. P2 Porous plug A

B

1

P1

P2

V1

V2 P Initial stage

2

Final stage

Fig. 13

If V1 is initial volume of gas at pressure P1 and V2 is the final volume of gas at pressure P2 work done on gas by piston 1 = P1V1 Work done on piston 1 by gas = P2V2 Net work done on piston 2 by gas = P2V2 − P1V1 Now, workdone from changing volume V1 to V2 =

V2

a

1

V2

∫V

dV =

a a − V1 V2

Now, total workdone by gas a a W = (P2V2 − P1V1) +  −   V1 V2  Using van der Waal’s equation a   P + 2  (V − b) = RT  V  PV +

ab a − Pb − 2 = RT V V

…(1)

H-141 PV = RT + Pb −

or

a V

 ab   V 2 is taken as negligible

 a P1V1 = RT + P1b −    V1 

So,

 a P2V2 = RT + P2B −    V2  a a P2V2 − P1V1 = (P2 − P1)b +  −   V1 V2  Putting this value into equation (1) a a W = (P2 − P1)b + 2 −   V1 V2 

We get, a a and are small quantities. V1 V2 Thus we can take

aP a aP1 a and = = 2 by using PV = RT r1 RT V2 RT W = b(P2 − P1) +

So,

2a (P1 − P2) RT

  2a = (P1 − P2) − b   RT If deficiency of temperature is −dT W = − JC P dT

So,

  2a − JC P dT = (P1 − P2)  − b   RT

i. e.

dT = −

1  2a  − b (P1 − P2)    JC P RT

This is the expression for cooling. Example 4. Calculate the power for refrigerating of 100 watt if the refrigenrator works at (i) 100% coefficient (ii) 60% coefficient. Solution : For exmple 1, (i)

k=3

For 100% coefficient, β actual = βmax = 3 But

k=

Q2 W

Therefore the work done (power required) for refrigerating effect of 100 watt. Q2 W= k actual =

100 = 33.3 watt 2

H-142 (ii)

60 60 × kmax = × 3 = 1.8 100 100 Q2 100 W= = = 55.5 watt k actual 1.8 k=

For 60% coefficient

Example 5. For one mole of hydrogen the van der Waal’s constants are, a = 0.245 ×

Litre2 × atoms mole2

b = 2.67 × 10 −2 Litre mole −1 and calculate inversion temperature. Given that R = 2 cal mole −1 K −1 and 1 atmosphere pressure = 10 6 dyne /cm 2 Solution : The temperature of inversion Ti is given by Ti =

2a Rb

a = 0.245 litre 2 atoms/mole 2

Here,

= 0.245 × 106 × 106 cm 4 dynes/mole 2 = 0.245 × 1012 cm 4 dyne/mole 2 b = 2.67 × 10−2 litre/mole = 26.7 cm 3/mole R = 2 cal/mole K = 2 × 4.2 × 107 erg/mole K Hence,

Ti =

2 × 0.245 × 1012 2 × 4.2 × 107 × 26.7

= 218.4 K

Example 6. A refrigerator works under a reversible cylce between the temperature 300 K and 400 K calculate (i) the thermal efficiency (ii) the coefficient of performance. Solution : (i) Thermal efficiency η=1− (ii)

T2 300 =1− = 0.25 or 25% T1 400

The coefficient of performance K=

Q2 T2 300 = = =3 W T1 − T2 400 − 300

Example 7. Calculate the least amount of work, expressed in kilowatt hours, necessary to convert 50 g of water at 30°C, into ice at −10 ° C using reservoir of cooling water at 30° C. Let that the pressure remains constant throughout the process given CP of ice = 0.5 K cal/kg °C and latent heat of ice = 3.4 × 10 5 Joule/kg. Solution : T1 = 30° = (273 + 30 ) = 303 K

H-143 T2 = − 10° C = 273 − 10 = 263 m = 50 g = 0.05 kg C P = 0.5 kilo, cal/kg °C and the quantity of heat extracted from the reservoir Q2 = 0.05 × 30 + 0.05 ×

3.4 × 105 4.2 × 103

+ 0.05 × 0.5 × 10 kilo cal

= 1.5 + 4 + 2.5 kilo cal = 5.75 kilo calories For ideal refrigerator, the coefficient of performance Q T2 K= 2 = W T1 − T2 Thus the required work W = Q2 ⋅

T1 − T2 303 − 263 40 kilo calories . × = 5.750 × = 575 T2 263 263

. × = 575

40 × 4.2 × 103 J or watt sec 263

= 5.75 ×

40 4.2 × 103 KWh × 263 3600 × 103

(‡ 1 kilo cal = 4.2 × 103 J)

= 1.018 × 10−3 KWh Example 8. Calculate the critical temperature and Boyle’s temperature of CO 2 for which van der Waal’s constants are, a = 0.0072 and b = 0.002 at NTP P = 1 and V = 1. Solution : (i) From van der Waal’s equation we have a   P + 2  (V − b) = RT  V  at NTP, P = 1 and V = 1 (given) 0.0072   (1 − 0.002) = R × 273 1 +  12  This gives,

R=

1.0072 × 0.998 273

The critical temperature is given by TC = =

8a 27 Rb 8 × 0.072  1.0072 ×0.998 27 ×   × 0.002   273

= 289.6 K = (289.6 − 273)°C = 16.6°C

H-144 (ii)

The Boyle temperature TB in terms of critical temperature is given by So,

Ti = 2TB = 675 . TC 6.75 6.75 TB = TC = × 289.6 K 2 2 = 977.4 K = (977.4 − 273) °C = 704.4 ° C

Example 9. Enthalpies of a certain gas before and after suffering Joule-Thomson expansion through a throttle value are 77.2 cal and 106.8 cals respectively. Deduce the fraction of the gas liquified if the enthalpy of emerging liquid be 55.4 calories. Solution : The fraction of the gas liquified in a regenerative machine is given by H f − He x= H f − HL H e = 77.2 cal, H f = 106.8 cal and H L = 55.4 calories 106.8 − 772 x= 106.8 − 55.4 29.6 x= 51.5 x = 0.57 Example 10. Helium gas expanded by Joule-Thomson’s expansion at −173 °C when the pressure difference is 20 atm at porous plug. For gas CP = 5 cal/mole K a = 0.0341 litre 2 -atm/mole 2 , b = 0.0237 litre/mole. Calculate the temperature change if R = 8.31 J/mole-K, J = 4.18 J/cal and 1 atmospheric pressure = 1.01 × 10 5 N/m 2 Solution :

1  2a dT  =− − b   dP JC P  RT

or

dT = −

Given that

1  2a  − b (P1 − P2)   JC P  RT

a = 0.0341 litre 2 atm/mole 2 = 0.0341 × (10−3 m 3)2(1.01 × 105 N/ m 2)/ mole2 = 3.44 × 10−3 Nm 4 /mole 2 b = 0.0237 litre/mole = 0.0237 × (10−3 m 3)/mole = 2.37 × 10−5 m 3/mole T = − 173 + 273 = 100 K 2a 2 × 3.44 × 10−3 = = 8.26 × 10−6 m 3/mole RT 8.31 × 100 2a − b = 8.28 × 10−6 − 23.7 × 10−6 RT = −15.4 × 10−6 m 3/mole

H-145 P1 − P2 = 20 atmosphere = 20 × (1.01 × 105) = 20.2 × 105 N/m 2 JC P = 4.18

calorie joule = 20.9 J/mole-K × 5.0 × mole-K calorie

  2a Putting  − b and (P1 − P2) and JC P into equation (1)   RT dT =

−15.4 × 10−6 × 20 .2 × 105 20.9

= +1.49 K Hence, helium temperature will increase. Example 11. Calculate the temperature of oxygen after expanding at pressure difference of 150 atmosphere. Given that CP = 7.0 cal/mole-K, a = 1.32 litre 2 -atmosphere/mole 2 , b = 3.12 × 10 −2 litre/mole. Initial temperature = 27°C and R = 8.31 × 10 −2 litre atm/mole

Solution :

  2a − − b (P1 − P2)   RT dT = JC P 2 × 1.32 2a = RT 8.31 × 10−2 × 300 K = 10.6 × 10−2 litre/mole 2a − b = 10.6 × 10−2 − 3.12 × 10−2 = 7.48 × 10−2 litre/mole RT = 7.48 × 10−2 × 10−3 m 3/mole = 7.48 × 10−5 m 3/mole P1 − P2 = 150 atm = 150 × (1.01 × 105) N/m 2 = 152 × 105 N/m 2 JC P = 4.18

Putting

cal joule ×7 = 29.3 joule/mole-K mole-K cal

2a − b, (P1 − P2) and JC P into equation (1), RT dT =

−7.48 × 10−5 × 152 × 105 = − 38.8°C 29.3

So, temperature of oxygen after expansion = 27 − 38.8 °C = − 11.8 °C Example 12. Calculate the inversion temperature helium Given that a = 0.0341 atm-litre 2 /mole 2 , b = 0.02137 litre/mole and R = 8.31 joule/mole-K Solution : We have inversion temperature,

H-146 T1 =

2a Rb

a = 0.0341 atm-litre 2/mole 2 = 0.0341 × (10−3 m 3)2 × (1.01 × 105 N/m 2) mole 2 = 3.44 × 10−3 Nm 4 /mole 2 b = 0.0237 litre/mole = 2.37 × (10−3 m 3)/mole Ti =

2 × 3.44 × 10−3 8.31 × 2.37 × 10−5

= 35 K = − 238°C

Example 13. Calculate the temperature produced by adiabatic throttling process in case of O 2 when the pressure is reduced by 50 atmosphere. Temperature of gas is 27°C, a = 1.32 × 1012 cm 4 dyne/mole 2 , b = 31.2 cm 3 /mole-K and CP = 7 cal/mole-K Solution :

1  2a dT  = − b dP C P  RT  dT =

dP  2a  − b  C P  RT 

dP = 50 atmosphere = 50 × 76 × 13.6 × 980 dynes/cm 2 C P = 7 cal/mole-K = 7 × 4.2 × 107 ergs/mole-K dT =

50 × 76 × 13.6 × 980 7 × 4.2 × 107

 2 × 1.32 × 1012  − 31.2 × 7  8.31 × 10 × 300 

dT = 4.935 K Example 14. For a gas a = 0.1328 N-m 4 /mole 2 and b = 32 × 10 −6 m 3 /mole then calculate the inversion temperature R = 8.31 (joule/mole-K). 2a 2 × 0.1328 Solution : Ti = = 999 K = 726°C = Rb 8.31 × 32 × 10−6 Example 15. A carnot refrigerator is working between 260 K and 300K. If it takes 500 calories heat from low temperature side calculate the heat given to high temperature. Also calculate the heat in one cycle of refrigerator. Solution : We have Q1 T1 = Q2 T2 Here,

Q2 = 500 cal T1 = 300 K T2 = 260 K

H-147 Q1 =

500 × 300 260

Q1 = 577 cal Thus heat in a complete cycle = Q1 − Q2 = 577 − 500 = 77 cal-K Example 16. A refrigerator operates on a reversed Carnot cycle whose coefficient of performance is 5. The evaporator is maintained at a temperature of −6 °C and power required to run the refrigerator is 3.5 kW. Calculate the refrigerating effect and the condenser temperature of the refrigerator. Solution : Given T2 = − 6 °C = 273 + L − b = 267 K, coefficient of performance k = 5 We have

k=

T2 T1 − T2

5=

267 T1 − 267

So, condenser temperature T1 = 267 +

267 5

T1 = 320.4 K Q2 Q k= = 2 Q1 − Q2 W

Again

Q2 = W × k = 3.5 × 5 = 17.5 kW Example 17. A domestic food refrigerator maintains a temperature of −10 °C, the air temperature is −30 °C the heat leakage into the freezer is estimated to be at the continuous rate of 2 kJ/s. Calculate least power needed to pump out this heat continuously. Solution : Given Q2 = 2 kJ/sec We have or

k=

T2 T1 − T2

T1 − T2 Q1 − Q2 = T2 Q2 Q1 =

and or

k=

Q2 Q2 = W Q1 − Q2

T1 Q1 = T2 Q2

T1 303 × Q2 = × 2 = 2.304 kJ/sec T2 263

W = Q1 − Q2 = 2.304 − 2 = 0.304 kJ/sec Example 18. A refrigerator maintained temperature −5 °C the ambient temperature is 25 °C and the heat transfer rate is 5 kW. Calculate the performance index and minimum power required to operate the refrigeration plant. Solution :

T1 = 25° C = 273 + 25 = 298 K T2 = − 5° C = 273 − 5 = 268 K

H-148 k=

T2 268 = = 8.93 T1 − T2 298 − 268

or

k=

Q2 Q1 − Q2

∴

Q1 =

So,

W = Q1 − Q2 = 5.56 − 5 = 0.56 kW

or

8.93 =

5 Q1 − 5

49.65 = 5.56 kW 8.93

Example 19. A compressor delivers 4 m 3 of the free air per minute at a pressure of 7 bar gauge. Assuming that the compression follows the law pV 1.3 = constant, Calculate work done. Given data n is the polytropic = 1.3 for air, V1 = 4 m 3 , p1 = 1 bar (absolute) and p2 = 7 bar (gauge) = 8 bar (absoulte) p1V11. 3 = p2V21. 3

Solution :

p  V2 = V1  1   p2   1 V2 = 4    8

1. 3

1. 3

= 0.807 m 3/min

Work done during compression =

n ( p2V2 − p1V1) n −1

Work done during compression =

1.3 (8 × 0.807 − 1 × 3) × 105 0.3

On solving we get workdone = 14.98 × 105 Nm/minute = 0.2496 × 105 Nm/s = 0.25 × 105 Watts = 25 kW Example 20. A single stage air compressor running at 80 RPM, compress air from a pressure of 1 bar and temperature of 15 °C to a pressure of 5 bar (see figure 13). The clearance volume is 5% of swept volume which is 0.42 m 3 . Assuming that the compression and expansion to follows the law pV 1.3 = constant, determine the power required to drive the compressor. 5 bar

p 1 bar N v Fig. 14

Given data n is the polytropic = 1.3 for air, N = 80 RPM, p2 = 5 bar (absolute) p1 = 1 bar (babsolute) and T1 = 15 °C.

H-149 V  5 = 0.05 k =  c = 100 V  s

Solution :

Vs = 0.42 m 3 Volumetric efficiency referred to the suction conditions, ηvolumetric

ηvolumetric

1 1    n   p  5 1.4   2  = 0.8775 = 1 + k −   = 1 + 0.05 − 0.05    1   p1        Actual volume of air taken referred to free air conditions = Swept volume of the compressor

0.8775 =

V1 0.42

V1 = 0.3685 m 3/cycle Mass of air =

p1V1 0.3685 × 1 × 100 = 0.287 × 288 RT1

= 0.445 kg/cycle 0.445 × 80 Mass of air = = 0.5933 kg/sec 60 p  T2 = T1  2   p1 

n−1 n

0.3

= 288(5)1.3 = 417.53 K

Power =

n mR(T2 − T1) n −1

Power =

1.3 × 0.5933 × 0.287 × (417.53 − 288) = 95.5 kW 1.3 − 1

1.

In example-I calculate the power for refrigerating effect of 100 watt if the refrigerator works at (i) 100% coefficient, (ii) 60% coefficient.

2.

For one mole of hydrogen, the van der Waal’s constants are, a = 0.245

litre 2 × atoms mole 2

, b = 2.67 × 10−2 litre mole −1 .

Calculate the temperature of inversion. Given that R = 2 cal mole −1 K −1 and 1 atomosphere pressure = 106 dyne/cm 2 . 3.

A refrigerator works under a reversible cycle between the temperatures 300 K and 400 K. Calculate (i) the thermal efficiency, (ii) the coefficient of performance.

4.

Calculate the least amount of work expressed in kilowatt hours, necessary to convert 50 g of water at 30°C into ice at −10°C using a reservoir of cooling water at 30°C. Assume that the pressure remains constant throughout the process. Given CP of ice = 0.5 kilo cal/kg°C and latent heat of ice = 3.4 × 105 Joule/kg m.

5.

Calculate the critical temperature and Boyle temperature of CO2 for which the van der Waal’s constant are given to be a = 0.0072 and b = 0.002 . The unit of pressure is atmosphere and the unit of value is that of a gram mole of the gas at NTP.

H-150 6.

The enthalpies of a certain gas before and after suffering Joule-Thomson expansion though a throttle value are 77.2 cals. and 106.8 cals respectively. Deduce the fraction of the gas liquefied if the enthalpy of emerging liquid be 55.4 calories.

7.

Helium gas expanded by Joule-Thomson’s expansion at −173°C when pressure difference is 20 atmosphere at porous plug. Given that CP = 5.0 cal/mole-K, a = 0.0341 litre 2 -atmosphere/mole 2 , b = 0.0237 litre/mole, R = 8.31 joule/mole-K, J = 418 . joule/cal and atmospheric pressure = 1.01 × 105 N/m 2 . Calculate the temperature change.

8.

Calculate the temperature of oxygen after expanding at pressure difference of 150 atmosphere. Given that CP = 7.0 cal/mole-K, a = 1.32 litre 2 -atmosphere/mole 2 , b = 312 . × 10−2 litre/mole, Initial temperature = 27 °C and R = 8.31 × 10−2 litre-atmosphere/mole-K.

9.

Calculate the inversion temperature of helium. Given that a = 0.0341 atmosphere-litre 2 /mole 2 , b = 0.0237

10.

Calculate the temperature produced by adiabatic throttling process in case of oxygen when the pressure is reduced by 50 atmosphere. Given, a = 1.32 × 1012 cm 4 dynes/mole 2 , b = 31.2 cm 3 /mole and CP = 7 Cal/mole-K.

11.

For a gas a = 01328 N-m 4 /mole 2 , b = 32 × 10−6 m 3 /mole. Calculate the inversion temperature if R = 8.3 .

litre/mole and R = 8.31 joule/mole-K.

joule/mole-K. 12.

A refrigerator works between temperature 300 K and 400 K. Calculate (i) Thermal efficiency, (ii) Coefficient of performance.

13.

A refrigerator works between temperature 1000 K and 2000 K. Calculate its coefficient of performance.

14.

A temperature operates on a reversed carnot cycle whose coefficient of performance is 5.0. The evaporator is maintained at a temperature of −6°C and power required to run the refrigerator is 3.5 kW. Calculate the refrigerating effect and the condenser temperature of the refrigerator.

15.

A refrigerator has coefficient of performance is 7.0 the required to operate the refrigerator is 7 kW. Calculate the refrigerating effect and condenser temperature.

16.

If a refrigerator maintains temperature −10 °C the temperature is +30 °C the heat power needed to pump out this head continuously.

17.

A refrigerator maintaining temperature −20 °C, the air temperature is + 40 °C, the heat leakage into the freezer is estimated to be at the continuous rate of 5 kJ/sec. Calculate the least power needed to pump out this heat continuously.

18.

A refrigerator maintained temperature −5 °C the ambient temperature is 25°C the heat transfer is 4 kW. Calculate the performance index and minimum power required to operate the refrigeration plant.

19.

A refrigerator maintained temperature −10 °C, the ambient temperature is 40°C the heat transfer rate is 15 kW. Calculate the performance index and minimum power requred to operate the refrigeratiion plant.

20.

A refrigerator operates between temperature −10 °C and 30°C. Calculate the coefficient of performance.

21.

Calculate the coefficient of performance of a refrigerator working between ice point and room temperature.

H-151

Long Answer Type Questions

6.

What are temperature of a gas inversion and Boyle temperature ?

1.

Write an essay on production of low temperature.

2.

Explain regenerative cooling and show how it can be 7. 8. used to liquefy hydrogen.

3.

Give an account of liquefaction of gases. Explain regenerative cooling method.

4.

Write an essay on the liquefaction of gases.

5.

Explain cooling due to adiabatic expansion of a gas. Explain the theory of Joule-Thomson regenerative 10. cooling. How can you obtain liquid helium by its application ? Very Short Answer Type Questions

6.

What is a refrigerator? Describe vapour compression 1. machine and deduce its coefficient of performance. 2. Show that Joule-Kelvin coefficient of ideal gas is 3. zero. 4. Discuss the Joule-Thomson effect and derive and expression for Joule-Thomson cooling. 5. 1  2a dT = − − b (P1 − P2 )  6.  JCP  RT 7. Show that enthalphy remains constant in the Joule-Kelvin effect and derive and expression for 8. Joule-Kelvin coefficient for van der Waal’s gas.

7. 8.

9.

10.

What is adiabatic demagnetization ?

11.

Discuss Hampson’s regenerative cooling machine.

12. 13. 14.

9.

9. How will you liquify Hydrogen gas ? Discuss in detial. 10. Which gas is liquified at last ? Discuss its liquefaction 11. in detail. Discuss the liquefaction and solidification of helium 12. gas. 13.

Short Answer Type Questions 1. 2.

3.

14. Explain if it is possible to liquefy a perfect gas using 15. the Joule-Thomson effect. 16. Deduce relationship between the efficiency of heat engine and coefficient of performance of a 17. refrigerator. What is refrigerator ? Explain its working principle 18. using refrigeration cycle.

4. 5.

19.

What is the importance of Joule-Thomson effect 20. from the point of view of liquefaction of gases ? What is the most essential condition for liquefaction 21. of gas ?

Explain the regenerative cooling method. What do you understand by inversion temperature in relation to Joule-Thomson effect ? Obtain its expression. Discuss the process of liquefaction of gases.

What is the value of inversion temperature ? Write two methods of cooling of gas. What is the value of 0°F in 0°C ? What is the value of Joule-Kelvin coefficient for ideal gas ? Which method is called adiabatic throttling ? Give the name of gases shows the heating effect. Which quantity remains constant under Joule-Thomson’s effect? 2a If Ti < for a gas the gas will show cooling or Rb heating ? Which branch of physics deals at very low temperature ? What is the value of Boyle’s temperature ? What is the value coefficient of performance of refrigerator ? Which gas is used for domestic refrigeration ? What is the relation between coefficient of performance and efficiency of heat engine? Which gas was last gas to be liquefied ? What was first commercial liquefier ? On what principle of regenerative cooling is based upon? Who developed the concept of regenerative cooling ? Which things you use to remove CO2 and water ? Who developed hydrogen ?

the

apparatus

for

liquefied

What is the relation between inversion temperature and critical temperature ? What is the relation among T1 , TB and TC ?

H-152 9.

Physics of production and use of low temperature is called : (a) cryogenics (b) refrigeration (c) regelation (d) freezing

10.

The ratio of coefficient of viscosities of liquid HeII and He I is : (a) 10−2 (b) 10−1 −3 (c) 10 (d) 103

11.

Which of the following statements is wrong for liquid He II? (a) The viscosity of He II is almost zero (b) Thermal conductivity of He II is very low (c) Density of He II is almost the same as that of He I (d) Flow of liquid He II through capillaries is independent of pressure difference.

12.

Solid helium : (a) can be obtained merely by lowering the temperature of the liquid (b) melts to form liquid (c) melts to form gas (d) can not exist above the critical temperature

13.

The (a) (b) (c) (d)

14.

Cryogenics is the : (a) physics of high temperature (b) physics of low temperature (c) physics of very low temperature (d) none of the above

Objective Type Questions Multiple Choice Questions 1.

2.

3.

The last gas to be liquefied was : (a) oxygen

(b) hydrogen

(c) nitrogen

(d) helium

If Ti is temperature of inversion, TB is Boyle temperature and TC is critical temperature, then : (a) Ti < TB < TC

(b) TC > TB < Ti

(c) TB > Ti > TC

(d) Ti = 2TB = 675 . TC

The Boyle temperature TB is given by the equation : 27 (a) TB =   TC  8 a (b) TB = bR (c) both the above (d) none of the above

4.

The critical temperature of a gas is 31°C and the room temperature is 27°C. Then the gas in the room : (a) can be liquefied (b) cannot be liquefied (c) will be solidified (d) will break into electrons

5.

The temperature of inversion (Ti) of a gas is : a 2a (b) (a) Rb Rb b 2b (d) (c) Rb Ra

6.

The coefficient of performance of a refrigerator working between ice point and room temperature 30°C is : (a) 2 (b) 9 (c) 10 (d) 6

7.

The (a) (b) (c) (d)

8.

second virial coefficient B : varies in a different manner for different gas is positive at low temperature is zero at Boyle temperature is higher than the first virial coefficient

Fill in the Blank(s) 1.

The minimum lowest temperature is ………

2.

Joule-Thomson effect is a method of ……… of gas.

The critical temperature of hydrogen is the same as 3. the temperature of inversion of helium. The temperature is : 4. (a) −118°C

(b) −240°C

(c) −268°C

(d) −273°C

viscosity of liquid He II is : greater than the viscosity of liquid He I negligible very large infinite

5.

……… is the minimum temperature of Fahrenheit scale. 2a If > b then gas will show ……… RT Joule-Kelvin coefficient for ideal gas is ………

H-153 6.

Hydrogen and helium gas shows ……… effect after passing through porous plug.

7.

Relation between inversion temperature is ……… .

8. 9. 10.

True/False 1.

2. Helium gas can becomes solid if gas temperature is 3. ……… and pressure is ……… . The efficiency of ……… air liquefier is greater than 4. ……… liquefier. 5. Linde’s air liquefier is based upon ……… .

Minimum possible temperature is −100 °C. Regenerative cooling is first given by Simens. Hydrogen and helium gases shows heating effect after passing porous plug. Helium gas can becomes solid at 1.1 K temperature. Relation between inversion temperature and Boyle’s temperature Ti = 2TB .

H-154

Objective Type Questions Multiple Choice Questions 1.

(d)

2.

(d)

3.

(c)

4.

(a)

5.

(b)

6.

(b)

7.

(c)

8.

(b)

9.

(a)

10.

(c)

11.

(b)

12.

(c)

13.

(b)

14.

(c)

Fill in the Blank(s) 1.

− 273.15 ° C

2.

cooling

3.

− 17.222° C

4.

cooling

6.

heating

7.

T1 = 2TB

8.

1.1 K and 25° C

9.

Hampson, Linde

2.

True

5. 10.

zero Thomson’s effect

True/False 1.

False

3.

True

4.

True

5.

True

H-155

H ints and Solutions Numerical Questions 1.

T2

k=

T1 − T2

See example-4

261 T1 − 261

7=

Ans. (i) W = 33.3 watts (ii) W = 55.5 watts 2.

T1 = 261 +

See example 5. Ans. Ti = 200 K

3.

Condensor temperatureT1 = 298.2 K Q2 Q = 2 k= Q1 − Q2 W

See example-6 Ans. (i) 25%, (ii) 3

4.

See example-7 Ans. 1.018 × 10−3 kWh

5. 6.

Q2 = W × k = 7 × 7 = 49 kW 16.

See example Ans. TC = 16.6 °C and TB = 664.4°C

17.

See example-9

See example-11 or or

Ans. −238 °C 10.

See example-13

∴

Ans. 4.935 °C 11.

T2 T1 − T2

See example-12

See example-14

13.

Ans. 25% and 3 T2 k= T1 − T2 k=

1000 1000 = 2000 − 1000 1000

k =1 See example-16. Ans. Ti = 320.4 K Q2 = 17.5 kW 15.

T2 = − 12°C = 273 − 12 = 261 K Given k = 7.0

= =

W

=

Q2 Q1 − Q2

Q1 − Q2 Q2 Q1 Q2

Q1 = Q2 ×

T1 T2

= 6×

313 253

= 7.422 kJ/sec

Ans. 726°C

14.

T1 T2

See example-14

12.

Q2

k=

and

Ans. −11.8 °C 9.

Given Q2 = 6 kJ/sec T2 k= T1 − T2

See example-10 Ans. +1.49 K

8.

See example 17. Ans. 0.304 kJ/sec

Ans. x = 0.57 7.

261 7

W = Q1 − Q2 = 7.422 − 6 = 1.422 kJ/sec 18.

See example-18. Ans. 0.56 kW

19.

T1 = 40°C + 273 = 313 K T2 = − 10 °C + 273 = 263 K T2 263 = = 7.26 k= T1 − T2 313 − 263 k=

Q2 Q1 − Q2 7.26 =

15 Q1 − 15

H-156 7.26Q1 − 15 × 7.26 = 15 7.26Q1 = 15 + 108.9

9.

See-6

10.

See-3

7.26Q1 = 123.9 Q1 = 17.06 kW

Very Short Answer Type Questions

W = Q1 − Q2

1.

Ti=

2.

(i) Joule-Thomson effect (ii) Adiobatic Expansion

3.

−18° C

4.

Zero

5.

Joule-Thomson’s effect

6.

Hydrogen and Helium

7.

Enthalpy

8.

Cooling

9.

Cryogenics a TB = Rb T2 K= T1 − T2

= 17.06 − 15 kW W = 2.06 kW 20.

T1 = 30° C + 273 = 303 K T2 = − 10°C + 273 = 263 T2 263 = k= T1 − T2 303 − 263 k = 6.575

21.

T2 = 0 ° C + 273 = 273 T1 = 30 ° C + 273 = 303 T2 273 = k= T1 − T2 303 − 273

10. 11.

k=9 12.

Short Answer Type Questions 1.

13. No, we cannot liquefly ideal gas by Joule-Kelvin 14. effect. See-2.2

2.

See-5.4

3. 4.

2a Rb

Freon gas 1− η K= η Helium

15.

Hampson’s liquefier

See-5

16.

Joule-Thomson effect

See-2.4

17.

Simens in 1857

5.

See-6

18.

KOH and KCl

6.

See-11

19.

Travers and Nernest

See-7

20.

Ti = 6.75TC

See-2

21.

Ti > TB > TC

7. 8.

mmm

Book-2 Physical Optics Unit-1 : Interference Unit-2 : Diffraction Unit-3 : Polarization Unit-4 : Associated Optical Instruments

P-3

Unit

1

Interference

1. Interference of Light Waves When two light waves of same frequency travel simultaneously in the same direction then, due to their superposition, the resultant intensity of light at any point in space is different from the sum of intensities of two waves. At some points the resultant intensity is maximum while at some other points it is minimum (nearly darkness). The redistribution of light intensity due to the superposition of two light waves is called ‘interference of light’. The interference is said to be ‘constructive’ at points where the resultant intensity is maximum and destructive at points where the resultant intensity is minimum or zero.

2. The Principle of Superposition Superposition principle says that when two or more wave motions travelling through a medium and superimpose on another then they loose their individual identity. A new wave is formed in which resultant displacement (y) at any instant is equal to the vector sum of displacements due to individual waves at that instant i.e. →

→

→

→

y = y 1 + y 2 + y 3 + ...

3. Young’s Experiment (Two Slit Interference) In 1801, Sir Thomas Young first demonstrated experimentally the phenomenon of interference of light. His two slit interference arrangement is shown in fig 1. L is a screen in which there is a narrow slit S. In front of this screen there is another screen M which carries two slits S1 and S2 quite close to each other.

P-4

X B (Bright) D (Dark) fringes

B

D S1 Mono chromatic light source

B

S S2

Screen

D B

Double slits Y

Fig. 1 : Young’s Double Slits Experiment (YDSE)

Fig. 2

The slits S1 and S2 are symmetrical with respect to the slit S and at the same distance from S. In front of M there is a third screen N at quite a large distance from M. When a monochromatic light beam is allowed to fall on the slit S, alternately bright and dark bands of equal width and running parallel to the slits S1 and S2 are seen on the screen in fig. 2. In fig. 1, the places of bright bands are shown by B and the places of dark bands by D. These bands are called ‘fringes’ and this array of fringes is called the interferenence pattern of the two slits.

4. Explanation of the Formation of Fringes by Wave Theory A cylindrical wavefront starts from the slit S. When this wavefront reaches the screen M then according to Huygen’s principle, the slits S1 and S2 act as a new light sources and send secondary wavelets in all directions. These wavelets meet each other in the space between the screen M and N. Fig. (1) is drawn continuous and dotted arcs with S1 and S2 as centres.

P-5 The continuous arcs represents the crests of the wavelets and the dotted arcs represents the troughs. At places where a crest of one wavelet meet a crest of the other, or a trough of one meets a trough of the other, the wavelets are in the same phase. Hence, the resultant amplitude at these places is equal to the sum of amplitudes of the two wavelets, and the intensity of light is maximum. In the figure, such places are shown by small circles. The lines joining these circles are called the ‘antinodal lines’. On the other hand, at places where the crests of the wavelets due to one slit meet the troughs of the wavelets due to the other slit, the wavelets are in opposite phases. Hence, the resultant amplitudes at these places is equal to the difference in amplitudes of the two wavelets. If the amplitudes of the wavelets are equal, then the resultant amplitudes of these places is zero and hence, the intensity of light is also zero. In fig .1 such places are shown by black spots. The lines joining these spots are called the ‘natal lines’. Hence, bright and dark bands are seen alternately on the screen N. Since, the interference is a characteristics property of waves, the interference of light supports the wave nature of light. If the slit S is illuminated by white light, fringes are coloured and of unequal widths. If one of slits S1 and S2 is covered, the fringes on the screen disappear. It shows that fringes arises due to interference of light waves from the two slits.

5. Theory of Interference Fringes or Positions of Maxima and Minima in Interference Pattern : Fringe-width (Determination of Wavelength of Light) Let S be a narrow slit illuminated by monochromatic light and S1 and S2 be two parallel narrow slits very close to each other and equidistant from S. The light waves from S reach S1 and S2 in the same phase. According to Huygen’s principle, S1 and S2 acts as new light source from which secondary wavelets originate. These wavelets form interference fringes on a distant screen. On measuring the width of these fringes, the wavelength of light λ can be calculated. P x

S1 θ S

d

(Screen) θ

C

O

A S2 D Fig. 3

Let a bright fringe be formed at the point P on the screen. Let the right bisector CO of the double slite S1S2 meet the screen at the point O and let the distance of point P from O be x. The wavelets starting from S1 and S2 travels different distances S1P and S2P to reach P. Thus, the path difference between them is (S2P − S1P). Let S1 A be a perpendicular drawn on S2P from S1 then, the path difference between the two waves at the point P is (S2P − S1P) = S2 A The ∆ S1S2 A and ∆ PCO are similar. Therefore,

P-6 S2 A OP = S1S2 CP The distance CO is long in comparison to S1S2 and so CP may be taken equal to CO. S2 A OP ∴ = S1S2 CO

∴

S2 A x = d D x ⋅d Path difference, S2 A = D

5.1 Position of Bright Fringes The intensity of light will be minimum at those places where the path difference between the interfering light wave is zero or an integral multiple of λ (0, λ , 2λ , ...). Hence, for maximum intensity (bright fringe), we have x ⋅d (where m = 0, 1, 2, ...) = mλ D Dλ …(1) x=m d Putting m = 0 in this formula, we get the position of the central bright fringe which is formed at O (where the path difference between the two wavelets is zero) and is called the zero order fringe. Putting m = 1, we get the position of the first bright fringe, m = 2 the position of the second bright fringe …, and so on.

5.2 Position of Dark Fringes The intensity of light will be minimum at those places where the path difference between the interfering light λ  λ 3λ 5λ  waves is an odd integral multiple of  , , , ... . Hence, for minimum intensity (dark fringe), we have  22 2 2 1 λ  xd = (2m − 1) = m −  λ , 2  2 D where, m = 1, 2, 3, ... or

1  Dλ  x = m −   2 d

…(2)

Putting m = 1 in this formula, we get the position of the first dark fringe and putting m = 2 the position of second dark fringe ...., and so on. A comparison of equation (2) with equation (1) show that dark fringes are situated in between bright fringes.

5.3 Fringe Width The distance between two consecutive bright or dark fringes is called ‘fringe- width’. Let x m and x m + 1 be the distance of mth and (m + 1)th bright fringes respectively from the central fringe O. λD Then, xm = m d

P-7 and

x m + 1 = (m + 1)

λD d

Therefore, the distance between the mth and the (m + 1) th bright fringes, that is, the fringe width is given by λD λD x m + 1 − x m = (m + 1) −m d d λD = d We see that the fringe width is free from m, that is, the width of all bright fringes is the same (provided the distance D of the screen from the slits is much larger than the separation d between the slits). Similarly, it can be shown that the width of λd . The fringe width is all dark fringes is also the same and is d denoted by W. Thus, λD …(3) W= d

Dλ/d Width of bright fringe

Measuring fringe width W, the wavelength of light λ can be calculated. According to equation (3), the fringe width is directly proportional to λ. Hence, the fringes of red light (larger wavelength) are broader than the fringes of blue light (shorter wavelength).

Dλ/d

Width of dark fringe

Fig. 4 : Position of dark and bright fringes

6. Angular Fringe Width The angular position of the mth bright fringe is  mλD    x m  d  mλ = = θm = D D d and that of the (m + 1)th bright fringe θm + 1 =

(m + 1) λ d

∴ Angular distance between two consecutive fringes, that is, the angular fringe width is given by θ = θm + 1 − θm θ= λ/d The angular fringe width is independent of the position of the screen.

7. Conditions for Sustained Interference of Light Waves 1.

The two sources of light should be coherent, i. e. they should emit light waves having hardly defined phase difference that remains constant with time.

2.

The waves emitted by two sources should have same frequency.

P-8 3.

The separation between the light sources (d) should be a small.

4.

The distance (D) of the screen from the two sources should be quite large.

5.

The amplitude of the interfering waves should be equal or nearly equal.

6.

The two sources should be quite narrow.

8. No Interference by Two Independent Light Sources The two independent light sources (such as two bulbs or two candles) cannot produce sustained interference. We know that every light source consist of a large number of atoms which are responsible to produce light waves. Each atom consists of electrons which are moving in different electronic orbits around positive charged nucleus. The energy of particular electron depends on its orbits. When the external energy is supplied at an atom, the electrons jump from inner orbit to outer orbit. This is called excited state which is not permanent. After a time interval of 10−8 sec, the electron came back to inner orbit. They radiates energy in the term of light radiation. It is important to mention here that the phase is changing randomly because the atoms which were responsible for light radiation in first step are now replaced by other atoms. So, the phase is change in haphazard manner. As a result, the intensity on the screen changes very rapidly with time. So, two independent sources cannot produce a stationary interference pattern on the screen.

9. Coherent Sources The sources of light, which emit continuous light waves of the same wavelength, same frequency and in same phase or having a constant phase difference are called coherent sources. Two sources are coherent when the waves they emit maintain a constant phase relation. Effectively, it means that the waves do not shift relative to one another as time passes. If wave patterns, on the average, neither constructive nor destructive interference would be observed because there would be no stable relation between two wave patterns. In practice, coherence in interference is obtained by two methods : 1.

By division of wavefront such as in Young’s double slit experiment, Lloyd’s mirror experiment, Fresnel’s biprism experiment.

2.

By division of amplitude by partial reflection and refraction of waves such as in interference of light in thin film, Newton’s ring and Michelson-Morley interferometer.

10. Condition for Constructive and Destructive Interference (Intensity Equation) Let the waves from two coherent sources of light be represented as y1 = a sin wt

…(1)

y 2 = b sin (wt + φ )

…(2)

where a and b are the respective amplitudes of the two waves and φ is the coherent phase angle by which second wave leads the first wave. According to superposition principle, the displacement (y) of the resultant wave at time (t) would be given by y = y1 + y 2

…(3)

P-9 or

y = a sin wt + b sin (wt + φ) y = a sin wt + b sin wt cos φ + b cos wt sin φ

or

y = a sin wt (a + b cos φ) + cos wt b sin φ

Put

…(4)

a + b cos φ = R cosθ

…(5)

b sin φ = R sin θ

…(6)

y = sin wt R cos θ + cos wt R sin θ = R [sin wt cos θ + cos wt sin θ] y = R sin (wt + θ)

…(7)

Thus, the resultant wave is a harmonic wave of amplitude R. Squaring equation (5) and (6) and adding, we get R 2(cos2 θ + sin 2 θ) = (a + b cos φ)2 + (b sin φ)2 R 2 = a 2 + b2 cos2 φ + 2ab cos φ + b2 sin 2 φ = a 2 + b2 (cos2 φ + sin 2 φ) + 2ab cos φ R = a 2 + b2 + 2ab cos φ

…(8)

As intensity is directly proportional to the square of amplitude of the wave, ∴

I1 = Ka 2, I 2 = Kb2

and

I R = KR 2 = K (a 2 + b2 + 2ab cos φ) I R = I1 + I 2 + 2 I1I 2 cos φ

this is called Intensity equation. For constructive interference

Intensity (I) should be max, for which cos φ = max = ± 1 ∴

φ = 0, 2π, 4 π, ... φ = 2nπ

where, n = 0, 1, 2, ... If ∆x is the path difference between the two waves reaching point P, corresponding to phase difference ∆φ then, λ λ ∆φ = (2nπ) ∆x = 2π 2π ∆x = nλ For destructive interference Intensity (I) should be minimum cos φ = minimum = − 1 φ = π, 3π, 5π, ... or

φ = (2n − 1) π

P-10 where, n = 1, 2, 3, ... Corresponding path difference between two waves λ ∆x = ∆φ 2π λ (2n − 1) π = 2π λ = (2n − 1) 2 λ ∆x = (2n − 1) 2

11. Interference by Division of Wavefront We know that two slits illuminated by light coming from a single slit narrow source and its virtual image or two virtual images of a narrow sources can be used as coherent sources of light produce interference. In this case, a wavefront is divided into two parts by utilising the phenomenon of reflection, refraction or diffraction in such a way that after traversing slightly different optical path, the light from the two coherent sources so produced finally unites to produce interference band.

12. Interference by Division of Amplitude In this condition the amplitude of the incoming wave is divided into two or more parts by partial reflection and refraction thereby produces two or more beams which produces interference. Interference of light in this film, Newton’s ring and Michelson’s interferometer are the examples of interference by division of amplitude.

13. Optical Path Retardation (Lateral Shift of Fringes) If a transparent sheet of refractive index M and thickness t is introduced in one of the paths of interfering wave, the optical path length of this path will become Mt instead of t, increasing by (M − 1) t. D Let present position of a particular fringe is y = (∆x ), the new position of the same fringe will be given by, d

S1

S1

S1

t y0

y0 S2

S2

S2 Fig. 5

y′ =

D [∆x + (M − 1) t] d

t

P-11 ∴

Lateral shift of fringe, y0 = y ′ − y =

D (M − 1) t d

λD d D β = d λ β y 0 = (M − 1) t λ β=

∴

14. Fresnel’s Biprism Fresnel’s biprism is a device used to obtain two coherent sources to produce sustained interference. It is a combination of two prisms of very small refracting angles placed base to base. In practice, the biprism is made from a single plate by grinding and polishing, so that it is a single prism with one of its angles about 179° and the other two about 30° each.

14.1 Production of Fringes The method of obtaining interference fringe is shown in figure 6. S is a narrow vertical slit illuminated by monochromatic light. The light from S is allowed to fall symmetrically on the biprism P, placed at a small distance from S and having refracting edge parallel to the slit. The light beams emerging from the upper and lower halves of the prism appear to start from two virtual images, namely S1 and S2, which act as coherent sources. Light beams BS1 E and AS2C, diverging from S1 and S2 are superimposed and the interference fringes are formed in the overlapping region BC. These fringes are non-localised and may be obtained on a screen or seen through on eyepiece. X S1

P

A α B

2d

S C S2 E

D Y Fig. 6 : Fresnel’s Biprism

14.2 Theory of Fringes Let S1 and S2 be the two virtual sources produced by the biprism and XY the screen on which the fringes are obtained. The point O on the screen is equidistant from S1 and S2 reach O in the same phase and reinforce each other. Hence, the point O is the centre of a bright fringe.

P-12 The illumination of any other point P can be obtained by calculating the path difference S2P − S1P and draw perpendiculars S1M 1 and S2M 2 on the screen. Let S1S2 = 2d, S1M 1 = S2M 2 = D and OP = x . Then (Precise measurement) X P Screen x

M1

S1

d S

2d C

O d

S2

M2 Y D Fig. 7

(S2P)2 = (S2M 2)2 + (PM 2)2  (x + d)2  = D 2 + (x + d)2 = D 2 1 +  D2   1

or

 (x + d)2  2 S2P = D 1 +  D2    1 (x + d)2  = D 1 + , as (x + d) > λ So

An = πbλ

  λ2 is negligible (2n − 1) 4  

This is the area of n th zone and it is constant for every zone. Now distance of n th zone from point P λ λ + b + (n − 1) 2 2 dn = 2 λ d n = b + (2n − 1) 4 b+n

…(2)

Hence, the effect of n th zone at point ‘P’ Rn ∝

An (1 + cos θ n) d

…(3)

 λ2  π bλ + (2 n − 1)  4 Rn =  (1 + cos θ n) λ b + (2 n − 1) 4 λ  πλ b + (2n − 1)  4  (1 + cos θ n) Rn = λ b + (2n − 1) 4 or Where θ n is the angle of n

Rn ≈ πλ (1 + cos θ n) th

…(4)

zone. Hence, it is clear as we increase angle of zone then effect of the zone becomes

decreases, i.e. But two wavelets has difference

R1 > R2 > R3 > ... > Rn λ so phase difference will be π. 2

P-50 Hence, if effect of any zone is positive then the effect of next one will be negative and next to this will be positive. i.e.

R = R1 − R2 + R3 − R4 − ... + Rn

and if n is odd

R = R1 − R2 + R3 ... + Rn

if n is even

R = R1 − R2 + R3 − ... − Rn

On solving this equation we get the result that the effect of whole wavefront will be R=

i.e.

R1 2

R1 2

4. Zone Plate As we know that from previous article that the effect of any wavefront at point P is R = R1 − R2 + R3 − R4 + R5 − ... + Rn If n is odd Then

R = R1 − R2 + R3 − R4 − ... + Rn

If we remove the effect of alternative zones then the effect at point P will be maximum. Either maximum positive or maximum negative i.e. if we remove the effect of even zones. We get,

R = R1 + R3 + R5 + ...

and if we remove the effect of even zones Hence,

R = − [R2 + R4 + R6 + ...]

In both case the intensity at any point will be maximum to obtain this condition we can use zone plate.

4.1 Construction of Zone Plate To construct a zone plate first of all draw the concentric circles on a white paper by taking the radius of square root of natural numbers. These circles are called zones. Now, black painted 2, 4, 6 ... zones and take the photograph in reduce size, now see the negative of this photograph.

Negative zone plate

Positive zone plate Fig. 3

In negative, we see that the zones those are black painted (2, 4, 6) were transparent while others ones are opaque (1, 3, 5) this zone plate is called negative zone plate, its opposite plate is positive zone plates.

P-51

4.2 Formula of Focal Length of Zone Plate Let XY is a zone plate, S is a point source at distance a from zone plate, P be the point at distance b from zone plate where we wish to find intensity. X Mn rn S

P

O a

b

Fig. 4 Working of zone plate

Let at point M n, the n th zone is passing through it, radius of n th zone is rn.

X Mn

From fig (4) In ∆SOM n

rn

SM n2 = SO2 + M nO2

S

P

SM n2 = a 2 + rn2  r2  SM n2 = a 2 1 + n2  a    r2  SM n = a 1 + n2  a  

1/ 2

 r2  SM n = a 1 + n  2a   SM n = a +

rn2 2a

Mn P = b +

rn2 2b

Y Fig. 5

[(1 + x )n = 1 + nx + ... binomial theorem]

Similarly M n P in ∆POM n

So

SM n P = SM n + M n P =a+

rn2 r2 + b+ n 2a 2b

= a+ b+

1 2  1 1 rn  +  2  a b

P-52 Now path difference between SM nP and SOP and

= SM nP − SOP = a+ b+ Path difference =

rn2 2

1 2 rn 2

1 1  a + b − a − b

1 1  a + b

But the phase difference between two consecutive wavelets is

…(1) λ nλ , so between n wavelets is . 2 2

rn2  1 1  nλ = + 2 2  a b

So

…(2)

nλab a+b

or

rn2 =

or

rn2 ∝ n

or

rn ∝ n

Hence, we construct the circles of square root of natural numbers. Now from equation (2) 1 1 nλ + = a b rn2

We get, Comparing it with lens formula

1 1 1 + = u v f

i.e. u → a = distance of object. v → b = distance of image. So f =

rn2 will be the focal length of zone plate. nλ

4.3 Multiple Foci of Zone Plate We know that the focal length of zone plate is f= If a = ∞ then f = b =

rn2 nλ

rn2 . nλ

Hence, many focal points of zone plates will be at Thus,

b b b , , ... for third zone, fifth zone etc. 3 5 7

f1 =

rn2 nλ

f2 =

rn2 3 nλ

P-53 f3 =

rn2 5 nλ

…………… ……………

fP =

rn2 (2 p − 1) nλ

where p = 1, 2, 3, ... .

4.4 Compare of Zone Plate with Convex Lens 1.

Zone plate and convex lens both forms the image of any object on the other side of object.

2.

Formula of focal length of zone plate is same as the lens formula

3.

Both have the fault of achromatic aberration.

4.

Convex lens has only one focal point while a zone plate has multiple focal points.

5.

Image formed by zone plate is less sharp while by convex lens is highly bright.

1 1 1 + = . u v f

5. Fraunhofer Diffraction In diffraction of Fraunhofer class both source and screen are at infinite distance from the object producing the diffraction, wavefront in this diffraction are plane wavefronts circular, cylindrical etc are not allowed to produce diffraction. The diffracted light collected at lens and seen on a telescope. The principle of this diffraction is based upon the resultant vibrations which are produced by light waves on the screen.

6. Resultants of ‘n’ Simple Harmonic We know that the diffraction of light is the result of superposition of waves from coherent sources on the same wavefront after the wavefront has been distored by some slit or object. Let us consider, there are ‘n’ harmonic waves on the wavefront having equal amplitude ‘a’ and a common phase difference ‘d’ between successive vibration (or harmonic waves) which is increasing in arithmetic progression. To obtain a resultant amplitude ‘R’, a polygon may be constructed as shown in figure. If the angle between first vibration and the resultant is θ, then it can be resolved as the cosine component of resultant. R cos θ = a + a cos d + a cos 2d + ... + a cos (n − 1) d

a

d

ad

R

a θ

d a Fig. 6

…(1)

and the sine component of resultant as R sin θ = a sin d + a sin 2d + a sin 3d + ... + a sin (n − 1) d

…(2)

P-54 d Multiplying both these equation by 2 sin , we have 2 The equation (1) becomes 2R cos θ ⋅ sin

d d d d d   = a 2 sin + 2 sin ⋅ cos d + 2 sin cos d + ... + 2 sin cos (n − 1) d 2 2 2 2 2  

But we know that

and

…(a)

2 cos A sin B = sin ( A + B) − sin ( A − B)  C + D  C − D sin C + sin D = 2 sin   cos    2   2 

…(b)

Applying these, we get 2R cos θ ⋅ sin

3d d  5d 3d  d d   − sin  = a 2 sin +  sin − sin  +  sin    2 2 2 2 2 2    n − 3  n − 1 + ... +  sin    d − sin   2   2   d   n − 1 = a a sin + sin    2  2 

 d  

 d 

 nd  n − 1  = 2a sin ⋅ cos   d  2   2  2a sin So

R cos θ =

nd  n − 1 cos   d  2  2 d 2 sin 2

…(3)

d and applying relation (a) and (b), we get 2 nd  n − 1 a sin ⋅ sin   d  2  2 R sin θ = d sin 2

Similarly, multiplying (2) by 2 sin

Squaring and adding equation (3) and (4), we get nd a 2 sin 2 2 2 2 2 sin 2  n − 1 d + cos2  n − 1 R (sin θ + cos θ) =   2   2  2 d  sin 2 nd a 2 sin 2 2 2 or R = 2 d sin 2 nd 2 R= d sin 2

…(4)

 d 

a sin

or

…(5)

P-55 where a is the amplitude of each harmonic waves or vibration and there are ‘n’ number of vibration and ‘d’ be the common phase difference between these harmonic waves. If 2α be the phase difference between first and last simple harmonic wave, then nd = 2α nd =α 2

or

 nd  a sin    2 ∴ R= d sin 2 a sin α R= α sin n a sin α or R= α n α α α (as n is large, so becomes small and for smaller sin = ) 2 n n na sin α R= α A sin α So R= α

…(6)

where n a = A is the resultant amplitude of n harmonic waves. Dividing equation (4) by equation (5), we get or

(n − 1) d 2 (n − 1) d θ= 2

tan θ = tan

This is the phase of n harmonics.

7. Fraunhofer Diffraction at a Single Slit Let a beam of monochromatic light from source S, of wavelength λ incident normaly upon narrow slit AB of width ‘e’. The diffracted light from the slit is focussed by a convex lens ‘L’ on the screen. The diffraction patterns having central bright fringe with alternating bright and dark fringes of decreasing intensity on either sides of central bright fringe O. According to concept of wave theory, a plane wavefront is incident normally on the slit AB, then each point of AB emits secondary wavelets in all directions. Wavelets in same direction focussed at O while wavelets diffracted at an angle ‘Q’ focussed at point P on screen.

P-56

P θ

A θ e

B

θ G θ

O

Fig. 7 : Single slit diffraction

To find out the resultant intensity at P. Let draw a perpendicular AG. Hence, path difference for the rays from points A and B of slit AB is given by ∆ = BG = AB sin θ Hence,

∆ = e sin θ 2n Phase difference = (e sin θ) λ

Let us suppose that AB is deviced into a large number of n equal parts. All these parts emits secondary wavelets 2n of amplitude ‘a’ and their phases will vary from zero to (e sin θ). Hence phase difference between two λ waves from sucessive part of slit 1 2n …(1) δ= ⋅ (e sin θ) n λ According to theory of composition of n simple harmonic motions of equal amplitudes ‘a’ and common phase   2π difference  (e sin θ) between successive vibrations the resultant amplitude at point P is given by n λ    πe sin θ   nδ  sin   sin    λ   2 R= a =a  δ  πe sin θ  sin   sin    2  nλ  R=

α= If n is very large then

e sin α α sin   n π e sin θ λ

α is very small hence, n α α sin = n n

…(2)

P-57 From equation (2) we have, sin α α A ⋅ sin α R= λ R = nα ⋅

…(3)

A = nα where, hence resultant intensity at ‘P ’ is given by  sin α  I = A2    α 

2

…(4)

Equation (3) given the expression for resultant amplitude and equation (4) for resultant intensity due to Fraunhofer diffraction at single slit.

7.1 Position of Central Maxima The resultant amplitude is given by equation (3) as A sin α R= α R =

 A α 3 α 5 α7 + − + ... α − α 3! 5! 7 ! 

  α3 α4 α6 + − + ... R = A 1 − 3! 4 ! 7 !   Hence, for bright fringe at O, R will be maximum and α = 0. Hence, π e sin θ = 0 ⇒ sin θ = 0 ⇒ θ = 0 λ So, the maximum value of intensity at C proportional to A 2.

7.2 Position of Secondary Minima Equation (4) given the intensity as  sin α  I = A2    α 

2

I will be minimum when sin α = 0 provided α ≠ 0 α = nπ where m = ± 1, ± 2, ± 3, ... πa sin θ = nπ λ e sin θ = nλ

…(5)

By putting m = ± 1, ± 2, ± 3, ... gives the position of first order, second order, third order, … of minima. But m ≠ 0 because it gives the position of principal maxima.

P-58

7.3 Position of Secondary Maxima To obtain the position of secondary maxima we have to obtain mathematical method for maxima and minima. Hence, differentiating equation (4) w.r.t. to α and equation zero. dT =0 dα 2 d  2  sin α   A   =0  α   dα    sin α α cos α sin α −     2 A2    =0  α    α2 α cos α − sin α

=0 α2 α cos α − sin α = 0 α = tan α

…(6)

Thus the condition for secondary maxima. The equation (6) can be solved graphically by plotting the curves for y = α and y = tan α. The point of intersection of these two curves gives the position of secondary maxima Hence, α1 = 1.43π, α 2 = 2.46π, α 3 = 3.47 π, α 4 = 4.471π + ... The direction of secondary maxima is given by the relation (2m + 1) α= ± π n where m = 1, 2, 3, 4, ... y = tan α

45°

Graph between y = α and y = tan α

π O – 4π –3π –2π π 2π 3π 4π Fig. 8 Intensity distribution curve

P-59 (2m + 1) πe sin θ =± π 2 λ (2m + 1) λ α sin θ = ± 2 For various values of m, we get α= ±

3π 5π 7π 9π ,± ,± ,± + ... 2 2 2 2

The intensity of principal maxima at α = 0 I = A02 = I 0 The intensity of first secondary maxima, α1 = ±

3π 2

 3π  sin    2 I 4 I1 = I 0 I0 = 0 = 22  3π  9π 2    2 I1 1 = = 4.5% of I 0 I 0 22

So,

The intensity of second secondary maxima, α 2 = ±

5π 2

 5π  sin    2 I 4I0 I2 = I0 + + 0 2 62  5π  25π    2 I2 1 = = 1.61% of I 0 I 0 62 Thus, relative intensities of successive maxima are in the ratio 4 4 4 : + ... 1: 2 : 2 9π 25π 49π 2 1 1 1 : : : ... 1: 22 61 121

7.4 Width of Principal Maxima Let us consider that the distance of first secondary minima from the centre of principal maxima be ‘x’ width of central maximum = 2 x. The direction of first minima is given by e sin θ = ± λ λ sin θ = ± e  λ θ = sin −1  ±   e

…(1)

P-60 λ  λ Here, θ is the angular half width of central maxima. Because it extends from sin −1   to sin −1  −  . If lens is  e  e very near to slit or screen is very far from the lens then x sin θ = ± D where D is the distance and D = f λ x =± e f λf e 2λf Width of central maxima = e x=±

Hence, width of CPM is directly proportional to wavelength of light and inversely proportional to its width.

8. Fraunhofer’s Diffraction Due to Double Slit Let us consider a plane wavefront XY of monochromatic light of wavelength ‘λ’ is incident on two parallel slits AB and CD which are perpendicular to the plane of paper having aperture ‘e’ and separation between them being ‘d’. L

P P'

Lens Fig. 9 : Double slit arrangement

The diffracted light is focussed on the screen MN with the help of a lens L. We observe a diffraction pattern having equally spaced bright and dark fringes due to interference of lights from slits AB and CD. A sin α We have studied that in case of single slit diffraction, the resultant amplitude is where α being the half of α the phase difference between first and last vibration at a single slit. A sin α originating from However, in case of double slits, there are two such resultant vibrations of amplitudes α S1 and S2 in a direction θ from the original path. Let both resultants have a phase difference of δ. Let us draw a normal from the second resultant C to the first resultant R1, the path difference between these two resultant waves is

P-61 AO = (e + d) sin θ From ∆ ACO Thus, the phase difference between these two resultants R1 and R2 2π (path difference) δ= λ 2π δ= (e + d) sin θ λ

…(1)

Thus the resultant of these resultant vibration of amplitudes A sin α A sin α and R2 = R1 = α α R 2 = R12 + R22 + 2R1R2 cosδ 2

2

 A sin α   A sin α   A sin α   A sin α  =  cos δ    + 2  +  α   α   α   α  2

 A sin α  = 2  (1 + cos δ)  α  =2

A 2 sin 2 α   2 δ − 1 1 + 2 cos 2   2 α

= 4⋅ or

A 2 sin 2 α α

I = R2 = 4 ⋅

2

cos

2

2

α

or

 A sin α  R= 2  cos β  α 

where

β=

R2=

δ 2

A sin δ 2

R

cos2 β …(2)

A sin α α

δ R1=

A sin α α Fig. 10

δ π = (e + d) cos θ 2 λ

Hence, we see that the resultant intensity contains two factors : 1.

 A sin α   , which gives the diffraction pattern due to single slits.   α 

2.

cosβ, which gives the interference pattern due to diffracted lights from two slits. A sin α Interpretation of first term : α A sin α . It is obvious form the earlier studies that we have central maxima at α = 0, while its minimas Let I 0 = α lie at α = ± mπ, where m is a non-zero integer i.e. m = 1, 2, 3, ... The secondary maximas are obtained at α= ±

3π 5π 7π ... ,± ,± 2 2 2

P-62 I0

–3π –5π –2π 2

–3π 2

–π

+π +3π 2 α α=0

+2π +5π +3π 2

Fig. 11 Intensity distribution curve

The position of diffraction minima can be obtained at α = ± mπ π or e sin θ = ± mπ λ or

e sin θ = ± mπ

…(1)

Interpretation of second term cos β : This term represents an interference pattern of equidistant bright and dark fringes. The maxima of interference term can be obtained when But or

cosβ = 1 or β = ± nπ n (e + d) sin θ = ± nπ λ (e + d) sin θ = ± nλ

…(2)

where n = 0, 1, 2, 3, ... and these correspond to zero order, Ist order and IInd order maxima respectively. Missing orders in diffraction patterns : We know that the intensity of diffraction pattern of two slits depends

upon the diffraction term of single slit, as well as on the interference term between the resultant vibrations due to two slits. Certain maximas will be missing in the diffraction pattern, due to overlapping of diffraction minima and interference maxima. As the position and direction of interference maxima is given by equation (2) (e + d) sin θ = ± nλ and the diffraction minima is given by equation (1) e sin θ = mλ So from these two equations e+d n = m e Case (i) when e = d i.e. the slit width is equal to the separation between two slits. n 2e So = =2 m e or

n = 2m

…(3)

P-63 When m = 1, 2, 3, ... n = 2, 4, 6 , ... That is 2nd, 4th, 6th interference maxima will be missing and the central diffraction will contain and two first order i.e. three interference maxima in itself

Fig. 12

Case (ii). If d = 2e i.e. the seperation between two slits is double of slit width, we have from equation (3) n 3e = =3 m e or

n = 3m

when m = 1, 2, 3, ... we have n = 3, 6, 9, ... which means that 3 rd , 6 th , 9 th orders of interference maxima will be missing in diffraction pattern. And the central diffraction maxima will have one zero order and two interference maxima on each side i.e. five interference maxima in all.

9. Diffraction Grating An arrangement which consists of a large number of equidistant slits is known as a diffraction grating and the corresponding diffraction pattern is known as grating spectrum. A good quality requires a large number of slits. Hence, grating is achieved by ruling grooves (lines) with a diamond point on an optically transparent sheet of material. The number of grooves on a good grating is about 15000 per inch. These grooves acts as opaque spaces after each groove is ruled machine lifts the diamond point and moves the sheet forward with the help of screw which drives the carriage carrying it. For a good quality grating lines should be equidistant hence pitch of the screw must be constant. Commercial gratings are produced by taking cost of an actual grating on a transparent film like that of cellulose acelate. Now, a day gratings are also produced holographically. For this one records the interference pattern between planes or spherical waves. In holographic grating number of lines per inch are much larger than ruled grating.

10. N-Slit Fraunhofer Diffraction Pattern (Diffraction Grating) Let us consider N-parallel slits, each of width ‘a’ and seperation ‘d’ is illuminated by monochromatic light of wavelength ‘d’ the diffracted light form N-slits is focused by a convex len s ‘L’ on the screen placed at focal plane of the lens.

P-64 L

S1

X P

S2 K1 S3

C

K2 S4 K3 Fig. 13

All the points within each slit become the source of secondary wavelets, which spread out in all directions. From the Fraunhofer diffraction at a single slit, the resultant amplitude R due to all waves diffracted from each slit in direction Q is gives by A sin α …(1) R= α ne sin θ …(2) α= λ where A is constant. So all the secondary waves may be treated as single waves of amplitude ‘R’ and phase α is the direction Q from normal. Thus the waves diffracted from all the N-slits in the direction Q are equivalent to N parallel waves, each starting from the middle points of slits. Hence, path difference for any two successive slits is given by ∆x = (e + d) sin θ

…(3)

Hence, corresponding phase difference is given by 2β = δ =

2π (e + d) sin θ λ

…(4)

Hence, as we more from one vibration (wave) to another path difference increase by (e + d) sin θ and phase 2n change by (e + d). Thus phase increases in arithmetic progression. Hence, the resultant amplitude in the λ direction θ is given by sin Nβ R1 = R sin β R1 =

A sin θ sin Nβ ⋅ α sin β

…(5)

The corresponding intensity is given by I = R12 =

A 2 sin 2 α sin 2 Nβ ⋅ α2 sin 2 β

…(6)

 A 2 sin 2 α   sin 2 Nβ  gives the diffraction due to a single slit whereas In the equation (6),    gives the  2 2   α  sin β  interference of the wave from all N slits in the distribution of intensity in the diffraction pattern.

P-65

10.1 Position of Principal Maxima and Minima 10.1.1 Maxima The position of principal maxima is given by condition as, sin β = 0 i.e. β = ± nπ where n = 0, 1, 2, 3 ... . Since sin β = 0. Hence,

sin Nβ will be indeterminate. Hence, sin β

lim β → ± nπ

d (sin Nβ) sin Nβ dβ = lim d β → ± nπ sin β (sin β) dβ =

lim β → ± nπ

N cos Nβ =N cos β

Hence, intensity of principal maxima is proportional to N 2 and increases as number of slits increases from equation (6), we have N=

A 2 sin 2 α α2

⋅ N2

…(7)

The positing of principal maxima in given by sin β = 0, i.e. β = ± nπ Where n = 0, 1, 2, 3, … π (e + d) sin θ = ± nπ λ (e + d) sin θ = ± nπ

…(8)

If n = 0 in equation we get, θ = 0 It is the position of zero order principal maxima. By putting n = 1, 2, 3, … in equation (8) we get first order, second order, third order, … principal maximum.

10.1.2 Minima The intensity in diffraction pattern will be maximum if sin Nβ = 0 but β ≠ 0 Hence, Nβ = ± mN π N (e + d) sin θ = ± mπ λ N (e + d) sin θ = ± mλ

…(9)

where ‘m’ have all integral values except 0, N, 2 N, 3 N, 4 N … because for these values sin β = 0. From equation (1) it is clear that for m = 0, we get zero order principal maxima and m = 1, 2, 3, ... ( N − 1) gives the position of minima. Again at m = N, we get first order principal maxima. Hence, there are ( N − 1) minima between two successive principal maxima.

P-66

10.2 Position of Secondary Maxima To find the position of secondary maxima we have to put dI =0 dβ From equation (6) we have I=

A 2 sin 2 α sin 2 Nβ ⋅ α2 sin 2 β

 sin Nβ   N cos Nβ ⋅ sin β − sin Nβ cos β  dI A 2 sin 2 α = ⋅2   dβ  sin β   α2  sin 2 β =0 N cos Nβ sin β − sin Nβ cos β sin 2 β

=0

N cos Nβ sin β − sin Nβ cos β = 0 …(10)

N2 ta n

2β

tan Nβ = N tan β

1

+

N tan β

Nβ Fig. 14

 sin 2 Nβ  To find the intensity of secondary maxima, we have to calculate the value of   and put into equation 2  sin β  (6). N tan β sin Nβ = 1 + ( N tan β)2 sin 2 Nβ = sin 2 Nβ sin 2 β sin 2 Nβ sin 2 β

= =

N 2 tan 2 β 1 + N 2 tan 2 β N 2 tan 2 β (1 + N 2 tan 2 β) sin 2 β N2 1 + ( N 2 − 1) sin 2 β

 sin 2 Nβ  Putting the values of   from equation (11) into equation (6) we get,  sin 2 β 

…(11)

P-67

I1 =

 A 2 sin 2 α  N2  2 2 2  α 1 + ( N − 1) sin β 

Hence, intensity of secondary maxima (I1) is proportional to

N2 2

1 + ( N − 1) sin 2 β

whereas intensity of principal

maxima (I ) is proportional to N 2. The diffraction pattern in shown as : Zero order principal maxima

First order principal maxima Secondary maxima

First order principal maxima

Secondary maxima

O Secondary minima Fig. 15 : Intensity distribution curve

11. Angular Width of CPM (Central Principal Maxima) We know that position of CPM of grating is …(1)

(e + d) sin θ n = nλ where e is the width of slit and d is separation between two consecutive slits.

X

If the directions of first secondary minima on both sides of n th order CPM is θ n ± dθ as shown in fig.

dθn

Then, we have (e + d) sin (θ n ± dθ) = nλ ±

λ N

where N are number of rulings.

dθn O

From equation (1) and equation (2) (e + d) sin (θ n + dθ n) = (e + d) sin θ n

…(2)

nλ ±

A B C

θn

C

λ N

nλ

sin θ n cos dθ n ± cos θ n sin dθ n 1 =1± N ⋅n sin θ n For small angle dθ n, sin dθ ≈ dθ n and cos dθ n ≈ 1

Y Fig. 16

P-68 So

1 ± cot θ ndθ n = 1 ± cot θ ndθ n = dθ n =

1 N ⋅n

1 N ⋅n 1 N ⋅ n cot θ n

This equation gives half the angular width of CPM.

12. Missing Orders of Absent Spectra in Diffraction Grating The principal maxima in the grating spectrum are obtained in the direction given by (e + d) sin θ = nλ

…(1)

where n = 0, 1, 2, 3, ... and (a + d) is grating element. In diffraction at single slit the directions of minima is given by e sin θ = mλ

…(2)

where m = 1, 2, 3, ... If equation (1) and equation (2) are satisfied simultaneously, the principal maximum of order n will not be present in grating spectrum. Dividing equation (1) by equation (2) we get, e+d n = m e Equation (3) is the required relation for missing order in grating spectrum. Case Ist → If d = e, then n = 2 m. Hence, for m = 1, 2, 3, ... n = 2, 4, 6, ... i.e. second order fourth order sixth order … maxima will absent because they will coincide with first order, second order, third order … diffraction minima. Hence, central maximum has three interference maxima. Case IInd → If d = 2e, then n = 2m for m = 1, 2, 3, … n = 3, 6, 9, … i.e. the third order, sixth order, ninth order … interference maxima will absent because they will coincide with first order, second order, third order … diffraction minima. Hence, central maximum have five interference maxima.

13. Determination of Wavelength of Light By Diffraction Grating We know that for diffracting grating condition for maximum intensity is given by (e + d) sin θ = nλ Hence, to calculate ‘λ’ first we have to calculate (a + d) and angle of diffraction θ for particular order n. Calculation for ( e + d) The grating element (a + d) is determined from the number of lines per inch on the grating. In N lines are there an grating then, N(e + d) = 1 inch = 2.54 cm 2.54 cm e+ d= N

P-69 Calculation of ‘θ’ To determine the angle of diffraction (θ) we used a spectrometer illuminated with a given light source. The eye piece of the telescope in focussed on the cross wires. The collimator and telescope is adjusted for parallel rays. The grating is adjusted on the prism table such that light from collimator falls normally on it and rulings of the grating are adjusted parallel to the axis of the spectrometer and slit. The spectral line, whose angle of diffraction is determined, is viewed by telescope in the first order on either side of direct image. The cross wire of telescope is adjusted an that spectral line and reading of both the verniers are taken. The difference between two readings of same vernier gives 2θ. This process is repeated for order and by making half, we get value of θ. Second order (n=2) First order (n=1) Zero order (n=0)

Grating

First order (n=1)

Collimator

Second order (n=2)

Fig. 17 Grating spectra

Hence, we can calculate λ by knowing (e + d), n and θ.

14. Maximum Order of Spectra of Grating We have that position of CPM of grating is

or For maximum

(e + d) sin θ = nλ (e + d) sin θ n= λ (e + d) nmax = λ

Hence, grating period (e + d) is must be less then twice of wavelength i.e.

(e + d) < 2λ n max or nmax < 2
µ E . So, velocity of E-ray is equal to velocity of O-ray along the optical axis but it is maximum at right angles to the direction of the optical axis.

(2)

For positive cystal velocity of E-ray is least in the direction at right angle to the optic axis. It is maximum and equal to the velocity of the ordinary ray along the optic axis. Optical axis E-rays S

O-ray

(a) Negative crystal

(b) Positive crystal Fig. 7

7. Circular, Elliptical and Plane Polarized Light In plane polarized light the orientation of light vector remains fixed while magnitude charges during vibrations. In case of circularly polarized light, two plane polarized lightwaves are allowed to superimpose then the resultant vector rotates. The magnitude of such a resultant vector (light) remains constant and its orientation varies continuously. If magnitude and orientation both varies continuously then the light is said to be elliptically polarized.

7.1 Analytical Method (Mathematical Method) Case I : Let us consider two electric vectors of wave are vibrating along the same axis, say x-axis. Then X1 = a sin ωt

…(1)

X 2 = b sin(ωt + δ)

…(2)

From the principal of superposition, the resultant of waves, is given as X = X1 + X 2 X = a sin ωt + b sin(ωt + δ) X = a sin(ωt + δ)

…(3)

The equation (3) represents the resultant of two waves having vibrations along same axis. So it is linearly polarized. Case II : Let two linearly polarized light waves vibrating along x and y-axis and propagating along z-axis, are allowed to superimpose. Hence the equations are, X = a sin(ωt + δ )

…(4)

Y = b sin ωt

…(5)

P-100 From equation (5), we have sin ωt =

y b

cosωt = 1 − sin 2 ωt = 1 −

y2 b2

…(6)

From equation (4), we have X = sin(ωt + δ) a X = sin ωt cos δ + cos ωt sin δ a Y2 X Y = cos δ + 1 − 2 sin δ b a b Y2  X Y  − cosδ = 1 − 2 ⋅ sin δ  a b b Squaring both sides, 2  Y 2  X Y 2  − cosδ = 1 − 2  sin δ  a b b  

X2 a

2

+

Y2 b

2

X2 a

2

cos2 δ − +

Y2 b

2

−

2XY Y2 cosδ = sin 2 δ − 2 sin 2 δ ab b 2 XY cosδ = sin 2δ ab

…(7)

Equation (7) is the general equation of an ellipse. Special Cases : When δ = 0, 2π, 4 π, ..., 2nπ then ∴

cosδ = 1 X a

2

2

+

Y b

2 2

−

2 XY b2

=0

2

 X Y  −  =0 a b  X Y  − =0 a b ±Y = ±

b X a

…(8)

Equation (7) we get, X2 a

2

+

Y2 b2

=1

Equation (9) is general equation of ellipse, so emergent light is elliptically polarized. If a = b, then from equation (9)

…(9)

P-101 X 2 + Y 2 = a2

…(10)

Equation (10) is equation of a circle, so emergent light is circularly polarized light.

8. Application of Bi-refringence 8.1 Quarter Wave Plate A plate of doubly refracting uniaxial crystal of uniform thickness whose refracting surfaces are parallel to optic  π λ axis and produces a phase difference of   or a path difference of   between the ordinary ray and  2  4 extra ordinary ray is called as quarter wave plate. Explanation : Let us consider the case of a calcite plate cut with optic axis parallel to the surface. It has been showed that when a plane polarized light falls normally on a thin plate of uniaxial crystal (here calcite plate) cut parallel to its optic axis the light splits up into ordinary ray and extraordinary ray of light. They travel along the same path but with different velocities. The velocity of E -ray is greater than the velocity of O-ray. As a result a phase difference is introduced between them. If the thickness of the crystal plate is such that it  π λ introduces a phase difference of   radian or a path difference of   then it is called a quarter wave plate.  2  4 Thickness of plate : To calculate the thickness of a plate we proceed as follows : Let t be the thickness required for such a plate. If µ 0 and µ e be the refractive indices of the crystal for O-ray and E-ray waves, then the path difference between O-ray and E-ray is [(µ 0 > µ e ) in calcite crystal ]

µ 0t − µ e t = (µ 0 − µ e ) t For a quarter wave plate λ 4 λ (µ 0 − µ e )t = 4

Path difference =

or

t=

λ 4( µ 0 − µ e )

8.2 Half Wave Plate A plate of doubly refracting crystal, cut with faces parallel to optic axis, such that it introduces a phase λ difference of π or a path difference of between E-ray 2 and O-ray, then it is called a half wave plate. λ For a half wave plate, path difference = 2 λ (µ 0 − µ e )t = 2 λ t= 2 (µ 0 − µ e )

Optical axis

t

O E-rays

Thickness of plate

Emergent light Fig. 8 : Quarter wave plate

P-102

9. Dichroism It is the property due to which a doubly refracting material absorbs one of the two refracted ray’s i. e., the ordinary and allows the other (extraordinary) to pass through it. The crystals or substances which possess this property are called dichroic. Light emerging out of a dichroic substances will be plane polarized. For example, tourmaline crystal is dichroic. Explanation : This phenomenon can be explained on the basis of a wire-grid polarizer as a plane of vibrations of ordinary and extraordinary ray through a doubly refracting crystal are perpendicular to each other. Molecular alignment of substance is such that it provides high electric conductivity, in one direction and completely obstructs it in perpendicular direction. Thus component is allowed to pass through it and other component is completely obstructed.

10. Optical Rotation Light is polarized when it passes through a nicol prism or polaroid. The passed light is in a certain oscillating plane. The polarized light (polarized by first polariod) passes through the second polaroid when it is placed in parallel with the first one, but the polarized light completely disappear when the second polaroid is placed perpendicular to the first one. Some compounds can rotate the plane of polarized light to a certain amount. This phenomenon is referred to as optical activity and compounds with such ability is named as optical active substance. Optically active substances exist as a pair (+) and (−) forms, of enantiomers. If the enantamers rotates the polarized plane clockwise, the other enantiomer rotates the same amount but to reverse direction. Clockwise rotation is ‘‘dextrorotary’’, while the counter clockwise rotation is “levorotary’’. For fig.9(a) dextrorotary, symbols d or (+) and for levorotary symbol l or (−) are frequency used in fig.9(b). The magnitude of rotation depends not only the type of compound, but also the concentration of the solution, the length of light path, the solvent, the wave length of the light employed and the temperature. Optical activity (Optical roation) Unpolarized Light (dextrorotary)

1st polaroid

2nd polaroid

Fig. 9(a) : Optical rotation

Unpolarized Light

(levorotary)

1st polaroid

2nd polaroid

P-103

Fig. 9(b) : An optical active compound

11. Fresnel’s Theory of Optical Rotation (Rotatory Polarization) Fresnel’s theory has following points : 1.

When a plane polarized light is passed through a crystal along the optical axis, it is split up into two components, i. e. one clockwise and another anti-clockwise circular motion.

2.

Both the components travels with same velocity in an optically inactive substance (medium).

3.

In optically active crystals two components travels with different speeds so that a relative phase difference is developed between them in transmission through the crystal.

4.

In dextra rotatary substances, v R > v L . In laevo-rotatory substances v L > v R .

5.

The circular vibrations of components after emergence, recombine to produce a plane polarized light.

6.

The plane of polarization of the result wave depends upon the phase difference between the two vibrations with in the crystal.

12. Mathematical Analysis Let a ray of plane polarized light be incident normally on a doubly reflecting crystal with its faces perpendicular to optical axis then equation of incident wave is given as Y = 2a sin ωt and X = 0

…(1)

Y = a sin ωt + a sin ωt

…(2)

X = cos ωt − a cos ωt

…(3)

These equation may be written as

∴ Hence

y1 = a sin ωt, y 2 = a sin ωt x 1 = a cos ωt, x 2 = − a cos ωt

These components of wave possesses circular motion. These circular motion travels along the optical axis of the quartz with unequal velocities. Hence during emergence there is phase difference of δ between them. Hence x 1 = a cos(ωt + δ)

(Clockwise) …4(a)

y1 = a sin(ωt + δ)

…4(b)

x 2 = − a cosωt y 2 = a sin ωt Resultant vibrations along two axis i. e. X-axis and Y-axis is, X = x1 + x 2

(Anticlockwise) …5(a) …5(b)

P-104 = a cos(ωt + δ) − a cos ωt = 2a sin

δ δ  sin ωt +   2 2

…(6)

Y = y1 + y 2 = a sin(ωt + δ) − a sin ωt = 2a cos

δ δ  sin ωt +  2  2

…(7)

Dividing equation (6) by equation (7), we get δ X …(8) = tan 2 Y δ This is a equation of straight line with slope tan with Y-axis. Let µ R and µ L are the refractive index of right 2 handed and left handed rotatory waves and t is the thickness of the crystal then path difference

∴

Phase difference

∴

Angle of rotation is

∆ = (µ R − µ L ) t 2π (µ R − µ L ) t δ= λ θ= =

δ π = (µ R − µ L ) t 2 λ π λ

C C − t  v v  R L

θ=

1 πC  1 − t  λ  vR vL 

θ=

π1 1 − t  T  vR vL 

(‡ T =

In case of left handed substances vL > vR θ=

π1 π1 1 1 − t=  − t  T  vR vL  λ  vR vL 

In case of right handed substances, vR > vL θ= =

π1 1  − t T  vL vR  1 πC  1  − t λ  vL vR 

For optically inactive substances VL = VR θ= 0

λ , time period of wave) C

P-105

13. Specific Rotation : (Optical Rotation in Liquid or Crystal) The specific rotation (α) is defined for a given compound as rotation (θ) induced by 1 g/ml solution 10 cm long under a specified condition (the temperature, wavelength of the light used, the solvent) as shown below. Where λ is the wavelength of the light and t is the temperature. Observed rotation(θ) [α]tλ = Length of sample tube (dm) × Conc (g / ml) θ l ⋅C 10θ α= l ⋅C

or

α=

(If ‘l’ is measured in cm) If ‘l’ is in cms, unit of specific rotation is deg/dm-gm/cc.

14. Polarimeters Halfshade Polarimeter : It is an instrument used for measuring the optical rotation of certain substances. Optical parts are shown in figure 10. Construction : Monochromatic light from a source S, usually a sodium lamp, after passing through a slit is rendered parallel by a convex lens L and falls on a polarizing Nicol prism N1. After passing through N1 the light becomes plane polarized. This light passes through a half-shade device H (called Laurent’s plate) and then through a tube D containing the optically active solution (say sugar solution), and then falls on the analyzing Nicol prism N 2. The emerging light viewed through a telescope T . The analyzing Nicol N 2 can be rotated about the light as axis, and its rotation can be measured on a circular degree scale C by means of a vernier. Slits H

L

Laurent's plate D

N1

N2

C T

S Mono chromatic light source

Nicol prism

Liqid solution

Telescope

Fig. 10 : Half shade polarimeter

Working : Let us consider for a moment that the half shade device H is not present. The position of the Nicol N 2 is adjusted so that the field of view is completely dark when the tube D is empty. In this position, N 2 is crossed with respect to N1. The reading of N 2 is taken. Now the tube D is filled with the experimental solution. This rotates the plane of polarization of the light coming from N1 through some angle, say θ1, so that some light is transmitted by N 2. Now, N 2 is rotated until the field of view again becomes dark and its reading is noted. This will happen when N 2 has been rotated through the angle θ in the direction of the optical rotation produced. Hence, the difference of the two reading of N 2 gives the rotation of the plane of polarization. Action of the Half Shade : The above method of measuring the optical rotation is not accurate because the eye cannot judge the position of complete darkness accuratory. The field appears dark over considerable range N 2 is rotated. Hence a pair of Nicols alone is not preferable in measuring optical rotation.

P-106 The difficulty is avoided by introducing a half shade device H immediately after the Nicol N1. It is a combination of two semicircular plates ACB and ADB. The plate ACB is of quartz and is cut parallel to the optic axis, while the plate ADB is of glass. The two are comented along the diameter AB. The thickness of the quartz plate is such that it introduces a phase difference of π between the O and E vibrations. In other words, it is a half wave plate. The thickness of the glass plate is such that it absorbs the same amount of light as the quartz plate.

A Q

P

C

D

O

Quartz

Glass

B Fig. 11 : Optical rotation in half shade plate

The plane polarized light from N1 falls normally on the half shade plate. Suppose this light has vibration along OP. It is transmitted through glass half as such and emerges with vibrations still along OP. Inside the quartz half, however, the light is divided into two components, one E-component parallel to the optic axis, (OA) of the quartz and other O component perpendicular to the optic axis, i. e. along OD. The component travels faster in quartz. Therefore, on emergence it gains a phase of π over the E-component. Hence, now the O component has vibrations along OC (instead of OD), the E-component still having vibrations along OA . Therefore, the light emerging from quartz has resultant vibrations along OQ, where ∠AOQ = ∠AOP. A

A

Q

A

Q

P

Q

P

N2

N1 C

F

O

E

D

C

F

D

O

C

E N2

(a)

P

(b) Fig. 12

O

D

F N1

(c)

Now, if the principal section N 2ON 2 of the analyzing Nicol N 2 is parallel to COD. Both halves of the field appears slightly but equally illumintated. This is because the component OE and OF of the emergent vibrations OP and OQ along the principal section of N 2 are small but equal. If N 2 is slightly rotated from this position in the clockwise direction. The component OE appreciably decreases while OF appreciably increases. Hence, the right half of the field becomes darker and the left-half brighter. Similarly, a slight rotation of N 2 in the anti-clockwise direction make the right half brighter and left half darker. Our eye judge for more accurately the position of equal illumination of the two halves then the position of complete darkness as is necessary in the absence of half shade. Thus, the half shade enable us to set the analyzer more accurately. A given half shade can, however be used only for one perpendicular wavelength for which it has been designed.

P-107

15. Biquartz Polarimeter Biquartz polarimeter is an accurate instrument which is more sensitive than a half shade polarimeter used for finding the angle of rotation produced by an optically active substance. The experimental part of biquartz polarimeter are the same as that of the half shade polarimeter except that the half shade device is replaced by biquartz plate and monochromatic light (sodium light) is replaced by white light. A P'

90°

P

90°

L

R

O Q'

Q

B Fig. 13 : Biquartz polarimeter

Example 1. If the refractive index for glass is 1.5, then calculate the angle of refraction. Solution : Given that Refractive index of glass µ = 1.5 From Brewster’s law µ = tan i P tan i P = 1.5 i P = 56.31° We have

r + i P = 90° r = 90° − 56.31° r = 33.69°

Example 2. Calculate the Brewster angle for a glass of refractive index 1.5 and it is immersed in water of refractive index 4 /3. Solution : Given that µ g = 1.5 and µ w = wng

By Brewster’s law

=

1.5 . = 1128 1.33

u = tan i P 1128 . = tan i P i P = 48.4 °

4 = 1.33 3

P-108 Example 3. If the refractive index of water is 1.33, then calculate the angle of polarization. Solution : Given that µ = 1.33 By Brewster’s law

µ = tan i P 1.33 = tan i P i P = 33°

Example 4. If the angle of polarizer and analyzer is 45°, what will be intensity of transmitted light for original intensity of incident light as I 0 ? Solution : Given that θ = 45° By Malus law, I = I 0 cos2 θ I = I 0 cos2 45°  1  I = I0   2 I=

2

I0 2

I = 0.5I 0 Example 5. Two nicol prisms are so arranged that the amount of light transmitted through them is maximum. What will be the percentage reduction in the intensity of the incident light when the analyzer is rotated through 60° ? Solution : From Malus law, I = I 0 cos2 θ The percentage reduction in the intensity of light =

I0 − I × 100 = I0

 I 0 − I 0 cos2 θ   × 100  I0  

= (1 − cos2 θ) × 100 But given θ = 60° ∴ Percentage reduction in intensity = (1 − cos2 60° ) × 100 3 1  = 1 −  × 100 = × 100 = 75%  4 4 Example 6. Two polaroids are adjusted to obtained maximum intensity. At what angle should one polaroid be rotated to reduce half the intensity ? Solution : Given that I 1 = I0 2 From Malus law,

I = I 0 cos2 θ

P-109 I0 = I 0 cos2 θ 2 1 cos2 θ = 2 θ = 45° Example 7. Calculate the thickness of quarter wave plate when the wavelength of light is 6000 Å and µ e = 1.55 and µ 0 = 1.54. Solution : Given that λ = 6000 Å µ 0 = 1.54, µ e = 1.55 ∴ Thickness of quarter wave plate

⇒

t=

λ 6000 × 10−10 = 4(µ e − µ 0) 4(1.55 − 1.54)

t=

6 × 10−7 4 × 0.01

t = 1.5 × 10−5 m

Example 8. Calculate the thickness of a quarter wave plate of quartz for sodium light of wavelength 5896 Å. The ordinary and extraordinary refractive indices for sodium are 1.54425 and 1.55336 respectively. Solution : Given that

Thickness,

µ 0 = 1.54425, λ = 5893 Å λ t= 4(µ e − µ 0) t= =

5893 × 10−10 4(1.55336 − 1.54425) 5893 × 10−7 4 × 911 × 10−5

= 1.62 × 10−3 cm Example 9. Calculate the thickness of a doubly refracting crystal which can produce path π difference of between O-ray and E-ray with light of wavelength 5893 Å (µ 0 = 1.53, µ e = 1.54). 4 Solution : Given that µ 0 = 1.53, λ = 5893 Å Thickness of plate,

t=

λ 4( µ e − µ 0 )

t=

5890 × 10−10 4( 1.54 − 1.53 )

= 1.47 × 10−5 m

P-110 Example 10. Calculate the thickness of quarter plate which would convert plane polarized light to circularized light to circularly polarized light ( λ = 5890 Å, µ 0 = 1.658, µ e = 1.486). Solution : A quarter wave plate converts plane polarized light to circularly polarized light. So, thickness of plate λ t= 4(µ e − µ 0) t= =

5890 × 10−10 2 (1.658 − 1.486) 5890 × 10−10 4 × 0168 .

= 8.56 × 10−7 m Example 11. Calculate the thickness of half wave plate when the wavelength of light is 5000 Å , and µ 0 = 1.55 and µ e = 1.54. Solution : Given that λ = 5000 Å, µ 0 = 1.55 µ e = 1.54 Thickness of plate,

t=

λ 2 (µ e − µ 0)

t=

5000 × 10−7 2 (1.55 − 1.54)

=

5 × 10−7 = 2.5 × 10−5 m 2 × 0.01

Example 12. Calculate the thickness of half wave plate which used light wavelength 5890 Å and µ 0 = 1.54 and µ e = 1.55. Solution : Given λ = 5890 Å, µ 0 = 1.54, µ e = 1.55 So, thickness of plate,

t=

λ 2(µ e − µ 0)

t=

5890 × 10−10 5.89 × 10−7 = 2 (1.55 − 1.54) 2 × 0.01

= 2.945 × 10−5 m Example 13. Calculate the specific rotation if the plane of polarization is turned through 26.4° transversing 20 cm length of 20% sugar solution. Solution : Given that θ = 26.4 ° , l = 20 cm, C = 20% = 0.2 g/cm 3 ∴ Specific rotation

α=

10θ l×C

α=

10 × 26.4 ° = 66° 20 × 0.2

P-111 Example 14. The specific solution of quartz at 5000 Å is 29.73 deg/mm. Calculate the difference in the refractive indices. Solution : Optical solution is given by πt (µ L ~ µ R ) λ θλ ∆µ = πt θ = 2973 . deg/mm t θ=

∴ Given

=

(where ∆µ = µ L − µ R )

2973 . × 5.000 × 10−4 180°

= 0.82583 × 10−4 = 0.258 × 10−5 Example 15. Calculate the specific rotation of 10% sugar solution if the plane of polarization is turned through 13.2°. The length of tube is 20 cm. Solution : Given λ = 20 cm, θ = 13.2° C = 10% = 01 . g/km 3 Specific rotation,

α=

10θ l×C

α=

10 × 13.2 20 × 01 .

α = 66°(dm −1) (g/cm −3) −1 Example 16. A sugar solution in a tube of length 20 cm produces optical solution of 13°. The solution is diluted to one third of its previous concentration. Find optical solution produced by 30 cm along tube containing the diluted solution. Solution : We know that Specific rotation, So, we can write for two stages as,

α=

10θ l×C

10 θ 2 10 θ1 = l 1 × C1 l 2 × C 2

Given, l 1 = 200 cm, C1 = C, θ1 = 13° C l 2 = 30 cm , C 2 = 3 10 × 13 10 × θ 2 × 3 = 30 × C 20 × C θ2 =

13 × 30 20 × 3

θ 2 = 6.5°

P-112 Example 17. 80 g of impure sugar is dissolved in a litre of water. The solution gives an optical rotation of 9.9 ° when placed in a tube of length 20 cm. If the specific rotation of sugar is 66° dm −1 (g /cc) −1 . Find the percentage purity of the sugar sample. θ Solution : Specific rotation λ = l×C Given, θ = 9.9°, α = 66° dm −1 (g /cc) −1, l = 20 cm = 2 dm 66 = C=

9.9 2× C 9.9 = 0.075 g /cc 2 × 66

= 75 g /litre This shows that only 75 g pure sugar is available in 1 litre solution. 75 = × 100 ∴ Percentage purity of sample 80 = 9375 . % Example 18. In an experiment following data is used length of tube containing sugar solution = 22 cm, volume of solution = 88 cc, amount of sugar in solution = 6 g and angle of rotation = 9 °54′. Calculate specific rotation of sugar solution. 6g Solution. l = 22 cm = 2.2 dm, C = 88 cm 3 θ = 9° 54 ′ = 9.9° Specific rotation,

α=

θ l×C

α=

9.9 × 88 2.2 × 6

α = 66° Example 19. A 5% solution of cane sugar placed in a tube of length 40 cm causes the optical rotation of 20°. How much length of 10% solution of the same substance will cause 35° rotation? Solution : We know that α=

θ1 θ2 × l 1 × C1 l 2 × C 2

Here, l 1 = 40 cm, C1 = 5%, θ1 = 20° l 2 = ?, C 2 = 10%, θ 2 = 35° ∴ ⇒

20° 35° = 40 × 5% l 2 × 10% l 2 = 35 cm

P-113 Example 20. The refractive indices of quartz for right handed and left handed circularly polarized light of wavelength 6300 Å are 1.53915 and 1.53921 respectively. Calculate the angle of rotation produced by quartz plate of thickness 0.5 mm. πt Solution : The rotation of plane polarized light is given by θ = (µ L − µ R ). λ Given that t = 0.5 mm = 5 × 10−4 m, λ = 6300 Å, µ L = 1.53921, µ R = 1.53915 So,

θ=

or

θ=

. × 5 × 10−4 314 6300 × 10−7

(1.53921 − 1.53915)

314 . × 5 × 0.0006 6.3 × 10−7

radian θ = 01495 . θ = 8.57 ° Example 21. A 20 cm long tube contain 48 cm 3 of sugar solution rotates the plane of polarization by 11°. If the specific rotation of sugar is 66°. Calculate the mass of sugar in solution. Solution : Let m is the mass of sugar. Give that α = 66° , V = 48 cm 3, l = 20 cm θ = 11° m m g/cm 3 = V 48 10θ α= l×C

C=

66 =

10 × 11 × 48 20 × m

m=

10 × 11 × 48 =4g 20 × 66

Example 22. The critical angle of light in a certain substance is 40°. What is the polarizing angle for it ? Solution : If C is the critical angle, then 1 1 = sin C sin 40° 1 = 0.4628

µ=

= 1.56 Now

tan i P = µ = 1.56 i P = tan −1(1.56) i P = 57.3°

P-114 Example 23. The polarizing angle of a piece for green light is 60°. What is the angle of minimum deviation for a 60° prism made of the same glass ? Solution : Given that i P = 60° Using Brewster’s law µ = tan i P = tan 60° = 1732 . For angle of deviation,  A + δm  sin    2  µ= A sin 2 Here µ = 1732 . , A = 60°, δ m = ?  60° + δ m   60° + δ m  sin    sin      2 2 = = 1732 . 60° sin 30 ° sin 2

∴

 60° + δ m      2 1732 . = sin 0.5

∴

 60° + δ m  . × 0.5 = 0.86605 sin   = 1732   2 60° × δ m = sin −1(0.86605) = 60° 2 or

60° + δ m = 120° δ m = 120° − 60° δ m = 60°

Example 24. Calculate the velocities of O-ray and E-ray in calcite crystal in a plane perpendicular to the optic axis. Given that µ 0 = 1.658, µ e = 1.486 and C = 3 × 1010 cm/sec. Solution : The velocity of the O-wave (ray) is =

C 3 × 1010 = = 1.81 cm/sec µ0 1.658

=

C 3 × 1010 = = 2.02 cm/sec µe 1.486

The velocity of the E-ray is

Example 25. Find the thickness of a quarter wave plate when the wavelength of light is equal to 5890 Å and µ 0 = 1.55, µ e = 1.54. λ Solution : The thickness of a quarter wave plate is, t = 4( µ 0 − µ e ) Here, λ = 5890 Å, µ 0 = 1.55, µ e = 1.54

P-115 ∴

t= =

5890 × 10−8 4(1.55 − 1.54) 5890 × 10−8 4 × 0.01

t = 1.47 × 10−3 cm Example 26. Two nicols are cross to each other. Now one of them is rotated through 60°. What percentage of incident unpolarized light will pass through the system ? Solution : The intensity I θ of the light emerging from the second nicol is Iθ = Here,

1 I cos2 θ 2 0

θ = 90° − 60° = 30° 2

 3   2 I θ cos 30°  2  3 = = = 2 2 8 I0 Percentage transmission is

Iθ 3 × 100 = × 100 = 37.5% 8 I0

Example 27. The value of µ e and µ 0 for quartz are 1.5508 and 1.5418 respectively. Calculate the phase retardation for λ = 5000 Å when the plate thickness is 0.032 mm. Solution : For a quartz crystal of thickness t, the path difference is

The phase retardation Here,

∆ = ( µ e − µ 0) t 2π 2π ( µ e − µ 0) t = ⋅∆ = λ λ µ e = 1.5508, µ 0 = 1.5418 t = 0.032 mm = 0.0032 cm λ = 5000 Å = 5000 × 10−8 cm Phase retardation =

2 × 314 . × (1.5508 − 1.5418) 5000 × 10−8

= 3.617 radian Example 28. Calculate the thickness of double refracting crystal required to introduce a path λ difference of between the O-ray and E-ray when λ = 6000 Å, µ 0 = 1.65 and µ e = 1.48. 2 Solution : The thickness of a doubly refracting crystal is given by λ t= 2 (µ 0 − µ e ) Here, µ 0 = 1.65, λ = 6000 Å = 6000 × 10−8 cm µ e = 1.48

P-116 t=

6000 × 10− 8 = 0.000176 cm 2 (1.65 − 1.48)

t = 176 . × 10−4 cm Example 29. Calculate the specific rotation of sugar solution from the following data. Length of the tube containing the solution = 22 cm, volume of the solution = 88 cc, amount of sugar in the solution = 6 cm and angle of rotation 95 ° 4′ . θ Solution : The specific rotation of the sugar solution is given by S = . lC Here, l = 22 cm = 2.2 dm C= S=

6g 88 cm 3

, θ = 9° 54 ′ = 9.9°

9.9° × 88 2.2 × 6

S = 66° Example 30. A 5% solution of cane sugar placed in a tube of length 40 cm, causes the optical rotation of 20°. How much length of 10% solution of the same substance will cause 35° rotation ? Solution : If l 2 be the length of solution of conc. C 2 of the same substance causes θ 2 rotation, then specific rotation is

or

S=

θ1 θ = 2 l 1C1 l 2C 2

l2 =

θ 2 l 1 C1 θ1 C 2

l2 =

35° × 40 × 5° 7 = × 40 20° × 10 8

Here, θ 2 = 35° , θ1 = 20° l 1 = 40 , C 2 = 10% C1 = 5% ∴

l 2 = 35 cm

1.

Calculate the polarizing angle for crown glass (µ = 1.520), flint glass (µ = 1.650) and water (µ = 1.333).

2.

If the polarizing angle for a piece of glass for green light is 60°, find the angle of minimum deviation for green light for its passing through 60° prism made of the same glass.

3.

The value of µ E and µ O for quartz are 1.5508 and 1.5418 respectively. Calculate the phase retardation for λ = 5000 Å, when the plate thickness is 0.032 mm.

P-117 4.

Calculate the thickness of a quarter wave plate of quartz for sodium light of wavelength 5893 Å. The ordinary and extraordinary refractive indices for sodium are 1.54425 and 1.55336 respectively.

5.

80 g of impure sugar is dissolved in a litre of water. The solution gives an optical rotation of 9.9° when placed in a tube of length 20 cm. If the specific rotation of pure sugar is 66° dm −1 (g /cc) −1 , calculate the percentage purity of the sugar sample.

6.

A sugar solution of specific rotation 52° per decimeter per g/c.c. causes rotation of 12° in column of 10 cm long. What is the concentration of solution ?

7.

If the refractive index for glass is 1.5, then calculate the angle of refraction.

8.

4 Calculate the Brewster angle for a glass of refractive index is 1.5 and it is immersed in water of refractive index . 3

9.

If the refractive index of water is 1.33, then calculate the angle of polarization.

10.

If the angle of polarizer and analyzer is 45°, what will be intensity of transmitted light for original intensity of incident light as I 0 ?

11.

Two nicol prisms are so arranged that the amount of light transmitted through them is maximum. What will be the percentage reduction in the intensity of the incident light when the analyzer is rotated through 60° ?

12.

Prove that i P + r = 90° where i P is the angle of polarization and r is the refracted angle.

13.

Two polaroids are adjusted to obtained maximum intensity. At what angle should one polaroid be rotated to reduce half intensity ?

14.

Calculate the thickness of quarter wave plate when the wavelength of light is 6000 Å and µ E = 1.55 and µ O = 1.54.

15.

Calculate the thickness of a quarter wave plate of wavelength of sodium light 5896Å. Given that (µ O = 1.5333, µ e = 1.54425)

16.

Calculate the thickness of doubly refracting crystal which can produce path difference of π / 4 between O-ray and E-ray with light of wavelength 5890 Å.

17.

Calculate the thickness of wave plate which would convert plane polarized light to circularized light to circularly polarized light (λ = 5890 Å, µ O = 1.658, µ E = 1.486).

18.

Calculate the thickness of half wave plate when the wavelength of light is 5890 Å and µ O = 1.55 and µ E = 1.54 .

19.

Calculate the specific rotation if the plane of polarization is turned through 26.4° transversing 20 cm length of 20% sugar solution.

P-118

Long Answer Type Questions 1. 2. 3. 4. 5. 6. 7.

What is polarized light ?

What do you understand by polarization of light ? 4. Discuss it. 5. Give Brewster’s law. 6. What do you understand double refraction ? Give an 7. example of double refraction. 8. What is polarized light ? How will you deflect plane, 9. elliptically and circularly polarized light ? What is meant by plane circularly and elliptically 10.

What is circularly polarized light ?

polarized light ? 8. 9. 10.

Short Answer Types Questions

How will distinguish between polarized and 1. unpolarized light ? 2. What is the nature of light ? Give an example to show the nature of light. 3.

11.

What do you understand by wave plates ? Explain 12. half wave plate and quarter wave plate. 13. Obtain expression for the minimum thickness of half 14. wave plate and quarter wave plate.

What is difference between polarized light and unpolarized light ? What is plane polarized light ? What do you mean by polarizer ? Define double refractive crystal ? What are ordinary and extraordinary rays ? What do you mean by negative and positive crystals? Explain the quarter wave plate. What is half wave plate ? Define it. How will we produce elliptically polarized light ? How will we produce circularly polarized light ? What you mean by optical activity ? What are dextro-rotatary and laevorotatory substances ?

What do you understand by retardation plates ? Give 15. its construction of retardation plates. 16. Discuss the Fresnel’s theory of optical rotation.

What is the effect of wavelength of light on angle of rotation?

12.

What is specific rotation ? Discuss it.

Give relative merits of half shade polarimeter.

13.

Discuss the phenomenon of optical activity.

11.

14. 15.

16.

17.

18. 19.

20.

17.

18. Define specific rotation. Describe the construction 19. and working of Laurent’s half shade polarimeter. 20. Discuss the phenomenon of rotation of plane of polarization of light by optical active materials. Give Very necessary theory. 1. Show that the rotation of plane of polarization is π 2. given by (µ L − µ R ). λ 3. Discuss theoretically, the superposition of two 4. linearly polarized sinusoidal light waves of the same 5. frequency travelling along same direction when their 6. vibration are parallel and mutually perpendicular. 7. Give the laws of rotatary polarization. Give the principle of half shade polarimeter and 8. explain how will you use it to determine the specific 9. rotation of cane sugar solution ? 10. Described the construction and working of biquartz 11. polarimeter. 12.

What is specific rotation?

What is ordinary light? Define Brewster’s law. What is Malus’s law?

Short Answer Types Questions What is double refraction ? What is uniaxial crystal ? Give the formula for thickness of quarter wave plate. What is optical rotation ? Write the formula of specific rotation. What do you mean by half shade polarimeter ? What is polarization of light ? Prove that µ = tani P . What do you mean by polarizer ? What do you mean by analyzer ? What is polarized light ? What is polaroid ?

P-119 13.

Determined the specific rotation.

14.

Give two difference between polarized light and unpolarized light.

15.

Can sound wave be polarized ?

16.

Explain the basic concept of optical activity.

8.

Which one is optically active substance ? (a) Calcite (b) Tourmaline (c) Quartz (d) NaCl

9.

Which one is not a uniaxial negative crystal ? (a) Tourmaline (b) Ironoxide (c) Ruby (d) Calcite

17.

What is plane of vibration and plane of polarization ?

18.

Show that at polarizing angle, reflected and refracted 10. rays are at right angles.

19.

What do you understand by double refraction ?

20.

What is Nicol prism ?

21.

What is polarimeter ?

22.

What do your mean by extraordinary ray ?

11.

Nicol prism works on the phenomenon of : (a) Reflection (b) Refraction (c) Scattering (d) Double refraction

12.

Biaxial crystal among the following is : (a) Tourmaline (b) Selenite (c) Quartz (d) Calcite

13.

The phenomenon of rotation of plane polarized light is called : (a) Kerr-effect (b) Double refraction (c) Optically activity (d) Dichroism

14.

If µ 0 and µ e are the refractive indices of crystal axis, we have : (b) µ 0 < µ e (a) µ 0 > µ e (c) µ 0 = µ e (d) µ e = 0

15.

A polarizer reduces the intensity of unpolarized light by : (a) 40% (b) 50% (c) 60% (d) 70%

16.

Polaroid glass is used in sun glasses because : (a) It reduces the light intensity to half (b) It is cheaper (c) It is fashionable (d) It has good colour

17.

In dextrarotatory substances, the relation between the refractive indices of right and left handed circularly polarized light is : (b) µ R = µ L (a) µ R = µ L (c) µ R < µ L (d) µ R > µ L

18.

The correct expression between polarizing angle and refractive index is : (a) µ sini P = 1 (b) µ tan i = 1 P (d) µ cosi P = 1 (c) µ coti P = 1

19.

(a) Plane polarized (b) Circularly polarized

The plane of vibration and depolarization are : (a) Orthogonal (b) Parallel

(c) Unpolarized

(c) Non-existent

23.

What do you mean by ordinary ray ?

24.

Give two examples of uniaxial crystals.

25.

Give two examples of biaxial crystals.

Objective Type Questions Multiple Choice Questions 1.

Polarization of light proves the : (a) Quantum nature of light (b) Corpuscular nature of light (c) Longitudinal nature of light (d) Transverse nature of light

2.

The (a) (b) (c) (d)

wave that cannot be polarized are : Light waves E.M. waves Transverse waves Longitudinal waves

3.

Transverse nature of light is proved by phenomenon of : (a) Interference (b) Diffraction (c) Polarization (d) Reflection

4.

Polarization cannot occurs in : (a) X-rays (b) Light waves (c) Radio waves (d) Sound waves

5.

Which one is not a uniaxial crystal ? (a) Calcite (b) Tourmaline (c) Topaz (d) Quartz

6.

Calcite crystal is a : (a) Uniaxial (c) Opaque

7.

Optical device which produced plane polarized light is : (a) Nicol prism (b) A biprism (c) A crystal (d) Calcite

(b) Biaxial (d) None of these

Light transmitted by a nicol prism is : (d) Elliptically polarized

(d) Both (a) and (b)

P-120 20.

21.

If light is polarized by reflection, the angle between 12. reflected and refracted light is : π 13. (a) π (b) 2 π (c) 2π (d) 14. 4

3.

1.

Double refracting crystal splits up in the incident ray 4. into ……… ray.

2.

In double refraction the ray which obey law of refraction is called ……… . 5. The crystal having two optic axes are called ……… 6.

4. 5. 6. 7. 8.

9. 10. 11.

Polarimeter is a device used to measure the ……… of the suger solution. The expression of solution of optically active substance is given as ……… .

The specific rotation of optically active substance of length 1 cm and conc. C with an optical rotation Q True/False is : 1. The light having vibrations only along a single l×C QlC (b) (a) direction perpendicular to the direction of 10 Q 10 Q propagation is said to be linearly polarized light. 10Q 10 (d) (d) 2. In double refraction crystal, ordinary ray does not lC QlC obey law of refraction.

Fill in the Blank(s)

3.

The property of rotating the plane of vibration of the light by any substance is known as ……… .

The velocity of O-ray is ……… in all directions with 7. in the crystal. 8. The velocity of E-ray is ……… in different directions with the crystal. 9. Polarization of light process ……… nature of light.

10. In side the quartz crystal, the E-ray travels ……… as 11. compared to O-ray.

In biaxial crystals, there are two optical axis angle which the velocities of refracted rays are same. sin i , varies inversely For extraordinary ray the ratio, sin r with the angle of incidence. Calcite crystal is a uniaxial doubly refracting crystal. Quartz is a biaxial doubly refracting crystal. In quartz crystal, O-ray travels faster than O-ray. In uniaxial −ive crystal, O-ray travels faster than O-ray. In ruby crystal, E-ray travels faster than O-ray. In Nicol prism, the O-ray is eliminated by T. I . R. Light emerging from Nicol is not uniformly plane polarized.

Polarized light is in general elliptically polarized the 12. ……… polarized and ……… polarized are its special cases. 13. In left handed elliptically polarized light the light vector rotates in ……… direction. 14. The thickness of the quarter wave plate is ……… .

A half wave plate is used in Laurentz half shade polarimeter for the measurement of optic rotation.

A half wave plate introduces a further path difference 15. of ……… .

A white light is used to measure angle of rotation by Laurent’s half shade polarimeter.

A quartz wave plate introduces a path difference of λ/ 2.

Quarter wave plate and half wave plates are true only for the one particular wavelength only.

P-121

Objective Type questions Multiple Choice Questions 1.

(d)

2.

(d)

3.

(c)

4.

(d)

5.

(c)

6.

(a)

7.

(a)

8.

(d)

9.

(c)

10.

(a)

11.

(d)

12.

(b)

13.

(c)

14.

(c)

15.

(b)

16.

(a)

17.

(d)

18.

(b)

19.

(a)

20.

(b)

21.

(c)

Fill in the Blank(s) 1.

two

2.

ordinary light

3.

biaxial crystal

4.

same

5.

different

6.

transverse

7.

slower

8.

plane, circularly

9.

clockwise

λ 4 (µ e − µ o )

10.

t=

13.

strength

11.

λ/2

14.

α=

12.

optical activity

Q l×c

True/False 1.

True

2.

False

3.

True

4.

False

6.

False

7.

False

8.

True

9.

True

10. True

14. True

15. False

11. True

12. False

13. True

5.

True

P-122

H ints and Solutions ∴ Phase retardation 2 × 314 . × (1.3308 − 1.5418) × 0.0032 = 5000 × 10−8

Numerical Questions 1.

We know that Brewster’s law µ = tani P

i P = tan −1 (µ )

or

= 3.62 rad 4.

For crown glass i P = tan −1 (1.520) = 56.66°

We know that thickness of quarter wave plate is given by λ t= 4(µ E − µ O ) Here, λ = 5893 × 10−8 cm = 5.893 × 10−5 cm

= 56° 39.6′

µ O = 1.54425

For flint glass

µ E = 1.55336

−1

i P = tan (1.650) ∴

= 5878 . °

t=

= 58° 46.8′ =

For water i P = tan −1 (1.33) = 53.06° u = tan i P (Brewster’s law) µ = tan 60° = Again

5.893 × 10−5 4 × 911 . × 10−3 cm

= 1.62 × 10−3 cm

= 53° 03.6′ 2.

5.893 × 10−5 4(1.5536 − 1.54425)

5.

We know that C = θ / lS

3

Here θ = 9.9°, l = 20 cm, = 2 dm, S = 66° dm −1

 A + δm  sin    2  µ= A sin 2

∴

C=

= 0.075 g cm −3

Here A = 60° , and µ =

3  60° + δ m      2 3 = − sin sin 30° 3  60° + δ m  = sin     2 2

 60° + δ m  sin 60° = sin     2

3.

or

60° + δ m = 60° 2

or

δ m = 60°

= 75 g/litre The sugar sample dissolved in a litre of water is 80 gm. In this sample, 75 g is pure water. 75 × 100 i. e. 9375 . % ∴ Purity is 80 6.

θ = 12°, l = 10 cm, C = ?

Phase retardation =

Here, µ E = 1.5508, µ O = 1.5418 and

t = 0.032 mm = 0.032 mm = 0.0032 mm

Given that [α] = 52° per decimeter/g cc By the specific rotation θ [α] = lC θ C= ⇒ [α ]l

We know that 2π (∆) λ 2π = (µ e − µ O ) t λ

9.9 g cm −1 2 × 66

On solving, we get answer as 23%. 7.

See example-1

8.

See example-2

Ans. r = 33.69° Ans. i P = 48.4°

P-123 9.

See example-3

5.

The first crystal which polarizes the light wave is called polarizer.

6.

There are certain crystals like calcite, quartz, tourmaline, etc such that when a ray of light is incident upon them then, on entering the crystal, the ray is splitted into two refracted rays. This phenomenon is called double refraction. One of the refracted ray obeys the law of refraction and is called ordinary ray.

Ans. i P = 53° 10.

See example-4 Ans. I = 0.5 I 0

11.

See example-5 Ans. = 75%

13.

See example-6

14.

See example-8

Ans. θ = 45°

The other refracted ray does not obey these laws, hence it is called the extraordinary ray. These crystals show the double refraction is called double refractive crystals.

Ans. t = 1.62 × 10−5 m 15.

See example-9 Ans. t = 1.47 × 10−5 cm

16.

7.

In double refracting crystal one refracted ray obeys the refraction law is called ordinary ray or O-ray and other ray does not obey the law of refraction completely is called extraordinary ray or E-ray.

8.

Please see the article no. 10 (optical rotation).

9.

Quarter wave plate : A plate of doubly refracting uniaxial crystal of uniform thickness whose refracting surfaces are parallel to optical axis and produces a π λ phase difference of   or a path difference of    2  2 between the ordinary ray and extraordinary ray is called as quarter waveplate.

10.

Half wave plate : A plate of doubly reflecting crystal, cut with its faces its parallel to optic axis, such that it introduces a phase difference of π or a path difference of (λ / 2) between ordinary and extraordinary waves, then it is called a half wave plate.

See example-9 Ans. t = 1.47 × 10−5 cm

17.

See example-10 −7

Ans. t = 8.56 × 10 18.

cm

See example-12 Ans. t = 2.945 × 10−5 m

19.

See example-13 Ans. α = 66°

Short Answer Type Questions 1.

2.

Polarized light : In polarized light the vibrations of the electric vector occur in a plane perpendicular to the direction of propagation of light and are confined to a single direction in the plane (do not occur 11. symmetrically in all possible directions). Difference between polarized light and unpolarized light : In poliarzed light the 12. vibrations of electric vector occur in a plane perpendicular to the direction of propagation of light and are confined to a single direction in the plane (do not occur symmetrically in all possible directions). 13. In unpolarized light or ordinary light the vibrations of the electric vector occur symmetrically in all possible directions in a plane perpendicular to the direction of propagation of light.

3.

Plane polarized light : If the light is plane 14. polarized the vibration of electric vector are along a single direction.

4.

If the light vector rotates along a circle, i. e. does not change magnitude while rotating, it will be circularly polarized light.

If the light vector rotates along ellipse. i. e. changes in magnitude while rotating it will be elliptically polarized light. Elliptically polarized light is the resultant of two waves of unequal amplitudes vibrating at right angles π to each other and having a phase difference of . 2 Optical activity : The property of rotating the plane of virbation of a plane polarized light about its direction of travel by some crystal is known as optical activity. The phenomenon is known as optical rotation. The substances which rotate the plane of polarization in clockwise (looking against the direction of light) are called as dexo-rotatory or right handed. The substances which rotate the plane of polarization in the anticlockwise direction are called a laevorotatory or left handed.

P-124 15.

We know that

Specific rotation : The specific rotation of a substance at a particular temperature and for a given wavelength of light used may be defined as the rotation (of plane of vibration of polarized light) produced by on decimeter length of its solution when the concentration is 1 g per cc. θ Thus specific rotation S = l×c

By Brewster’s law, µ = tani p

where i p = Angle of polarization from snell’s law sin i p µ= ( i p = 1)…(2) sin r

‡

Comparing equation (1) and (2) sin i p tan i p = sin r sin i p sin i p = cos i p sin r

where θ is angle of rotation in degrees, l the length of solution in decimeter and C is the concentration of solution in g per cc. 16.

Angle of rotation (θ) is the inversely proportional to 1 the square of wavelength (λ) of light i. e. θ ∝ 2 λ

sin r = cos i p = sin(90°− i p ) r = 90° − i p

Thus it is least for red and greatest for violet. 17.

Half shade polarimeters are instruments signed to measure the angle of rotation produced by a 20. substance. With the help of polarimeter we calculate the with the help of polarimeter we calculate the specific rotation of sugar solution and concentration of solution is also calculated.

19.

The ordinary light also called as unpolarized light, consits of a very large number of vibration in all planes with equal probability at right angles to the direction of propagation. Hence, the unpolarized is represented by a star.

i p + r = 90°

Malus’s law : In 1909, Malus discovered a law regarding the intensity of light transmitted by the analyzer. According to Malus when a completely plane polarized light transmitted through the analyzer varies as the square of cosine of the angle between the plane of transmission of the analyzer and plane of polarizer. I θ ∝ cos 2 θ or

Brewster’s law : According to this law the tangent of the angle of polarization (i p ) is numerically equal to the refractive index µ of the medium, i. e.

I θ = I 0 cos 2 θ

where I 0 be the intensity of incident polarized light

Plane of analyzer

µ = tani p

Plane of polarizer

Brewster also prove that the reflected and refracted ray are perpendicular to each other.

A a cos θ θ

i p + r = 90°

i. e. Proof :

Unpolarized light

Air glass

N

A

ip i p

O

90°

O

Hence proved

i. e. Intensity of transmitted light through the analyzer

Pl l ig ane ht p ol ar ize d

18.

…(1)

a

a sin θ

Fig. 15 : Malus law

r

Very Short Answer Type Questions B (Brewster law)

N' Fig. 14 : Polarization by reflection

1.

There are certain crystals like calcite, quartz, tourmaline, etc such that when a ray of light is incident upon them then entering the crystal the ray

P-125 splitted into two refracted rays. This phenomenon is called double refraction. One of the refracted ray obeys the law of refraction and is called ordinary ray or O-ray. And the other refracted ray does not obey these laws hence it is called extra ordinary ray. 2.

3.

direction of propagation of the light is called plane of vibration.

The crystals which show the phenomenon of double refraction is called uniaxial crystals. The examples of 18. uniaxial crystals are calcite crystal, quartz, tourmaline etc. Thickness of quarter wave plate is given by λ t= 4(µ E − µ O )

Plane of polarization : The plane containing the direction of propagation of light and perpendicular to the plane of vibration is called the plane of polarization. Angle between reflected and refracted ray : Suppose a beam of ordinary is incident on glass surface at polarizing angle i p as shown in fig. Let r be the angle of refraction from Brewster’s law

N

where µ E = Refractive index of extra ordinary ray or (E-ray)

5.

The property of rotating the plane of vibration of plane polarized light about its direction of travel by some crystal is known as optical activity. The phenonomenon is known as optical rotation. θ Specific rotation is given by S = l×C

7.

µ = tani p

µ=

The property of one sideness of electric vector of light is called polarization of light. See the answer of short answer type question no. 4.

9.

The first crystal which polarizes the light wave is called polarizer.

10.

The second crystal which analyzes the light wave is called analyzer.

11.

See the short answer of question no. 7.

12.

Polaroid is a device which measure the intensity of plane polarized light and polaroid used both as polarizer and analyzer.

13.

See the answer of short answer type question no. 15.

14.

See the answer of short answer type question no. 2.

15.

Sound waves never can be polarized.

16.

See the answer of short answer question no. 4.

17.

Plane of vibration : The plane containing the direction of vibration of the electric vector and the

…(1)

From snell’s law

Half shade polarimeter is an instrument to measure specific rotation of sugar solution.

8.

r

Fig. 16 : Brewster’s law

where θ is the angle of specific rotation in degree l is the length of solution in decimeter and C is the conc. of solution in g per cc. 6.

90°

O

glass

λ = Wavelength of incident light. 4.

ip ip

Air

µ O = Refractive index of ordinary ray or O-ray)

sin i p

…(2)

cos r

Comparing equations (1) and (2), sin i p tan i p = sin r cos i p sin i p = cos i p sin r

( ip = i )

‡

sin r = cos i p = sin(90° − i p ) r = 90° − i p r + i p = 90° This shows that at the polarizing angle, reflected ray and refracted rays are at right angle. Hence proved. 19.

See the answer of question no. 1.

20.

Nicol prism is made of calcite crystal. Which shows the phenomenon of double refraction and gives the polarized light.

P-126

22.

Polarimeter : Polarimeters are used to measure the 24. angle of rotation of plane of polarization caused by optically active substance. 25. See the answer of short answer type question no. 6.

23.

See the answer of short answer type question no. 6.

21.

The examples of uniaxial crystals are calcite, quartz etc. Example of biaxial Herpathite crystal etc.

crystal

are

Tourmaline, mmm

P-127

Unit

4

Associated Optical Instruments

1. Interferometer It is an optical instrument which is based upon the principle of interference with the help of interferometer. We can determine the wavelength of light, the difference between two wavelengths and refractive index of any medium.

2. Michelson-Morle’s Interferometer This is a famous interferometer. We can use this to determine the wavelength of light and width of spectral lines. This interferometer is devloped by Michelson and Morle’s in 1881 and it is based upon the principle of interference.

2.1 Construction S is a monochromatic light source (Sodium lamp) L is a convex lens. Screw 10–5 cm

M1 1

Half silvered

1 S

45° Convex lens L

2

1 G1

45° 2

Screw 2

2

'T ' Telescope Fig. 1 : Michelson-Morle’s interferometer

M2

P-128 M 1 and M 2 are the two mirrors having screws on back side with the help of them both mirrors can be rotated. G1 and G2 are the two plates of same material and same refractive index with same thickness both plates are inclined at 45° angle. Plate G1 is a half-silvered plate and its side towards G2 is half-silvered. So, it can be used as a beam splitter. Mirror M 1 can be move in the direction of arrow and the least count of its movement is 10−5 cm. ‘T ’ is the telescope where interference between two waves occurs.

2.2 Working As light emits from source S, it falls upon lens ‘L’ which converts this light into parallel beams. This beam is made to full on glass plate G1 which splits this beam into two beams. One of them moves towards mirror M 1 ray (1) and other moves towards mirror M 2 ray (2). Both rays strikes the mirror M 1 and M 2 and reflects back towards G1 meets together and proceed towards telescope T1 where phenomenon of interferences is observed.

2.3 Working of Plate G 2 Plate G2 is similar to plate G1, this is also called compensating plate. If this plate is not placed in the path of ray (2) then ray (2) will travel in completely air until it reached at M 2 and reflects back at G1. While in compare at this, ray (1) cross plate G1 two times. First time when moving towards M 1 and second time when it moving towards T after reflecting mirror M 1. Hence, extra path difference 2 (µ − 1) t will produce in ray (1) where µ-refractive index of plate and t is the thickness. At this condition, both ray (1) and ray (2) cannot in same phase. For making them into same phase, we need a glass plate G2 of same refractive index µ and thickness in the path of ray (2). Hence, now ray (2) has extra path difference 2 ( µ − 1) t and hence, constant phase difference between ray(1) and ray (2) observed which is the necessary condition of interference.

2.4 Types of Fringes The fringes in this interferometer depends upon the inclination of M 1 and M 2. Let M 2′ be the image of M 2 formed by the reflection at half silvered surface of the plate P, so that G1M 2 = G1M 2′ . The interference fringes may be regarded as formed by the light reflected from M 1 and M 2′ . Hence, an air film develops between M 1 and M 1′ . But the path difference between two rays is 2 ( µ − 1) t. This path difference can be varied by moving M 1 backward or forward. Now, we see the fringes in telescope and the shape of the fringes depends upon the inclination of M 1 and M 2. If M 1 and M 2 are at right angle to each other i. e. mirror M 1 and M 2′ (image of M 2) are parallel. Thus, we get circular fringes. When the distance between the mirrors M 1 and M 2 or between M 1 and M 2′ decreases, the circular fringes shrinks and vanishes at the centre. A ring disappears each time when the path 2(µ − 1)t decreases by λ. If M 1 and M 2 are not perfectly perpendicular a wedge-shaped film will be formed between M 1 and M 2′ then we get almost straight line fringes of equal thickness in the field of view of telescope, as the radius of fringes is very large. Fig. 2 : Types of interference fringes

P-129

3. Applications of Michelson-Morle’s Interferometer 3.1 To Determine the Wavelength of Sodium Light First of all set up both mirror M 1 and M 2 perpendicular to each other and obtain the circular fringes. For this we put a pinhole between lens L and plate G1. We get four images of this pinhole, two of them are fade and other two are sharp. Now, mirror M 1 is moved away by working the micrometer screw. So, both fade images coincides with two sharp images in such condition that both mirrors are perpendicular. Now, remove pinhole. Let mirror M 1 is moved with distance d and then n th rings displaced. So,

2d = nλ

…(1)

or

2d + λ = λ(n + 1)

…(2)

λ  Hence, if d becomes  d +  then (n + 1 ) fringes displaced.  2 Now, if mirror M 1 is moved with distance x and N fringes becomes displaced. So,

2x = Nλ 2x λ= N

We can determine the wavelength of light. This method is very accurate method.

3.2 Determination of Difference between Two Close Wavelength and Width of Spectral Lines Michelson-Morle’s interferometer is adjusted in order to obtain the circular fringes. Some sources such as sodium lamp are not perfect monochromatic source. Sodium lamp has two very near wavelengths λ 1 and λ 2 are called sodium D1 and D 2 line. The two wavelengths form their separate fringes patterns but as λ 2 and λ 2 are very close to each other and thickness of air film is very small, the two patterns practically coincide with each other. As the mirror M 1 is moved slowly, the two patterns of circular fringes separate slowly and when the thickness of air film is such that the dark ring of λ 1 coincides with bright ring of λ 2 and vice-versa. This is called maximum in distinctness. Now keep moving mirror M 1 a position comes when bright ring of λ 1 coincides with bright ring of λ 2 and similarly for dark ring. This condition is called maximum distinctness. Let mirror M 1 is moved with x distance between two maximum distinctness or indistinctness. If n fringes of λ 1 wavelength and (n + 1) of λ 2 wavelength is appear at centre. So, for λ 1 wavelength 2x = nλ 1

or n =

2x λ1

…(1)

For λ 2 wavelength 2x = (n + 1)λ 2 2x (n + 1) = λ2 Subtracting equation (2) − equation (1) n +1−n=

2x 2x − λ 2 λ1

…(2)

P-130 λ − λ 2 1 = 2x  1   λ 1λ 2 

or or

λ1 − λ 2 =

λ 1λ 2 2x

…(3)

If λ 1 ≈ λ 2 so, λ 1 − λ 2 ≈ λ 2 λ=

∴

λ1 + λ 2 2

The equation (3) gives the difference of two near wavelengths called width of spectral lines.

4. Twyman-Green Interferometer This interferometer is same as Michelson-Morle’s interferometer, we can test the glass plate, prism or lens etc. with the help of this interferometer. This interferometer was developed by Twyman and Green in 1916.

4.1 Construction and Working The apparatus of Twyman-Green interferometer is shown in figure (3). S is a light source, L is a lens, M 1 and M 2 are two glass mirrors and P1 and P2 are the two glass plates where P1 is at 45° angle and P2 is the plate to be tested. As light rays fall upon lens L1 after emitting from source S and lens L1 converts these rays into parallel light rays. These light rays falls upon half silvered plate which divides these rays into two light rays. One will move towards mirror M 1 and other will move towards M 2. After reflecting from mirror reached at telescope passing through P1. In telescope interference occurs between the fringes. Screws

1

M1

1

1 1

2

2

2

2

S

L

1 2

1 2

Testing plate Telescope

Fig. 3 : Twyman-Green interferometer

M2

P-131 Now put a glass plate to be tested in the path of light ray (2). If plate is not completely plain at a point then a wedge shape film will develop at this place and hence circular rings will formed in telescope. Now take off this plate and polished this area. Now put for again testing. If the plate is completely plain then no circular fringe will form. If we test a convex lens then we use convex mirror in place of mirror M 2 and for fine testing we use LASER light in place of light used in experiment.

5. Fabry-Perot Interferometer This is another famous interferometer developed by C. Fabry and A. Perot in 1899. Interferometer is based upon the principle of interference of infinite light rays.

5.1 Construction and Working Different parts of Fabry-Perot interferometer are shown in figure 4. It has two glass plates P1 and P2 both are silvered at inner side and both plates reflected 75% to 85% part of light. Plate P2 is stable and some screws are mounted on back side of this plate with the help of these screws. We can rotate plate P2. Plate P1 can be moved in the direction of arrow and we can measure this distance upto 10−5 cm accuracy. A thin film of air is now developed between P1 and P2 we can vary its thickness by sliding P1. S is a broad light source, L is a convex lens light rays falls upon P1 after crossing L. Now between P1 and P2, multiple reflection occurs and finally we get many transmitted rays. These rays now focused on screen by adjusting the lens of telescope. S2

t

S1 90° θ

S θ

Broad source L

Plates system

Telescope 'T ' Fig. 4 : Working of F-P interferometer P1

P2

Screen

If light ray incident at P1 makes angle θ then the transmitted rays also have the same angle ‘θ’. That is why fringes are called Haidinger fringes i. e. fringes of equal inclination.

P-132

5.2 Derivation for Coefficient of Finesse SA1 is the incident light ray, incident at plate P1, P1 t

R4 R3

A4

R2

A3

R1

A2

S

P2 P B4 B3 B2 B1

A1

T Fig. 5

Some part of its amplitude is reflected ( A1R1) and some part is transmitted A1B1. Thus, there will be multiple reflection occurs between P1 and P2. Hence, we get some reflected rays A1, R1, A2R2, ... and some transmitted rays B1T1, B2T2 ... Let R is reflected part of amplitude and T is the transmitted part of amplitude. If amplitude of incident ray is a then amplitude of transmitted rays B1T1 = a T T = aT B2T2 = a T R R T = aTR B3T3 = a T R R R R T = aTR 2 …………………………………… …………………………………… Hence, amplitudes are aT , aTR, aTR 2 ……… Now, displacements y = ae iwt y1 = aTe iwt y 2 = aTRe i(wt − δ) y 3 = aTR 2e i(wt − 2δ) …………………… …………………… Thus, amplitude of resultant wave A = aT + aTRe − iδ + aTR 2e −2iδ + ...... A = aT [1 + Re − iδ + R 2e −2iδ + ...] This series can be written as   1 A = aT  − iδ  1 − Re 

P-133 Its complex conjugate   1 A* = aT  + iδ  1 − Re  Now intensity I = AA* So,

  1   1 × aT  I = aT  − iδ  iδ  1 Re −  1 − Re  

or

  1 I = a 2T 2  − iδ iδ  (1 − Re )(1 − Re )

or

  1 I = a 2T 2  iδ − iδ 2 1 − Re − Re + R 

or

  1 I = a 2T 2  2 iδ − iδ  1 + R − R(e + e )

or

  1 I = a 2T 2   2 1 + R − 2R cosδ 

or

  1 I = a 2T 2   2 + R − R + R − R 1 2 2 2 cosδ  

or

  1 I = a 2T 2   2 ([1 − R) + 2R(1 − cos δ)

But

cos δ = 1 − 2 sin 2

 e iδ + e − iδ  ‡ cosδ =   2  

δ 2

or

    1  I = a 2T 2     2 2 δ (1 − R) + 2R 1 − 1 − 2 sin 2     

or

 1 I = a 2T 2  2 − R + R sin 2 δ / ( 1 ) 4 

or

   1 a 2T 2  I=   (1 − R)2 1 + 4 R sin 2 δ   2  (1 − R)2

or

I=

1 a 2T 2   (1 − R)2 [1 + F sin 2 δ /

  2 …(1)

  2

Where ‘F’ is known as coefficient of finesse and

F=

4R (1 − R)2

…(2)

P-134 Now, it is clear from equation (1) that intensity depends upon

δ 2

i. e. for maximum intensity δ = 2nπ δ sin ≈ 0 2 So,

Imax =

a 2T 2

…(3)

(1 − R)2

For minimum intensity δ = (2n + 1)π δ sin = 1 2     aT  1  = (1 − R)2  4 R2  1 +  (1 − R)2   2

Imin

2

  1 Imin = a 2T 2   2 − + ( 1 ) 4 R R   Imin = Imin =

a 2T 2 1 + R 2 − 2R + 4 R a 2T 2

…(4)

(1 + R)2

5.3 Visibility of Fringes The visibility of fringes is given as V=

Imax − Imin Imax + Imin a 2T 2

V=

(1 − R)2 a 2T 2 (1 − R)2

− +

a 2T 2 (1 + R)2 a 2T 2 (1 + R)2

a 2T 2[(1 + R)2 − (1 − R)2] V=

(1 − R)2(1 + R)2 a 2T 2[(1 + R)2 + (1 − R)2] (1 − R)2(1 + R)2

V=

(1 + R)2 − (1 − R)2 (1 + R)2 + (1 − R)2

P-135 V= V= V=

1 + R 2 + 2R − 1 − R 2 + 2R 1 + R 2 + 2R + 1 + R 2 − 2R 4R 2 + 2R 2 2R (1 + R)2

This is called visibility of fringes.

5.4 Fabry-Perot (Etalon) The biggest difference between Fabry-Perot interference and Etalon is that in interferometer the distance between P1 and P2 is not fixed while it is fixed in Etalon. Thus the thin film of air formed between P1 and P2. To fix both plates arranged both plates in a rigid from the distance between two plates is nearly 100 mm for best use. The resolving power of Etalon is very high.

6. Tolansky Fringes In 1945, Tolansky used white light in place of monochromatic light in Fabry-Perot interferometer and found interference fringes called equal chromatic fringes. Such kind of fringes are called Tolansky fringes. We know that, Resultant intensity in, Fabry-Perot interferometer I= where F =

4R 2

(1 − R)

and phase difference δ =

Imax

1 + F sin 2 δ / 2

2π (2 µ + cos θ). λ

Hence, intensity I depends upon phase difference δ and δ depends upon µ , t, θ and λ. Here, µand t are kept constant and θ = 0 for perpendicular incident. Hence, δ depends upon λ i. e. for the value of λ the intensity I can be maximum and minimum. Hence, the locus of points of equal wavelengths are called Tolansky fringes.

Example 1. A Michelson interferometer is set for the white straight fringes. When a mica sheet of thickness 0.005 cm is put in front of the fixed mirror then in order to bring back the coloured fringes to their original position, the movable mirror is moved by 0.0025 cm. Calculate the refractive index of mica. Solution : Given, x = 2.5 × 10−5 m t = 5 × 10−5 m Formula used is 2x = 2 ( µ − 1) t

P-136 where t is the thickness of mica sheet and µ is the refractive index. x µ= +1 t µ=

2.5 × 10−5 5 × 10−5

+ 1 = 1.5

µ = 1.5 Example 2. If a movable mirror of Michelson’s interferometer is moved through a distance 0.06 mm, 200 fringes crossed the field of view. Find the wavelength of light. Solution : Given, x = 6 × 10−5 m

or

N = 200 fringes 2x Formula used = λ = N Nλ x= 2

Where x is the separation of movable mirror from the fixed mirror then 6 × 10−5 = 200 ×

λ or λ = 6000 Å 2

Example 3. In Michelson’s interferometer, a thin plate is introduced in the path of one of the beams and it is found that 50 band crosses the line of observation. If the wavelength of light used is 5896 Å and µ = 1.4, determine the thickness of the plate. Solution : Given, n = 50 λ = 5896 × 10−10 m µ = 1.4 Formula used 2(µ − 1)t = nλ =

nλ 50 × 5896 × 10−10 = 2 (µ − 1) 2 (1.4 − 1)

t = 3.68 × 10−5 m Example 4. In Michelson’s interferometer 100 fringes crossed the field of view when the movable mirror is displaced through 0.02948 mm. Calculate the wavelength of monochromatic light used. Solution : Given, x = 0.02948 × 10−3 m n = 100 Formula used λ=

2x N

P-137 =

2 × 2.948 × 10−5 m 100

= 5.896 × 10−7 m = 5896 Å Example 5. Calculate the distance between successive positions of the mirror M1 of Michelson’s interferometer giving best fringes in case of a sodium source having wavelength 5896 Å and 5890Å. What will be the change in path difference between two successive reappearances of the interference pattern ? Solution : Given, λ 1 = 5.896 × 10−7 m and

10−7 × λ 2 = 5.89 × 10−7 m

Formula used ∆λ = (λ 1 − λ 2) = or

x= =

λ 1λ 2 2x

λ 1λ 2 2(λ 1 − λ 2) 5.896 × 5.89 × 10−14 2 × 6 × 10−10

= 0 .289 mm

The path difference will be equal to = 2x = 2 × 0.289 mm = 0.5788 mm Example 6. The wavelength of two components of D-lines of sodium are 5890 Å and 5896 Å. By how much distance one of the mirror of Michelson’s interferometer be moved so as to obtain consecutive position of maximum distinctness? Solution : Given, λ 1 = 5.896 × 10−7 m λ 2 = 5.89 × 10−7 m Formula used ∆λ = λ 1 − λ 2 =

λ 1λ 2 2x

where x is distance through which the movable mirror is moved from one position of maxima to the next, then we have λ 1λ 2 x= 2(λ 1 − λ 2) x=

5.896 × 5.89 × 10−14 2 × 6 × 10−10

x = 0.289 mm

P-138 Example 7. In an experiment with Michelson’s interferometer, the distance travelled by the mirror for two successive position of maximum distinctness was 0.2945 mm. If the mean wavelength for the two component of sodium D-line is 5896 Å. Calculate the difference between the two wavelengths. Solution : Given, x = 0.2945 × 10−3 m and

λ aV = 5.896 × 10−7 m

Formula used δλ = λ 1 − λ 2 = − =

λ 2av 2x

(5.893 × 10−7 )2 2 × 0.2945 × 10−3

∆λ = 5.896 Å Example 8. In an experiment for determining the refractive index of a gas using Michelson’s interferometer a shift of 140 fringes is observed when all the gas is removed from the tube. If the wavelength of light used is 5460 Å and the length of the tube is 20 cm. Calculate the refractive index of the gas. Solution : Given, λ = 5.46 × 10−7 m t = 0.2 m n = 140 Formula used

or

2 (µ − 1)t = nλ nλ µ= +1 2t =

140 × 5.46 × 10−7 + 1 = 0.00019 + 1 2 × 0.2

µ = 1 .00019 Example 9. In moving one mirror in a Michelson interferometer through a distance of 0.1474 mm, 500 fringes cross the center of the field of view. What is the wavelength of light ? Solution : Let x be the distance moved by the mirror when N fringes cross the field of view, then λ 2x x = N or λ = 2 N In the given problem x = 014 . 74 mm or

x = 0.01474 cm

and

N = 500

P-139 Therefore,

λ=

2 × 0.01474 500

λ = 5896 × 10−8cm or

λ = 5896 Å

Example 10. For a sodium lamp, the distance traversed by the mirror between two successive disappearances is 0.289 mm. Calculate the difference in the wavelengths of the D1 and D2 lines of the sodium lamp. Assuming λ = 5890 Å. Solution : We have, λ1 − λ 2 =

λ2 2x

In the given problem x = 0.289 mm or

x = 0.0289 cm

and

λ = 5890 Å

or

λ = 5890 × 10−8 cm

Therefore,

λ1 − λ 2 =

5890 × 10− 8 × 5890 × 10− 8 2 × 0.0289

λ 1 − λ 2 = 6 × 10−8 cm Example 11. When the movable mirror of Michelson interferometer is moved through 0.05896 mm, a shift of 200 fringes is observed, what is the wavelength of light used ? Solution : This distance x moved by the mirror when N fringes cross the field of view is given by Nλ x= 2 2x ∴ λ= n Here, x = 0.05896 mm = 5.896 × 10−2 mm and N = 200 ∴

λ=

2 × 5.896 × 10−2 200

= 5.896 × 10−4 mm λ = 5896 Å Example 12. In Michelson’s interferometer the scale readings for two successive maximum indistinctness of fringes were found to be 0.6939 mm and 0.9884 mm. If the mean wavelength of the two components of light be 5896 Å, deduce the difference between the wavelengths of the components. Solution : We have, ∆λ =

λ 1 × λ 2 λ2 = 2x 2x

P-140 Where λ is the average of λ 1 and λ 2. Here,

λ = 5893Å = 5893 × 10−10 m

and

x = 0.9884 − 0.6939 = 0.2945 mm = 2.945 × 10−4 m ∆λ =

∴

(5893 × 10−10 m )2 2 × 2.94 × 10

−4

m

= 5.896 × 10−10

∆λ = 5.896 Å Example 13. Calculate the visibility of the fringes for a reflection of 80% multiple-beam interferometery. Solution : The visibility of fringes V, in terms of reflectivity R is given by V= Here,

Imax − Imin 2R = Imax + Imin 1 + R 2

R = 80% = 0.8 2 × 0.8 V= = 0.976 1 + (0.8)2

∴

Example 14. Two F– P interferometers have equal plate separations but coefficients of intensity reflection are 0.80 and 0.90. Deduce the relative width of the maxima in the two cases. Imax Solution : I= δ 1 + F sin 2 2 4R and R is the reflecting power of the plates. Where F = (1 − R)2 1 1 = 2 1 + F sin 2 δ ′ 2 δ′ F sin 2 = 1 2 δ′ 1 sin = 2 F For large R, 1/ F is so small that we can write sin ∴ For R = 0.80, F = ∴

4R (1 − R)2

δ′ δ ′ = 2 2 2 δ′ = F

= 80 and for R = 0.90, F = 360 δ1 = δ2

F2 = F1

360 = 80

9 2

P-141 Example 15. For a source radiating at mean Å the coherence time τ is 2 × 10 −10 sec. Deduce the order of magnitude of (i) coherence length L, (ii) the spectral width of the line, (iii) purity factor Q. Solution : (i) The coherence length L = τC = 2 × 10−10 × 3 × 106 L = 6 × 10−2 m (ii)

The spectral width ∆λ =

λ 2 (6000 × 10−10)2 = 3 × 10−12 m = 0.03 Å = −2 2L 2 × 6 × 10 6000 × 10−10 m = 2 × 105 0.03

(iii)

The purity factor

1.

If plate separations in F − P interferometer is same and coefficients of intensity reflection are 0.60 and 0.90, deduce the relative width of maxima in two cases.

2.

For a source radiation at mean wavelength λ = 5896Å, the coherence length is 6 × 10−2 m. Calculate the width of

Q=

spectral line. 3.

Calculate the visibility of the fringes for a reflection of 90% in multiple beam interferometer.

4.

If mirror M1 is moved through 0.05896 m and 150 fringes displaced from the central point, what is the wavelength of light used ?

5

If λ = 6000Å and x = 0.0289 cm, calculate λ1 − λ 2 = ?

6.

A Michelson’s interferometer is set for the straight fringes when a mica sheet of thickness 0.005 cm is put in front of fixed mirror then in corder to bring back the coloured fringes to their original position the mirror M1 is moved by 0.0025 cm. Calculate refractive index of mica.

7.

If mirror M1 is moved through a distance 0.06 mm, 200 fringes crossed the field view, find the wavelength of light.

8.

In Michelson’s experiment a thin plate is introduced in the path of one of the beams and it is found that 50 band crosses the line of observations. If λ = 5896Å and µ = 1.4, determine the thickness of plate.

9.

In Michelson’s experiment 100 fringes crossed the field of view when mirror M1 is halved through 0.02948 mm. Calculate the wavelength of light.

10.

Calculate the distance between two successive position of the mirror M1 of Michelson’s interferometer giving best fringes in case of a sodium source having wavelength 5896 Å and 5890Å. What will be the change in path difference between two successive reappearances of interference pattern ?

11.

The wavelength of two compounds of D-lines of sodium are 5890Å and 5896 Å. By how much distance one of the mirror of Michelson’s interferometer be moved so as to obtain consecutive position of distinctness?

12.

In Michelson’s experiment the distance travelled by mirror M1 in two consecutive position of maximum distinctness was 0.2945 mm. If the mean wavelength for the two components of sodium D-line is 5893 Å. Calculate the difference between the two wavelengths.

13.

In an experiment for determining the refractive index of a gas using Michelson’s experiment a shift of 140 fringes is observed when all the gas is removed from the tube. If the wavelength of light used is 5460 Å and the length of the tube is 20 cm, calculate the refractive index of the gas.

P-142 14.

In moving one mirror in Michelson’s experiment through a distance of 0.1474 mm and 500 fringes cross the centre of the field of view. What is the wavelength of light ?

15.

For sodium light, the distance travelled by mirror M1 between two successive disappearances is 0.289 mm, calculate the difference in the wavelength of D1 and D 2 lines of sodium lamp λ = 5890 Å.

16.

When the mirror M1 of Michelson’s interferometer is moved through 0.05896 mm a shift of 200 fringes is observed. What is the wavelength of light B used?

17.

In Michelson’s interferometer the scale reading for two successive maximum indistinctness of fringes were found to be 0.6939 mm and 0.9884 mm. If average wavelength of D1 and D 2 line is 5893 Å then calculate the difference between D1 and D 2 .

18.

Calculate the visibility of fringes for reflections of 80% multiple beam interferometers.

19.

F − P interferometer have equal plate separations but coefficients of intensity reflection are 0.80 and 0.90. Calculate the relative width of maxima in two cases.

20.

For a source radiating at mean Å the coherence time τ is 2 × 10−10 sec. Deduce the order of magnitude of (i)

Coherence length

(iii) Purity factor

(ii) Spectral width of line

P-143

Long Answer Type Questions

13.

Discuss the brief study of some interferometer.

1.

What do you mean by an interferometer ? Discuss 14. Michelson-Morle’s interferometer in detail.

2.

3.

Give the construction and working of 15. Discuss the F.P. interferometer and F.P. Etalon. Michelson-Morle’s interferometer. Discuss the working of compensating plate. What kind of fringes Short Answer Type Questions we can produce by moving mirror M1 and M 2 ? What is interferometer ? Decribe the construction and working of a 1.

4.

Michelson’s interferometer and explain the use of 2. compensating plate. Describe how the interferometer may be used to obtain (i) circular 3. fringes (ii) straight lined fringes. Discuss the construction and working of Michelson’s 4. interferometer. Give its one application.

5.

Explain the Michelson’s interferometer is used to determine. (i)

Wavelength of light

(ii) Difference between two wavelengths 6.

7.

8.

9.

5.

interferometer

What is interferometer ? Discuss the work of compansatory plate in Michelson’s interferometer ? Draw neat and clean of Michelson’s interferometer. 2x . Derive the formula λ = N Give the applications of Michelson’s interferometer to determine the wavelength of sodium light. Give one application of Michelson’s interferometer. Give the applications of Michelson’s interferometer to determine the difference between two wavelengths. Determine the difference between two close wavelengths or width of spectral lines. Write a short interferometer.

note

on

Twyman-Green

Write a short note on F.P. Etalon. Write a short note on Tolansky fringes.

interferometer. Show that coefficient of finesse is 4R F= (1 − R)2

12.

Discuss maximum and minimum intensities.

Very Short Answer Type Questions

Describe the theory and working of Fabry-Perot 1. interferometer. Show that visibility of fringes 2. 2R V= (1 + R 2 ) 3.

10.

11.

Write short note on : Tolansky fringes

(i)

What is interferometer ? What is the work of plate G2 in Michelson’s interferometer ? On what principle interferometer is based on ? What is the least count of mirror M1 in Michelson’s interferometer ? What are the basic applications of Michelson’s interferometer ? What are the types of fringes in Michelson’s interferometer ?

7.

At what angle we can find curved fringes in Michelson’s interferometer ?

8.

At what angle we can find circular fringes in Michelson’s interferometer ?

Wavelength of light

(ii) Width of spectral lines

Write a short note on fringes produced by Michelson’s interferometer.

6.

(ii) Twyman-Green interferometer Write short note :

one

6.

Describe the construction and working of 8. Michelson’s interferometer and give its applications to determined the wavelength of sodium light. 9. Explain the principle of Fabry-Perot interferometer. Obtain an expression for the intensity distribution 10. and discuss sharpness of fringes. Describe theory and working of Fabry-Perot 11.

(i)

with

7.

Discuss Fabry-Perot interferometer in details. What is 4. the difference between Fabry-Perot interferometer and F.P. Etalon. 5.

12.

Discuss Michelson’s application.

P-144 9.

What is the distance travelled by Mirror M1 if one ring is displaced in Michelson’s experiment ?

(c) Centimeter order (d) None of the above

10.

What is the formula of visibility of fringes in F.P. 6. interferometer ?

11.

What is the angle of glass plate in Michelson’s interferometer ?

12.

What is the formula of difference between two near wavelengths ?

The instrument which is capable of forming the fringes of equal thickness and of equal inclination is : (a) Michelson interferometer (b) Fabry-Perot interferometer (c) Newton's rings (d) None of the above

13.

What is the use of Twyman-Green interferometer ?

14.

What is the path difference is produced in a light ray in Michelson’s interferometer ?

15.

Give the example of light source having two nearest wavelengths.

7.

16.

Who developed F.P. interferometer ?

17.

What is the main feature of Tolansky fringes ?

18.

Which interferometer is very near to Michelson’s 8. interferometer ?

19.

What are Haidinger fringes ?

20.

What is the difference between Tolansky and F.P. interferometer ?

Objective Type Questions 9.

Multiple Choice Questions 1.

Michelson interferometer with monochromatic light, the central fringe is : (a) Bright (b) Dark (c) White (d) May be dark or bright

2.

In Michelson interferometer with monochromatic 10. light, when mirrors are exactly perpendicular, the fringes are : (a) Straight line (c) Circular

3.

In Michelson interferometer the order of central fringes is n = 1 : 11. (a) Maximum (b) 0 (c) 1

4.

5.

(b) Elliptical (d) Parabolic

(d) 2

The least count of the micrometer attached with the Michelson interferometer is of the order of : (a) 10−5 m

(b) 10−5 cm

(c) 10−10 cm

(d) 105 cm

In the Michelson interferometer, two waves interfere after travelling a distance of : (a) Wavelength order (b) Kilometer order

12.

In the Michelson interferometer the shape of Fringes depends upon : (a) Distance of source from compensating plate (b) Distance of semi-silvered plate from compensating plate (c) The inclination between two mirrors (d) Distance of semi-silvered plate from the telescope. In the Michelson interferometer, if a uniform sheet of thickness t and refractive index µ is introduced in one of the paths, then the additional one-way path introduced in that beam is : D (a) 2(µ − 1) t (b) (µ − 1) t 2d (c) (µ − 1)t (d) µ t In the Fabry-Perot interferometer, the sharpness of maxima depends upon : (a) The inclination between the two plates (b) The reflectivity of the glass plates (c) The resolving power of a telescope (d) The distance between the glass plates The most suitable method for the determination of the refractive index of gases and liquid is : (a) The Jamin refractometer (b) The Fabry-Perot interferometer (c) Newton’s rings (d) The Michelson interferometer Which of the following is not suitable for a non-reflection coating ? (a) Zinc sulphide (b) Magnesium fluoride (c) Sodium aluminum fluoride (d) None of the above In Fabry-Perot interferometer : (a) Interference between infinite waves (b) Interference between two waves (c) Interference does not occur (d) None of the above

P-145 13.

14.

One ring will displace in Michelson-Morle’s 6. interferometer is mirror M1 is travelled with distance : λ (a) λ (b) 7. 2 8. λ λ (c) (d) 4 6 9. The fringes in Michelson-Morle’s experiment will be circular if the angel between two mirror is : 10. (a) 0° (b) 90° (c) 180°

15.

(d) 45°

I max − I min I max + I min

is used for ……… of fringes.

The fringes of equal inclination are ……… fringes. Glass plates are at angle ……… in Michelson-Morle’s interferometer. When mirror M1 and M 2 are the perpendicular to each other the fringes will be ……… Tolansky used ……… light in place of ……… light.

Fringes are very sharp for :

True/False

(a) Michelson interferometer

1.

We cannot measure wavelength of light by Michelson interferometer.

2.

We use monochromatic light in Michelson-Morle’s interferometer.

3.

Accuracy of measuring wavelength is very high for Michelson interferometer.

4.

Glass plate G2 is called compensation plate.

(b) Fabry-Perot interferometer (c) Tolansky interferometer (d) Newton’s rings 16.

Formula

Interferometer depends upon the principle of : (a) Interferometer

(b) Diffraction

(c) Polarization

(d) Double refraction

Fill in the Blank(s)

5.

Movable mirror is M1 .

6.

Straightlined fringes formed when mirror M1 and M 2 has any angle between them.

1.

The least count of mirror M1 in Michelson-Morle’s 7. interferometer is ……… cm.

2. 3.

The formula for measuring wavelength by Michelson 8. Morle’s experiment is ……… 9. Movable mirror is ……… mirror.

4.

Glass plate is inclined at ……… angle.

5.

In Michelson-Morle’s interferometer the central fringe order is ……….

10.

Twyman-Green interferometer cannot test the glass plates. Tolansky fringes are equal chromatic fringes. Least count of scale in Michelson-Morle’s interferometer is 10−10 m. − I min I is the formula of sharpness of fringes. V = max I max + I min

P-146

Objective Type Questions Multiple Choice Questions 1.

(a)

2.

(b)

3.

(a)

4.

(b)

5.

(c)

6.

(a)

7.

(c)

8.

(a)

9.

(c)

10.

(a)

11.

(a)

12.

(a)

13.

(b)

14.

(b)

15.

(b)

16.

(a)

Fill in the Blank(s) 1.

10 −5

2.

λ=

2x N

3.

M1

4.

45°

5.

maximum

6.

sharpness of fringes

7.

Haidinger

8.

45°

9.

circular

10.

white light, monochromatic light

True/False 1.

False

2.

True

3.

True

4.

True

6.

False

7.

False

8.

True

9.

False

5.

True

10. True

P-147

H ints and Solutions numerical Questions 1.

We know that intensity distribution I max I = δ 1 + F sin 2 2 4R where, F = (1 − R)2

3.

Visibility of fringes − I min I 2R = V = max I max + I min 1 + R 2 Here

4.

x=

Nλ 2

λ=

2x 2 × 5.896 × 10−2 = 7861Å = N 150

Here, R is the reflecting power of the plates. Let δ′ is the half width of fringes 1 So, I = I max 2 1 1 = 2 1 + F sin 2 δ′ 2 δ′ 1 2 δ′ = 1 or sin 2 = F sin 2 2 F 1 For large R, is so small, F δ′ δ′ Thus we can write sin = 2 2 For R = 0.60

F=

λ2 6000 × 10−8 × 6000 × 10−8 = 2x 2 × 0.0289 6 × 10−5 × 6 × 10−5 0.0578

= 622.837 × 10−10 = 6.2 × 10−8 cm 6.

See example-1. Ans. 1.5

7.

See example-2 Ans. 6000 Å

4R

See example-3 Ans. 3.68 × 10−5 m

(1 − R)2 4 × 0.60 (1 − 0.60)2

9. = 15

See example-4 Ans. 5896 Å

10.

For R = 0.90

See example-6 Ans. 0.5788 mm

F = 360 δ1 δ2

=

360 = 15

11.

24 1

12. 13. 14.

2 × 6 × 10−2

dλ = 0.028 Å

m

See example-9 Ans. 5896 Å

2

= 2896901.33 × 10−18 m = 0.028 × 10

See example-8 Ans. 1.00019

(5896 × 10−10 )2

−10

See example-7 Ans. 5.896Å

λ2 dλ = 2L

= 2896901.33 × 10−20 +

See example-6 Ans. 0.289 mm

We know that, width of spectral line

dλ =

λ1 − λ 2 = =

8. F=

2.

5.

R = 90% = 0.9 2 × 0.9 1.8 = = 0.994 V= 1 + 0.92 1.81

15.

See example-10 Ans. λ = 6 × 10−8 cm

16.

See example-11 Ans. λ = 5896 Å

P-148 17.

See example-12

(ii) To determine the difference between two nearest wavelengths.

Ans. ∆λ = 5.896 Å

Short Answer Type Questions

6.

(i)

7.

At any angle

Straight lined or curved (ii) Circular

1.

See article 1

8.

90°

2.

See 1 and 2.3

9.

λ /2

3.

See diagram in article 2.1

10.

V=

4.

See article 3.1

5.

See article 3.1

11.

45°

6.

See article 3.1

12.

7.

See article 3.2

λ1 − λ 2 =

8.

See article 3.2

13.

To test the glass plate

9.

See complete article 4

14.

10.

See article 5.4

11.

See article 6

12.

See article 2.4

Very Short Answer Type Questions

λ2 2x

2(µ − 1)t µ → refractive index t → thickness of plate

15.

Sodium light (5896 Å and 5890 Å)

16.

Fabry and Perot.

17.

There are equal chromatic fringes.

2.

It is an optical instrument which is based upon the 18. principle of interference. 19. To compensate the path difference. 20.

3.

Interference principle

4.

10−5 cm

5.

(i) To determine the wavelength of sodium light.

1.

2R (1 + R 2 )

Tolansky interferometer. The fringes of equal thickness. In F.P. interferometer we use monochromatic light and while we use white light in Tolansky interferometer. mmm

Book-3 Statistical Physics Unit-1 : Basic Concepts Unit-2 : Ensembles and Thermodynamic Connections Unit-3 : Classical Statistics Unit-4 : Quantum Statistics (BES & FDS)

S-3

Unit

1

Basic Concepts

1. Basic Postulates of Statistical Physics To study the statistical physics the macroscopic or microscopic properties of a system such as : (i) total energy of a system at temperature T and volume V (ii) pressure exerted on the wall of the container by the particles of the system (iii) distribution of the particle in different energy ranges or different velocity ranges. Following postulates are made : 1.

To consider a phase space : We consider the phase space to represent each microstate of the system by a point in it. If a system hf-degree of freedom, the phase space is 2f dimensional with f -position axes and f momentum axes.

2.

To construct a statistical ensemble : We calculate the number accessible microstates of the system under the constrains, such as number of particles of the system is constant, energy is constant and volume is constant. The accessible microstates constitute a statistical ensemble.

3.

To divide phase space into phase cells : The phase space is divided into small phase cells. Each has the same volume. Each phase cell corresponds to a microstate of that system. If a system has f degree of freedom, the volume of each phase cell is h 0f , where h 0 is a constant in classical statistic, where h 0 is the Planck’s constant in quantum statistics.

4.

To find probability of a microstate by the postulate of a priori-probability : According to the postulate of equal a priori-probability each of the accessible microstate is equally probable. Then probability for finding the system in any of the accessible microstate in the energy range E and E + dE is P( E ) = CΩ( E ) where C is the proportionality constant.

5.

To find the most probable microstate : We determine the equilibrium state of the system. Statistically, the equilibrium state of the system is the most probable state. Then the equilibrium state is obtained by finding the microstate for which the probability is maximum.

6.

To find the number of accessible microstates : The microstates are grouped energywise. The number of microstates in the unit volume of the phase space in energy range E to E + dE is represented by Ω( E ) .

7.

To determine the average value of property : The property of the system is explained by finding the average value of that property in the equilibrium state of the system.

S-4 Example : The mean energy of the system is < E > = Σ E i P(E i ) i

where P(E i) is the probability of occurrence of microstate corresponding to the energy E i.

2. Specification of States 2.1 Microstates and Macrostates of a System of Particles Let us consider a system consisting of a large number of gas molecules in the phase space which has been divided into tiny cells. Each cell represents small region of position and momentum. Each molecule may be specified by a point (phase point) lying same where inside one of these cells. The microstate of the system at a particular instant can be defined when we specify as to which particular cell each molecule of the as to which particular cell each molecule of the system belongs at the instant. This keep intermating is, however, unnecessary to determine the observable properties of the system (gas). For example the density is uniform if the number of molecules in each cell is same, regardless of which particular molecule lies in which particular cell. A macrostate of the system, on the other hand, can be defined by just giving the molecules in each cell, such as n1 molecules are in cell1, n 2 are in cell 2, and so on. There may be a large number of microstates corresponding to the same macrostate. As an example, let us consider a system of the molecules only ( for simplicity) named A, B C : which are to be distributed in two halves of a box, the left half L and the right half R. There are four possible distributions. (i)

3 molecules in L and 0 in R

(ii)

2 molecules in L and 1 in R and

(iii)

0 molecules in L and 33 in R

Let us call these distributions as (3, 0), (2, 1), (1, 2), (0, 3) respectively. Now, the distributions (3, 0) and (0, 3) can occur in one way only; while the distributions (2, 1) and (1, 2) can each occur in three ways as illustrated in the following table : Distribution

Left (L)

Right (R)

(3, 0)

ABC

—

(2, 1)

AB

C

AC

B

BC

A

A

BC

B

AC

C

AB

—

ABC

(1, 2)

(0, 3)

Thus, the total number of ways in which three molecules can occupy two halves of the box are 1 + 3 + 3 + 1 = 8 = (23) corresponding to four different distributions. Each way of arrangement of molecules is a microstate of the system, while each different distribution of molecules is a macrostate. Thus there are eight

S-5 microstate and four macrostates of the system. There is only one microstate ( ABC,–) corresponding to the macrostate (3, 0), three microstates ( AB, C) ; ( AC, B) ; (BC, A) corresponding the macrostate (2, 1) and so on. The number of microstates corresponding to a particular macrostate is called the thermodynamic probability of that macrostate. Obviously, the equilibrium macrostate of a system is one for which the number of microstates is maximum.

3. Phase Cell The concept of a point in the phase space (position-momentum) is to be considered in the light of uncertainty principle. This phase space is divided into tiny six cells whose side are ∆X, ∆Y, ∆Z, ∆px , ∆py , ∆pz such are called phase cells. The volume of each of these cells is ∆τ = ∆x ∆y ∆z ∆px ∆py ∆pz Classically, there is no restriction on the volume of the phase cell. It may be reduced to any extent, tending to zero and also without affecting the classical results. However, according to uncertainty principle h ∆x∆px ≥ 2π h ∆y∆py ≥ 2π h ∆z∆pz ≥ 2π From this, we have  h  ∆τ ≥    2π 

3

Thus, a ‘point’ in the µ-phase space is actually a cell whose minimum volume is of the order of h 3. It means that a particle in phase space cannot be considered exactly located at the point x, y, z, px , py , pz but can only be found somewhere within a phase cell of the order of h 3 centered at that point. In other words, the volume of a quantum state in phase space is of the order of h 3. The unit of volume in phase space is ( Js)3. In a similar way, the volume of phase cell in 6-N dimensional space can be written ∆Γ =

N

∏ ∆x i ∆yi ∆z i ∆pxi ∆pyi ∆pzi = ∏ ∆Γi i=1

i=1

Thus, Γ-space is built up as the produce of N-µ space.

4. Phase Space (µ-space and Γ-space) In classical mechanics, the instantaneous dynamical state of a particle is completely specified by its three position coordinates x , y, z and the corresponding momentum components px , py , pz . Thus, six coordinates are needed to specify a one particle system completely. Gibbs suggested that any instantaneous state (position and momentum) of the particle may be conveniently represented by some point in an imaginary x , y, z , px , py , pz are marked along six mutually perpendicular axes in space. This six dimensional space is known as phase-space or µ-space. The point in phase-space representing in the instantaneous state of the particles is

S-6 called the phase point. As the time progresses, the phase point moves in the phase space. The path of the phase point represents the trajectory of the particles. Let us now consider a system consisting of N-particles such as a gas in a container. The instantaneous coordinates q1, q2,…, q3N and 3N momentum coordinates p1, p2,...,p3N , that is by a total number of six-N parameters. An imaginary 6N-dimensional space in which the state of the system can be represented by a point having 6N co-ordinates (q1, q2, ..., q3N , p1, p2, ..., p3N ) is called E-space. Both the 6-dimensional µ -space and 6N dimensional Γ-space are hypothetical. The Γ-space may be taken as a superposition of µ-space.

4.1 Representation of Phase Space and Division of Phase Space If the position and momentum co-ordinates can assume continuously all the possible values, there will be infinite points possible in the phase to represent the different states of a molecules and hence there will be infinite number of microstates accessible in a finite volume element of phase space. Thus in classical statistics, the phase space can be divided into small cells of definite size. Since the position and momentum of a particle can be measured with precision upto a definite unit, therefore assuming ∆x ⋅ ∆px = h 0 ∆y ⋅ ∆py = h 0 ∆z ⋅ ∆px = h 0 where h is a constant, whose dimensions are dimension of energy x times, i. e. J-sec. The value of phase cell in 6-dimensional phase space is dΓ = ∆x ∆px × ∆y ∆py × ∆z ∆pz = h 03 Hence, the number of one phase cell in volume element dx dy dz dpx dpy dpz of 6-dimensional phase-space is dx dy dz dpx dpy dpz dΩ = h 03 The dimension of volume of phase cell in six-dimensional phase space is not equal to that of (length) 3, but is equal to that of (length X momentum) 3. Its SI unit is (J-s) 3. In general, in a 2f dimensional phase (M-space), the volume of a phase cell is h 0f and the number of phase cell in the phase volume element dΓ = dq1 dq2 ... dq f dp1 dp2 ... dp f is dΓ dΩ = f h0 or

dΩ =

dq1 dq2 ... dq f dp1 dp2 ... dp f h 0f

Similarly, in 2Nf dimensional space (Γ-space) the volume of one phase cell is h 0Nf and number of phase cells in a phase volume dΓ = dq1 dq2 ... dqNf dp1 dp2 ... dpNf dΓ dΩ = Nf h0 or

dΩ =

dq1 dq2 ... dqNf dp1 dp2 ... dpNf h 0Nf

S-7

4.2 Density of Distribution in the Phase Space Ensemble Density The use of ensembles in statistical mechanics is guided by the following points : 1.

There is no need to distinguish between different constituting an ensemble because the laws of statisticals mechanics aim to tell us only the number of systems or elements which would be found in different states, that is in different regions of phase space at any time.

2.

The number of system is an ensemble is so large that there is an ensemble is so large that there is continuous change in their number in passing from one region of phase space to another.

Keeping these points in mind, the condition of an ensemble at any time can be suitably specified by a density function ρ with which the phase points are distributed in the phase space. This density function is called the density of probability distribution or the ‘ensemble density’. In an ensemble of f degrees of freedom, the density of distribution ρ is a function of f position coordinates q1, q2, ... q f and f momentum coordinates p1, p2, … p f , corresponding to 2f combined position momentum axes in the phase space. The density of distribution ρ is also function of time t because due to the motion of phase points the density of distribution will change with time at any given point. Hence we can write ρ = ρ (q1, q2, ... q f , p1, p2, p f ... p f , t) or briefly,

ρ = ρ (q, p, t)

Obviously, the density of distribution ρ denotes the number of systems (or element) dN which are found at any given time in a given infinitesimal region of Γ-space. If the region chosen in such that the position coordinates lie between qi and qi + dqi and momenta lie between (i = 1, 2, 3, … f ), then the hyper volume dΓ of this region will be given by f

dΓ = dq1 dq2 ... dq f dp1 dp2 ... dp f =

∏

dqi dpi

i=1 f

where ∏ stands for the product over all values of i = 1 to i = f . i=1

The number of systems dN lying in the specified infinitesimal region can be obtained by multiply the density of distribution p with this hyper volume in the phase space, that is, f

dN = ρdΓ =

∏ dqi dpi i=1

Integrating over the entire phase space we can obtain total number of systems under consideration, that is, N = ∫ ρ dΓ = ∫ ...∫ ρ dq1dq2 ... dqi dp1 ... dp f

5. Statistical Average Value To find the average property of a system statically, we divide the property of the system into the convenient internals number of system at an instant having the same value of that property. Then the average value of that property is obtained by dividing the sum of products of all properties with the number of systems of all property with the number of systems corresponding to that property by the total number of systems. Let, we are to find the mean value of a property X of system. For this, we consider an ensemble of N such identical systems and take different value X1, X 2, X 3, ... of that property. Let the number of systems in ensemble corresponding to the value of property X1, X 2, X 3 ... be n1, n 2, n 3, ... respectively, then

S-8 Total number of systems in ensemble N = n1 + n 2 + n 3 + ... and mean (or-expectation) value of X is n X + n 2 X 2 + n 3 X 3 + ... X (or) < x > = 1 1 n1 + n 2 + n 3 n1 X1 + n 2 X 2 + n 3 X 3 + ... N n1 n2 n X= X1 + X 2 + 3 X 3 + ... N N N X=

Now, if P1 =

n1 n n , P2 = 2 , P3 = 3 ... gives the probability for the values of X to be X1, X 2, X 3 ... respectively, N N N

then < x > = P1 X1 + P2 X 2 + P3 X 3 + ... < x > = ΣPr X r where Pr =

nr and r = 1, 2, 3 ... N

Hence, the average value of property is equal to the sum of products of different values of that property with their properties.

6. Liouville Theorem If the state of an ensemble changes with time, the position of phase points in the phase will change,and so the density of states ρ will change with time. The rate of change of density ρ in the phase space is governed by an equation known as ‘Liouville theorem’. This theorem is expressed as f  ∂ρ ∂ρ   ∂ ρ ɺ i qɺ i + p = −∑     ∂t  q, p ∂pi  i =1  ∂qi

6.1 Condition of Equilibrium or Statistical Equilibrium An ensemble is said to be in statistical equilibrium if the probabilities of finding the phase points in the various regions of the phase space, and the average value of the properties of the systems in the ensemble are  ∂ ρ independent of time. Mathematically, the condition for an ensemble to be in statistical equilibrium   = 0,  ∂t  q, p for all values of q and p. This means that the density ρ must be independent of time at all points in the phase space. Derivation : For any ensemble, ρ is a function of some property, which in turn can be expressed as a function of q’s and p’s. Let this property be represented by E. Then This gives

ρ = ρ(ε) ∂ρ ∂ρ ∂ε and = ∂qi ∂ε ∂qi

∂ρ ∂ρ d ε = ∂pi ∂ε dpi

Now, the Liouville’s theorem is f  ∂ρ  ∂ρ  ∂ ρ ɺ i qɺ i + p =− Σ   i = 1  ∂q  ∂t  q, p pi  ∂ i

Substituting value from equation (1), we have

...(1)

S-9  ∂p ∂ε ∂p ∂ε   ∂ ρ ɺ i p qɺ i + =−Σ   i  ∂ε ∂q  ∂t  q, p dl ε ∂pi  i

...(2)

Since ε would itself be a function of q’s (position coordinates) and p′s (momentum coordinates) such that its value for any given system in statistical equilibrium would not change with time, we may write  ∂ε ∂ε  dε ɺ i = 0 = Σ qɺ i + p i  ∂q ∂pi  dt i  ∂ ρ =0    ∂t  q, p

Hence equation (2) gives

We conclude that if ρ is a function of a some property of the ensemble (say, energy for conservative system) which depends on the q’s and p’s and is independent of time, the ensemble will be in statistical equilibrium.

7. Stirling’s Approximation For a very large value of n , the calculation n ! becomes very trouble some Stirling’s formula gives a suitable analytic approximation of log n ! for large value of n. We note that n ! = n(n − 1)(n − 2) ... 4 × 3 × 2 × 1 log n ! = log n + log(n − 1) + log(n − 2) + ... + log 4 + log 3 + log 2 + log 1

Thus

=

...(1)

n

∑ log x (since log 1 = 0)

x =2

Let us now plot log x versus x for value of x = 2, 3 ... n as shown in figure. y-axis y = log x

log 5 log 4 log x

log 3 log 2

0

1

3

5

4

6

7

8

9 x Fig. 1 : Stirling’s approximation 2

x-axis

The area under the stepped curve is

∑ log x = log(n)!

x =2

For large value of x, the stepped curve and the smooth curve of log x become indistinguishable. So we can approximate log n ! by integration of log x for x = 1 to n. This approximation is quite satisfactory for large value

S-10 of n since log x in the range of large values of x varies only slightly when x is increased by unity. Thus for large value of n. log n ! =

n

∫1 log x dx

= [x log x ]1n −

n

∫1

1 ⋅ x dx = [x log x ]1n − [x ]1n x

log n ! = n log n − n + 1 As n is very large, we may neglect 1 in the above result, so we obtain. log n ! = n log n − n This equation is called Stirling formula and this approximation is known as Stirling approximation.

8. Entropy ‘‘Entropy is a measure of disorderness of molecules of system’’. i (i) (ii)

P

f O

V Fig. 2 : P-V diagram for change in state (entropy)

This figure represents two different path (i) and path (ii) on the P-V diagram for the change in state of a system from initial equilibrium state i to the final equilibrium state f . The process is reversible for both the paths. Then we get a cyclic process. According to Clausius theorem for the cyclic process, dQ ∫ T =0 or

 f dQ +   ∫i  T  along path (i)

i

 dQ =0   along path (ii)

∫ f  T

...(1)

Since the process (ii) is reversible ∴

i

 dQ =−   along path (ii)

∫i

∫ f  T

f

 dQ    T  along path (i)

…(2)

Hence from equation (1) and equation (2), we have

∫i or

∫i

f

f

 dQ −    T  along path (i)  dQ    T 

= along path (i)

∫i

∫i

f

f

 dQ    T 

 dQ    T 

=0 along path (ii)

…(3) along path (ii)

S-11 Thus, we conclude that the quantity ∫

i

f

dT remains constant between two states. Its value depends only on the T

initial and final states.

8.1 Entropy as a Function Consider a function S such that the difference in its value between the final state f and initial state i is (S f − Si) f dQ equal to ∫ if the change from state i to f is reversible. In mathematical form, i T f dQ …(4) S f − Si = ∫ i T The function S is called entropy and quantity (S f − Si) is called change in entropy. If the initial and final equilibrium states are very close, the change in entropy is very small then dT (for a reversible process) dS = T ⇒ Entropy is the unique function or the point function of the state of the system then dT dS = T

8.2 Change in Entropy of Universe in a Reversible Process Consider a reversible process in which the system absorbs heat dQ from its surrounding a temperature T . The increase in entropy of the system is dQ dS1 = T and there is a simultaneously decrease in heat dQ of the surrounding at temperature T1 therefore the decrease in entropy of the surrounding is dQ dS2 = − T ∴

Total change in entropy of the universe (i. e. system and surroundings) is dS = dS1 + dS2 dQ dQ − dS = T T dS = 0

Thus ‘‘there is no change in entropy of universe in a reversible process’’. In other words, ‘‘the entropy of universe remains constant in a reversible process.’’

9. Thermodynamic Probability For a given macrostate of a system in thermal equilibrium there may be large number of accessible microstates. ‘‘The total number of microstates corresponding to a given macrostate is called the thermodynamic probability of that particular macrostate’’.

S-12 Let us consider a system containing a large number N of molecules. A macrostate of the system can be expressed in terms of distribution of the N molecules among varies cells in the phase space. Suppose, in a particular macrostate, there are n1, n 2, n 3 ... n r molecules in the Ist, IInd, IIIrd … rth cells respectively. The number of ways this can be done is N! n1! n 2! n 3! ... n r! Each different way corresponds to a different microstate. Thus

N! is the thermodynamic n1! n 2! n 3! ... n r!

probability of the particular macrostate.

9.1 Entropy and Thermodynamic Probability : (S = K ln Ω) Boltzmann Entropy Relation We know that a given system so alter itself that in the state of thermal equilibrium its entropy is maximum. On other hand, statistically, the system alter in the direction of increasing probability and the equilibrium state of the system is the most probable state, that is, the state of maximum thermodynamic probability. Boltzmann conclude that the entropy of system is a function of the probability of the state of the system. That is S = f (Ω)

…(1)

where S is the entropy and Ω is the thermodynamic probability of the state of the system. To find out the nature of the function, let us consider two independent system having entropies S1 and S2 and probabilities Ω1 and Ω 2 Then

S1 = f (Ω1) and S2 = f (Ω 2)

The entropy of the combined system is S = S1 + S2

(entropy is additive)

S = f (Ω1) + f (Ω 2)

…(2)

But the probability, being multiplicative for the combined system is Ω1Ω 2. Hence equation (i), the entropy of the combined system should be S = f (Ω1Ω 2)

…(3)

Comparing equation (2) and equation (3), we get f (Ω1Ω 2) = f (Ω1) + f (Ω 2)

…(4)

The solution of equation (4) determines f . To solve it, we first differentiate in partially with respect to Ω1 and then with respect to Ω 2. This gives Ωf ′ (Ω1Ω 2) = f ′ (Ω1) and Dividing

Ω1 f ′ (Ω1Ω 2) = f ′ (Ω 2) Ω 2 f ′ (Ω1) = Ω1 f ′ (Ω 2)

Generalizing this relation, we can write Ωf ′ (Ω) = constant = k (say) k f ′ (Ω) = Ω

∂f    where f ′ is equal to   ∂Ω 

S-13 Integration gives f (Ω) = k log e Ω + C Where C is a integration constant. Substituting this result in equation (1), we get S = k log e Ω + C

…(5)

Now according to Nernst heat theorem, the entropy of a thermodynamic system tends to zero as its temperature tends to zero (as T → 0, S → 0). Further near absolute zero, the thermodynamic probability Ω → 1. Thus 0 = k log e 1 + C or

C= 0

∴

S = k log e Ω

This is the Boltzmann’s entropy probability relation. The constant k appearing in the relation has been identified as the Boltzmann’s universal constant.

Example 1. Using the statistical interpretation of entropy, show that the difference in entropy V  in two state of volumes V1 and V2 of n-mole of an ideal gas is nR loge  2  , where the  V1  temperature and number of molecules remains constant. Solution : Let V1 and V2 are the volume of the system in the initial and final state respectively. Consider a rectangular box of total volume V2. Let it is divided in two parts by a wall X which can be shifted to any place. Let the wall divides the box in volumes V1 and V2 − V1. V1

V2 – V1

X Total volume = V2

Let there is a single molecule of an ideal gas inside the box. The probability of finding this molecule in volume V V1 is 1 . V2 If there are two molecules in the box, the probability for finding both the molecules simultaneously in volume 2

V  V1 will be  1  .  V2  V  Similarly, for n-mole of gas, the probability of finding all the nN molecules in volume V1 will be  1   V2  V  Here, N is Avogadro’s number and W1 =  1   V2 

nN

nN

Now if the wall in the box is completely removed, the probability of finding of the molecules simultaneously in volume V2 will be i. e.

S-14 W2 = 1 ∴ or

W1  V1  =  W2  V2 

nN

W  V  log e  1  = n N log e  1  W  2  V2 

…(1)

From the statistical interpretation of energy, the entropy in the initial state is S1 = k log e W1 The entropy in the final state is S2 = k log e W2 ∴ Change in entropy is S2 − S1 = k log e W2 − K log e W1 or

W  S2 − S1 = k log e  2   W1 

…(2)

From equation (1) and equation (2), we get V  S2 − S1 = − Nnk log e  1   V2  or

V  S2 − S1 = n Nk log e  2   V1 

or

V  S2 − S1 = nR log e  2   V1  R = Nk (Gas constants)

Here

Example 2. A particle in a system has two possible energy states at E = 0 and E > 0. The system is in equilibrium at an absolute temperature. Calculate the mean energy of the particle. Solution : According to Boltzmann–Canonical law, the mean energy of the particle is < E>=

ΣE i e − E i / kT Σe − E i / kT E e − E i / kT

or

< E>= 0+

or

< E>= 0+ E

e − E i / kT

< E>= E Example 3. The entropy of a system in equilibrium is 1.5 × 10 −23 cal/K. Calculate the number of accessible macrostates of the system. Solution :

S = 1.5 × 10−23 cal/K = 1.5 × 10−23 × 4.2 J/K k = 1.38 × 10−23 J/K

From relation

S = k log e W

S-15 1 .5 × 10−23 × 4.2 = 1.38 × 10−23 log e W

Then

1.5 × 4.2 = 1.38 × 2.38 × 2.3026 log10 W 1.5 × 4.2 log10 W = 1.38 × 2.3026

or

log10 W = 1.9826 W = Antilog 1.9826 W = 98.34 Example 4. A system has only the three energy states ε1 = 0 J, ε 2 = 1.38 × 10 21 J and ε 3 = 2.76 × 10 −21 J which are obtained in 2,5 and 4 forms respectively. Calculate the probability for the system at 100K temperature to be (i) in the ground state ε1 and (ii) in a macrostate of the energy ε 2 . Given Boltzmann’s constant ( k) = 1.38 × 10 −23 J/K. Solution : Given g1 = 2 ε1 = 0 g2 = 5 ε 2 = 1.38 × 10−21 J g 3 = 4, ε 3 = 276 . × 10−21 J T = 100 K k = 1.38 × 10−23 J/K The probability for the system to be in ground state ε1, P1 = g 1C ⋅ e − ε1 / kT = 2C ⋅ e −0 = 2C

…(1)

The probability for the system in a macrostate of energy ε 2, P2 = g 2C ⋅ e − ε 2 / kT = 5C ⋅ e −1. 38 × 10

−21

/ 1. 38 × 10−23 × 100

= 5Ce −1

…(2)

The probability for the system in a macrostate of energy ε 2, P3 = g 3Ce − ε 3 / kT = 4 C ⋅ e −2.76 × 10

−21

/ 1. 38 × 10−23 × 100

= 4 Ce −2

Since, total probability is 1 P1 + P2 + P3 = 1

Hence From equations (1), (2) and (3), we have

2C + 5e −1 + 4 Ce −2 = 1 or

2C + 5C × 0.368 + 4 C × 0135 . = 1 or

4.38C = 1 or

C = 0.228 (i)

The probability for the system to be in ground energy state P1 = 2C = 2 × 0.228

C=

1 4.38

…(3)

S-16 P1 = 0.456 (ii)

The probability for the system to be in ground energy state =

P3 g3

4 Ce −2 = Ce −2 = 0.228 × 0135 . = 0.031 4 Example 5. The relative probability for a system at temperature T to be found in two states with an energy difference of 4.8 × 10 −21 J is e2 . Calculate the temperature T. (Given (k = 1.38 × 10 −23 J /k) Solution : According to the Boltzmann-Canonical Law, the probabilities for system to have energies E1 and E 2 at temperature T are respectively. P1 = Ae − E1 / kT P2 = Ae − E 2 / kT P1 = e( E 2 P2

− E1 ) / kT

 P  E − E1 log e  1  = 2 kT  P2  T=

Given

E 2 − E1 P  k log  1   P2 

E 2 − E1 = 4.8 × 10−21 J k = 1.38 × 10−23 J/K P1 / P2 = e 2

∴

P  log e  1  = 2 log e = 2  P2  T=

Hence

4.8 × 10−21 1.38 × 10−23 × 2

T = 173.9 K

Example 6. A monoatomic ideal gas is enclosed in a vessel of volume V at an absolute Vb 4 temperature T. The partition junction of the molecule in the gaseous system is 3 . Prove that β /2 1 3 the average energy of a molecule is kT. Hence b is a constant and β = , where k is the kT 2 Boltzmann constant. Solution : The relation between the average energy of molecule with the portion junction Z ix is, < E>= −

∂ (log e Z )V ∂β

S-17 where β =

1 kT Z=

Given

Vb4 β 3/ 2

< E>= − < E>= =

∂ ∂β

∂ ∂β

  Vb4   log e  3/ 2   V  β   

3   log e V + 4 log e b − 2 log e β

3 1 3 3 × = × kT = kT 2 β 2 2

Example 7. 100 particles are distributed in 10 cells, each of same energy. Find the thermodynamical probability for the maximum and minimum probable macrostates. Solution : Thermodynamical probability is n! Ω = n ! n 2! ... n10! For the most probable state, n1 = n 2 = ... = n10 = 10 The minimum probable state, n1 = 100 and n 2 = n 3 = ... = n10 = 0 Then, the maximum probable macrostate is Ωmax =

100 ! 100 ! = (10)!(10)!(10)!(10)!(10)!(10)!(10)!(10)!(10)!(10)! (10 !)10

The minimum probable state is Ωmin =

100 ! 100 ! =1 = (100)!(0)!(0)!(0)!(0)!(0)!(0)!(0)!(0)!(0)! 100 !

Example 8. A system can have the four energy states E1 = 0, E 2 = 1.38 × 10 −21 J, E 3 = 2.76 × 10 −21 J and E 4 = 4.14 × 10 −21 J. The number of accessible macrostates in these energy states are 3, 5, 7 and 9. Calculate the probability at 100 K, when (i)

The system is in energy state E 2

(ii)

The system is in energy state E 3

(iii) The system is in energy states E 4 (iv)

Also calculate the mean energy of the system ( k = 1.38 × 10 −23 J/K).

Solution : Given

g1 = 3 g2 = 5 g3 = 7 g4 = 9

S-18 E1 = 0 J E 2 = 1.38 × 10−21 J E 3 = 276 . × 10−21 J E 4 = 4.14 × 10−21 J T = 100 K k = 1.38 × 10−23 J/K The probability of the system to be in energy state E1 P1 = Cg 1e − E1 / kT = 3Ce −0 P1 = 3C

…(1)

The probability of the system to be in energy states E 2 P2 = Cg 2e − E 2 / kT = 5 Ce −1. 38 × 10−21 /1.38 × 10−23 × 100 = 5 Ce −1

…(2)

The probability of the system to be in energy state E 3 P3 = Cg 3e − E 3 / kT = 7 C. e −2.76 × 10−21 / 1.38 × 10−23 × 100

…(3)

= 7 Ce −2 The probability of the system to be in energy state E 4 P4 = Cg 4 e − E 4 / kT P4 = 9C ⋅ e −4.14 × 10−21 / 1.38 × 10−23 × 100 = 9Ce −3 Total energy of the system is equal to 1 then, P1 + P2 + P3 + P4 = 1 From equations (1), (2), (3) and (4), we get 3C + 5Ce −1 + 7 Ce −2 + 9Ce −3 = 1 C(3 + 5e −1 + 7 e −2 + 9e −3) = 1

or or

C[3 + 5 × 0.368 + 7 × 0135 . + 9 × 0.049) = 1

or

C[3 + 1.840 + 0.945 + 0.441) = 1

or

C × 6.226 = 1

(i)

The probability of the system in the energy range E 2 is P2 = 5Ce −1

(ii)

or

P2 = 5 × 0161 . × 0.368

or

P2 = 0.29624

The probability of the system in the energy range E 3 is P3 = 7 Ce −2 = 7 × 0161 . × 0135 . = 0152145 .

(iii)

The probability of the system in the energy range E 4 is P4 = 9Ce −3 = 9.0161 × 0.049 = 0.071001

…(4)

S-19 (iv)

Mean energy of the system is E = P1E1 + P2E 2 + P3E 3 + P4 E 4 E = 3C × 0 + P2E 2 + P3E 3 + P4 E 4 = P2E 2 + P3E 3 + P4 E 4 = 0.29624 × 1.38 × 10−21 + 015145 . × 276 . × 10−21 + 0.071001 × 4.14 × 10−21 = 1.38 × 10−21[0.29624 + 0152145 . × 2 + 0.071001 × 3] = 1.38 × 10−21 × (0.29624 + 0.304290 + 0.213003) = 1.38 × 10−21 × 0.813533 < E > = 1123 . × 10−21 J

Example 9. The molar specific heat at constant for silver in the range of temperature 40K to 120 K is expressed as CP = (0.076 T − 0.00026 T 2 − 0.15) cal/mol-K where T is the Kelvin temperature. Calculate the change in entropy, if 5 mole of silver are heated from 40 K to 120 K (log10 3 = 0.4771). Solution : Heat required to increase the temperature of 5 mole of silver by dT is dQ = 5 × C P × dT ∴

Increase in entropy of 5 mole of silver when temperature is increased 120 5 ⋅ C dT 120 dQ P S − Si = ∫ =∫ 40 40 T T =

120

∫40

5 (0.076T − 0.0026T 2 − 015 . )dT T 120

  T2 . log T  = 5 0.076T − 0.00026 × − 015 2  40 

  120  = 5 0.076(120 − 40) − 0.00013 {(120)2 − (40)2} − 015 . log e    40    = 5 [0.076 × 80 − 0.00013 × 160 × 80 − 015 . × 2.3026 × 0.4771] = 5 [6.08 − 1.664 − 016478557 . ] = 5 × 4.25 = 21.25 cal/K Example 10. 100 g steam at 100°C is converted into water at the same temperature. The latent heat of steam is 540 cal/kg. Calculate the change in entropy. Solution : Given M = 100 g = 01 . kg T = 100°C = 373 K L = 540 kcal/kg Decrease in entropy is ∆S =

∫

dQ Q mL 01 . × 540 = = = T T T 373

kcal/K ∆S = 0145 .

S-20 Example 11. 1 kg water at 0°C is mixed with 1 kg water at 100°C. Calculate the change in entropy of the mixture. Given : Specific heat of water = 1 kcal/kg °C log 3.23 = 0.5092,

log 3.73 = 0.5717,

log 2.73 = 0.4362

Solution : If t °C is the temperature of mixture, then heat given by water at 100°C = heat taken by water at 0°C. 1 × 1(100 − t) = 1 × 1 × (t − 0) 100 − t = t 100 = 2 t t = 50°C Increase in entropy in heating 1 kg water at 0°C (273 K) to water at 50°C (= 323 K) Tf   323 ∆S1 = mC ⋅ log e   = 1 × 1 × 2.3026 log10    273  Ti  kcal/K = 2.3026 (2.5092 − 2.4362) = 0168 . The decrease in entropy in cooling 1 kg water at 100°C (= 373 K) to 50°C = (323 K), Tf   323 ∆S2 = mC log e   = 1 × 1 × 2.3026 log10    273 T  i = 2.3026 × (2.5092 − 2.5717) kcal/K = − 0144 . The change in entropy of the system is . + (−0144 . ) ∆S = ∆S1 + ∆S2 = 0168 = 0168 . − 0144 . ∆S = 0.024 kcal/K Example 12. 0.1 kg steam at 100°C is converted in to ice at − 10 °C. Calculate the change in entropy. Given :

Latent heat of ice = 80 cal/g Latent heat of steam = 540 cal/g Specific heat of ice = 0.5 cal/g °C Specific heat of water = 1 cal/g °C log 2.363 = 0.4200 log 2.73 = 0.4363 log 3.73 = 0.5717

Solution : This process is done in four steps : (i)

Steam at 100°C converts into water at 100°C

(ii)

Water at 100°C converts into water at 0°C

S-21 (iii)

Water at 0°C converts into ice at 0°C

(iv)

Ice at 0°C converts into ice at −10°C

Given, mass of steam, m = 01 . kg = 100 g In first step : The change in entropy dQ Q mL1 100 × 540 ∆S1 = ∫ = = = T1 T T 373 = 145 cal/K (decrease)

[‡ T1 = 100 ° C = 373 K]

In second step : The change in entropy Tf  ∆S2 = mc1 log e    Ti   273 = 2.3026 × 100 × 1 × log10    373 = 2.3026 × 100 × (2.4363 − 2.5717) cal/K = 3118 = − 31177 . . cal/K (decrease) In third step : The change in entropy dQ ∫ T Q mL2 100 × 80 = 29.30 cal/K (decrease) ∆S3 = = = = T2 T T 273 In fourth step : The change in entropy Tf  ∆S4 = mC 2 log e    Ti   263 = 2.3026 × 100 × 0.5 × log10    273 = 2.3026 × 100 × 0.5 × (2.4200 − 2.4363) = − 1.88 cal/K ≈ 1.88 cal/K (decrease) ∴ Total decrease in entropy ∆S = ∆S1 + ∆S2 + ∆S3 + ∆S4 ∆S = 145 + 3118 . + 29.30 + 1.88 ∆S = 207.36 cal/K Example 13. 0.2 kg water at 100°C is changed into ice at 0°C. Calculate the change in entropy (Latent heat of ice = 80 cal/g). Solution : The process is done in two processes : (i)

Water at 100°C converts into water 0°C

(ii)

Water at 0°C converts into ice at 0°C

Given, mass of water (m) = 0.2 kg m = 200 g In first step : Change in entropy

S-22 Tf   273 ∆S1 = mC log e   = 200 × 2.3026 × log10    373  Ti  = 200 × 1 × 2.3026 × (2.4363 − 2.5717) = − 62.36 cal/K ≈ 62.36 cal/K (decrease) In second step : Change in entropy dQ Q mL 200 × 80 = = = = 58.60 cal/K (decrease) ∆S2 = ∫ T T T 273 The net change in entropy is ∆S = ∆S1 + ∆S2 = 62.36 + 58.60 = 120.96 cal/K Example 14. 10 g of steam at 100°C is converted into water at the same temperature. Find the change in entropy. Solution : The change in entropy to convert steam at 100°C into water at 100°C is dQ Q = ∆S = ∫ T T mL ∆S = T Given that

m = 10 g L = 540 cal/g T = 100°C = 373 K ∆S =

10 × 540 373

∆S = 14.5 cal/K (decrease) Example 15. 1 kg of ice of at 0°C is converted into water at the same temperature. What is the change in entropy ? Solution : The change in entropy to convert ice at 0°C into water at 0°C is dQ Q mL ∆S = ∫ = = T T T 1000 × 80 = 293.04 cal/K ∆S = 273 Example 16. 100 g steam at 100°C is convert into water at the same temperature. The latent heat of steam is 540 k cal/kg. Calculate the change in entropy. Solution : Given that n = 100 g = 01 . kg T = 100°C = 373 K L = 540 kcal/kg Decrease in entropy is ∆S =

∫

dQ Q mL = = T T T

S-23 01 . × 540 373

or

∆S =

∴

kcal/K ∆S = 0145 .

Example 17. A system of n particles has only two allowed state A and B. The probability for state A is P and that for state B is (1 − P). What is the probability for system to be in macrostate defind by distribution ( r, n − r)? Solution : The probability for the distribution of n particles in states A and B in macrostate (r , n − r ) is given by P( r, n − r) =

n! P r (1 − P)n − r r !(n − r )!

Example 18. Calculate the percentage error made in Stirling’s formula loge( n !) = n loge n − n when n = 5. Solution : For n = 5 log e (n !) = log e (5 × 4 × 3 × 2 × 1) = log e 120 = log e 1.20 + log e 102 = 01823 . + 2 × 2.3026 = 4.7875 n log e n − n = 5 log e (5) − 5 = 5 × 1.6094 − 5 = 3.047 Difference = 4.7875 − 3.047 = 17405 . . 17405 % error = × 100 = 36.38% 4.7875 Example 19. 100 molecules of a gas are enclosed in a cubical volume. Imagine the volume to be divided into two equal halves. Calculate the ratio of the time spent by the system in the most probable macrostate and the macrostate : (i) ( 40 : 60), (ii) ( 45 : 55). Solution : Number of particles n = 100 Number of compartment C = 2 As the number of particles is 100, the most probable macrostate is (50 : 50) 100 T( 50, 50) C (i) For the macrostate (50 : 50), = P( 50, 50) = 10050 T 2 (ii)

For the macrostate (40 : 60), T( 50, 50) T( 40, 60)

T( 40, 60) T

100

= P( 40, 60) =

C 40

2100

100 ! 50 ! 50 ! = 40 ! 60 ! = 100 = 100 ! 50 ! 50 ! C 40 40 ! 60 ! 60 × 59 × 58 × 57 × 56 × 55 × 54 × 53 × 52 × 51 = 50 × 49 × 48 × 47 × 46 × 45 × 44 × 43 × 42 × 41 100

= 7.3

C 50

S-24 (iii)

For the macrostate (45, 55) 100 T( 45, 55) C = P( 45, 55) = 10045 T 2 T( 50, 50) 100 C 50 45 ! 55 ! (55 × 54 × 53 × 52 × 51) = 1.64 = = = T( 45, 55) 100 C 45 50 ! 50 ! 50 × 40 × 49 × 48 × 47 × 46

Example 20. 80 molecules of a gas are enclosed in a cubical volume. Let this volume be divided into two equal halves by means of an imaginary partition. Calculate the ratio of the time spent by the system in the most probable macrostate ( 40, 40) and (38, 42). Solution : Given that n = 8, C = 2 Most probable state is (40, 40) T( 40, 40) T( 38, 42)

=

80

C 40

80

C 38

80 ! 40 ! 40 ! = 42 × 41 = 1722 = 1104 = . 80 ! 40 × 39 1560 38 ! 42 !

Example 21. From a well shuffled deck of cards, what is the probability that three cards drawn in succession are all aces ? Solution : The probability will be

Here

P = p1 × p2 × p3 m 4 1 p1 = 1 = = n1 52 13 p3 =

and p2 =

m2 3 1 = = n 2 51 17

m3 2 1 = = n 3 50 25

P = p1 × p2 × p3 =

1 1 1 1 × × = 13 17 25 5525

1.

A dice is thrown, calculate the probability of face with two dots coming up.

2.

Pair of six faced dice are thrown simultaneously. Calculate the probability that face 4 comes up in both dices ?

3.

There are two boxes with weights ratio 4 : 1 between which 10 particles are distributed. What is the probability for a distribution (2, 8) ? Which distribution has the maximum probability ?

4.

Four distinguishable particles are to be distributed in two compartment. First compartment is divided into 3 cells and other is divided into 2 cells and there is no restriction on the number of particles in any cell. Calculate thermodynamics probability for the distribution (4, 0), (2, 2) and (1, 3).

5.

Let an assembly of N = 10 identical particles having a fixed amount of energy E = 20 J. The particles are distributed in three cell in such away that n1 = 5, n2 = 3 and n3 = 2 with energies E1 = 0, E 2 = 2 and E 3 = 4 J per particles. If δn2 = 2, Calculate δn1 and δn3 assuming N and E remains constant. Also calculate the initial and final probability.

6.

A particle in a system has two possible energy states at E = 0 and E > 0. The system is in equilibrium at absolute temperature. Calculate the mean energy of the particle.

S-25 7.

The entropy of system in equilibrium is 1.5 × 10−23 cal/K. Calculate the number of accessible macrostates of the system.

8.

The relative probability for a system at temperature T to be found in two states with an energy difference of 4.8 × 10−25 J is e 2 . Calculate the temperature T (Given k = 1.38 × 10−23 J/K).

9.

1 kg water at 1°C is mixed with 1 kg water at 100°C. Calculate the change in entropy of the mixture.

10.

0.2 kg water at 100°C is changed into ice at 0°C. Calculate the change in entropy. (Latent heat of ice = 80 cal/g) V  Prove that S2 − S1 = nR log e  2  .  V1 

11.

3 kT. 2

12.

Prove that the average energy of a molecule is

13.

10 g of steam at 100 °C is converted into water at the same temperature. Find the change in entropy.

14.

1 kg of ice at 0°C is converted into water at the same temperature. What is the change in entropy ?

15.

100 particles are distributed in 10 cells each of same energy. Find the probability for maximum and minimum probable macrostates.

16.

Using the statistical interpretation of entropy, show that the difference in entropy in two states of volumes V1 and V2 V  of n-moles of an ideal gas is nR log e  2  .  V1 

17.

The molar specific heat at constant pressure for silver in the range of temperature 40K to 120K is expressed as CP = (0.076T − 0.00026T 2 − 015 . ) cal/mol-K. Calculate the change in entropy if 5 mole of silver are heated from 40K to 120K (Given log10 3 = 0.4771).

18.

A system has only the three energy states E1 = 0 J, E 2 = 1.38 × 10−21 J and E 3 = 276 . × 10−21 J, which are obtained in 2, 5 and 4 form respectively. Calculate probability for the system at 100 K temperature to be (i) in the ground state E1 and (ii) in a macrostate of energy E 2 . Given that Boltzmann’s constant k = 1.38 × 10−23 J/K.

S-26

Long Answer Type Questions

3.

What do you mean by probability ?

1.

What do you mean by the phase space ?

4.

Write down equilibrium condition.

2.

How is the phase space divided into phase cells of 5. arbitrary size? 6. Define thermodynamic probability in energy range E to E + dE for a macroscopic system. 7.

What is entropy ?

A simple particle of mass m is enclosed in a vessel of 8. volume V. Find the number of accessible macrostates 9. in the energy range (i) 0 + 0 E (ii) E to E + dE. 10. State the fundamental postulates of statistical physics or mechanics.

What is phase cell ?

3. 4.

5. 6.

Write down the equilibrium condition and prove it.

7.

What do you mean by entropy of a function ?

8. 9. 10.

Give the statistical interpretation of second law of thermodynamics. Give in physical significance of energy. Define constraint. Define microconical state.

Objective Type Questions

Prove that the change in entropy of of the universe Multiple Choice Questions 1. β-parameter is : remains constant. ∂ ∂ State and derive Boltzmann’s entropy probability (b) (a) (Ω) (log Ω) ∂E ∂E relation. 1 (c) (d) none of these (log Ω) Give the statistical interpretation of entropy. E

Short Answer Type Questions 1.

Define degree of freedom of a system.

2.

Explain µ-space.

3.

What is Γ-space or V-space?

4.

Give the number of microstates in a phase volume.

5.

Give the number of microstates in energy range E to E + dE.

6.

State the postulates of statistical physics.

7.

Use statistical mechanics to determine the average value of any property.

8.

Described the macrostate and microstates of any system.

9.

Define a system.

10.

Define thermal equilibrium

11.

What is the principle of increase of entropy ?

12.

State the postulates to construct a statistical ensemble.

Very Short Answer Type Questions

2.

100 molecules are divided in 10 cells each of same energy. The incorrect statement is : (a) Most probable distribution is in which one cell contain all the molecules and rest of the cells are empty (b) The number of accessible microstages of the 100 most probable distribution are (10!)2 (c) The value of log Ω is zero for the last probable distribution where W are the number of accessible microstate (d) In the least probable distribution, all the molecules will occupy one cell and the rest of the cells will be empty

3.

The hypothesis of equal a priori-probability for an isolated physical system tells that : (a) All the microstates of a macrostate are equally probable (b) The macrostates of different energies are equally probable

1.

Find the change in entropy of a gas.

(c) The macrostates of mean energy is most probable

2.

Define phase space.

(d) The state of lowest energy is most probable

S-27 4.

A system consist of 10 distinguishable particles which 12. are distributed in two equal cells of a box. Then :

The number of most probable macrostates for a system having odd number of particles :

(a) Total number of macrostates are 210

(a) 4

(b) 3

(b) Total number of microstates are 11

(c) 2

(d) 1

(c) Only 1-microstate is possible for the macrostate 5.

6.

7.

(a) 1

(b) maximum

If Pr and PS are the probabilities for the occurrence of r and s-events respectively the probability for the 14. occurrence of both r and S-events simultaneously will be :

(c) zero

(d) minimum but not 1

Probability is given by :

(a) Pr − PS

(b) Pr + PS

(c) Pr /PS

(d) Pr × PS

(a) (0, 4)

(b) (1, 3)

(c) (2, 2)

(d) (3, 1)

11.

16.

n

Cr

(b)

2 n Cr

(d)

n

Cr ⋅ 2n

n

Cr

(d) All of these

(a) −∞ to +∞

(b) 0 to 1

(c) −∞ to 1

(d) −1 to 1

Constraints imposed on a system :

number

of

inaccessible

(d) None of the above 17.

A coin is tossed 10 times. The probability for getting head each time is : 19. 252 1 (a) (b) 1024 2 1 120 (d) (c) 1024 1024 Three identical dice A, B, C each with six faces are simultaneously tossed. The probability of getting the 20. same number uppermost in each dice is : 1 1 (a) (b) 36 216 1 1 (d) (c) 18 72

(d) negative

(c) By ratio

The value of probability lie :

(c) Decrease the microstates

Three identical coins are tossed several times. The probability to get head in two coins and tail in coin uppermost is : 1 3 (a) (b) 8 8 18. 2 1 (d) (c) 64 64

(c) zero

(b) By percentage

(b) Increase the number of inaccessible microstates

2n

The value of probability of an event cannot be : 1 (a) 1 (b) 2

(a) By same ratio

(a) Have no effect

If n distinguishable particles are distributed in two identical cells the probability for macrostate (r, n − r) is :

(c)

10.

15.

Four distinguishable coins are tossed many times. The most probable macrostate will be :

n

9.

The thermodynamic probability of a system in equilibrium :

(d) (5, 5) is the least probable macrostate (10, 0)

(a)

8.

13.

21.

Statistical method give greater accuracy when the number of observation is : (a) Very small (b) Very large (c) Negative (d) Neither very small nor very large The probability of occurrence of two independent event is equal to their : (a) Sum

(b) Difference

(c) Product (d) Ratio Px falls rapidly with deviation x the ratio is Pmax maximum when : (a) n = 102

(b) n = 104

6

(d) n = 108

(c) n = 10

A card is drawn from a pack of 52. The probability of its being an ace or king is : 2 8 (a) (b) 52 52 4× 3 4 (d) (c) 52 × 51 52 Same events has : (a) Even probability (b) Odd probability (c) Zero probability (d) Differ probability

S-28 22.

23.

24.

A coin and a six faced dice are thrown probability 5. that coin shows tail and dice shows five is : 6. 7 1 (b) (a) 12 8 2 1 (c) (d) 7. 12 12 8. Pmin (minimum probable macrostate) is given by : 9. 1 1 (b) (a) n 2n 2 10. (c) 2n (d) 1 Pmax (most probable state) is given : n! 1 (a) Pmax = ⋅  n  ! n  ! 2n     2  2 (b) Pmax

In thermodynamic, there are two boundaries for each accessible state, there boundaries are called ……… β-parameter is given by ……… Partition function is represented by Z = ……… ……… is a measure of disorderness of molecules of the system. Change in entropy of universe remains constant in a ………

True/False 1 hν . 2

1.

Zero point energy of an oscillator is < E > =

n! = n!(n − 1)!

2.

n! (n − 1)!

It is impossible to construct the perpetual machine of second kind.

3.

Boltzmann entropy probability is S = k log Ω( E) .

4.

The value of Boltzmann’s constant

(c) Pmax =

(d) none of the above 25.

The entropy of universe remains ………

= 1.88 × 10−23 J/K.

Entropy of universe : (a) Remains constant

(b) Changed

5.

Phase space is denoted by µ-space.

(c) Decreases

(d) None of these

6.

The entropy of the system in equilibrium is minimum.

Fill in the Blank

7.

The internal energy of universe increases.

1.

The total probability of any event is equal to ………

8.

2.

Minimum probable macrostate Pmin is given by 9. ………

The potential energy of molecules of a perfact gas is zero.

3.

The macrostate is the state which represents the 10. ……… properties of a system.

4.

Maximum or most probable state is given by ………

The point function of the state of a system is work. dQ Clausius theorem is ∫ > 0. T

S-29

Objective Type Questions Multiple Choice Questions 1.

(b)

2.

(a)

3.

(a)

4.

(a)

5.

(d)

6.

(c)

7.

(d)

8.

(b)

9.

(c)

10.

(b)

11.

(d)

12.

(c)

13.

(b)

14.

(d)

15.

(c)

16.

(b)

17.

(b)

18.

(c)

19.

(a)

20.

(b)

21.

(a)

22.

(d)

23.

(a)

24.

(a)

25.

(a)

Fill in the Blank 1.

One

2.

1 2n

3.

microscopic

4.

n! 1 ⋅  n  n 2n  ! !  2  2

5.

constant

6.

constraint

7.

1 / kT

8.

z = e−βEi

9.

Entropy

10. reversible process

True/False 1.

True

2.

True

3.

True

4.

False

5.

True

6.

False

7.

True

8.

True

9.

False

10.

False

S-30

H ints and Solutions 4.

Numerical Questions 1.

Let a dice has six faces. Hence the total numbers of cases with which a dice can thrown n = 6

The thermodynamic probability Ω can be written as N Ω(n1 , n2 ) = (g1 )n1 (g 2 )n2 n1 ! n2 ! (i)

n2 = 0 4! (3)4 (2)0 = 34 = 81 Ω(4, 0) = 4! 0!

The face with two dots will come up only once. Therefore number of favourable cases i. e. m = 1 m 1 Priori-probability P = = ∴ n 6 2.

Probability of falling of first dice with face 4 up P1 =

(ii) Here, N = 4, g1 = 3, g 2 = 2, n1 = 2, n2 = 2 4! Ω(2, 2) = (3)2 (2)2 = 6 × 9 × 4 = 216 2! 2!

1 6

Probability of falling of second dice with face 4 up 1 6

(iii) Here, N = 4, g1 = 3, g 2 = 2,

P2 =

n1 = 1, n2 = 3 4! (3)1 × (2)3 Ω(1, 3) = 1! 3!

Probability of both the dices simultaneously with face 4 up P = P1 × P2 1 1 1 P= × = 6 6 36 3.

= 4 × 3 × 8 = 96 5.

The probability of distribution in which r particles are in one box and ( N − r) in other box. N! br CN − r P = (r, N − r) = r !(n − r)!

The expression of total number of particles and energy is given by N = n1 + n2 + n3 N = 5 + 3 + 2 = 10

…(1)

and

E = n, E1 + n2 E 2 + n3 E 3

…(2)

or

E = 5 × 0 + 3 × 2 + 2 × 4 = 14 J

If

N = n1 + n2 + n3 (∵ N is constant)

Where b is probability of going a particle in one box and C that of going in the other box.

δN = δn1 + δn2 + δn3 δn1 + δn2 + δn3 = 0

We know b + C = 1

nE1 + n2 E 2 + n3 E 3 = 14

b = 0.8 ∵

C = 0.2 Hence, for the distribution (2, 8), r = 2, N − r = 8, the probability is 10! (0.8)2 (0.2)2 P(2, 8) = 2! 8!

that

Hence, for the distribution (2, 8), r = 2, N − r = 8, the probability is 10! P(2, 8) = (0.8)2 (0.2)8 2! 8!

that

−2

−8

= 45 × (64 × 10 ) × (256 × 10 ) = 7.37 × 10−5

(∵ δn2 = 2)

δn1 + δn3 = − 2

The statistical weights of the boxes are given 4 : 1 or

Here N = 4, g1 = 3, g 2 = 2, n1 = 4

is

E1 = 0, E 2 = 2 and E 3 = 4 2n2 + 4n3 = 14 2δn2 + 4δn3 = 0

Putting the value of δn2 = 2 in above equation 4δn3 = − 2δn2 = − 2 × 2 = 4 is

δn3 = − 1 Using equation δn1 + δn3 = − 2 δn1 − 1 = − 2

⇒ δn1 = − 1

The new distribution of particle n1′ = n1 + δn1 = 5 − 1 = 4

S-31 n2′ = n2 + δn2 = 3 + 2 = 5

T=

n3′ = n3 + δn3 = 2 − 1 = 1 Initial thermodynamical probability N! 10! = = 2520 n1 ! n2 ! n3 ! 5! 3! 2!

Given, E 2 − E1 = 4.8 × 10−21 J k = 1.38 × 10−23 J/K

Final probability n! 10! = = 1260 n1 ! n2 ! n3 ! 4! 5!1! 6.

=

< E> = 0+

or

−

T=

Ei

9.

E − i kT

< E> = 0+ E

or

=E

log 2.73 = 0.4362 Let t ° C be the temperature of mixture, then heat given by water at 100° C = Heat taken by water at 0° C

= 1.5 × 10−23 × 4.2 J/K K = 1.38 × 10

1 × 1 × (100 − t) = 1 × 1 × (t − 0) 100 − t = t

J/K

100 = 2t

S = k log e W

t = 50 ° C

From relation −23

1.5 × 10

−23

× 4.2 = 1.38 × 10

log e W

1.5 × 4.2 = 1.38 × 2.38 × 2.3026 log10 W 1.5 × 4.2 log10 W = 1.38 × 2.3026 log10 W = 1.9826 W = Antilog 1.9826 ⇒ 8.

W = 98.34

According to the Boltzmann-canonical law, the probabilities for system to have energies E1 and E 2 at temperature T are respectively P1 = Ae

E − 1 kT

P2 = Ae P1 P2

Given that log 3.23 = 0.5092, log 3.73 = 0.5717

S = 1.5 × 10−23 cal/K −23

1.38 × 10−23 × 2

Specific heat of water = 1 kcal/kg ° C

E − i e kT

or

4.8 × 10−21

T = 173.9K

kT

Ee

= e2

P  log e  1  = 2 log e = 2  P2 

E − i kT

Σ Ei e

7.

P1 P2

According to Boltzmann-canonical law, the mean energy of the particle is Σ E ie

E 2 − E1 P  k log  1   P2 

E − 2 kT

( E 2 − E1 )

=e

kT

P  E − E1 log e  1  = 2 P kT  2

Increase in entropy in heating 1 kg water at (0° C = 273), T  ∆S1 = mC log e  f   Ti  323  = 1 × 1 × 2.3026 log10    273  = 2.3026 (2.5092 − 2.4362) = 0.168 kg-cal/K The decrease in entropy in cooling 1 kg water at 100 ° C (= 373 K) to 50 ° C(= 323 K), T  ∆S2 = mC log e  f   Ti  323  = 1 × 1 × 2.3026 log10    373  = 2.3026 × (2.5092 − 2.5717) = − 0.144 kilo-cal/K The change in entropy of the system is ∆S = ∆S1 + ∆S2

S-32 = 0.168 + (− 0.144)

=

= 0.168 − 0.144 13.

∆S = 0.024 kilo-cal/K 10.

3 kT 2

Given that m = 10 g

This process is done in two processes :

L = 540 cal/g

(i)

T = 100° C = 273 K

water at 100° C convert into water 0° C

Then change in entropy dQ Q mL = = ∆S = ∫ T T T 10 × 540 ∆S = 373

(ii) water at 0° C converts into ice at 0° C Given that, mass of water (m) = 0.2 kg m = 200 g In first step : Change in entropy T  ∆S1 = mC log e  f   Ti 

∆S = 14.5 cal/k 14.

273  = 200 × 2.3026 × log e    373  ∆S1 = 200 × 1 × 2.3026 × (2.4363 − 2.5717)

∆S = 293.04 cal/k

= − 62.36 cal/k ≈ 62.36 cal/k (decrease)

15.

In second step : Change in entropy dQ Q mL ∆S2 = ∫ = = T T T 200 × 80 = 273

n1 = n2 = .... n10 = 10 The minimum probable state, n1 = 100

∴ The net change in entropy is

n2 = n3 = … n10 = 0

∆S = ∆S1 + ∆S2

Then, the maximum probable macrostates is 100! Ωmax = (10)! (10)! (10)! (10)! (10)! (10)! (10)! (10)! (10)! (10)!

∆S = 62.36 + 58.60 = 120.96 cal/k Please see the examples no. (1) on page no 11.

12.

The relation between the average energy of molecule with the portion junction z is ∂ (log e z) V =− ∂β where β =

=

=

Ωminimum = 1

=

 Vb4   ∂   log e  3 / 2   V ∂β  β 

3 ∂  log e V + 4 log e b − log e β   2 ∂β  3 1 × 2 β

3 = × kT 2

(100)! (10!)10

Similarly, least probable macrostate is 100! Ωmin = (100)! (0)! (0)! (0)! (0)! (0)! (0)! (0)! (0)! (0)!

1 Vb4 and z = 3 / 2 kT β

Given < E > = −

Thermodynamic probability is given by n! Ω= n1 ! n2 ! n3 ! ... n10 ! For the most probable state,

∆S2 = 58.60 cal/k (decrease)

11.

The change in entropy to convert ice at 0° C into water at 0° C is dQ Q mL 1000 × 80 = = = ∆S = ∫ T T T 273

=

100! (100)! Ωminimum = 1

16.

Please see the example no. 1 on page no. 13

17.

Please see the example no. 9 on page no. 19

18.

Please see the example no. 8 on page no. 17 mmm

S-33

Unit

2

Ensembles and Thermodynamic Connections

1. Ensemble A system is defined as a collection of number of particles. An ensemble is defined as a collection of number of particles. An ensemble is defined as a collection of a large number of systems which are essentially independent (non-interacting) of one another but are macroscopically identical. The number of ensemble are known as ‘elements’. Thus, the elements of an ensemble are identical in macroscopic parameters (like volume, pressure, energy, total number of particles), but differ in position and momentum coordinates with one another, that is differ in their (unobservable) microscopic states. The various elements being imaginary do not interact with one another. Each element behaves independently, following the law of mechanics. There is a clear difference between the actual system under study and an element of the ensemble. The system is the physical object, while the elements are imaginary copies of it to enable us to use the probability theory. The concept of ensemble helps in understanding the actual system more completely. It represents at one time the (microscopic)) states of the actual system which would occur in the course of time. It is easier to predict the statistical behaviour of a chosen ensemble than to study the behaviour of any particular system. By the knowledge of the behaviour of the ensemble, we can predict the probable behaviour of the system under consideration.

2. Classification of Ensembles The macroscopic identity of the systems constituting an ensemble may be achieved by choosing the same values of some set of macroscopic parameter which uniquely determine the state of the system. Thus, there may be many types of ensembles. Most widely used of them are micro-canonical, canonical and grand-canonical ensembles. 1.

Micro-canonical ensemble : It is the collection of a large number of essentially independent system having the same energy E, some volume V and some number of particles N for simplicity, we assume that all the particles are identical. The individual system of a micro-canonical ensemble are separated by rigid, impermeable cable and well-insulated walls in figure. Such that the value of E, V and N for a particular system are not affected by the presence of system are not affected by the presence of other systems.

S-34

E, V, N

E, V, N

E, V, N

E, V, N

E, V, N

E, V, N

E, V, N

E, V, N

E, V, N

Fig. 1 : Micro-canonical ensemble

2.

3.

Canonical Ensemble : It is the collection of a large number of essentially independent system having the same temperature T , same volume V and the same number of identical particles N. The equality of temperature of all systems can be achieved by bringing each in thermal contact with a large heat reservoir at constant temperature T or bringing all of the system in normal contact with one another. The individual systems of a canonical ensemble are separated by rigid, impermeable and conducting walls in figure (2). As the separating walls are conducting, heat can be exchanged betwen the systems. As a result all the systems will arrive at the common temperature T . Grand-canonical Ensemble : It is the collection of large number of essentially independent system having the same temperature T , same volume V and same chemical potential µ. The individual systems of a grandcanonical ensemble are separated by rigid, permeable and conducting wall in fig (3). As the separating walls and conducting and permeable, the exchange of heat as well as that of particles between the system takes place in such a way that all the system arrive at a common temperature T and common chemical potential µ.

T, V, N T, V, N T, V, N T, V, N T, V, N T, V, N T, V, N T, V, N T, V, N

Fig. 2 : Canonical ensemble

T, V, µ T, V, µ T, V, µ T, V, µ T, V, µ T, V, µ T, V, µ T, V, µ T, V, µ

Fig. 3 : Grand-canonical ensemble

3. Thermodynamic Connections 3.1 Ideal Gas Equation Suppose a vessel of volume V1 contains 1-mole of an ideal gas. The number of molecules in the gas is N (Avogadro number). The probability of a molecule of being found in a small part of volume V of the vessel is 2

V V   . The probability of both the molecules being found simultaneously in the same part is   , because  V1   V1  probability is multiplicative. Hence, the probability of all the N-molecule of being found in the volume V is V Ω=   V1 

N

Putting this value of Ω in Boltzmann’s entropy probability relation,

S-35 S = k log e (Ω) V S = k log e    V1 

We have

N

S = kN (log e V − log e V1) Differentiating it with respect to V (V1 → is constant), we have 1 ∂S = kN V ∂V

…(1)

The Ist and IInd laws of thermodynamics dQ = dU + dW dQ = dU + PdV

or and dQ = TdS, gives

dU + PdV = TdS Differentiating it with respect to V at constant temperature, we have  ∂S   ∂U   + P =T    ∂V  T  ∂V  T But, for an ideal gas  ∂U   = 0 (Joule’s law)   ∂V  T ∴

P  ∂S    =  ∂V  T T

…(2)

Comparing equation (1) and equation (2) we get kN P = V T PV = kNT But kN = R (universal gas constant) ∴

PV = RT

This is an ideal gas equation.

3.2 Change in Entropy of an Ideal Gas in Isothermal Expansion Let a vessel of volume Vf has µ-mole of an ideal gas. The number of molecules in the gas is µN, where N is Avogadro’s number. Suppose the vessel is divided by an imaginary wall and its volume on one side of the wall is Vi. The probability of all the µN molecules of being found in the volume Vi is V Ω i =  i   Vf 

µN

If the wall be removed, then the probability of all the molecules of being found in the whole volume Vf of the vessel will be 1 that is Ω f = 1. Therefore,

S-36 Ω i  Vi  =  Ω f  Vf 

µN

…(1)

S = k log e Ω

We have

S f − Si = k log e Ω f − k log e Ω Ωf = k log e Ωi Ωf

 Vf  But =  Ω i  Vi 

µN

, by equation (1).  Vf  S f − Si = k log    Vi 

or

µN

 Vf  = k µN log e    Vi  But kN = R S f − Si = µR log e

∴

Vf Vi

since Vf > Vi, the entropy of the gas increases.

4. Entropy of Mixing and Gibb’s Paradox The partition function of a perfect gas is Z=

V h3

(2πmkT )3/ 2

…(1)

The entropy of a perfect gas is S = Nk log Z +

3 Nk 2

…(2)

Substituting value of Z from equation (1), the expression for the entropy of a perfect gas becomes V  3 S = Nk log  3 (2πmkT )3/ 2 + Nk h  2

…3(a)

3 3   S = Nk log V + log m + log T + C 2 2  

…3(b)

This may be written as

where C is a constant term including constant factors h, k . The entropy given by above equation does not satisfy the additive property thus giving paradoxial result. To explain this, consider two system denoted by indices a and b held at the same temperature Ta = Tb = T (say) and partitioned by a barrier as Rn figure.

S-37 Barrier (a)

(b)

Na, Va T, ma Nb, Vb T, mb Sa

Sb

Fig. 4 : Entropy of gas : Gibb’s Paradox

If the particles of the two systems different, then using equation (3b), the entropies of the system ‘a’ and ‘b’ are given by 3 3   …4(a) Sa = N a k log Va + log ma + log T + C 2 2   3 3   Sb = N b k log Vb + log mb + log T + C 2 2  

…4(b)

where N a , ma and Va refer respectively to the number of particles, mass of each particle and the volume of system ‘a’ and N b, mb and Vb refer to the corresponding quantities for system ‘b’. As the entropy is an extensive quantity. It must satisfy the additive property. The entropy of the joined system would have been given by 3 3 3 3     …(5) Sab = Sa + Sb = N ak log Va + ma + log T + C + N bk log Vb + mb + log T + C 2 2 2 2     However, if the particles of the two systems are same and if for convenience we take Va = Vb = V and N a = N b = N the entropy of each of the individual system would be 3 3   …(6) Sa = Sb = Nk log V + log m + log T + C 2 2   Then entropy of the combined system would have been Sab = Sa + Sb 3 3   Sab = 2Nk log V + log m + log T + C 2 2  

…(7)

Let the partition be removed to allow the molecules of the gas to mix freely. We have to consider the mixed system having 2N particles in a volume 2V then, 3 3   Sab = 2Nk log 2V + log m + log T + C 2 2   3 3   = 2 Nk log V + log m + log T + C + 2 Nk log e 2 2 2   Sab = Sa + Sb + 2 Nk log e 2

…(8)

which is not in (7), but contains an extra factor 2 Nk log e 2. This indicates that by mixing two different gases each containing same number of molecules N, by removing a partition between them. The entropy of joined system increases by an unaccounted amount 2 Nk log e 2. This additional entropy is called the entropy of mixing. Thus we use equation (3) for entropy, we get paradox results because entropy being as extensive thermodynamic function. The total entropy of the joined system would have been given by equation (7). This peculiar behaviour of the entropy is called Gibb’s paradox.

S-38

5. Partition Function The partition function ‘Z’ of a system is the sum of all the Boltzmann factor over all energy states of the system Z=

∑ e − βE

i

i

Let us consider a dilute ideal gas in equilibrium at an absolute temperature T . Let us focus attention on a particular molecule of the gas and regard it as a small system in thermal contact with a heat reservoir (at temperature T ) consisting of all the other molecules of the gas. The probability of finding the molecule in any one of its quantum states i where its energy is E i is given by the canonical distribution. Pi =

e − βE i

∑ e − βE i

, where β =

1 kT

i

Now, if a molecule is found with probability Pi in a state i of energy E i, then its mean energy is given by E=

∑ E i e − βE

∑ Pi E i =

i

i

∑ e − βE

i

…(1) i

i

where summation is over all possible state i of the molecule. Let us consider the numerator of equation (1)

∑ E i e − βE

i

=−

i

∂

∑ ∂β (e −βE ) = i

i

−∂ − βE  ∑ e i ∂β  i 

Making this substitution in equation (1), we get − E=

∂  − βE  ∑ e i ∂β  i 

∑ e − βE

i

i

Let us now define the quantity

∑ e − βE

i

= Z . Then we get

i

−∂Z 1 ∂Z ∂β E= =− Z Z ∂β or

E=−

∂ log g Z ∂β

The calculation of mean energy E thus requires only the evaluation of the sum Z. The sum Z over all the molecule is the ‘partition function’ of the molecule. Again, let us consider an assembly of molecules in equilibrium at absolute temperature T . The probability of finding a molecule in a particular microstate of energy E i is Pi =

e − βE i Σ e − βE i i

where β =

1 kT

S-39 Now, the denominator D r Σ e −βE i is the partition function Z of the molecule. i

−Ei

∴

Pi =

e

kT

Z

By probability concept, we can write Pi = n i / N where n i is the number of molecules having energy E i and N is the total number of molecules. −Ei

or

n i = Pi N =

∴

ni =

e

kT

Z

N

Ei

N − kT e Z

6. Statistical Definition of Temperature According to Boltzmann’s canonical law, the probalitity of finding the system in an energy state E i in thermal − E1

equilibrium with a heat source at an absolute temperature T , is P1 = Ae system in energy state E 2 at the same temperature T is P2 = Ae

kT

and the probablity of finding the

.

− E1 / kT

∴

P1 e = P2 e − E 2 / kt

or

P1 = e −( E1 P2

or

 P  −(E1 − E 2) log e  1  = kT  P2 

or

− E 2 / kT

T=

− E 2 ) / kT

E 2 − E1 P  k log e  1   P2 

where T is called statistical interpretation of temperature. This equation defines the temperature according to Boltzmann’s canonical law according to which knowing the probabilities P1 and P2 of the system in state E1 and E 2 respectively at a temperature can be determined.

6.1 Interpretation of Second Law of Thermodynamics According to second law of thermodynamics, an irreversible process always occurs in a direction in which the entropy of universe increases. In reversible process the entropy always remains unchanged. Thus,

dS ≥ 0

Statistically, S = k log Ω, therefore when entropy S increases , the thermodynamic probability Ω also increases. Thus, statistically in irreversible processes, the change in the system occurs such that it takes the most probable state.

S-40 Equipartition of energy : According to the law of equipartition of energy, ‘‘If a system made up number of particles is in thermal equilibrium at an absolute temperature T , its total kinetic energy is distributed equally 1 amongst its all degrees of freedom and the mean kinetic energy per degree of freedom is kT where k is 2 Boltzmann’s constant = 1.38 × 10−23 J/K. Proof : Consider a thermodynamical system with degrees of f freedom. We require f position co-ordinates q1, q2, q3 ... q f and f -momenta coordinates p1, p2, p3 ..., p f to express the position and motion of the system. The total energy E of the system will be a function of these f -position co-ordinates and momentum coordinates i. e. E = E(q1, q2, q3 ..., q f , p1, p2, p3 ... p f ). The above expression of energy can be written in the following form : E = E i( pi) + E ′ (q1, q2, q3 ... q f , p1, p2, p3 ... p f ) where E i is the function of pi only, while E′ is the function of f position coordinates and ( f − 1) momentum coordinates, i. e. does not depond on pi. According to Botzmann’s Canonical law, the probability for molecules to be with energy E at temperature T in the range qi ɺ 2 is P = Ae − E / kT dq1dq2 ... dq f dp1dp2 ... dp f ɺ 1, p2 and p2 + dp q1 + dqɺ 1, q2 + dqɺ 2 ..., p1 and p + dp Hence mean value of E i in thermal equilibrium is ∞

∞

− E / kT

∞

∞

−( E i + E ′) / kt

∫ ... ∫−∞ E ie dq1dq2 ... dq f dp1 ... dp f < E i > = ∞−∞ ∞ − E / kT ∫−∞ ... ∫−∞e dq1dq2 ... dq f dp1dp2 ... dp f or

∫ ... ∫ E ie < E i > = −∞∞ −∞∞ −( E ∫−∞ ... ∫−∞e ∞

or

< Ei > =

− E i / kT

dq1dq2 ... dq f dp1dp2 ... dp f ∞

∞

− E ′ / kt

dpi ∫ ... ∫ e dq1 ... dq f dp1 ... dp f ∫−∞ ... ∫−∞ E ie −∞ −∞ ∞ − E / kT ∞ − E ′ / kT ∞ dp1 ∫ ... ∫ e dq1 ... dq f dp1 ... dp f ∫−∞e −∞ −∞ i

∞

or

∞

i + E ′) / kT

dq1dq2 ... dq f dp1dp2 ... dp f

− E i / kT

dpi ∫ E ie < E i > = −∞∞ − E / kT dpi ∫−∞e i

E i = αp12

But

∞

∴

2 2 − αp i / kT

∫ αp1 e < E i > = −∞ ∞ − αp ∫−∞e

2 / kT i

dpi

dpi

But from standard integrals, ∞

∫−∞x

2 − αx 2

e

=

1 2α α×

From

< Ei > =

π and 2 1  2α     kT  π /α kT

∞

∫−∞e π /α kT

− αx 2

dx =

π α

S-41 < Ei > =

or

kT 2

1 kT which is same for each position and each 2 momenta coordinate. This is the law of equipartition of energy. Hence proved. Thus, the mean energy corresponding to the ith momenta is

6.2 Statistical Interpretation of Second Law of Thermodynamics The second law of thermodynamics is concerned with the direction in which any natural process occurs in a system. According to the entropy interpretation of law ‘‘Any natural process in a system occurs in such a direction that result in an increase in the entropy of the system.’’ (In the equilibrium state the entropy remain unchanged). Statistically, an isolated system tends to approach a state (macrostate) of large probability where the number of microstates accessible to it is larger than initially. (If the system is already in its most probable state, it remains in equilibrium). This means that the direction of natural processes is controlled by the laws of probability. Hence, statistically, the second law should be modified as : A natural process occurring in a system has a large probability of producing a net increase in the entropy of the system and surroundings. The probability of decrease in entropy is very small, but not zero. Under very unusual conditions, the entropy may be decrease. Such event are known as ‘fluctuations’. Some examples to demonstrate the statistical nature of the second law of the thermodynamics are following : 1.

Let us consider the conduction of heat from a hot stove to a kettle of water. According to the statistical interpretation of the second law, we assert, not that heat must flow from the stove to the water. It is to be understood that there is always a chance although an extremely small one, that the reverse may take place.

2.

Suppose that a vessel containing a gas is divided into two equal compartment by a partition in which there is a small trap door. On the average, the molecules of the gas will be equally divided between the two compartments, but occasionally there may be more molecules in one compartment than in the other. Similarly there are chances, although extremely small, when all the molecules are confined to one compartment only. According to the statistical interpretation of second law of thermodynamics we would say that there is small probability of all the molecules being conform to the one compartment but a much larger probability of an even distribution of the molecules between the two compartments.

The second law tells not only the possible events, it tells which events are most probable.

7. Entropy, Pressure and Chemical Potential 7.1 Entropy Entropy is a physical quantity which shows the random motion of molecules. If dQ heat is given to the system at temperature T then entropy of system will be dQ dS = T Its unit is calorie/kelvin.

S-42

7.2 Chemical Potential As we know that about term electric potential is the amount of work done in bringing unit charge W V= q If q = 1 coulomb then V = W. Thus, the work done in bringing unit charge q to a point from infinity. Here, q is the capacity factor. Similarly, we can define the capacity and potential quantities for a chemical system in which ‘mole’ is taken as the capacity factor and molal free energy as the potential. This molal free energy is known as chemical potential. The amount of work done in transferring n mole of a substance from a state of molal free energy F1 to another state of molal free energy F2 (likewise V1 to V2) will be n(F2 − F1). Chemical potential of a system is denoted by µ  ∂F  where µ =   .  ∂N 

7.3 Pressure Pressure is a force exerted by the substance per unit area on a another substance. The pressure of a gas is the force that the gas exerts on the walls of its container. When you blow air into a balloon, the baloon expands because the pressure of air molecules is greater on inside of the balloon then the outside.

8. Physical Significance of Various Statistical Quantities 8.1 Temperature Temperature is a quantity we are all familiar with.We know that it measures the microscopic jiggling of atoms : the faster they are moving, the hotter an object feels. So when defined it in terms of a derivative of the density of states, a seemingly unrelated concept you were probably a little surprised. To resolve this seeming paradox, I am going to prove a very important theorem called the equipartition theorem. Once you reach the end, the significance of temperature will become clear. In general, the energy of a system can depend in a arbitrary way on all of its microscopic variables, which we will call x 1, x 2, etc. That is, the energy is a function E(x 1, x 2, ...). But there are some very important special cases where it takes particular form. Let us make three assumptions : 1.

The energy of the system can be written E(x 1, x 2, x 3, ... 0 = E1(x 1) + E 2(x 2, x 3, ...) That is, the energy separates into two terms. The first term depends only on x 1, while the second term does not depend on x 1 at all.

2.

x 1 can be treated as a continuous variable, and can take on any value from − ∞ to ∞.

3.

The energy is quadratic in x 1 is E1(x 1)= Cx 12 where C is a constant.

S-43

8.2 Thermodynamic Potentials as a Measure of Probability Thermodynamic potentials are confusing. Not only are they an unfamiliar concept, but as soon as you start trying to do calculations with them, it is very easy to get all tangled up in a net of subtle mistakes. In the following sections, I will approach thermodynamic potentials from several different directions. I will point out some of the particular issues that make them confusing, and try to help you form a clear understanding of the essential concepts. The first thing that makes them confusing is that they mix together quantities related to two different systems. Recall that we started with a single isolated system, then spilt it into two pieces called A and B. A is the system of interest, the thing we really care about B is the heat bath. Consider the Gibbs free energy as an example. Let me restate the definition, including a script on every variable to indicate which one it refers to : G A = E A + PB VA − TB S A is very easy to get confused about which system each variable relates to. For example easy to see the term TS and remember that T and S are each defined in terms of log(Ω) forget that their definitions refer to completely different ΩS. T is related to the density of states of the heat bath, while S measures the density of states of the system of interest. A thermodynamic potential, is first and foremost, a measure of how probable it is for a system to be in a particular state. It appears in the exponent of the Maxwell-Boltzmann distribution. Lower values of the potential indicate more likely states, while higher values indicate less likely states.

8.3 Thermodynamic Potentials and Thermodynamic Forces According to classical mechanics, if the potential energy U of a system depends on a variable the system experiences a generalized force ∂U Q= − ∂x When the value of x changes by an amount ∆x, the system performs work equal to W = Q∆ x Since we have defined physical quantities called thermodynamic potentials and thermodynamic forces, you may be wondering how closely they are related to potentials and forces of the conventional, non-thermodynamic kind.

8.4 Intensive and Extensive Variables Macroscopic variables that are independent of the size of the system are called intensive variables. Temperature, pressure, and chemical potential are all intensive variables. For example, the temperature of a system has nothing to do with how large that system is. Energy is the relevent thermodynamic potential. Physically speaking, T and P are the quantities you control, with E and V vary in response to term.

8.5 The Mechanics of Thermodynamic Forces Consider a balloon filled with gas. It is subject to three different forces : the outward pressure of the gas inside the balloon, the inward pressure of the surrounding air, and the elastic tension of the balloon itself that resists expansion. At equilibrium these three forces exactly balance each other out. If they do not balance, the balloon will expand or contract until they do. But what is actually happening at a microscopic level ?

S-44 Molecules of gas are constantly striking the surface of the balloon. How do we know that ? Because the density of states of the gas increases with increasing volume. If its volume were greater, there would be more microstates available to it. The balloon is restricting it from visiting those microstates, and if the balloon were not there, it would not remain contained in such a restricted volume. That is what pressure really is : the mechanical force exerted by the gas molecules as they strike against the balloon (or any other object that restricts their motion). So pressure is not merely like a force. It is a force. It can do all the same things other forces can do, including producing accelerations and performing work. Chemical potential can be understood in exactly the same way. Imagine a box with a small hole in it, so that air moelcules can diffuse in and out. The chemical potential is essentially a measure of the density of air molecules. In equilibrium, we expect the density to be the same inside and outside. If that is not the case, we expect to find a net flow of molecules one way or the other unit equilibrium is achieved. On the other hand, if there is another force involved (such as one that repels molecules away from the interior of the box), then we expect to find different densities inside and outside. We can find the equilibrium distribution by looking for the values that minimize the thermodynamic potential. Here is one more critical observation about thermodynamic forces, they are always proportional to the temperature. Given the microscopic dscription above, this is now easier to understand. For example, pressure is the force of particles randomly striking the walls of a container. The faster they ar moving, the harder they strike it. The average velocity of each particle is proportional to the temperature. If the temperature were exactly zero so the particles were not moving at all, there would be no pressure. All thermodynamic forces would disappear, and all thermodynamic potentials would simply become equal to the energy.

8.6 Thermal Equilibrium We took the derivative of G with respect to volume, and derived a condition for the system to be in equilibrium. Let’s repeat the same calculation, only instead taking the derivative with respect to energy ∂S A ∂G A = 0 = 1 − TB ∂E A ∂E A = 1 − kTB

∂ log(Ω A) T =1− B ∂E A TA

From which we conclude T A = TB Unsurprisingly, the requirement is that both subsystems must have the same temperature. Two system whose temperatures are equal said to be in thermal equilibrium. If you bring them into contact with each other, no energy will flow between them. On the other hand, if the systems have different temperatures, there will be a net flow of energy until their temperatures become equal. This energy transfer is different from the ones seen in the previous sections, in that if does not involve any mechanical work. There is no change to any independent macroscopic variable other than energy. It is simply the result of random collisions between molecules that transfer kinetic energy from one subsystem to the other. This type of energy transfer is known as heat.

8.7 A Brief Rant : Internal Energy We have written E for energy, they instead write U for internal energy. At best, internal energy is a useless distinction, and at worst it can be actively misleading.

S-45 The idea is that the total energy of a system can be divided into three parts : the kinetic energy of the system as a whole, the potential energy of the system as a whole, and the internal energy. Only the last of these affects the internal dynamics of the system. For example, if you put your experimental apparatus on an airplane flying at 1000 km/hour, that greatly increases its kinetic energy, but has no effect on you results. Likewise if you take the apparatus to type top of not. Everest, that greatly increases its potential energy but again does not affect your results. The problem with this ideal is that it simply wrong. Every time the airplane encounters turbulence, the bouncing will add heat to the system, thus affective you results. The same thing happens to a lesser extent every time it speeds up, slows down, turns, or accelerates in any other ways,the only case where the internal energy is fully decoupled from the overall motion of the system is when it moves at a constant speed in a straight line with no acceleration at all. But in that case, the only difference between internal energy and total energy is what reference frame you calculate it in. Relativity tells us that all your results must be independent of what reference frame you use, so the choice of whether to use internal energy or total energy is irrelevant. Similarly, the gravity at the top of Mt. Everest is slightly weaker than at sea level, and that does have the potential to change your results. In many (but not all) cases, the difference is negligible, and in that event the only difference between internal energy and total energy is that they are offset by a constant. But the zero point of energy is always arbitrary; adding a constant never affects behavior. So once again, it is irrelevant which one you use. In summary, the choice to use internal energy instead of total energy by definition is only correct if does not affect any of your results. If it does affect them, that proves you have defined the internal energy incorrectly. But it is very easy to define it incorrectly, such as by neglecting a contribution that actually does matter. And the only way to make sure is to repeat your calculation using the total energy and verify that the results do not change.

8.8 Physical Significance of Entropy dQ is called T the increase in entropy of that system and if ∆Q is rejected by the system at constant temperature T , the dQ quantity is called the decrease in entropy of the system. T ∆Q dQ Thus change in entropy ∆S = or, for an infinitesimal change dS = . T T If dQ heat is absorbed by a system at a constant temperature in a reversible process, the quantity

It is clear that from the above equation that the unit of entropy is cal/kelvin or J/kelvin. The dimension of heat energy are equal to the product of dimensions of energy and that of absolute temperature. Similarly as the dimension of potential energy of a body in the earth’s gravitational field are equal to the product of dimensions of mass of the body that of the above the earth’s surface. If height is considered to corresponding temperature the mass will correspond to entropy. Thus the energy of a system is the quantity which is reated to the flow of heat in a similar way as mass is related to the vertical motion under gravity. All natural processes are irreversible process in which th entropy of universe (system or surrounding) increases only in ideal reversible process the entropy remain unchanged. Thus dimension form dS ≥ 0.

S-46

Example 1. What is the probability of drawing (i) a king (ii) king of diamond, (iii) any card of diamond from a deck of 52 cards ? Total number of cards = 52

Solution : (i)

Number of kings in the deck = 4 4 1 ∵ Probability of getting a king in a drawing = = 52 13 1 52

(ii)

There is only 1 king of diamond. Hence, the probability of getting it =

(iii)

There are 13 cards of diamond. Hence, the probability of getting any card of diamond =

13 1 = 52 4

Example 2. A bag contains 6 green balls, 8 white balls and 10 black balls. If a ball is drawn from the bag, what is the probability of its being either white or black ? 8 and that Solution : The total number of balls is 6 + 8 + 10 = 24. The probability of getting a white ball is 24 10 the getting a black ball is . Therefore, by addition, the probability of getting either a white or a black ball is 24 8 10 18 3 = = + 24 24 24 4 Example 3. Find the probability that from two dice either 7 or 11 is obtained. Solution : A die has six faces numbered 1, 2, 3, 4, 5, 6. From two dice there are 6 × 6 = 36 alternatives in all. The number of cases in which 7 is obtained are : Die 1 :

1

2

3

4

5

6

Die 2 :

6

5

4

3

2

1

Thus, there are 6 cases in which the sum of two dice comes out be 7. Hence, the probability is P(7) =

6 36

Similarly, there are only 2 cases (6, 5) and (5, 6) in which the sum comes out to be 11. Hence the corresponding probability is 2 P(11) = 36 Therefore, by addition theorem, the probability of getting either 7 or 11 is 6 2 2 P(7) or 11 = P(7) + P(11) = + = 36 36 9 Example 4. A bag contains 3 white and 4 black balls. If two balls are drawn out one after the other, what is the probability that the first ball will be black? The first ball is returned to the bag after it is taken out. Solution : The total number of ball is 7. The total number of ways of drawing 2 balls from the bag is

S-47 7

C2 =

7! 7 × 6 × 5! = = 21 2 !(7 − 2)! 2 × 5!

The number of ways in which one white ball may be drawn out of the 3 is = 3C1 = 4

C1 =

3! =3 1! 2!

4! 4 × 3! =4 = 1! 3! 1 × 3!

The probability of obtaining one white ball is 3

C1

7

C2

=

3 1 = 21 7

and the probability of obtaining one black ball is 4 7

C1 C2

=

4 21

By multiplication theorem, the probability of obtaining one white and one black ball is

1 4 4 × = 7 21 147

Example 5. From a pack of 52 cards two are drawn at random. Find the chance that one is a king and the other a queen. Solution : The total number of ways of drawing 2 cards out of 52 is 52 C 2 and there are 4 kings and 4 queens. Therefore, the number of ways of having a king and a queen in 4 C1 × 4 C1 Required probability =

∴

4

C1 × 4 C1 52

C2

4! 4! × 1! 3! 1! 3! = 4 × 4 = 8 52 ! 52 × 51 663 2 ! 50 ! 2 Example 6. Find the probability that is tossing a coin 12 times we get (i) 4 heads, 8 tails (ii) 6 heads, 6 tails. N! Solution : The probability of getting r heads out of N tosses of a coin is given by P(r ) = P r qN − r r !( N − r ) where P is the probability of getting a ‘head’ and q is the probability of getting a ‘tail’ in a single toss a ‘head’ up 1 1 a ‘tail’ up and each is equally likely to occur, thus the probability of each is that is P = q = 2 2 The last expression thus reduces to P(r ) =

r!  1  N r !( N − r )!  2

Here N = 12 (i)

The probabiloity of getting 4 heads (r = 4) is P(4) =

12 ! = 4 !(12 − 4)!

 1    2

12

=

12 !  1    4 ! 8 !  12

12

S-48

=

(12 × 11 × 10 × 9) × 8 !  1    (4 × 3 × 2 × 1) × 8 !  2

 1 = 495   2 (ii)

12

=

12

495 = 0121 . 4096

The probability of getting 6 heads (r = 6 ) is P(6) = =

12 !  1   6 !(12 − 6)!  2

12

=

12 !  1    6 ! 6 !  2

12

(12 × 11 × 10 × 9 × 8 × 7) × 6 !  1    (6 × 5 × 4 × 3 × 2 × 1) × 6 !  2

 1 = 924    2

12

=

12

924 = 0.225 4096

Example 7. Identical coins are tossed simultaneously. Calculate probability to get (i) the heads of 5 coins up, (ii) the tails of 4 coins up. Also, calculate the probabilities of maximum and minimum possible combinations. Solution : Given n = 12 (i)

To get the heads of 5 coins up r=5 Macrostate = (r , n − r ) = (5, 7) From relation n

P(r , n − r ) =

2n n

P(5, 7) = (ii)

Cr

C5

212

12 ! 212 7 ! = 29 = 0.07132 = 5 !12 2

To get the tails of 4 coins up n −r= 4 ∵

Macrostate = (8 , 4) 12

P(8, 4) =

C8 12

2 12 ! 8 4 ! = 495 = 0121 = !12 . 2 212

The maximum probability of combination n 12 =6 r= = 2 2 12 ! 12 C 6 6 ! 6 ! 924 = 12 = 12 = 12 2 2 2

S-49 Pmax = 0.2256 The minimum probability of combination r = 12, n = 12 Pmin = P(12, 0 ) 12

=

C12

212

=

1 212

= 0.000244 Example 8. Ten particles are distributed in two identical cells. Calculate : (i) the probability for (3, 7) distribution (ii) the most probable macrostate (iii) ratio of the probability for (3, 7) macrostate with the maximum probability. n

Solution : Probability for (r , n − r ) = (i)

Cr

2n

For (3, 7) distribution n = 10 and r = 3 10 ! 3 7! P(3, 7) = 10 = !10 2 2 10 × 9 × 8 1 × = 3× 2×1 20 10

=

C3

120 210

P(3, 7) = 0117 . (ii)

For the most probable macrostate or or

r=n −r 2r = n

∴ (5, 5) is the most probable macrostate (iii)

Ratio of the probability for (3, 7) macrostate with the maximum probability 10

C3 10 P (3, 7) 2 = P (5, 5) 10 C 5 210 10 ! 10 C3 3 ! 7 ! 5 ! 5 ! = = 10 10 ! 3! 7 ! C5 5! 5! =

5! × 5 × 4 × 3! 3! × 7 × 6 × 5!

=

10 = 0 .476 21

S-50 Example 9. Assuming that in a six faced die, the probability of getting any face uppermost is equal. Five dies are tossed simultaneously. Calculate the probability of getting the number ‘2’ uppermost (i) in any one die (ii) in all 5 dice (ii) in at least 1 die. 1 Solution : The probability of getting the number (2) uppermost in a dice is and the probability of not getting 6 in uppermost is 1 5 1− = 6 6 1 5 Then, and PB = PA = 6 6 (i)

If out of the 5 dice, the number 2 comes uppermost only in one dice and does not come in the rest 4 dice the macrostate is (1, 4) i. e. r=1 n −r= 4 n= 5

and From the relation

P(r , n − r ) = nC r (PA)r (PB )n − r 1

 1   5 P(1, 4) = 5C1      6  6 5! × 1! 4 !

1

4

4

5× 5× 5× 5× 5  5  1 = 0.48   ×  =  6  6 6× 6× 6× 6× 6 P(1, 4) = 0.48

(ii)

To get 2 uppermost in all five dice and not to get 2 in any dice the macrostate is (5, 0) i. e. r − 5 and n − r = 0 5

 1   5 Probability P( 5, 0) = 5C 5      6  6

0

5

=

0

5!  5  1  1 ×  ×  =   6  6 5 ! 0 !  6

5

P( 5, 0) = 1.286 × 10−4 (iii)

5 Since the probability of not getting 2 in a die is , therefore the probability of not getting 2 uppermost in 6 any 5 dice is 5

 5 = 1 −   = 0.48  6 Hence, the probability of getting 2 uppermost at least in 1 die 5

 5 1 =   = 1 − 0.48 = 0.52  6

S-51 Example 10. 14 particles are distributed in two boxes A and B. The ratio of probabilities to occupy the boxes A and B by a particle is 4 : 1. Calculate the probability for (i) (6, 8) distribution (ii) (8, 6) distribution. Solution : The probabilities for finding the particle in the box is A and B is PA and PB respectively then PA : PB = 4 : 1 PA = 0.8 and PB = 0.2 Now for (r , n − r ) distribution probability P(r , n − r ) = nC r (PA)r (PB )n − r (i)

For (6, 8 ) distribution, probability 14 ! × (0.8)6 × (0.2)8 6! 8! 14 × 13 × 12 × 11 × 10 × 9 × (0.8)6 × (0.2)8 = 6× 5× 4 × 3× 2×1

P(6, 8) =

14

C 6(PA)6 (PB )8 =

= 0.002015 (ii)

For (8, 6) distribution, probability P( 8, 6) =

14

C 8(PA)8(PB )6

14 ! × (0.8)8 × (0.2)6 8! 6! 14 × 13 × 12 × 11 × 10 × 9 × (0.8)8 × (0.2)6 = 6× 5× 4 × 3× 2×1 =

= 0.03224 Example 11. 20 particles are distributed in 5 identical cells. Calculate the thermodynamic probability for the macrostate (10, 5, 2, 3, 0). Solution : Given n = 20 Macrostate = (10, 5, 2, 3, 0) From the relation W=

n! n1 ! n 2 ! n 3 ! n 4 ! n 5 !

Thermodynamic probability W=

20 × 19 × 18 × 17 × 16 × 15 × 14 × 13 × 12 × 11 × 10 20 = 10 ! × 5 × 4 × 3 × 2 × 1 × 2 × 1 × 3 × 2 × 1 × 1 10 ! 5 ! 2 ! 3 ! 0 !

= 465585120 Example 12. Calculate the probability that is tossing a coin 10 times we get (i) all heads (ii) 4 heads, 6 tails (iii) 2 heads, 8 tails (iv) 7 heads, 3 tails (v) 1 head, 9 tails. Solution : the probability of getting r heads out of N tosses of a coin is given by P(r ) =

N!  1   r !( N − r )!  2

N

S-52 Here N = 10 (i)

The probability of getting all heads (r = 10 ) is P(10) =

10 !  1   10 !(10 − 10)!  2

 1 = 1   2 (ii)

=

1 = 0.00976 1024

The probability of getting 4 heads (r = 4) is P(4) = = =

(iii)

10

10

P(2) =

Similarly,

= (iv)

P(7) =

10 !  1    7 ! 31  2

(v)

P(1) =

10 !  1    1 ! 9 !  2

10

=

10

=

10 !  1    4 ! 6 !  2

10

(10 × 9 × 8 × 7) × 6 ! 1 × (4 × 3 × 2 × 1) × 6 ! 210 210 10

2

=

210 = 0.205 1024

10 !  1    2 ! 8 !  2 45 210

=

10

=

(10 × 9) × 8 ! 1 × 10 (2 × 2) × 8 ! 2

45 = 0.0439 1024

(10 × 9 × 8) × 7 ! 1 120 120 × 10 = 10 = . = 0117 7 !(3 × 2 × 1) 1024 2 2

10 × 9 !  1    1 ! 9 !  2

10

=

10 × 9 !  1    1 × 9 !  2

10

=

10 10

2

=

10 = 0.00976 1024

Example 13. Three similar dice A, B and C each having six equally likely faces marked as 1, 2, 3, 4, 5, 6, are thrown simultaneously and in a toss all the faces have equal probability of appearing up. Calculate the probability of getting the faces of all the dice up marked with 1 number. 1 Solution : Probability of dice A to shown face marked up P1 = 6 1 Probability of dice B to show face marked up P2 = 6 1 Probability of dice C to show face marked up P3 = 6 ∴

Probability of all the three dice A, B and C to show face marked 1 up, 1 1 1 1 P = P1 × P2 × P3 = × × = 6 6 6 216

Example 14. Three particles are distributed in three compartment of equal size. Find the number of microstates in (i) macrostates (0, 0, 3) and (ii) macrostates (1, 2, 0). Solution : (i) For macrostates (0, 0, 3)

S-53 3! =1 0! 0! 3!

Number of macrostates = (ii)

For macrostates (1, 2, 0) Number of macrostates =

3! 1! 2! 0!

Example 15. Calculate the thermodynamic probability of various macrostates corresponding to distribution of 5 particles in two compartment. Solution : Number of particles n = 5 Number of compartment c = 2 Total number of macrostates = n + 1 = 6 The various macrostates are (0, 5), (1, 4), (2, 3), (3, 2)(4, 1) and (5, 0) n! Thermodynamic probability W( n , n = 1 2) n1 ! n 2 ! (i)

For the macrostates (0, 5) thermodynamic probability 5! =1 W( 0, 5) = n1 ! 5 !

(ii)

For the macrostates (1, 4) thermodynamic probability 5! W(1, 4) = =5 1! 4 !

(iii)

For the macrostates (2, 3) thermodynamic probability 5! W( 2, 3) = = 10 2! 3! Similarly for the macrostates (3, 2) and (5, 0), the thermodynamic probabilities respectively are W( 3, 2) = 10, W( 4, 1) = 5, W( 5, 0) = 1

Example 16. In a system of 8 distinguishable particles distributed in two equal sized compartments, calculate the probability of the macrostate (3, 9), ( 4, 4) and ( 2, 6). Solution : Total number of particles n = 8; Total number of compartment C = 2 ∴ Total number of microstate C n = 28 = 256 (i)

Number of microstates in the macrostates (3, 5 ) = ∴ Probability of microstates (3, 5 ) =

8! 8×7 × 6 = 56 = 3! 5! 3 × 2 × 1

56 7 = 256 32

Number of macrostates in the macrostates (4, 4) = ∴ Probability of the macrostates (4, 4) =

8! 8×7 × 6× 5 = = 70 4 !4 ! 4 × 3 × 2 × 1

70 35 = 256 128

S-54 (ii)

Number of microstates in the microstate (2, 6) =

8! 8×7 = = 28 2! 6! 2 × 1

28 7 = 256 64

∴ Probability of the macrostate (2, 6) =

Example 17. Eight distinguishable particles are distributed among three compartments of equal size. Find the probability of the macrostate (i) ( 4, 3, 1) and (ii) (3, 3, 2). Solution : The probability of occurance of the macrostate (n1, n 2, n 3, ... n k ) corresponding to the distributed of n distinguishable particles into k compartment of equal size is given by 1 n! P( n , n , n , ... nk) = × 1 2 3 n1 ! n 2 ! ... n k ! k n Here, total number of particles, n = 8 and the total number of compartment, k = 3. (i)

For the macrostates (4, 3, 1); n1 = 4, n 2 = 3, n 3 = 1 8! 1 × 8 P( 4, 3, 1) = 4 ! 3 !1 ! 3 8×7 × 6× 5 1 280 = × 8 = 3× 2 6581 3 = 0.0421

(ii)

For the macrostate (3, 3, 2), n1 = 3, n 2 = 3, n 3 = 2 8! 1 P( 3, 3, 2) = × 3 ! 3 ! 2 ! 38 8×7 × 6× 5 1 460 × 8 = = 3× 2× 2 6561 3 = 0.0854

Example 18. What is the maximum and minimum value of the probability of occurence of a macrostate for the case of n distinguishable particles distributed in two compartment ? Prove. Solution :The total number of terms representing all possible combinations is Σ n C r = 2n Probability of distribution (r , n − r ), is given by P( r, n − r) = Pmax =

=

Pmin =

1 n! × r !(n − r )! 2n

…(1)

1 n! × n n n     2   !  !  2  2 n!

1

n  n ! 2    2 2

n! 1 1 × n = n n !(n − n)! 2 2

…(2)

…(3)

S-55 Example 19. An excited state of an atom is 1.38 eV above the ground state. Calculate the number of atoms in this excited state relative to the ground state at 16000 K. ( k = 1.38 × 10 −23 J/K −1 ) Solution : Let E1 and E 2 be the energies of the ground state and of the excited strate respectively. By canonical distribution the probability of finding an atom in these states are P1 = Ce −βE i , P2 = Ce −βE i where C is a constant and β =

1 . kT P2 e −βE i = = e −β[ E 2 P1 e −βE i

∴

= e −( E 2

− E1 ]

− E1 ) / kt

 −1.38 × 1.6 × 10−19  P2 = exp   −23 P1 (1.38 × 10 × 16000) = e −1 = 0.368 Example 20. Calculate the number of phase cells in a given energy range for a one dimensional oscillator. Solution : For 1-dimensional oscillator, the phase space is two-dimensional (x − px ). The area of a phase cell in this phase space is δ x δpx = h (by uncertainty principle). The (total) energy of the oscillator of mass m is given by E=

px2 1 2 + kx 2m 2

which may be written as p2 x2 + x =1  2E  2mE    k  This represents an ellipse of semi-major axis a = 2E / k and semi-minor axis b = 2mE . Thus, the area of the phase space of the oscillator having charging between 0 and E is πab = π

2E m 2mE = 2πE k k

Frequency of the oscillator is x and so x= ∴

1 k 2π m

Area of phase space = 2πE

1 E = 2π x x

Since the area of each phase cell is h, the number of phase cells (in energy range 0 to E) is E E/ x = hx h

S-56

1.

A system is in thermodynamic equilibrium at absolute temperature T. The energy difference between two states is 4.83 × 10−21 J and relative probability of these states is e 2 . Calculate the value of temperature ? (Given that k = 1.38 × 10−23 J/K)

2.

Find the entropy of 1 mole of chlorine gas at standard scale at 27°C.

3.

Calculate the ratio of microstates of 1 g water at 400 K temperature. Given that : Specific heat = 4.2 joule/g and k = 1.38 × 10−23 joule/Kelvin.

4.

For a system entropy = 15 cal / K. Calculate the number of microstates if 1 cal = 4.2 × 107 erg, k = 1.38 × 10−23 erg/K.

5.

If β parameter of a body is 2 × 1020 joule and temperature is 362 K. Calculate Boltzmann’s constants.

6.

If β = 1.85 × 1020 Joule −1 and K = 1.38 × 10−23 J/K then calculate the temperature of body.

7.

What is the probability of drawing (i) a king (ii) of diamond from a deck of 52 cards.

8.

Calculate the number of phase cells in a given energy range for one dimensional oscillator.

9.

An excited state of an electron is 1.38 eV above the ground state. Calculate the number of atoms in this excited state.

10.

A bag contains 6 green balls, 8 white balls and 10 black balls. If a ball is drawn the bag, what is the probability of its being either white or black ?

11.

A bag contains 3 white and 4 black balls if two balls are drawn out one after other. What is the probability that first ball will be white and second ball will be black ? The first ball is returned to the bag after it is taken out.

12.

From a pack of 52 cards two are drawn at random. Find the chance that one is a king and other a queen.

13.

Find the probability that is tossing a coin 12 times we get (ii) 4 heads, 8 tails (ii) 6 heads, 6 tails.

14.

Identical coins are tossed simultaneously. Calculate probability to get (i) the head of 5 coins up, (ii) The tails of 4 coins up. Also, calculate the probability of maximum or minimum possible combinations.

15.

Ten particles are distributed in two identical cells. Calculate the probability for (3, 7) distribution.

16.

14 particles are distributed in two boxes A and B. The ratio of probabilities of occupy the boxes A and B by a particle is 4 : 1. Calculate the probability for (i) (6, 8) distribution (ii) (8, 6) distribution.

17.

20 particles are distributed in 5 identical cells. Calculate the thermodynamic probability for the macrostate (10, 5, 2, 3, 0).

18.

Calculate the probability that tossing a coin 10 times we get (i) all heads (ii) 4 heads, 6 tails.

19.

Calculate the thermodynamic probability of various macrostates corresponding to distribution of 5 particles in two compartment.

S-57

Long Answer Type Questions 1.

2. 3.

5.

What is meant by an ensemble ? Give a brief 6. comparative account of micro-canonical, canonical 7. and grand-canonical ensembles ? 8. Discuss the density of distribution in the phase space. 9. Discuss statistical equilibrium and derive the 10. neccessary condition for it.

4.

What is Louville theorem ? Prove it.

5.

Give the interpretation thermodynamic.

6.

What is entropy ? Discuss it.

7.

Using Boltzmann’s entropy-probability relation, 1. obtain the ideal gas equation.

8.

What do you understand by the term ‘partition function’ ? Obtain expression for the mean energy of molecules in a system in term of partition function.

of

second

law

of

What is canonical ensemble ? Give the definition of probability. Entropy of a universe always increases. Why ? Give physical signifance of entropy. What is the difference between micro-canonical, ensembe and grand-canonical ensemble ?

Objective Type Questions Multiple Choice Questions

9.

Explain the statistical meaning of entropy.

10.

What is Gibb’s Paradox ? How this paradox is solved ?

2.

(a) h03

(b) h06

(c) h0−3

(d) h0−6

The correct statement is :

(b) Only the probability of equilibrium is maximum

1.

What is density of states ?

2.

Describe statistical ensembles.

3.

Define partition function.

4.

What is relation between thermodynamic probability and entropy ? 3. What is the Boltzmann’s canonical partition function ?

the

system

in

(c) Both the probability and entropy of the system in equilibrium are maximum

6.

What is Boltzmann’s Canonical distribution law ?

7.

What is the law of equipartition energy ?

8.

Find statistical entropy thermodynamics.

9.

Write the Clasius theorem.

10.

Give the statement of entropy.

of

The volume of an elementary cell in the six dimensional phase space is :

(a) The entropy of the system in equilibrium is minimum.

Short Answer Type Questions

5.

Define phase space.

second

Very Short Answer Type Questions 1.

What is Grand-canonical ensemble ?

2.

Define the term entropy.

3.

What do you mean by Gibb’s paradox ?

4.

Write the second law of thermodynamics.

4. law

of

5.

(d) Only the entropy of the system in equilibrium is maximum In a two dimensional phase space, one dimensional oscillator can be represented by : (a) A straight line

(b) A parabola

(c) A circle

(d) An ellipse

The number of accessible microstates Ω in the energy range E and E + dE for a classical macro-system (with degrees of freedom 3) is proportional to : (a) E1 / 2

(b) E

(c) E −3

(d) E −1

If the probabilities of a molecule of a gas to be in energy states E and 2E respectively at temperature T and T0 are equal, then T and T0 are related as : T (b) T0 = (a) T0 = T 2 T (d) T0 = (c) T0 = 2T 4

S-58 6.

100 molecules are divided in 10 cells each of same energy. The incorrect statement is : (a) The most probable distribution is in which one cell contains all the molecules and rest of the cells are empty (b) The number of accessible microstates of the 100! 13. most probable distribution are (10!)10

7.

(d) In the least probable distribution, all the 14. molecules will occupy one cells and the rest of the cells will be empty

The point function of the state of a system is :

In a micro-canonical ensemble, each system has the 15. same :

According to Clausius theorem : dQ dQ (a) ∫ (b) ∫ >0 > ni ni EF Ei =

1 e∞ + 1

=0

(e + ∞ = ∞)

Similarly, at T = 0 K and E i = E F Ei = =

1 e° + 1 1 1 = 1+1 2

Example 14. Calculate the average speed of oxygen molecules at 300 K temperature if mass of oxygen molecule is 5 .31 × 10 −26 kg and k = 1 .38 × 10 −23 J/K. Solution : We know that average speed of molecules

v = 1 .596 v = 1 .596

kT m 1 .38 × 10− 23 × 300 5 .31 × 10−26

v = 1 .596 × 2 .8 × 102 v = 447 m/sec

S-86 Example 15. If temperature of O 2 molecules is 300 K, mO = 5 .31 × 10 −26 kg and Boltzmann’s constant K = 1 .38 × 10 −23 J/K. v rms = v 2

Solution :

= 1 .732

kT mO

= 1 .732 ×

1 .38 × 10−23 × 300 5 .31 × 10−26

= 1 .732 × 2 .8 × 102 = 485 m/sec Example 16. Calculate most probable speed v p for oxygen molecules. If mass of oxygen molecule is 5 .31 × 10 −26 kg. Solution : v p = 1 .414

kT = 1 .414 × 2 .8 × 102 = 0 .396 m/sec m

Example 17. Gas molecules has different speeds. Average speed v, rms speed rrms and probable speed v p which energy is correspondence to kinetic energy. Solution :

Average kinetic energy =

1 v12 + v 22 + v 32 + ... + v n2  m  N 2  

1 m v2 2 1 2 E = m v rms 2 =

∴

v rms = v 2

Hence, rrms is correspondence to kinetic energy of molecules. Example 18. First excited state of H 2 atom is 10.2 eV. How much temperature is needed to excite H 2 atom to first excited state from ground state? Solution : The kinetic energy of H 2 gas molecule

E= Now, or

3 1 KT = mv 2 2 2

1 mv 2 = 10.2 eV = 10 .2 × 1 .6 × 10−19 J 2 3 × 1 .38 × 10−23 × T = 10 .2 × 1 .6 × 10−19 2 T=

2 × 10 .2 × 1 .6 × 10− 19 3 × 1 .38 × 10− 23

T = 7 .88 × 104 K

S-87 Example 19. If a gas density is 1 .49 g / m 3 at pressure 10 2 N / m 3 . Then calculate rms speed of molecules. ρ = 1 .4 g/ m 3

Solution :

P = 105 N/ m 2 v rms =

3P = ρ

3 × 105 1 .4

v rms = 4 .6 × 102 m/sec

1.

If there are so many molecules in a system at 600 K. Each molecule has mass 1 × 10−3 kg. Calculate the probability for the molecule that any one molecule will rise up to 1 Å height itself [k = 1.38 × 10−23 J/K].

2.

Calculate the rms speed of a molecule of a gas at 300 K while atomic mass of gas is 221 and R = 8.3 J/mole K.

3.

Calculate the rms speed of a molecule of a gas at 300 K if atomic mass of gas is 32 and R = 8.3 J/mole K.

4.

Calculate speed of nitrogen molecules at 127°C.

5.

Calculate the probability that speed of oxygen molecule at 200 K lies between 99.5 m/sec and 100.5 m/sec.

6.

Calculate vp and rms speed of molecules for hydrogen gas at N.T.P., if k = 1.38 × 10−23 J/K and N = 6 × 1023 mole.

7.

Calculate the value of vx for which defficiency in the probability is (i)

8.

Calculate the probability that the molecules speed of oxygen gas lies between 99 and 100 m/sec at temperature 200 K.

9.

Calculate the probability of that the speed of oxygen molecules lies between 100 and 101 m/sec at 200 K temperature.

10.

If there are many molecules obeying classical statistics in a box at 300 K mass of each molecule is 0.19 m. Calculate that any molecule that will fly to height of 1 Å.

11.

Calculate the fraction of O2 molecules within 1% of the most probable speed at N.T.P. Calculate the effect of changing :

1 1 times (ii) times. 5 10

(i) the gas to CO2 (ii) the temperature to 600 °C 12.

If number of molecules and the corresponding velocities are given in table then calculate average speed v , vrms and most probable speed of the molecules : S.N.

ni

vi m/sec

1

3

4.0

2

2

2.0

3

6

3.0

4

6

4.0

5

8

3.0

S-88 13.

14.

If number of molecules and corresponding velocities are given in table then, calculate average speed v, rrms and most probable speed of the molecules. S.N.

ni

vi m/sec

1

4

1.0

2

2

2.0

3

8

3.0

4

6

4.0

5

5

5.0

10 molecules are moving with speed 1.0, 2.0, 3.0, 3.0, 3.0, 4.0, 4.0, 5.0 and 6.0 m/sec Calculate : (i)

average speed v

(ii) rms speed vrms

(iii) most probable speed 15.

For the given data calculate : (i)

verage speed v

(ii) rms speed vrms

(iii) most probable speed vp S.N.

ni

vi (m/sec)

1

2

1.0

2

4

2.0

3

8

3.0

4

6

4.0

5

3

5.0

16.

Calculate rms speed of H 2 gas molecule at N.T.P.

17.

Mass of oxygen molecule is 5.28 × 10−2 kg and its average speed at N.T.P. is 425 × 102 m/sec. Calculate the average kinetic energy.

18.

At what temperature the speed of H 2 molecules is same for N 2 molecules at 35°C given N = 28 and H = 2.

19.

At what temperature the average speed of H 2 molecules is same as the speed of N 2 molecules at 27°C MH = 2, M N = 28.

20.

Calculate the average speed of oxygen molecules at 300 K temperature if mass of oxygen molecule is 5.31 × 10−26 kg and K = 1.38 × 10−23 J/K.

21.

If temperature of O2 molecules is 300 K, k = 1.38 × 10−23 J/K. Calculate most probable speed vp for oxygen molecule if mass of O2 molecule is 5.31 × 10−26 kg.

22.

First excited state of H 2 atom is 10.2 eV. How much temperature is needed to excite H 2 atom to first excited state.

23.

If a gas density is 1.4 g/m 3 at pressure 10 5 N/m 2 , then calculate rms speed of molecules.

24.

In previous question if gas density ρ = 1.499 g/m 3 and pressure P = 105 N /m 3 . Calculate most probable speed vp .

S-89

Long Answer Type Questions

17.

1.

What is classical statistics ? Discuss its assumptions. Derive its distribution law.

2.

Discuss the classical statistic and hence derive its 18. distribution law.

3.

Give the assumptions of Maxwell−Boltzmann’s statistics and derive its distribution law, ni = e −βE i

4.

5. 6.

7.

8. 9. 10.

19.

Derive an expression for Maxwell–Boltzmann’s 20. statistics distribution law and hence, show that 1 β= 21. kT Give the assumptions of Maxwell’s distribution law and derive an expression for energy distribution law. 22. What are the basic assumptions of Maxwell–Boltzmann’s statistics? Derive energy distribution law. Derive for Maxwell–Boltzmann’s statistics g Number of molecules ni = ( α + iβE ) i e

Show that maximum probability function 8m  Pmax (v) =    πkT 

1/ 2

e −1

Given the experimental verification for Maxwell’s velocity distribution law. Discuss : (i) Zartman’s and K 0 experiment (ii) Doppler broadening of spectral line What is Doppler’s broadning of spectral lines ? Discuss in details? Discuss the postulates of classical statistics and hence derive an expression of energy of molecules. Discuss limitations of classical mechanics.

Short Answer Type Questions 1.

Write a short note on classical statistics.

2. What is Maxwell’s–Boltzmann’s velocity distribution 3. law? Explain Maxwell’s–Boltzmann’s distribution law of 4.

What is Maxwell–Boltzmann’s statistics?

velocity distribution.

Derive an expression φ

5.

Derive an expression for the number of molecules within velocity range v and v + dv m  i.e. n(v) dv ≈ 4 πN    2 πkT 

3/ 2

v2 e

−

1 mv2 2 kT dv

Derive energy distribution Maxwell–Boltzmann’s statistics.

7.

Derive expression

12.

On the basis of Maxwell–Boltzmann’s statistics derive 8. the formulas of average speed, rms speed and most probable speed. 9. Derive an expression for vrms , v and vp where all are 10. having their usual meaning. What is Maxwell’s velocity distribution law ? Derive an expression for vp , v and vrms .

15.

Prove that vrms > v > vp .

16.

Show that by using Maxwell–Boltzmann’s statistics vp : v : vrms = 1 : 1.129 : 1.225

Derive Maxwell’s-Boltzmann’s distribution law.

6.

What is Maxwell’s velocity law of distribution? Discuss in detail?

14.

Write postulates of classical statistics.

ni = λe − βE i .

11.

13.

Calculate the number of molecules corresponding to the most probable speed and the maximum value of the probability function.

ni =

for

gi α + βE i (e )

Define mean or average speed, rms speed, most probable speed. Distinguish among v , vrms and vp . Show that the formula of average speed v = 1.59

11.

function

kT m

Show that the formula of rms speed vrms = 1.73

kT m

S-90 12.

Show that mean energy of a particle is : 3 E = kT 2

Show that the formula of probable speed kT . vp = 1.44 m 27.

13.

Show that vrms > v > vp .

14.

Calculate the number of molecules corresponding to the most probable speed and the maximum value of 28. probable function. 1/ 2

15.

8m  Show that Pmax (v) =    πkT 

16.

Discuss Zartman and Ko experiment.

17.

Discuss the width of spectral line.

18. 19.

20.

21.

e −1 .

23.

24.

25.

Calculate the value of vx for which defficiency in the 3. 1 1 times. probability is : (i) times (ii) 5 10 4. 4kT for Show that the component of velocity vx is m which probability f (x) will becomes its maximum 5. value. Write a short note on limitations of classical 6. 7.

Discuss the limitations of classical mechanics.

Which statistics is called classical statistics ? What kind of spin a molecule Maxwell–Boltzmann’s statistics ?

has

in

What are the examples of Maxwell–Boltzmann’s statistics? What kind of particles Maxwell–Boltzmann’s statistics?

we

study

in

What are the restrictions on the number of molecules in a classical statistics? What is Stirling formula ? Give the formula for thermodynamical probability.

8. If a particle of mass m is in a phase cell of volume V 9. then show that number of phase cells in energy range are 10. 4 π V (2mE)3 / 2 n= 11. 3 h3

Can phonon obeys Maxwell–Boltzmann’s statistics?

If a particle of mass m is in a phase cell of volume V 12. then show that number of microstates in energy 13. range E and E + δE are 14. 4 2 πV 3 / 2 1 / 2 m E dE φ(E) dE = 15. h3 Number of molecules in energy range E and E + dE 16.

Give the formula of velocity distribution law.

1  n(E) dE = 2 πN   πkT 

3/ 2

E1 / 2 − e Ε / kT dE

Then show that most probable energy is 1 E = kT 2 26.

As we know that gas molecules can have different kind of velocities as, v , vrms and vp . Which velocity is correspondance to energy?

Very Short Answer Type Questions

1. What is Zortman and Ko’s experiment ? Why it has 2. been performed ?

mechanics. 22.

Show that that the probability of finding of a particle 1 at Fermi level is . 2

Using the formula 1  n(E) dE = 2 πN   πkT 

3/ 2

E1 / 2 e − E / kT dE

What is the formula for Maxwell–Boltzmann’s distribution law ? What is the value of β-parameter ? Give the distribution Maxwell–Boltzmann’s statistics.

function

for

What is the volume of a phase cell ? What is the formula for average speed ? What is the formula for rms speed ? What is the formula for most probable speed ?

17.

What is the relation between the velocities of molecules ?

18.

Give the name of two experiments can verify the Maxwell−Boltzmann’s velocity distribution law.

19.

Zartman and Ko experiment is based upon.

20.

What was the important limitation of classical mechanics?

21.

What is the probability of electron at Fermi law?

S-91 10.

Multiple Choice Question

For one - dimensional oscillator of mass m frequency v and energy between O to E, the number of phase cell in two dimensional phase space is : E (a) (b) Ehv v

1.

(c)

Objective Type Questions The formula for the most probable speed of the molecules in a gas is : 3kT m 11. (a) VmP = (b) VmP = 2kT m (c) VmP =

2kT m

(d) VmP =

m 3kT

12.

2.

The rms speed of an proportional to the square (a) Mass (b) (c) Both (a) & (b) (d)

3.

In Zartman-Ko experiment distance from the plate is a measure of : (a) Molecular speed (b) Electron speed 13. (c) Electron density (d) None of these

4.

Five particles have speeds 1, 3, 4, 5 and 7 m/sec respectively. Their rms speed is : 10 (a) (b) 10 5 (c) 5

5.

(d) None of these

(c) T

7.

8.

9.

2

E2 hv

Classical statistics is : (a) M.B.

(b) F.D.

(c) B.E.

(d) None of these

β-parameter is : ∂ (a) log e Ω ∂E ∂ (b) log e Ω 2 ∂E (c) log n! = n log n − n ∂Ω (d) log e ∂E For classical statistics. Which is not true? (a) Molecules are distinguishable (c) In a cell there is only one molecule (d) None of the above

14.

For classical statistics. Which one is true? (a) No restriction on molecules in a cell (b) Molecules obeys Pauli’s law strictly g (c) Distribution law is ni = α + βEi + 1 i (e )

(b) T1 / 3 (d) T

The number of coordinate in the phase space of a single particle is : 15. (a) 2 (b) 3 (c) 5 (d) 6 Five particles are distributed in two phase cells. Then the number of macrostate is : Fill (a) 10 (b) 6 1. (c) 32 (d) 32 2. Stirling formula for large n is : (a) log n! = n log n 3. (b) log n! = n log n − n 4. (c) log 2n! = 2 log n − n (d) log n! = n log n + n 5. Phase space is divided into : (a) Groups (b) Sub-groups (c) Sets (d) Cells

(d)

(b) Molecules are identical

The most probable speed of a gas molecule varies with temperature as : (a) T1 / 2

6.

ideal gas is inversely root of its : Temperature None of these

E hv

6.

(d) None of the above Which one is true? (a) vp < v < vrms

(b) vrms < v < vp

(c) vp < v < vrms

(d) vp = v > vrms

in the Blank Classical statistics is ……… statistics. Molecules are ……… in Maxwell–Boltzmann’s statistics. ………… follows Maxwell–Boltzmann’s statistics. In M.B. statistics we use ……… value for the molecules. Molecules can have ……… Maxwell–Boltzmann’s statistics.

spin

in

The number of molecules remains ……… in Maxwell–Boltzmann’s statistics.

S-92 7.

……… is statistics formula.

8.

……… is known as Maxwell’s distribution law.

9.

The value of β is ……… .

10.

……… is known as number of molecules in Maxwell–Boltzmann’s statistics.

11.

Number of ……… cells is h3 .

12.

Average speed v is ……… .

13.

rms speed vrms is ……… .

14.

Probable speed vp is ……… .

15.

Relation between vp , v and vrms is ……… .

16.

……… is the method for verification of M.B. velocity 6. distribution law. 7. According to Dulong Petit’s law CV is ……… .

17.

True/False 1.

Classical statistics is Maxwell–Boltzmann’s statistics.

2.

Molecules in Maxwell–Boltzmann’s statistics are identical.

3.

Electrons follows Maxwell–Boltzmann’s statistics.

4.

Molecules following Maxwell–Boltzmann’s statistics 1 has Spin. 2

5.

Indistinguishable particles Boltzmann’s statistics.

18.

……… of photon could not be explained by 8. Maxwell–Boltzmann’s statistics.

19.

The probability of finding of electron at Fermi level is 9. ……… .

20.

The speed related to kinetic energy is ……… speed.

21.

10. The volume of a phase cell is six dimensional phase 11. space is ……… .

22.

In two dimensional phase space, a one dimensional 12. harmonic oscillator can be expressed as ……… .

23.

At absolute temperature T, the number of atoms in a 13. cell of energy E i as equilibrium is ……… . The minimum size of elementary phase cell in 14. quantum statistics is ……… .

24. 25. 26.

The ratio of vmp to vrms to a gas molecules is ……… . 15. The number of molecules having speed between v and v + dv is proportional to ……… power of 16. velocity.

27.

For small values of speed, the fraction of number of 17. molecules having velocity rises ……… .

28.

For hydrogen gas at N.T.P. the ……… speed is 18. maximum.

29.

The ratio of average speed to rms speed of a gas 19. molecule is ……… .

30.

The expression of average speed of gas molecule is 20. ……… .

follows

Maxwell–

log n! = n log n − n is Stirling approximation. Relation between entropy and thermodynamical probability is S = k log e Ω. g ni = α + iβE are number of molecules. i) (e Maxwell’s velocity distribution law is obeyed in F.D. statistics. Volume of phase cell is h3 . Most of the molecules moves with most probable speed. The valid relation is Vrms < vp < v . The value of rms speed is 173 .

KT . m

Zartman & Ko experiment Maxwell–Boltzmann’s velocity law. The ratio of vp and v in a gas is

can

verify

π . 2

Kinetic energy of molecules is related with v average speed. Average energy of the molecules of monoatomic gas is kT. kT Energy of harmonic oscillation is . 2 Maximum number of molecules with most probable speed. 1 Probability of finding of electrons at Fermi level is . 2

S-93

Objective Type Questions Multiple Choice Questions 1.

(c)

2.

(a)

3.

(a)

4.

(a)

5.

(a)

6.

(d)

7.

(d)

8.

(b)

9.

(d)

10.

(c)

11.

(a)

12.

(a)

13.

(c)

14.

(a)

15.

(c)

Fill in the Blank 1.

M.B.

2.

identical

3.

Gases

4.

average

5.

ny

6.

constant

7.

log n ! = n log n − n

8.

ni = λ e − β E i

9.

β = 1/kT

gi α + βE i

11.

phase

12.

1.59 kT /m

10.

ni =

(e

)

13.

1.73 kT /m

14.

1.44 kT / m

15.

v rms > v > v p

16.

Zartman and Ko

17.

3R

18.

Frequency distribution of photon

19.

1/2

20.

rms

21.

h3

22.

ellipse

23.

ni = e− Ei / kT

24.

h3

25.

0.81

26.

double

27.

parabolically

28.

rms

29.

0.92

30.

8 kT m

True/False 1.

True

2.

True

3.

False

4.

False

6.

True

7.

True

8.

True

9.

False

10. True

5.

False

11. True

12. False

13. True

14. True

15. True

16. False

17. False

18. False

19. True

20. True

S-94

H ints and Solutions k = 1.38 × 10−23 J/K

Numerical Questions 1.

T = 200 K

We know that probability of a particle to have energy E 1  P(E) = 2 π   πkT 

i.e.

3/ 2

Pv = 4 × 3.14 32    6 × 1026 × 3 ⋅ 14 × 1 ⋅ 38 × 1023 × 200 

1 / 2 − E / kT

E

e

−

Energy required to lift up to 1 Å −3

×e

−10

E = mgh = 1 × 10

× 9.8 × 10 −14

≈ 9.8 × 10

or

J

6.

Now, energy

7.

= 8.28 × 10−21 J

e − E / kT ≈ e − 1.18 × 10

≈ Very small or i.e. m zero Hence, the probability of that particle is very small. vrms =

3RT = M

3 × 8.3 × 300

3.

vrms =

3RT = M

3 × 8.3 × 300

=

221 × 10−3

= 183.3 m/sec

8.

vp =

9. 10. 11.

233437.5

12.

See example-6 See example-7 See example See example-10 Ans. v = 3.28 m/sec, vrms = 9.68 m/sec,

−23

1.38 × 10

× 400 −27

28 × 1.66 × 10

3/ 2

v 2 e − mv

32 6 × 1026

vp = 4.0 m/sec 13.

See example-10 Ans. v = 3.24 m/sec = 3.49 m/sec and vp = 3 m/sec

We have probability formula between velocity range v and v + dv

m = 32 amu =

See example-6

Ans. 0.0451 e −1

2kT kT = 1.44 m m

m  Pv = 4 π    2 πkT 

2kT log e 10 m

Ans. Zero

32 × 10

= 496.12 m/sec 5.

2kT log e 5 m

Ans. p(v) = 6.13 × 10− 4

−3

= 1.44

See example-3

Ans. P(v) = 6.3 × 10−4

= 483.15 m/sec 4.

See example−2

(ii) vx =

7

2.

×1

Pv = 6.11 × 10−4

Ans. (i) vx =

9.8 × 10−14 E = 1.18 × 10 7 = kT 8.28 × 10−21

or

32 × ( 100) 2 6 × 10 26 × 2 × 1 ⋅38 × 10 −26 × 200

(100)2

Ans. vp = 1503 ⋅ 48 m/see, 1841.603

E = kT = 1.38 × 10−23 × 600

So,

3/ 2

2

/ 2 kT

dv

14.

Ans. v = 3.3 m/sec, vrms = 3.6 m/sec and vp = 3.0 m/sec 15.

dv ~ − 100.5 − 99.5 = 1 m/sec

See example-10 Ans. v = 6.17 m/sec, vrms = 3.36 m/sec

kg

v = 100 m/sec

See example-10

vb = 3.0 m/sec 16.

See example-2 Ans. 1841.603 m/sec

S-95 17.

8kT m

9.

See article-5

10.

See article-6.1

11.

See article-6.2

12.

See article-6.3

13.

See article-6

14.

See example-1

15.

See example-1

16.

See article-7.1

17.

See article-7.2

18.

See article-7.1

19.

See example-3

See example-11

20.

See example-4

Ans. − 250.6 °C 3 = × 3.14 × (5.28 × 10−26 ) × (4.25 × 10 2 )2 16

21.

See article-8

22.

See article-8

23.

See example-5

Avg. speed v =

3kT m

rms speed vrms = thus

vrms 2 vrms

…(1) …(2)

3π = v 8 3π = + v2 8 1 2 m vrms 2 1 3π 2 v = m× 2 8

K.E. of O2 molecule K.E. =

18.

= 5.61 × 10−21 Joule. 19. 20. 21. 22. 23.

24.

24.

See example-5

See example-12

25.

See example-8

Ans. − 251.57 °C

26.

See example-8

See example-14

27.

See example-13

Ans. 447 m/sec

28.

See example-17

See example-16 Ans. 0.396 m/sec

Very Short Answer Type Questions

See example-18

1.

Ans. 7.88 × 104 K

2.

Any spin

See example-19

3.

Gases

Ans. 4.6 × 102 m/sec

4.

Identical and distinguishable

5.

No restrictions on number of molecule and energy of molecules.

6. 7.

log n! = n log n − n n! Ω= n1 ! n2 ! n3 ! ...

8.

No

9.

ni = λe − βE i

10.

β=

11.

ni =

12.

m  4 πN    2 πkT 

13.

h3

14.

v = 1.59

vp = vp =

2 vrms 3 2 × 4.6 × 102 3

= 3.77 × 102 m/sec

Short Answer Type Questions 1.

See article-1

2.

See article-1

3.

See article-1

4.

See article-2

5.

See article-2

6.

See article-3

7.

See article-3

8.

See article-5

Maxwell-Boltzmann’s statistics

1 kT gi

e( α +

βE i )

kT m

3/ 2

v2 e

−

1 mv2 2 kT dv

S-96 kT m

15.

vrms = 1.73

16.

vp = 1.44

17.

vrms > v > vp

18.

(i) Zartman and Ko experiment (ii) Width of spectral line

kT m

19.

Fizeau modes

20.

In this statistics we assume that free electrons and contribute to specific heat. 1 2

21.

mmm

S-97

Unit

4

Quantum Statistics (BES & FDS)

1. Equilibrium between Two Systems in Thermal Contact (Microscopic View) If two systems are capable of exchanging energy of thermal agitation between one another then they both are called in thermal contact. Let two systems X and Y with specified volumes VX and VY and having number of particles n x and n y . The total energy of system is E. Now we will find that how energy E will be distributed in both systems. We will use maximum probability condition for this. Let energy E x of system X which is the part of total energy E. Thus, by using energy conservation principle E = EX + EY X nX VX

Conducting wall

nY VY

Y Fig. 1

or

EY = E − EX

…(1)

Now probability of energy E X P(E X ) ∝ Ω (E X ) or

P(E X ) = kΩ (E X )

…(2)

where Ω (E X ) are the number of microstates in energy range E and E + δE and k is any constant. Similarly probability of energy E of system Y P(E Y ) = k ′ Ω (E Y ) where k ′ is any constant.

…(3)

S-98 Now total energy of system P(E X , E Y ) = kk ′ Ω (E X ) Ω ′ (E Y )

…(4)

when both system X and Y comes in contact. Hence they can exchange their energy and comes into equilibrium state soon. So from equation (4) Taking log of both sides of equation (4) We get, log e P(E X , E Y ) = log e kk ′ + log e Ω (E X ) + log e Ω ′ (E Y )

…(5)

Differentiating equation (5) w.r.t. E X and taking for maximum ∂ log e P [E X , E Y ] = 0 ∂E X We get,

∂ ∂ log e Ω (E X ) + log e Ω ′ (E Y ) ∂E X ∂E X

0= =

∂E ∂ ∂ log e Ω (E x ) + log Ω ′ (E y ) Y ∂E X ∂E Y ∂E X

…(6)

From equation (1), we get, ∂ ∂ (E − E X ) EY = ∂E X ∂E X ∂E Y = −1 ∂E X

…(7)

∂ ∂ log e Ω (E X ) = log e Ω ′ (E Y ) ∂E X ∂E Y

…(8)

⇒ Using this value into equation (6) We get,

Hence, if two systems are in thermal contact then they will be in equilibrium if function

∂ log e Ω is constant. ∂E

2. Bridge with Macroscopic Physics Let

β parameter = β=

Similarly,

β′ =

∂ log e Ω ∂E X 1 ∂Ω Ω ∂E X

…(1)

1 ∂Ω ′ ∂ log e Ω ′ = Ω ′ ∂E Y ∂E Y

…(2)

β and β′ has the dimensions reciprocal to energy. But from equation (8) from previous article we get β = β ′. Thus, according to temperature two systems are in thermal contact if both has same temperature i. e. T = T ′.

S-99 Hence, macroscopic quantity temperature is a constant and has maximum value. It the temperature of two bodies are same then it means β parameters are also same for both particles 1 1 or β′ = i. e. β= kT kT ′

3. Indistinguishability of Particles and its Consequences In classical statistics (or M.B statistics), the particle are distinguishable and there is no restriction on the

number of particle in any one cell. The size or volume of the cell can be made as small as we want. Hence, the number of phase cell (i.e. g where i denotes i th compartment) in phase space can be made as large as possible. i

Each particle has equal priori-probability of going into any cell. If n i and g i are number of particles and number of elementary phase cell, respectively in i th compartment, then, g i > > n i (according to classical statistics)

…(1)

and occupation index f (M.B) =

ni 1 so we can neglect 1 in comparison to (n i − g i). For the most probable or the equilibrium state Ω B.E , the thermodynamic probablity and hence, ln Ω B.E is maximum and therefore, δ (ln Ω B.E ) = 0 k

∑ [ln (n i + g i) − ln n i] δn i = 0

i. e.

…5(a)

i=1

Apart from this, the system must obey the following two constraints : (i)

Due to conservation of the total number of particles i.e. N = a constant.

∑ n i = constant

N=

∵

…5(b)

i

So (ii)

δN =

∑ δn i = 0 i

Due to conservation of energy of the system i. e. E = a constant ∴ So

E=

∑ n i E i = constant

δ∑ ni Ei = i

…(6)

i

∑ ε i δn i = 0 i

Now using Lagranging undetermined multipliers technique we multiply equation 5(b) by − α, equation (6) by − β and adding to equation we get,

S-103

∑ [ln (n i + g i) − ln n i − α − β E ] δn i = 0

…(7)

i

i

As the variations δn i are independent of each other, the sum given in above equation will be zero only if the expression within the bracket vanishes i.e. ln (n i + g i) − ln n i − α −

εi =0 kT

1  ‡ β = kT 

n + g i ε ln  i  =α + i kT  ni  or

1+

or

gi = e α ⋅ e ε i / kT ni ni =

gi α ε i / kT e e

…(8)

−1

which is known as Bose − Einstein’s distribution law. The occupation index f (M.B) of B.E statistics can  n  1 be expressed f (M.B) = i  α ε / kT . gi  e ⋅ e i − 1 6.

Fermi-Dirac (F.D) distribution : Basic characteristics or assumptions of F.D statistics are : (i) The particles are identical and indistinguishable. (ii) The particles follow Pauli’s exclusion principle i. e. there can be either no particle or only one particle in a given cell. Hence, the number of cells must be larger than the number of particle (g i > > n i). (iii) The size of or volume of elementary phase cell cannot be less than h 3. (iv) The particles which obey F.D statistics are known as fermions. They possess half integral spin. Consider a system having N indistinguishable particles which obey F.D statistics. To determine energy distribution of these particles in the most probable state. We divide available volume in phase space into large number of compartments or energy levels (say k). Each compartment represents a small interval of energy or momentum. Due to larger size as compared to elementary phase cell. each compartment contains a very large number of elementary phase cells (N compartment has g 1 cells, 2 nd compartment has g 2 cells ...i th compartment has g 1 cells and so on). Let n1, n 2, ... n i ... n k be the particles having mean energy E1, E 2 ... E i ... E k , respectively in compartments numbered 1, 2, ... i, k, then, N = n1 + n 2 + n 3 + ... n i + ... n k =

k

∑ ni

i=1

= Total number of particles of given system Consider the i

th

compartment in which n i identical and indistinguishable particles distributed among its

g i phase cells. According to Pauli’s exclusion particle, if we place n i particle among g i phase cells, the first particle can be placed in any one of the available g i cells. Thus, the first particle can be distributed in g i different ways and remaining (g i − 1) cells will remain vacant. The second particle can be placed in (g i − 1) different ways and so on. The total number of arrangements of arranging n i particles among available g i cells or states with energy level E i is given by

S-104 = g i (g i − 1) ... [g i − (n i − 1)] gi ! = (g i − n i)! Due to indistinguishable nature of particles n i ! way of arranging n i particles among themselves are meaningless. Therefore, the total number of distinguishable and different arrangement or ways is gi ! …(1) (Ω i) = (g i − n i)! n i ! If now, we consider all the possible compartments or energy level, the total number of different and distinguish arrangements or ways for distribution of N particles of the system into k compartments gives thermodynamic probability. Hence, gi ! k …(2) Ω (n i, n 2, ... n k ) = Π (g i − n i)! n i ! i=1

k stands for the product of all terms with various values of i from 1 to k. Taking natural where Π i=1

logarithms on both sides, we get, k

ln Ω =

∑ [ln g i ! − ln (g i − 1)! − ln n i !]

ln Ω =

∑ [g i ln g i − g i] − (g i − n i) ln (g i − n i) + (g i − n i) n i ln n i] n i]

Using Stirling formula, we get, k

i=1

=

k

∑ [g i ln g i − (g i − n i) ln (g i − n i) − n i ln n i]

i=1

To find the state of maximum thermodynamic probability or equilibrium state. We differentiate both sides keeping in mind that g i is a constant and not a variable like n i δ [ln Ω] =



∑  − (g i − n i) (g i

=



1 1 (− δn i) + δn i ln (g i − n i) − n i δn i − δn i ln n i] ni i − n i)

∑ [ln (g i − n i) − ln n i] δn i

…(3)

i

For a system to be in equilibrium, the thermodynamic probability Ω F.D (and subsequently ln Ω F.D) is maximum. So that δ (ln Ω) = 0 i. e.

∑ [ln (g i − n i) − ln n i] δn i = 0

…(4)

i

We also know that the system must satisfy two auxiliary conditions. Conservation of the total number of particles i. e. N = constant δN =

k

∑ δn i = 0

i=1

Conservation of total energy of the system i.e. E = constant

…(5)

S-105 δE =

k

∑ ε i δn i = 0

…(6)

i+1

Now we will apply Lagrangian undetermined multipliers technique and multiply equation (5) by − α and equation (6) by − β and adding to equation (3), we get, k

∑ [ln (g i − n i) − ln n i − α − β εi] δn i = 0

…(7)

i=1

Since the various terms for different values of i are independent of each other or (δn i ≠ 0), the quantity within the bracket in above equation must vanish for all values of i,  g − ni ln  i  = α + β εi  ni 

i. e.

g i − ni = eα + β ε i ni gi = eα + β ε i + 1 n! ni 1 = α +β ε i + 1 gi e

or

or

ni =

Substituting the value of β = We get,

gi α + β εi e

+1

1 , where k is Boltzmann’s constant. kT gi n i = α ε / kT e e i +1

…8(a)

This above equation is referred as Fermi-Dirac distribution law. The occupation index f (F.D) of F.D statistics can be expressed as n 1 …8(b) f (F.D) = i = ( α + ε / kT) i gi e +1

4. Comparison of M.B, B.E and F.D Statistics The expressions of the occupation index for M.B, B.E and F.D are as follows : n 1 f (M.B) = i = α ε / kT gi e e i

…9(a)

f (B.E.) =

ni 1 = g i e α e ε i / kT − 1

…9(b)

f (F.D) =

ni 1 = α ε / kT i gi e +1

…9(c)

When g i > > n i irrespective of any value of temperature, the three satistics coincide. When g i > > n i for B.E statistics. We get,

S-106 ε  gi  = exp. α + i  ni kT   where we have neglected − 1 in comparison to e

εi    α +  kT 

…(10)

because g i > > n i

Similarly for F.D statistics 

α + gi  =e ni

εi   kT 

…(11)

where we have neglected + 1 It is clear from equations 9(a), 10 and 11, B.E and F.D statistics reduce to M.B statistics when number of phase cell is large in comparison to number of particles. Hence, for small value of occupation index, the two quantum statistics (B.E and F.D) yield M.B statistics or classical statistics. Thus, classical statistical mechanics is a special case of quantum statistical mechanics.

4.1 Fermi-Dirac Gas Fermi-Dirac statistics follows these assumptions : 1.

Molecules are identical

2.

Independent to each other

3.

Non-interacting molecules

4.

Spin of molecules is half-integral

5.

Molecules follows Pauli’s principle

6.

Molecules are named as Fermions example electron, proton and neutrons etc.

We know that distribution law for the molecules of obeying Fermi–Dirac statistics is g n i = α + βEi i + 1 e

…(1)

Let A = e α So

ni =

gi βE i

Ae

+1

Here factor + 1 indicates that there is no restriction on α that is should be positive only. Total number of phase cells between momentum range p and p + dp Momentum range between p and p + dp are g ( p) dp = g S Volume of one phase cells g ( p)dp = g S But

So

∫ ∫ ∫ ∫ ∫ ∫ dx dy dz dpx dpy dpz h3

∫ ∫ ∫ dx dy dz = V 4 4 ∫ ∫ ∫ dpx dpy dpz = π ( p + dp)3 − πp3 = 4 πp2dp 3 3 g ( p)dp = g S

4 πVp2dp h3

…(2)

S-107 Where g S = 2S + 1 is known as spin degeneracy factor which is arising because of spin S of the Fermions. Now we have relation between energy and momentum p2 = 2 mE or

2 p dp = 2 mdE

or

dp =

1 2m 2 E

Thus, number of states in the energy range between E and E + dE 4 πV

g (E) dE = g S

h3

m dE 2E

, 2mE

…(3)

where g (E) is density of states function. Using equation (1) and equation (3) we get number of molecules within range E and E + dE n(E) dE = dn(E) = g S

4 πmV h

2m

3

E1 / 2 Ae

E / kT

+1

dE

…(4)

1 kT E x= kT dE dx = kT

where,

β=

Let variable

Then using equation (4) dn = g S dn = g S

4 πmV h3

2m

x 1/ 2dx (kT )3/ 2 Ae x + 1

2  2πm kT    π  h2 

3/ 2

V⋅

x 1/ 2 Ae x + 1

dx

As we discuss earlier partition function is  2πm kT  Z=   h2  So

dn =

3/ 2

V

2g S Z x 1/ 2dx π Ae x + 1

Thus number of particles n= n=

∫ dn = 2g S Z π

∞

∫

2g S Z π

0

∞

∫

0

x 1/ 2 Ae x + 1

Now total energy of system E=

=

∫ E n(E) dE

x 1/ 2dx Ae x + 1 dx

…(5)

S-108

∫ E dn = kT ∫ x dx =

E = kT

∞

2g S Z

∫

π

0

x 3/ 2dx

…(6)

Ae x + 1

Degeneracy : We can evaluate the integrals (5) and (6) for two cases when α < 0 or when α > 0. It α > 0 then

case is weak degeneracy while if α < 0 then case is strong degeneracy. Case I : For α is positive i.e. A is greater than 1. Using equation (5) ∞

∫

0

x 1/ 2dx Ae x + 1

∞

=

∫

0

∞

=

x 1/ 2dx = 1   Ae x 1 + Ae x  

∫x

1/ 2

dx

0

e− x A

∞

1/ 2 ∫ x dx 0

=

1 1 x 1/ 2e − x dx − 2 ∫ A0 A π 2A

 e− x  1 +   A  

−1

  e − x e −2 x + − ... 1 − 2 A A  

∞

=

e− x A

∞

∫x

1 / 2 −2 x

e

dx +

0

1 A

3

∞

∫0

x 1/ 2e −3x dx ...

1 1   1 − 23/ 2 A + 33/ 2 A 2 − ...

So, total number of particles using equation (5) g Z 1 1  n = S 1 − 3/ 2 + 3/ 2 2 − ... A  2 A 3 A 

…(7)

Now using equation (6) We have integral ∞

∫

0

x 3/ 2dx x

Ae + 1

∞

= =

1 1 x 3/ 2e − x dx − 2 A ∫0 A 3 π 4A

∞

∫x

3 / 2 −2 x

e

dx + ...

0

1 1   1 − 25/ 2 A + 35/ 2 A 2 − ...

Hence, total energy E=

3g S Z 1 1   kT 1 − 5/ 2 + 5/ 2 2 − ... 4A 2 A 3 A  

Using the value of g S Z from equation (7) into this equation, we get,

Putting

E=

3 n kT 2

1 1   1 − 25/ 2 A + 35/ 2 A 2 − ...

=

3 n kT 2

1 1   1 + 25/ 2 A − 35/ 2 A 2 + ...

A=

gS Z n

1 1   1 − 23/ 2 A + 33/ 2 A 2 

−1

S-109

So

E=

 3 1 n kT 1 + 5/ 2 2 2  

2   n    + ...  g SZ   

 n  1   − 5/ 2 g Z  S  3

Now pressure of gas 2   1  n  1  n   dE  nkT  1 + 5/ 2  P=−  = −    + ... 5 / 2  dV  V  2  g SZ  3  g SZ    

or

P=

RT  1 1 + 5/ 2 V  2 

 n  1   − 5/ 2 g Z  S  3

2   n    + ...  g SZ   

Case II : When α is negative e − α > > 1 so here A is much less than one. Thus the value of

1 will increase A

n 1 = A gS Z or

n h3 = g S (2πm kT )3/ 2 V

…(8)

at absolute zero i. e. at T = 0 K, A = 0. So using equation (5), we get n= If A = 0, then we can replace upper limit by So, we get

π

∞

∫x

1/ 2

dx

0

1 A n= n=

or we get

2g S Z

1 = A

2g S Z π 2g S Z π

1/ A

∫x

1/ 2

dx

0

×

1/ A

2 3A

 3nπ1/ 2     4 g SZ 

3/ 2

∫

x 1/ 2dx =

0

2 3 A 3/ 2

2/ 3

But partition function is given as  2π mkT  Z=   h2  So

3/ 2

V

h 2  3n  1 =   A 2mkT  4πVg S 

This is the degeneracy of Ideal Fermi gas at T = 0 K. Now we will evaluate zero point energy E 0. Now putting E = E 0 and A = 0 into equation (6), we get

2/ 3

…(9)

S-110 E 0 = kT Putting

2g S Z π

∞

∫x

3/ 2

dx

0

1 for upper limits A E 0 = kT E 0 = kT

Putting the value of

2g S Z π 2g S Z π

1/ A

∫x

3/ 2

dx

0

×

2 5 A 5/ 2

1 from equation (9) we get, A 2/ 3 2g S Z 2  h 2  3n   ⋅  E 0 = kT    π 5  2m kT  4πVg S    

2g S  2π mkT  = kT   π  h2  E0 =

3 nh 2  3 n    10 m  4 πg S V 

3/ 2

5/ 2

2  h2 V×  3  2mkT 

 3n     4 π Vg S 

2/ 3

Similarly, zero point pressure P0 =

2 E0 3 V

n h 2  3n  P0 =   V 5 m  4π g S V 

2/ 3

5. Bose-Einstein Gas We have some postulates for Bose–Einstein statistics : 1.

All the molecules are identical

2.

− Molecules has spin integral mulitiple of h

3.

Phase space volume is h 3

4.

Number of distinguishable ways are π

5.

Distribution law is

6.

At high temperature this distribution approaches to M.B statistics.

7.

There is no restriction on the number of molecules in a cell.

i

(e

gi α + β ∈i

(n i + g i − 1)! n i ! (g i − 1i)!

− 1)

2/ 3

  

5/ 2

S-111 As the distribution function is ni =

(e

gi α + β ∈i

− 1)

An assemble of identical indistinguishable particles with spin zero or integral spin is known as Bose–Einstein gas or an assembly of bosons is known as Bose–Einstein gas. Let us consider a Bose–Einstein gas consisting of N boson and these bosons are distributed in different energy states. Such that n number of bosons are in the energy state E i (i th energy state) and its degeneracy is g i. n i = no. of particles in the i th state ni =

gi

…(1)

e α + β ∈i − 1

α= −

µ kT

A = e− α ni =

…(2)

gi 1 β ∈i e −1 A

As the number of particles in a given state cannot be negative i.e. n i ≥ 0 1 β ∈i So e ≥1 A

…(3)

A = e− α e α = e − µ / kT =

1 A

…(4)

If E i = 0 i. e. the energy of the lowest energy state (ground state) is taken to be zero. 1 Then from equation (3) ≥ 1 A So ‘A’ should lie between 0 and 1 …(5)

0≤ A ≤ 1 The constant α can be determined by the condition n= =

gi ∑ ni = eα ∑ + β∈ −1 i

∑ gi 1 β ∈i e −1 A

…(6)

It is known that for the particles in a box of normal size, the translation levels are closely spaced. Thus we can replaces summation by integration. The number of quantum states lying in the momentum range P and P + dp. g ( p) dp = g s where g s is the spin degeneracy.

V ⋅ απp2dp h3

…(7)

S-112 Let g s = 1 for S = 0 (‡ g s = 2S + 1)

V ⋅ 4 πp2dp

g ( p) dp =

…(8)

h3

According to the most probable distribution the number of particles in the momentum range P and P + dp is given by g (P) dp …(9) d n(P) = α + βP 2 / 2 m −1 e P2 m

ε=

…(10)

The number of particles in the energy range ε and ε + dε then 4 π mV 2m ε1/ 2dε

dn(ε) = Now let us substitute

…(11)

h 3 e α + ε / kT − 1

ε kT dε dx = kT

…(12)

X=

dn(ε) =

…(13) x 1/ 2

V

2

h

π εα × x − 1

(2πm kT )3/ 2 3

…(14)

∫ dn (ε)

n=

V

n=

h3

(2πm kT )3/ 2 F1(α) ∞

2

F1(α) =

…(15)

x 1/ 2

∫ e α + x − 1 dx

π

…(16)

0

The total energy of the system or Bose-Einstein gas

∫ ε dn(ε) = ∫ kTx dn(ε)

E=

=

V

2

h

π

(2πm kT )3/ 2kT 3

∞

x 3/ 2dx

∫ eα + x − 1 0

Using equations (13) and (14) = F2(α) =

3 V (2πm kT )3/ 2kT f 2(α) 2 h3 ∞

4 3 π

x 3/ 2dx

∫ eα + x − 1

…(17) …(18)

0

For A < 1 F1(α) and F2(α) can be written F1(α) =

∞

Ar

∑ r 3/ 2

r =1

…(19)

S-113 F2(α) =

∞

Ar

∑ r 5/ 2

…(20)

r =1

Therefore, the expression for n and E is given by ∞

V

n=

h

E=

(2πm kT )3/ 2 ∑ 3

Ar

r = 1r

…(21)

3/ 2

3 V (2πm kT )3/ 2kT 2 h3

Ar

∑ r 5/ 2

   A2 A3 A2 A3 E 3 = kT  A + 5/ 2 + 5/ 2 + ...  A + 5/ 2 + 3/ 2 + ... n 2 2 3 2 3    E=

…(22) −1

3 A A   nkT 1 − 3/ 2 + 5/ 2 + ...   2 2 3

…(23)

h h3 V (2πm kT )3/ 2

…(24)

From equation (21) f1(α) = Thus, f1(α) is proportional to

n − the particle density and it is inversely proportional to temperature and low V

density i.e. the classical limit. For A < < 1 : Degeneracy

A = e− α =

h h3 V (2πm kT )3/ 2

…(25)

A= 0 ∫ εP(ε) dε N 0

 1  = 2π   πkT 

3/ 2 ∞

 1  < ε > = 2π    πkT 

3/ 2 ∞

 1  = 4π   πkT 

3/ 2 ∞

∫ε

3/ 2 − ε / kT

e

dε

0

Substituting ε1/ 2 = x we get

∞

∫x

4 − ax 2

e

dx =

0

∫x

e

x dx

0

∫x

4 − x 2 / kT

e

⋅ dx

0

3 π 8 a5

=

3 π 8 (1 + kT )5

=

3 x ⋅ (kT )5/ 2 8

 1  < ε > = 4x    πkT  < ε>=

3 − x 2 / kT

3 kT 2

3/ 2

⋅

3 π (kT )3/ 2 8

S-127 Example 10. If electron density in a crystal is 0.5 electron/Å and h = 6.62 × 10 −34 J-S, me = 9.1 × 10 −31 kg then calculate Fermi energy. Solution : We have Fermi energy per unit length

or

EF =

h2  2 N  3π  2m  V

EF =

h2  N    2m  4 L

2/ 3

2

Planck’s constant h = 6.62 × 10−34 J-sec me = 9.1 × 10−31 kg 5 electrons N 0.5 = = m L A ° 10−10 electrons = 5 × 109 m EF =

(6.62 × 10−34 )2 2 × 9.1 × 10−31

 5 × 109  ×   4 

2

= 3.76 × 10−19 Joule =

3.76 × 10−19 1.6 × 10−19

eV

= 2.35 eV Example 11. Fermi energy for lithium is 4.70 eV and electron density is 4.6 × 10 28 /m 3 . Calculate the electron density for Fermi energy 2.35 eV. Solution : We have Fermi-energy EF =

h 2  3π 2 N    2m  V 

 N EF ∝    V EF

1

EF Here, E F = 4.70 eV, E F 1

2

2

2/ 3

2/ 3

 N / V1  =   N / V2 

2/ 3

N = 2.35 eV and = 4.6 × 1028 / m 3. V

Putting the value   4.70  4.6 × 1028   = 2.35   N       V  2 

2/ 3

S-128  2.35 = 4.6 × 1028 ×    4.70

3/ 2

= 4.6 × 1028 × 0.35 = 1.6 × 1028 / m 3 Example 12. Calculate the Fermi energy for the sodium. If sodium has 2.5 × 10 28 electron/m 2 . Solution :

Given that

2/ 3

EF =

h 2  3π 2 N    2m  V 

EF =

h2  2 N 3π V  8π 2m 

EF =

h2  3 N ⋅ 8m  π V 

2/ 3

2/ 3

N = 2.5 × 1028 / m 3 V

Planck’s constant h = 6.62 10−34 Js Mass of electron m = 9.1 × 10−31 kg  6.62 × 10−34   3 × 25 × 1028  EF =   × 3.14   8 × 9.1 × 10−31  

2/ 3

= 0.6 × 1037 × 8.29 × 1018 =

4.97 × 10−19 1.6 × 10−19

= 3.1 eV

Example 13. Classify the following particles according to statistics e.g. M.B, B.E and F.D : (i) electrons, (ii) mesons, (iii) photons, (iv) neutrons, (v) holes, (vi) phonons, (vii) protons, (viii)He4 atom, (ix) H 2 molecule, (x) positron, (xi) hydrogen atom, (xii) α-particle, (xiii) O 2 molecule, (xiv) gas molecule. Solution : Particles

Statistics

(i)

Electron

F–D

(ii)

Mesons

B–E

(iii)

Photons

B–E

(iv)

Neutrons

F–D

(v)

Holes

F–D

(vi)

Phonons

B–E

S-129 (vii)

Protons

F–D

(viii)

He4 atom

B–E

(ix)

H 2 molecule

B–E

(x)

Positron

F–D

(xi)

H 2 atom

F–D

(xii)

α-particle

B–E

(xiii)

O2 molecule

M–B

(xiv)

Gas molecules

MB

Example 14. Show that the average energy of electron at 0K is →

E=

3 E where E F is the Fermi energy. 5 F

Solution : The number of electrons is an electron gas having energies between E and E + dE is given by n(ε) dε =

8π mV h3

(2m)

ε1/ 2dε e( ε − ε F ) / kT + 1

…(1)

where m is mass of electrons, the Fermi energy ε F is EF = V

∴

h

3

=

h2  3N   8 m  πV 

2/ 3

3N 8π m ⋅ 2 (2m)

[E F ]3/ 2

Substituting this in equation (1), we get. n(ε) dε =

3 ε1/ 2dε N { ε F (0)} −3/ 2 ( ε − ε F ) / kT 0 2 e +1

At T = 0K all of the electrons have energies less than or equal to ε F (i.e. ε ≤ ε F ) So that at T = 0 K we have e( ε − ε F ) / kT = 0 ∴ At absolute zero i.e. at T = 0 K Number of electrons, n(ε) nε =

3N [ε F (0)]−3/ 2 ε1/ 2dε 2

…(2)

Now, total energy at absolute zero EF

E0 =

∫ ε n(ε) dE 0

=

3N [ε ]−3/ 2 2 F

ε F ( 0)

∫ε 0

3/ 2

dε =

3 Nε F 5

S-130 ∴ Average energy E at absolute zero is, E=

E0 3 = εF N 5

Example 15. If sodium atom has one electron in atom and density of sodium is 0.97 g/cm 3 then calculate the Fermi energy of sodium. Atomic weight of Na is 23. Solution : We know that

Fermi energy is as εF =

h 2  3n    8m  πv 

If sodium has one electron per atom the electron density

2/ 3

n is given by v

n Nρ = v W where Avogadro number, N = 6 × 1026 atoms/kg ⋅ mole Density of sodium, ρ = 0.97 g/cm 3 = 0.97 × 103 kg/m 3 Atomic wight of sodium W = 23 ∴

Electron density,

n (6 × 1026 atoms / kg ⋅ mole) (0.97 × 103 kg/ m 3) = 23 v = 2.53 × 1028 electrons/m 3

Planck’s constant, h = 6.62 × 10−34 J-sec Mass of the electron, m = 9.1 × 10−31 kg Thus, Fermi energy EF =

h2  3 n  ⋅  8m  π v 

2/ 3

2/ 3 (6.62 × 10−34 joule − sec)2 3 3 28 ( 2.53 electron / m ) 10 = × ×  3.14 8 × (9 × 10−31 kg) 

= 5.032 × 10−19 joule =

5.032 × 10−19 1.6 × 10−19

eV

= 3.145 eV Example 16. Fermi energy of an atom is 5.52 eV. Calculate (i) average energy of free electrons at 0K, (ii) calculate the temperature at which a classical particle will have this energy. Given Boltzmann’s constant k = 1.38 × 10 −23 J/K.

S-131 Solution : (i) We know that average energy electrons is 3 E = EF 5 Given that,

E F = 5.52 eV 3 E = × 5.52 5 E = 3.312 eV

(ii)

We know that kinetic energy to classical particle at temp. T is 3 E = kT 2 As per question

E = 3 .312 eV

i.e. or

E = 3.312 × 16 × 10−19 Joule 3.312 × 1.6 × 10−19 = T= T=

3 × 1.38 × 10−23 × T 2 3.312 × 1.6 × 10−19 × 2 1.38 × 10−23 × 3 10.5984 × 10−19 4.14 × 10−23

T = 2.56 × 104 K Example 17. The Fermi energy of a metal is 10 eV. What is the corresponding classical energy. Solution : At temperature T the amount of energy a particle has is kT i. e.,

E = kT EF = EF =

Given,

1.38 × 10−23 1.6 × 10−19

T

T eV 11594

E F = 10 eV

So, temperature T = 11594 × 10 T = 115940 K Example 18. Calculate the Fermi energy of copper on the assumption that such atom of copper contributes one free electron to the electron gas, density of copper = 8 .94 × 10 3 kg/m 3 . Solution : First we calculate the number of electrons per unit volume kg mass 8.94 × 103 3 3 atoms atoms N m = = m = = 8 .48 × 1028 −27 3 mass V m 63.5 × 1.66 × 10 kg m3 atom

S-132

Now, Fermi energy

EF = =

h2  3N   8m  πV 

2/ 3

(6.63 × 10−34 )2 8 × 91 . × 10−31

 3 × 8.48 × 1028  ×  . 314  

2/ 3

= 7.06 eV Example 19. Assuming atomic nucleus as an ideal gas (Fermion). Show that the Fermi energy of gas is 28 MeV. Solution : Let an atom Where

Z

XA

A → Atomic weight Z → Atomic number

N = ( A − Z ) → Number of neutrons Concentration of neutrons =

A−Z A−Z = 4 V πR 3 3

Where R → nucleus rating R = R0 A1/ 3

But,

R0 = 1.2 × 10−15 m

Where

R = (1 .1 × 10−15) A 1/ 3m

So, So,

Concentration of neutrons = ≈

3 4 π × (1 .2 × 10−15)3

 A − Z ×   A 

3 × 0 .5 4 π (1 .2 × 10−15)3

≈ 0 .5432 × 1044 Here

A−Z = 0.5 for light nucleus A

Now Fermi energy h 2  3π 2 N  EF =   2m  V  = or

2/ 3

(1.05 × 10−34 )2 2 × 1.6 × 10−27

× 3π 2 × 0.5432 × 1044

E F = 4 .5 × 10−12 J E F = 28 MeV

Example 20. Three particles are to be distributed in four energy states a, b, c and d. Write down all the possible ways for such a distribution of the particles are (a) Fermions (b) Bosons (c) Classical particles.

S-133 Solution :

(i)

Fermions : The fermions or the Fermi–Dirac particles are identical and indistinguishable and half

integral spin particles. The number of possible ways of distribution of three particles in four energy states a, b, c and d being gi ! 4! = =4 n i ! (g i − n i)! 3 ! (4 − 3)! Their four ways are given below : a

b

c

d

a

b

c

d

a

b

c

d

a

b

c

d

1.

2.

3.

4.

(ii)

Bosons : The bosons are identical and indistinguishable and integral spin particles. Each energy state

can contain any number of particles. The number of possible ways in which three bosons may be distributed in four possible states a, b, c and d being : g i(n i + g i − 1)! 4 (4 + 3 − 1)! = 3! 4 ! ni ! g i ! =

4 × (6)! 4 × 5 × 6 = 20 = 3! 4 ! 2× 3

These 20 ways are given below a 1. 2. 3. 4.

b

c

d

S-134

5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

Thus, there are twenty possible ways in which there bosons can be distributed in four energy states. (iii)

Classical particles : The classical particles are identical but distinguishable. Each energy state can

contain any number of particles. Let the particles be denoted by p, q, r which are to be distributed in four energy state a, b, c and d. The number of possible ways being g ini = (4)3 = 64. These 64 ways are given below.

S-135

1.

pqr pqr

2.

pqr

3.

pqr

4. 5.

pq

6.

pq

7.

pq

8.

qr

9.

qr

10.

qr

11.

pr

12.

pr

13.

pr

14.

r

r r r p p p q q q pq

15.

pq

16.

pq

17.

p

r r

qr

18.

qr

19.

qr

p p

S-136

20.

q

pr

21.

pr

22.

pr

23.

r

p

q

r

p

qr p

q

pq qr

p

37. 38.

pq r

36.

q pq

r

34. 35.

pr pr

33.

p

pr

q

31. 32.

qr qr

28.

30.

r

qr p

27.

29.

pq pq

25. 26.

q pq

r

24.

q

qr pr

S-137

q

39.

pr q

40.

pr

41.

p

q

42.

p

q

43.

p

r

44.

p

r

45.

p

q

r

46.

p

r

q

47.

q

p

48.

q

p

49.

q

r

50.

q

r

p

51.

q

r

p

52.

q

53.

r

p

54.

r

p

55.

r

q

56.

r

q

57.

r

r r q q

r r p

p

r

q q p p p

q

S-138

58.

r

q

p

59.

p

q

r

60.

p

r

q

61.

q

p

r

62.

q

r

p

63.

r

p

q

64.

r

q

p

Example 21. A system consist of 5 particles arranged in two compartments, the first compartment is divided into 6 cells and second into 8 cells. The cells are of equal size. Calculate the number of microstates in the microstate (2, 3) if the particles obey B.E. statistics. Solution : n1 = 2

gi = 6

n2 = 3

g2 = 8

According to B.E. statistics, the thermodynamic probability Ω (n1, n 2) for a microstate (n1, n 2) is given by (n − g 1 − 1) (n 2 + g 2 − 1)! Ω (n1, n 2) = 1 × n1 ! (g 1 − 1)! (n 2 !) (g 2 − 1)! =

(2 + 6 − 1)!  3 × 8 − 1 ×  2! 5!  3 !7 ! 

=

7 10 ! × = 2520 2! 5! 3! 7 !

1.

If Fermi energy of a metal is 3.14 eV then c alculate average energy of electron.

2.

At absolute zero Fermi energy of lithium is 4.70 eV and electron density is 4.6 × 1028 /m 2 . If Fermi energy of any metal is 2.35 eV. Calculate the electron density in per meter cube.

3.

How can we divide three molecules into four cells if they followed Fermi−Dirac statistics?

4.

Calculate the rms speed of nitrogen at 27°C. Given N = 6 × 1023 molecule/mole and k = 1.38 × 10−16 erg K.

5.

Calculate most probable speed for a gas whose density is 1.4 g/litre at pressure of 106 dynes/cm 2 .

6.

Calculate the probability that speed of N 2 molecules at 280 K lies between 149.5 m/sec and 150.5 m/sec.

S-139 7.

At what temperature will average speed of molecule of H 2 gas be double the average speed of O2 at 300 K?

8.

There are 3 × 1027 free electrons per cubic metre of Na. Calculate the Fermi energy.

9.

Calculate the different number of ways of arranging 6 Fermions in 10 phase cells.

10.

Compute the possible number of arrangements of two particles in three phase cells according to : (i) M.B. statistics (ii) B.E statistics, (iii) F.D statistics.

11.

Compute the possible number of arrangement of two particles in four cells. According to : (i) M.B (ii) B.E (iii) F.D statistics.

12.

Three particles are distributed into four phase cells. Calculate the distributions for then according to : (i) M.B statistics, (ii) F.D. statistics, (iii) B.E. statistics.

13.

Calculate the number of different arrangements of 8 particles among seven cells by using B.E. statistics.

14.

Fermi energy of certain metal M1 is 5 eV. A second metal M 2 has an electron density which is 6% greater than that of M 2 free. Calculate the Fermi energy of M 2 .

15.

The number of molecules in energy range E and E + dE is given by 1  n(E) dE = 2 πN    πkT 

3/ 2

E E1 / 2 × Exp  −  dE  kT 

Derive the expression for the most probable and the number of molecules having most probable energy. Also calculate mean energy. 16.

For a crystal there are 0.5 elections/Å in per unit length. Given h = 6.6 × 10−34 J-sec, mass of electron . × 10−31 kg me = 91

17.

Fermi energy of lithium is 4.70 eV and electron density is 4.6 × 10−28 / m 3 . Calculate the electron density for Fermi

18.

Calculate the Fermi energy for the sodium atom if sodium has 2.5 × 10−8 electrons in per m 3 .

19.

If sodium has 1 electron in atom and density of Na is 0.979 g/cm 3 then calculate the Fermi energy of sodium

energy is 2.35 eV.

Na = 23. 20.

Fermi energy of an atom is 5.52 eV. Calculate average energy of free electrons at 0 K.

21.

If E F = 5.52 and average energy E = 3.312 eV then calculate the temperature at which classical energy is same as average energy. Given Boltzman’s constant k = 1.38 × 10−23 J/K

22.

If E F = 10 eV for a metal then what is the classical energy of a metal?

23.

If Fermi energy of lithium is 4 ⋅ 70 eV what will be the classical energy of lithium?

24.

Calculate the Fermi energy of Cu on the assumptions that such atom of Cu contributes one electron to the electron gas, given ρ = 8.94 × 103 kg/m 3 .

25.

Show that Fermi energy of ideal gas is 28 MeV.

26.

Calculate the number of ways in which 8 particles obeying Bose−Einstein statistics are distributed into six phase cells.

S-140

Long Answer Type Questions 1.

2.

20.

Discuss the equilibrium between two systems in thermal contact. How it is related with macroscopic 21. physics? Show that for two system in an equilibrium in condition β = β ′ 22. 1 1 i.e. = kT kT ′

3.

How two systems can be in equilibrium if they are placed with connecting each other?

4.

What is classical statistics ? Discuss it in detail.

5.

What do you mean by indistinguishability of particle and its consequences?

6.

Explain the indistinguishability and discase its consequences distribution function.

Discuss the assumptions for M.B., B.E. and F.D. statistics and give the comparison between them. Derive expression for distribution functions corresponding to three statistics. Mention with examples the condition for the states of degeneracy of gas. Discuss the free electrons in metals and Fermi level. Derive an expression for Fermi energy EF =

h2  3π 2 N    2m  V 

2/ 3

23.

What is Fermi energy ? Give an article for the Fermi energy.

24.

Discuss on the following : (i)

Photons in blackbody chamber

7.

Derive an expression for the of M.B. statistics.

(ii) Planck’s constant and its implications

8.

Discuss the Bose−Einstein statistics.

(iii) Particle in one dimensional box

9.

Write the postulates of Bose−Einstein statistics and 25. hence give its distribution function. 26.

10.

11. 12.

13. 14. 15. 16. 17. 18.

19.

Write the basic assumptions of Bose−Einstein statistics and derive expression for distribution  n  1 functions fM.B. = i   g i  e α e E i / kT − 1 

Discuss the one dimensional harmonics oscillator. Give the derivation for Fermi energy and hence 3 show that the average energy is of Fermi energy. 5

Short Answer Type Questions

1. Derive the distribution function for the particle 2. obeying Fermi−Dirac statistics. Write the basic assumptions of F–D statistics and derive the expression for distribution functions 3. n 1 fBE = i E i / kT α g i (e e + 1) 4. Write the postulates of Fermi-Dirac distribution law and hence give the distribution functions. 5. What is Fermi gas? Discuss its degeneracy. Write the postulates of F.D. statistics. Discuss Fermi 6. gas. 7. What is Bose−Einstein gas ? Discuss in detail.

How can two systems comes in equilibrium? Show that two systems can come in the thermal ∂ contact if function log e Ω = constant ∂E What is the condition for Bridging for macroscopic physics? Show that if temperature of two bodies are same then β parameters for both systems becomes equal. Discuss indistinguishability of particles and its consequence. Discuss consequence of indistinguishability. How indistinguishable molecule can be divided into (i) M.B. (ii) B.E. (iii) F.D. statistics.

Write the postulates of Bose−Einstein. Discuss 8. Boson. 9. Derive an expression for number of particles and total energy for a F.D. statistics and gives its 10. degeneracy. 11.

Discuss the comparision between all three statistics.

Discuss liquid helium problem on the basis of 12. statistics.

Derive an expression for number of molecules in Fermi−Dirac gas.

Write a short note on quantum statistics. Write basic postulates of Bose−Einstein statistics. Write basic postulates of F.D. statistics.

S-141 2g s Z

∞

x1 / 2

13.

What is the distribution formula for Bose–Einstein statistics into the energy levels?

13.

Give the application of quantum statistics for liquid 14. helium.

Give the formula of distribution of molecules into energy levels according to F.D. statistics. gi what indicates ‘+1’ ? In formula ni = α βE i e e +1

i.e. n =

π

∫ Aex + 1 dx 0

14.

What are helium-I and helium-II. Discuss them.

15.

Give the theory of free electrons in metal.

16.

What do you mean Fermi level and Fermi energy ? 16. 17. Derive an expression for Fermi energy

17.

E=

h2  3π 2 N    2m  V 

2/ 3

15.

What is the value of β parameter ? What is the energy of a photon ?

18.

What is the average energy in terns of Fermi energy?

19.

What is Fermi energy ?

20.

How many types of quantum statistics?

18.

Discuss for photons in a blackbody chamber.

19.

What is Planck constant? Given its implications.

20.

Discuss the application for a particle in one dimension box.

21.

Discuss the one dimensional harmonic oscillator.

Multiple Choice Questions

22.

Classify the following particles according to :

1.

Objective Type Questions

(i) M.B. statistics (ii) F.D. statistics (iii) B.E. statistics (a) mesons (b) O2 molecules (c) photons (d) holes (e) phonons (f) He 4 atoms (g) H 2 molecule. 2. 23.

Show that the average energy of free electrons at 0 K 3 is E = E F . 5

24.

Assuming atomic numbers as an ideal gas and following Fermi−Dirac statistics. Show that the Fermi energy of gas is 28 MeV.

How many types of quantum statistics? (a) 1

(b) 2

(c) 3

(d) 4

What is the equilibrium condition for two systems? ∂ (a) log e Ω = constant ∂E ∂ (b) log e Ω 2 = constant ∂E ∂ (c) log e Ω = 0 ∂E (d) None of the above

Very Short Answer Type Questions

3.

1.

When two systems are said to be in equilibrium ?

2.

What is the condition of bridging for macroscopic physics?

3.

Which gas was liquified at last?

4.

Give the condition at which quantum mechanics can reduces to classical mechanics. 4.

5.

Give four postulates of B.E. statistics. 1 The molecules having spin h are called. 2

6.

The mean energy of Planck’s oscillator is : 1 (a) hν (b)  n +  hν  2 hν (d) None of there (c)   hν  e kT − 1     Quantum mechanics is applied to a system of particles : (a) If wave function of particles is highly overlaped

7.

Define Bosons.

(b) If particles has same masses

8.

In which statistics Paulie’s law obeyed ?

(c) If particles has same charge

9.

Give the distribution law for Bose–Einstein statistics.

(d) None of these

10.

Give the distribution law for Fermi−Dirac statistics.

11.

What is the formula for Fermi energy?

(a)

7

(b) H 2 atom

12.

What type of volume of a phase cell has ?

(c) H 2 molecule

(d) Electron

5.

Which one is a Boson among these? 3 Li

S-142 6.

7.

15.

Spin of photon is : 1− (a) h 2

(b) −h

(c) Zero

(d) 2 −h

What types of indistinguishable?

molecules

are

compartment or energy level : gi! n !(g − ni)! (b) i i (a) ni !(g i − ni)! gi

called

(a) If the distance between two molecules is greater than de Broglie’s length. 16. (b) If the distance between two molecules is lesser than de Broglie’s length. (c) Both (a) and (b)

17.

(d) None of these 8.

At Fermi level what is the probability of electrons? 1 (a) 1 (b) 2 (d) ∞

(c) 0 9.

18.

Fermi–Dirac particles has : (a) Spin integral multiple ℏ (b) Spin integral multiple of

10.

13.

14.

(g i − ni)! ni !

(d) None of these

Planck's radiation law follows : (a) F.D statistics

(b) B.E estatistics

(c) M.D statistics

(d) None of these

How many molecules can exist in a phase cell, as per F.D statistics? (a) 1

(b) 0

(c) 3

(d) 4

According to which statistics the energy of a system is not zero at absolute zero i. e. T = 0 K : (a) F.D

(b) B.C

(c) M.B

(d) None of these

The Bose–Einstein condensation is the rapid increase in the population of the ground state : (a) Bellow critical temperature

(d) None of these

(b) At critical temperature

Fermi-Dirac distribution law is : βE i

(b) ni =

g i(e α + βE i

(c) ni =

gi α + βE i

(d) ni =

12.

19.

(c)

(c) Both (a) and (b)

(a) ni = g i(e α +

11.

1 ℏ 2

The number of distinct arrangements of distributing ni particles (or Fermions) into g i phase cell of i th

(e

(d) None of these 20.

− 1)

βE i

Which particle statistics?

What are alphas particles 2 He 4 ? (a) Fermions

+ 1)

(b) Bosons

(c) Classical particle (d) None of these 21.

gi (e α +

(c) Above critical temperature

+ 1)

− 1) follows

Maxwell–Boltzmann’s

(a) Phonon

(b) O 2 molecule

(c) Electron

(d) Photon

He 4 follows : (a) F.D statistics

(b) M.B statistics

(c) B.E statistics

(d) None of these

Fill in the Blank 1.

Two systems are said to be in equilibrium if ………

The number of ways of distribution 3 Bosons in 3 2. 3. phase cells :

The gas which is liquified at last is ……… gas.

(a) 8

(b) 100

4.

(c) 300

(d) 15

According to F.D statistics there are only ……… particles in each cells.

In Maxwell-Boltzmann’s statistics the particles are :

5.

Paulie’s law is followed in ……… statistics.

(a) Indistinguishable (b) Distinguishable

6.

Photon obeys .........statistics.

(c) Both (a) and (b) (d) None of these

7.

……… is the minimum size of phase cell.

The particles which obey BE statistics are known as :

8.

At absolute zero ……… is the probability of finding of electrons at Fermi level.

9.

Fermions has ……... .

(a) Fermions

(b) Muons

(c) Bosons

(d) None of these

2 He

4

is ……… .

S-143 10.

In blackbody chamber number of photons ……… . 7.

11.

Electron is a ……… .

8.

12.

The electron in metal follows ……… statistics.

13.

Molecules in quantum statistics are ……… .

14.

Fermi–Dirac distribution law is ……… .

15.

Zero point energy of harmonic oscillator is ……… .

16.

The molecules has any spin are called ……… 11. particles.

17.

Molecules obeying classical statistics ……… .

18.

The average energy is ……… .

19.

The energy of photon is ……… .

20.

Holes are ……… .

True/False

Planck’s law follows B.E statistics. The average energy of electron in free electron is 3 EF . 5

9.

Alpha particle is Fermion.

10.

Helium was last gas to be liquified. g If i >> 1 so, quantum statistics can change into ni classical statistics.

12.

The number of arrangement of ni particles in g i cell gi! . by F.D. statistics are ni !(g i − ni)!

13.

Gas molecules follows M.B or classical statistics.

14.

If two particles are distributed into two cells then cell be like that

1.

Phonon is a Fermion.

2.

Electron follows Fermi-Dirac statistics.

3.

There can be only one particles in a cell as per 15. Fermi-Dirac statistics.

4.

Paulie’s law is not valid for B.E statistic.

16.

Energy of a quanta or photon is E = hν.

5.

Bose-Einsteins distribution function is gi ni = (e α e E i / kT + 1)

17.

The volume of phase cell is h3 .

18.

The electron gas has very less repulsion forces.

19.

The energy of electrons in metal is discrete.

6.

Quantum statistics can change into classical statistics 20. g if i = 1. ni

The zero point energy of harmonic oscillator is zero.

Neutron is Boson.

S-144

Objective Type Questions Multiple Choice Questions 1.

(b)

2.

(a)

3.

(c)

4.

(a)

5.

(c)

6.

(c)

7.

(a)

8.

(b)

9.

(b)

10.

(c)

11.

(b)

12.

(d)

13.

(b)

14.

(a)

15.

(a)

16.

(b)

17.

(a)

18.

(a)

19.

(a)

20.

(b)

21.

(a)

Fill in the Blank 1.

∂ loge Ω = constant ∂E

2.

helium

3.

Boson

4.

one

5.

Fermi–Dirac

6.

Bose–Einstein

7.

h3

8.

1/2

9.

spin integral multiple of −h

10.

continuously varies

11.

Fermion

13.

Indistinguishable

14.

ni =

16.

Classical

17.

gas molecules

19.

E = hν

20.

Fermions

gf (e eβEi + 1) α

12.

Fermi-Dirac

15.

1 hν 2

18.

3 E 5 F

True/False 1.

False

2.

True

3.

True

4.

True

6.

False

7.

True

8.

True

9.

False

10. True

5.

False

11. True

12. True

13. True

14. True

15. False

16. True

17. True

18. True

19. True

20. False

S-145

H ints and Solutions Numerical Questions 1.

We know that Average energy of free electron 3 E0 = EF 5 3 = × 3.14 = 1.844 eV 5

2.

5.

Ans. vmp = 377 . × 104 cm/sec 6.

7.

Ans. T = 75 K 8. 9.

EF =

2

h  3N  2m  8πV 

2/ 3

N  2m  3/ 2 8π =  EF 3 V  h2  N 3/ 2 ∝ EF V N = n′ (for metal) V N = n (for Lithium) V n′  E F′ = n  E F

  

 E′ n′ =  F  EF

  

3/ 2

See example-20

11.

See example-20

12.

See example-20

13.

See example-7

Ans. (i) 9 (ii) 6 (iii) 3 Ans. (i) 16 (ii) 10 (iii) 6 Ans. (i) 4 (ii) 20 (iii) 64 Ans. 3003

15.

See example-9 3 Ans. < ε > kT 2

16.

See example-10

×h 3/ 2

× 4.6 × 1028

Ans. 2.35 V 17.

Number of ways of distribution of molecules into 18. energy levels gi! = ni(g i − ni)! 19. 4! = 3! (4 − 3)! 4× 3× 2×1 = 3 × 2 × 1 × 1!

Ans. 516 . × 104 cm/sec

See example-11 Ans. 1.6 × 1028 /m 3 See example-12 Ans. 3.1 eV See example-15 Ans. 3.14 eV

20.

See example-16 Ans. First part 3.31 eV

21.

=4 See example-1

See example-8 Ans. 5.2 eV

≈ 1.6 × 1028 / m 3

4.

See example-6 Ans. 210

14.

3/ 2

2.35  n′ =    470 . 

3.

See example-5

10.

3/ 2

Let

See example-4

Ans. E F = 3.499 eV

Fermi–Energy

or

See example-3 Ans. P(v)dv = 7.4 × 10−3

We know that,

or

See example-2

See example-16 Ans. Second part 2.56 × 104 K

22.

See example-17 Ans. T = 115940 K

S-146 23.

We have,

Very Short Answer Type Questions

E = kT or

EF = EF =

Given

1.38 × 10−23 1.6 × 10−19 T eV 11594

2.

β = β′

3. 4.

Helium gi >> 1 ni

5.

(i)

T = 4.70 × 11594 T = 54491.8 K

24.

Two systems can comes into equilibrium is function ∂ log e Ω is constant for both. ∂E

T

E F = 4.70eV

So,

1.

Molecules are identical

See example-18

(ii) Molecules are indistinguishable

Ans. 7.06 eV 25.

See example-19

(iii) Molecules does not obeys Paulie’s law i. e. only one molecule in one phase cell

26.

See example-20

(iv) Each phase cell has volume h3

Ans. 28 MeV

Short Answer Type Questions

6.

Fermions i. e. obeying Fermi-Dirac statistics.

7.

If Boson’s are the molecule obeying Bose–Einstein statistics and having spins integral multiple of ℏ.

8. 9.

Fermi–Dirac statistics gi ni = α E i / kT (e e − 1)

10.

ni =

1.

See article-1

2.

See article-1

3.

See article-2

4.

See article-2

5.

See article-3

6.

See article-3.1

7.

See article-3.2

8.

See article-4

9.

See article-5

10.

See article-6

11.

See article-7

12.

See article-8

13.

See article-9

14.

See article-9

15.

See article-10

16.

See article-11

17.

See article-11

18.

See article-12

19.

See article-13

20.

See article-14

21.

See article-15

22.

See article-13

(i)

23.

See article-14

(ii) Fermi-Dirac statistic

24.

See article-19

gi α E i / kT

(e e

+ 1)

2

h  2 N  3π  2m  V

11.

EF =

12.

h3 -type (n + g i − 1)! Ωi = i n1 !(g i − 1)!

13.

2/ 3

(g i)!

14.

Ωi =

15.

Factor ‘+1’ indicates that there is no restriction in α that it should be. 1 β= kT

16. 17. 18.

(g i − ni)! ni)!

E = hν 3 E = EF 5

19.

It is the energy where probability of finding of 1 electrons is 2

20.

Two Bose–Einstein statistics mmm