2,339 320 25MB
English Pages 611 Year 1969
Table of contents :
FOREWORD......Page 7
CONTENTS......Page 9
PART 1 QUANTUM PHYSICS......Page 13
1  The Foundations of Quantum Physics......Page 15
2  Quantum Mechanics......Page 64
3  Atoms with One Electron......Page 120
4  Atoms with Many Electrons......Page 161
5  Molecules......Page 194
6  Solids......Page 242
7  Nuclear Structure......Page 294
8  Nuclear Processes......Page 340
9  Fundamental Particles......Page 388
PART 2 STATISTICAL PHYSICS......Page 443
10  Classical Statistical Mechanics......Page 445
11  Thermodynamics......Page 473
12  THERMAL PROPERTIES OF GASES......Page 505
13  Quantum Statistics......Page 530
APPENDIX......Page 561
SUPPLEMENTARY NOTES......Page 563
Tables......Page 587
LIST OF TABLES......Page 593
ANSWERS TO ODDNUMBERED PROBLEMS......Page 595
Theatomic masses, based on the exact number 12.00000 as the assigned atomic mass of the'prir.';* cipal isotope of carbon, 12C, are the most recent (1961) values adopted by the International Union of Pure and Applied Chemistry. The unit of mass used in this table is called atomic mass Group—*
I
II
ill
IY
Period
Series'
I
I
IH 1.00797
2
2
3 Li 6.939
4 Be 9.0122
5B 10.811
6C 12.01115
3
3
11 Na 22.9898
12 Mg 24.312
13 Al 26.9815
14 Si 28.086
4
19 K 39.102
20 Ca 40.08
4
29 Cu 63.54
5
37 Rb 85.47
6 5 I* r^.
55 Cs 132.905
m
30 Zn 65.37
7
87 Fr [223] ,
10
‘ Lanthanide series: ** Actinide series:
Table A—2
57 La 138.91 89 Ac [227]
5771 Lanthanide series*
80 Hg 200.59
59 Pr 140.907 91'Pa [231] ,
50 Sn 118.69 *72 Hf 178.49
81 Tl 204.37
88 Ra [226.05] 58 0». 140.12:' 90 Th ' * 232.038
40 Zr 91.22
49 In 114.82
48 Cd 112.40 56 Ba 137.34
79 Au 196.967
32 Ge 72.59
39 Y 88.905
6 9
22 Ti 47.90
31 Ga 69.72
38 Sr 87.62
47 Ag 107.870
7
21 Sc 44.956
82 Pb 207.19
89Actinide series** 60 Nd 144.24 92 U 238.03
61 Pm [147] 93 Np [237]
62 Sm 150.35 94 Pu [242]
Fundamental Constants
Constant
Symbol
Value
Velocity of light
C
2.9979 X IO8 m s” 1
Elementary charge
€
1.6021 X 10~19 C
Electron rest mass
me
9.1091 X 10“ 31 kg
Proton rest mass
mp
1.6725 X IO"27 kg
Neutron rest mass
Wn
1.6748 X IO27 kg
h
6.6256 X IO *4 J s
h = h/2ir
1.0545 X IO"34 Js
Planck constant Chargetomass ratio for electron
e/me
Quantum charge ratio
h/e
Bohr radius ,Compton wavelength: of electron
I of Protoni Rydberg constant
oo v \c,e ^C,P B
t 1.7588 X IO11 kg” 1 *
4.1356 X IO "15 J s ( 5.2917 X I0~u m 2.4262 X IO 12 m 1.3214 X 10“ 15m 1.0974 X IO7 Hi1
unit (amu); I amii “ 1.6604 X 10~27 kg. The atomic mass of carbon is 12.01115 on this scale because it is the average of the different isotopes naturally present in carbon. (For artificially produced elements, the approximate atomic mass of the most stable isotope is given in brackets.) y vi V iI y in o 2 He 4.0026 7N 14.0067
80 15,9994
9F ' 18.9984
10 Ne 20.183
15 P 30.9738
16 S 32.064
17 Cl 35.453
18 A 39,948
33 As 74.9216 : 41 Nb * 92,906
34 Se 78.96 42 Mo 95.94
83 Bi 208.980
44 Ru 101.07
75 Re 186.2
65 Tb 158.924 97 Bk [249]
28 Ni 58,71 '
45 Rh 102.905
46 Pd 106.4
53 1 126.9044
84 Po [210]
64 Gd 157.25 96 Cm [245]
27 Co 58.9332
36 Kr 83.80
43 Tc [99]
74 W 183.85
73 Ta 180.948
26 Fe 55.847
35 Br 79.909
52 Te 127.60
51 Sb 121.75
63 Eu 151.96 95 Am [243]
25 Mn 54.9380
24 Cr 51.996
23 V 50.942
54 Xe 131.30 76 Os 190.2
77 Ir 192.2
78 Pt 195.09
85 At [210]
66 Dy 162.50 98 Cf [249]
86 Rn [222]
67 Ho 164.930 99 Es [253]
Constant
68 Er 167.26 100 Fm [255]
69 Tm 168.934 101 Md [256]
70 Yb 173.04 102 No
Value
Symbol
Bohr magneton
MB
Avogadro constant
Na
Boltzmann constant
k
1.3805 X IO 23J 0K " 1
Gas constant
E
8.3143 J 0K " 1 mol" 1
Ideal gas normal volume (STP)
V0
2.2414 X IO" 2 m3 mol” 1
9.2732 X IO" 24 J T " 1 6.0225 X IO23 mol" 1
Faraday constant
F
9.6487 X IO4 C mol” 1
Coulomb constant
Xe
8.9874 X IO9 N m2 C ” 2
Vacuum permittivity
Co
8.8544 X IO 12N " 1 m" 2 C2
Magnetic constant
Xm
1.0000 X 10” 7 m kg C " 2
Vacuum permeability
MO
1.3566 X IO" 9 m kg C " 2
Gravitational constant
7
6.670
Acceleration of gravity at sea level and at equator
Q
9.7805 m s” 2
Numerical constants:
*
*• 3.1416;
e
» 2,7183;
71 Lu 174.97 103
X l O " 11 N m 2 kg” 2
\/2 — 1.4142:
\/ 3
1.7320
FUNDAMENTAL
ADDISONWESLEY PUBLISHING C O M PA N Y
U N IV E R S IT Y PHYSICS
VO LU M E II I
Q U A N T U M A N D STATISTICAL P H Y S IC S
M A R CE LO ALO NSO
Department o/ Scientific Affairs, Organization of Am erican States
E D W A R D J. F IN N
Department of Physics, Georgetown University
R e a d in g , M a ss a c h u s e tts · M en lo P a r k , C a lifo rn ia D o n M ills, O n ta rio · W o k in g h a m , F.ngland · A m s te rd a m · B o n n · S y d n e y S in g a p o re · T o k y o · M a d rid · B o g o ta · S a n tia g o · S a n J u a n
T H IS
BOOK
W O RLD
IS A N
A D D IS O N W E S L E Y
STUDENT
S E R IE S
E D IT IO N
A com plete and unabridged reprint of the original Am erican textbook, this W orld Student Series edition m ay be sold only in those countries to which it is consigned b y AddisonW esley or its authorized trade distributors. I t may not be reexported from the country to which it has been consigned, and it m ay not be sold in the United States of Am erica or its possessions.
T h e cover design o f this volum e is adapted from the August 1966 issue o f the m agazine Am ericas, published b y the Pan Am erican U nion in Spanish, English, and Portuguese; used b y special permission.
C op yrigh t (c) 1968 by AddisonW esley Publishing Com pany, Inc. AU rights reserved. N o part of this publication m ay be reproduced, stored in a retrieval system, or transm itted, in any form or b y any means, electronic, mechanical, photocopying, recording, or otherwise, w ithout the prior w ritten permission of the publisher. O r i g i n a l e d i t i o n p u b lis h e d in th e U n it e d States o f A m e r i c a . P u b lis h e d s im u lt a n e o u s ly in C a n a d a . Philippines C opyright 1968.
Q R S T U V D A 8 9 8 7 6
FO RE W O RD
This book is the third and last volum e of a series published under the general title of Fundamental University Physics. T h e purpose of this series is to offer to students of science and engineering a logical and unified presentation of physics at the introductory under graduate level, w ith emphasis on the basic ideas which form the core of physics: the conservation laws, the interrelation between particles and fields, and the atom ic vie w of m atter. W e have tried to present physical concepts in such a w ay th at the student w ill attain a clear understanding of their theoretical m eaning and recognize their experim ental foundations, noting the close interrelation between th eory and experim ent. W e also have tried to develop in the student the ab ility to manipulate the mathem atics required for the expression o f such concepts. T h e three volum es cover the equivalent of a twosem ester course in general physics plus a one (or tw o ) semester course in modern physics. Volum e I treats mechanics and the gravitation al interaction. Volum e I I deals w ith electrom agnetic interactions and waves. V olu m e I I I covers quantum and statistical physics (including therm odynam ics).
A lth ou gh the three volum es are closely related and follo w a logical
sequence, each one is selfcontained and can be used independently of the others. T his is particularly true of Volum e I I I , which covers m ost of the subject m atter usually in cluded in an introdu ctory modern physics course. T h e curricula for all sciences are under great pressure to incorporate new subjects th at are becom ing more relevant. W e expect th at this series w ill relieve this pressure b y raising the level of the student’s understanding of physical concepts and his ab ility to apply them to concrete situations. T his w ill perm it m any interm ediate courses presently offered in the undergraduate curriculum to be upgraded. T h e tradition al undergraduate courses in mechanics, electrom agnetism , and modern physics w ill benefit m ost from this upgrading. Thus the student w ill finish his undergraduate career at a higher level of knowledge than form erly: an im portant benefit for those who term inate their form al education at this point. A lso there w ill now be room for newer and m ore exciting courses at the graduate level. T his same trend is found in the more recent basic textbooks in other sciences for freshman and sophomore courses. T h e first part of this volum e is called Quantum Physics. Quantum ideas are the essence of to d a y ’s physics. U n fortunately, except for a brief introduction to B ohr’ s ideas and to w aveparticle du ality in the introdu ctory general physics course, there has often been a delay in exposing students to quantum m echanical concepts and their applications. T ra d ition a lly on ly the physics and chem istry m ajors learned quantum mechanics, and then rarely before the senior year. H ow ever, physics and chem istry majors should acquire a w orking knowledge of quantum ideas as early in their curricula as possible so th at
Pt
Foreword
they may utilize this knowledge in subsequent undergraduate courses. This procedure is strongly endorsed by the Commission on College Physics. Present trends in biology and engineering demand that students in these fields also have a basic understanding of the solid state and of molecular structure. Therefore we have been careful to introduce the student to quantum mechanics in a way which, although elementary, allows him to apply quantum concepts to different situations. Chapter I is an introduction to the foundation of quantum ideas. This is followed in Chapter 2 by the necessary background in quantum mechanics; here we emphasize the way in which physical information about a system is extracted from the shape of the potentialenergy function and a knowledge of the general nature of wave functions. In the succeeding chapters, 3 through 9, quantal concepts and techniques are applied to the analysis of atoms, molecules, solids, nuclei, and fundamental particles. In the second part of the text (designated Statistical Physics), we use statistical methods to consider the properties of matter in bulk. Like quantum mechanics, statistical physics is a wellfounded, powerful tool, to which the student should be introduced as early as possible. After discussing classical statistical mechanics in Chapter 10, we present thermo dynamics from a statistical point of view in Chapter 11 and apply it to both ideal and real gases in Chapter 12. We are firmly convinced that this is the most appropriate method to follow in introducing the student to the concepts of thermodynamics. The text ends with a brief introduction to quantum statistics in Chapter 13. Since many students now learn the basic ideas of relativity in their general physics course, the special theory of relativity is discussed in the appendix. (A more complete discussion of relativity appears in Volumes I and I I of the series.) Several collateral topics, such as group velocity and the methods of particle detection, are also discussed in the appendix. We have kept the mathematical requirements within the topics covered by a standard calculus course. We have also often either omitted or relegated to the problem sections those mathematical calculations which are not essential to an understanding of the main trend of physical ideas; one example of such calculations is the sometimes boresome task of finding certain solutions to Schrodinger’s equation. Many applications of the fundamental principles, as well as the discussion of a few more advanced topics, appear in the form of worked examples. The text has been written so that the student may omit all examples at his first reading. During a second reading the student should consider those examples chosen by the instructor. The instructor may discuss these examples at his convenience or propose them on a selective basis. Certain sections of the text may be omitted without loss of continuity. The problems at the end of each chapter follow the sequence of the chapter, with a few more difficult problems at the end. The large number of varied problems means that the instructor can choose problems to match the abilities of his students. Hence by proper selection of the material in this text, the instructor can adapt the text to a one or twosemester course and at the same time give the student both a challenge and a motivation to meet that challenge. We want to express our gratitude to all those who, by their assistance and encourage ment, have made this work possible. We recognize in particular Professor David Lazarus, whose comments and criticisms helped to improve many aspects of the text. Last— but not least—we thank our wives, who have so patiently stood by us. Washington, D. C. January 1968
Marcelo Alonso Edward J. Finn
CONTENTS
P A R T I Q U A N T U M P H Y S IC S
^
C h a p te r
I
T h e F o u n d a tio n s o f Q u a n tu m Physics
Introduction 4 □ Electromagnetic radiation 4 (5 Blackbody radiation 7 □ Photoelectric emission 11 O ' Scattering of radiation by free electrons 14 □ Photgns 18 □ Stationarystates 22 □ Experimental evidence of stationary states 26 □ Interaction of radiation with matter 29 □ Particles and fields 33 □ Particles and wave packets 38 □ Heisenberg’s uncertainty principle for position and momentum 39 □ The uncertainty relation for time and energy 43 C h a p te r
2
Q u a n tu m M ech an ics
Introduction 53 □ W a vefu n ction a n d p rob ab ilityd en sity 53 □ Schrodinger’s equation 56 □ Potential step 59 □ ' Particle in a potential box 63 □ The harmonic oscillator 71 □ Energy levels and wave functions in general 75 □ Potentialbarrierpenetration 80 □ Symmetry, wave functions, and parity 88 □ The timedependent Schrodinger equation 90 □ Transition probabilities and selection rules 94 □ The formal theory of quantum mechanics 96 C h a p te r
3
A to m s w ith O n e Electron
Introduction 109 □ T h eh yd rogen atom 109 □ T h esp ectru m of hydrogen 115 □ Quantizationofangularm om entum 117 □ Oneelectron wave functions under central forces 121 □ The Zeeman effect 132 □ Electron spin 135 □ Addition of angular momenta 137 □ Spinorbit interaction 139 j
Cha pt e r
4
A to m s w ith M a n y E lectrons
Introduction 150 □ T h eh eliu m ato m 150 □ Theexclusion principle 158 □ E lectronicstru ctureofatom s 161 □ LS coupling 164 □ Atoms with one or two valence electrons 171 □ Xray spectra 176 titi
v iii
Conlenls
C h a p te r
5
M olecules
Introduction 183 □ The hydrogen molecule ion 183 □ Molecular orbitals of diatomic molecules 191 □ Electronic configuration of some diatomic molecules 194 □ Polyatomic molecules 202 □ Conjugated molecules 208 □ Molecular rotations 212 □ Molecular vibrations 215 □ Electronic transitions in molecules 222 □ Conclusion 225 C h a p te r
6
Solids
Introduction 231 □ Types of solids 231 □ Band theory of solids 243 □ Freeelectron model of a solid 246 □ Electron motion in a periodic structure 251 □ Conductors, insulators, and semiconductors 261 □ Quantum theory of electrical conductivity 268 □ Radiative transitions in solids 274 C h a p te r
7
N u c le a r S tru c tu re
Introduction 283 □ Isotopes, isotones, and isobars 283 □ The atomic mass unit 286 □ Properties of the nucleus 286 □ Nuclear binding energy 293 □ Nuclearforces 298 □ T h e g ro u n d s ta te o f the deuteron 301 □ Neutronproton scattering at low energies 303 □ The shell model 310 □ Nuclear radiative transitions 319 C h a p te r
8
N u c le a r Processes
Introduction 329 □ Radioactive decay 329 □ Alpha decay 335 □ Beta decay 340 □ Nuclear reactions 348 □ Nuclear fission 357 □ Nuclear fusion 363 □ The origin of the elements 367 C h ap ter
9
F u n d a m e n ta l P articles
Introduction 377 □ Particle genealogy 378 □ Particles and antiparticles 379 □ Particle instability 386 □ The conservation laws 397 □ Invariance, symmetry, and conservation laws 403 □ Resonances 414 □ What is a fundamental particle? 419 P A R T 2 S T A T I S T I C A L P H Y S IC S C h a p te r IO
E lassical S tatistical M ech an ics
Introduction 434 Statistical equilibrium 434 @ The MaxwellBoltzmann distribution law 436 Jg Temperature 443 □ Thermal equilibrium 448 @ Application to the ideal gas 450 C h a p te r 11
T h e rm o d y n am ics
Introduction 462 □ Conservation of energy of a system of particles 462 □ Manyparticle systems: work 464 □
Contents
ix
Manyparticle systems: heat 466 □ The first law of thermodynamics 467 □ Graphical representation of processes 469 □ Special processes 473 □ Entropy and the second law of thermodynamics 475 □ Entropy and heat 4S0 □ Discussion of processes in terms of entropy 484 C h a p te r 12
T h e rm a l Properties o f G ases
Introduction 494 □ The equation of state of an ideal gas 494 □ Equation of state for real gases 497 □ Heat capacity of an ideal monntomic gas 504 □ Heat capacities of an ideal polyatomic gas 506 □ The principle of equipartition of energy 512 C h ap ter 13
Q u a n tu m Statistics
Introduction 519 □ FermiDirac distribution law 519 □ The electron gas 522 □ Application of FcrmiDirac statistics to electrons in metals 526 □ BoscEinstein distribution law 528 □ The photon gas 531 □ Heat capacity of solids 536 □ The ideal gas in quantum statistics 540 Appendixes
I Relativistic mechanics 551 □ I I Collisions 555 □ I I I Group velocity 560 □ IV Some useful integrals 562 □ V Stirling’s formula 563 □ V I Lagrange’s undetermined multipliers 564 □ V I I The detection of particles 564
T ab les
577
L i s t o f TahIcs
581
A n sw ers to O d d N u m b e re d P ro b le m s Index
589
583
PART I QUANTUM PHYSICS
i
The Foundations o f Quantum Physics 2 3 U
Quantum Mechanics
Atoms with One Electron
Atoms with M any Electrons 5
Molecules 6
9
Solids
7
Nuclear Structure
8
Nuclear Processes
Fundamental Particles
2 One o f the fundam ental ob jectives o f physics is to an alyze the properties o f th e basic com ponents o f m atter and the processes th at occur am ong them as a result o f th eir interactions. T hese basic com ponents— called fundam ental or elem en ta ry p articles— are electrons, protons, neutrons (and others) which group to geth er to form nuclei, atoms, and molecules. These groups, in turn, com bine to form m a tte r in bulk.
A lth ou gh the m otion o f fundam ental particles com plies w ith the prin
ciples o f conservation o f m om entum , angular m om entum , and energy, th e a n a ly sis o f this m otion requires a fram ew ork d ifferent in several respects from the one developed in classical (o r N ew to n ia n ) mechanics fo r the analysis o f m acroscopic m otion. T h is m ore refined th eory is called quantum mechanics. W e m ust under stand it w ell before we em bark on a discussion o f atom s, molecules, and nuclei. F ortu n ately, atom s and m olecules are essentially the result o f electromagnetic in ter actions betw een the p ositively charged nuclei and the n eg a tively charged electrons. Thus w e can discuss atom s and molecules, w ith ou t havin g to appeal to oth er lesswellunderstood forces, b y com bining the laws o f electrom agnetism w ith those o f quantum mechanics. T h e sam e technique m ay also be used for gases, liquids, and solids. On the other hand, nuclei are basically the result o f a new ty p e o f force, the socalled strong or nuclear interaction.
Since the strong interaction is not y e t well
understood, its analysis is much m ore involved .
C onsequently, in this te x t our
discussion o f nuclei must be o f a m ore d escriptive nature. Perhaps the m ost dynam ic and stim ulating field o f con tem porary physics is the study o f the fundam ental particles. T h e interactions observed betw een these par ticles require the introduction o f another ty p e o f force, in ad dition to th e stron g interaction.
T h is force is called the weak interaction.
A n oth er force, the gravita
tional interaction, w hich is the w eakest o f all interactions, plays a lesser role insofar as the basic structure o f m atter is concerned. T h e rela tive value o f the four interactions is: Strong
I
E lectrom agnetic
IO2
W eak
ΙΟ" 13
G ravitatio n al
1038
T h e processes in v o lvin g fundam ental particles h ave m otivated a new form alism , som ewhat d ifferent from quantum mechanics, called quantum field theory. theory, how ever, is too com plex to be considered in this text.
T h is
I THE FOUNDATIONS OF QUANTUM PHYSICS
1.1 Introduction 1.2
Electromagnetic Radiation 1.3 IA
1.5
Blackbody Radiation
Photoelectric Em ission
Scattering o f Radiation by Free Electrons 1.6
Photons
1.7 1.8 E xperim ental Evidence o f Stationary States 1.9
Interaction o f Radiation with M atter 1.10 1.11
1.12
Particles and Fields
Particles and Wave Packets Heisenberg's Uncertainty P rin c ip le Position and Mom entum
1.13
The Uncertainly Relation fo r T im e and Energy
«Ρ —
U
The foundations o f quantum physics
( 1.2
1.1 ln trod u vtion B y the end o f the nineteenth century, and during the first quarter o f the tw entieth, experim ental evidence began to accum ulate which indicated th at the interaction o f electrom agnetic radiation w ith m atter was n ot en tirely in accordance w ith the IqMis o f electrom agnetism . T h ese laws w ere the result o f the w ork o f Am pbre, Laplace, F araday, H en ry, M axw ell, and m any others, and are synthesized in M a x w e ll’s equations for the electrom agnetic field. A t the same tim e the th eory o f the atom ic structure o f m atter was developin g, m ainly as a result o f the discovery o f the elec tron and the confirm ation o f the nuclear m odel o f the atom . S till another series o f experim ents forced the physicist to review his concepts o f the m otion o f subatomic particles, since th ey app aren tly did not m ove precisely in accordance w ith the assumptions o f N ew ton ian mechanics. T o explain the new observations, a sequence o f new ideas, introduced in a m ore or less acl hoc fashion, were incorporated b y sev eral physicists. W ith the passage o f tim e, and by the efforts o f m any brillian t men, these ideas e v olved until th ey becam e w hat is now known as the quantum theory, a th eory which is, perhaps, the essence o f con tem porary physics. In this chapter we shall review the m ore im portant experim ental bases o f quantum physics. Y
F ig. 11.
Electric field of a charge at rest.
F ig . 12. E lectrican dm agneticfieldsof a uniformly moving charge.
1.2 E ivvt rom a gn vtiv H ndiation T h e electrom agnetic interaction between tw o charged particles can best be de scribed in terms o f the concepts o f electric and m agnetic fields produced b y the charges.
W h en a charged p article is a t rest rela tiv e to an inertial observer, the
observer measures a field which is called the electric field o f the charge (F ig . 11). H ow ever, if the charge is in m otion rela tive to the observer, he observes a different field, called the electrom agnetic field o f the charge (F ig . 12).
One com ponent o f
the field is still called electric, while the rem aining com ponent is called the m agnetic field. Such fields depend on the v e lo c ity and acceleration o f the charge rela tive to the observer.
Since the separation o f the field produced b y a charge into an elec
tric and a m agnetic part depends on the rela tive m otion o f the charge and the ob server, w e should speak on ly o f the electrom agnetic field o f the charged particle. C onversely, when a particle m oves through the electrom agnetic field produced by
Electromagnetic radiation
1.2)
5
other charges, it experiences a force given by F =
g (€ + I· x (B),
where ε and ® are the electric and m agnetic fields, respectively, as measured b y an observer w h o measures the v e lo c ity o f the particle as v. In this w ay w e can describe the electrom agnetic interaction o f charged particles in terms o f fields. E n ergy is required to set up an electrom agnetic field. T h e energy per unit volum e o f an electrom agnetic field in vacuum is
( 1. 1) where « 0 and g o arc the vacuum p e rm ittiv ity and perm eab ility, respectively. T h e energy o f a static electrom agnetic field (th a t is, a field th at does not change w ith tim e) ob viou sly remains constant. H o w ever, when the field is time dependent, the electrom agnetic energy at each point changes w ith tim e. T h e tim e variation s o f an electrom agnetic field g iv e rise to an electrom agnetic w ave which propagates w ith a v e lo c ity c =
! / V i 0Mo «
3 X IO8 m s ~ \
(1.2)
which is the same as the v e lo c ity o f light in vacuum. W c m ay say th at the w a v e carries the energy o f the electrom agnetic field. T h is energy which is carried b y an electrom agnetic w a v e is som etimes called electromagnetic, radiation. Since a charge a t rest relative to an observer produces a static field, the charge does not radiate electrom agnetic energy. A lso it can be shown th at a charge which is in uniform rectilin ear m otion does not radiate electrom agnetic energy because the total energy o f its electrom agnetic field remains constant. A v e ry different situation exists for a charge which is in accelerated m otion. T h e total energy o f the electrom agnetic field o f an accelerated charge varies w ith tim e. T h erefore an accelerated charge radiates electromagnetic energy. T h e rale o f energy radiation by a charge q m ovin g w ith v e lo c ity i> and acceleration a, when the v e lo c ity is small rela tive to the v e lo c ity o f light, is (1.3) One im portant conclusion is that, if a charge is to be m aintained in accelerated m otion, energy must be supplied to com pensate for the energy transferred as radia tion. T h is means th at when an ion is accelerated, for exam ple in a V an de G raaff accelerator or in a cyclotron, a fraction o f the energy supplied to the ion is lost as electrom agnetic radiation. tivistic energies.
T h is en ergy loss, however, is negligible except at. rela
C harged particles trapped in the earth ’s m agnetic field, in sun
spots, or in distant celestial bodies such as the Crab nebula also em it radiation, called synchrotron radiation. T h is radiation extends from radio frequencies to the extrem e ultraviolet.
6
The foundations of quantum physics
(1.2
I f the p article is decelerated instead o f being accelerated, E q . (1.3) still holds and the energy radiated is th at excess which th e electrom agnetic field has as a result o f th e decrease in the v e lo c ity o f the charge. F o r exam ple, when a fast charge such as an electron or a proton hits a target and is stopped, a substantial p art o f its total energy goes o ff as radiation (F ig . 1 3). T h is radiation is called deceleration radia tion, or m ore com m only bremsstraldung (from the Germ an words Brem sung (decel eration ) and Strahlung (rad iation )]. T h is is the chief m echanism by which radiation is produced in the xray tubes which are used in physical, m edical, and industrial applications.
T h e energy radiated b y a charged particle m ay be absorbed b y oth er charged particles which are subject to the action o f the electrom agnetic field produced by the first particle.
H en ce w e m ay describe the interaction o f tw o charged particles
as an exchange o f energy b y means o f emission and absorption o f radiation.
For
example, the oscillating electrons in the antenna o f a radio broadcasting station radiate energy. P a rt o f this energy is absorbed b y the electrons in th e antenna o f a radio receiver, which results in a signal a t the receivin g station. A n analysis o f the processes o f emission and absorption o f radiation (th a t is, the interaction o f radiation and m a tte r) is fundam ental for understanding the behavior o f m atter. A s w e shall see in th e follow ing sections, quantum physics evolved as a result o f the analysis o f such processes. E X A M P L E 1.1.
The rate at which energy is radiated by an oscillating electric dipole.
S o lu tio n : Consider a charge q moving along the Zaxis in such a way that at any time its position is given by z = z0 cos cot. This corresponds to an oscillatory motion of ampli tude zo and angular frequency ω. Thus the charge is equivalent to an oscillating electric dipole. The acceleration of the particle is a = —ω2ζ. Substituting this value of a in Eq. (1.3), we have
Iilackbody radiation
7
The rate of energy radiation oscillates because of the variation of 2 with time. T o obtain the average rate of energy radiation, we recall that (22) » vο = £20· Thus (d lA
=
\ dl ) avo
q2Zpoii 12xtoC3
.
.
We may say that an oscillating electric dipole radiates energy at an average rate given by Eq. (1.5) and that the radiation corresponds to an electromagnetic field oscillating with the same frequency as the dipole. K ( x)
F ig. 1t. Monochromatic energy density of blackbody radiation at different temperatures as a function of the frequency. !•'requeney v
Φ # ..?
Itla r lib o tly
H a tlia iio n
Consider a c a v ity whose walls are"at a certain tem perature. T h e atom s com posing the w alls are em ittin g electrom agnetic radiation; at. the same tim e th ey absorb radiation em itted b y other atom s o f the walls. T h e electrom agnetic radiation field occupies the w hole ca vity .
W hen the radiation trapped within the c a v ity reaches
equilibrium w ith the atom s o f the walls, the am ount o f energy em itted by the atom s per unit tim e is equal to the am ount absorbed b y them. Hence, when the radi ation in the c a v ity is a t equilibrium w ith the walls, the energy density o f the electrom agnetic field is constant.
E xperim ent has shown that, at equilibrium ,
the trapped electrom agnetic radiation has a welldefined energy distribution; that is, to each frequency there corresponds an energy density which depends solely on the tem perature o f the walls and is independent o f th eir m aterial.
T h e energy
density corresponding to radiation w ith frequency between v and v  f dv is w ritten e (i') dv, where E (g) is the energy density per unit frequency range, som etimes called monochromatic energy density. T h e observed variation o f E(V) w ith the fre quency v is illustrated in Fig. 14 for tw o temperatures.'
C u r v e s lik e th e s e w e r e
first obtained experim entally by Lum m er and Pringsheim in 189!).
It m ay be
seen from th e curves th at for each tem perature the energy density shows a p ro nounced m axim um a t a certain frequency.
N o te also th at the frequency a t which
the energy d ensity is maxim um increases as the tem perature increases. plains the change in color o f a radiating b od y as its tem perature varies.
T h is ex
8
The foundations of quantum physics
U .3
I f a sm all hole is opened in one o f the walls o f the ca v ity , some o f the radiation escapes and m ay be analyzed. T h e hole appears v e ry bright when the body is a t high tem peratures and the inten sity o f the equilibrium radiation within the c a v ity is high, but it appears com pletely black at low tem peratures, when the intensity o f the equilibrium radiation is negligible in the visible region o f the spectrum . F o r th at reason the radiation com ing ou t o f the c a v ity was called blackbody radia tion by those who analyzed it in the nineteenth century. T h e problem o f finding w h at mechanism causes radiating atom s to produce the observed energy distribution o f blackbody radiation led to the birth o f quantum physics. B y the end o f the last century all attem pts to explain this energy distribu tion using the concepts availab le a t th at tim e had failed com pletely. T h e Germ an physicist .Max P lanck ( 18581947) suggested, abyut 1900, th at if the radiation in th e c a v ity was in equilibrium w ith the atom s o f the walls, there should be a cor respondence between the energy distribution in the radiation and the energies of the atom s in the ca vity . A s a m odel for the radiating atom s, Planck assumed that atom s behave as harmonic oscillators, and th at each one oscillates w ith a given frequency v. A s a second assumption Planck suggested that each oscillator can absorb or emit radiation energy only in an amount proportional to its frequency v. T h is la tter condition is not required b y the classical theory o f electrom agnetism (as expressed by M a x w e ll’s equations), which perm its a continuous emission or absorption
o f energy. G iven th at E is the energy
process o f interaction assumption states that E =
absorbed or em itted in a single
o f an oscillator w ith electrom agnetic radiation, P lan ck ’s
hv,
( 1.6)
where h is a p rop ortion ality constant assumed to be the same for all oscillators. Hence, when an oscillator absorbs or emits electrom agnetic radiation, its energy increases or decreases b y an am ount hv.
E quation (1.6) then im plies that
the energy o f atomic oscillators is quantized. T h a t is, the energy o f an oscillator o f frequency v can attain on ly certain values, which are (assuming th at the m inim um energy o f the oscillator is zero) 0 , hv, 2hv, 3hv, . . . .
Thus, in general, the possible values o f the energy o f an oscillator o f fre
quency v are ® En =
nhv,
where n is a p ositive integer.
(1.7) A s w e know, the energy o f an oscillator is propor
tional to the square o f its am plitude and, a p rio ri, by properly adjusting the am pli tude o f the oscillations, w e can m ake an oscillator o f a given frequency have any a rb itra rily chosen energy. T h erefo re P lan ck ’s idea was an ad hoc assumption which could not be explained b y means o f classical concepts; it w as justified on ly because
Blackbody radiation
1.3)
.9
it “w o rk e d ,” and because physicists a t the tim e lack ed a b e tte r explanation. W e still do n o t liave a b etter explanation; w e must a c c e p t the quantization o f some physical quantities as a fundam ental fact o f nature. B y ap p lyin g some considerations o f a statistical n a tu re, togeth er w ith E q . (1.6), Planck obtained, for th e energy density in b la c k b o d y radiation, an expression o f the form
e
W
where k is B oltzm an n ’s constant. w ith the experim ental values o f
e
(L8)
T h is expression, w h ich agrees surprisingly well ( c ) a t d ifferent tem peratures, has been accepted
’ as the correct expression fo r blackbody radiation. It is called P la n ck ’s radiation law. A n interesting aspect is th at P lan ck ’s derivation c a n n o t presently be considered as physically sound (which is the reason w e have o m itte d it). In other words, the problem which precipitated the birth o f the qu an tu m th eo ry was first solved by means o f an unsatisfactory m ethod.
T h e problem had to w a it several years until
the quantum th eory was developed along other lines o f th ou ght before an adequate m ethod o f calculation was found.
T h is revised d e riv a tio n w ill be given in Sec
tion 13.6. H o w ever, P lan ck ’s ideas, especially E qs. ( 1.6 ) and (1 .7), prom pted new thinking b y m any other physicists w ho w ere w orking on th e interpretation o f other related phenom ena; this led to rapid developm en t o f quantum theory. In E q . (1.6) w e introduced an arb itrary constant li, called P la n ck ’s constant. Its value, obtained b y m aking E q. ( 1 . 8 ) f i t the exp erim ental results for I, =
6.6256 X IO" 34 J s.
e
( c ) , is
(1.9)
P lanck’s constant is one o f the m ost im portant con stants in physics. E X A M P L E 1.2. Express the ,monochromatic energy density of blackbody radiation in terms of wavelength. S o lu tio n : Sometimes it is preferable to express the monochromatic energy density in terms of wavelength instead of frequency. We define Ε ( λ ) according to the relation E (X ) d\ =  E (V ) d r. We introduce the minus sign because d\ and d v have opposite signs, although E (r ) and E (X ) are both positive. Thus, since v = c/ λ , we have dv/dk
cfk 2
and E (X ) =
— E (r ) dv/dk = E(i„,,„ = Pphoton, which again is equivalent to Eq. (1.30). So, when we use the same approximation as above, Eq. (1.34) becomes

I B, 
B,) ( )  5 ^ ) " ‘ 
or
(ft 
ft) ( l + ^
i)
2 hv = E j — E i +
^
^ c2 1 ■·
(1.35)
Therefore, for absorption to take place, the energy of the absorbed photon must be slightly greater than the energy difference between the two levels of the absorber to account for the kinetic energy of the recoiling absorber. A consequence of this analysis is that a photon emitted by a system (atom, molecule, or nucleus) in the transition a —> 6 cannot be absorbed by another identical system in order to undergo the reverse transition b —> a, and therefore the emission spectrum is not identical to the absorption spectrum. W e shall come to this matter again in Example 1.10. For atomic and molecular transitions in which E j — E i is of the order of few electron volts and M c 2 is of the order of 1 0 " eV, the correction term in Eqs. (1.33) and (1.35) is about IO10 eV and thus is negligible. On the other hand, for nuclear transitions, E j — Ei may be of the order of IOfi eV. Since M c 2 is of the same order as in atomic transitions, the cor rective term is about 10 eV, which is relatively more important.
I . it K xpvrinivnta l K riilvn vv o f S tationa ry Statvn So far w e h a ve introduced the idea o f station ary states as a convenient concept to explain the discrete spectrum o f atom ic systems. H ow ever, the existence o f transi tions betw een stationary states is am p ly corroborated by m any experim ents.
The
m ost characteristic is th at o f inelastic collisions, in which p art o f the kinetic energy
Experimental evidence of stationary stales of the p rojectile is transferred as internal energy to the target.
27
T h ese are called
inelastic collisions o f the first kind. Inelastic collisions o f the second kind correspond to the reverse process. Suppose th at a fast particle q collides w ith an oth er system A (w hich m ay be an atom, molecule, or nucleus) in its ground state o f energy E i .
A s a result o f the
projectilesystem interaction (w hich m ay be electrom agnetic or nuclear), th ere is an exchange o f energy. L e t E 2 be the energy o f the iim t excited state o f the system . The collision w ill be elastic (i.e., the kinetic en ergy w ill be conserved) unless the projectile has enough kinetic energy to transfer the excitation energy E 2  E
i to
the target. W h en this happens the collision is inelastic, and w e m ay express it by
A }~ t/fast
*
E
T 9»low
W lien the mass o f the p rojectile q.is v e ry sm all com pared w ith th at o f the target A , as happens fo r the case o f an electron colliding w ith an atom , the con dition for in elastic collision (see E xam p le 1.7) is Ek > E2 
E 1,
(1.36)
where E k = \mv2 is the kinetic energy o f the p rojectile before the collision. T h e kinetic energy o f the p ro jectile a fte r the collision is then E 1 k = E k — ( E 2 — E i ), since the energy lost b y the p rojectile in the collision is E 2 — E i . T o g iv e a concrete exam ple, suppose th at an electron o f kinetic energy E k m oves through a substance, let us say mercury vapor.
P rovided th at E k is sm aller than
the first excitation energy o f m ercury, E 2  E it the collisions are all elastic and the electron m oves through the vapor, losing en ergy v e ry slow ly, since the m axi mum kinetic energy lost in each collision (see P ro b lem 1.55) is ap p roxim ately AEk «
— 4 (m . J M )E k ~ 5
How ever, i f E k is larger than E 2  E
X
IO 0J?*.
it the collision m ay be inelastic and the elec
tron m ay lose the energy E 2 — E i in a single encounter. energy o f the electron was not much larger than E 2  E
I f the initial kinetic
it the energy o f the elec
tron a fter the inelastic collision is insufficient to excite other atom s. the successive collisions o f the electron w ill be elastic.
T h e rea fter
But if the k inetic energy o f
the electron was in itia lly v e r y large, it m ay still suffer a few m ore inelastic colli sions, losing the energy E 2 — E i a t each collision and producing m ore excited atoms b efore being slowed down below the threshold fo r inelastic collisions. T h is process was observed for the first tim e in 1914 b y Franck and H ertz. T h e ir experim ental arrangem ent is indicated schem atically in Fig. 117.
A heated fila
m ent F em its electrons which arc accelerated tow ard the grid (7 b y a variab le potential V . T h e space between F and G is filled w ith m ercury vapor. B etw een the grid G and the collecting plate P a small retard in g potential F ', o f ap p roxim ately 0.5 vo lt, is applied so th a t those electrons which are left w ith v e ry little kinetic energy a fte r one or m ore inelastic collisions cannot reach the plate and are not registered b y the galvanom eter.
A s V is increased, the plate current I fluctuates
28
The foundations of quantum physics
+ 
F ig. 117. Franck and Hertz experimental arrangement for analyzing inelastic collisions of the second kind.
Fig. 118. Electron current versus ac celerating potential in the FranckHertz experiment.
as shown in F ig. 118, th e peaks occurring a t a spacing o f abou t 4.9 volts. T h e first dip corresponds to electrons th at lose all th eir kinetic energy a fte r one inelastic collision w ith a m ercury atom , which is then le ft in an excited state.
T h e second
d ip corresponds to those electrons th at suffered tw o inelastic collisions w ith tw o m ercury atoms, losing all th eir kinetic energy, and so on.
T h e excited m ercury
atom s return to their ground state b y emission o f a photon, according to H g * —> H g ) hv w ith hv = E q — E 1. F rom spectroscopic evidence w e know th at m ercury vap or, when excited, em its radiation whose w avelen gth is 2.536 X IO 7 m (o r 2536 A ), corresponding to a photon o f energy hv equal to 4.86 eV. R a d ia tio n o f this w avelength is observed com ing from the m ercury v a p or during the passage o f the electron beam through the vapor.
T h u s this sim ple experim ent is one o f the
m ost striking proofs o f the existence o f stationary states. A n oth er sim ilar experim ent is the coulomb excitation o f nuclei. F o r exam ple, w hen a proton passes near a nucleus, the electrical interaction betw een the tw o m ay g iv e rise to an inelastic collision, resulting in the excitation o f the nucleus to one o f the lowest excited states. T h e nucleus returns to its ground state, em ittin g ganunaray photons which have an energy o f the order o f several keV.
F o r that
reason coulom b excitation is one o f the m ost im portant experim ental m ethods for d etectin g and an alyzin g the low lyin g stationary states o f nuclei. E X A M I’ L E 1.7. Calculation of the threshold kinetic energy required for the excitation of the target in an inelastic collision of the first kind. S o lu tio n : Let us designate the mass of the projectile and the target by m and M , respec tively. W e assume that the target is initially at rest in the laboratory or /.frame of reference. Given that p is the momentum of the projectile before the collision and p ' and I’ the momenta of projectile and target after the collision, the conservation of mo mentum requires that p = p
I’
(1.37)
\ j
Interaction of radiation with matter
29
Similarly, if £ i is the energy of the initial stationary state of the target and £'2 the energy of the final state, the conservation of energy requires that I
2
2 P
I
I
p
/2 ,
+ * 1 2
I
j\2
I
JI
+ 2MP + E i
or, if wc set A E = £2 — E i , then
^ / = ^ ' 2+ ύ ϊρ2+ΕΕ·
( 1 ·3 8 )
The minimum kinetic energy of the projectile required for the process is that in which both projectile and target are at rest in the center of mass or Cframe of reference after the collision, so that all the kinetic energy in the Cframe is used in the excitation of the target. In this case both target and projectile, after the collision, move in the Cframe with the same velocity ccm of the center of mass of the system. Therefore p ' = mvcm and P = M v c m  But if v is the velocity of the projectile before the collision, we have that ecu =
m+ M
And therefore , P
m e = m" i+ M
mp = '„ m~+f "Ir M ’
mM v P = m + M
Mp m+ M
equations which are, of course, compatible with Eq. (1.37). Substituting these values in Eq. (1.38) after a straightforward simplification, we obtain M m+ M
( i p2)
“
AE
&  ά " ’  ( Ι + ϊ ) ΔΕ·
('·39>
This equation gives the threshold kinetic energy which the projectile must have for excit ing the target to its first excited level. I f the projectile is much lighter than the target, m ater, in the developin g process, the film is treated in such a w ay that a perm anent im age is formed. W h en the photon has enough energy, its absorption b y an atom o r a m olecule m ay result in the ejection o f an electron. W h a t is le ft is an ionized atom or m ol
Interaction o} radiation with matter
1.9)
31
ecule. W e m ay w rite the process as A + hv  * A + + e . T h is process, called photoionization, is the eq u iva len t o f the photoelectric effect in m etals discussed in Section 1.3. F o r th at reason it is also called the atomic photo electric effect.
A s a result o f photoionization, w h en a beam o f ultraviolet, x or 7
radiation passes through m atter, it produces ion ization along its path.
L e t us
designate the energy required to extract an electron from an atom or m olecule b y /; this energy is called the ionization potential.
T hen the kinetic energy o f the
ejected electron is g ive n b y Ek
(1 .4 0 )
 hv — I ,
an equation analogous to E q. (1.14). neglected the recoil energy o f the ion.)
(In this equation, incidentally, w e h ave E q u a tio n (1.40) shows that, in order to
produce photoionization, the energy o f the photon must be equal to or larger than I.
T h e value o f I depends on the stationary sta te initially occupied b y the ejected
electrons. For exam ple, if an electron is to be ejected from the ground state in a hydrogen atom , the m inimum energy o f the photon must be 13.6 eV. B u t if the electron is in the first excited state, on ly 3.4 e V a re required. F o r helium atom s the ionization energy needed to rem ove an electron from the ground state is 24.6 eV. In the region o f the upper atm osphere called the ionosphere, the large concen tration o f ions and free electrons (abou t 10"
per m 3) is due m ostly to the photo
electric effect in atom s and molecules produced b y u ltraviolet and xradiation from the sun. Som e o f the reactions that occur m ore frequ en tly are N O + hv > N O + F e ~ N 2 + Iiv  *
(5.3 e V ),
N / + e_
(7.4 e V ),
O 2 F hv —> O f + e ~
(5.1 e V ),
H e + hv —» H e + + c ~
(24.6 e V ).
T h e ionization potentials are indicated in parentheses.
M a n y other secondary
reactions take place in the atm osphere as a result o f these ionizations. A process which is the reverse o f p h otoion ization is radiative capture.
In radia
tiv e capture, a free electron w ith kinetic energy Ek is captured into a bound state by an ion w ith the emission o f a photon. T h a t is, A + fe
—* A F hv.
T h e energy o f the photon is Iiv =
Ek +
I.
(1.41)
W h en the energy o f a photon is much larger than the binding energy o f the electron in an atom or molecule, the electron can be considered as free. In th at case C om pton scattering (Section 1.5) is a m ore probable process than the photo
■
32
The Joundalions of quantum physics
( 1.9
K Absorption edge
K r2
10 *
I
10
IO2
P h o to n en ergy, M e V
F ig. 119. Macroscopic cross sections for the absorption of photons in aluminum and lead. T lie solid curves are the total cross sections. Those labeled I are the partial cross sections due to the photoelectric effect; those labeled I I are partial cross sections due to the Compton effect; those labeled I I I are partial cross sections due to pair production. electric effect. I f the energy o f the photon is larger than 2mcc2 ( =
102 M e V , which
is tw ice th e rest energy o f the electron), y e t another process m ay occur: the crea tion o f an electronpositron pair. (T h e positron is a particle havin g the same mass as the electron, but p ositive charge.) T h e creation o f an electronpositron pair, called p a ir production, m ay be writ ten as Iiv —> e + + e~.
(1.42)
T h is process w ill be discussed in detail in Section 9.3. A t present, let us say on ly that it consists in the transform ation o f a photon into an electron plus a positron. Since the energy associated w ith the rest mass o f an electron o r a positron is mcc2, the minimum energy o f the photon needed to produce an electronpositron pair must obviou sly be 2m,.c2. A t high energy, electrom agnetic radiation m ay also interact w ith atom ic nuclei, either raising them to an excited state, ejectin g a nuclear particle (such as a proton ) in a socalled photonuclear reaction, or even breaking the nucleus apart.
These
nuclear processes w ill be discussed in C h ap ter 8. W hen a beam o f radiation passes through m atter, its energy is gradu ally absorbed by the various processes we have m entioned. Thus, given th at I o is the inten sity o f the radiation before it enters the substance, its intensity a fte r it has traversed a
Particles and fields
1:10)
33
thickness x o f the substance is given b y I =
Io e ~ Sz,
(1.43)
where S (expressed in m eters1 if x is expressed in m eters) is a qu a n tity character istic o f each substance and o f each process; it is called the coefficient o f linear ab sorption or the macroscopic cross section. I t is a function o f the energy o f the photons (fo r the d erivation o f E q. 1.43, see E xam ple 8 .8). F o r each substance, there is one m acroscopic cross section for each possible process, such as photoelectric effect, C om pton scattering, pair production, etc. T h e to ta l cross section o f a substance is the sum o f all partial cross sections.
F ig
ure 119 shows the individual and total m acroscopic cross sections for aluminum and lead for the three processes m entioned.
N o te that a t low energy the photo
electric effect is the m ost im portant, a t m edium energies the C om pton effect dom inates, and a t high energies pair production is the main process.
I.
IO
I tUrtivIes and Fields
A n oth er im portant revolution in physical concepts took place a t the end o f the first quarter o f this century, and radically changed our approach to the description o f m otion o f a particle. Our sensory experience tells us th at the objects w e touch and see have a w elldefined shape and size and therefore are localized in space. W e thus tend to extrap olate and think o f the fundamental particles (i.e., electrons, protons, neutrons, etc.) as havin g shape and size, and so w e tend to im agine them as being som ewhat like small spheres, w ith a characteristic radius, as w ell as mass and charge. Th is, how ever, is an extrapolation beyond our direct sensory experience and w e must analyze it carefully before w e accept it. E xperim ents have shown that our extrapolated sensory picture o f the basic con stituents o f m atter is erroneous.
T h e dynam ical beh avior o f elem entary particles
requires that we associate w ith each p article a field— a mailer field— in the same w a y that, in th e reverse manner, we associate a photon (which is equivalent to a particle) w ith an electrom agnetic field. T h is m atter field describes the dynam ical condition o f a particle in the same sense th at the electrom agnetic field corresponds to photons which h ave precise m om entum and energy. In discussing the connec tion betw een the m atter field and the dynam ical properties o f the particle (i.e., m om entum and en ergy), we m ay be guided b y the relations previously found for the photon. W ritin g the relations (1.28) in reverse, w e m ay assume that the w a ve length λ and the frequency v o f the m onochrom atic field associated w ith a particle o f m om entum p and energy E are given b y
where h, as before, is P la n ck ’s constant. These relations were first proposed in 1924 b y the French physicist Louis de B ro glie (1892
), and for th at reason λ =
is som etim es called the de B roglie wavelength o f a particle. num ber k =
2ir/\ and the angular frequency ω =
h/p
Introducing the w ave
2κν, w e m ay w rite the rcla
35
The Joundalions of quantum physics
tions (1.44) in the m ore sym m etric form
or, defining a new constant designated b y h and called hbar, h =
h /'lir =
p =
ΛAr,
1.0.544 X IO34 J s,
w e have E 
Ιιω.
(1.4.5)
I f our assumption, as expressed b y Eqs. (1.44) or (1.4.5), is correct, w e m ay expect th a t w h en ever the m otion o f a p article is disturbed in such a w a y th at the field associated w ith it cannot propagate freely, interference and diffraction phenomena should be observed, as is the case for elastic and electrom agnetic waves.
T h is is
indeed what happens.
F ig. 120. Experimental arrangem for observing electron diffraction throi crystalline material.
B efo re w e discuss the experim ents that reveal interference and diffraction o f th e m atter field, le t us try to estim ate the value o f the de B roglie w avelength λ asso ciated w ith a particle.
Electrons accelerated b y an electric p oten tial V gain an
energy e V ; hence their kinetic energy is p /2mr = e V so that p =
\/:2 )itL.eV.
T h erefore, introducing the values o f e, mc, and Λ, w e obtain the de B roglie w a ve length o f such electrons λ =
h / V 2 m ec V =
where V is expressed in volts.
(1.40)
1.23 X IO V v I 7 m,
T h is form ula can also be used when the kinetic
energy o f the electron is expressed in electron volts.
F o r V ~ IO4 V (w hich is in
the range o f v o lta g e used in T V tubes), the w avelength is abou t IO 1 1 no, com parable to the w avelength o f xrays.
T h is means th at if w e send a beam o f fast
electrons through a crystal, w e should obtain diffraction patterns which result from scattering o f the m atter field. T h ese diffraction patterns, corresponding to the in com ing electrons, should be sim ilar to those observed for xrays.
Particles and fields
UO)
35
F ig. 121. Diffraction of electrons by crystal powder (courtesy of Dr. Lester Germer).
In 1927 the B ritish scientist G. P. T h om son (1892
) began a series o f
experim ents whose purpose was to study the passage o f a beam o f electrons through a thin film o f crystalline m aterial. A fte r the electrons passed through the film , th ey struck a photographic plate, as shown in Fig. 120. I f the electrons had behaved as particles in the m acroscopic sense, a blurred im age would have been observed because each electron would undergo, in general, a different scattering b y the atom s in the crystal.
H o w e v e r, the result obtained was identical to the D eb ye
Scherrer patterns for xray d iffra c tio n b y a p oly crystal line substance, as indicated in the photograph o f Fig. 121. Sim ilarly, when an electron beam passes through a single crystal, one obtains L a u e spot patterns (also observed w ith xrays) as seen in the photograph o f F ig. 122.
From the structure o f these patterns one
36
The foundations of quantum physics
( 1.10
Electron source
S ingle crystal
F ig . 123.
Davisson and Germer arrangement for observing liragg scattering of
electrons. can com pute the tie B roglie w avelength λ if one knows the spacing between the crystal planes and if one applies the formulas that have been d erived for xrays. T h e resulting values o f λ can be compared w ith those obtained from Etp (1.46). T h e result is com p lete agreem ent, w ith in the lim its o f experim ental error. In the celebrated experim ents b y C. Davisson and L. G erm er (m ade at about the sam e tim e as those o f Th om son ), a beam o f electrons was sent at an angle to the face o f a crystal. T h e diffracted electrons w ere observed b y means o f a d etector sym m etrically located, as indicated in F ig. 123. rangem ent fo r observing xray diffraction.
T h is is sim ilar to the B ra g g ar
I t was found th at the electron current
registered b y the detector was a maximum e v ery tim e the B ra g g condition, derived fo r xrays, was fulfilled. T h e B ra g g condition is expressed b y * 2d sin θ =
nX,
where d is the separation o f successive atom ic layers in the crystal, and λ is g iven b y E q. (1.46).
R ea cto r shielding M on oen ergetie
neutn R ea cto r
Fig. 124.
Neutron crystal spectrometer.
R o lyen ergetie neutron Iieam
T h e same phenomenon o f B ra g g diffraction has been observed in experim ents w ith protons and neutrons. N eu tron d iffraction is especially useful, since it is one o f the most pow erful means o f studying crystal structure.
E xperim enters use
m onoenergetic beams o f neutrons and an alyze th eir passage through the crystal. T h e neutrons em erging from a nuclear reactor through a porthole (see Fig. 124) * See, for example, Fundamental University Physics, Volume II, Section 23.8.
Particles and fields
37
liave a w id e spectrum o f energy (in other words, th ey v a r y w id e ly in m om entum ), or, which is equivalent, the neutron beam is not m onochrom atic; rather it contains a spectrum composed o f m any de B roglie wavelengths.
W h en the neutron beam
from the reactor falls on a crystal, o f L iF fo r example, the neutrons observed in the sym m etric direction correspond on ly to the w avelength X given b y B ra g g ’s condition (1.47). T h erefo re th ey have a welldefined energy and m omentum. T h e crystal then acts as an energy fille r or monochromator. T h e m onoenergetic neutron beam is in turn used to study other m aterials, b y diffraction, or to analyze nuclear reactions in vo lvin g neutrons o f definite energy. E X A M P L E 1.8. ture of 25°C?
What is the de Broglie wavelength of thermal neutrons at a tempera
S o lu tio n : Ily thermal neutrons we mean neutrons which are in thermal equilibrium with matter at a given temperature. Thus the neutrons have an average kinetic energy identi cal to that of the molecules of an ideal gas at the same temperature. Therefore the average kinetic energy' of thermal neutrons is Ziove = §kT, where T is the absolute temperature and k is Boltzmann’s constant (see Eq. 10.41). Given that the temperature is 25°C, we have T = 298°K and therefore Eove = J k T _ = 6.17 X IO" 21 J = 3.85 X IO2 eV. The corresponding momentum is p = V 2mnEave = 4.55 X IO2 '1 m kg s1 . Then, using Eq. (1.44), we find that the average de Broglie wavelength of the thermal neutrons is X = 1.85 X IO10 m. (Incidentally', noting that the separation of the planes in a NaCl crystal is d = 2.82 X IO10 m, we see that the first diffraction maximum for neutrons of this wavelength occurs at an angle Θ — 19°.) E X A M P L E 1.9. An electron is released at a great distance from a proton. Find the wavelength of the electron when it is (a) I m from the proton, (b) 0.5 X IO10 m from the proton (this distance is of the order of magnitude of the radius of the orbit of an electron in the ground state of a hydrogen atom). S o lu tio n : The potential through which the electron has moved, when it is at a distance r from the proton, is V = e/4ir3
Classical and quantal path of a particle in phase space.
In quantum mechanics the picture is different.
L e t us d iv id e the phase space in to
cells, each cell having sides Δ χ and Δ/> so that Δ χ Ap =
A. Th en th e m ost w e can
s a y is that a t each m om ent the represen tative point o f the particle lies w ith in one such cell.
A s tim e goes on, the path o f the represen tative p oin t falls w ith in a
rib b o n lik e path form ed b y patching to geth er a series o f cells (see Fig. l2 9 b ).
f.1.7
Tliv I 'n vvrta in ty Itvla lion fo r T im v anti Envrfiy
In a d d itio n to the uncertainty relation Δ χ Ap ~ Ii betw een a coordinate and the corresp on din g m om entum o f a m ovin g particle, there is an uncertainty relation b e tw e e n tim e and energy.
Suppose that w e w ant to m easure not on ly the en ergy
o f a particle but also tbe tim e a t which the p article has such energy. I f Δ< and A E a r e th e uncertainties in the values o f these quantities, the follo w in g relation holds:
(1.49)
A tA E Ii. W e can understand this relation in the follow in g w ay.
I f w e want to define the
t im e a t which a particle passes through a g iven point w e m ust represent the par tic le b y a pulse or w a ve packet h a vin g a v e ry short duration Al. B ut to build such a p u lse it is necessary to superpose fields which h ave d ifferen t frequencies, w ith an a m p litu d e appreciable on ly in a frequency range Δ ω centered around the frequency ω a n d such that, according to the th eory o f F ou rier analysis, At Δ ω ~ 2π. M u ltip ly in g b y ft and recalling from E q. (1.4.5) th at E — Ιιω and that 2Trh = h, w e ob ta in relation (1.49). R elation (1.49) g ives the optim um relation am ong the u n certain ties Al and AE. H ow ever, in m ost cases, t and E are known with less ac c u ra c y , so th at w e m ust w rite instead o f relation (1.49) the m ore general expression At A E > h.
44
The foundations of quantum physics
(1.13
T h e uncertainty relation (1.49) requires that w e revise our concept o f stationary states.
Let us consider an electron in an excited stationary state in an atom . T h e
electron a fte r a certain tim e will suffer a rad iative transition into another station a ry state o f less energy.
H ow ever, w e have no means o f p redicting w ith certa in ty
how long the electron will remain in the stationary state before m aking the transi tion. A s w ill be seen in the next chapter, the most w e can talk about is the prob ab ility p er unit tim e that the electron w ill jum p into a low er en ergy state. T h ere fore the avera ge length o f tim e the electron is in the stationary state, also called the lifetim e o f tin; state, and which is inversely proportional to the transition p rob ab ility w e have m entioned, is known within an uncertainty At. H ence the energy o f the stationary state o f the electron is not known precisely but has an uncertainty Al'J, such that relation (1.49) holds.
O ften A E is designated as the
energy width o f the state whose energy is m ost probably between E — \ A E and E T h A E (F ig . 130). W e m ay assume that Δ/ is o f the order o f m agnitude o f the lifetim e o f the excited state. Thus, the shorter the lifetim e o f an excited state, the larger the uncertainty in the energy o f the state.
F o r the ground state, whose
life t im e is in fin ite hecAtiso n s vsto m w h ich is A t its Around s ta te C a n n o t s u ffe r a
 cc.
T h is yield s
ccurately.
G round state
F ig. 130.
Width of energy levels.
Because o f the energy w idth o f station ary states, the energy released or absorbed in a transition is not well defined.
Thus in a transition between states o f
en ergy E i anil E 2, the photons em itted or absorbed fall in the energy range E 2 — L· I ±
h Al·.,
w here A E is the total energy w idth o f both states.
V i.13)
The uncertainly relation for lime and energy
45
A further broadening o f the energy o f the stationary states is caused b y the electrom agnetic D op p ler shift.
T h e atom s o f a substance are in m otion, some
advancing tow ard and some retreating from the observer, and the observed fre quency o f the radiation em itted b y the atom differs according to the direction o f relative m otion.
In m ost atom ic and m olecular transitions, the broadening due
to the D op p ler effect is much greater than the broadening due to the uncertainty principle, but is usually sm aller for nuclear transitions. E X A M P L E 1.10. The possibility of resonance absorption in atomic and nuclear transi tions as a result of the energy width of the stationary states. S o lu tio n : In Example 1.6 we saw that in a transition between states of energies E i and # 2, the photon emitted or absorbed has an energy which is smaller or larger, respectively, than Eo — E i by an amount (E 2 — E i ) 2/2M c2. Therefore, if the energy width X E is larger than (Eo — E i ) 2/2.1/c2, photons emitted by one system can be absorbed by an other of the same kind; but if X E is smaller, the absorption does not necessarily occur. The first situation is normally found in atomic and molecular systems and the second is more common in nuclei. For example, in the atomic case of the 4.86 eV transition in mercury, mentioned in Section 1.8, the lifetime of the excited state of mercury is about 10~s s. A mercury atom, whose atomic mass is about 200 amu or 3.34 X IO25 kg, has a rest energy M c2 = 1.86 X IOn eV. Therefore the term accounting for the atomic recoil with Eg — E i = 4.86 eV is
^
7 ^
= 7 .1 5 X 1 0  , l eV.
The uncertainty in the energy of the mercury atom excited level, which has a lifetime Xl ~ IO" 8 s, is X E ~ h/Xt = 4.14 X IO7 eV. This is much larger than the correction term, by a factor of about IO4. From this typical example we may thus conclude that recoil effects due to momentum conservation in atomic and molecular transitions do not hinder resonance absorption. As an example of a nuclear transition, consider the case of the 1.33 M eV gamma ray emitted by 00N i. A nickel atom, whose atomic mass is about 60 amu or I X IO25 kg, has a rest energy M c2 = 5.61 X IO10 eV. Also E? — E i = 1.33 X 10® eV. Therefore the correction term which accounts for the nuclear recoil is (E 2 — E i ) 2/2M c2 = 15.8 eV, which is relatively larger (compared with E 2 — E i) than in the atomic ease. The Iifetinte of the nuclear excited state is Xt ~ IO14 s. Hence the uncertainty in the energy of the excited level is X E ~ h/Xt = 0.414 eV, which is smaller than the correction term by a factor of about 38. Therefore recoil effects due to momentum conservation in nuclear transitions are important. These effects make it impossible, in many cases, for a nucleus to absorb the gamma photon emitted by another similar nucleus. E X A M P L E 1.11.
Discussion of the Mossbauer effect.
S o lu tio n : In the previous example we saw that in nuclear transitions the energy uncer tainty due to the finite life of an excited nuclear state is much less than the recoil energy of the nucleus, thereby making it impossible for a Ύray photon emitted by a nucleus to be absorbed by another identical nucleus. However, under special circumstances, recoil effects can be reduced by a large factor. This is possible when the emitting and absorbing
46
The foundations of quantum physics
nuclei are bound in a crystal lattice and the conditions are such that the whole crystal recoils, instead of just a single atom recoiling. Then the mass is so large that the recoil energy is very small compared with E 2 — E j. Thus resonance absorption may occur, resulting in the socalled Mossbauer effect, which was observed for the first time in 1958 by the German physicist R. L. Mossbauer.
j
t j
·
Iψ
Experimental arrange ment for measuring the width of an energy level. F ig . 131.
Absorber
'V
"ιγ
_
/r^I)etector
The Mossbauer effect has been used to investigate several important physical proper ties. For example, by means of the arrangement of Fig. 131, the natural energy width A E of a nuclear state can be determined. A 7ray source is mounted on the rim of a turn table whose velocity can be adjusted. When the source is at .1 its radiation passes through the hole in the shielding and falls on an absorber, which is composed of atoms of the same material as the source. Both emitter and absorber are embedded in crystals to essentially eliminate recoil effects. When the emitter is at rest relative to the absorber, resonance absorption is observed. But if the turntable is set in motion resonance absorption becomes impossible. This is due to the Doppler shift in the frequency of the emitted 7ray as a result of the motion of the source relative to the absorber.
Absorption curve
Fig. 132. Detector current as a function of the turntable rim velocity. 8
6
4
2
0
+2
+4
+G
+8
I), cm s“ 1
Figure 132 shows the intensity of the transmitted radiation. Note that maximum absorption occurs at zero relative velocity, and that the absorption decreases when the rel ative velocity increases in either direction. For a relative velocity of about 4 cm s1 , corre sponding to a change in the Doppler frequency of about Av ~ v(v/c) = 1.33 X IO l l V, or a change of energy A E ~ 1.33 X IO l o Zi1absorption is practically negligible, indicating that the energy width of the state is about half as great.
Problems
47
K e fe re n c e e 1. “ E arly Work in Electron Diffraction,” G. Thomson, Am. J . Phys. 29, 821 (1961) 2. “ T h e Scattering of Xrays as Particles,” A. Compton, Am. J . Phys. 29, 817 (1961) 3. "E instein’s Proposal of the Photon Concept,” A. Arons and M . Peppard, Am. J. Phys. 33, 367 (1965) 4. “ 60 Years of Quantum Physics,” E. Condon, Physics Today, October 1962, page 37 5. “ Paths to Quantum Theory Historically Viewed,” F. Hund, Physics Today, August 1966, page 23 6. “ T h e Mossbauer Effect,” S. de Benedetti, Sci. Am., April 1960, page 72 7. Introduction to Modern Physics, F. Richtmyer, E. Kennard, and T . Lauritsen. New Y o r k : McGrawHill, 1955, Chapter 3, Sections 43, 44, 4955; Chapter 4; Chapter 5, Sections 86 and 87; Chapter 6, Sections 9095 8. Great Experiments in Physics, Morris Shamos, editor. N ew Y ork : Holt, Rinehart, and Winston, 1959, Chapter 17 (Einstein); Appendix 2 (Planck); Appendix 5 (Compton) 9. The Feynman Lectures on Physics, Volume I, R. Feynman, R. Leighton, and M . Sands. Reading, M ass.: AddisonWesley, 1963, Chapter 37 10. Foundations of Modern Physical Science, G. Holton and D. H. D. Roller. M ass.: AddisonWesley, 1958, Chapters 31 and 32
Reading,
i*r o b le m s 1.1 When an electron is accelerated through a potential difference of I volt its gain in kinetic energy is called I elec tron vo lt (eV ). (a) Show that I eV = 1.602 X IO l o J. (b) W hat is the energy increase of an electron when it is acceler ated through 10 V, 50 kV, and I M V ? (c) Assuming that the electron starts from rest, calculate the final velocity. 1.2 A gaseous source emits light of wave length 5 X IO7 m. Assume that each molecule acts as an oscillator of charge e and amplitude IO10 m. (a) Compute the average rate of energy radiation per mol ecule. (b) I f the total rate of energy radi ation of the source is I W, how many molecules are emitting simultaneously? 1.3 Estimate the value of (dE/dl)aye as given by Eq. (1.5) for a proton in a nu cleus. Assume Zn of the order of IO15 m and ω about 5 X IO20 Ilz for lowenergy gamma rays.
1.4 I t can be shown from M axwell’s equations that the electric and magnetic fields in a plane electromagnetic wave are related by 8 = cffi. Show that the energy density in the wave may be written as E = «o8 2. The intensity I of the wave is equal to the energy flowing, per unit time, across a unit area perpendicular to the direction of propagation. Show that I = ceo62. 1.5 The momentum density in an elec tromagnetic wave may be written /> = eo£ X (B. Show that this expression has the units of momentum per unit volume. Also show that for a plane wave, E = pc. [M int: Recall the relation between 8 and a, Schrddinger’s equation with Ep(x) = 0 becomes r dV „. — 2m dx2 =
°r
dV d ^ 
where a2 = Im E Ih
2
= ImEb/h2.
2
^
.
.. = 0’
78
(2.7
Quantum mechanics
When we recall the discussion of the potential step in Section 2.4, we see that the solution of this equation is Ψ ί(χ) = De " . N ext we must satisfy the continuity conditions of the wave function at x = a; that is, ^ t = Ψ2
and
(/ψι/dx = d^/dx.
T hey yield CsinAqa — De00
and
AqCcosAqa = —a D e ~ °a.
Dividing these two equations to eliminate the constants C and D, we obtain Aq cot Aqa = —a
(2.23)
or, introducing the expressions for Aq and a, we have [2m(Co 
Ε ύ /h] 1' 2 cot [2m(A?o 
& )/ A l1/2a =  [ 2 mEb/H]'12.
In this equation the only quantity that has been left arbitrary so far is the energy Eby Thus it expresses a condition for the possible energy levels inside the well. However, since it is a transcendental equation, it is difficult to obtain Eb in a closed form, as we did for the potential box. The situation is very similar to that described before in connection with the square well (Example 2.5). Thus, depending on the depth Eo of the well, there may be none, one, two, etc., possible energy levels. I t can be verified that if E o a 2 < π 2ft2/Sm ,
there is no bound state; if ■K2 h 2 /8 m < Eoa 2 < Qn2H2/ 8m , there is only one bound state; if Qtc2H2/Sm
0). VVe note that they do not correspond to bound states, since a particle may now move between x = 0 and i = oo. Again using k f = 2m(Eo + E)/h2, so that p, = hki is the momentum in side the well, we may write the solution for x < a as before; that is, φ\(χ) = C sin kiX. For x > a, where E p — 0, Schrodinger's equation is now h~ drf/
2m djfi ~
ά 'φ
°Γ
2.
n
d ^ + * * = 0
where k2 = 2m E / fr and p — hk is the free particle momentum outside the well. The solution to this equation is of the type given by Eq. (2.7). However, in this case it is more convenient to write it in the equivalent form* ψ ί(χ ) = O sin (At + 6). The quantity δ is called the phase shift; its physical meaning will be explained shortly. Applying the continuity conditions of the wave function at x = a, we obtain ki cot k,a = k cot (ka 4 5).
(2.24)
This equation differs from Eq. (2.23) in that it contains, among other things, an arbitrary quantity 5, the phase shift. Thus we can always satisfy Eq. (2.24) for any energy E (or momentum hk) by properly choosing the phase shift 5; therefore the positive energy spectrum is continuous, as expected. We can understand the origin of the phase shift in the following manner. I f the poten tial energy were of the type considered in Fig. 25, and shown (reversed) in Fig. 222(a), the wave function would have been ψ ~ sin kx ~ e'kx — e ~ 'kx which contains the incident and reflected beams of particles, both having a wavelength λ = 2ir/k. But when we in troduce the potential well (Fig. 222b), the wave function is distorted in the region 0 < x < a, and, although the wavelength at z > a is still 2ir/k, the curve sin kx must be moved along the Yaxis the distance δ/k so that it smoothly joins with the wave func tion inside the well at x = a (which has a different wavelength λ, = 2π/kf). In other words, a local modification of the potential energy between x = 0 and x = a affects the whole wave function. This is expressed by a phase shift δ for x > a. * W e may see that the two forms are equivalent if we recall that Ciik l — cos kx ± i sin kx. Then E 0
Classical mechanics requires th at a p article com ing from the le ft w ith
2.8)
Potential barrier penetration
81
*(x)
F ig. 224. W ave function corresponding to the potential barrier of Fig. 223 for an energy less than the height of the barrier.
an energy E < E
0
should be reflected back a t x =
0, as discussed before in Sec
tion 2.4. H o w ever, when w e consider the problem according to quantum mechanics b y obtaining the solution o f Schrodinger’s equation for regions ( I ) , ( I I ) , and ( I I I ) , w e find th at the w ave function has, in general, th e form illustrated in Fig. 224. Its com ponents are o f the form ψι =
A e ikx + B e ~ ikz,
φ2 =
Ceax + D e ~ ax,
where k and a arc the same as in Section 2.4; th at is, k 2 = 2 m (E 0 — E )/ h 2.
φ3 =
A 'e ikx,
2m E /h 2 and a 2 —
T h e w a v e function ψχ, as in previous cases, contains the inci
dent and the reflected particles; ψ2 decays exponentially, but it must also contain the p ositive exponential, since the barrier extends on ly up to x — a and the posi tiv e exponential is not necessarily to be excluded, as it was in the case o f the potential step.
Because f
2
is not y e t zero a t x =
a, the w a ve function continues
into region ( I I I ) w ith the oscillating form ψ2, which represents the transm itted particles which have the same energy as the incident, particles, but an am plitude A ' which, in general, is different from A . Since ψ 3 is not zero, there is a finite p rob a b ility o f finding the particle in region ( I I I ) .
In other words, it is possible f o r
a particle to go through the potential barrier even i f Hs kinetic energy is less than the height o f the potential barrier. W h en E > E 0, the classical description o f th e process indicates that all particles should cross the poten tial barrier and reach the righthand side.
H ow ever, in
quantum mechanics, for the same reason as in th e case o f the potential step, som e particles are reflected a t x =
0 and a t x =
a.
H ence the w a ve functions in the
three regions are now ψι = w here k ' 2 =
A e ikx + B e ~ ikz,
φ2 =
Ceik'x +
D e ^ 'f ,
φ3 =
A 'e ikx,
2m (E — E 0 )/ h 2 and hk' is the m om entum o f the particles while they
are crossing the poten tial barrier.
82
Quantum mechanics
(.2.8
E/Eo
F ig. 225.
Transmission coefficient of a rectangular potential barrier. I
B y ap p lyin g the boundary conditions a t x =
0 and x =
a, we can determ ine the
coefficients B 1 C , D , and A ' in terms o f A both in the case o f E < E 0 and in the case o f E > E 0. T h e transparency o f the barrier is defined as T = \A'\2/\A\2. I t varies w ith the energy E , as shown in Fig. 225. N o te th at for E > E 0 theref is p erfect transmission (T =
I ) for certain values o f E f E 0 which correspond to a
w avelen gth o f the particle while it is crossing the potential barrier given b y λ ' = '2ic/k\ which is equal to a m ultiple o f 2a. T h is is considered a resonance effect. F igu re 226 illustrates a m ore general case. A ccord in g to classical mechanics, if the p article has an energy E it m ay m ove between A and B or between C and D . I f the particle was in itia lly between A and B it w ill never be found between C and D , and conversely.
B u t again in quantum mechanics the logic is different.
W hen
w e solve Schrodinger’s equation for this potential, w e obtain a w a v e function φ,
F ig. 226.
General case of a potential barrier.
Potential barrier penetration
2.8)
F ig. 227.
83
Shape of the wave function in the general ease of a potential barrier.
as shown in F ig. 227.
W e expect, front our previous examples, th at between A
and B the w a ve function w ill be oscillatory, but to the right o f /f it, must v a ry exponentially.
H o w ev e r at C the w a v e function still has a finite value.
B etw een
C and D , which is a classically allowed region, the w a ve function becomes oscilla to ry again. T o the righ t o f I ) the w a ve function decreases once m ore exponentially. T h erefo re w e conclude th at there is a finite p rob ab ility that the particle will be found either between A and B or between C and D .
In other words, if the p article
is in itially between A and B, a t a later tim e it m ay be between C and D , and con versely.
F o r this to happen, the p article m ust penetrate the potential barrier be
tween B and C. P en etration o f a poten tial barrier has no analog in classical mechanics because it corresponds to a situation in w hich a particle has a negative kinetic energy or an im agin ary m om entum . T h is should not present any difficulty in quantum m echan ics, how ever, since w e m ust n ot insist on using the logic o f classical mechanics in the atom ic dom ain. B arrier penetration has been observed in m any situations, o f which w e shall discuss two. T h e m olecule o f am monia, N H 3, is a pyram id w ith the N atom a t the v e rte x and the three H atom s at the base, as shown in F ig. 2 2 8 (a ).
O bviou sly the N
F ig. 228. (a) Inversion motion of the nitrogen atom in the ammonia molecule, Potential energy for the inversion motion.
(b)
84
Quantum mechanics
(2.8
F ig. 229. (a) Potential energy for an electron in a metal with no external electric field, (b) Resultant potential energy when an external electric field is applied.
atom m ay be at one o f the tw o sym m etric equilibrium positions N and N ' on either side o f the base o f the pyram id.
Since both N and N ' must be equilibrium posi
tions, the potential energy for the m otion o f the N atom along the axis o f the pyram id must have tw o m inim a and have the sym m etric shape indicated in F ig. 2 2 8 (b ), w ith a potential barrier between N and N '. I f the N atom is initially a t N , it m ay even tually leak through the potential barrier and appear a t N '.
If
the energy o f this m otion is less than the height o f the potential barrier, such as th e energy level E in the figure, the m otion o f the N atom is composed o f an oscil lato ry m otion between R and S or between T and U , depending on which side o f the plane it happens to be, plus a much slower oscillatory m otion between the tw o classical regions passing through the potential barrier. T h e frequency o f this second m otion is 2.3786 X IO 10 I I z for the ground state o f X H 3. I t is this second ty p e o f m otion which w e use to define a tim e standard w ith atom ic clocks. A s another example, let us consider the case o f the electrons in a m etal.
T h e ir
poten tial energy is shown in F ig. 2 2 9 (a ). I f an electron at the highest energy level is to escape from the metal, an am ount o f energy at least equal to φη must be sup plied to the electron. T h is am ount o f energy m ay be provided b y heating the m etal or b y the absorption o f a photon (as in the photoelectric effect).
B ut the electron
m a y also escape if one applies an external electric field whose potential is shown by the dashed straight line in Fig. 2 2 9 (b ). T h e resultant poten tial energy is illus trated by the solid line in Fig. 2 2 9 (b ), and is equ ivalent to a potential barrier. T h e m ost energetic electrons in the m etal can leak ou t through the barrier, result ing in what is called field electron emission. In Section 8.3, w e shall discuss another barrier penetration effect, the emission o f alpha particles b y nuclei. E X A M P L E 2.7. Analysis of wave functions and energy levels for the inversion motion of nitrogen in the XtH 3 molecule. S o lu tio n : T o show how much useful information we can derive from a qualitative analysis of the wave functions, we shall now discuss the inversion motion of nitrogen in the N H 3 molecule in more detail. Figure 230 reproduces the potential energy shown in Fig. 228(b). Let us look at this potential energy as if it were the potential energy of a
2.8)
Potential barrier penetration
85
F ig. 230. Potential energy for the in version motion in N H 3.
F ig. 231. W ave functions corresponding to the four lowest energy levels of the inver sion motion in N H 3.
86
Quantum mechanics
(2.8
State la
State Oa
F ig. 232.
Probability densities corresponding to the wave functions shown in Fig. 231.
simple harmonic oscillator with a bump or barrier at the center. The effect of this central bump is a perturbation that affects the motion of the particle, mainly as it passes through the center. As indicated before, this effect— for particles with energies such as E 1 smaller than the height of the potential barrier— is to decrease the probability of finding the particle in the central region. VVe can translate this into quantummechanical language by saying that the bump distorts the wave functions of the harmonic oscillator potential in the central region, decreasing their amplitude in that region. Figure 231 shows the first four wave functions for the harmonic oscillator potential without the bump as dashed lines. The actual wave functions, when the effect of the central barrier is taken into account, are shown as solid lines. Figure 232 indicates the corresponding probability densities, \ψ(χ)\2. An analysis of these probability densities shows that ψο[2 and \ T im vIlvp on ilfn t Svh roilinqvr Kquntion
So far w e have discussed on ly the space distribution o f the m atter field, a n d have com puted its am plitude or w a v e function f i x ) . N o th in g has been said about the tim e variation o f the field. T o discuss tim e dependence, w e must have an equation containing the tim e; this equation is called Schrodinger’s limedependent equation. I t plays a role in quantum mechanics sim ilar to M a x w e ll’s equations in electro magnetism. Schrodinger’s tim edependent equation is: A2 O2V
, „
. ,T
2m d x 2
.. 3 * —
Ot ’
^
^
where w e now w rite T (x , I) instead o f ψ (χ ) for the w ave function, so that w e in clude both tim e and space dependence.
W e see th at this equation for the m atter
field, Ί^ χ , t ), differs radically from the w a v e equation O2 Zfd t 2 =
v 2 O2 ZfO x2, which
describes elastic or electrom agnetic waves, because it contains on ly the firstorder tim e d e riv a tiv e ΟΨ/ΟΙ and not the secondorder tim e d erivative. In a sense Ecp (2.27) resembles the equation fo r transport phenomena, OZfOt =
a 2 O2 ZfO x2,
which is also o f first order in the tim e d e riv a tiv e and describes heat conduction and diffusion. i =
H ow ever, there is again an im portant difference: the im aginary factor
\/— I, which m ultiplies the tim e d e riv a tiv e in Schrodinger’s equation. T h e re
fore the field described b y E q. (2.27) is different from elastic or electrom agnetic w aves on the one hand and diffusion or therm al w aves on the other. W e m ust re frain from carryin g the an alogy between quantum m echanical fields and elastic and electrom agnetic w aves too far. N o te th at w e have on ly w ritten down Schrodinger’s tim edependent equation and have not derived it in the same w ay that w e can establish the w a ve equation fo r elastic w aves in a string or for electrom agnetic waves, starting from m ore basic principles.
W e could now say th at E q. (2.27) is a fundam ental law o f nature, in
the same w ay as the F aradajrH en ry law o f electrom agnetic induction or the prin ciple o f conservation o f m om entum are fundam ental laws o f nature, and that its m athem atical form is the result o f a careful analysis o f experim ental facts and in tu itiv e thought.
T h is statem ent is essentially correct.
H ow ever, w e can in fact
obtain Schrodinger’s equation if w e start from certain basic assumptions.
T h is is
done in the form al th eory o f quantum mechanics, whose discussion is b ejrond this book; however, Section 2.12 does g iv e an introdu ctory analysis. W c m ay point out here th at the fact th at Schrodinger’s tim edependent equation is o f first order in the tim e d e riv a tiv e and o f second order in the position d e riv a tiv e m ay be d irectly
The limedependent Schrodinger equation
2.10)
91
related to the fact th at the energy o f the particle depends on the square o f the mom entum through the relation E =
p 2 / 2 m + E 1,(x ).
N e x t w e shall tr y to obtain a solution o f E q . (2.27) which is adequate to describe stationary states.
F rom our knowledge o f standing waves, w e m ay assume that
such a solution must Iiave the position and tim e variables separated.
W e shall
try a solution o f the form * { x , t) =
ψ ( χ ) β  ίΕ ιι\
(2.28)
N e x t w e m ust p ro ve th at E is the energy and ψ (χ ) is the am plitude satisfyin g Eq. (2.3). T o sec this w e note that θΦ _ _
=
_
_
i E*
W
,e  iE im
.
O2V _ =
d2\p e —
iEtin
Substituting these tw o expressions in Eq. (2.27) and canceling the com m on factor eiE tir,' w e 0btain p;() ( 2.3) for ψ {χ ), and therefore E m ust be the total energy o f the system. T h erefore the m atter field given b y E q. (2.28) oscillates w ith an angular frequency ω =
E/h
or
E =
Ιιω 
hv,
in accordance w ith Eqs. (1.44) and (1.45). A s indicated before, the field expressed b y E q. (2.28) is typical o f standing w aves because the space and tim e parts are included in d ifferent factors.
H ow ever, the im portant difference, when com pared
w ith standing w aves on strings, in air columns, or in m etallic cavities, is that the tim e part cannot be w ritten as sin ωί o r cos oil, but on ly as e ~ 'wt = e ~ /Λ, and this is because E q. (2.27) is o f first order in time. T h a t is, quantum m echanical w aves always have a com plex tim e dependence. F o r a free particle m ovin g in the + Y d ir e c tio n w ith m om entum p = have ψ (χ ) =
hk, we
.4e‘*x. Thus the tim edependent w a v e function is * ( x , 0 = i { x ) e  iE,!h =
A e iikl * 0 ,
(2.29)
which describes a w a ve p ropagating in the K Y direction w ith a phase v e lo c ity Vp
 ω/k =
E / p — \v, as w e discussed in Section 1.11.
T h is corroborates the
statem ent m ade in Section 2.3 th at th e w a v e function ψ (χ ) = e‘kjr corresponds to a particle m ovin g in the K Ydirection. S im ilarly, the tim edependent w a v e function corresponding to ψ (χ ) = e ~ 'kx is '!'(x , t) = A e  ,{kz+ut), which describes a w a ve propagating in the n egative Y direction. H o w ever, w e cannot use cos {kx — ω ί) or sin {kx — ω ί) instead o f E q. (2.29) to express V (x , I) for a free particle, because these functions are not solutions o f E q . (2.27). F o r a particle in a poten tial box, . / \
V (X ) =
j
·
„ TlTTX
,1 sm —
=
a / in *xla A \B
C
— inr x/ a\ )
 2Ϊ
and Ψ(χ, I) =
A sin
e ~ iE " K = j£. e * " " ' 
E t/ fi ) __
* (n x x / a  f Et/ ti)
.92
Quanlum mechanics
(.2.10
T h is equation corresponds to tw o w aves tra velin g in opposite directions which result in standing waves, as in the case o f a v ib ratin g string w ith fixed ends. A solution o f the form given b y E q. (2.28) describes a particle m ovin g w ith a welldefined energy, corresponding to one o f the energy levels o r stationary states. In this case,  *(x , O l2 =
[ r ( x ) e iE ll* M x ) e  iE" * ] =
*(*)2,
(2.30)
so that the p rob ab ility d ensity Φ(χ, , we have that the rate of energy radiation is
d li
r,
r
ε ~ω >
i t = h u T it = w
I
I2
A fter comparing this with the result of Example 1.1 for a classical oscillator, we conclude that during the transition the electron behaves as an electrical dipole of angular frequency ω and an amplitude 2[r,/.
R p fo rv m O H 1. “ QuantumMechanical Tunneling," J. Former, Am. J . Phys. 34, 1168 (1966) 2. “ The Philosophical Hasis of Bohr’s Interpretation of Quantum Mechanics,” K. IIall, Am. J. Phys. 33, 624 (1965) 3. “ Quantum Mechanics as a M odel,” T . Bucht Am. J . Phys. 34, 653 (1966) 4. “ Interpretation of Quantum Mechanics and the Future of Physics,” E. Witmer, .lm. J. Phys. 35, 40 (1967) 5. Elements of ll'ave Mechanics, N. M ott. 1962, Chapters I, 2, 3, and 4
Cambridge: Cambridge University Press,
6. Introduction to Modern Physics, F. Richtmyer, E. Kennard, and T. Lauritsen. Y ork: McGrawHill, 1955, Chapter 6, Sections 96102
New
7. Elementary Wave Mechanics, W. Heitler, second edition. Oxford: Oxford University Press, 1956, Chapters I and 2 8. Atomic Spectra and Atomic Structure, G. Herzberg. Chapter I
New Y ork: Dover Press, 1944,
9. Structure of Matter, W. Finkelnburg. N ew Y ork : Academic Press, 1964, Cha))ter 4 10. Quantum Theory of Matter, J. Slater. N ew Y ork: McGrawHill, 1951, Chapters 1,2, and 3 '
P r o M o n tH
2.1 A particle is represented by the wave function ψ (χ) = sin kx. Plot the wave function φ (χ ) and the probability distribution ^(.r)2. Estimate the uncer tainty in the position and in the momentum of the particle. 2.2 Show that RJT = I for a stream of particles impinging on a potential step of height Eo when E > Eo (refer to Ex ample 2.2).
2.3 W hat is the effect on the energy levels of a onedimensional potential box as the size of the box (a) decreases? (b) increases? 2.4 Consider an electron in a onedimen sional potential box of width 2.0 A. Cal culate the zeropoint energy. Using the uncertainty principle, discuss the effect of incident radiation designed to locate the electron with a 1% accuracy (that is, Ax = 0.2 A).
Problems 2.5 Estimate the zeropoint energy of an electron confined inside a region of size IO14 m, which is the order of magnitude of nuclear dimensions. Compare this energy with both the gravitational potential energy and the coulomb potential energy of an electron and a proton separated the same distance. On the basis of this comparison, discuss the possibility that an electron can exist within a nucleus. 2.6 Calculate the zeropoint energy of a neutron which is confined within a nucleus which has a size IO15 m. 2.7 Show that the fractional energy dif ference A E / E between any two adjacent levels of a particle in a box is given by ( 2n + I )/n2. 2.8 Normalize the wave functions for a particle in a potential box, given by Eq. (2.18), showing that C = V 8 /abc = V 8/F, where V is the volume of the box. 2.9 Prove that the threedimensional wave functions given in Eq. (2.18) are orthogonal. 2.10 Plot the function g(p) given in Ex ample 2.4 versus p, and compare with Fig. 212. Explain the difference. 2.11 Show that the volume of phase space corresponding to a particle moving within a region of volume V and having a momen tum between p and p + dp is 4π \'ρ 2 dp. Recalling that the minimum size of a cell in phase space within which a particle can be localized is h3, find the number of quan tum states with momentum between p and p + dp accessible to the particle. Compare with the results given in Examjjle 2.4. 2.12 Show that the energy levels and wave functions of a particle moving in the X T plane within a twodimensional potential box of sides a and b are E = (π 2 h2 /2 m) (n 2 /a 2 ) n\/b2) and Φ = C sin (ηιττχ/α) sin (nziry/b). Discuss the degeneracy of levels when a = 6. Find the normalization constant C. 2.13 Referring to Problem 2.12, if the twodimensional box is very large, but has
103
equal sides, find the number of quantum states (a) with energy between E and E T dE, (b) with momentum between p and p f dp. 2.14 Verify by direct integration that wave functions φο and \f/\ for the harmonic oscillator (Table 22) are normalized. Also verify that ψι> is orthogonal to ψ\ and ψ2. 2.15 Calculate the zeropoint energy and the spacing for the energy levels (a) in a 1dimensional harmonic oscillator, with an oscillatory frequency of 400 Hz, (b) in a threedimensional harmonic oscillator with an oscillatory frequency of 400 H z1 (c) in the CO molecule, if the two atoms oscillate with a frequency of 6.43 X IO11 Hz. 2.16 The general expression of the solu tions of Schrodinger’s equation for the harmonic oscillator is = N nH n(OX)e * * '2' 2, where N n = \Ζα/πΐ'2 2"ηΙ is the normali zation constant, a = s/moi/fi, and the functions H n(ax) are called Hermite poly nomials, defined by / / „ (* )
=
( — l ) " e {2
( e £2) .
Write the first four wave functions (n — 0, I, 2, 3) and compare with the expressions given in Table 22. 2.17 The wave functions for a threedimensional harmonic oscillator can be written as ψη, η2 η$(%, V, Z) ~ N niN niA n^ X H ni(OX)H ni(a y )H n3 (Ca)eS S ^ , where the different quantities are as de fined in Problem 2.16. The energy of the oscillator is E = (η + §)Λω, where n = m + «2 + « 3. Show that the degen eracy of each state is g = £ (n f l )(n f 2). Fully write out the wave functions corre sponding to n = 0, I, and 2. Note that the wave function for w = 0 is spherically symmetric. .Vnalyze the shape of the wave functions for n = I.
IOb
Quantum mechanics
Ep
F ig u re 239 Kp
F ig u re 2—10
F ig u re 2—11
Kp
F ig u re 2—12
Problems 2.18 W ia t is the effect on the energy levels of a onedimensional potential well as the depth of the well (a) decreases? (b) increases? Repeat the analysis, con sidering that the width changes but the depth remains constant. 2.19 Discuss the effect on the energy levels and wave functions of a potential well of range a when a hard core of range b is added, as shown in Fig. 238. [H int: See Examples 2.5 and 2.6.) 2.20 Given a potential well of depth E o and width a (Example 2.6) draw the possible wave functions for the bound states if v (a) E oa 2 < Tr2E2/Sm, (b) TT2 fi2/Sm < E oa 2 < 9ir2A2/8m, (c) Ihr2 Ii2 /Sm < E oa 2 < 25rr2h2/Sm. 2.21 Estimate the well depth E o for a neutron in a onedimensional rectangular well of width 3 X IO15 m, given that its binding energy Ei, is 2.0 MeV, and assum ing that only one energy level is possible. 2.22 Equation (2.23), Zr, cot £,a = —a, may be solved graphically in the following way: Define £ = k,a, η — a a, and P = (V 2 m E o / h )a . Therefore £2 + η2 = P2. Also from Eq. (2.23), η = — ξ cot ξ. With £ as abscissa and η as ordinate, plot η = — £ cot £ and show that the circle £2 jT V2 = p 2 intersects the curve once, twice, three times, etc., depending on the value of p. The points of intersection cor respond to the allowed values of E o a 2. Show that these values are consistent with the values of E oa 2 given in the text.
105
2.25 Repeat the preceding problem for a case in which the particles are initially in cident from the right. 2.26 Consider the potential energy shown in Fig. 240. Discuss the general mathe matical expression and the shape of the wave function for a particle incident from the right when its energy is (a) E < Eo, (b) E 0 < E < E'o, and (c) E > E'o. R e peat the last two cases for a particle which is incident from the left. 2.27 Write the wave function in each of the regions of the potential energy shown in Fig. 241. Also sketch the wave func tions. Consider a particle incident from the left, first with E < Eo, and then with E > EoIs there a possible stationary state for a particle initially in region (3)? 2.28 Write the wave function in each of the regions of the potential energy shown in Fig. 242. Also sketch the wave func tion. 2.29 A particle moves rectilinearly under the action of a uniform electric field S so that its potential energy is E p = — Sx. (a) W'rite the Schrodinger equation for this motion, (b) Make a sketch of the wave functions for an energy E larger and smaller than zero. Is the energy quantized?
2.23 For the square potential barrier of Fig. 223, determine the coefficients B, C, D, and .1' in terms of the coefficient A for (a) E < E o and (b) E > Eo
2.30 Given the potential energy shown in Fig. 243, sketch the wave functions for a particle coming from the right and having a total energy that is (a) negative, (b) be tween zero and Eo, (c) larger than Eo. H ow does the wavelength of the particle change as the particle moves in the regions x > 0 and x < 0, if the total energy is larger than E o l Repeat for a case in which the particle comes from the left. This situation corresponds to the potential energy of an electron in a metal when an external electric field G is applied.
2.24 Write the wave functions in each of the regions of the potential energy shown in Fig. 239. Also sketch the wave func tions. Consider that the incident particles come from the left and discuss the three distinct cases in which E < Eo, E o < E < E'o, and E > E'0.
2.31 A particle moves under the potential energy E p(x ) = — Foe " ' 2, (a) Plot E p(x ). (b) Make a sketch of the wave functions when the total energy is both negative and positive, (c) Do you expect to have quan tized energy levels for certain ranges of the energy? (d) Estimate the zeropoint energy
i06
Quantum mechanics 'j I'
E„
F ig u re 245
of the particle. (H in t: N ote that for ener gies close to — Eo the particle oscillates with a frequency ω = [(d 2 E p/dx2 )/ m ]U2.) 2.32 Show that for the Hermite poly nomials defined in Problem 2.16, / / .( ί) = if n is even and ΙΙΛ ΰ
= U n it)
if n is odd. Conclude from this that the wave functions for the onedimensional harmonic oscillator have even or odd parity depending on whether n is even or odd. Also show that the parity of the threedimensional harmonic oscillator of Problem 2.17 is even (odd) depending on whether n is even (odd). 2.33 Show that the wave functions of a particle in a potential box of Eq. (2.18) have even (odd) parity if n = ni + «2 + «3 is odd (even). 2.34 Given the potential energy shown in Fig. 244, write the wave function for each region and sketch the wave functions and energy levels for E < Eo and E > EoN ote that the wave functions are symmet ric (even) and antisymmetric (odd). Dis cuss the effect on adjacent energy levels as either Eo or b becomes very large. 2.35 A particle moves with the potential energy shown in Fig. 245. Plot the wave
functions and the probability distributions for the three lowest energy levels. Recog nize that the three levels have very close energies. Make a general sketch of the energy levels. 2.36 Consider the timedependent wave function for a beam of free particles as given by Eq. (2.29), where \A\2 is the average number of particles per unit volume. Show that the number of particles crossing a unit area per second (i.e., the particle flux) is h\A\2 k/m. 2.37 Given that Φ (*, i) =
+ C2’/'2(z )c _ ‘a'2'/a,
where the proper functions φι (x ) and φ/ζ(χ) are orthogonal and normalized, show that Ci = J ^ t(Z )1POr, 0) dx and ·
Co = j φ * (χ )* (χ , 0) dx. This shows that the wave function at any time can be found in terms of the initial wave function Ψ{χ, 0). Extend the result to the case in which ΊΌ τ, t) = ^
c„\l/n(x )e ~ 'En‘" ‘· n
2.38 Show that the frceparticle wave functions φ (χ ) = e±,k* are proper func tions of the momentum operator corre
Problems spending to respectively.
the
proper
values
±hk,
2.39 Indicate which of the following func tions are proper functions of the operator d/dx: (a) Cirt, (b) eat, (c) sin kx. Indicate in each case the proper value. Repeat for the operator d~/dx2. 2.40 Show that the momentum operator — ifid/dx is hermitian. [H in t: Integrate the expression on the left of Eq. (2.40) by parts, with A replaced by the momentum operator, and take into account the be havior of wave functions at d b » .] 2.41 Find the average or expectation values of x, x2, p, and p 2 for the n = 0 and n = I states of the linear harmonic oscillator. 2.42 Find the average or expectation values of x, x 2, p, and p 2 for the ground and first excited states of a particle in a onedimensional potential box. 2.43 Write out fully the operators corre sponding to the three rectangular com ponents of the orbital angular momentum L = r X p. 2.44 The commutator of two operators A and B is A B — BA and is designated by the symbol [A , B], Show that [x, ρχ\ψ = ilop,
[y, PxWr = o, and [z, p M = 0, where \p is an arbitrary function. Usually the wave function is not explicitly written and the relations are written as [x, px] = Hi, [l/, Pil = 0, and [z, px] = 0. (Commuta tors play a very important role in quantum mechanics.) 2.45 Find the matrix elements 201, * 02, pot, and po2 for the coordinates and
momenta of a linear harmonic oscillator.
107
(onedimensional)
2.46 A particle moves in a onedimen sional potential box of width a. From the normalized wave functions find the matrix elements x„m and p nm. Show also that Pnm = im (XnmI nm, where ω„„, = (E n — E m)/h. 2.47 Find the selection rule for the elec tric dipole transition of a particle in a one dimensional potential box. [H in t: Recall Example 2.10.) 2.48 Show that if of these three functions that are normalized and orthog onal to ψι. 2.49 The curvature of a function ψ (χ) is proportional to d2 \p/dx2. Using Schrodin ger’s equation, show that if E < E p(x ), the wave function \p(x) is concave toward the χΥaxis, and that if E > E p(x ), the wave function \p(x) is convex toward the .Yaxis. Check this result with the plots of wave functions considered in this chapter.
3
ATOMS WITH ONE ELECTRON
3.1 3.2
The Hydrogen Atom
3.3 3A
Introduction
The Spectru
Quantization o f A ngular M om entum
3.5 OneElectron Wave Functions Under Central Forces 3.6
The Zeeman Effect 3.7 Electron S p in
3.8 A d d ition o f A ngular Mom enta 3.9 S p in O rb il Interaction
The hydrogen atom
3.2)
3.1
109
In trod u ction
W e shall begin our study o f atom s b y sum m arizing our fundam ental ideas about atom ic structure.
E v e r y atom has an overall dimension o f abou t IO 9 m.
I t is
composed o f a rela tively m assive nucleus (whose dimensions are o f the order o f IO 14 m ) about which m ove a number o f electrons, each o f charge — e, occupying the rest o f the atom ic volum e. T h e nucleus is composed o f A particles ( A is the mass number) called nucleons, o f which Z are protons {Z is the atomic number), each o f charge + e , and N ( =
A — Z ) are neutrons, which have no electric charge. T h e re
fore the nucleus possesses a positive charge o f + Z e . T h e number o f electrons in an y atom is equal to the number o f protons (th a t is, Z o f them ) and th erefore an atom is an electrically neutral system. H ow ever, in certain instances an atom m ay gain or lose some electrons, so th at it becomes n egatively or p ositively charged; in this case it is called an ion.
T h e mass o f the nucleon is about 1850 tim es the electron
mass. Thus the mass o f an atom is practically equal to that o f its nucleus. H ow ever, the Z electrons o f an atom are responsible for m ost o f the atom ic properties which are reflected in the properties o f m atter in bulk, such as the elastic and electrom agnetic properties of different materials.
E lectrom agn etic
interactions between electrons and nuclei o f different atom s play the basic role in the binding togeth er o f atom s to form molecules, in chemical reactions, and in practically all the properties o f m atter in bulk. W e can explain the m otion o f the electrons around the nucleus if w e consider on ly the electrom agnetic interactions between the electrons and the com ponents o f the nucleus (protons and neutrons). Since electrom agnetic interactions are w ell understood, it has been possible to d evelop an accurate description o f the electronic m otion.
T h e corresponding problem fo r the nucleus, on the other hand, is m ore
complex, since other interactions enter which are not so w ell understood. W hen w e analyze electronic m otion, we must use the m ethods o f quantum mechanics dis cussed in the previous chapter. In this chapter we shall discuss the properties o f atom s and ions having just one electron (o f which the sim plest is the hydrogen ato m ) and in the follow ing chapter w e shall consider the problem o f m anyelectron atoms. T h e oneelectron atom w ill help us to understand the basic problem s related to atom ic structure.
3.2
T he H y d ro g en A to m
T h e sim plest o f all atom s is the hydrogen atom. Its nucleus is composed o f on ly one particle, a proton, so that it has /1 =
1 and Z =
I.
Around the proton, a single
electron moves. So th at our calculation w ill be applicable to other atom s w e shall assume, however, th at the nucleus contains Z protons w ith a to ta l p ositive charge equal to + Z e (F ig . 3 1 ). A t this m om ent w e shall make tw o approxim ations. First, we shall consider the nucleus to be a t rest in an inertial system. T h is is a reasonable assumption because the nucleus, being m ore massive than the electron, practically coincides w ith the center o f mass o f the atom , which certainly is a t rest in an inertial system so lon g as no external forces act on the atom. Second, w e shall assume that the electric field o f the nucleus is th at o f a point charge.
T h is is also reasonable,
HO
(.3.2
Atoms with one electron
since the nucleus has a v e ry small size (abou t IO14 m ), com pared w ith the average distance o f the electron from the nucleus (abou t IO10 m ).
H ow ever, in a m ore
refined analysis, the size and shape o f the nucleus would h a ve to be taken into account.
Electron Fig. 31.
An electron moving around a nucleus.
T h e m otion o f the electron relative to the nucleus is determ ined b y the coulomb interaction between the two. T h is interaction is expressed b y an a ttra c tiv e inversesquare central force acting on the electron, given b y
*■ = 
4
^

T h e poten tial energy o f the electronnucleus system is then


Π 5 Γ
 9 7 ! I' £ ) 3
I
2
* ■ ■ < '>  m
I' S
* “ r, K — Θ, ir + φ. Analyze the behavior of the angular wave functions given in 'fable 34 and confirm the rule given in the text that the parity of the F im, functions is ( — 1)'. Repeat for the wave functions appearing in Table 35. 3.16 (a) Show that the angular wave functions of Table 35 may be obtained by a proper linear combination of the wave functions in Table 34. (b) Express the wave functions of Table 35 in terms of x, y, z, and r and justify the notation used for their description. 3.17 Express the operators L1 and Lv in spherical coordinates. Using these values, together with Eq. (3.21) for L1, obtain expression (3.23) for L2. 3.18 W rite the radial equation (3.24) for a free particle (E p = 0). Show, by direct substitution into the equation, that the solutions for I = 0 and I = I satisfying the requirement that u = 0 for r = 0 are u = sin At and u = (sin kr)/kr — cos Ar, where A2 = 2mE/h2. Write the complete free particle solution φχ„ (r ), including the angular part, for I = 0 and 1 = 1. Inter pret the wave function I = 0 as a combina tion of an incoming and an outgoing spherical wave. [.Vote; I t can be shown that the general wave function for a free particle of momentum p = hk and orbital angular momentum L — +1(1 + I )h is φ ι„ = j i ( k r ) Y tm(e, φ), where Ji(Ar) is called the spherical Hessel function of order I. The student should consult a mathematical handbook to investigate the Bessel functions and obtain
146
Atoms with one electron
jo (k r) and ji( k r ), comparing these with the results obtained in the first part of this problem.) 3.19 I t can be shown that the wave function for a free particle moving along the ifaxis with momentum hk may be expressed as
I
ψ = e
3.23 Show, by direct substitution, that R io is a solution of Schrodinger’s radial equation.
ikt
= Y
j
t IV 4 ir (2 f + D j d k r ) Y 10(B).
I W hy are only mj = 0 wave functions allowed? W hy does the solution contain many angular momentum functions? Write the first two terms in the summation in full. (See the previous problem for the definition of j I.) 3.20 The radial wave functions for the hydrogen atom are given by R nM
3.22 From the information given in Problem 3.20, show that the radial wave functions for hydrogen behave as p' for small values of r and that they behave as Pne"'2 for large values of r. Therefore conclude that the larger the value of I, the less penetrating the orbit.
=
N n ie  " ' 2 P1 L f t i 1 (p ),
where p = 2Zr/nao, the IJt are the associ ated Laguerre polynomials, which are de fined as L \(p) = — dp
[L i(P )J ,
where L i ( p ) = e’ — ' [p‘ e~p\ , dp
3.24 Verify that the angular momentum wave functions Y imi(0, φ) given in Table 34 are orthogonal and normalized. (The condition of orthogonality and normaliza tion is that
So S0 * Y*mt Y Pml dil = Si p where dLl = sin Θ d$ άφ and Satl is I if a = b, and zero otherwise.) 3.25 The volume element coordinates (Fig. 326) is
The probability of finding an electron within that volume element is ψ2Γ2 dr dil. Verify that the probability of finding the electron within the spherical shell of radii r and r + dr, regardless of the angular posi tion, is given by /?„i2r2 dr. [H in t: Replace Φ by its expression (3.18) and integrate over all angles, using the normalization condi tion for Τι™, given in Problem 3.24.) Z
v
= _ Γ [\naoJ
( »  * 
2n[(n
Ο ’Γ 2 1) I]3 J
W rite all the radial functions for n = 1,2, and 3, and compare with the expressions given in Table 36. 3.21 Show that the associated Laguerre polynomials, defined in Problem 3.20, are of degree t — s. This shows that s < I. Thus verify the rule that for a given n the maximum value of ( i s n — I.
spherical
dV = dr dS = T2 dr dtt.
and N „ i is the normalization constant given by
in
Problems 3.26 Using the result of the preceding problem, find (a) the most probable dis tance, and (b) the average distance of the electron from the proton in the lsstate in hydrogen. Compare with the Rohr radius of the orbit when n = I. 3.27 (a) Using the volume element given in Problem 3.25, show that the normaliza tion condition of the ψ „ι„, wave functions requires that $o\Rni\2r 2 dr = I. (b) R e ferring tp Fig. 313, show that the area under each curve is equal to one, as required by the normalization condition. [H int: Consider each lobe as a triangle.] 3.28 The average value of r hydrogenlike atoms is given by
in
the
Compute r»„c for all states with n  1,2, and 3. Compare these values with the corresponding Rohr radii. Using ί"ftvo BS fl< measure of the size of the orbit, arrange the nl states according to increasing average distances from the nucleus. 3.29 The radial wave functions have one or more nodes, i.e., regions in which the probability of finding the electron is zero. Find the value of r at which this happens for the 2swave function in Table 36. 3.30 Using Eq. (3.24), show that, regard less of the value of the angular momentum, the radial wave function of a free particle at large distances from the origin of coordi nates is proportional to e iilcrZr1 where k2 = 2mE/h2. Hence the complete wave function can be written as / W e iikr/r, where }(θ ) depends on the angular momen tum. Interpret both signs. 3.31 Write the radial and angular integrals required to compute the matrix elements of z = r cos Θ between states nlmi and n'l'm'i. Show that the matrix elements are zero unless m i = m\. 3.32 Estimate the fractional relativistic correction A ErfEn for the n = 2 energy levels in the hydrogen atom.
147
3.33 Analyze the splitting of p, d, and flevels of a oneelectron atom as a result of the relativistic effect. [H in t: See Example 3.6.] 3.34 Determine the electric current of the electron in the first three Rohr orbits ( n = I, 2, 3). Also calculate the magnetic dipole moment of the electron for each case. 3.35 Draw an energylevel diagram for the 4f and 3dstates of hydrogen in the pres ence of a magnetic field. Show that in the 4f —* 3d transition, the number of spectral lines is three. I f the magnetic field is 0.5 T , would the lines be observable, given that the resolution of a spectrometer is 10_1 1 m? 3.36 The value of e/me may be obtained experimentally by observing the Zeeman effect. Find the value of e/m* if the separa tion between two lines in a field of 0.450 T is 0.629 X IO10Hz. What is the wave length separation for the hydrogen line in the transition n = 2 t o n = l ? Is the wavelength separation for the H a line (n = 3 to n = 2) greater than, less than, or the same? 3.37 The force exerted on a magnetic dipole of moment M by an inhomogenous magnetic field (B of gradient d(&fdz is F — ± M [dR/dz). Given that the gradi ent in a certain region is 1.5 X IO2 T m ~ l , calculate the force exerted on an electron due to its spin magnetic dipole moment. If a hydrogen atom moves I m in a direction perpendicular to such a field, calculate the vertical displacement, given that the velocity of the hydrogen atom is IO5 m s_ I and the electron spin is either parallel or antiparallel to the magnetic field. 3.38 A beam of silver atoms with an average velocity of 7 X IO2 m s 1 passes through an inhomogenous magnetic field 0.1 m long which has a gradient of 3 X IO2 T m 1 in a direction perpendicular to the motion of the atoms. Find the maximum separation of the two beams which emerge from the field region. Assume that the net magnetic moment of each atom is I Rohr magneton.
ihK
Atoms with one electron
3.39 What radiofrequency signal will in duce electron spin transitions to change from parallel to antiparallel orientation (or vice versa) in a magnetic field of IO1 T ? 3.40 Using Eq. (3.39) verify that the numbers given for the spinorbit splitting of the energy levels for hydrogen shown in Fig. 323 are correct. Calculate the same splitting energies for H e +. 3.41 It was indicated in Example 3.8 that the spinorbit interaction causes a preces sion of L and S about their resultant J , which is constant. Show that, in this case, L z and S z cannot have welldefined values even though their sum J z is constant. Consequently, m i and m , are not good quantum numbers, wbile m is a good number. 3.42 Analyze the splitting of the 3dlevel in hydrogen due to a magnetic field (a) when the magnetic field is weak and (b) when it is strong, compared with the spinorbit interaction. 3.43 Discuss the splitting of the lines in the 3d —» 2p transition in the presence of a magnetic field, when the magnetic field is weak compared with the spinorbit inter action. 3.44 The relativistic expression for the energy of a free particle is E 2 = riigc4 + P 2C2 . Write the corresponding relativistic Schro«> o f electron I b y a and the quantum numbers o f electron 2 * This is also a straightforward conclusion from Schrodinger’s equation for a system of independent particles.
152
Atoms with many electrons
b y b, w e must then w rite ^atom =
(4.3)
t a W t b i 2),
resulting in a p rob a b ility distribution l*atom2 =
lAaO¥/.(2)2 =
(4.4)
 ^ ( 1)2^6( 2)2.
D ue to the m otion o f the electrons, the avera ge central field th at one electron produces on the other deviates from the 1/r coulom b field produced b y the nucleus. T h is ob viou sly requires a slight m odification o f the w a ve functions ^ a( I ) and i/'t( 2), which are no longer identical to the hydrogen like w a ve functions.
T h e change
affects the radial part R „ i but not the angular p art Y imi o f the w a ve function be cause the resultant force on each electron is still a central force. U sing appropriate m athem atical techniques, we can optim ize the electron w ave function, and so ob tain the energy levels o f the atom w ith re la tiv ely good accuracy. H ow ever, even if the functions appearing in E q. (4.3) are so optim ized, this ex pression for the w ave function o f the atom cannot be correct. W a v e function (4.3) says th at electron I is in state a and electron 2 is in state b. B ut the w ave function (4.5)
'/'atom = Φα( 2 )ψί>( 1 )
corresponding to electron 2 in state a and electron I in state b, must represent a state o f the same energy as the w ave function o f E q. (4.3) and should describe the state o f the atom ju st as well as E q. (4.3). T h e fact th at the w a ve functions given b y E qs. (4.3) and (4.5) correspond to the same energy is called exchancje degeneracy. N o w , electrons are identical and indistinguishable, and the m ost we can say is th at in the atom one electron is in state a and the oth er is in state b. T h is requires th at the w ave function ψ,Μι,η be such th at i/'»,OJn2 (which gives the p ro b ab ility dis tribution o f both electrons) be sym m etric w ith respect to the tw o electrons, in such a w ay th at both electrons p lay the same role.
N eith e r the atom ic w a v e function
given b y E q. (4.3) nor the one given b y E q. (4.5) m eets this requirem ent. B u t we m ay obtain an adequate atom ic w a ve function (incorporatin g the fact th at the electrons are indistinguishable) b y m aking ap propriate linear com binations o f Eqs. (4.3) and (4.5), which happen to be the functions ’/'atom —
 M l ¥ 6 (2 )
±
M
(4.6)
2 )^ (1 ).
T h e student can see th at in both cases the expression ^at0m2 is sym m etric w ith respect to both electrons.
T h e inform ation contained in the functions g ive n by
Eqs. (4.3), (4.5), and (4.6) is expressed schem atically in F ig. 42.
S ta te b
O
O ·>
2 State a
—
o —
I'  S zi. I f S and M s are the quantum
numbers associated w ith S 2 and S z, then S2 =
S (S + I )ft2,
Sz =
M s h,
M s = ±S, M S  1 ) , . · · , (4.18)
with M s =
D .' msi· Again , to a given configuration there m ay correspond several
values o f S, depending on the relative orientation o f the Si’s. Once L and S arc found, the total angular m om entum for the configuration is given b y J =
L + S. T h e possible values o f the quantum number J (w hich depend
on the rela tive orientation o f L and S ), J =
arc
L + S , L + S — I , . . . , \L — 0 is forbidden b y the law o f conservation o f angular
m om entum , since it can be shown that a photon has an intrinsic angular m om en tum or spin o f one unit (see Section 9.2). transition ./ =
0 —* ./ ' =
H ence it is impossible fo r there to be a
0 with emission (o r absorption) o f a photon carrying
L S coupling
*#.1/
2s 2p3 ( 2P) HS
Hp
 18.50
(  150.3(H) cm” ' )  ^
150
Zs2 2ICt (2I)) 140
HS
Hp
(>— ' I ) 5 ' I )
130
4
120
3D 2s 2p·’
'I )
4=
 3P S p
. ,  3S
1P 3 ___
90
13.01 3D
5D
, —
1D
— 3I)
3D
5I)
jS i
'I··
ο—
6
100
%
3= ·,—
nd
_ _ _ m _____ m . L1H9^ o . ?
'/////. 7 
«5 =%
3D
'l)
2s2 2p3 (4S) Hp
10.93  1S  1D Ip
I" 4
HO
H(I
'M .___ ^ ___ ^ ____ O7 1O
( —136,540 e m ‘)
, 3
3P
Sp
80 .. — 3S 70
IS (2s2 2p4) 30
20 1D (2s2 2p4)
10 0
3P (2s2 2p4)
F ig. 112.
Energy levels of the oxygen atom.
Oil· J unit o f angular m om entum w ith ou t viola tin g the conservation o f angul;
me imentum. A ls o A M = 0 is n ot possible w ith A J = 0, because in this case aga the i conservation o f angular m om entum requires a change in the direction of ich im plies a change in J 2, w ith a corresponding change in M .
«■V
170
Aloms with many electrons T A B L E 1—3
{0.0
Possible mi a n d m, V alu es for the np2 C o n figu ratio n
—I
+1
0
Ml \ +2 + 1
0
(I,+ ) ( ! ,  ) (I , — ) (0,  )
(I , — ) ( — I, — )
(l,+ )(0 ,  )
( I , + K O ,+ )
(I ,  M O , + )
(l,+ )( l,  ) (0 , + ) ( 0 ,  )
(I,+ K  I , + )
(I, K  I , + ) —I
( 0 , — ) ( — I,  )
—2
E X A M P L E 4.3.
( 0 , + ) ( — I,  ) (0 ,  K  I , + )
(0, + ) ( — ! , + )
( — ! , + ) ( — I,  )
Analysis of the configuration np2.
S o l u t i o n : In this configuration, corresponding to I = I, the quantum numbers of the electrons are (n, I, mi, m,) and (n, I, m\, m',). T o comply with the exclusion principle, the two sets of quantum numbers must be different and therefore must differ either in mi or m „ or both. The possible values for mi and m\ are + 1 , 0, — I, and those of m, and m'„ are + J or — J. T o simplify the notation, the spin values will be designated as plus ( + ) or minus ( — ) and the common quantum numbers » and / ( = I) will be omitted. Table 43 presents the 15 combinations of quantum numbers compatible with the exclusion prin ciple. Each entry gives a possible combination (mi, m,) and (m'i, m',). They are arranged according to the common values of M l — m/ + m\ and M s — m, m',. Each set of quantum numbers corresponds to a wave function of the type given by Eq. (4.13). Figure 4—13(a), in which a method developed by J. C. Slater for analyzing configura tions is used, shows the number of different wave functions corresponding to each value of M l and M s From the table or the figure it may be seen that the largest value of M l
M,.
M,.
M,.
O
(d) Fig. 413.
—
L = O1S = D
Slater method for obtaining the terms of the np2configuration.
U s
4.6)
Atoms with one or two valence electrons
F ig . 414.
171
Binding energy of inner electrons as a function of the atomic number.
is 2, and this must correspond to L = 2 or a Dterm. The value M l = 2 is associated with M s — 0 only, and must correspond to S = 0, making up a ’ Dterm. This state requires five wave functions with M l = 2, I, 0, — I, — 2, and M s = 0. These wave func tions are indicated in Fig. 413(b). The remaining ten wave functions are properly grouped in Figs. 413(c) and (d ). Arrangement (c) contains all nine combinations of M i = + 1 , 0, — I , with M s = + 1 , 0, — I, and therefore corresponds to a term with L = I and S = I or 3P. Finally, in (d ), we have the last wave function, with M l — 0 and M s = 0, which must correspond to a term with L — 0 and S = 0 or ’ S. Thus the only possible terms of the np2 configuration are 1S, 1D 1 and 3P.
4 .Ii AiontH irith O n v o r T iro V a lv n v v E lv v tro n s T h e general th eo ry o f a tom ic structure developed in the previous sections can be illustrated in a v e ry sim ple w a y in th e case o f atom s com posed o f com plete shells plus one or tw o electrons. T h e electrons fillin g the com plete shells constitute the core or kernel and the rem aining electrons are called the valence electrons. T h e binding en ergy o f the kernel electrons is higher than th at o f the valence electrons and increases rap id ly w ith the atom ic number, as shown in F ig. 414. T h e figure
172
Atoms with many electrons
gives the binding energy o f Is, 2s, 2p, etc., electrons as a function o f the atom ic number.
Hence, the kernel electrons are rather tig h tly bound and remain prac
tically undisturbed in m ost o f the processes in which the atom participates. I t is the valence electrons th at are m ostly responsible for the chemical properties o f the ato m ; these are the ones that participate in chem ical reactions and chem ical binding. I t can be easily shown th at a com plete shell, filled w ith its full quota o f electrons, necessarily has L =
0 and S = O .
In other words a com plete shell, and b y exten
sion the kernel, does not contribute either to the orb ital angular m om entum or to the spin o f the atom .
H ence the orbital angular m om entum and the spin o f the
atom are both due entirely to the valence electrons.
F o r exam ple, in the case
o f onevalenceelectron atoms, the spin o f the ato m is S = £ and all energy levels in which on ly the valence electron is excited are doublets (2 S + 1 = 2). F o r these atom s L = I , where I is the orbital angular m om entum o f the valence electron. T h e sim plest singlevalenceelectron atom is lith ium ( Z = ou ter or valence electron and Z — 1 =
3), which has an
2 electrons in the kernel. T h e tw o kernel
electrons occupy the level w ith n = I , havin g a configuration I s 2. I f the valence electron m oves in such a w a y th at it does not penetrate the kernel, the effec tiv e field it perceives is th at o f the + 3 e charges o f the nucleus and the — 2e charges o f the kernel, resulting in an effective charge o f + e . T h erefo re the m otion and the energy levels o f the valence electron w ould be sim ilar to those found for the h yd ro gen atom. T h is, according to our discussion o f the w a v e functions in Section 3.5, occurs on ly app roxim ately w h enever the angular m om entum is large. B u t for low values o f the angular m omentum, and especially for the sstates (I =
0) the orb it
o f the valence electron is penetrating and extends through the kernel, even reach ing v e ry close to the nucleus. Figure 415 shows the radial charge distribution o f the lithium kernel, togeth er w ith the radial charge distribution o f the valence electron for the states 2s and 2p. (T h e radial charge distribution is proportional to r 2[I? (r )]2; see F ig. 3 1 3 .)
We
thus see th a t the valence electron in its m otion spends some tim e w ith in the kernel, the penetration being greater for an sorbit than for a porbit. T h e effective charge which the valence electron feels during its m otion in a penetrating o rb it goes from
r2[/f(r) I2
Fig. 415. Radial charge distribution of the kernel and of the valence electron in the 2s and 2pstates in the lithium atom.
4.5)
Atoms with one or two valence electrons I Iy d ro gcn
F ig. 116.
Lithiu m
173
Sodium
Energy levels of hydrogen, lithium, and sodium.
+ e when it is far out up to + 3 c , when it has penetrated deep into th e kernel. T h e corresponding energy is interm ediate between th at o f hydrogen ( Z = I ) and o f L i 2+ ( Z = 3). Since the penetration depends on the angular mom entum , the energy o f the valence electron also depends on the angular momentum.
T h e sm aller the
angular m omentum, the low er (m ore n egative) the energy; i.e., the greater the penetration. So instead o f the le v el scheme shown in Fig. 38, in which all states having the same n have the same energy irrespective o f their angular m omentum, w e have the situation d epicted in F ig. 416, fo r the onevalenceelectron atom s lithium and sodium. T h e en ergy levels o f hydrogen are also shown for comparison. W e see th at the larger the value o f n and o f the angular m om entum I, the closer the energy levels o f lithium and sodium are to those o f hydrogen, because the orbits are less penetrating and the effective coulom b field corresponds to Z =
I.
H ow ever, the m ore penetrating orbits h ave energy levels th at are qu ite distinct from those o f hydrogen. T h e possible transitions o f the valence electron are regulated by the selection rules (3.17), resulting in the follo w in g spectral series: Sharp series: Principal series:
ns —» ?i'p, np —* n's,
D iffuse series:
nd —> n'p,
B ergm ann series:
» f —» » 'd ,
17U
Atoms with many electrons
(4.S 7 X 1 0 + V
4—
1= 0
I
5S/2
£24 2 ±
3Ds/2 I
4P>'2 '
3 l ^ T 6> I ), could be expressed by ... RhcfZ  I ) 2 h — ' R h c Z 2 — xn This expression assumes that the electron in the ground state fully screens one nu clear charge. Discuss the plausibility of this expression. Compute the energy levels for helium when n = 2, 3, and 4, and compare with the experimental result. W hy does the accuracy of the above ex pression for E increase when n increases?
4.2 Write, in determinantal form, the three wave functions ^ s Xa of Eq. (4.10). N ote that the wave function for M s = 0 consists of two determinants. 4.3 The ground state of lithium has the electron configuration Is 2 2s. Write out the wave function in determinantal form for the state when M s = 4.4 For the p2 configuration (see Table 42), write the determinantal wave functions corresponding to M l = 2, M s = 0 and M t = I, M s = I. 4.5 Consider a threeelectron system in an sp2 configuration in the oneelectron approximation. Write the total wave func
180
Atoms with many electrons
tion in determinantal form so that the wave function corresponds to M l = I and Ms = i 4.6 Given that the electron configuration of an atom is 4s 4p 3d, write all possible wave functions which correspond to M l I and M s = \. Use the determinantal form. 4.7 Find the groundstate configuration for the following atoms: (a) Si; (b) M n; (c) Rb; (d) N i. Also write their groundstate term. 4.8 Determine the groundstate configura tion and the number of unpaired electrons in the following atoms: (a) S, (b) Ca, (c) Fe, (d) Br. 4.9 (a) Show, on a diagram similar to Fig. 47, the occupied states for the atoms of Si, Cl, and As when they are in their groundstate configuration. (b) Write the electron configuration and the ground terms of each. 4.10 Calculate the angle between the total and orbital angular momentum for the 4D 3/2 state. 4.11 Find the terms for the following con figurations and indicate in each case which term has the lowest energy: (a) ns, (b) np3, (c) (np2)(n 's), (d) np5, (e) (nd2)(n 'p ), (f) (nd)(n 'd ). 4.12 List all the possible radiative transi tions for oxygen (see Fig. 412) when the excited configuration is 2s2 2p3 ( 2D) 3d ( 3G). 4.13 Look at Figs. 411 and 412. Draw some of the possible transitions compatible with the selection rules. Determine the wave numbers of these transitions and check with experimental values, which may be found, for example, in the Hand book of Chemistry and Physics. 4.14 Calculate the terms for np3 con figuration. 4.15 Give the S, L, and 7values for the terms 1S0, 2S p 2, 'P r , 3P 2, 3F 4, 5D 1, 1D 2, and 0Faz2.
4.16 Find the terms corresponding to the configuration nd2. A pply your result to determine the ground and first excited state of titanium. 4.17 Show that a complete nl shell must necessarily have L = S = O. 4.18 T w o equivalent pelectrons have strong spinorbit coupling. Find the pos sible values of the resultant angular momentum J if the problem is considered as j  j coupled. Repeat the problem, con sidering L S coupling. Does the same value of J appear the same number of times in each case? 4.19 In L S coupling, one can obtain the magnetic moment of an atom as IWave = — (e/2mc)g j, where J refers to the total angular momentum of the atom and g is given by Eq. (3.40), with I, s, and j re placed by L, S, and ./. Find g for calcium and aluminum. Discuss the splitting of a 3p term under the action of a weak mag netic field. Find the splitting if the mag netic field is strong. 4.20 Discuss the Zeeman effect under weak and strong magnetic fields for the transi tions 3F —> 3D and 1F —> 1D in calcium. 4.21 The 4 1D 2 —►4 l Pi transition in cal cium yields a single line at 6439 A. What wavelengths are observed if cadmium is placed in a magnetic field of 1.40 T ? 4.22 The relative separation between the different levels of an L S J multiplet due to the spinorbit interaction can be con sidered as proportional to S  L . Apply the relation to the 3F and 3D multiplets. Draw the energy levels of each multiplet and in dicate, by arrows, the allowed 3F —» 3D transitions. 4.23 Repeat the preceding problem for the case of 4D —* 4P and 4P —> 4S tran sitions. 4.24 Using the values given in Fig. 417, calculate the wavelength separation of the sodium Dlines. From the result, estimate the constant a in E s l — aS · L for the 3 2P states.
Problems
181
4.25 Analyze the 3 2D —» 3 2P transition in sodium. Determine the wavelength separation, if any, for this transition.
4.30 The following AVlines have been measured;
4.26 Compute the energy, in electron volts, of “red” (6500 A ) and of “ blue” (4000 A ) photons. Using the energy scale on Fig. 418, determine which of the sin glet and triplet transitions for calcium are in the visible range of the spectrum. Com pare your rough estimates with a table of wavelengths.
magnesium: 9.87 A ;
4.27 An expression which fits the energy levels of the valence electron for onevalenceelectron atoms is E n =  R h c ( Z — S )2/(n 
δ)2,
where S is the screening constant and δ is the quantum defect (which depends on the n and I values of the particular valence electron). Values of δ for lithium and sodium are:
0.40
P 0.04
0.00
1.37
0.88
0.01
S
Li (Z = 3) N a (Z = 1 1 )
d
sulfur:
5.36 A ;
calcium:
3.35 A ;
chromium: 2.29 A ;
cobalt:
1.79 A ;
copper:
1.54 A ;
rubidium:
0.93 A ;
tungsten:
0.21 A .
Plot the square root of the AVfrequency against the atomic number of the element. A young graduate student of Ernest Rutherford’s, by the name of H. G. Mose ley found, in 1912, an empirical relation ship of the form V v = A (Z — c). (This relation served to clarify the concept of atomic number.) From your plot, verify this relation and estimate the values of A and c. Develop an explanation for this relation of Moseley’s. 4.31 Calculate the wavelengths and ener gies for the K a xray lines of aluminum, potassium, iron, nickel, zinc, molybdenum, and silver. Compare the energy values with those shown in Fig. 414. Use the Moseley function of the previous problem.
For S, one takes a value equal to the num ber of electrons in the kernel. Find the energy of the ground state and the first two excited states of the valence elec tron in lithium and sodium. [H in t: See Fig. 416.]
4.32 The AV lin e for cobalt is 1.785 A . What is the energy difference between the Is and 2porbitals in cobalt? Compare with the energy difference between the Isand 2porbitals in hydrogen (i.e., the first Lyman line). W hy is the difference much larger for cobalt than for hydrogen?
4.28 The transition from the 3)>level to the 3slevel of sodium results in a line with a wavelength of 5890 A . (W e ignore the doublet structure.) Compute the wave length, using the values of the energy levels given by the expression in Problem 4.27, and compare with the experimental value. Repeat for the 2p —* 2s transition in lithium.
4.33 (a) The Aabsorption edge for tung sten is 0.178 A and the average wavelengths of the Aseries lines are K a = 0.210 A , Κβ = 0.184 A , and K y = 0.179 A . Con struct the xray energy level diagram for tungsten similar to that in Fig. 420. (b) What minimum energy is needed to excite the Aseries for tungsten? Deter mine the wavelength of the A0Iine.
4.29 From the information contained in Fig. 414, estimate the energy of the A and /.absorption edges for aluminum and oxygen. Plot the xray absorption curve for these two substances in that region.
4.34 The Αχabsorption edge in tungsten is 1.02 A . Assume that a A„photon is absorbed by one of the 2selectrons in an Auger process. Determine the velocity of the ejected photoelectron.
5 MOLECULES
5.1 5.2 5.3 5A
Introduction
The Hydrogen M olecule Io n
M olecular Orbitals o f D ia tom ic Mo
Electronic Configuration o f Some D iatom ic Molecules 5.5
Polyatom ic Molecules
5.6 Conjugated Molecules 5.7
M olecular Rotations
5.8
M olecular
5.9 Electronic Transitions in Molecules 5.10
Conclusion
The hydrogen molecule ion
5.2)
5.1
183
In tro d u ctio n
An im portant problem related to the structure o f m atter is the structure o f m ol ecules. W h y do carbon atom s com bine w ith four hydrogen atoms, form ing m eth ane, but never w ith three or five hydrogen atom s? W h y is the C O 2 m olecule linear, while H 2O is bent and N H 3 is a pyram id ?
W h y does benzene ( C 6H 6) have the
form o f a hexagon w ith all hydrogen atom s in the plane? H ow can hydrogen atoms join togeth er to form the m olecule H 2 but never form H 3? W h y are the spectra of molecules so com plex when com pared w ith atom ic spectra, ranging from the m icro w ave up to the u ltraviolet? These and m any other questions could not be answered satisfactorily before the quantum th eory was developed.
One o f the m ost spec
tacular successes o f the quantum th eo ry is the answers it provides to m any o f the questions physicists and chemists h ave been asking about m olecular structure. W e cannot say, however, th at our v ie w o f m olecular structure is a closed chapter of physics. L e t us begin our study o f m olecules b y asking ourselves: W h a t is a m olecule? Th e first impulse is to say th at a m olecule is a group o f atom s bound togeth er by some kind o f interaction. B u t this im m ediately raises new questions. D o the atoms conserve their in d ivid u ality? molecule is form ed?
H o w is the m otion o f the electrons affected when a
W h a t kind o f interaction gives rise to a m olecule?
W e m ay
also adopt an opposite v iew b y saying th at a m olecule is a group o f nuclei sur rounded b y electrons in such a w a y th at a stable arrangem ent is produced. T h is is the natural extension o f the concept o f an atom, which is a nucleus surrounded by electrons. T hen instead o f looking at the H 2 m olecule as tw o hydrogen atoms, w e m ay consider it as tw o protons and tw o electrons arranged according to the laws of quantum mechanics and held togeth er b y electrom agnetic forces. F ro m a struc tural point o f v ie w this second approach is more fundam ental than the first. H ow ever, as it is in m any cases, an interm ediate position is the m ost convenient. W hen tw o atom s com bine to form a m olecule, the m ore tig h tly bound, or inner, electrons o f each atom (w hich fill the com plete shells o f their respective kernels) are p ractically undisturbed, rem aining attached to th eir original nuclei. O n ly the outermost, or valence, electrons in the unfilled shells are affected and th ey m ove under the resultant forces o f the ions, com posed o f the nuclei and kernels, as w ell as their m utual repulsion.
T h ese valence electrons are responsible for chem ical
binding and for m ost physical properties o f the molecule. T h is interm ediate posi tion is the m odel w e shall ad op t in this chapter; it is an excellent approxim ation for analyzing m olecular structure.
5.2
T h e H y d r o g e n H loieeule Io n
Th e simplest o f all m olecules is the hydrogen m olecule ion H 2", consisting of tw o protons and one electron.
Chem ists w rite the form ation o f this m olecule as
H + H + —* H 2, where H + is just a proton. In other words, the H 2" m olecule is formed when a hydrogen atom captures a proton. B u t once the H 2" m olecule is formed, it is no longer possible to tell w hich is the hydrogen atom and which is the
18k
Molecules
(5.2 —e ,Λ
\
\
\
\ r2 \ \ \ \
F ig . 51.
The H j molecule ion.
^
r
proton. Our picture o f this m olecule is th at shown in F ig. 5 1 : an electron m ovin g in the electric field o f tw o protons w hich are separated a distance r. W e m igh t say th at the electron does not rem em ber which proton it in itially be longed to.
T h is lack o f m em ory is clearly revealed b y chargeexchange collisions.
Suppose th at the electron is in itia lly associated w ith proton P 1, and proton p 2 approaches the hydrogen atom from the righ t from a large distance (F ig . 5 2 ). In the region o f closest approach the breaks apart.
m olecule lasts on ly a short tim e and then
W h en this happens there is a certain p ro b ab ility th at the incoming
proton w ill tak e the electron w ith it. Thus w e m ay w rite the process as H+ + H
(elastic collision),
H
(chargeexchange collision).
/ H + + H —» H 2\ + H +
T h e rela tive prob ab ility o f the tw o processes shown ab ove can be calculated as a function o f the energy o f the incom ing proton. T h e result checks well w ith experi m ental evidence. H
H+
HJ
II
I>2
1>2
(c)
(d )
F ig. 52. Collision of a hydrogen atom (H ) and a proton( H +), as seen in the CM frame of reference, (c) Elastic collision, (d ) Chargeexchange collision.
5.2)
The hydrogen molecule ion
185
T o discuss the stationary states o f the H J m olecule, we first assume th a t the two protons are a t rela tive rest, and set up Schrodinger’s equation w ith a potential energy Ep = ”
/
 ( 
Aire 0 \
rl
r2
(5.1)
'
r)
Th e first tw o term s in the parentheses g iv e the a ttra ctive poten tial energy between the electron and the tw o protons, and the third term is the repulsive poten tial energy between the tw o protons. T h e variation o f the electronic poten tial energy along the line jo in in g the tw o protons is shown in Fig. 53. I t resembles the double potential w ell considered in Section 2.8 in connection w ith the N H 3 molecule.
Fig. 33. Potential energy along the line joining the two nuclei in H j and H 2.
T h e solution o f Schrodinger’s equation for H 3, w hich is rather com plex, w ill be sketched in E xam ple 5.2. A t the m om ent, however, we shall follow a m ore intu itive reasoning.
F irst w e shall consider the situation in which the electron is initially
orbiting around proton P 1, form ing a hydrogen atom in its ground state Is, and the proton p 2 (o r H + ) is a t the right, v e ry far away.
T h e w a ve function o f the
electron is practically undisturbed by proton p 2. Thus it coincides w ith the hydro gen lsfunction drawn in Fig. 39. T herefore, if we draw the electron’s w a ve func tion along the line join ing the tw o protons as a function o f the distance from p i, we obtain the curve shown in Fig. 5 4 (a ), w hich shows that the electron is to be found predom inantly around proton p i. Suppose now th at the electron is ini tially orb iting around proton p 2, form in g a hydrogen atom in the lsstate, and that proton p i is far to the left. T h en the w ave function o f the electron is as shown in Fig. 5 4 (b ). As the separation between the tw o protons decreases, the w a v e function o f the electron is disturbed because the approaching proton (p 2 in F ig. 5 4 (a ) and P 1 in Fig. 5  4 (b )) tries to pull the electron aw a y from the other proton. T h e sym m etry of the electron’ s potential energy shown in Fig. 53 suggests th at the prob ab ility distribution o f the electron m ust exh ibit the same sym m etry. Thus, once the H 2 molecule is form ed, the electron w a v e function must have pronounced peaks around each proton, where the poten tial energy is less, according to F ig. 53.
T h e tw o
possible w ave functions for I I J which have the required sym m etry for their prob ab ility distribution are shown in F ig. 5 4 (c ) and 5—4 (d ).
T h e w ave function in
(c ) is even and in (d ) it is odd relative to the center, 0 , o f the m olecule. w ave functions w ill be designated as ^ ovon and Ψ,,μ , respectively.
These
W c can express
186
Molecules
(5.2
1A1 ( a) I >2
< l> )
IM
F ig. 5—i.
Even and odd molecular orbitals in H j .
either w a v e function in an approxim ate w a y b y linearly com bining the hydrogen w a ve functions (o r atomic orbitals) Ψι and Ψ2 corresponding to an electron orbiting around either proton, designated by I and 2. Thus lAcvcn
~
+
Φ
2.
^odd
~
Ψ ΐ
—
(5.2)
Ϊ 2 ·
T hese w a v e functions are called molecular orbitals, ab b reviated M O .
T h e theory
we are d evelopin g for w ritin g the m olecular orbitals is called the linear combination o f atomic orbitals, abb reviated L C A O . T h e states corresponding to m olecular orbitals and Ψ,,μ o f F ig. 54 are designated, for reasons to be given later, a ^ sls and σ „ Is. T h ese states must have different energies, as we can see from the follow in g explanation.
B oth w a v e func
tions g iv e m axim um p ro b ab ility for finding the electron near either proton, but ^even has an appreciable valu e in the region between the protons, while Ψ,,μ is v e ry small (o r zero) in that region, as depicted in F ig. 55.
F igu re 5 5 (a ) shows the
variation o f the p rob ab ility density o f the electron along the line join ing the tw o protons; Fig. 5 5 (b ) shows the contour lines o f equal p rob a b ility on a plane passing through the tw o protons, and in F ig. 5 5 (c ) the shading indicates the relative p rob ab ility o f finding the electron at d ifferent places. W h en the electron is in the region between the tw o protons, it pulls the protons together, offsetting their electric repulsion. helps to separate the tw o protons.
W hen the electron is on either side, it
In other words, when the electron is in the
region between the protons, it acts as a "c e m e n t” holding them and results in a stable configuration.
In term s o f energy, we m a y say that the state σ ^ ΐζ cor
responding to ^cvcn has a low er energy than σ „ 1β corresponding to ψ0j, and p. wave functions
F ig. 518.
(b) sp3 hybridized wave functions
W a v e functions resulting from sp3 hybridization.
204
Molecules
F ig. 519.
(5.5
Localized sp3 molecular orbital bonds in (a) methane, (b ) ethane.
o f its lack o f d irectio n a lity ) w ould not produce a bond o f the same strength, m ak ing it difficult to explain m olecules such as m ethane, C H 4. I t is possible how ever, b y means o f the technique called hybridization o f wave functions, to produce new atom ic w a v e functions oriented in the desired directions. T h e one 2s and the three 2p w a v e functions o f carbon do not h a ve exa ctly the same energy, but the differ ence in their energy is v e ry small.
B y m aking proper linear com binations o f the
four functions, w e m ay obtain new or h ybrid w a ve functions, all corresponding to the same energy and having a pronounced m axim um in any desired direction.
A
possible set o f linear com binations o f the s, p*, p„, and P 1 functions g iv e four new hybrid w a ve functions w ith m axim a w hich p oin t tow ard the vertexes o f a tetra hedron, as shown in F ig. 5 1 8 (b ). T h e directions along which the new w a v e func tions h a ve their m axim a form angles o f 109°28'.
T h e y are expressed in terms o f
the s and p w a v e functions as Ψι =
i(s +
Pi +
P v + Pz),
=
£ (S +
P i —
p„ 
P z ),
1^3 =
$ (S
P x H P » 
P z ),
φ4 =
£(s —
~
P i 
P „
+
,
.
P z ).
T h is ty p e o f h ybridization is designated b y sp3. N o te th at because s and p w a ve functions correspond to d ifferent values o f the angular m om entum , the hybrid w a ve functions do n ot describe states o f w elldefined angular m omentum. N o w let us consider the m ethane m olecule: C H 4. W e conclude th at the m axi mum strength o f the bonds is obtained when the Iselectrons o f each hydrogen atom achieve m axim um overlap pin g w ith each o f the four sp3 h ybrid w a v e func tions o f carbon.
T h is results in the tetrahedral m olecule shown in F ig. 5 1 9 (a ).
Polyatomic molecules
5.5)
205
7
}’
X
X
( b ) ρ ζ w a ve function
(a ) sp2 h yb rid ized w a ve functions
F ig. 520.
W ave functions resulting from sp2 hybridization.
Sim ilarly, th e ethane (o r H 3C — C H 3) m olecule has the structure indicated in Fig. 5 1 9 (b ). In this m olecule the tw o carbon atoms are held togeth er b y the o v e r lapping o f tw o hybrid sp3 w a v e functions. T h is is called a σ bond, due to its sim ila rity to the situation o f σ orbitals in d iatom ic molecules. I t is im portant to recognize th at hyb rid ization applies not on ly to carbon but to any atom or ion w hich has the same electronic configuration. F o r exam ple, the ion N + has the same electronic arrangem ent as carbon and the structure o f the radical N + H 4 is similar to th at o f C H 4, w ith N + replacing the carbon atom. Besides sp3 hybridization, there are several other possibilities o f h ybridization including w a v e functions whose angular m om enta h ave values higher than one. W e shall discuss on ly sp2 and sp hyb rid ization because o f th eir im portance. In sp2 hyb rid ization the s, p*, and p„ w a ve functions com bine to produce three w a v e functions in the X T p la n e , w ith th eir m axim a pointing in directions m aking angles o f 120°, as indicated in F ig. 520. These w a ve functions are expressed in term s o f the s and p w a v e functions as
(5.10)
T h e fourth electron occupies a p., state.
T h is is the typ e o f hyb rid ization re
quired to explain molecules such as ethylene H 2C =
C H 2, shown in Fig. 521.
T h e double bond between the carbon atom s results from the overlap pin g o f one
206
Molecules
(5.5 λ
(b)
F ig . 521.
E lectronic distribution in ethylene using sp2 h yb ridization : (a ) σbond,
(b ) 7rbond.
sp 2 hybrid w a v e function from each carbon atom along the line C — C (th a t is, a σ bon d) and the overlappin g o f p2 w a v e functions, which constitute a πbond, so called because o f its sim ilarity to the case o f 7rorbitals in d iatom ic molecules. T h e hydrogen atoms are attached to the rem aining sp2 hybrid w a v e functions. A s a result, the ethylene m olecule is planar. T h e double σττbond has a certain rigid ity which m akes it difficult to tw ist the m olecule around the C — C axis. T h is rig id ity does not occur when there is a single bond, such as there is in ethane, and it has som e effect on the m olecular properties.
N o te th at the degree o f overlap pin g in a
πbond is m uch less than in a σ bond, and therefore πbonds contribute less to the binding energy than σ bonds. F o r example, 6.33 e V are required to break a single σ bond, but on ly 3.98 e V are needed to break a double σ π bond into a single σ bond.
Polyatomic molecules
5.5)
207
s+p.F ig. 522.
W ave functions resulting from sp hybridization.
F ig. 523. Electronic distribu tion in acetylene using sp hybrid ization. (a) W ave functions with atoms separated; (b ) a and 7rbonds.
T h e third ty p e o f h yb rid iza tion w ith s and p atom ic orbitals is called sp, and corresponds to w a v e functions s ± p „ w hich have pronounced m axim a in the ± Z d ire ctio n s . These h yb rid w a v e functions are shown in F ig. 522. H y b rid sp w a ve functions are required to explain m olecules such as acetylene, H C = C H (F ig. 523). T h e trip le bond betw een the tw o carbon atom s results from the over lapping o f one sp h y b rid w a v e function from each carbon atom (th a t is, a σ bon d) and the overlap pin g o f p* and p y w a v e functions, resulting in tw o 7rbonds. hydrogen atom s are attached to the rem aining sp w a ve functions. acetylene is a linear molecule.
The
A s a result,
208
Molecules
F ig . 524. πbonds.
ii.G
(.5.6
Electronic distribution in butadiene, (a) Localized σbonds; (b) unlocalized
C on ju g a ted M o le c u le »
L e t us now consider a class o f compounds, called conjugated b y organic chemists, o f which butadiene (C 4H 6) is an exam ple.
T h e electronic structure o f butadiene
is indicated in Fig. 524. T h e carbon atom s along the chain are bonded b y σ bonds using sp2 hybrid w a ve functions. T h e hydrogen atom s are attached to the carbon atom s using the rem aining sp2 w a v e functions. B u t in addition there are four p2 electrons form ing 7rbonds along the carbon atom chain. In conjugated molecules the π bonds behave in a special w ay.
T h e π bonding electrons— one fo r each
carbon atom — instead o f being localized in particular regions o f the m olecule, as are the electrons in σ bonds, are m ore or less free to m ove along the m olecule, as indicated in Fig. 5 2 4 (b ) by their prob ab ility distribution. T h e π electrons accord
Conjugated molecules
5.6)
1
F ig. 525.
2
3
209
4
Molecular orbitals for irelectrons in butadiene.
ingly act in a v e ry special w a y which requires further consideration. F o r example, the p olarizab ility o f the conjugated molecules in the direction o f the carbon chain is much greater than the p o larizab ility o f saturated hydrocarbons which h ave on ly localized σ bonds and sp3 w a v e functions. A ls o the τ electrons introduce a certain rigid ity into the m olecular structure, so that all carbon atom s lie in a plane. L e t us now discuss the w a v e functions or m olecular orbitals o f the four πelectrons. A t the b ottom o f Fig. 525, w e have schem atically indicated the radial w ave function o f each πbonding electron when the carbon atom s are all v e ry distant from one another. W h en the atoms are v e r y close together, as th ey are in a molecule, the possible w a v e functions are obtained b y m aking proper linear com binations o f the individual or atom ic w a v e functions. same as th at used for
T h e technique is the
and H 2, but now w e have four atom s instead o f two.
Because o f the sym m etry o f the m olecule the m olecular orbitals m ust be either sym m etric or an tisym m etric rela tive to the center o f the molecule. T h e four pos sible m olecular w a ve functions o f the π electrons are designated b y ψι, ψ2, ψ3ι and in F ig. 525, and each corresponds to a differen t energy, which increases w ith the number o f nodes o f the w a v e function.
T h erefore, as the carbon atoms are
drawn closer, the energy o f the system splits into four closely spaced levels instead
210
Molecules
(5.6
F ig. .>26. Electronic potential energy of irelectrons in butadiene as a function of internuclear separation.
o f tw o, as in
and H 2.
T h e four energy curves are shown q u a lita tiv e ly in
F ig. 526, as functions o f the internuclear distance. E ach energy' level can accept tw o electrons w ith opposite spins.
Thus, in the ground state o f butadiene, the
four ττelectrons occupy the states corresponding to φ ι and φ 2. T h e next tw o nearby states, corresponding to φ3 and φΛ, are em p ty. T h e resulting p rob ab ility distribu tion o f the electron is shown in F ig. 527.
M olecu lar orbital φ\ is o f the bonding
ty p e for each pair o f carbon atoms, while φ 2 *s bonding for the pairs 12 and 34 o f carbon atom s and antibonding for the pair 23. F o r th at reason th e total prob a b ility distribution shows a dip a t the center o f the molecule. T h is means th at the strength o f the bond between the pair 23 o f carbon atom s must be less than for the pairs 12 and 34. E xp erim en ta lly th e length o f the bond 23 is 1.46 X IO 10m, w h ile bonds 12 and 3 4 have a length o f 1.35 X IO10 m. Our discussion can easily be generalized to include a polyene or conjugate com pound consisting o f a carbon chain o f 2 n atom s; in the classical valence m odel this would be w ritten as: ·
C = C — C = C — C = · · · In addition to the σ bonds
betw een pairs o f adjacent carbon atoms, there are 2>i 7relectrons spread along the molecule, as previou sly shown for butadiene in F ig. 524. T h ere are then 2 n closely spaced energy levels available, w ith cap acity for 4n electrons, to be filled b y the 2n irelectrons.
H ence when these m olecules are in the ground state on ly
the low er h a lf o f the en ergy levels are occupied. E lectronic m otion in these m ole cules can be excited b y a rela tiv ely small am ount o f energy because o f the nearby em p ty levels.
W hen there are 4n 7relectrons, all energy levels are occupied and,
Ci
F ig. 527.
C2
C3
Ci
Total probability distribution of τ electrons in butadiene.
Conjugated molecules
5.6)
F ig. 528.
211
Benzene molecular orbitals, (a) Localized σbonds; (b ) unlocalized 7rbonds.
in order to excite the electronic m otion, the m olecule m ust undergo a transition to another electronic configuration; this requires much m ore energy. T h e avail ab ility o f low lyin g e m p ty energy levels is responsible for the fact th at m any o f these conjugate molecules absorb photons in the visible region and therefore have characteristic colors. As a last exam ple, w e shall consider the cyclic conjugate m olecule benzene, C e H 6.
T h e carbon atom s are located at th e vertexes o f a regular hexagon and
joined b y σ bonds using sp2 hybrid w ave functions along each C — C line, as shown in F ig. 5 2 8 (a ). T h e hydrogen atom s are attached at the rem aining sp2 orbital o f each carbon atom .
T h e re are also six 7relectrons, one from each carbon atom, in
2/2
Molecules
(5.7
p , orbitals (th e Zaxis is taken perpendicular to the plane o f the m olecule). These 7Telectrons are free to m ove along the hexagon, constituting a sort o f closed cur rent, as indicated in F ig. 5 2 8 (b ). T h is accounts for the strong diam agnetism o f benzene and other cyclic con jugate molecules. T h e analysis o f the w a ve functions and the energy levels o f the 7relectrons is sim ilar to th at for butadiene, but the fact th at the carbon chain is closed introduces som e new features w ith respect to open carbon chains, which w e shall not discuss (see P ro b lem 6.20). W e shall not pursue the subject o f m olecular structure any further.
T h is b rief
discussion o f chemical binding is enough to illustrate the m ore fundam ental ideas and to forewarn the student o f the difficulties w hich arise when one attem pts to use the classical concept o f valen cy in too rigid a form.
T h e form ation o f a m ol
ecule is a m anyparticle problem in vo lvin g electrom agnetic interactions, and the problem m ust be solved according to the rules o f quantum mechanics. E X A M P L E 0.6. molecule.
Estimate the first excitation energy of a πelectron in a conjugate
S o lu tio n : One can estimate the excitation energy of a πelectron by considering that πelectrons move independently of each other and that their motion within the molecule resembles motion in a potential box (Section 2.5) whose dimension is of the order of the length of the carbon chain. The wave functions would then correspond to the dashed lines in Fig. 525. I f the carbon chain has 2n atoms, there are 2 « πelectrons and the last occupied electronic level corresponds to the nth level in Eq. (2.14). The energy difference between the nth and ( n + I) th levels is the first excitation energy. Then 2, 2 ,
Δ£ = Η
I
^
,.2
2 ,2
  ι ϋ
2

2 ,2
= ^
(2'ι + 1 ) 
(5n )
where a is the length of the region in which the πelectrons move, taken as the molecular length extended half a bond length on each end. For example, in the case of butadiene, with a about 5.6 X IO10 m and n = 2, we get A E = 5.86 eV, which corresponds to photons of wavelength 2.12 X IO7 m. I t has been observed that butadiene has strong absorption for radiation of wavelength 2.17 X IO7 m. Thus our crude model shows a reasonable agreement with the experiment.
.7 .7
X lolvvu ia r R ota tion s
N o w th at we have considered electronic m otion in molecules which have fixed nuclei, the next step in our analysis o f m olecular structure is to stu d y nuclear m o tion.
T h e simplest nuclear m otion is th at o f rotation o f the m olecule around its
center o f mass as if the m olecule w ere a rigid body. W e shall first discuss the rota tion o f diatom ic molecules. W e can study the m otion o f a rigid bod y m ost easily b y using a fram e of reference attached to the bod y and com posed o f the principal axes o f inertia. In the case of diatom ic molecules (F ig. 5 2 9 ), the principal axes are the line join ing th e tw o nuclei, N i and N 2, or the Z 0 axis, and any line perpendicular to it through the center o f mass. Because their mass is so sm all, w e can neglect the m om ent o f inertia
5.7)
Molecular rotations
Fig. 529.
Rotation of a diatomic molecule about its center of mass.
N1
213
N2 CM,
due to the electrons, and th erefore the m om ent o f inertia o f the m olecule rela tive to the Z 0 axis is zero.
Thus the angular m om entum o f the molecule, for rotation
around the Z 0 axis, is also zero and the to ta l angular m om entum L o f the m olecule is perpendicular to the m olecular axis. G iven th at r 0 is the equilibrium separation o f the nuclei and μ the reduced mass o f the m olecule, the m om ent o f inertia about an axis perpendicular to Z 0 and passing through the center o f mass o f the m olecule is / = μ τ 0. W e m ay then w rite the kinetic energy o f rotation o f the m olecule as Er =
L 2/21.
B y virtu e o f the quantization o f the angular m om entum w e m ay
write, according to E q. (3.15), L 2 =
h 2l ( l +
I ) , w here Z = 0, 1, 2, 3, . . . , (th a t is,
a positive in teger). T h erefo re the kinetic energy o f rotation o f the molecule becomes
(5.12)
Er = h W + !) = Dhcl{·1 + 1^where it is custom ary to set h 2/21 =
Jihc. T h e quantities h 2/21 and B for several
molecules are given in T a b le 53, expressed in e V and in cm  1 , respectively.
By
givin g successive values to I, w e can obtain the rotational energy levels o f the m ol ecule (F ig. 5 3 0 ). Successive energy levels, corresponding to I and I ' =
I +
I, are
separated by the amount AEr =
T A B L E 53
2 B h c(l +
I).
(5.13)
K o tatio n al an d V ib ra tio n a l C o n stan ts o f Som e D iatom ic M olecules
Λωο, eV
v0, cm 1
H2 C l2 N2 L i2 O2 CO HF HCl HBr BeF
8.0 X IO 3 3.1 X ΙΟ" 5 2.48 X 10—*
60.809 0.244 2.010 0.673 1.446 1.931 20.94 10.59 8.47 1.488
0.543 0.0698 0.292 0.0434 0.194 0.268 0.510 0.369 0.326 0.151
4395 565 2360 351 1580 2170 4138 2990 2650 1266
1.78 2.38 2.48 1.31 1.05 1.84
X X X X X X
I O
Ii = h/M rlc cm 1
X
h2/2I, eV
CO CO
Molecule
IO 4 10~4 IO 3 ΙΟ" 3 IO" 3 ΙΟ4
2/4
Molecules
(.5.7
Parity
F ig. 530. Rotational energy states of a diatomic molecule.
■t +
Because o f the sm all valu e o f h 2/ 2 I (~ 1 0  4 e V ) when com pared w ith the trans lational kinetic energy (w hich is o f the order o f k T ~
eV =
2.5 X IO 2 e V at
room tem perature), m any molecules are found in excited rotational states at room tem perature.
A lth ou gh the rela tiv e equilibrium population o f th e d ifferent rota
tional en ergy levels is an im portan t problem , its analysis requires the use o f statistical m ethods, and w e shall th erefore defer it until Section 12.5. T h e w a ve functions corresponding to the rotational m otion are the F (m,(0, Φ) introduced in Section 3.5, and the angles (0, φ) define the direction o f the m olecular axis Z 0 rel a tiv e to the coordinate axes X , Y, Z , fixed in the laboratory.
H ence the p a rity
o f each le v el is ( — I ) 1, and thus successive energy levels have opposite p arity, since th ey are sim ilar to the s, p, d, . . . atom ic orbitals. radiation, the allowed transitions are those for which A f=
F o r electric dipole
±1,
(5.14)
a selection rule sim ilar to th at found in the atom ic case. Thus the o n ly transitions possible are those betw een adjacent levels. Som e o f the transitions, corresponding to absorption, are indicated in Fig. 530. W h en w e use B oh r’s rule, E 2 — E x = hv, the frequency o f the radiation em itted or absorbed in a rotational transition is v =
A E /h =
2 B c (l + 1 )
or
v =
2 B (l + I ) ,
(5.15)
w here I refers to the angular m om entum o f the low est level in vo lved in the transi tions and P is the w a ve number expressed in cm  1 . Thus th e rotation al spectrum o f d iatom ic m olecules consists o f a series o f lines equ ally spaced an am ount A? =
2B cm  1 .
B y measuring AP, w e can com pute B and thus the m om en t o f inertia, and from it estim ate the nuclear separation r 0. Pure rotational spectra fall in the m icrow ave or farinfrared regions o f the spec trum . In addition, for a m olecule to exh ibit a pure rotational spectrum , it must
Molecular vibrations
5.8)
2/5
v, pm 1
Fig. 531.
Rotational absorption spectrum of IIC l in the gaseous phase.
possess a perm anent electric d ipole m om ent. In the process o f absorption o f radia tion b y the m olecule, the perm anent electric dipole m om ent interacts w ith the electric field o f the incom ing w ave. In the process o f emission o f radiation, the rotation o f the dipole is responsible for the radiation.
T h erefore, homonuclear
diatomic m olecules (w h ich do not have a perm anent electric d ip ole) d o n ot show pure rotation spectra. F igu re 531 illustrates the absorption rotation spectrum o f H C l in the gas phase. Each d ip corresponds to a maxim um or resonant, absorption. W e can see that, in agreem ent w ith the theory, as given b y E q. (5.15), the fre quencies are even ly spaced. Equation (5.12) holds for the case o f a rigid molecule.
H o w ev e r, when the
rotational energy increases, there is a stretching o f the m olecule due to a centrif ugal effect; this results in an increase in the m om ent o f inertia and a correction to Eq. (5.12) is required. Our discussion applies also to the rotational spectra of linear molecules such as COo or IIC N t. T hose m olecules— which, in general, have relatively larger m om ents o f in ertia— have much closer rotational energy levels than most diatom ic molecules, and correspondingly low er frequencies in the rota tional spectrum.
5./1 M o l w u l a r Vibra tion s So far w e have considered th e nuclei as fixed rela tiv e to each oth er; how ever, the shape o f the poten tial energy curve for a diatom ic m olecule (such as those shown in Figs. 56, 513, and 5 14) suggests th a t the nuclei in a m olecule are in rela tive oscillatory m otion. L e t us consider a d iatom ic m olecule which has a poten tial energy for a given electronic configuration, as shown in F ig. 532, where r0 is the equilibrium separa tion.
G iven that the nuclear m otion corresponds to an energy E , the tw o nuclei
oscillate so that, classically, th eir separation varies between Oa and Ob.
The
motion o f the nuclei must, how ever, be described according to quantum mechan1CS. I f the poten tial energy w ere th at o f a sim ple harm onic oscillator, represented by the parabola %k(r — r 0) 2, the rela tive oscillatory m otion o f the tw o nuclei would be sim ple harmonic, and the results o f Section 2.6 w ould be applicable.
216
Molecules
·
(5.8
E
T lic angular frequ en cy o f the oscillations is ω 0 = V k /μ, w h ere μ is the reduced mass o f the molecule. In Section 2.6 it was shown th at the energy o f the oscillatory m otion is quantized and g iven b y E q. (2.21).
H en ce the vib ration a l energy levels
o f a diatom ic m olecule are expressed app roxim ately b y Ev = where v =
( ο \ £)Λω0,
(δ . 16)
0, I, 2, 3, . . . (p o s itiv e integer). T h erefo re the vib ration a l energy levels
o f the m olecule are e qu ally spaced an am ount Iicon and th e m olecule has a zerop o in l vibrational energy equal to iheo0. Because o f th e zeropoin t energy, the dis sociation energy o f a d iatom ic m olecule is D e = D — Ishco0, as m a y be inferred from F ig. 532. T h e selection rule for electric dip ole transitions am ong vibrational levels is the same as E q. (2 .34 ); th at is,* Av=
±1.
(5.17)
Since the on ly allowed change o f v is to a neighboring energy level, the on ly fre quency absorbed or em itted in a v ib ration a l transition is equal to th e natural classical frequency V0 =
ω 0/2ττ. T a b le 53 give s the values o f Iieo0 and V0 =
V0I c
* This selection rule is not rigorous because the potential energy, as shown in Fig. 532, is not that of a harmonic oscillator, and higher values of Av are also possible, although less probable.
5.8)
Molecular vibrations
217
for several diatom ic molecules. T h e vib ration al frequencies o f m ost d iatom ic molecules fall in the infrared region o f the spectrum. F o r a v ib ration a l transition to occur, either in emission or absorption, the d iatom ic m olecule must have a permanent electric dipole m om ent.
T h u s hom onuclear m olecules such as H 2 or
N 2 do not show pure vib ration a l transitions, since they have no perm anent electric dipole m om ents and they do not exh ibit an infrared spectrum . B ut polar molecules, such as H C l, show strong vib ration a l transitions. T h e m olecular energy due to both rotation and vib ra tion can be expressed by combining Eqs. (5.12) and (5.10), so th at w e have
E =
Ev+ Er =
( v + i)ha>0 + ^ j l ( l +
I).
(5.18)
In general, the qu a n tity Λ2/2/ is much sm aller ( ~ I 0 4 e V ) than Iicon (~ 1 0 1 e V ), and we m ay say that, to each v ib ration a l level, there correspond several rotational levels, as shown in Fig. 533.
W h en the selection rules
=
Av
±
I
and
Al
—
± 1
are taken into account for a transition between tw o rotational levels belonging to two adjacent vib ration a l levels, the rota tion vib ra tion spectrum results.
V ib ra tio n a l ievels
Iii
Th is
R o ta tio n a l levels
i ’r= £/(102)_ JQ22)= ( (121V (220)“ (041)
(201 y
(300) 
10,000
(012) 
9000 .
(210)“
“ 8762
(031)(130)“
8374 "8274
(101)( 021) ( 120) =
7251 6871 =6775
(0 U )_ (U O )
_5331 = 5235
(Ill)
( 002)· ( 200) 
7445
“ 10284 9834
(030)
(100)
(001)3657 ( 020) 
(010)
M ixc d m odes
F ig. 537. Vibrational levels of HaO in the gaseous phase. The energies are expressed in cm 1 . The numbers in parentheses correspond to the vibrational mode (niri2» 3)·
On the other hand, ω 3 induces a dipole m om ent parallel to the axis o f the m olecule and Co2 a dipole m om ent perpendicular to the axis o f the m olecule; therefore both are a c tiv e spectroscopically. In cid en tally, co2 is doubly degenerate, since it consists o f the m ode shown plus one in which the atom s oscillate perpendicular to the page. A p olyato m ic m olecule therefore has several sets o f vibration al energy levels. Som e are illustrated in F ig. 537 for H 2O. Each o f these levels is associated w ith corresponding rotational levels, th ereby m aking the spectra rather complex.
An
analysis o f rotation vib ration spectra provides valu able inform ation about m olec ular dimensions, geom etry, dissociation energy, and so on. A n interesting process, called the Ram an e je c t— after the In d ian physicist C. V . R am an (1888
) — is related to m olecular vib ration s and rotations, al
Molecular vibrations
5.8)
221
1η term ed in te o r virtu a l state
 In it ia l state
^scattered —v
rscaltorcd ~ v Stokes
R a y le ig li scatterin g
Cr
esrattrird = VHelA ntiS tokes
Ram an scatterin g
F ig. 538.
Rayleigh and Raman scattering.
thougli it is essentially a scattering process.
Suppose that w e illum inate a gas
sample w ith m onochrom atic radiation o f frequ en cy v. I f w e observe the radiation scattered in a direction at right angles to th at o f incidence, w e find th at the radia tion contains, in addition to the frequ en cy v equal to th at o f the prim ary or incident radiation, which is the coherent or Rayleigh scattering, radiation o f frequ en cy v ±
vv
(where vv corresponds to a frequ en cy o f the vib ration al spectrum o f the m olecule). Th is is called Raman scattering. W e can interpret Ram an scattering in the follow in g manner: suppose th at a m olecule is in itia lly in the v ib ration a l state v (F ig . 538). When the m olecule absorbs radiation o f frequency v, it m ay go to an excited state. I f the excited state is not a station ary one, there is an im m ediate reradiation o f the absorbed energy. T h e m olecule m ay return to the initial state, em ittin g radiation of the same frequ en cy as the incident lig h t; this is R a y le ig h scattering. T h e m ol ecule m ay also return to another vib ration al level which, due to the selection rule Av =
± 1 , must be one im m ed iately ab ove or b elow the initial le v e l; this is Ram an
scattering.
Thus the em itted radiation has a frequ en cy v + vD or v — vv.
The
line o f frequency v — vv is called the Stokes lin e and th at o f frequency v j νΏ the antiStokes line.
T h e R am an effect has also been observed in the rotational spec
trum. In this case the selection rule is A l =
0, ± 2 .
E X A M P L E 5.7. Relation between the constants in the Morse potential (Example 5.3) and the vibrational frequency of a diatomic molecule. S o lu tio n : Given that E p = D [ I — βαo where x , = Λωο, 4 D.
.7. U
E iv e ir o n ir
T r n n M iiio n s
in
M o Ie e u ie H
A given m olecule m ay have several electronic configurations or stationary states. L e t us consider, for sim plicity, a d iatom ic m olecule; to each electronic state there corresponds a potential energy sim ilar to the electronic states o f I I 2*" shown in Fig. 57.
T h e tw o nuclei in the excited electronic states L iv e , in general, equ ilib
rium distances which are different from the one for the ground state. T w o such poten tial energy curves arc shown in F ig. 539. T h e separation o f these energy curves is o f the order o f I to 10 cV. Thus when a m olecule experiences an elec tron ic transition, ju m ping from one electronic configuration to another, the radia tion in vo lved falls in the visible or the u ltraviolet regions o f the spectrum.
F ig. 539.
Vibrational and rotational energy levels associated with two electronic states.
Electronic Iransilions in molecules
5.9)
'223
T o a given electronic state there correspond m any vib ration al states and to each vibrational state there correspond several rotational states. A s a first approxim a tion, w e m ay w rite the energy o f the m olecule in the form
E = Ec + E v+ E r =
E c + [p +
ft)Aw0 +
/(/ +
I),
(5.21)
where E c refers to the electronic energy at the m inimun of the poten tial energy curve (see Fig. 539).
In an electronic transition all three energies m ay change.
T h e frequency ω 0 and the m om ent o f inertia I are, in general, differen t for both electronic states. T h e refo re wc must w rite the energy change in the electronic transition as ΔΛ = where A E c =
E" — E' 
A E c i A E v t A E r
E c — E c is the change in electronic energy, given b y tb c energy d if
ference o f the m inimum o f the tw o electronic states, AEv =
E" 
E 'v =
(v " +
J)*Wo 
(" ' + 4 )M i
is the change in vib rational energy, and
.
AEr =
EV 
Ei =
I " (I" +
is the change in rotational energy.
I) 
~
I '(I ' +
D
T h e terms w ith double primes correspond to
the state o f higher electronic energy and those w ith single primes to the state of lower electronic energy.
T h e frequency o f the radiation em itted or absorbed in
an electronic transition is therefore the sum o f three terms,
" = X = ^ T
V ) + V rK ", I '),
where vc — A E ,/ h is due to the change in electronic energy, and vv(v " , ν’) and vr( l " , I ') correspond to the changes in vib ration a l and rotational energies, respec tively. T h e frequency vc is the largest. For a given electronic transition the spectra consist o f a scries o f bands; each band corresponds to a given value o f v " and ν' and all the possible values o f I " and I'. T o determ ine the possible values of (υ ", ν ') and ( I " , I ') w e need selection rules. W hen w e consider electric dipole transitions, the rotational selection rule is Al = N o te that Al =
0, ± 1 ,
no I " =
0 v " =
0.
and
T h e reason for this is th at the m axim um o f the w a v e function for the ground state o f an oscillator occurs at the center (see F ig. 219).
W e have considered
absorption transitions; the same logic applies to emission transitions.
T h erefore
vib ration a l transitions associated w ith electronic transitions do not com ply w ith the selection rule Av =
± I and v m ay suffer a change o f several units.
T h e rule w e have explained for determ ining the vib ration al transition in an electronic transition is known as the FranckC oiulon p rin ciple.
I t was form ulated
first in classical term s b y James Franck and reform ulated according to quantum m echanics b y E dw ard U. Condon about 1926.
Conclusion
5.10)
225
E
F ig. 3—40.
5.10
Electronic transitions allowed by the FranckCondon principle.
I onclusion
In our discussion o f m olecular structure w e h ave considered on ly the most im portant aspects. T h e re are m any details th at w e have not m entioned. F o r example, nuclear spins h ave an influence on m olecular energy levels. In the simplest m ol ecules, H f and I I 2, the tw o protons m ay have th eir spins parallel or antiparallel, resulting in orthohydrogen and parahydrogen. M olecu lar sym m etry plays a v e ry im portant role in determ ining th e possible normal vibrations as w ell as the allowed transitions; how ever, a discussion o f m olecular sym m etry is beyond the level o f this book. In the case o f polyatom ic molecules the spectra m ay be so com plex th at it m ay be impossible to classify them as w e have done in Section 5.9 for the spectra of diatom ic molecules. H o w ev e r, this com p lexity serves a useful purpose, in th at it perm its us to id en tify a g iven com pound by observing its spectrum. On the other hand, certain atom ic groups exhibit w elldefined electronic and vib ration a l spectra. Observation of the corresponding frequencies serves as an indication of the presence o f such groups in a molecule. W e m ay conclude by saying th at one o f the most pow erful tools for determ ining molecular structure is the analysis o f m olecular spectra.
226
Molecules
RoierpnveH 1. "T h e Force Between Molecules,” B. Derjaguin1Sci. Am., .Iuly 1960, page 47 2. “ Some Aspects of Molecular Physics,” J. Levelt1 .lm. J. I 'In/s. 28, 192 (1960) 3. “ Chemistry of the Noble Gases,” H. Selig, J. M alm .^nd H. Claassen, Sci. .Dn., M ay 1964, page 66 4. 5.
“ The Zeeman Effect in Molecules,” C. Quailc,
bn. J. Phys. 32, 634 (1964)
The Scientific Endeavor. New Y ork: Rockefeller Institute Press, 1965; “ The Architec ture of Molecules," by L. Pauling
6. Symmetry in Chemistry, I’ . Dorain. Reading, Mass.: AddisonWesley, 1965 7. The Structure of Molecules, G. Barrow. New Y ork: Benjamin, 1964 8. Electrons and the Covalent Bond, II. Gray. New Y o r k : Benjamin, 1965 9. Elements of Wave Mechanics, N. M ott. Cambridge, England: Cambridge University Press, 1962, Chapter 5, Sections 2.3, 3, 4.2, 4.3, 4.4, 4.5, and 4.6 10.
Structure of Matter, W . Finkelnburg. Section 7; Chapter 6
New Y o rk : Academic Press, 1964, Chapter 4,
11. Quantum Theory of Matter, J. Slater. and 9
New Y ork: M cGrawHill, 1951, Chapters 8
12. Elementary Introduction to Molecular Spectra, B. Bak. N ew Y ork : Interscience, 1960 13. Chemical Binding and Structure, J. Spice. N ew Y ork: Macmillan, 1964 14. Wave Mechanics and Valency, J. Linnett. London: Methuen, 1960 15. The Chemical Bond, third edition, L. Pauling. Ithaca, N .Y .: Cornell University Press, 1967
P r o b Io n iM 5.1 The groundstate energy of H f rela tive to the system composed of hydrogen in its ground state and H + at infinite sep aration is — 2.65 eV. Compute: (a) the energy of I l / relative to the system H++ H++ e at infinite separation; (b) the energy of the system Ho' + e at infinite separation rela tive to the system H + H, again at infinite separation, and with both atoms in their ground states; (c) the ionization energy of Ho if the dissociation energy of this mol ecule into two hydrogen atoms in their ground states is 4.48 eV. Compare the result in (c) with the ionization energy of atomic hydrogen.
5.2 Calculate the contribution of the coulomb repulsion of the two nuclei to the energy of Ho in its ground state. Compute the energy due to the interaction of the electron with the nuclei. [H in t: Use the data of the preceding problem and recall that the equilibrium separation of nuclei in H ^ is 1.06 A.] 5.3 Dissociation and ionization energies are frequently expressed in kcal mole1 . Show that one kcal mole1 is equal to 4.338 X IO2 eV. Expressthedissociation energy of H 2 in kcal mole1 . 5.4 Explain why the bond length of H j is 1.06 A, while H j has a shorter bond length equal to.0.74 A. Also explain why the dis sociation energy of LIj (103.2 kcal mole1 )
Problems is less than twice the dissociation energy of (61.06 kcal mole1 ). 5.5 Explain why the ion H j" is less stable than the ion H e i if both have the same electronic configuration. Which ion should have the lafger internudear separation? I
5.6 Find the energy of L ij relative to the system 2 L i+ + 2 c , with all four particles at infinite separation. The ionization energy of lithium is 5.39 eV. Calculate the coulomb repulsion energy of the Li + ions at the equilibrium separation, 2.67 A . Then compute the energy of the two val ence electrons. 5.7 Discuss the electronic configuration and bond structure of (a) A I2, (b) S2, and (c) CI2. Write, in each case, the symbol for the groundstate term. 5.8 Given the molecule NO , discuss its electronic configuration and bond struc ture. Compare with the N 2 molecule and state. Decide which molecule is more stable. 5.9 Write the electronic configurations for the following heteronuelear diatomic mol ecules: (a) L iH 1 (b) C N 1 (c) SO, (d) C lF 1 and (e) H I. Determine, whether the mol ecules arc polar and, in the affirmative case, identify which end of the molecule is positive and which end is negative. 5.10 Calculate the energy released when AlCl and A lIlr dissociate into neutral atoms. The equilibrium separations for these molecules are 2.13 A and 2.30 A , respectively. The respective experimental values are 71.5 kcal mole1 and 55.3 kcal mole 1 . [H int: Sec Example 5.5 for a sim ilar calculation for NaCl.) 5.11 When a hydrogen atom captures an electron to become I I  , an energy equal to 0.. 49 eV is released. The ionization energy of lithium is 5.39 eV. Compute the dis sociation energy of LiH , given that the nuclear equilibrium separation is 1.60 A. 5.12 An acceptable empirical expression for the potential energy in ionic bonding is
227
given by E , = —v2e2/(4 rtur) + be~33xlu '° rt where v is the charge of the ions and b is a constant determined from experimental values for the dissociation energy. Find 6 for K C l, given that the equilibrium dis tance for this molecule is 2.79 X IO10 m. Plot the potential energy. Find the dis sociation energy of K C l into K + and C l . Calculate the energy required for its dis sociation into neutral products. 5.13 A fairly precise empirical expression for calculating the potential energy of two ions in an ionic bond is E , (r ) = — e2e2/(4jre0r)
b e ~ " — d/r®,
where v is the charge of the ions and a, 6, and d are constants obtained experi mentally. For L iF the constants arc: a = 3.25 A  ’ 1, 6 = 895 eV, and d = 2.68 eV A®. Determine the energy required to disso ciate a L iF molecule into its two ions. (The bond length of L iF is 1.54 A .) Also com pute the energy required for dissociation into two neutral atoms. The ionization potential of lithium is 5.39 eV and the electron affinity of fluorine is 3.45 eV. 5.14 Show that the wave function i
= N ( s + Oip1 + /Spv + Tp1)
has its maximum along the direction of the vector a = UxO + u j 3 + u,7, and its minimum in the opposite direction. IIcnce two hybrid wave functions corresponding to vectors a and a ' have their maximum in directions making the same angle as a and o'. Write the vector a for each of the sp3 and sp2 hybrid wave functions introduced in Section 5.5. Show that in the sp3 case the vectors point toward the vertexes of a tetrahedron, making angles of 109° 28', and in the sp2 case they are in the XTplane, making angles of 120°. [H in t: Note that the vector uxpx + uvpv + u,p, is propor tional to r.]
228
Molecules
5.15 Show that the four sp3 hybrid wave functions are orthogonal. Repeat for the sp2 wave functions. 5.16 Discuss the bond structure (a) C3H 8, (b) C 3H 0, and (c) C3H j.
in
5.17 Analyze the bond structure of the following molecules: (a) C O 2, (b) CS2, (c) C S T e, (d) Cs, (e) C dC l2l (f) O C l2, (g) O N C l1 (h) SnCl2, (i) S§, and (j) C N 3  . Indicate which molecules are linear and which are bent. 5.18 Analyze the bond structure of the following molecules: (a) AsH 3l (b) SbF3, (c) I lO i , (d) 0 H 3+ , (e) P C l3, (f) SeO §(g) HrO3", (h) SiFi, and (i) P F 3. Indicate which molecules are planar and which are pyramidal. 5.19 Analyze the bond structure of H II3 and B H i", indicating whether the mol ecules are planar or pyramidal. 5.20 Discuss the H 2O and N H 3 mol ecules, using sp3hybridized wave func tions for the oxygen and the nitrogen atoms, respectively. Do you think that this is a better description of these mol ecules than the one presented in the text? 5.21 Analyze the bond structure of C H 3" and I i 3O +. 5.22 Discuss the bond structure of the molecule H 2CO. Is the molecule planar? 5.23 The electric dipole moment of H 2O is 6.2 X IO 30 m C. Find the dipole mo ment corresponding to each O— H bond. Given that the bond length of O— H is 0.958 A, what fraction of the hydrogen electron has been transferred to the oxy gen atom? 5.24 The effective total length of the con jugate molecule C H 3— (C H = C H — JiC H 3 is 9.8 A. Plot the energy levels occupied by the πelectrons. Find the energy and wavelength of the photons absorbed when one of the uppermost πelectrons is excited. 5.25 The /3carotene molecule is a conju gated molecule having 22 πelectrons. It has been found that this molecule shows
strong absorption of radiation at 4510 A. Estimate the total length of the molecule. 5.26 Show that the moment of inertia of a diatomic molecule about an axis perpen dicular to the line of nuclei and passing through the center of mass is I = pro, where μ is the reduced mass of the mol ecule and ro is the internuclear distance. 5.27 The adjacent lines in the pure rota tional spectrum of 35C l19F are separated by a frequency of 1.12 X IO10 Hz. AMiat is the interatomic distance of this molecule? 5.28 (a) Calculate the energy and wave length of the photon absorbed when a 200H g35Cl molecule (ro = 2.23 A ) makes the rotational transitions I = 0 —» I = I and I = I —> I = 2. (b) In what region of the electromagnetic spectrum are these lines found? 5.29 Suppose that the equilibrium sep aration in the H 35Cl and H 37Cl molecules is the same and equal to 1.27 A. Compute for each molecule (a) the constant B for the rotational levels, (b) the energy of the first two excited rotational leifels, (c) the frequencies and wavelengths correspond ing to the transitions I = 0 —* I = I and I = I —> I = 2, and (d) the frequency difference for successive lines. Compare your results with Fig. 531. 5.30 Compute the energy of the first three excited rotational states in CO and CO2. Determine the wavelength of the photons absorbed in transitions among such energy levels. 5.31 The rotational energy levels of a molecule having two equal principal mo ments of inertia are given by _
rot
I 2/i
,2
, I
Λ
+ 2 \/2
Λ Z1/
,2
*’
where I i corresponds to axes X and I' and 12 corresponds to the Zaxis, and L 2 = 1 (1 + I )Λ2,
L 1 = Ui h .
Problems Using this equation, estimate the relative position of the rotational energy levels of a molecule having (a) 12 = Ο.8 /1, (b) 12 = 1.2/i. Plot the levels as multiples of
h2/2h . 5.32 A diatomic molecule is not rigorously rigid and, due to a centrifugal effect, the internuclear distance increases as the angu lar momentum of the molecules increases. (a) IIow should this molecular stretching affect the energy levels, in comparison with the values given by Eq. (5.12)? (b) An empirical expression for the rota tional energy is Ε ,ο ι
=
{ h 2/ 2 1 ) {1(1
+ I) 
51/(1+ I)]2},
where b is the stretching constant. Obtain an expression for the frequencies due to transitions among rotational levels. Com pare with Eq. (5.15) and see if the stretch ing effect is recognizable in Fig. 531.
m1
m2 F ig. 341
5.33 T w o bodies of masses mi and m2, respectively, are joined by a spring with an elastic constant k (Fig. 541). Show that if the spring is stretched and released the masses will oscillate with a frequency v = ( 1/2jt)V k /μ, where μ is the reduced mass of the system. 5.34 Calculate the energy of the three lowest vibrational levels in H F, given that
229
the force constant is 9.7 X IO2 N m 1 . Find the wave number of the radiation absorbed in the transition e = 0 —* e = I. 5.35 The infrared spectrum of CO, at low resolution, shows an absorption band cen tered at 2170 cm 1 . Find the force con stant in CO. Plot, to scale, the potential energy curve. 5.36 Calculate the ratio of the vibra tional frequencies of 1I I 35Cl and 2H 35Cl molecules, assuming that the force con stant is the same for both molecules. 5.37 What is the force constant for the HCl molecule, given that the vibrational frequency is 9 X IO13 Hz? Also find the zeropoint energy. 5.38 Find the vibrational frequencies of IID and D 2, given that the Ho molecule absorbs infrared radiation of wavelength 2.3 X 10G m corresponding to a vibra tional transition with Ac = I and no change in rotational or electronic energies. Assume the same force constant for the three molecules. Compare with the experi mental values. 5.39 The three vibrational frequencies of CO2 are 667.3 cm 1 , 1388.4 cm 1 , and 2349.4 cm 1 . Make a sketch of the first few vibrational energy levels of this molecule. 5.40 The molecule CO has an electronic transition that produces several bands in the visible region (around 6000 A). Esti mate the separation among adjacent rotational lines in each band. Do you expect the band to appear continuous in a lowresolution spectroscope?
6 SOLIDS
6 .I 6.2 6.3 6A
Introduclion
Types o f Solids Band T
FreeEleelron M odel o f a Solid
6.5 Electron M o tio n in a Period ic Structure 6.6 6.7
Conductors, Insulators, and Semiconductors Quantum Theory o f Electrical Conductivity 6.8
Radiative Transitions in Solids
Types of solids
6.2)
S.1
231
In tro d u c tio n
M a tte r in bulk, in the w ay it affects our senses, is an aggregate o f a v e r y large number o f atoms. G rossly speaking, these aggregates appear to be in three physical states or phases: gases, liquids, and solids. In gases the average distance between molecules is much greater than the size o f the m olecules and the interm olecular forces are much weaker than the forces which hold the atom s in the m olecule together.
Thus in gases the m olecules retain their ind ivid u ality.
A t the other
extreme, in a solid, the atom s (o r m olecules) are tig h tly packed and held in m ore or less fixed positions by forces, o f electrom agnetic origin, which are o f the same order o f m agnitude as those in volved in m olecular binding. T h u s the shape and volum e o f a solid remain essentially constant so long as the physical conditions, such as pressure and tem perature, do n ot undergo any appreciable change. Liquids fall in between gases and solids. H o w ever, the th eory o f liquids is still incom plete and w ill not be considered in this text. In m ost solids the atom s (o r m olecules) do not exist as isolated entities; rather their properties are m odified b y the nearby atoms. T h e regular arrangem ent o f the atom s or groups o f atom s is one o f the m ost im portant features o f solids; that is, the structure o f solids exhibits a regularity or p eriodicity constituting w hat is called a crystal lattice; therefore, to understand the structure o f a solid, it is neces sary to study on ly the basic unit or cell o f the lattice, since all properties repeat from cell to cell.
One o f the m ost effective m ethods o f analyzing a crystal struc
ture is b y means o f xray or neutron diffraction. F rom the quantum m echanical p oin t o f view , determ ining the structure o f a solid does not differ, fundam entally from determ ining th at o f a molecule.
I t con
sists in finding a'stab le configuration o f nuclei find electrons which are subject to their electronic interactions and which m ove according to the laws o f quantum mechanics. T h e tw o main differences between the structure o f a solid and that o f a m olecule are the large number o f atoms in volved and the regu larity in their arrangem ent.
Several typ es o f approxim ations are used to study the structure o f
a solid, depending on the dom inant factors in volved in each solid. In this chapter w e shall an alyze som e o f these m ethods and ap p ly t hem in order to explain certain properties o f solids, leavin g some considerations o f a statistical nature for C hap ter 13.
β.2
Typott o f Solidtt
Solids m ay be classified according to the predom inant ty p e o f binding.
In this
section w e shall b riefly discuss the m ost representative typ es o f solids. (I)
C o v a le n t, s o lid s .
In a covalent solid the atom s are bound togeth er b y local
ized directional bonds sim ilar to those found in our discussion o f the H 2 molecule. T h e crystal lattice is determ ined by the orientation and nature o f the directional bonds. A typ ica l case is diam ond, in which the four bonding electrons o f each carbon atom are oriented as shown previou sly in F ig. 518 b y means o f sp3 hybrid
232
Solids
(6.2
F ig. 61. Diamond lattice.
w a v e functions, resulting in the crystal structure indicated in F ig. 61. Each ball represents a carbon nucleus and each bar a pair o f localized bonding electrons. T h e separation between tw o carbon atoms is 1.54 X IO10 m. C ovalen t solids, because o f their rigid electronic structure, exhibit several com m on m acroscopic features.
T h e y are extrem ely hard and difficult to deform .
AU
arc poor conductors o f heat and electricity because there are no free electrons to carry energy or charge from one place to another. A lso a rela tiv ely high energy is required to excite wholecrystal vib ration s in a covalent solid due to the rigid ity o f the bonds. W h olecrystal vib ration s therefore have a high frequency. Sim ilarly, electronic excitation energies o f covalent solids are o f the order o f a few electron v o lts (fo r exam ple, the first electronic excitation energy o f diam ond is about G eV ). T h is electronic excitation energy is rela tiv ely large com pared w ith the average I therm al energy (o f the order o f k T ) , which at room tem perature (298 cK ) is about 2.4
X IO 2 e V ; hence covalent soljjls are n orm ally in their electronic ground state.
M a n y covalen t solids are transparent, especially diam ond, because their first electronic state is higher than the photon energies in the visible spectrum, which lie between 1.8 and 3.1 eV. (2 )
I o n i c c r y s ta ls .
A t the other extrem e are ionic crystals, which consist o f a
regular array o f p ositive and negative ions resulting from the transfer o f one eleci tron (o r m ore) from one kind o f atom to another.
T h is is, for exam ple, the case
for N a C l and C sC l, whose crystal structures are shown in Fig. 62. T h e separation between the N a and C l atom s is about 2.81 X 10~ 10 m, while the shortest distance between identical atoms is 3.97 X IO1 0 m. T h e ions are so arranged th at a stable configuration is produced under their m utual electronic interactions.
Types o} solids
6.2)
233
(a) F ig . 62. (a) N aC l lattice; (b ) CsCl lattice.
T hese solids, because th ey have no free electrons, are also poor conductors of heat and electricity.
H ow ever, at high tem peratures the ions m ay gain som e m ob ility,
resulting in better electrical con du ctivity.
Io n ic crystals are usually hard, b rittle,
and have a high m elting point due to the rela tiv ely strong electrostatic forces be tween the ions. Som e ionic crystals stron gly absorb electrom agnetic radiation in the far infrared region o f the spectrum. T h is p rop erty is associated w ith the energy needed for exciting lattice vibrations. T h is energy is gen erally low er for ionic than for covalent crystals, due to their rela tiv ely weaker binding force.
F igu re 63
shows the transniission o f infrared radiation through a thin (1.7 X IO 7 m ) sodium 7 chloride film ; m axim um absorption occurs at a w avelen gth o f 6.11 X IO 5 m or a ' frequency
2.03
X
of
4.91 X IO 12 H z,
corresponding
to
photons
w ith
I t r 2eV.
Fig. 63. Transmission of infrared radiation through a N aC l film. [Adapted from R. Barnes, Z . Physik 73, 723 (1932)]
L
λ, IO
an
energy
of
234
Solids
(6.2
F ig . 64. Electronic configura tions of L i + and F  .
K
N
H H
11 H
Lr
M o s t ionic crystals are diam agnetic because the ions, having a com plete shell structure w ith all electrons paired, have no net m agnetic m om ent.
F o r exam ple,
in L iF , the ions L i + and F  have the configurations shown in F ig. 64, w hich resemble, resp ectively, the configurations o f the inert gases H e and N e (rem em ber F ig. 4 7 ). T h is sim ilarity is also reflected in another im portant p rop erty. T h e ions are spherically sym m etric, and thus their binding does not show directional preference, as do those o f covalent solids. T h e refo re the ions are arranged in the crystal as if th ey w ere closely packed spheres. T h e re are an infinite number o f w ays o f packing spheres.
T w o com m on ones are shown in F ig. 6 5 : the cubic
closepacked (ccp ) and the hexagonalclosepacked (hep ).
F ig . 65. (a) Cubic closepacking (ccp) of spheres; (b) exploded view of ccp; (c) hexagonal closepacking (hep) of spheres; (d) exploded view of hep.
(d)
Types of solids
6.2) (3) H y d r o g e n  b o n d s o lid s .
235
C losely related to ionic crystals are the hydrogen
bond solids, w hich are characterized b y stron gly polar molecules havin g one or m ore hyd rogen atoms, such as w ater, H 2O, and hydrofluoric acid, H F .
T h e posi
tiv e hydrogen ions, since th ey are rela tiv ely small, m ay attract the n egative end o f other molecules, form in g chains such as (H + F  )
(H + F  )
(H + F  ) . . .
Th is is p articularly interesting in the case o f ice, in which the w ater m olecules have the tetrahedral arrangem ent shown in Fig. 6 6. T h e rela tiv ely open structure o f ice accounts fo r the larger volu m e which ice has b y com parison w ith w ater in the liquid phase. (4 ) M o le c u la r s o lid s . These solids are m ade o f substances whose m olecules are not polar. A ll electrons in these molecules are paired, so th at no covalen t bonds between atom s o f tw o differen t molecules m ay be formed. o f solid retain their ind ivid u ality.
M olecules in this typ e
T h e y are bound b y the same interm olecular
forces that exist between m olecules in a gas or a liquid: van der Waals forces, which are v e ry weak, and which correspond roughly to the forces between tw o electric dipoles. T h is m ay be explained as follow s: Although, on the average, these m ole cules do not have a perm anent electric dipole m om ent, their electronic configura tion at each instant m ay g iv e rise to an instantaneous electric dipole. T h e van der W aals forces result from the interaction o f these instantaneous electric dipoles. F or the ab ove reasons m olecular solids are not conductors o f heat and electricity, have a v e ry low m elting point, and are v e ry com pressible and deform able. E xam ples o f m olecular solids are C H 4, C l2, I 2, C O 2, C eH e, etc., in their solid state. In e rt gases, whose outer shells are com plete, solid ify as m olecular sol ids. F igu re 67 shows the potential energy curves describing the inter action between tw o atom s o f an inert gas in the solid state.
F ig. 6 6. Arrangement of water molecules in ice. [From L. Pauling, The Nature of the Chemical Bond. Ithaca, N . Y . : Cornell University Press, 1960; Iiy permission of the publisher.]
236
Solids
(l6.2
F ig. 67. Van der Waals interatomic potential energy for inert gases.
(5 )
M e t a ls .
F in ally, there are the metals, which are solids o f great practical and
theoretical im portance. M eta ls are elements which have rela tiv ely small ionization energies, and whose atom s have on ly a few w eak ly bound electrons in their outer m ost incom plete shells.
These outerm ost w eak ly bound electrons are easily set
free using the energy released when the crystal is form ed. A m etal thus has a regu lar lattice o f spherically sym m etric positive ions th at rem ain when the outer most electrons are set free. T hroughout this structure there is an electronic “g a s " form ed b y the released electrons which are responsible for the bonding. These electrons m ove, m ore or less freely, through the crystal lattice and th erefore are not localized. M e ta llic solids exhibit excellent therm al and electrical con du ctivity, for which the free electrons are m ainly responsible, the reason being th at the free electrons easily absorb any energy from electrom agnetic radiation or lattice vib ration s— no m atter how sm all— and thus increase their kinetic energy and their m ob ility. F o r this same reason m etals are also opaque, since the free electrons can absorb the photons in the visible region and be excited to one o f the m any eloseb’y quantum states availab le to them.
T h e free electrons are also largely responsible for an
other characteristic exhibited by m etals: their high reflection coefficient for elec trom agnetic waves, especially in the radiofrequency and infrared regions.
Forces
holding the m etal lattice togeth er are spherically sym m etric; therefore these lat tices resem ble the closely packed spheres discussed for ionic crystals. T h e preceding classification o f types o f solids should not be taken too s trictly; som e solids are a m ixture o f m ore than one type. A n interesting exam ple is graphite
Types of solids
6.2)
237
F ig. 68. Graphite lattice. (th e graph ite lattice is shown in F ig. 6 8 ), which consists o f layers o f carbon atoms arranged in the form o f a hexagon. T h e atom s in a layer are bonded by localized covalent σ bonds which use sp2 h ybrid w ave functions and nonlocalized Tbonds, as in benzene (Section 5.6). T h e nonlocalized rrbonding electrons are free to m ove parallel to the layers, which explains the electrical con d u ctivity o f graphite parallel to the layers but n ot perpendicular to them.
Successive layers o f atom s act as
macromolecules. These layers are held togeth er b y weak van der W a als forces, just as m olecular crystals are, which accounts for the flaky, slippery nature o f graphite. In fact, it is because o f this structure that graphite is used as a lubricant. T h e length o f the sp 2 bonds in graphite is I .42 X I O ' 10 m and 'the separation between thei layers is 3.35 X I 0 IO m. O
O
O
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Ό (a)
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O (b)
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O (c)
O
O
O
O
O (d)
O
O
° n °
° n °
O O
O
O
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O O
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(e)
F ig. 69. Imperfections in a crystal lattice.

238
Solids
(6.2
T h e structure o f each o f the types o f solid th at w e have described is determ ined by the electronic structure o f the com ponent atoms, which indicates the ty p e and number o f electrons available for bonding, as w ell as the energy required to adjust their m otion to the conditions prevailin g in the lattice (conditions which are d if ferent from those in the isolated atom s). In general, when a solid is form ed, a given am ount o f energy is required in order to m od ify the m otion o f the electrons; how ever, much m ore energy than this is released in the process o f lattice form ation. C rystal lattices are not perfect, and the im perfections in them m ay be due to several causes.
F igu re 69 illustrates som e o f the m ost typ ical im perfections.
In
(a ) w e have a vacancy left by a missing atom, w h ile in (b ) w e have a substitutional im p u rity replacing an atom o f the lattice.
Im p erfection s due to an interstitial
atom— either o f the same ty p e or o f an im pu rity— are shown in (c ) and (d ). F in ally (e ) shows an edge dislocation, w hich m ay be looked upon as if an extra layer o f atom s had been pushed halfw ay through the top o f the lattice.
L a ttic e im perfec
tions have a m ost im portant effect on m any electric, elastic, and optical properties o f solids. E X A M P L E 6.1.
Calculation of the internal potential energy of an ionic crystal.
S o lu tio n : Since, in a first approximation, we may consider an ionic crystal as a regular array of positive and negative point charges, we may expect that the internal potential energy of an ionic crystal is essentially due to the electric interaction between the ions. Let us consider, as an example, the case of NaCl, and calculate the electric interaction of a N a + ion with all other ions in the crystal. As we can see from Fig. 62, each N a + has 6 C l ions as nearest neighbors, all at the same distance. Thus, since both N a + and C lhave charges which are equal in absolute value to e, the attractive electric potential energy of N a + due to the 6 nearest C l ions is
where R is the distance between nearest neighbors. The next series of ions are 12 N a + ions at a distance s/2 R, resulting in a repulsive potential energy
N ext follows a group of 8 Cl  ions at a distance \Ζ3 R, producing an attractive potential energy
The next layer contains 6 N a + ions at a distance 2R, producing a repulsive potential
Types of solids
6.2)
239
This process continues until all the crystal is included. The resultant potential energy is their sum or
" '   S £ » ( "  ί  + ά  3+
)   ώ
·
where a is the value of the sum inside the parentheses, and is called Modelling's constant. In the case of a facecenteredcubic lattice such as NaCl, the Madelung constant is a = 1.7476. In general, a depends only on the geometry of the crystal; for a bodycentered lattice, such as CsCl, its value is 1.7627. Since a is positive, the potential energy is negative and the interionic force is attractive at all distances, with no minimum. Therefore the crystal should coalesce into a closely packed structure with no stable configuration. However, we know that this is not the case. The conclusion results from considering the ions as point charges. When two atoms come very close together the nuclear repulsion (partially screened by the electron shells) and the repulsion among the filled electron shells enter into effect. Therefore we must add a shortrange repulsive term to the potential energy. W e did this before, in Sec tion 5.4, when we were discussing ionic bonds. T o obtain this repulsive term we assume, on an empirical basis, a reasonable expression for the potential energy. Several expres sions have been proposed, but we shall consider only the simple one suggested by the German physicist, Max Born (1882):
K
F
“ 47Γ£οβ ” ’
where β and n are two constants to be determined. When we use R n with n > I, the repulsion is a shortrange force which is negligible at large distances in comparison with the coulomb ionic attraction. When this potential energy is added to Eq. (6.1), the effec tive potential energy is
£’   ά ( ΐ  £ ) '
E —> F —* C (F ig . 629).
256
Solids
(6.5
F ig. 629. Electronic motion in first and second Brillouin zones in a linear lattice. Under the action of an external force, the wave number and the energy of the electron vary within a zone, as shown by the arrows.
F ig. 630. Velocity and acceleration as a function of k in the second Brillouin zone of a linear lattice.
Electron motion in a periodic structure
6.5)
257
T h e variation o f the v e lo c ity and the acceleration o f the electron in the second Brillouin zone is shown in Fig. 630.
Sim ilar logic applies to the other Brillouin
zones. A n im portant conclusion to be derived from th e ab ove kinem atical description is that an external force cannot rem ove an electron from a BrilIouin zone; there fore the electron remains in the same energy band unless enough energy is gained in a single process (as b y absorbing a p hoton) to cross the energy gap and pass to the next zone.
(T h e re is, how ever, a certain p rob a b ility th at an electron under an
applied force w ill bridge the gap at k — ή ζη π / α ; this is called the Zener effect. H ow ever, w e shall not discuss it here.) I t has been found con ven ien t to define an effective mass m * of the electron accord ing to the relation m * =
5*
F /a. H e re F is
the external force applied to the electron and o the actual acceleration due both to F and to the lattice interaction. Thus we cannot expect th at m * w ill be the same as the electron mass TOc, nor should we expect it to be a constant. C onsidering that a =
dv/dl =
k
(dv/dk) dk/dt,
and using Eqs. (6.22) and (6.23), we have
m
Fig. 631.
(6.24)
d 2E / d k 2
Effective mass as a func
tion of k.
N o te th at when the electron is free and its energy is given b y E q. (6.11), w e have m * = TOe. O bviou sly to* is a function o f the param eters o f the lattice and o f the electron’s lattice m om entum hk. F ro m the graphs o f /iversusk (F ig . 62 5), we see that to* is p ositive at the b ottom o f an energy band and n egative at the top. I t becomes v e ry large, actually infinite, at the inflection point o f the energy curve; that is, at the m axim um o f v in Fig. 628. T h e variation o f to* w ith k in the first Brillouin zone is shown schem atically in Fig. 631. T h e values o f in * a t the bottom o f the energy band (a t k =
0) are given for certain m etals in T a b le 62.
T A B L E 6—2
Effective M ass
M ctal
Li
Na
K
Rb
Cs
m*/me
1.40
0.98
0.94
0.87
0.83
W e shall com plete our description o f electronic m otion in a periodic lattice b y considering the density o f states Q (E ) — dn/dE.
W e have already discussed this
density for the freeelectron m odel [E q. (6.15) and Fig. 620].
A t the bottom of
258
(6.5
Solids
the band th e d ensity o f states closely resembles the parabolic curve o f the freeelectron m odel, but instead o f increasing steadily, the curve decreases alm ost p arabolically at the top o f the band. T h is is shown in F ig. 632, w hich m ust be considered as on ly qu a litative. T h e actual shape o f dn/dE depends on the struc ture o f the lattice and the position o f the band. T h is m ore or less sym m etric shape o f dn/dE can be easily understood.
Suppose th at a band is com pletely filled.
If
one electron is rem oved (perhaps to the conduction band), it is possible to say th at a hole has been created, since there is now a vacan t s tale in the band. W h en an external force (such as an electric field) is applied, some electron m ay be m oved into the vacan t state, th ereby filling the hole. B u t in so doing, this electron leaves a new hole in the band corresponding to the state it previou sly occupied. W e can say th a t the hole m oves in a m anner exa ctly opposite to th at o f the electrons under the applied force, and thus acts as a p ositively charged particle. T h e zero o f energy for the holes is at the top o f th e band and th eir energy is measured d ow n w ard ; th a t is, Fmax — E .
T h erefo re d n/dE a t the to p o f the band is v e ry sim ilar
to E q. (6.15), but w ith E replaced by Fmax
F.
dn dE
F ig. 632. in a band.
Density of energy states
0
«K P
T h e student m a y realize th at in our d escriptive analysis o f electron m otion in a periodic lattice, w e have blended the quantum and classical pictures o f the elec tron. W e h ave done this in order to obtain the basic results desired w ith ou t be com ing in vo lved in com plex quantum m echanical calculations which are beyond the scope o f this book.
E X A M P L E 6.4. Calculation of the average value of the momentum of an electron described by the wave function φ = e'klu (x ). S o lu tio n : Assuming that the wave function φ (χ ) is normalized to unity, we have j φ*φ dx  J u * u dx = I.
Applying Eq. (2.50), we may write the average momentum of the electron as
6 .5 )
Kledrori mol ion in a periodic structure
259
But
and, since Φ* = ce ikxu ,kxu *(x * (x),), we obtain
Because the normalization condition makes the first integral on the right equal to I, we may write
where p„,aVe corresponds to the second term, and may be considered as the average momentum of the electron as a result of its interaction with the ions forming the lat tice, since this is the physical meaning we have attributed to « ( z ) . In this way we con clude that the average momentum of the electrons is composed of tw’o parts: the lattice, or quasifree, particle momentum hk and the momentum due to the interaction with the lattice. E X A M P L E 6.5. Proof of Bloch's theorem, which states that u (x ) in Eq. (6.19) is a peri odic function with the same period as the lattice spacing. S o lu tio n : Let us consider a linear lattice of spacing a, such that E p(X) = E p(x + a). Then, since the probability distribution of the electrons must show the same periodicity as the potential energy, we may write \i(x)\2 = \ψ(χ + α1φ[χ — (η — l)e ], n I f the number N of atoms which constitute thelattice isvery large sothat we may dis regard theend effects, the summations appearing inbothexpressions of u are identical and u (x ) = u (x + a), as required by Bloch’s theorem. T o find the average energy of an electron described by the wave function (6.26), we apply Eq. (2.51); that is,
£‘ n = % 4 d x ’
(6 27)
where H is the hamiltonian operator of the electron, given by
and E p(x ) is the periodic potential energy of the electron in the lattice. A straightforward calculation, which we omit, gives E ave = Eat — ot — 2/3 cos ka
(6.28)
where Ent is essentially the energy of the atomic state associated with wave functions φ(χ — na) and a and /3 are appropriate constants. Equation (6.28) shows that the values of Eavc are between E l l — a — 2/3 and E„t — a + 2/3, depending on the value of k. In other words, the width of the band is 4/3. The values of E, a, and β depend on the atomic state through the wave functions φ; therefore there are a series of energy bands, each correlated with an atomic state. Figure 633 shows the expression (6.28) plotted for several bands. The portions corresponding to each of the Brillouin zones are indicated by heavier lines. The frceparticle energy is also shown by the dashed parabola. Note the similarity to Fig. 625. Instead of choosing several ranges of k to express the Brillouin zones, we could have limited k to the range —π/β < k < ττ/α for all bands and used only the central portions of the curves for the different zones.
6.6)
Conductors, insulators, and semiconductors
261
F ig . 633. Allowed energies in the tightbinding approximation.
li.ti
t'om iuctorH, I nsnlalors. anti Svm U'onduetorH
An interesting property o f solids is their electrical conductivity. Som e materials, traditionally called insulators, are extremely poor conductors o f electricity (examples are diamond and quartz and, in general, m ost covalent and ionic solids). Other solids are exceedingly good conductors o f electricity; in this group fall the metals, such as copper and silver. (T o get a quantitative idea, consider copper, whose electrical conductivity at room temperature is IO20 times greater than that of quartz.) Intermediate between these tw o extreme groups is a third class of solids, called semiconductors. Although sem iconductors arc much poorer electrical conductors than the metals, their condu ctivity increases w ith the temperature, while that o f metals decreases w ith the temperature. T ypical sem iconductors are germanium and silicon. One of the most im portant reasons for the initial success o f the band theory of solids was that it offered a simple explanation of this markedly different electrical behavior o f solids. W e shall make our first analysis by means o f the frecclectron model, which we shall later refine by taking into account the periodic structure of the lattice.
262
Solids
(6.6
L e t us consider a m etal havin g the band structure shown in Fig. 634, which m igh t correspond, for exam ple, to the energy levels o f sodium (Z =
11).
Bands
corresponding to the Is, 2s, and 2p atom ic levels are com pletely filled because the respective atom ic shells are also com plete.
B u t the 3s band, which can accom
m odate up to tw o electrons per atom , is on ly half filled, since the 3s level in each sodium atom has on ly one electron. D u e to therm al excitation, some electrons in the 3s band, which have an energy o f the order o f the F erm i energy for this band, pick up energies of the order o f k T (abou t 0.025 e V at room tem perature) and their distribution am ong th e energy states o f the band resembles th at o f F ig. 622.
U n der the action o f an external electric field, these electrons m ay,
w ith ou t vio la tin g the exclusion principle, pick up additional small amounts o f energy and pass to any o f the m any nearby e m p ty states within the band. In sharp distinction to disordered therm al excitation, the electrons excited b y the electric field gain m om entum in the direction opposite to the field; this results in a collective m otion through the crystal, which constitutes an electric current. T h erefo re w e conclude that a substance having a band structure such as th at o f F ig. 634 should be a good conductor o f electricity, and for the same reason it should be a good therm al conductor, w ith the electrons in the upperm ost par tia lly occupied band being responsible for both processes. In other words, the good conductors o f electricity (also called metals) are those solids in which the upper m ost occupied band is n ot com pletely filled.
Empty " band
I
F ig. 634. ductor.
H a lf
Energy bands in a con
In n er
2s: Is 
filled bands
A c tu a lly the situation is sligh tly m ore com plex because o f the possible super position o f the upperm ost bands. sodium.
F igu re 635 shows the actual band structure of
A t the equilibrium distance r 0 in the m etal, about 3.67 X IO1 0 m, the
2p level remains p ractically undisturbed, but the bands corresponding to the 3s and 3p atom ic levels overlap.
T h erefo re the conduction electrons have m any
m ore states availab le than those corresponding to the 3s band alone. In fact, this overlap pin g o f the outerm ost bands is the com mon situation for m ost m etals or conductors. F o r exam ple, consider the case o f magnesium ( Z = 12). T h e m ag nesium atom has the configuration I s 2 2s2 2 p e 3s2, and therefore all the atom ic shells are filled.
H ow ever, the first excited level, 3p, is rather close to 3s.
In the
Conductors, insulators, and semiconductors
6.6)
263
D 3d
10
s
2 0 
0
3.67 5
10
In tern u clea r distance, IO10 I
F ig. 635. Energy bands of sodium.
solid state the band structure is similar to th at o f F ig. 635 for sodium. and 3p bands are indicated schem atically in Fig. 636.
T h e 3s
N o rm a lly , w ith no over
lapping, the 3s band should be filled and the 3p band em pty, and magnesium should be an insulator, as explained below. B u t because o f the overlappin g, the upper m ost electrons o f the 3s band have the low est energy states o f the 3p band a v a il able. Thus some 3s electrons m ove to occupy som e low 3plevels until an equilib rium energy level for both bands is established. Since the total number o f energy levels availab le from the 3s and 3p bands is 2.V ) 6iV = 8.V and w e h ave on ly 2N electrons, there are 6N accessible em p ty states.
T h erefo re m agnesium should
be a good conductor; this is in agreem ent w ith the experim ental facts. T h ose sub stances whose atom s have com plete shells but which, in the solid state, are con ductors because o f the overlap pin g o f a filled band and an e m p ty band are often called semimetals.
■3p
3s
F ig. 636. Overlapping of energy bands in a conductor. "2 p
264
Solids
(.6.6
Con du ction M m nd
J (em pty) _
Ω
Im rge en erg y gap
V alen ce M m nd J (fille d )
dn dE
F ig. 637.
Energy bands of an insulator.
F ig. 638. Energy band structure for diamond (C ) 1 silicon (Si), and germanium (G e).
I n the transition m etals group, such as iron, the overlap pin g bands are 3d, 4s, and 4p, and the number o f electrons is insufficient to fill these bands.
Sim ilarly,
in the rareearth group, the overlap pin g bands in vo lved are 4f, 5d, 6s, and 6p. H ence these elements, when in the solid state, are conductors. L e t us now consider the case o f a substance in w h ich the uppermost or valence band is com p letely filled and does not overlap the next band, which is to ta lly e m p ty (F ig . 6 3 7 ). Since all states o f the valence b a w f are occupied, the electron energy is “ fro zen ,” th at is, the electrons cannot change their state w ith in the band w ith ou t vio la tin g the exclusion principle. T h e on ly possibility for exciting an elec tron is to transfer it to the e m p ty conduction band; but this m a y require an energy o f a few electron volts.
H ence an applied electric field cannot accelerate th e elec
trons in the valence band, and thus cannot produce a net electric current. substance is therefore an insulator.
T h is
(O f course, at a su fficiently high tem perature
or under v e r y strong electric fields, some electrons m ay be excited to the conduc tion band, and then an electric current is possible.) M o s t covalen t solids, w hich are composed o f atom s h avin g an even number o f valence electrons, are insulators. Figure 638 shows a sim plified band scheme of diam ond ( C ) . T h e bands correspond to the atom ic 2s and 2p levels in diam ond, which can accom m odate up to 8 elec trons.
H o w ever, the carbon atom has on ly 4 electrons available for these levels
(rem em ber Figs. 47 and 4 1 0 ).
A s the atom s g e t closer, the 2s and 2p bands
begin to overlap. A t smaller interatom ic distances, th ey sp lit again into tw o bands, each accom m odating up to 4 electrons per atom. (T h is conclusion is arrived at by a detailed calculation.)
H ence the 4 electrons from each atom are norm ally in
the low er or valen ce band, w h ile th e upper band is em pty.
A t the equilibrium
distance in diam ond, about 1.5 X IO10 m (ind icated by C in Fig. 6 3 8 ), the gap separating the low est or valence band from the upper e m p ty band is about 5 eV. T h is m ay be considered as a r e la tiv e ly large energy gap ; it explains w h y diam ond is such a good insulator.
I,
j
Conductors, insulators, and semiconductors
6.6)
265
Con du ction band Sm all en ergy gap
Fig. 639. Energy bands and electron distribution in a semi conductor.
V alen ce band
T h e sameafaand scheme also applies to silicon and germ anium (excep t th a t the bands correspond to differen t atom ic energy levels and en ergies); the equilibrium separation o f the atom s in their solid states is also shown in F ig. 638. N o w , how ever, the gap between the valence and conduction bands a t the equilibrium separa tion o f the atom s is much sm aller (1.1 e V in silicon and 0.7 e V in germ anium ), and this makes it much easier to excite the upperm ost electrons in the valence band into the conduction band. T h e situation is illustrated in F ig. 639. A s the tem pera ture increases, m ore electrons are able to ju m p into the next band.
T h is has tw o
results: T h e few electrons in the upper or conduction band act as th ey would in a m etal, and the em p ty states, or holes, left in the low er or valence band act in a similar w ay, but as if th ey w ere p ositive electrons (also, their effective mass m ay be d ifferent because th ey are in a differen t energy band).
Thus w e h ave electric
conduction from the excited electrons in the conduction band and from the holes in the valence band; the con d u ctivity increases rap id ly w ith the tem perature b f cause m ore electrons are excited to the conduction band.
F o r exam ple, in silicon,
the number o f excited electrons is increased b y a factor o f IO6 when the tem pera ture is raised from 250 0K to 450 °K . H ence semiconductors are insulators in which the energy gap betw een the valence band and the conduction band is about one e V or less, so th a t it is re la tiv ely easy to th erm ally excite electrons from the valence to the conduction band. T h e energy gaps o f som e insulators and semi conductors are g iven in T a b le 63. T A B L E 63
E n ergy G a p s (e V )
Insulators
eV
Semiconductors
eV
Diamond Zinc oxide Silver chloride Cadmium sulfide
5.33 3.2 3.2 2.42
Silicon Germanium Tellurium Indium antimonide
1.14 0.67 0.33 0.23
T h e electrical conduction in sem iconductors w hich w e have described is called in trin sic conductivity.
T h e con d u ctivity o f a sem iconductor can also be enhanced
b y the ad dition o f certain im purities. Suppose th at w e replace som e o f the atom s o f the sem iconductor b y atom s o f a different substance (these atom s then constitute
266
Solids
II
III I I
II II
J T___
I
111
III
I I
III
•w
t
(.6.6
■*—
LConduction
}
I band Impurity levels — — — f———
L Valence J band
ί
——f— — 4 I
Conduction band
L Valence
J band (b)
(a)
F ig . 6—tO. Impurities in a semiconductor: (a) donors, or ntype, (b) acceptors, or ptype.
the im p u rity ), and suppose th at these im pu rity atom s have more electrons than those o f the semiconductor.
F o r example, if to silicon or germ anium , w hich con
trib u te four electrons per atom to the valence band, vve add a few atom s o f phos phorus or arsenic, each o f w hich contributes five electrons per atom to the valence band, w e have an extra electron per im pu rity atom. These additional electrons (w hich cannot be accom m odated in the valence band o f the original lattice) occupy some discrete energy levels just below the conduction band; the separation m ay be a few tenths o f an eV (F ig. 6 4 0 (a )l· These excess electrons are easily released by the im pu rity atom s and excited into the conduction band.
T h e excited electrons
then contribute to the electrical con d u ctivity of the semiconductor. Such im pu rity atoms are called donors; the semiconductor is called an n type (o r n egative) semiconductor. C onversely, the im pu rity m ay consist o f atoms havin g fewer electrons than those o f the semiconductor.
F o r the cases in which silicon and germ anium are the host
substances, the im pu rity atoms could be boron or aluminum, each o f w hich con tributes on ly three electrons.
In this situation the im pu rity introduces vacant
discrete energy levels, v e ry close to th e top o f the valence band (F ig . 6  4 0 (b )). T h e refo re it is easy to excite some o f the more energetic electrons in the valence band into the im pu rity levels. the valence band. trons.
T h is process produces v acan t states, or holes, in
A s explained previou sly, these holes then act as p ositive elec
Such im pu rity atom s are called acceptors; the sem iconductor is called a
p typ e (o r p ositive) semiconductor. T o produce significant changes in the con d u ctivity o f a semiconductor, it is sufficient to have about one im pu rity atom per m illion sem iconductor atoms. Sem iconductors have a w ide industrial application as rectifiers, m odulators, detec tors, photocells, transistors, etc. E X A M P L E 6.7.
Discussion of the pn junction.
S o lu tio n : One important application of semiconductors to modern electric circuitry is the pn junction. Suppose that we have two samples of the same semiconductor— say germanium— one of ptype and the other of ntype (F ig. 6—41 fa )) . I f the two samples are placed in contact (F ig. 6 4 1(b )), there is a diffusion or flow of holes from the left to the right and of electrons from the right to the left. This double flow produces a double layer of positive and negative charges on both sides of the junction, setting up a potential
Conductors, insulators, and semiconductors
6.6)
(a)
O
0
0
o : holes (positive) • : electrons (negative)
·
·
• 0 • 0
0 0
·
Oo
267
0
·
0 ·ο o ° .s O o ·
°
·
·
° 0· · — In «O* eο L · 0 · ··
(b)
(c)
f (d)
F ig. 641. The pn junction.
difference across the junction [as shown on the right in part (b )] which, when equilibrium is reached, opposes the further flow of holes and electrons across the junction. In the succeeding discussion we shall concentrate on the holes; the situation for the electrons is equal and opposite. Due to the recombination of holes and electrons, the number of holes in the ntype semiconductor tends to decrease, which allows a small hole current l\ to flow contin uously from the pside to the nside. A t the same time, due to thermal excitation, holeelectron pairs are produced in the ntype semiconductor, and these excess holes can flow very readily across the junction into the pside with a current 12 · A t equilibrium both hole currents are identical; that is, /1 = 12 (similar logic can be applied to the electrons).
268
{6.7
Solids
I f a potential difference V is applied, as in Fig. 6—41(c ), with the pside joined to the positive terminal and the nsidc to the negative terminal of the source of V, the height of the potential difference across the junction decreases. This allows a larger current I i to the right, without actually changing the thermally generated current /2 to the left. Thus a net hole current I i — /2 results across the junction to the right, and this current in creases very rapidly with V, due to the large supply of holes from the pside. On the other hand, if the potential difference V is reversed, as in Fig. 641 (d), the potential dif ference across the junction increases. This reduces the value of 11, again without sub stantially affecting 12 , since the supply of holes from the nside is temperature limited. Thus a net current to the left will exist across the junction which will approach the constant value 12 with increasing V . I = I i I 2
F ig. 642. Current as a function of voltage across a pn junction. The voltage V is considered positive when applied in the direction p —> n and negative when applied in the opposite direction.
V
Figure 642 shows the graph of the net current across the junction as a function of V, with V considered positive when applied as in Fig. 641 (c) and negative otherwise. The net current is expressed fairly accurately by the expression I = /1 
/2 = I 2(eVlkT 
I).
W e conclude that a pn junction acts as a rectifier or a detector device favoring the passage of a current in the direction p —» n. This is the same function performed by diode and triode electron tubes, but the pn junction may perform it with an expenditure of considerably less energy.
6 .7
H u a n tu m Thpwry o f E le e trira l C o n d u etirity
In the preceding section we have discussed electrical conduction in solids from the p oin t o f v ie w o f the freeelectron model. T h e periodic structure o f the solid must be taken into account, how ever, and the results o f Section C.5 must be used in order to obtain qu a n tita tive results. on a onedim ensional model.
F or sim plicity, w e shall base our discussion
L e t us consider the solid to be in its ground state,
Quantum theory of electrical conductivity
6.7)
E
E
O o tio n to le ft
269
a M o tio n to right
a M o tio n to le ft
(a)
O
k a
M o tio n to righ t (b )
F ig. 643. Occupation of energy levels in the first Brillouin zone: (a) no electric field applied, (b) external electric field applied from right to left.
w ith no electric field applied, and assume th at the electrons are in the first Brillouin zone (th e same logic, how ever, applies to any other zone). T h e electrons are occupy ing the low est states w ith in the band, in a sym m etric form , so th at no net current exists (F ig . 6  4 3 (a )). I f an electric field is applied, all electrons experience a force in a direction opposite to th e field and, according to E q. (6.23), th eir fcvalues in crease in the direction o f the force. T h e result is an asym m etric distribution o f the electrons w ith in the m etal, as shown in F ig. 6 4 3 (b ). T h is g ives rise to a net electric current in the m etal, since m ore electrons m ov e in one direction than in the opposite direction. A s long as the electric field is applied, the occupation o f states w ith k parallel to the force increases w ith tim e and the occupation o f those states w ith k opposite decreases w ith tim e.
In other words, the current increases continuously w ith
time, even if the electric field is constant, due to the continuous acceleration o f the electrons.
(W e shall ignore th e com plex effects th at m ay result when som e elec
trons even tu ally reach the boundary o f the B rillouin zone a t fc =
ττ/α; this problem
has previou sly been discussed in Section 6.5.) T h e ab ove result contradicts experi ence.
M o s t conductors ob ey O hm ’s law, w hich m ay be w ritten either in the
fam iliar form V =
R l or in the m ore convenient form
j = σ δ,
(6.29)
where j is the current density, expressed in A m  2 , 6 is the applied electric field, expressed in N C  1 , and σ is the conductivity o f the substance, expressed in Ω1 m —'. T h e reciprocal o f the con du ctivity, P =
Ι/σ',
(6.30)
is called the resistivity, and is expressed in Ω m. In either of the tw o forms, O hm ’s law states th at a constant electric field produces a constant electric current; th at is, when an electric field is applied, the conduction
270
Solids
(6.7
electrons in a m etal acquire an average and constant d rift velocity.
T h erefo re we
conclude th at there must be some mechanism which prevents the electrons under the applied electric field from acceleratin g indefinitely up to the top o f the con duction band. In the classical theory, form ulated by D rude and L oren tz before the advent o f quantum theory, this constant average v e lo c ity o f the electron resulted from the frequ en t collisions o f the electrons w ith the p ositive ions which constitute the m etal lattice.
In such collisions, an electron was supposed to transfer the m om en
tum it had gained from the electric field since the previous collision to the ion, and this prevented the electron from being accelerated continuously as it drifted in the direction opposite to the applied electric field. T h is mechanism also explained the Joule effect as being due to the energy gained b y the ions from the electronic collisions.
In the D ru d eLoren tz theory, the con d u ctivity is given by σ =
ne2T/mc,
(6.31)
w here n is the number o f electrons per unit volu m e and r is a param eter called the relaxation time. T h is expression can be derived as follows. T h e acceleration o f an electron due to the applied electric field is a = — eS/?nc. I f I is the tim e between tw o successive collisions o f the electron w ith the lattice ions, the average d rift v e lo c ity o f the electron is Vt v e = j =
C T ivtve =
(n e 2l/2me)S ,
E q . (6.31) shows that r =
ja l =
which
— e(6)t/2m,..
yield s
σ =
T h e current density is then
ne2t/2m0.
Com parison
w ith
%t. H ence in the D ru d eLoren tz th eory the relaxation
tim e r is assumed to be o f the same order o f m agnitude as the tim e between tw o successive collisions o f the electron w ith the ions o f the lattice. F o r most m etals at room tem perature, the valu e o f r, com puted from the measured valu e o f σ, is o f the order o f IO14 s. G iven th at I is the avera ge separation o f the ions and v, is th e average therm al v e lo c ity o f the electrons in the absence o f the electric field, w e can assume that the relaxation tim e is o f the order o f m agnitude o f l/vt. F o r most solids I is o f the order o f IO 9 m. Using for th e electrons the same relation d erived for gas molecules, vt =
\/3kT/m e (see Section 10.6), w e find th a t at room
tem perature v, is o f the order o f IO5 m s  1 . w ith the experim ental valu e o f r.
T h en I fv l ~
IO14 s, in agreem ent
H o w ever, at tem peratures th at are low (but
n ot too close to absolute zero), the con du ctivity o f m etals varies approxim ately as the reciprocal o f the absolute tem perature (th a t is, σ ~ l / T or p ~ T ) , as shown in F ig. 6—14 for sodium.
T h is means th at the relaxation tim e also varies
reciprocally w ith th e tem perature. W hen w e com pare the experim ental valu e o f r at low tem peratures w ith th at of l/vi, in order to obtain a qu a n tita tive agreem ent w e would have to assume values fo r I m any times larger than the interionic separation.
T h is w as one o f the first
clues th at the D ru d eLoren tz theory was not correct. H en ce w e shall first try, using the quantum theory, to explain w h y the elec trons m aintain a constant average v elocity , and next obtain a qu a n tita tive expres sion for the con du ctivity, sim ilar to E q. (6.31). In the quantum theory, an electron is represented by a w a ve packet.
T h e re is
an en ergy spread centered about the energy E and an associated wavenum ber
Quantum theory of electrical conductivity
6.7)
271
spread centered about the w a v e number A. T h e m otion o f a w a ve packet m ay be hindered by scattering.
In itia lly the w a ve packet is m ovin g in a particular direc
tion w ith the w a ve number A ; a fter the scattering it is m ovin g w ith a different w ave number, say A·', in a different direction. In other words, scattering produces a transition A· —> A'. In the transition, some m om entum and energy are transferred to the scatterer. Th&^effect o f the applied electric field is to accelerate the electrons in a certain direction; the effect o f the scattering is to disarray the electron m otion, hindering the accelerating effect of the electric field.
S cattering in to e m p ty states
Fig. 6—14. Variation of resistivity with temperature.
F ig. 645. Scattering of electrons in first Brilloiiin zone by impurities.
A steady state is produced when these tw o effects balance each other, in a sta tistical sense, resulting in a constant average v e lo c ity o f the conduction electrons. F or this situation to exist, the scattering m ust occur at any Avalue within the conduction band.
Consider, for example, the onedimensional case shown in
Fig. 645, which is sim ilar to F ig. G 43(b).
Since scattering cannot v io la te the
exclusion principle, the electrons must be scattered into va c a n t states. Thus the elect rons that are ,scattered are t he most energetic ones. These elect rons are scattered into va c a n t states which, in a linear lattice, have opposite m om entum and (in gen eral) low er en ergy; the figure indicates this schem atically. T h e energy lost by the electron is absorbed by the scatterer. T h e result is that the most ener getic electrons frequ en tly reverse their m otion, thereby checking the accelerating effect o f the electric field. T h e number o f electrons m ovin g to the right and to the le ft is m aintained at a statistically constant difference, g ivin g rise to a steady cur rent.
A lth ou gh w e h a ve used a onedim ensional m odel, a sim ilar situation exists
for the threedim ensional case. W e t hen see that the idea o f scattering o f the elec tron w a v e packet accounts for both Ohm ’s law and the Joule effect. T h e latter results since energy and m om entum are transferred to the scatterers.
272
Solids
(6.7
Since w e h ave previou sly seen (Section 6.5) th a t an electron can m ove freely through a crystal lattice (excep t when k = ± 7 r / a ), our next task is to id en tify the sources o f the assumed scattering.
T h e general answer is the follo w in g: any
irregu larity in the p eriodicity o f a lattice disturbs the otherw ise free m otion o f the electron, and the disturbance can be considered as a scattering.
These lattice
irregularities are due to tw o factors: ( I ) I mperfections in the solid, such as vacant spaces, interstitial and displaced atoms, dislocations, and im purities; for exam ple, if small amounts o f im pu rity atom s are added, and these are un iform ly distributed throughout the solid, the con du ctivity is m odified. T h e contribution to the con d u c tiv ity due to scattering by im perfections in the lattice is essentially independent o f tem perature.
(2 ) Therm al oscillatory m otion o f the ions w hich constitute the
lattice. Since the ions do not all oscillate in phase, their vibrations g iv e rise to small fluctuations in the lattice spacing; although these fluctuations are small, th ey are spread over all the lattice.
In addition, because the oscillations increase the effec
tiv e cross section which the ions present to the m otion o f the electrons, the prob ab ility o f scattering is prop ortion ately larger.
C lea rly the latticevib ration effect
is tem peraturedependent, since the am plitude o f the vibrations depends on the vibration al energy and this in turn depends on the tem perature. T o be m ore qu a n tita tive, w e m ay still use E q. (6.31) to express the conduc tiv ity , replacing the electron mass mc by its effective mass m * ; th at is, σ =
(6.32)
ne2T/m*.
In general w e can calculate m * at the F erm i en ergy «κ, since the energy o f the conduction electrons is not v e ry differen t from e F. A lso n is not the to ta l number o f electrons per unit volu m e in the conduction band; rather it is the effective number o f electrons which p articipate in the conduction.
T h is number is smaller
than the total number of electrons in the conduction band, due to the restrictions imposed by the exclusion principle. G iv e n th at P , is the prob ab ility o f scattering o f the electrons per unit tim e (thus it is expressed in s 1 ), the relaxation tim e is given by r =
(6.33)
I / P ..
In other words, the larger the scattering p rob ab ility, the sm aller the relaxation tim e and the sm aller the con du ctivity, as our physical m odel requires.
L e t us
designate the scattering p ro b ab ility per unit length b y Σ , (thus it is expressed in m ~ ‘ ).
T h is q u a n tity is also called the macroscopic scattering cross section.
The
conduction electrons m ove w ith a v e lo c ity v e ry close to th at corresponding to the F erm i energy eF.
D esignating this v e lo c ity b y e F, w e have that P , =
t’F2 s, and
E q . (6.33) becomes T =
l/ c F2,.
(6.34)
T h e problem has thus becom e one o f calculating 2,. T h is is a difficult problem w h ich is beyond the scope o f this text. H ow ever, we can m ake certain estimates.
Quantum theory of electrical conductivity
6.7)
273
I t is natural to assume th at Σ , is proportional to the number o f scattering centers per unit volum e, designated b y n , (expressed in m ~ 3), and so w e m ay w rite 2, =
η ,σ ,, where σ , (n o t to be confused w ith t he con d u ctivity) is called the scat
tering cross section per center (it is expressed in m 2). W e can calculate σ„ b y using the techniques o f quantum mechanics if w e know the interaction between the electron and the scatterer.
T h e F erm i v e lo c ity Vp is,
by definition, tem perature independent. Also, in the case o f lattice im perfections, the term s appearing in 2 , are basically tem perature independent. Thus we v e rify that the con du ctivity due to lattice im perfections is independent o f the tem pera ture.
On the other hand, it is reasonable to assume (and it can be shown theo
retica lly) that for lattice v ib ra t ions σ » is proportional to the square o f the am plitude o f the ion oscillations; th at is, a s ~ A 2. B ut the energy o f an oscillator is propor tional to the square o f the am plitude, and we m ay w rite a s ~ vib ration a l energy. A t tem peratures not too close to absolute zero, the average vib ration al energy of a solid is proportional to the absolute tem perature T . T h erefo re w e conclude th at σ , ~ T , which, according to E q. (6.25), means th at τ ~ I / T .
T herefore, in v ie w
o f E q. (6.23), w e have that σ ~ I / T . W e thus v e r ify th at the con d u ctivity due to lattice vib ra t ion has the proper tem perature dependence. I f 2,.,· is the m acroscopic cross section due to the im purities and 2 , i( is th at due to the th erm al vib ration s o f the lattice, w e have 2 , =
2,.,· + 2 ,,t) so long as
we m ay assume th at the tw o scattering probabilities are ad ditive. I/ r =
νρΣ, =
T h erefore
AfQSj,i + £,.
T h is density is about IO 15 tim es greater than the
density o f m atter in bulk, and gives us an idea o f the degree o f compactness o f the nucleons in a nucleus.
I t also shows th at m atter in bulk is essentially em pty,
since m ost o f the mass is concentrated in the nuclei. (2 )
A n g u la r m o m e n t u m .
T h e resultant angular mom entum o f a nucleus is
called (fo r historical reasons) the nuclear spin. I t is designated by I. B oth protons and neutrons, like electrons, have spin
In addition, protons and neutrons
possess orbital angular m om entum associated w ith their m otion in the nucleus. T h e resultant nuclear angular m om entum (o r spin) is obtained by com bining, in a proper w ay, the orbital angular m om enta and the spins o f the nucleons com posing the nucleus.
T h e nuclear spin is designated b y a quantum number I such
th a t the m agnitude o f the nuclear spin is h\/7(7 +
I ).
T h e com ponent o f the
nuclear spin in a g ive n direction is given b y m/ft, where m/ =
± 7 , ± ( 7 — I ), . . . ,
or 0, depending on w hether 7 is a halfinteger or an integer. are 27 +
I possible orientations o f the nuclear spin.
T h erefore there
A s explained in Section 34,
these same rules are va lid for all angular m om enta in quantum mechanics.
E x
p erim entally the values o f I are integers (if .4 is even ) or halfintegers (if A is odd) ranging from zero, as in 4H e and 12C, up to 7, as in 176Lu.
T a b le 71 presents
some nuclear spins. I t has been noted that p ractically all eveneven nuclei (i.e., nuclei that have an even number o f neutrons and protons) have 7 = 0, which indicates th at identical nucleons tend to pair th eir angular m om enta in opposite directions. T h is is called the pa irin g effect. E ven od d nuclei (i.e., nuclei that have
Properties of the nucleus
289
an odd number o f either protons or neutrons) all have halfintegral angular mo menta, and it is reasonable to assume th at the nuclear spin coincides w ith the angular m om entum o f the last or unpaired nucleon, a result which seems to hold in m any cases.
Oddodd nuclei have tw o unpaired nucleons (one neutron and one
proton) and the experim ental results are a little m ore difficult to predict, but their angular m om enta are integers, as one would expect, since there is an even total number o f particles.
A n y theory o f nuclear forces, to be satisfactory, must
account for the experim ental values o f I. (3)
M a g n e t ic d ip o le m o m e n t .
W e m ay recall (see Section 3.0) that a m oving
charge possesses an orbital m agnetic dipole m om ent M t, which is proportional to its orbital angular m om entum L and which is g iven by M
l
=
(q / 2 m )L .
F o r the case o f protons, the charge is q = M
l
=
e, and therefore
(e/2mp) L .
(7.3)
O bviously neutrons have no charge and do not have an orbital m agnetic dipole moment. T h e com ponent o f the m agnetic mom ent along the Zaxis is
M l ., = (c /2 mp)L 2 = {ch/2mv)m i = μ κιη t, where the constant μκ =
eh/2mp =
5.0504 X IO27 J T 1
(7.4)
is called a nuclear magneton. I f the particle has spin S, it m ay have an additional spin m agnetic dipole m om ent given by Ms =
gs {e/2mp) S
or
M s ., =
g.sp\ms ,
(7.5)
where gs is a constant characteristic o f the particle, called the spin gyromagnetic ratio.* T h e value for the proton is gs .p = +5.5855, indicating that M s is parallel to S (as corresponds to a positive charge).
I t has been observed th at the neutron,
although it has no electric charge, has a spin m agnetic m om ent corresponding to
gs.n =
— 3.8263.
T h e negative sign indicates th at M s is antiparallel to S, as cor
responds to a n egative charge.
These results suggest th at the proton and the
neutron have a com plex structure which w e have not y e t been able to determ ine exactly. W e can obtain the resultant m agnetic dipole m om ent o f a nucleus b y combining, in a proper w a y , the m agnetic dipole mom ents o f all the nucleons. O bviously, the resultant m agnetic dipole m om ent must be related to the nuclear spin I . * Recall that for the electron gs = —2.0 (see Section 3.7).
The
290
Nuclear structure
(7.4
relation, which is an extension o f E q . (7.5), is
where = 87‘J M eV . We >ec that, because of the Z  term, the coulomb repulsion energy increases very rapidly with the number of protons. Fnless the nuclear interaction is strong enough and attractive, the nucleus cannot be stable. We must compare these electric energies with those of atoms and molecules, which are of the order of a few electron volts. The reason for the big difference is that the protons in a nucleus are much more closely packed than electrons in an atom, their separation being about one hundredthousandth as much as the average separation of electrons in atoms or molecules, resulting in an energy roughly 100,000 times greater.
? ..*
X u v lv a r H ind in y Iinvrfiy
W e can g et an idea o f the strength o f the nuclear interaction by studying the bind ing energy o f nuclei; th a t is, the energy required to separate the nucleons com posing a nucleus.
T h e mass o f a nucleus is appreciably less than the sum o f the
masses o f its com ponent nucleons. W0 =
F or exam ple, in the case o f the deuteron,
2.014102 umu. T h e masses o f the proton and the neutron arc W1, =
1.007825 anni
and
w„ =
1.008605 antu.
2,9'/
(7.5
Xurlear structure
T h c ir sum is Wip H m n = 2.016490 amu, w hich is larger than wz,j b y the am ount 0.002388 amu.
T h e explanation o f this loss o f mass lies in the binding energy o f
the system. C onsider one proton and one neutron a t rest and sufficiently far apart so that there is no interaction between them. T h e ir total energy is then the sum o f their rest masses; th at is, E =
WipC2 + WinC2 =
(wip Λ wi,Jc2.
N o w suppose th a t the tw o particles are close enough to form a deutcron nucleus as a result o f th eir nuclear interaction. energy.
L e t us call E p th eir nuclear potential
T h e y are m ovin g about their center o f mass and thus also h a ve som e
kinetic energy E k. T h erefo re th eir total energy is now E' =
(wip  f wi„)c2 + E k  f Ep.
But if the nuclear forces are a ttra c tiv e and the zero o f poten tial energy is chosen at infinity, E p is negative. A lso, since the deutcron is a stable system, E p must b e larger than E k in absolute valu e so that E k T E p is n egative (rem em ber th at in the case o f th e hydrogen atom, as w e ll as in all bound systems, th e total energy, excluding the mass energy, is n egative). T h erefo re, E ' is sm aller than E , or in other words, energy is liberated when the system o f tw o bound nucleons is form ed. T h e energy liberated is Eb =
E 
E 1,
(7.9)
w hich is called th e binding energy o f the system .
T h is am ount o f energy m ust be
give n back to the deuteron to separate the neutron and the proton.
W c can m ore
rea d ily understand these relations if we look a t Fig. 75. T h e nuclear binding energy in the case o f the deuteron is the equ ivalen t o f the ionization energy in the case o f t he hydrogen atom. E n ergy a t in finite separation E n erg y in Ixnind state
by = m.ic2
Ep
F ig. 73. Binding energy of deuteron.
A n "e ffe c tiv e ” mass o f the np system can be defined b y the relation E ' = resulting in a value for wia sm aller than mp T wt„. deuteron is then Eb = =
( » i p T Wi1J c 2 — WiaC2 = 931.48(wip T
Win
(wip T wi„ — wi.Jc2
— wia) M cV .
m,ic2,
T h e binding energy o f the
Nuclear binding energy
7.5)
295
In th e last expression th e masses must be expressed in amu. Since »ip + » i „ — wi,i = 0.002388 amu, th e binding energy o f the deuteron is E t =
2.224 M e V .
T h is is therefore the
energy required to separate the proton and the neutron in a deuteron, or the energy released when a proton and a neutron at rest are com bined to form the deuteron. C om parin g this energy w ith the equivalent result for hydrogen, whose ionization (o r binding) energy is o n ly 13.0 eV, w e conclude th at the nuclear about
IO'1 tim es stronger than electrom agnetic interactions.
interaction is
Incid en tally, no
measurable change in mass is observed in chemical reactions because th e energy in vo lved in most electrom agnetic processes is rela tiv ely small, o f the order o f a few eV , equ ivalent to a change in mass o f a few billionths o f mass units per atom. F o r a nucleus o f mass M composed o f A nucleons, o f w hich Z are protons and A
 Z
are neutrons, the binding energy (i.e., the energy required to separate all
nucleons) is E b= =
IZn ip + ( A — Z )m „ — M } c 2 931.48[Zmp + {A 
Z )m n 
M ) M eV .
(7.10)
In the second expression all masses must be expressed in amu. A good indication o f the stab ility o f a nucleus is the average binding energy per nucleon, Ei,/A . Its valu e fo r various nuclei is shown in Fig. 76. One can see that it is m axim um for nuclei in the region o f mass number A — CO.
Therefore,
if tw o light nuclei are join ed togeth er to form a mediummass nucleus (a process
F ig. 76. Binding energy per nucleon as a function of mass number.
296
Nuclear structure
(7.5
called fu s io n ), energy is liberated, and if a h eavy nucleus is d ivid ed into tw o m ediummass fragm ents (a process called fission ), energy is also liberated. T h e fact th at the binding energy per nucleon is practically constant (a b ove A =
10, the v ariation is on ly 10% ) suggests th at each nucleon interacts on ly
w ith its im m ediate neighbors, independently o f th e total number o f nucleons; this p rop erty is called saturation. T h is saturation effect is also found in molecules, in w hich the binding energy o f tw o atom s is practically independent o f the other atom s in the molecules. T h e same situation occurs in liquids and solids. A n em pirical expression for the binding energy o f a nucleus was proposed around 1935 b y the G erm an physicist C. V. W eiszacker.
A s w e have indicated, the bind
ing energy per nucleon is approxim ately constant and therefore the total binding energy o f a nucleus should be proportional to the total number o f nucleons or the mass number A ; that is, Es =
Oi A , where a, is a constant o f proportion ality.
H o w ev e r, the nucleons close to the nuclear surface are less tigh tly bound than those inside because they have few er neighbors wi th which to interact. T h u s w e have to correct our first estim ate o f Es by adding a n egative term. Since this is a surface effect, the new term must be proportional to the square o f the nuclear radius or, in view o f E q . (7.2), proportional to A 213. Thus w e m ay w rite Es =
U1A 
a 2A 213.
N e x t w e must consider the repulsive coulom b energy o f the protons, which also tends to decrease the binding energy.
F ro m E xam ple 7.1 w e conclude that the
coulom b correction must be proportional to Z 2A ~ ' 13. Thus w e m ay w rite Es
 α ι I — a 2A 213 — a3Z 2A ~ 113.
A n oth er term w hich tends to decrease the binding energy is the kinetic energy o f the nucleons. A s w e shall see in E xam ple 13.3, the kinetic energy o f the nucleons is p roportion al to .4 + f (. V — Z ) 2A ~ l . T h e first part, which is proportional to .1, can be included in the a%A term.
T h u s the kinetic energy correction am ounts to
adding a term o f the form — a4( N — Z ) 2A ~ ', so th at Es =
U1A 
U2A 213 
O3Z 2A  ' 13 
Ci4(.V 
Z ) 2A ~ '.
(7.11)
W e can obtain the four constants by fittin g the equation to the experim ental values o f Es, which w e get b y using E q . (7.10).
I f w e choose the constants to fit the
binding energies o f nuclei w ith odd .4 (th a t is, Z od d and Ar even, or the reverse), we find th at the binding energy o f nuclei w ith .4 even fall system atically on either side o f the values g iven by E q. (7.11), w ith even even nuclei havin g a larger bind ing energy and oddodd nuclei having a sm aller binding energy. the best fit o f E q. (7.11) fo r all nuclei w ith .4 >
Thus to obtain
10, we add a corrective term 5,
such that N
A
E ven
E ven
Even
E ven
Odd I
Odd
E ven I
Odd
Odd
Odd E ven
7.5)
Nuclear binding energy
T h e origin o f this term is related to the p a irin g energy o f nuclei.
297
T h a t is, as a
result o f the pairing effect m entioned in Section 7.4, the nuclear interaction seems to depend on the rela tive orientation o f the angular m om enta o f the nucleons; therefore the binding energy depends on how m any pairs o f nucleons have their angular m om enta in opposite directions, or are paired.
T h is number is rela tiv ely
sm aller in oddodd nuclei and larger in eveneven nuclei, resulting in a larger binding energy for the latter. T h e best values o f the constants, expressed in M c V , are O1 =
15.700,
a2 =
17.810,
a4 =
23.702,
a5 =
34.
a3 
0.711,
I t is suggested th at the student check the v a lid ity o f E q . (7.11) by ap plyin g it to several nuclei. N o te that, from E q. (7.10), w e m ay express the mass o f a nucleus in the form Al
= Zm p + =
Zm p + + a \ (N 
(/I — Z )m „ — (A 
Z ) m n
Z ) 2A ~ x 
Eb/c2 a\A + a'2A 213 + a'3Z 2A ~ 113 S'.
(7.12)
T h e values o f the constants appearing in E q. (7.12), when expressed in amu, are a\ = a’A
1.69 X IO 2 , =
2.54 X IO 2 ,
a'2 = 1.91 X IO 2 , o'j =
o'3 =
7.63 X IO 4 ,
3.6 X 10 ~ 2.
E X A M P L E 7.2. Difference in mass of “ mirror” nuclei; i.e., nuclei having the same odd A but with N and Z interchanged. One pair of mirror nuclei is 2H and 2He. Another pair is u N a and 22MgSolution: Consider odd.4 mirror nuclei which differ by one unit in N and in Z, with one having (N , Z ) and the other (N — I 1Z j I). Disregarding the pairing energy term in Eq. (7.12), since Λ is odd for both nuclei, and noting that N  Z = ± 1 , so that the kincticenergy term cancels out, we have Atz+ 1 — Atz —>'ip — m „ j a'3[ ( Z  f I ) 2 — Z 2JA = 7«p 
m„ + a3( 2 Z +
1 )A
1/2
~ 1/3.
Actually, since mirror nuclei are light nuclei, with low Z , the coulomb term must be used with Z ( Z — I ) instead of 7 2, as indicated in Example 7.1. Thus A fz + 1 
AIz
= mp 
m„ + 2 a ^ r 1/3
= (  1 . 3 9 0 + 1.526Z.11 ' 3) X IO3 amu. W c can use this equation to obtain a3 from the observed value of the masses of the mirror nuclei; from the measured value of a3, we can obtain the constant ro for the nuclear radius. I t is suggested that the student apply this formula to the two cases mentioned above.
298
(7.6
Xuctear structure
E X A M P L E 7.3. Calculation of Ihc atomic number of the most stable nucleus for a given mass number .4. S o lu tio n : maximum stant, and N — 7j =
The most stable nucleus with a given mass number .1 is that which has the value of the binding energy. Thus we have to compute OEb/dZ with Λ con equate it to zero. From the expression for Eb given in Eq. (7.11), and since .1 — 2Z, we have that OEb/dZ
2α3Ζ/1_1/3+ 4 α 4(Λ  2 Z )A ~ 1 = 0,
or, introducing the numerical values of β3 and a4, we have
Z = 2 + 0.0157Λ 2' 3 ‘
(7' 13)
For light nuclei, having small .1, we can neglect the second term in the denominator, and we have approximately Z « £.1, a result confirmed experimentally. It is suggested that the student make a plot of Z as a function of .1 up to 240 (go in steps of 20) and compare with Fig. 71.
7.H
X u r lv u r Fttrvvtt
L e t us now sum m arize the m ain properties o f nuclear forces, some o f which we h ave already m entioned. (I)
T h e n u c le a r f o r c e is o f s h o r t ra n g e .
Short range means th at the nuclear
force is appreciable on ly when the interactin g particles are v e r y close, at a sepa ration o f the order o f 10 l s m or less. A t greater distances the nuclear force is negligible. W e m ay infer that the nuclear force is o f short range because a t dis tances greater than IO14 m, corresponding to nuclear dimensions, th e interaction regulating the scattering o f nucleons and the grouping o f atom s into molecules is electrom agnetic.
I f the nuclear force w ere o f long range, the nuclear interaction
between the atom ic nuclei would be fundam ental in discussing m olecular form a tion, d om inating the w eaker electrom agnetic forces in the same w ay th at the electrom agnetic interaction dom inates the even w eaker g ravita tion al interaction in the form ation o f atom s and molecules. T h e range o f nuclear forces m ay be determ ined d irectly b y perform ing scattering experim ents. Suppose, for example, th at w e send a proton against a nucleus. T h e proton, on approaching the nucleus, is subject both to the electric repulsion and the nuclear force. I f the nuclear force is o f a range com parable to th at o f the elec tric force, the m otion o f the proton, no m atter how close or how distant it is when it passes the nucleus, would be affected b y both typ es o f force and the angular dis tribution o f the scattered protons would d iffer appreciably from the results for pure electric (o r coulom b) scattering. On the other hand, if the range of the nuclear force is small, those protons passing at a distance from the nucleus greater than the range o f the nuclear force essentially experience on ly a pure electric force.
O nly those
protons w ith enough kinetic energy to overcom e the coulom b repulsion and pass close to the nucleus are affected b y the nuclear force, and their scattering is differen t
7.6) from pure coulom b scattering.
299
Nuclear forces
T h is la tter case is the situation observed experi
m entally, confirm ing the short range o f the nuclear force. (2) T h e n u c le a r f o r c e s e e m s I o b e in d e p e n d e n t o f e le c t r ic c h a rg e .
T h is
means th at the nuclear interactions betw een tw o protons, tw o neutrons, or one proton and one neutron are basically the same.
F o r exam ple, from the analysis
o f protonproton and neutronproton scattering, scientists have concluded th at the nuclear p art is essentially the same in both cases. Also th e facts that (a ) light nuclei arc com posed o f equal numbers o f protons and neutrons, (b ) the binding energy per nucleon is app roxim ately constant, and (c ) the mass difference o f m irror nuclei (E xam p le 7.2) can be accounted for by the difference in coulomb energy alone, indicate th a t the nuclear interaction is charge independent.
B e
cause o f this p rop erty, protons and neutrons are considered equ ivalen t insofar as the nuclear force is concerned. F or that reason, as indicated before, th ey are designated by the com mon name o f nucleons. (3 ) T h e nuclea r f o r c e depen ds on the relative orien ta tion o f the spins o f th e in tera ctin g nu cleon s. T h is fact has been confirm ed b y scattering experi ments and by analysis o f the nuclear energy levels. I t has been found that the energy o f a twonucleon system in which the tw o nucleons have their spins parallel is different from the energy o f such a system in which one has spin up and the other down. In fact, the neutronproton system has a bound state, the deuteron, in which the tw o nucleons have their spins parallel (S = to exist if the spins are antiparallel (.S' =
I ), but no such bound state seems
0).
(4 ) T h e nuclea r f o r c e is not c o m p le te ly centred; it depen d s on th e o rie n tation o f th e spins relative to th e line jo in in g the tw o n u cleon s.
Scien
tists have concluded this b y noting th at even in the sim plest nucleus (th e deuteron), the orbital angular m om entum o f the tw o nucleons rela tive to their center o f mass is not constant, con trary to the situation when forces are central. T h erefore, to explain the properties o f the ground state o f the deuteron, such as the m agnetic dipole and electric quadrupole m om ents, w e m ust use a linear com bination o f s (o r I — 0) and d (o r 1 = 2 ) w a ve functions. P a r t of the nuclear force seems to be due to a rela tiv ely strong spinorbit interaction.
A n oth er part, called the tensor
force, closely resembles the interaction between tw o dipoles. (5) T h e nuclea r f o r c e has a rep u lsive core.
T h is means that at v e ry short
distances, much sm aller than the range, the nuclear force becomes repulsive. T h is assumption has been introduced to explain the constant average separation of nucleons, resulting in a nuclear volu m e p roportion al to the total number o f nu cleons, as w ell as to account fo r certain features o f nueleonnueleon scattering. In spite o f all this inform ation about nuclear forces, the correct expression for the potential energy fo r the nuclear interaction between tw o nucleons is not y e t well known. Several expressions have been proposed. One is the Y u la w a potential,
300
C7.6
Nuclear structure
Ep
Ep
Ep
F ig. 77. Empirical shapes for the nuclear potential energy.
proposed about 193S by th e Japanese physicist H id e k i Y u k aw a, and given by ο  τ >τ
E p( r ) =
o
— E 0r 0 — ^r  »
w here E 0 and r 0 are tw o em pirical constants. T h e constant r 0 is the range o f the nuclear force and E 0 gives the strength o f the interaction. T h e decreasing expo nential factor e ~ rlr° drops the Y u k a w a poten tial energy to zero faster than the electric poten tial energy, which varies as I /r.
H o w ever, there are even some
doubts that the nuclear interaction can be described in term s o f a poten tial energy function in the same w a y that we have been able to explain the g ravita tion al and electrom agnetic interactions. In any case, for m ost problem s the nuclear interaction at low energies m ay be represented schem atically by a potential energy as shown in Fig. 7 7 (a ). B eyond a certain distance the poten tial energy is practically constant, or (which is equ iva len t) the force is zero. A t v e ry short distances a repulsive core m igh t be added (F ig . 7  7 (b )). T h is is the ty p e of potential to be included in Schrodinger’ s equation w hen w e discuss nn and np interactions. F o r pp interactions, however, w e must also include the coulomb repulsion e2/ ln t0r, which is im portant on ly outside the range o f the nuclear forces (F ig . 7  7 (c )).
T h e solution o f S clu odinger’s equation
then g ives the station ary states o f twonucleon systems.
Sometimes, to obtain
sem iqu an titative inform ation, one can replace the nuclear potential b y a square poten tial well. AVe cannot discuss nuclear structure w ith the same thoroughness as w e dis cussed atom s and molecules, since w e do not know the exact form o f the nuclear poten tial energy to be incorporated in the Schrodinger equation.
N evertheless,
the know ledge w e do have about nuclear forces is sufficient to p rovid e a sound basis for discussing nuclear structure.
7.7)
7.7
The ground slate of the deuteron
301
T h e G ro u n d State o f the O eu teron
T h e deuteron, com posed o f a proton and a neutron, is the sim plest o f all nuclei (if we exclude the triv ia l case o f the nucleus o f hydrogen which is a single p roton ). B asically, on ly the nuclear interaction is op erative in binding a neutron and a proton togeth er (w e m ay neglect the small m agnetic interaction resulting from their m agnetic m om ents).
T h erefore, from a detailed analysis o f the properties
o f the deuteron, w e can obtain valu able inform ation about the nature o f nuclear forces. T h e deuteron has on ly one stationary state (w hich ob viou sly is its ground state) w ith an energy E = I =
E f1=
— 2.224 M e V . Also the spin o f the deuteron is
I. W e m ay then assume th at the proton and neutron have th eir spins parallel
(th at is, S =
I ) and th at the orb ital angular m om entum o f their rela tive m otion
around the center o f mass is zero (th a t is, L =
0 ), resulting in a state sS ].
In a
state 3S i there is no orbital m agnetic m om ent, and the m agnetic m om ent o f the deuteron should be equal to μρ + μ„ =
2.7927 — 1.9131 =
0.8796 nuclear m ag
netons, which is v e r y close to the experim ental valu e μ(ι =
0.8574 nuclear m ag
netons. T h erefo re our assumption about the ground state seems to be well founded. Considering th at the nuclear forces are central, w e m ay express the w a v e function o f the deuteron in the same form as was done for an electron in E q. (3.18).
In an
sstate the angular p art is constant (see T a b le 3 5 ) and w e h ave to consider on ly the radial p art R ( r ) = u (r )/ r , w here u (r) is obtained from E q. (3.24). S etting I =
0 in th at equation, which corresponds to a 3S i state, and recalling th at m is
the reduced mass o f the neutronproton system (th a t is, m ~ %mp), w e have h*_ (E u 2m dr2
+ E p(r )u =
Eu.
(7.14)
L e t us next, for sim p licity, express the nuclear poten tial energy b y the square w ell represented in F ig. 7 8 (a ), which has a depth E 0 and a radius a *
T h e problem
(a) * The inclusion of a repulsive core would not substantially change the results we shall obtain in this section, and for that reason we omit its consideration.
302
(7.7
.Xuclear structure
then becomes identical to th at discussed in E xam p le 2.0, since the Schrodinger equation for u (r ), E q . (7.14), is the same as for ^(.e) in that exam ple.
W e conclude
that the form o f u (r ) is . . U (r) =
( A sin k,r,
r < a,
I f t  ',
r >
a,
where kf =
2 m (E 0 
a2 =
2 m E 0/h2.
E b)/ h 2
and
T h is w a v e function is shown in Fig. 7 8 (b ). T h e con tinuity o f the proper function at r =
a requires th at Eq. (2.23) hold, nam ely k~i cot k,a =
—a
(7.15)
or [2 m ( E u 
E b)/ h 2\ '12 cot [2 m (E 0 
E b)/ h 2] U2a =
 [ 2 m E 0/h2] ' 12.
In our case w e know E b and we m ay use this equation to obtain inform ation about E 0 and a.
Since there is on ly one station ary state, w e recall from E xam ple 2.6
that n 2h 2/8m < E 0a 2 < Dtt2Ii 2/8m.
T h e ground state o f the deuteron deter
mines on ly the valu e o f E 0a 2, w hich is about 1.48 X IO28 McVr m 2. T o obtain E 0 and a separately, we must h ave another relation. F rom scattering experi m ents we can estim ate the range a (see Section 7.8). Thus if a is o f the order o f 2 X IO15 m, w e have th at E 0 is abou t 37 M e V .
I t m ust be noted that if we
use som e oth er ty p e o f shortrange poten tial instead o f a square w ell, w e obtain differen t values o f E 0 and a. W e indicated at the beginning th at there is a sm all discrepancy between the values of Mu and μ 0 f Mn which must be accounted for. electric quadrupole m om ent Q0 =
2.82 X IO31 m 2 C.
A lso the deuteron has an B ut a 3S j state is spheric
a lly sym m etric and does not have an electric quadrupole m om ent. T h u s w e m ay suspect th at our assumption o f a 3S j ground state for the deutcron is n ot com p letely correct. N o w a 3S 1 state is not the on ly state com patible w ith 1 = 1 . h a ve the states 1P 1 ( L =
I 1S =
0), 3P i ( L =
I,S =
W e m ay also
I ), and 3D 1 ( L =
2, S =
I).
I f the values o f mu and Qll are com puted for such states, th e results arc all v e r y d if ferent from the experim ental values. T o solve this dilem m a, w e m ay assume th at the w a v e function o f the ground state o f the deuteron is a linear com bination o f the four possible w a v e functions w e have m entioned.
But the 3S i and 3D ] states
have even p a rity and the 1P 1 and 3P i states h a ve odd p arity.
T h erefore, if the
ground state o f the deuteron has a welldefined p arity (as is to be expected if the force is sym m etric in the tw o interacting particles), w e m ust com bine on ly the 3S i and 3D 1 states.
Λ d etailed calculation, w h ich w ill be om itted, shows that to
reproduce the experim ental values o f mu and Qll w e must h a ve * d ,u „.r o n =
( U ) S 1K 3S 1) +
0 . 2 0 v K :!D , ) .
Neulronprolon scattering at low energies
7.8)
T h e fact th a t w e h a ve to m ix .states w ith L =
0 and L =
303
2 shows th at the orb ital
angular m om entum o f the deuteron does n o t have a w elldefin ed valu e and that therefore the nuclear force is n o t rigorously central.
In the deuteron, the non
central part is attribu ted to th e tensor force.
7M
X r u t r o n I tVoton S ru llv rin g at Loir E n r r g iv s
A n oth er valu ab le source o f inform ation about the nuclear force betw een tw o nu cleons is scattering experim ents.
T o p erform protonproton scattering experi
m ents, a beam o f protons from an accelerator is m ade to im pin ge on a target containing hydrogen atoms, and the scattered protons arc analyzed. T h e d evia tion from pure coulom b scattering gives inform ation about the nuclear force. Sim ilarly, in ncutronproton scattering, a beam o f neutrons from a nuclear reactor or oth er neutron source is p rojected on to a hydrogen target and the scattered neutrons are observed.
N eu tronneu tron scattering experim ents are m ore d iffi
cult, since it is impossible to have a target com posed on ly o f neutrons; therefore some indirect m ethods are necessary. L e t us now consider np scattering. Suppose th at a beam o f neutrons o f m om en tum p =
hk m oves along the Zaxis, as in F ig. (7 9 ).
m a y be described b y the w a v e function ^ mc = responding to these particles I a j mc =
«#in«2 =
Thus the incident particles
c 'k:.
T h e current density cor
v. W h en the neutrons pass close
to the target they are subject to the nuclear forces resulting from th eir interaction w ith the protons and are d evia ted from their initial direction o f m otion.
T h e scat
tered neutrons at a great distance from the ta rg et are represented b y a w a v e func tion resem bling spherical w aves; th at is, th eir w a ve function Iuis th e form e'kr/r (see P rob lem 3.30). T h e inten sity o f the scattering is not necessarily the same in all directions, and in general depends on the angle 0 which the direction o f scat
Scattcred neutrons
F ig. 79. Scattering of particles.
304
Nuclear structure
{7.8
tering makes w ith the Zaxis. Thus w e m ay w rite ^ scat = }{ ff)c 'k'/ r, where J(O) is called the scattering amplitude. T h e total w a ve function is
(7.16) T h e flux o f particles scattered through the surface dS per unit tim e (F ig. 7 9 ) is i# sc»t 2 =
H /(0)12 d S /r2.
B u t d S/r2 is equal to the solid angle dQ subtended b y dS. Thus F lu x o f particles scattered per unit tim e through dS =
v\f(0)\2 dU.
T h e differential cross section σ (β ) is defined as
σ
flux o f particles scattered per unit tim e within dil
(7.17) I f we observe σ {θ ) experim entally, w e m ay obtain J(O) b y ap p lyin g E q. (7.17). T h en w e m ay look for the nuclear forces which g iv e the observed function /((?). T h is provides a means o f guessing the nature o f the nuclear interaction. W h en the energy o f the incom ing neutrons is v e ry low, w e m ay obtain a sim ple expression for f (6 ).
W e note first th at the incom ing beam is composed o f particles
o f m om entum hk, parallel to the Zaxis but m ovin g at differen t distances from the axis and thus having differen t angular m om enta w ith respect to the scattcrer. T h a t is, th e w a ve function e'kz does n ot g iv e inform ation about the position o f the particles and w e m a y speak on ly o f the p rob a b ility th at a g iven incom ing p article w ill h ave a certain angular m om entum . T h u s w e m ay say th a t the w a v e function e'kz describes a beam o f particles having the same energy and m om entum but d if feren t values o f angular m om entum . In other words, w e m ay express e'kl as a sum mation o f angular m om entum w a v e functions according to
Σ Ί'1'
(7.18)
w here each ψι (called a partial w a v e ) corresponds to a g ive n angular m om entum and has the sam e angular dependence as the functions Y io o f T a b le 35.
(W e
m ust have mj = 0 because, if the particles m ove p arallel to the Zaxis, their angular m om entum rela tive to the origin o f coordinates m ust be perpendicular to th e Zaxis; th at is L z =
0.)
U sing a classical picture, w e m a y define the im pact parameter b as the perpendic ular distance from the Zaxis when the particle is still v e ry far to the left, so that L =
bp =
bhk. B u t L =
ftV T ( T F T ) ~ hk T h e refo re bk ~ I. Thus, according to
their angular m om entum , the particles fall w ith in the differen t zones shown in Fig. 710. W h en the nuclear force is o f short range, there is appreciable scattering
Neutronproton scattering at Ioio energies
7.8)
305
F ig. 710. Angular momentum zones.
on ly if b is o f the order o f the range, or smaller. T h a t is, the on ly particles which w ill be scattered out o f the incom ing beam are those having I < k X range.
For
v e ry low energies, k is v e ry small and on ly particles w ith I = 0 are scattered. W e say then th at w e have sscattering. In quantal language w e m ay say that in the presence o f a scatterer o f short range on ly th e I =
0 part o f the w a v e function
^ inc in E q. (7.18) (o r the partial w ave ^ u) o f a beam o f v e ry low energy neutrons is m odified, becom ing ψ'η, w h ile for I > 0 the ψι are the same as in E q. (7.18). Thus the w a v e function m odified b y the scattering must be expressed as i =
Ψ'ο + Σ
Φι·
I t can be shown th at the partial w a ve ip0 o f zero angular m om entum , corresponding to the freeparticle w a v e function ψ„ Kinc =
eikz, is
sin kr kr T h e function λτψ 0 =
sin kr is represented b y the dashed line in Fig. 7—11(b).
Er
ύ
Q
J f
r
£
(a) F ig. 711. (a) Rectangular deuteron potential well, (b) Positive energy wave functions.
306
Xurlear slructure
(,7.8
T o obtain the m odified zero angular m om entum w a ve function ψ'0, we note th at in the case o f a rectangular poten tial w ell, shown in F ig .·7 1 1 (a), the p article is free when r > a, but its w a v e function for r < a depends on the nuclear poten tial energy.
T h u s the w a v e function for r > a is still a frccparticle solution, but
shifted along the radial direction so th at it joins sm ooth ly w ith the solution for r < a (recall E xam p le 2.(5). W e must therefore w rite ,, φ° =
sin( A r  M 0) C
Fr
T h e qu a n tity δ0 is the phase shift. T h e w a v e function tr\p'0 = C sin (Ar f 50) is represented by the solid line in Fig. 7  1 1(b). N o t e th at the am plitude o f ψ0 is also differen t from that o f ψ0· T h e n it can be shown (see E xam ple 7.4) th at the differential scattering cross section is given by (7.19) Since σ (θ ) is independent o f the angle 0, the scattering is spherically sym m etric. T h e to ta l scattering cross section is (7.20) because the to ta l solid angle around the target is 4 π. F rom the observed valu e of σ, we m ay obtain the phase shift A0 experim entally.
On the other hand, if one
uses E q. (7.15), it can be shown th at a t v e ry low energies hk?/‘2 m
H ence E q . (7.20) m a y be transform ed into
In scattering experim ents th e neutron and proton m ay h a ve their spins parallel (trip let state) or antiparallel (single state) and th erefore we m ay determ ine whether the force is or is not spindependent b y exam ining the cross section in each case. E xperim ental results show th at the tw o cross sections are different, thus confirm ing the spindependence o f the force.
T h e depth and range for the trip let state,
when w e assume a square potential, have been found to be I i 0 = a =
2.25 X IO 15 m, w h ile for the singlet state w e m ust use I i 0 =
26.2 M eV , 17.8 M e V
and η = 2.51 X IO15 m. A s the energy o f the incom ing neutrons increases, the scattering affects partial w aves o f angular m om entum higher than zero.
T h e effect on each partial w ave
ψι is described b y a corresponding phase sh ift 5/, so that the to ta l cross section is given by (7.22)
Neutronproton scalterinij at low eneri/ies
7.8)
307
O bviou sly E q . (7.20) corresponds to the case in which the on ly nonzero phase shift is S0. T h e phase shifts Si are functions o f the energy. B y m easuring σ at different energies and an alyzin g the angular distribution o f th e scattered p arti cles, we m ay obtain the Si’s experim entally.
On the other hand, w e can com pute
the Si's th eoretically by assuming a reasonable poten tial energy fo r the interaction between the incom ing particle and the scatterer. T h e refo re scattering experim ents are v e ry useful in our search for the form o f the nuclear interaction. E X A M P L E 7.1.
Calculation of the cross section for sscattering.
Solution: Combining Eqs. (7.16) and (7.18), \vc may write I Arr
* = I which, when compared with Eq. (7.19), gives ikr
Va = \h + m
— · T
From the result of Problem 3.18 we know that = I (sin kr)/r. A more detailed analysis (which we omit) based on the normalization of the wave function indicates that we must make A = I /k. Thus ,
sin kr = ^ T '
Also, as indicated previously in the text, „ sin (kr + Sn) Vo = C FrHence we have that c sin (kr + 5p) _ sin kr kr
kr
e^_ r
Recalling that sin a = (e'a — e ~ ‘a)/2i, after a simple algebraic manipulation we obtain the scattering amplitude as e’>0 sin Sn m
=
k*
Using Eq. (7.17), we may then write the differential cross section as sin“ So VO) = Im \ ~ which is the expression given in Eq. (7.19) for / = 0 or sscattering.
308
!Vuclear structure
(.7.8
db Z
F ig. 712. Coulomb scattering of a charged particle. E X A M P L E 7.5. Calculation of the cross section for coulomb scattering of a charged particle by a nucleus. S o lu tio n : When a charged particle, let us say an electron, a proton, or an aparticle (or He nucleus), passes near a nucleus it experiences an electric force which is repulsive or attractive depending on the charge of the incoming particle. As a result, the incom ing particle is scattered or deviated from its original path. Let O be the position of the nucleus and A the initial position of the projectile, very far from O (Fig. 712). The ini tial velocity of the particle, at A 1 is vo. The impact parameter b is the perpendicular dis tance from .4 to a line OZ passing through 0 and parallel to ro. Thus when the particle is at .4, its ^ngular momentum relative to 0 is L  mvob. I f the force is central, the angu lar momentum remains constant during the process. Given that re is the charge of the particle (r = — I for electrons,  r I for protons and deutcrons, and + 2 for He nuclei) and Ze is the charge of the nucleus, the repulsive force on the particle is
Under the action of this force the incoming particle describes a hyperbolic path, suffering a deviation (or scattering) given by the angle Θ, which is a function of vo and 6, or (which is the same) of the energy and angular momentum of the projectile. When the particle is at any position, such as .1/, its angular momentum is mr2(d / = f + and is an E3 transi tion w ith an energy o f 0.032 M e V . T h e lifetim e o f the / =
i state is about 114 min.
N u clei w hich are in excited states and which have a reasonably long lifetim e are called isom ers. Transitions for which A l > 3 are designated as is o m e ric tra n s ition s . F igure 725 shows the distribution o f nuclei having isomeric states. T h e y appear in groups called isla nds o f is om erism , which fall just below the m agic numbers, according to the predictions o f the shell model.
In fact, the explanation o f
these islands o f isom erism is one o f the successes o f the shell model.
324
Nuclear structure
U e ferfin e v tt 1. “The Nuclear Force,” R. Marshak, Sci. Am ., March 1960, page 98 2. “ Nuclear Models,” It. van Wageningen, Am. J. Phys. 28, 425 (1960) 3. “ Pauli and Nuclear Spin,” S. Goudsmit, Physics Today, June 1961, page 18 4. “ The Nucleus T od ay,” D. Bromley, The Physics Teacher 2, 260 and 320 (1964) 5. “ Problem of Nuclear Structure,” V. Weisskopf, Physics Today, July 1961, page 18 6. Nuclear Forces, D. Brink. N ew Y ork: Pergamon Press, 1965 7. Structure of Atomic Nuclei, C. Cook. N ew Y ork: Van Nostrand, MomentumBooks, 1964 8. Structure of Matter, W. Finkelnburg. Sections 18
New Y ork: Academic Press, 1964, Chapter 5,
9. Foundations of Modern Physical Science, G. Holton and D. H. D. Roller. Mass.: AddisonWesley, 1958, Chapters 36, 37, 38 10. Principles of Modern Physics, R. Leighton. p ters 13, 14, 16, 18 and 19
Reading,
New Y ork: McGrawHill, 1959, Chap
11. Nuclear Physics, I. Kaplan. Reading, Mass : AddisonWesley, 1963 12. Introduction to the Atomic Nucleus, J. Cuninghamc. Co., 1964
Amsterdam: Elsevier Publishing
Problfitns 7.1 Examine Fig. 71 (or the chart of nuclides provided with the text) carefully to verify the statement that adjacent stable isobaric pairs do not occur in nature. 7.2 M ake a list of all adjacent stable iso topic sets larger than 4. For example, titanium (Z = 22) has five adjacent stable isotopes. Use Fig. 71 or the chart of nu clides provided with the text. 7.3 With the aid of Fig. 71 (or the chart of nuclides), investigate stable isotonic sets. W rite the neutron number N for sets of more than four stable isotones. Are any of these adjacent stable isotones? 7.4 Ordinary boron is a mixture of the 10B and 11B isotopes. The composite (or chemical) atomic mass is 10.811 amu. W hat percentage of each isotope is present in natural boron (a) by number and (b) by mass?
7.5 Calculate the nuclear radius of 10O, 120Sn, and 208Pb. 7.6 Choosing the most abundant isotope, compute the nuclear radius for nuclides with Z equal to 10, 30, 50, 70, and 90. Also compute the radius of the inner elec tronic orbit, using Eq. (3.11). Plot both radii on the same graph. Draw conclusions about the interaction of inner electrons with the nucleus. 7.7 What nuclei have a radius equal to onehalf the radius of 236U? 7.8 Estimate the kinetic energy of a nu cleon inside a nucleus, both by using the quantummechanical picture of a particle in a potential box of width IO15 m and by considering the de Broglie wavelength X of the nucleon, which is of the order of 2irr, where r is IO15 m.
Problems 7.9 The interaction between the electron and the nuclear magnetic moments gives rise to a hyperfine splitting of the atomic energy levels. The energy due to this inter action is proportional to ( μ ο / 4 ι τ ) μ χ μ Β ( ί ·~ 3)a ve,
where r is the distance of the electron from the nucleus. Show that the hyperfine split ting is of the order of IO5 eV in energy, giving rise to a hyperfine structure of spec tral lines which is of the order of IO2 A when the lines are in the visible range. 7.10 The hyperfine splitting of atomic energy levels discussed in the previous problem is also proportional to I ■ J, where I is the nuclear spin and ./ the electronic angular momentum. Analyze the hyper fine splitting of the electronic states 2S 1/2, 2P 1/2, and 2P 3/2 of 23N a (/ = § ). In each case determine the number of levels and their relative spacing. Show your results on an energy level diagram. 7.11 The electric quadrupole moment of an ellipsoidal charge distribution which has axial symmetry is Q = §Z (a  — 62), where a is measured along the symmetry axis and b is measured along the perpendic ular direction. The mean nuclear radius R is defined by R  = ab2. (This guarantees that a sphere of radius R has the same volume as the ellipsoid.) Show that if a = R\~ A R , then b = R — $ A R and Q = ^ Z R A R , to the first order of approxi mation. Also note that Q /ZR2, which is plotted in Fig. 74, is proportional to A R /R . Determine whether this plot is more informative than a plot of the qua drupole moment alone, as far as information about nuclear deformation goes. 7.12 Using the data of Table 71 and the results of Problem 7.11, estimate A R /R for 176Lu1 which is one of the most highly deformed nuclei. 7.13 Estimate the coulomb repulsion energy of the two protons in 3IIe (assume that they are 1.7 X 10 ~ ’ 5 m apart).
325
Compare this energy with the difference in the binding energies of 3H and 3He. Is the result compatible with the assump tion that nuclear forces are charge inde pendent? 7.14 How much energy is required to re move (a) a proton and (b) a neutron from 16O and from 17O? Is the result an indica tion of the existence of pairing forces? 7.15 From the atomic masses given in Table 71, calculate the total binding energy and the binding energy per nucleon for 'L i, 16O, o7Fe, and 176Lu. 7.16 The empirical Weiszacker formula, Eq. (7.11), contains four terms. Plot each of the terms as a function of .1. For the appropriate Z and N values for the fourth term, use those corresponding to the most stable isobar and choose only selected A values. Determine the relative importance of each term in the different regions of the periodic table. 7.17 With the aid of Eq. (7.13), determine the atomic number of the stable nuclide for .1 = 27, 64, 82, 125, and 180. Compare these values with the data in Fig. 71. 7.18 Using the Weiszacker formula, cal culate the atomic masses of 12C, 27Al, 88Sr, 185Re, and 232U. Compare with the values listed in Table 71 and judge the accuracy of the formula. 7.19 Using the Weiszacker formula, cal culate the mass difference between the mirror nuclei 23N a and 23M g. Compare with the result using the experimental values of the masses, which are 22.98977 and 22.99412 amu, respectively. 7.20 (a) Calculate the binding energies for 14O, 13O, lfiO, 17O1 18O, and 19O. (b) Repeat for 11C 1 13N 1 16O, 17F, and 18Ne. (c) From the results of (a) and (b), cxirlain the variation in binding energy as a neutron or a proton is added to a nucleus. The masses of those nuclei not listed in Table 71 are (in amu): 14C, 14.00324; 14O 1 14.00860; 13O, 15.00307; 18O, 17.99916; 190 , 19.00358; 17F 117.00209; 18Ne, 18.00571.
326
Nuclear structure
7.21 The following mass differences have been found experimentally: 1I I 2 — 2H
= 1.5434
3 2II — i 12C
= 4.2300
12C 1I I 4 
16O
10 " 27.26 amu. Through what angle in the laboratory frame of reference is a 4M cV aparticle On the basis of 12O = 12.000 amu, calcu scattered when it approaches a gold nu late the atomic masses of 1I I 1 2H, and lfiO. cleus with an impact parameter which is Compare with the experimental values. 2.6 X I O  13 m? 7.22 A simple empirical approximation for 7.27 What is the impact parameter of a the nuclear potential is the Yukawa poten 4M eV α particle, given that. it. is scattered tial Jip = — /ioroer/ro r , where through an angle of 15° by a gold nucleus? Eo = 50 M eV
and
= 3.6364
X X
scattered more than 90° in the laboratory frame of reference. (Assume equal masses.) [H int: Use the nonrelativistic formulas 1 0 3 amu, relating the scattering angle in the C and lO ^Λframes ’ amu, of reference given in the appendix.)
X
ro = 1.5 X 10~l a m.
Plot the Yukawa potential for r = O.lro, O.oro, ro, 1.5ro, 2ro, and 3ro, and compare with the electric potential energy of two protons at the same separations. 7.23 Show how the 3S, 1P 1 3P 1 and 3D states of a neutronproton system are the only ones compatible with a nuclear spin equal to one. 7.24 It can be shown that, when all phase shifts are considered, the scattering ampli tude /(0) of neutron scattering is given by ](θ )
ν'4 π (2 / + k
D
t V
i i Sin
Sl Y l0l
I
where Θ is the scattering angle and the l ‘io are angular momentum functions. When the energies are not too great, only the first two phase shifts, ίο and Si, are needed to describe the neutron scattering and all other phase shifts arc zero, (a) Write /(0) and the differential cross section α(θ) for this approximation, (b) Show that the scattering is not spherically symmetric. How do you correlate the asymmetry with the signs of the phase shifts? (c) Compute the total cross section by integration and verify that it coincides with the first two terms in Eq. (7.22). [H in t: Use the expres sions for Poo and P io given in Table 3 I.] 7.25 Show that, in an elastic scattering experiment involving protons on protons or neutrons on protons, no particles are
7.28 Find the distance of closest ap proach for (a) IOMeV and (b) 80MeV protons, incident headon on a gold nu cleus, and compare with the nuclear radius, (c) In which ease will the proton “ hit” the nucleus? Find the kinetic energy of the proton when it “hits” the nucleus. 7.29 An aluminum foil scatters IO3 aparlieles per second in a given direction. I f the aluminum foil is replaced by a gold foil of identical thickness, how many aparticles will be scattered per second in the same direction? 7.30 The number of αparticles scattered by a given foil is IO6 per second at a scat tering angle of 10°. Calculate the number of αparticlcs scattered at each 10° interval up to 180°. Plot your results for σ(0). 7.31 A beam of 12.75MeV αparticlcs is scattered by an aluminum foil. It is found that the number of particles scattered in a given direction begins to deviate from the predicted coulomb scattering value at ap proximately 54°. I f the αparticles are assumed to have a radius of 2 X 10_ , ° m, estimate the radius of the aluminum nu cleus. [H in t: It can be shown that the dis tance of closest approach is given by r — ( vZe2 4xeo»H 'o)(lesc 20)·] 7.32 A t what neutron energy docs the relation k X range of nuclear force = 0.5 hold? Could this result be used to estimate the upper limit of the energy at which only
Problems sscattering of neutrons by protons need be considered ? 7.33 Calculate the nuclconnucleon scat tering cross section in the limit of low energies, E  * 0, for the triplet potential IEo = 36.2 M eV ) and the singlet poten tial (E 0 = 17.8 M cV ). What cross section would be measured experimentally if the spins of the particles were to be oriented randomly? 7.34 Calculate the cross section for nucleonnucleon scattering in the triplet state (E o = 36.2 M eV ) for E = 0.1, 0.5, I, and 5 M eV. Do you conclude then that at low energies the cross section for nucleonnucleon scattering is not sensitive to the energy? 7.35 (a) Using the shellmodel scheme given in Fig. 716, find the nucleon con figuration for 2H 1 3H 1 3He, 4H e1 7Li, 12C, 13C, and 30Cl. (b) Predict the nuclear spin for these nuclides. Compare with the values listed in Table 71. 7.36 Compute the magnet dipole moments of the nuclei listed in the previous problem, using the method of Example 7.6, and compare with the experimental values given in Table 71. 7.37 The moment of inertia of a nucleus of mass M and average radius R, if con
327
sidered a solid sphere, is β = f . l / Λ 2. W ith a mass number .4 equal to 50, 100, and 150, estimate the energy (in M eV ) of a Ύray emitted in a transition from a rotational energy level with I  2 to one with I = 0. Compare with the energy of Ύrays emitted by eveneven nuclei in those regions. What do you conclude? 7.38 From the rotational energy levels shown in Fig. 720, estimate the moment of inertia of 180Hf. The moment of inertia about the symmetry axis of an ellipsoid of revolution is S = §.l/62, where 6 is the length of the other semiaxes. Estimate the deformation A R / R of l80Hf, and compare with the value obtained from the electric quadrupole moment. [H int: Recall from Problem 7.11 that 6 = R — % A R .[ 7.39 Referring to Fig. 722, determine the energy and multipole order of the 7transi tions shown for each of the three nuclei. 7.40 Refer to Fig. 722(a) and (b ), and assume that each of the levels shown is due to a single particle transition; with the aid of the shellmodel level scheme of Fig. 716, write the configuration for the ground state and each of the excited states. 7.41 Referring to Fig. 7—22(c), deter mine whether or not the level scheme can be attributed to rotational collective excitations.
NUCLEAR PROCESSES
8.1 In lrod ud ion 8.2 Radioadive Decay 8.3 A lpha Decay 8A
Bela Decay
8.5 Nuclear Reaclions 8.6 Nuclear Fission 8.7 Nuclear Fusion 8.8 Tlie O rigin of Ihe Elemenls
8.2)
II. I
Radioactive decay
329
In troiIin sIion
In the previous chapter w e an alyzed the basic properties o f nuclei, considered as stable assemblies o f nucleons, and proposed a m odel which, in a m ore or less sat isfactory w ay, accounts fo r such properties. A n oth er source o f inform ation about nuclear structure is the analysis o f processes, such as nuclear disintegrations and nuclear reactions, in which there is a rearrangem ent in the energy or configuration o f the nucleons. In this chapter w e shall analyze som e nuclear processes. M a n y o f them occur naturally, but others are produced artificially in the lab oratory using different typ es o f accelerating machines or nuclear reactors. T h e experim ental and theoretical discussion o f nuclear processes is still an unfinished chapter o f con tem porary physics, in which a large am ount o f research is being done.
it.'2
Ila ilion rtin · H o r n y
W e rem ind the student that, when we were discussing Fig. 71, w e indicated that some nuclei have a com bination o f protons and neutrons which does not lead to a stable configuration. T h ese nuclei are therefore unstable or radioactive.
U nstable
nuclei tend to approach a stable configuration by releasing certain particles. These particles, when th ey were first observed a t the end o f the last century by J. B ecquerel, Pierre and M a rie Curie, and others, were designated as a  and /3particles. A lp h a particles are helium nuclei, com posed o f tw o protons and tw o neutrons as m a y be verified by m easuring their charge and mass. T h u s when a nucleus emits an alpha p article its atom ic number Z decreases by tw o units and its mass number A decreases by four units.
T h e new nucleus therefore corresponds to a
different chemical elem ent. F o r exam ple, when the rad ioactive nucleus 2g?U em its an αparlicle, the residual nucleus is 2ISTh. B eta particles are electrons which carry a charge — e.
Hence, when a nucleus
emits a /3particle, the atom ic number o f the nucleus increases b y one unit, but the mass number does not change.
F or exam ple, when the rad ioactive nucleus 2SjjTh
emits a /3particle, the residual nucleus is 2StPa. Som e nuclei, instead o f em ittin g electrons, release positrons which carry a charge fe ; therefore the residual nucleus, after positron emission, has an atom ic number sm aller by one unit.
F o r example,
when the nucleus 12N em its a positron, the residual nucleus is 1C. T h e tw o types o f /3decay are designated β ~ and β + , respectively.
T h e residual or daughter
nucleus is som etim es left in an excited state and in the transition to its ground state em its gam m a rays, as explained in Section 7.10. M a n y isotopes o f elem ents w ith Z > SI (o r A > 20G) are n aturally radioactive. A few other naturally existing lighter nuclei, such as u C and 40Iv, are also radio active. M a n y m ore rad ioactive nuclei have been produced in the lab oratory using nuclear reactors and particle accelerators. F igu re 81 shows one o f the three natural rad ioactive chains, the socalled u ra nium series, and T a b le 81 g ives the relevan t inform ation about this series. T h e mass number o f nuclei in this series is given b y 4 « f 2, where n is an integer.
T h e tw o other natural rad ioactive chains are the
actinium series and the thorium series, com posed resp ectively o f nuclei o f ty p e
330
(£.2
Nuclear processes
Atom ic number
F ig. 81. The naturally radioactive uranium or 4n + 2 series. 4 n + 3 and 4n. T h e heavier nuclides in these tw o series are 2^ U and 2goTh. is suggested th at the student m ake a p lot o f these
It
tw o series, sim ilar to Fig. 81,
using the chart o f nuclides p rovided w ith the text. In
the
next
tw o
sections
we
shall
present a m ore detailed analysis o f a  and /3decay. B u t let us now discuss certain features th at are com mon to both radio active processes.
I t has been observed
that all rad ioactive processes follow an exponential law. T h erefore, if N 0 is the initial number o f unstable nuclei,
the
number o f nuclei rem aining after a tim e I is given by N =
N 0e ~ u ,
(8.1)
where λ is a constant characteristic of each
nuclide,
called
the
disintegration
constant. I t is expressed in s1 (o r in the reciprocal o f an y other tim e un it). E qu a tion (8.1) is represented in Fig. 82.
For
each radioactive nuclide there is a fixed
F ig. 8 2. Radioactive decay as a function of time.
Hadioaclive decay
8.2)
331
TABLE 8—1 The Uranium Scries
Radioactive species
Uranium I (U I) Uranium X i (U X i ) Uranium X 2 (U X 2) Uranium Z (U Z ) Uranium I I ( U l I ) Thorium (T h ) Radium (R a) Radon (R n) Radium A (R a A )
Nuclide
238T 92UT 2311.1 90 I n 234Pa 91* “
iItPa 2IiU 2IoTh 225Ra 2IlR n
2ItPo
Radium 11 (R ah) Astatine218 (2l8A t) Radium C (R aC )
2I 4 2Pb 2I 8A t
Radium C ' (R a C ') Radium C " (R a C ") Radium D (R a D ) Radium E (R a E ) Radium F '(R a F ) Thallium206 ( 206T l) Radium G (R aG )
2IJPo 2ItT I 2ISPb
2IlOi
T yp e of disinte gration
SI 1 1
Particle energy, M eV 4.19 0.19 2.31 0.5 4.768 4.68 4.777 5.486 a: 5.998
4.51 X IO9 y 24.1 d 1.18 m 6.66 h 2.48 X IO5 y 8.0 X IO4 y 1620 y 3.82 d 3.05 m
4.88 3.33 9.77 2.88 8.80 2.75 1.36 2.10 3.78
β a a, β
26.8 m 1.3s 19.7 m
4.31 X 104 0.4 5.86 X 104
1.64 X IO 4 s 1.32 m 21 y 5.0 d 138.4 d 4.2 m
4.23 8.75 1.13 1.60 5.80 2,75
a
a β β β a
210IU S3**1
20eP 82* h 13
Disintegration constant, s1
β β β a a a a Of, β
210Pn si1 0
200TM
Halflife
β Stable
X X X X X X X X X
X X X X X X
i o  18 ΙΟ "7 IO3 10~5 IO 14 I O  '3 IO  11 IO "6 10~3
IO3 IO 4 109 10° 10“ 8 103
0 :? 0.7 6.70 a : 5.51 /3:3.17 7.683 1.96 0.0185 1.155 5.300 1.51
tim e in terval T , called the halflife, during which the number o f nuclei a t the be ginning o f the interval is reduced, by the end o f the interval, to onehalf.
So if
w e in itia lly had N 0 nuclei (or atom s), a fter tim e T on ly N 0/2 are left, a fte r tim e 2T on ly N 0/4 remain, and so on. T o find this tim e T, we set N = in E q . (8.1). \T =
In 2 =
Then $ N 0 
N 0e ~ XT or ex r =
2.
^ N 0 and I =
T
T a k in g logarithm s, w e have
0.693 or T = 0.693/X,
(8.2)
which relates T and λ. T h e recorded halflives g o from a great m any years— such as the h a lflife o f the αdecay o f 209Bi, which is abou t 2 X IO 18 years, and the β ~·decay o f 115In , which is about 6 X IO u years— down to fractions o f a second ( 8B e has ;ui α decay halflife o f the order o f IO 10 s). F rom E q. (8.1) we can find the rate a t which nuclei disintegrate: d N /d t =
— \ N 0e ~ u =
\ N .
(8.3)
332
Nuclear processes
(8.2
T h is indicates th a t the disintegration rate dN/dt is proportional to the number o f nuclei present.
T h erefo re the disintegration rate dN/dt decreases in the same
proportion and w ith the same halflife as the number o f nuclei N .
T h e absolute
value o f the disintegration rate, \dN/dt\, is called the activity o f the substance. D isintegration rates are usually expressed in curies, abbreviated C i, in honor o f P ierre and M a rie Curie, discoverers o f polonium and radium. T h e curie is defined as the a c tiv ity o f a substance in which 3.700 X IO 10 nuclei disintegrate per second.* Subm ultiples o f the basic unit are the m illicurie ( I m C i =
IO3 C i) and the micro
curie ( I p C i = 10B C i). E quations (8.1) and (8.3) are both statistical laws which are valid on ly when the number o f nuclei is v e ry large, and which m ay be interpreted in the follow in g way.
T h e re is a decay probability per u n it time X th at an unstable nucleus w ill
disintegrate according to a specific process.
T h e p ro b ab ility that the nucleus w ill
disintegrate in the tim e interval dl is X dt. I f there are N nuclei present (w h ere N is v e ry large), then we m ay expect a number o f nuclei equal to (λ d t)N to disinte grate during dt. T h erefo re we m ay w rite dN =
— (X d t )N
or
dN/dt =
— XN,
which is E q . (8.3).
T h e minus sign appears because N decreases w ith tim e as a
result o f the decay.
T h e calculation o f the decay p ro b ab ility per unit tim e X for
each decay process is an im portant problem for which refined techniques o f quan tum mechanics must be used. N o te from the ab ove discussion th at w e cannot speak o f the halflife o f a single nucleus or predict w ith certainty when a given nucleus will disintegrate, and we repeat th at Eqs. (8.1) and (8.3) are correct on ly in a statistical sense. E X A M P L E 8.1.
Compute the mass of 1.00 Ci of 14C'. The halflife of 14C is 5570 years.
S o lu tio n : S in c e T = 5570 y r = (5.570 X IO3 yr) X (2.156X 107 s/yr) = 1.758X IOn s, the disintegration constant is λ = 0.693/T = 3.94 X 10_ , 2 s_ I . Also IA V M l = I Ci = 3.70 X IO7 s ~ l . Therefore, using Eq. (8.3) with absolute values, we find that
N =
= 9.38 X 10IS nuclei of 14C 1
which is also the number of carbon atoms present. The atomic mass of 14C is 14.0077 amu. Thus the mass of the above number of carbon atoms is .1/ = (14.0077 X 1.6604 X IO27 kg a t o m  ') X (9.38 X IO18 atoms) = 2.18 X IO 7 kg. * This rate is approximately equal to the activity of I g of 11a.
8.2)
Radioaclire decay
333
E X A M P L E 8.2. One method of producing a radioactive nuclide is to place a sample of a given substance inside a nuclear reactor. Radioactive nuclides are produced as a result of neutron capture by the nuclei of the substance. For example, when we bombard 59Co with neutrons, we find that we obtain 60Co, which is ^radioactive with a halflife of 5.27 years. Another method of obtaining radioactive nuclides is to bombard the sub stance with charged particles, such as protons or deuterons, using accelerators to energize the projectiles. In both cases the new nuclide is produced at the constant rate of g nuclei per second. Calculate the number of nuclei of the radioactive nuclide produced in terms of time. X/V1I1UX
F ig. 83. Production of a radioactive nuclide as a function of time.
S o lu tio n : The radioactive nuclide is fabricated at the rate of g nuclei per second, but at the same time, according to Eq. (8.3), the nuclide disintegrates at the rate of λ .V nuclei per second, where N is the number of nuclei present at that instant. Thus the net rate of increase of nuclei per second* is rf.v Ti
=
3“
λΛ ·
Separating the variables and integrating, we have
Γ L 0T T ^
, [ ‘ „ =  x L dt
or
, .V  g/\ =  λ ί. 1o r T T = I T x
Assuming that the number of nuclei of the substance was initially zero (that is, N o = 0), we have, at a later time t, N = f ( I — β_λ n + e + + v.
(8.15)
In the case o f electron capture, the process is w ritten as p f~ e
—►n
v.
(8.16)
In this w ay a nucleus m ay get rid o f its excess neutrons or protons w ithout actually e m ittin g either o f these particles. T h is th eory o f j8decay was proposed in 1934 by Ferm i.
E quations (8.14), (8.15), and (8.16) are com patible w ith Eqs. (8.11),
(8.12), and (8.13), and express in a m ore fundam ental w a y w hat happens in /Jdecay. T h e ab ove processes allow us to infer another p rop erty o f the neutrino: its spin. T h e neutron, proton, electron, and positron each h ave spin
T herefore, in the
/J+ process, p —> n + e + + v, the angular m om entum on the left, in units o f h, is J, while the angular m om entum o f n + e + m ust be £ ±
J =
I or 0, depending
on whether the neutron and the positron have their spins parallel or antiparallel. T h u s the neutrino must have spin j , and be so oriented that the total spin or angu lar m om entum o f the three particles on the right add to j . conservation o f angular m om entum is also saved!
In this w a y the law o f
T h e same logic applies to the
tw o other processes: (8.14) and (8.16). T h is also explains the observed changes o f spin Δ / o f nuclei in /Jdecay. I f the electron (o r positron) and the neutrino have th eir spins parallel (trip le t states), then we h ave Δ / = ± 1 or 0, but if their spins are an tiparallel (singlet states), then A l = 0. Process (8.14) has been observed w ith free neutrons, since the reaction is exoergic. In fact, the energy availab le is Q =
931.48 [m„ — (m p + m e  f m f )] M e V =
0.7834 M e V .
Free neutrons decay according to E q. (8.14), w ith a halflife o f 13 min. T h is is one o f the reasons w h y free neutrons do not exist.
Free neutrons, sh ortly a fter they
are produced, are either captured by other nuclei or disintegrate into protons, electrons, and neutrinos. On the oth er hand, processes (8.15) and (8.16) are endoergic, as the student can v e rify by com puting the Q value for each process; therefore
Beta decay
365
free protons are stable to /3decay, which accounts for the existence o f hydrogen. Otherwise all hydrogen w ould have disappeared, either by capture o f the orb ita l electron or b y disintegration o f the nuclear proton.
Processes (8.1.5) and (8.16),
however, can occur in m ore com plex nuclei, when the required energy is supplied by the difference in the binding energies o f the parent and the daughter nuclei. H ow ever, neutrons bound in nuclei do not in general disintegrate spontaneously because the presence o f the other nucleons m ay m ake the process en ergetically impossible. F o r th a t reason m ost neutrons in nuclei are stable. B eta decay illustrates tw o fundam ental facts o f physics. One is the im portance o f conservation laws in an alyzing processes th at occur in nature. T h e other is the variable nature o f the fundam ental particles.
T h a t is, fundam ental particles, al
though th ey have welldefined properties, are not perm anent structures, and one particle m ay change into sets o f the others, w ith in the lim itations imposed by the conservation laws. T h is is a radically new concept which was not con tem plated either in classical or quantum mechanics. T h e interrelational character o f the fundam ental particles opens an en tirely new v ista o f ideas about the structure o f matter.
In C hap ter 9 we shall explore this new situation in greater detail.
A com plete and satisfactory theory which can account for all the transform a tions am ong fundam ental particles has not y e t been developed. I t is assumed th at processes such as (8.14), (8.15), (8.16), and others (which w ill be described in C h ap ter 9) arc the result o f a differen t interaction, called the weak interaction. From the analysis o f the halflives and energy distribution o f /3emitters, it has been estim ated that the strength o f the w eak interaction is o f the order o f IO13 when com pared w ith that o f the strong or nuclear interaction, or about 10—11 when com pared w ith the electrom agnetic interaction. A t present both strong and w eak interactions p rovide challenging and excitin g areas o f research for the physicist, and, presum ably, th ey w ill continue to do so fo r m any years to come.
E X A M P L E 8.6.
Analysis o f the CowanReines neutrinodetection experiment.
S o lu tio n : When an antineutrino is captured by a proton, a neutron and a positron are produced; that is, p + P —* n + e +.
(8.17)
This process o f antineutrino capture may be related to Eq. (8.15) in the same way as the electron capture process, Eq. (8.16), is related to Eq. (8.15). Nuclear reactors produce large numbers of /3“ emitters, which are the products of uranium fission (see Section 8.5), and thus nuclear reactors are copious sources of anti neutrinos. I f the antineutrinos from a reactor fall on hydrogenous material, process (8.17) may take place. The positron can be detected because it may collide with an electron, resulting in the annihilation of both particles and the emission of 7rays (Section 9.3), which can be easily detected by means of scintillation counters. (See Appendix V II.) One can detect the neutron by adding cadmium to the hydrogenous substance. Neutrons— after being slowed down in their motion through the substance on account of collisions with other nuclei— are captured by the cadmium nuclei with the emission of some Ύrays. This chain of events is
346
Nuclear processes Incident antineutrino
F
Scintillation counter Target tank with cadmium
Scintillation counter
F ig. 813. Schematic diagram of the neutrino detection experiment. The incoming neutrino reacts with a proton at I. The ejected positron annihilates with an electron at 2 and the neutron is captured by a cadmium nucleus at 3.
depicted in Fig. 813. There is, of course, a delay of several microseconds between the production of the 7rays resulting from the electronpositron annihilation and the emission of the 7rays resulting from the capture of the neutron by cadmium. B y means of proper electronic circuitry, the two sets of 7rays are identified. An analysis of the experimental results provides convincing evidence that the assumed chain of events has taken place and that antineutrinos are actually coming from the reactor. Several other experiments have confirmed the neutrino hypothesis. In the CowanReines experiment, performed in an underground room below one of the reactors at Savannah River, the antincutrino flux was about 4 X IO10 cm2 s1 and the number of events registered was about three per hour. I t is considered that the neutrino flux on the earth, coming mainly from the sun, is about 4 X IO10 cm 2 s1 . E X A M P L E 8.7. Determination of the energy available for each of the βdecay processes described by Eqs. (8.11), (8.12), and (8.13). S o lu tio n : L e t us first consider β “ decay, Eq. (8.11). Initially we have a nucleus of atomic number Z and mass viz.. A t the end of the process we have a nucleus of atomic number Z I and mass m z + i, plus an electron (mass mc) and a neutrino (zero mass). Thus the energy available for the process is Q s
=
I ' » Z — ( ’» 2 + 1 +
m .)]c 2 =
( v i z — Vl2 Jr I — vie)c 2.
Normally one uses the atomic masses M z instead of the nuclear masses, such that M z = viz + Zm 9. Making this substitution, we get Q f  = (M z 
il/ z+ i)c2.
For/3+decay (Eq. 8.32), we have
(8.18)
Bela decay
347
or, transforming into atomic masses, Q i+ = ( M z ~ M z  i 
2m.)c2.
(8.19)
Finally, for electron capture (Eq. 8.13), Qeo = ( M z 
M z  i ) c 2.
(8.20)
Thus, whenever the atomic mass of a given atom is larger than that of either of the two neighboring isobars, it will decay by either β ~ or electron capture. However, for β + decay, the masses must differ by at least 2me (about 1.097X IO3 amu or 1.022 M eV ). I t is suggested that the student check these rules by looking atseveralgroups of isobars in a table of nuclides. In writing these equations, we have neglected the effect due to the binding energy of the electrons in the atoms. The energy Q is shared (as kinetic energy) by the decay products. I f we neglect the recoil energy of the daughter nucleus, then Q also gives the maximum kinetic energy of the electron or the positron in Eqs. (8.11) and (8.12). Let us illustrate these important rules with some examples. The nucleus 14C disin tegrates into 14N according to the scheme of Eq. (8.11), giving off an electron and a neutrino. The masses of the atoms are M z ( 14C) = 14.007682 amu and M z + 1 ( 14N ) = 14.007515 amu. Therefore Eq. (8.18) gives Qg = 0.000167 amu or 0.1556 M eV. The observed maximum kinetic energy of the ejected electrons is 0.155 M eV, which is in excellent agreement. N ext let us consider the decay of 11C into 11B1 according to the scheme of Eq. (8.12). The masses of the two atoms involved are M z ( 11C ) = 11.01492 amu and M z  i ( 11B) = 11.01279 amu. Thus their mass difference is 0.00213 amu or 1.985 M eV 1 which is larger than 2»ie or 1.022 M eV. Therefore positron emission is possible with Qi + = 1.985 M eV 
1.022 M eV = 0.963 M e V .
The measured maximum kinetic energy of the positrons observed in this decay is 0.96 M eV, again in excellent agreement. Finally we consider the decay of "Be, whose mass is M z ( 7Be) = 7.01915 amu. As explained before, 7Be is found to decay into 7L i1whose mass is M z  1 ( 7L i) = 7.01822 amu. Their mass difference is 0.00093 amu or 0.866 M eV. This is less than 1.022 M eV, and therefore positron emission is impossible. The decay therefore occurs via electron capture, with Qec = 0.866 M eV. E X A M P L E 8.8.
Calculation of the shape of the energy spectrum in /3decay.
S o lu tio n : Consider a large number of /3radioactive nuclei. VVe observe the emitted electrons (or positrons) during a certain time interval. Let us designate by dN the number of electrons (or positrons) emitted with a kinetic energy between E ke and E ke 4 dEke. Our purpose is to calculate dN/dEke, which is the number of electrons (or positrons) per unit energy range. W e shall designate the kinetic energy available for the electron and the neutrino by Eo which, if we neglect the recoil energy of the daughter nucleus, is prac tically equal to Q. Then Eo — E ke + E k,. Obviously Eo must be equal to the maximum kinetic energy of the electron. When the kinetic energy of the electron falls within the range dEke, that of the neutrino falls in the interval dEkr = — dEke. A fter the decay we
308
(8.5
Nuclear processes
may treat the electron and the neutrino as free particles enclosed in a very large potential box. Then dN /d Eu must be proportional to the number of electronic states per unit energy range [that is, ge(E ke)] and the number of neutrino states per unit energy range (Jy(E ky). That is, dN/dEu ~ ge(E u )g y(E ky). From Problem 2.11, we have g(p ) dp ~ p2 dp, and therefore g (E ) = g(p ) dp/dE ~ p 2 dp/dE. For a neutrino having zero rest mass, the energymomentum relation is p, = E ky/c. IIcnce Qy(E ky) ~
E2 ky = (E 0 
E u ) 2.
For an electron (which must be treated relativistically), the energymomentum relation is E u = c V m jfc2 + p2 — ViyC2■ Hence g J E u ) ~ (E u V VieC2) ( E ky + ZmeP 2E u ) 1' 2. Therefore we have dN / d E ke = C (E u +
wteC2) ( E l  f ZmeC2E u ) 1/2(£ o 
E u ) 2,
(8.21)
where C is a constant of proportionality which depends on several other factors involved in the 0decay, such as the atomic number Z of the decaying nucleus and the strength of the weak interaction responsible for the decay. It also has some dependence on the electron’s energy. Plotting dN/dEu against Eu , we may compare with the experimental results shown in Fig. 8 1 1(a) and (b). The agreement in general is rather satisfactory. W e may note that Eq. (8.21) has been derived on the assumption that the neutrino has a zero mass. However, if the rest mass had not been zero, we would have obtained a dif ferent result. Therefore the experimental confirmation of Eq. (8.21) is an indirect proof that the neutrino has a negligible rest mass. Given the degree of accuracy of present ex periments, we can say that the rest mass of the neutrino must be less than OOOlm0, and therefore it can safely be taken as zero.
tt.S X u v iv a r Itv a v lio n s W h en tw o nuclei, overcom ing th eir coulom b repulsion, com e v e ry close together (w ith in the range o f the nuclear force), a rearrangem ent o f nucleons m ay occur. T h is m ay result in a nuclear reaction, sim ilar to the rearrangem ent o f atom s in reacting m olecules in a chemical reaction. N u c lea r reactions are usually produced by bom barding a target nucleus (Λ/,·) w ith a nuclear projectile (m,·), in most cases a nucleon (neutron or proton) or a ligh t nucleus such as a deuteron or an alpha particle.
H e a v ie r nuclei are not generally used because the electric repulsion be
tween h e a v y nuclei requires a projectile w ith a large kinetic energy. Som etim es photons are used as projectiles. M o s t o f the reactions result in the same particle, or another particle (ni/) being ejected and a residual or final nucleus (.1//) being le ft in its ground state or in an excited state. the sym bols .!/,(w,·,
T h e reaction is designated by
where the initial and final nuclei are to the le ft and
8.5)
Nuclear reactions
359
to the righ t o f the parentheses and the incom ing and ou tgoing ligh t particles are inside the parentheses.
F o r exam ple, when w e bom bard
1^ N w ith α particles
(o r ^ H e), the result m ay be a proton (o r [ I I ) and a residual ' 0 nucleus. process m ay be w ritten in the form
T h is
14
IN + 2H e ►
or in the m ore ab b reviated notation mX (a, P ) 17O. In general, when the energies o f the particles in vo lved are not too high, a nuclear reaction supposedly occurs in tw o steps.
F irs t: an incom ing particle or
projectile is captured, resulting in the form ation o f an intermediate or compound nucleus which is in a highly excited state.
In the second step, the com pound
nucleus m ay be deexcited, either b y emission o f the same incom ing particle or by some other means. T h e ab ove exam ple can thus be w ritten in the form
th at the nucleus is in an excited state. G enerally speaking, to a g iven first step in a nuclear reaction, there are several modes o f deexcitation fo r the com pound nucleus. E ach m ode is called a “channel. ” F o r exam ple, when 27A l is bom barded w ith protons, several products result, some o f which arc listed below :
IJ S i +
7
i l N a f 3 JH + n Som e nuclear reactions cannot be described by means o f the com pound nucleus model. O ne exam ple is the strippin g reaction, i M g + 2H —* JH + or f M g (d , p ) 22
S
i iM g ,
E xp erim en tal evidence indicates th at in this process, when
the deuteron comes v e ry close to the nucleus, the neutron is captured and the pro ton is repelled w ith ou t the form ation o f a com pound nucleus. N u clea r reactions are essentially collision processes in which energy, m om entum , angular m om entum , number o f nucleons, and charge must be conserved; the m eth ods o f newtonian and relativistic mechanics are used to com pute some o f these quantities (see Ap p en d ix I I ) . I f the incom ing and ou tgoing particles are the same, the process is called scattering. Scatterin g is elastic if the nucleus is le ft in the same state, so that kinetic energy is conserved, and inelastic if the nucleus is le ft in a different state.
In the inelastic case, the kinetic energy o f the ou tgoing particle
differs from the kinetic energy o f the incom ing particle.
350
Nuclear processes
C8.5
T h e Q o f a nuclear reaction A/,(m,·, m f ) M / is given b y the expression
Q = [(M 1 + m,) 
(M t + m/)]c2.
(8 .22)
In the ab ove equation atom ic masses arc alw ays used. I f Q is positive, the reaction occurs at all values o f the kinetic energy o f the incom ing p ro jectile; but if Q is neg ative, v ii m ust h a ve a t least a threshold kinetic energy to produce the reaction. A s w e shall see in E xam ple 9.8, the threshold kinetic energy o f the incom in g par ticle in the L fra m e o f reference is
, M i I m, + mj I Mj if the particles m ust be treated rela tivistically.
H o w ever, if the particles can be
treated n on relativistically, so that Q is sm all and m /
M / can be replaced in the
num erator b y Λ/,■ f m,·, the threshold kinetic en ergy o f the p ro jectile in the L fram e o f reference is
T h is is the expression used in most cases. In some cases the p rojectile is captured but no new particle is em itted. Instead, a gam m a ra y (o r p h oton ) is em itted, whose energy depends on several factors, such as the state o f the resulting nucleus and the binding energy plus the kinetic energy o f the captured particle. A n exam ple o f a capture reaction is 27A l(n , 7 ) 28A l. A n oth er is the capture o f a neutron b y hydrogen in the reaction 'I I ( n , 7 ) 2H ; the resulting atom is deuterium. T h e reverse process m ay also occur: a nucleus m ay absorb a photon or 7ray whose energy is sufficient for a particle to be ejected. T h is process, which is equ ivalent to the photoelectric effect in atoms, is called a pholonuclear reaction. E xam ples are 25M g (7 , p ) 24N a and 21I(7, n )'H . A nuclear reaction m ay be described in term s of a cross section.
T h e concept
o f cross section was introduced in Sections 1.9 and 7.8 fo r certain processes.
Of
course, for each particular nuclear reaction, there is one cross section, which is ex pressed as a function o f the en ergy o f the p rojectile or incom ing particle. sections are defined experim entally in the follow in g w ay.
Cross
Suppose th at a sam ple
o f thickness A x (w here Ax is sm all) and area Λ , containing n, target nuclei per unit volum e, is exposed to a current density nav o f incom ing particles (o f ty p e a ) per unit area and unit tim e, where na is the number of incom ing particles per unit volu m e and v th eir v e lo c ity . G iv e n that N lj ou tgoing particles (o f ty p e b) are ob served leavin g the sam ple per unit tim e, the cross section for the reaction (a, b) is
σ (α’ ^
(n uv )(n ,A Δ χ )
(8.23)
T h is definition coincides w ith E q. (7.10), since n tA A x is the total number o f target nuclei and thus N h Zfn lA A x ) gives the total flux per unit tim e o f ou tgoing particles
Nuclear reactions
8.5) per target nucleus.
351
W e note th a t Ni, is expressed in s_ l , th at nav is expressed in
in 3 ms1 or m ~ 2 s—1, th at n tA Ax is a unitless number, and th at σ is expressed in m 2 as in previous cases. M o s t cross sections are o f the order o f R 2 ~ IO28 in2, where R is the nuclear radius. T h e qu a n tity IO28 in 2 is called a bam , ab b revi ated b, as m entioned in Section 7.3. cross sections o f nuclear processes.
I t is a unit com m only used for expressing
Λ subm ultiple is the m illibarn, mb, equal to
IO31 m 2. F o r a g iven incom ing particle a, there m ay result several d ifferent ou tgoin g par ticles b, b ', b " , corresponding to the several reaction channels, each with its own cross section σ (α , b), σ (α , b '), etc. T h e total reaction cross section fo r p article a is then σ (α ) = σ (α , 6) + [ ‘ «C l* 
'iiC + 7.
(8.27)
The Q of this reaction is 10.4 M eV. Since the kinetic energy of the αparticle is about 5 M eV 1 the total energy available is 14 M eV in the cm frame of reference. This energy must be shared by the '«C atom and the 7photon. Thus the 7rays should have an energy slightly less than 14 M eV. From absorption measurements it was estimated that the photon energy should be about 7 M eV. Tw o years later (1932) Curie and Joliot ob served that when the radiation from the above reaction passed through a hydrogeneous material, highly energetic protons were produced, with a maximum energy of about 7.5 M e V . The natural interpretation was to assume that the protons had been knocked out by collision with 7photons produced in the above reaction, resulting in a (7, p) nuclear Compton effect. The most energetic protons result from a headon collision in which the photons recoil or are deflected 180°. Given that E r and pr = E r/c are the energy and momentum of the incident photon, E r and p'r = E r/c those of the recoil photon, and Eic and p = V 2m p/s’* those of the proton (which may be treated nonrelativistically), the conservation of energy and momentum give Er = E1 r + Es,
E r/c =  E'r/c+ V 2 m pE t .
from which we may get E r = i ( E t + V 2 (m , Inserting the maximum value of Eu, 7.5 M eV , and recalling that mvc is about 938 M eV, we then get E r ~ 64 M eV. This value of the 7ray photons is much larger than the energy available from the reaction, given above as 14 M eV , or that deduced from absorp tion measurements. What is worse, considering the effect of the products of the BeHe reaction on other substances, we obtain other values of E r , in some cases as high as 90 M eV . Therefore we can obtain no consistent results for E r from Eq. (8.27), com patible with energy and momentum conservation. In 1932 the British physicist J. Chadwick showed that all these difficulties disappeared and the conservation laws were restored if. instead of 7rays, neutral particles were emitted, having a mass close to that of protons. These neutral particles were called neu trons, and the process can now be written as 4Be + e lle 
l ‘ 2C]* 
'oC + !,n.
(8.28)
Chadwick made careful measurements of the kinetic energy of protons and nitrogen atoms knocked out by the neutrons when they were passing through a substance containing hydrogen and nitrogen, respectively. This allowed the mass of the neutron to be calculated,
Xuclear reactions
8.5)
355
resulting in a value between 1.005 and I.OOS amu, which is consistent with the energies involved in the above, reaction. Later on, more precise measurements yielded Vin = 1.008665 amu. Chadwick’s experiments were the foundation of our present model of the nucleus, a model which assumes that the nucleus is composed of protons and neutrons. Prior to that time scientists considered nuclei as being composed of protons and electrons (.I protons and .1 — Z electrons, or a total of 2.1 — Z particles). But that assumption resulted in insurmountable difficulties: electrons were too big compared with the nuclear radius, their magnetic moments were IO3 times larger than nuclear magnetic moments, and their presence inside a nucleus made it impossible to account for the observed values of the nuclear spins. Thus the timely recognition of the neutron was more than welcome. E X A M P L E 8.11. A sheet of gold 0.3 mm thick is exposed to a slow neutron current density of IO7 neutrons cm 2 s1 . The capture cross section of 107Au for thermal neu trons is 94 X IO28 m2. The density of gold is 19.3 X IOi kg m 3 and its atomic mass i s 197.2 amu. Find the number of nuclei of 1 0 8 A u formed per second and per cm 2 of the sheet, according to the reaction 197Aufn1T I 108Au. Solution: Let us designate the capture cross section of 107Au for thermal neutrons by σ(η, T) and the number of atoms of gold per unit volume by nAu. Then the macroscopic cross section for neutron capture, according to Eq. (8.24), is Σ = « Αι,σ(η, 7 ). Given that I 0 is the incident neutron current density, the number of neutrons that have not been captured after a distance x, according to Eq. (8.26), is I = I Oe L The number of neu trons that have been captured in the distance x is thus I0 
I = /o(l 
e~z).
This is equal to the number of 108Au nuclei formed. In our case, _________ 19.3 X IO3 kg m3_______________________ μ ” Au
—3
197.2 amu X 1.66 X 10~27 kg a m u '1
is the number of 197Au atoms per m3. Thus Σ = nAucr(n, I ) = 562 m 1 . Setting x = 0.3 mm = 3 X IO4 m, we have S r = 0.1686 and e~tx = 0.845. Therefore I0 
I = IO7 n cm2 s l ( l 
0.845) = 1.55 X IO0 Ocm 2 S1 .
This gives the number of neutrons absorbed and of 198Au atoms formed per cm2 and per second. E X A M P L E 8.12. Determination of the energy levels of 28Si by use of the nuclear reac tion 27Al(p , a ) 24 M g and its inverse, 24M g fa 1p )27Al. Solution: The compound nucleus of the reaction 27Alfp, a ) 24M g Ls 28Si1 so that we may write 27A l + 1H  » [28Si]*  » 4H e + 24Mg. I t is expected that the cross section for this reaction will show marked peaks (i.e., reso nances) w'hen the total energy in the Cframe of reference of the system 27A l + 1H minus the restmass energy of 28Si coincides with the energy required to excite 28Si to
356
Nuclear processes
(,8.5
I 11.588 lit,.
'7A l + 9 .990
t 2
4M
r
+
*
p
'! I
10 4M c 9 I I
I I
Kinetic energy o f incoming particle plus target nucleus in the Cframc o f reference (It)
F ig. 816. Evidence for the compoundnucleus concept by use of the reactions 27A I(p ,o ()24M g and 24M g fa 1 p )27A l The compound nucleus is 28Si. (a) Energy rela tions, (b) cross section for both reactions. Data from Kaufmann1Goldberg, Koester1and Mooring, Pliys. Rcv., 88, 673 (1952)] one of its stationary states. This energy is [ M ( 27A l) + »ip 
JI/(28Si)]c2 + E kp = (1 1 .5 8 8 + Ekl,) M e V 1
where AJp is the total kinetic energy of the system (p, 27A l) in the Cframe. Once the compound nucleus is formed, it may break into the system 24M g + a. The 24M g may be in its ground state or in an excited state; the state of the 24M g nucleus determines the kinetic energy of the αparticle and of the 7rays observed when the 24M g nucleus proceeds to its ground state. The inverse reaction, that is, 24M g + 4H e > [28Si]4 —> Ή + 27A l1 produces a 28Si nucleus with an excitation energy given by [.U (24M g) + m (JHe) 
.U (28Si)Jc2 + E ka = (9.990 + E ka) M c V 1
8.6)
Nuclear fission
357
where Eka is now the total kinetic energy of the system (a, 24M g) in the Cframe. The different energy relations are sketched in Fig. 816(a). From our discussion it is clear that the cross sections for the reaction 27Alfp, a )24M g and its inverse, 24M g fa 1p )27Al, must show the same set of resonances. However, for each resonance the kinetic energy of the system (a, 24M g) is uniformly larger than the kinetic energy of the system (p, 27A l) by the amount (11.588 — 9.990) M eV = 1.598 M eV when the energies are referred to the respective Cframes. This conclusion is con firmed by the experimental results shown in Fig. 816(b), which gives the cross sections for protons with energies from 1.1 to 1.8 M eV and for αparticles with energies from 2.7 to 3.4 M eV . The result constitutes a satisfying verification of the compound nucleus concept, and serves to show the origin of resonances in nuclear reactions. The energy levels of 28Si, in the energy region considered, can be determined from the energies of the peaks shown in the figure.
0 .6 ' X u e lv a r JFittnion A n im portant nuclear process is fission.
I t consists in the division o f a h eavy nu
cleus, such as uranium or thorium, into tw o fragm ents o f com parable sizes. Fission as a natural process is v e ry rare ( 238U is believed to fission spontaneously w ith a halflife o f app roxim ately 1 0 '0 years). T h e usual m ethod o f producing fission arti ficially is to excite the nucleus.
T h e threshold or m inim um a c tivatio n energy re
quired fo r fission o f a heavy nucleus is from 4 to 6 M e V . One o f the m ost effective means o f inducing fission is by neutron capture.
T h e binding energy o f the cap
tured neutron is, in som e cases, enough to excite the nucleus a b o v e the threshold energy, so that division into tw o fragm ents takes place. T h is is, for example, the case o f the nucleus 2U , which undergoes fission after capturing a slow (o r therm al) neutron. T h e process m ay be expressed by the equation 23D2 5 U  F n  * [2gU ]*  » X + Y . For other cases, in order for fission to take place, the neutrons must have some kinetic energy— o f the order o f I M e V — in addition to the binding energy. T h is is what occurs with 2gU, which fissions on ly after capturing a fast neutron. T h e reason for this differen t b ehavior lies in some details o f the structure o f the different nuclei related to the pairing term in the mass form ula given in E q. (7.1‘2 ).
The
nucleus 2Q8U is evenodd, w ith 143 neutrons, and when a neutron is captured, an eveneven nucleus, 22U , is formed. T h e captured neutron is paired w ith the last odd neutron o f 22U , releasing the additional pairing energy δ ~ 0.57 M e V .
On
the other hand, 2g lU is an eveneven nucleus, w ith 146 neutrons, all paired, and when a neutron is captured, an evenodd nucleus, 2g 2U, results, with no extra pairing energy available. F o r the same reason 2gPu, w ith 145 neutrons, undergoes fission b y slow neutron capture.
T a b le 8 3 gives the excitation energy o f some
nuclei resulting from neutron capture and their fission activation en ergy; from this inform ation it is possible to deduce which nuclei are fissionable by thermal neutrons.
358
8.6
Nuclear processes T A B L E 8—3
C K issionahility
o f H eavy
N u clei w ith T h e r m a l
N e u tro n s
Target nucleus 23SU 235 p; 238 ( ;
232Th 23lPa 237Np 23aPu
Compound nucleus
Excitation energy, M eV
Threshold energy, M eV
I231U] [236U] I239U] [233Th] I232Pa] (238Np) [240Pu]
6.6 6.4 4.9 5.1 5.4 5.0 6.4
4.6 5.3 5.5 6.5 5.0 4.2 4.0
T A B L E 8—1
T h re sh old
E nergies
fo r
Bhotofission
Nuclide
238 u 235t . 233 j 
239Pu 232Th
Photofission threshold, M eV 5.08 5.31 5.18 5.31 5.40
± 0 .1 5 ± 0.25 ± 0.27 ± 0.27 ± 0.22
A nucleus m ay also be excited enough to suffer fission by absorbing 7rays o f energy equal to or larger than the threshold energy required for fission. T h is proc ess is called photofission.
T h e threshold energies for photofission are g iven in
T a b le 84 for some nuclei.
O O
O F ig . 817. Deformation of a nucleus according to the excitation energy.
(a)
(b)
(c)
(d)
F igu re 817 depicts the supposed mechanism o f nuclear fission. nucleus whose equilibrium shape is spherical, as in Fig. 8 1 7 (a ).
C onsider a
I f p roperly ex
cited, it m a y experience collective vibrations, as explained in Section 7.10. W h en the excitation energy is low, the oscillations about the spherical shape are small, so th at at its m axim um deform ation the nucleus adopts the ellipsoidal shape shown
8.6)
Nuclear fission
359
N 1,
F ig. 8 1 8 .
Potential energy function for a nucleus under deformation.
in F ig. S—17 (b ). O n c e the excitation energy has been released in the form o f 7rays, the nucleus r e tu r n s t o the equilibrium shape. ra d iative neutron
cap tu re.
T h e process has therefore been a
W h en the excitation energy is larger, the nucleus is
deform ed more, a s in F ig . 8 1 7 (c ). E v en in this case there is some prob ab ility th at it m ay return to t h e original shape a fte r deexcitation by 7 ray emission.
B u t if
the excitation e n e r g y is large enough, the deform ation m ay be so large th at the electrical rep u lsio n b etw een the tw o halves is larger than the shortrange nuclear interaction, and t h e r e is a greater prob ab ility that the nucleus, instead o f return ing to the s p h e r ic a l shape by releasing 7rays, deform s m ore and m ore until it divides into tw o fr a g m e n t s — as indicated in Fig. 8 1 7 (d )— resulting in fission. F igu re 818 s h o w s schem atically the potential energy o f a nucleus versus its deform ation. F o r d e fo rm a tio n s sm aller than a certain critical value, designated I ) in the figure, the n u c le a r forces dom inate the electrical forces and the potential energy increases w i t h th e deform ation, resulting in the curve A B . T h is is the region o f stable os c illa tio n s . F o r deform ations larger than D , the electrical forces dom inate the nuclear forces,
a n d the poten tial energy decreases w ith further deform ation,
resulting in curve B C , which corresponds to the division into tw o fragm ents; these fragm ents fly a p a r t because o f their coulom b repulsion. I f E 0 refers te the ground state o f the n u cleu s, t h e fission threshold energy is E/. T h is curve should be taken on ly in a q u a lit a t iv e sense. I f the nucleus i s
in it ia lly in a state w ith energy less than th at a t B, such as
E 0 or E , it u n d e r g o e s oscillations w ith out fission. I t is prevented from spontaneous fission b y the p o t e n t i a l barrier.
T h e nucleus m ay, how ever, tunnel through the
poten tial barrier, r e s u lt in g in spontaneous fission. T h e p rob a b ility o f tunneling is extrem ely low w h e n t h e nucleus is in the ground state, so th at spontaneous fission is a rather im p r o b a b le even t.
H ow ever, the pen etrab ility o f the barrier increases
360
(8.6
Muclear processes
w ith the excitation energy o f the nucleus, m aking fission m ore probable.
I f the
nucleus has an excitation energy larger than E i , it ob viou sly undergoes fission. T h erefo re fission by neutron capture w ill occur if the binding energy plus the kinetic energy o f the captured neutron are enough to take the nucleus ov e r the poten tial barrier. As we said before, if the binding energy alone is enough, slow neutron fission occurs. F or 218U 1 the height, o f the potential barrier is about 5.3 M e V ; thus 5.3 M e V is the critical energy required to produce the fission o f 235U b y neutron capture. I t is im portant to note, however, that there is not a single threshold energy or a unique critical deform ation ; these quantities depend on the excitation m ode o f the nucleus and the initial state o f the nucleus. T h e capture o f a neutron does not necessarily lead to fission even if the energy is available, because (as explained p reviou sly ) before the nucleus has tim e to split, it m ay relc.asc its excitation energy in the form o f 7rays, resulting in ra d ia tiv e capture instead o f fission. T h erefore, when neutron capture takes place, tw o com p eting processes enter into p lay: rad iative capture (n, 7 ) and fission (n, f).
Each
o f flic tw o processes is characterized by its ow n cross section— designated by σ (η , 7 ) and σ (η , f), resp ectively— which depends on the energy o f the neutron. T h e values o f these cross sections are given in T a b le 8 5 fo r neutrons h a vin g a v e lo c ity o f 2200 m s  1 , which corresponds to the average v e lo c ity o f thermal neutrons at room tem perature.
T h e last column lists the average number o f
neutrons released per fission.
T A B L E 85
Properties o f S om e F issionable M a te ria ls fo r T h e r m a l N e u tro n s
Nucleus
σ(η, f), barns
σ(η, 7 ), barns
V
2 33 LT
525
53
2.51
2 3 5 jj
577
101
2.44
239Pu
742
286
2.89
Fission is not a sym m etric process; in general the tw o fragm ents have unequal mass numbers.
T h e m ost probable division is into fragm ents w ith mass numbers
around 95 and 135, as shown in Fig. 819, which gives the fission y ield o f 233U, 235U, and 239Pu for therm al neutrons.
T h is can also be seen in F ig. 71, where
the rad ioactive nuclides above the s ta b ility region are particularly abundant in the regions around A =
90 and /I =
135, associated w ith N =
50 and N =
82,
respectively. T h e reason for this asym m etry seems to be the tendency o f a h eavy nucleus to split into fragm ents h a vin g closed neutron shells around the m agic numbers 50 and 82, respectively. T w o properties o f fission m ake it a v e ry im p o rtan t process for practical applica tions: One is th at neutrons are released in fission and the other is that energy is released in fission.
Nuclear fission
8.6)
Mass number
Mass number
(a)
(b)
361
Fig. 819. (a) Thermal and fast fission yield of 235U. (b) Thermal fission yield of 233U and 239Pu. [Data from S. KatcofT, Nucleonics 16, No. 4, 78 (April 1958)]
W e can see th at neutrons are released in fission by exam ining the stab ility region in F ig. 71.
F o r the heaviest nuclei, such as uranium, the ratio o f neutrons to
protons is N / Z ~ 1.55.
Th is, o f course, w ill be app roxim ately the ratio for the
resulting fragm ents. H o w ever, from the same figure w e see that for mediummass stable nuclei the ratio is N / Z ~ 1.30. T h is means th at the resulting fragm ents have too m any neutrons, and thus lie ab ove the stab ility region. T h erefore they are β ~ ■ radioactive.
In fact, uranium fission is one o f the m ethods o f producing
β ~ radioactive isotopes. H o w ever, the neutron excess is so large that a few neutrons are released at the tim e o f fission.
T h e average number o f neutrons released per
fission (designated as v) is given in T a b le 8 5 for some nuclides, assuming that fission is induced b y slow neutrons. T h a t energy is released in nuclear fission m ay be seen from the binding energy per nucleon, represented in Fig. 75.
F o r a h eavy nucleus, the binding energy is
about 7.5 M e V per nucleon, but fo r mediummass nuclei, corresponding to the tw o fragm ents, it is about 8.4 .McV per nucleon, resulting in an increase o f binding energy per nucleon o f abou t 0.9 M e V 1 o r a total o f about 200 M e V for all nucleons in a uranium nucleus. This, then, is the order o f m agnitude o f the energy liberated in the fission o f a uranium atom. T h e energy released appears as kinetic energy o f the fragm ents, o f the released neutrons, and o f the disintegration products (th a t is, electrons, photons, and neutrinos) resulting from the ddeeay o f the rad ioactive
362
P1 JucIear processes
(8.6
fragm ents. Since the neutrinos em itted in the /3decay (and a fe w photons as w e ll) n orm ally escape from the m aterial in which fission takes place, on ly about 180 M e V per atom can be retained, an energy still considerably larger than the energy lib erated in a chem ical reaction (which is o f the order o f 3 to 10 e V per atom ). F o r exam ple, the energy released in the fission o f 235U is distributed, on the average, as follow s: K in e tic energy o f fragm ents
1G7 M e V
K in e tic energy o f fission neutrons
.■>
E n ergy o f 7rays (radiated at the tim e o f fission)
7
E n e rgy o f /3_ decay electrons E n ergy o f 7decay o f fragm ents E n e rgy o f β ~ decay neutrinos T o ta l energy
200 M e V
T h e exact energy distribution varies, o f course, from one case to another. T h e fact th at for each neutron absorbed in order to produce one fission, m ore than tw o new neutrons are em itted (on the avera ge) suggests the possibility o f a chain reaction. T h a t is, if things are arranged in such a w ay that, after each fission, at least one o f the new neutrons produces another fission, and o f the neutrons released in this fission, again at least one produces a fission, and so on, then a selfsustaining process or chain reaction results.
(C h ain reactions are v e ry com mon in chem istry.
C om bustion is a chain reaction.
B urning requires th at a m olecule have a certain
excitation energy so that it can com bine w ith an oxygen molecule.
B u t once the
first m olecules are excited and com bine w ith oxygen, the energy liberated is enough to excite m ore molecules o f the fuel, and burning results.) I f in each stage o f the process more than one neutron per fission produces a new fission, the number of fissions increases exponentially and a d ivergen t chain reaction results. what happens in an atom ic bomb.
T h is is
B ut if, on the average, on ly one neutron of
each fission produces a new fission, a steady chain reaction is m aintained under controlled conditions. T h is is what happens in a nuclear reactor. In fast nuclear reactors the neutrons arc used a t the same energies ( I to 2 M e V ) a t which they are released in the fission process.
But in lhennal nuclear reactors
the neutrons arc first slowed down b y allow ing them to collide w ith the atom s o f som e other substance, called a moderator, until they com e to therm al equilibrium w ith the substance.
T h e neutrons are then called thermal.
T h e m oderator must
be a substance which has small mass number and a small neutron capture cross section.
W a ter, heavy water, and graphite are the substances m ost used as
moderators. T h e energy released in a nuclear reactor is extracted by means o f a circulating fluid called a coolant.
In pow er reactors this energy is used for heating or for the
generation o f electric power.
In research reactors the neutrons are used for d if
ferent kinds o f experim ents, or for isotope production. E X A M P L E 8.13. Determination of the energy released when a neutron is captured by 235U and by 238U and the feasibility of fission of 23sU and 238U by thermal neutrons. *
Nuclear fusion
8.7)
363
S o lu tio n : The Q for the capture reaction of a neutron by an atom of mass number Λ is Q = (Μ
λ
+ m„ 
.V .i+ i)c 2 = 931.48(37,, + m„ 
Μ λ + ύ M eV,
where the masses in the second expression must be expressed in amu, and all refer to atomic masses. Considering the case of 235U 1 which becomes 230U after capturing a neutron, we have AZa(235U) = 235.1170 amu and .l/a + ι ( 230U) = 236.1191 amu. Using mn = 1.0090 amu, we find that Q = 6.13 M eV. On the other hand, the potential barrier of 230U for fission is about 5.3 M eV. Therefore the excitation energy of 230U formed as a result of the capture of a neutron by 235U is larger than the height of the potential barrier for fission. W e conclude then that the 238U nucleus resulting from the neutron capture by 235U undergoes fission even if the neutron is so slow that it contributes a negligible kinetic energy. Considering the case of 238I 11 which becomes 239U after capturing a neutron, we have AZa(238U) = 238.1249 amu and M λ +1 ( 239U) = 239.1287 amu. Thus the Q of the cap ture reaction is 4.85 M eV 1 which is smaller than the height of the fission barrier of 239U by about 0.6 M cV. Therefore, unless the captured neutron has sufficient kinetic energy to make up for the energy required, fission is not produced. Experimentally it is found that the minimum laboratory kinetic energy of the neutrons must be about I M eV to fission 238U.
//. 7
\ u r l f a r F u sion
A process which is the reverse o f nuclear fission is nuclear fusion . I t consists in the coalescence o f tw o colliding nuclei into a larger nucleus.
Because o f the coulom b
repulsion between nuclei, they must have a certain kinetic energy to overcom e the coulom b poten tial barrier and g et close enough so that nuclear forces produce the necessary consolidating action.
T h is problem is not present in nuclear fission be
cause the neutron does not have an electric charge, and thus can approach the nucleus even if its k inetic energy is v e ry sm all or p ractically zero.
Since the cou
lom b barrier increases w ith atom ic number, nuclear fusion occurs at reasonable kinetic energies on ly for ve ry light nuclei w ith low atom ic number or nuclear charge. W e shall now estim ate the kinetic energy required to place tw o nuclei o f atom ic numbers Z 1 and Z 2 in contact. L e t r in the electric poten tial energy o f tw o nuclei ( E p = Z \Z2e~/A7Γ€/)>') be equal to the sum o f the nuclear radii, or about IO 14 m. T h en w e obtain E p ~ 2.4 X I O  14Z 1Z 2 J =
1.5 X IO5Z 1Z 2 e V =
O Io Z 1Z 2 M e V .
T h is gives the height o f the poten tial barrier and therefore the m inim um initial relative kinetic energy o f the tw o nuclei necessary for fusion to occur.
I f the col
liding particles do not have a kinetic energy equal to or larger than E p, fusion cannot occur. H ow ever, at energies sligh tly low er than E p there is some p rob ab ility o f fusion by penetration o f the coulom b barrier.
T h e average kinetic energy o f a
system o f particles havin g a tem perature T is o f the order o f k T , or about 8.6
X IO 5 T eV,
where T is in absolute degrees. T h u s the energy o f IO5 e V corresponds to a tem pera ture o f about IO9 0K , which is much higher than the tem peratures believed to exist a t the center o f the sun.
E ven so, fusion is one o f the m ost im portant processes
364
Nuclear processes
(8.7
occurring in the sun, and is its main source o f energy.
T h e fusion takes place
am ong the re la tiv ely small number o f light nuclei which have energies w ell above the average energy at the sun’s tem perature. W e conclude that, if nuclear fusion o f a large number o f nuclei is to take place, it is necessary fo r the reacting nuclei to be a t tem peratures much higher than those generated by even the m ost exoergic chemical reaction. T h e extrem e tem perature creates a problem o f containm ent o f the reacting particles, since no known m aterial can sustain such tem perature. A lso, a t these tem peratures, the nuclei are stripped o f all their surrounding electrons (because o f collisions) and the substance con sists o f a neutral m ixture o f p ositively charged nuclei and negative electrons called a plasma.
C ontainm ent has been attem p ted by means of m agnetic fields.
Also,
when the inten sity o f the m agnetic fields is rap idly increased, the plasma is com pressed adiab atically and its tem perature increases until fusion begins.
S everal
ingenious devices have been built th at perform these tw o functions o f containm ent and heating. A s we m ay see by referring to Fig. 76, energy is liberated in nuclear fusion o f light nuclei (A < 20). W hen tw o ligh t nuclei coalesce in to a heavier one, the bind ing energy o f the product nucleus is greater than the sum o f the binding energies of the tw o lighter nuclei, and this results in a liberation o f energy.
I f the conditions
are appropriate, the energy liberated in fusion is enough to excite other nuclei, and a chain reaction results.
T h e chain reaction becomes a nuclear explosion by a
mechanism sim ilar to a chemical explosion, but in this case the explosion is due to nuclear instead o f electrical forces. T h e chain reaction m ay also occur under con trolled conditions, although no fu lly satisfactory fusion reactor has yet been built. T h e sim plest fusion react ion is the capture o f a neutron by a proton (o r hydrogen nucleus) to form a deuteron: (8.29)
} H + n — ?H + 2.226 M e V .
T h e great ad van tage o f this reaction is th at there is no electrical potential barrier to be overcom e.
A s will be seen in the next section, scientists assume th a t this
fusion reaction played an im portant role in the ea rly stages o f the evolution o f the universe.
A t present, however, this fusion reaction is rela tiv ely unim portant due
to the lack o f a sufficient number o f free neutrons.
N evertheless, reaction (8.29)
occurs when neutrons from a nuclear reactor diffuse through a hydrogeneous sub stance, such as w ater or paraffin. A n o th er sim ple fusion reaction is that between tw o protons.
Since a diproton
nucleus does not exist, the process is accom panied by the conversion o f one o f the protons into a neutron a t the expense o f the binding energy o f the resulting deu teron, and the emission o f a positron and a neutrino. T h a t is, H +
} I I  > ?H I e + + v +
1.35 M e V .
(8.30)
A th ird im portant fusion reaction is th at between hydrogen and deuterium, re sulting in a tritium nucleus: jH + * H  » iH + e + + v + 4.6 M e V .
(8.31)
8.7)
Nuclear fusion
365
A fusion reaction which has a large cross section and which liberates a great am ount o f en ergy is th at between deuterium and tritiu m : TH + ?H  » He + n +
17.6 M e V .
(8.32)
A n oth er possible deuterium fusion reaction, in v o lvin g 3I I e 1 is I H + I H e > !H e + } H + H ow ever,
18.3 M e V .
since tritium and 3H e are n o t so readily available, th e fusion
(8.33) o f tw o
deuterium nuclei or deuterons is o f m ore practical im portance. T h is hasthe addi tional ad vantage o f using on ly one class o f nuclei. T h e resulting products m ay be d ifferent;
tw o
possibilities
occur
with
about th e same p ro b ab ility: jjH e +
η + 3.2 M e V
/ i H + ?H
\ ? II
+
} H + 4.2 M e V (8.34)
T h e cross sections for processes (8.32) and (8.34) are shown in F ig. 820.
F ig. 820. Cross sections of the deuterondeuteron and deuterontriton fusion re actions as a function of deuteron energy. [Data from A. Bishop, Project Sherwood, Reading, Mass.: AddisonWesley, 1958]
A lth ou gh the energy released in a single fusion reaction is much less than th at released in a single fission reaction, the energy per unit mass is larger (because deuterium is a v e ry light fuel). energy is about 2
X
F o r the deuterium deuterium fusion reaction, the
IO m J per kilogram o f fuel. T h is is m ore than double the value
for uranium fission. Due to the rela tiv e abundance o f deuterium (abou t one deute rium atom fo r about 7000 hydrogen atom s), and the rela tiv ely low cost o f extracting deuterium from w ater (ab ou t 300 per gram ), scientists predict th at— once con trol ledfusion devices becom e practical— the fusion process will provide an alm ost unlim ited source o f energy. Fusion reactions are the source o f the energy o f the sun and the other stars. One o f the m ost im portant fusion processes is the Bethe or carbon cycle, which is
366
(8.7
Nuclear processes
equ ivalen t to the fusion o f four protons into a helium nucleus.
T h e cycle occurs
in the follow in g steps: In
+
12o C  > * ? N
' I n  » 1JJc + e+ + v }H +
1GC  » 1JN
Ih + 'I n  » 1I o
Iη +
(83o)
1Io
 > ‘ ?N + e + + V
1I n
*
1I c
+ !H e .
A d d in g all the reactions and canceling com m on terms, we have 4 iH  > H e + 2e+ + 2v + 26.7 M e V . The
(8.36)
net energy lib erated in the process is 26.7 M e V , or about 6.6 X IO 14 J per kg
o f JH consumed.
N o te that the carbon atom is a sort o f catalyzer, since it is re
generated a t the end o f the cycle. T h e tim e required for a carbon atom to go through this cycle in the sun is about 6 X IOe years. T h e cycle is shown schemat ically in Fig. 821.
Fig. 821. The carbon cycle.
A n oth er fusion reaction in stars is the Critchfield or prolon prolon cycle, which consists o f the fo llo w in g steps:
Jh
+ J ii » ? h
Jh
+ TH
+ e+ + *
— IH e
(8.37)
I l I e + He  » He + 2 }H . P ro p erly com bined, these reactions also y ield E q. (8.36). T h e period o f the protonproton cycle in the sun is about 3 X IO0 years. T h e cycle is shown schem atically in F ig . 822. A s explained in the next section, astrophysicists b elieve th at the protonproton cycle presently predom inates ov e r the carbon cycle in the sun and stars o f sim ilar structure, but th at in m any youn ger stars the situation is the reverse and the carbon cycle is m ore im portant.
In older stars that have much higher tem pera
tures, other types o f fusion reactions are also b elieved to take place.
8.8)
The origin of the elements
367
I t is estim ated that process (8.36) is occurring in the sun a t the rate o f 5.64 X 1 0 " kg per second o f hydrogen fusing into helium, w ith a release o f 3.7 X IO25 W . O f this, o n ly about 1.8 X IO 14 W falls on the earth, m ostly in the form o f electrom agnetic radiation; however, this is still IO5 tim es greater than all the industrial pow er generated on the earth.
S M T h v O rig in o f th v E iv m v n t* W e shall now consider the interesting and im portant question o f the origin o f the elements. B y origin we m ean the mechanism b y which the present, com position o f the universe was achieved, assuming the existence o f certain prim eval m atter. T h e statem ents w e shall m ake arc m erely speculative, since there is on ly incom plete evidence available. One o f the clues w e use for our speculation is the rela tive abundance in the universe o f the different chem ical elements, as shown in F ig. 823, and the other is their isotopic com position.
A s w e can see from the figure, hydrogen is the most
abundant, follow ed b y helium. A fte r the sudden drop in abundance corresponding to lithium , beryllium , and boron, the abundance follow s a regularly decreasing trend, levelin g off for Z > 35 or A > 80, but w ith some pronounced maxima, especially for iron and neighboring nuclides.
T h e elem ents stop a t Z =
92, since
the am ount o f existing nuclei w ith Z > 92 is essentially zero, although they have been produced a rtificia lly in the laboratory up to Z =
103.
A s to isotopic com
position, the lighter elements are richer in those stable isotopes w ith low er neutron content, while the heavier nuclei are richer in isotopes w ith higher neutron content. Also, e v en A nuclei are m ore abundant than odd.4 nuclei.
A n oth er interesting
feature is th at no nuclei w ith A = 5 or 8 are found in nature. T h e relative abundance and isotopic com position have been found to be the same not on ly in samples taken from various parts and depths o f the earth’s crust but also in samples taken from m eteorites which h ave fallen on the earth from outer space.
T h is constant com position suggests that, at least in our galaxy, the
elem ents w ere all form ed b y app roxim ately the same process.
T h e “ zero tim e ”
368
(8.8
Xuclear processes
F ig. 823. Cosmic abundance of the elements. Rlnjs. Rev. 88, 248 ( 1952) J
[Data from H. Urey and H. Brown,
o f the process is estim ated to have occurred a t about (8 ± 2) X IO9 years ago. T h is figure was obtained b y tak in g into account several astrophysical considera tions, such as the v e lo c ity o f receding galaxies, and by an alyzing the natural radio active chains. M a n y theories h a v e been advanced to explain this general com position o f the universe, but no single th eory is com pletely satisfactory. T h e m ost acceptable th eory at this tim e (attrib u ta b le to Fow ler, H oy le, B urbridge, and Greenstein, am on g others) postulates that the elem ents are synthesized in stars under v aryin g conditions.
T h e sequence o f even ts contem plated by this theory is as follow s:
Suppose th at in itially (i.e., a t “zero tim e ") there is a v e ry large mass o f gaseous hydrogen (w ith , perhaps, som e free neutrons).
Due to statistical fluctuations and
g ravita tion al interaction, some o f this hydrogen m ay condense in to clusters, or stars, reaching a density o f the order o f IO5 kg m  3 . In the process o f condensation there is a transform ation o f gravitation al potential energy into kinetic energy, resulting in an increase o f tem perature (to about IO7 ° K ) of the gas. tem peratures the protonproton cycle o f E q . (8.37) is possible. gins to be form ed.
A t such
H elium thus be
I t is possible that, as byproduct reactions o f the proton 
proton cycle, rela tiv ely small quantities o f nuclei w ith higher atom ic mass m ay be form ed.
F o r exam ple, 7Li is produced by the process OlIe j 2H e —* JBe F Ύ,
^Be F c
3Ei F v.
The origin of Ihe elements
8.8)
369
P a rt o f the lithium reverts back to helium through the reaction
I L i + {Η — He + !H e . O ther fusion reactions, such as (8.29), (8.31), and (8.33), m ay also tak e place, but in much less amount. Since helium is more m assive than hydrogen, the helium nuclei produced in the protonproton cycle are carried to the center (o r core) o f the star by gravita tion al action.
T h e density at the core m ay then becom e as high as IO8 kg m ~ 3.
The
resultant gain in kinetic energy o f the helium nuclei a t the core increases the core tem perature (u p to values o f IO8 ° K ). T h e surface tem perature o f a star in whose interior a large qu a n tity o f helium nuclei has accum ulated increases and its color consequently changes.
T h e increase in tem perature and density o f helium nuclei
at the core increases the possibility for the production o f fB e by the helium fusion reaction ! H e + ! H e  » fB e , which is follow ed, in about IO 16 s, b y the decay process fB e  > !H e + !H e . H o w ever, due to the rela tiv ely large helium concentration, it is possible for another helium nucleus to be captured before this d ecay occurs, resulting in the reaction fB e + ! H e  » ‘ iC . A lte rn a tive ly , the interm ediate products in the protonproton chain f H and He m ay be captured by the beryllium , form in g 1IJB and 1^C 1 respectively, although in much sm aller am ounts th an 1§C.
A lso m an y other lessprobable reactions can
take place which result in other light elements. T h e chain o f events just described explains how the gaps a t A = be bypassed.
δ and 8 can
T h is mechanism also makes the scarcity o f lithium , beryllium , and
boron understandable, as well as explaining the rela tiv ely great abundance o f ‘ C. B j' the same process o f helium capture, the production o f successively m ore m assive nuclei such as 'fO , fJjNe, anc^ others is possible. T h e on ly lim itation to helium capture is the am ount o f energy required by the !H e nuclei to overcom e the cou lom b repulsion o f the h eavier nucleus.
F igu re 824 is a schem atic representation
o f the series o f events proposed by this th eory o f helium capture (o r burning).
F ig. 824. The helium burning process.
370
8.8
(.
Nuclear processes
W ith the production o f heavier nuclei great quantities o f helium are used up and a further g ravita tion al contraction o f the star should take place, leading to core densities o f the order o f IO9 kg m 3 (w ith a corresponding increase in the kinetic energy o f the nuclei and a core tem perature which approaches 10° ° K ). U n der such conditions o f v e ry high density and extrem e tem perature, other nuclear reactions are possible which m ay produce nuclei o f higher mass number, up to the iron group (ab ou t A = CO), but not heavier, due to the energy lim itations men tioned above. F ree neutrons m ay be produced in som e instances through (a, n) reactions, such as 1I C
+
's O
f 2 lie —» io N e ( n,
I oK e +
!H e —> 19O
+ n,
jH e —* Ia M g I n.
T hese neutrons, togeth er with som e o f the original (o r prim eval) neutrons which have not decayed into protons, arc availab le for exten din g the production o f nuclei to higher mass numbers— that is, nuclei beyond the iron group— by neutron capture rath er than by the capture o f charged particles. In m any instances neu tron capture reactions result in /3~decay (as w e have seen, every tim e a nucleus acquires too m any neutrons, it decays by electron emission).
T h e resultant nu
cleus has a higher atom ic number a fte r the /3“ decay and the chain can thus ad vance tow ard higher Zvalucs.
O f course, as tim e passes, the number o f free
neutrons decreases. T h is then is an explanation for the fact th at the h eavier elem ents are rela tiv ely less abundant since the production o f the heavier elem ents is alm ost entirely dependent on neutron capture. Since the in itial “ firin gu p” o f a star is based on a statistical fluctuation, the stars in the universe cannot all be expected to follo w the same sequence a t the sam e rate, and the stars are presently in differen t stages o f evolution. T h e sun itself is still m ostly in the first stage o f evolu tio n ; its com position is assumed to be 81.70% hydrogen, 18.17% helium, and 0.07% fo r the rest o f the elements.
Stars
In which the protonproton cycle presently appears to be the dom inan t process are called m ain sequence stars. T h ose stars in which a t present it appears th a t the m ost im portant process is helium burning are called red (jiants because o f their color.
In m an y stars the three stages o f nucleosynthesis proposed by this theory
are assumed to be tak in g place sim ultaneously, w ith hydrogenburning occurring a t the surface, helium reactions prevailin g a t an interm ed iate (and h o tte r) layer, and h eavy elem ents being produced in the much h o tter core o f the star. Stars that have e v o lv e d in the w ay predicted by this th eo ry are called firstgeneration stars. In som e instances instabilities which arise during the evolu tion and aging o f a star m ay result in the ejection o f som e o f the m aterial from its in terior into inter stellar space, where it mixes w ith uncondensed hydrogen and the dust o f outer space.
C ondensation o f some o f this m ixture a t som e later tim e results in second
hand later) generation stars.
I t is in these youn ger stars th at the carbon cycle,
E q . (8.3.5), plays an im portant role.
Problems
371
T h e theory o f the origin o f the elements is much m ore com plex than our lim ited review can possibly indicate, but w e cannot explore this m atter in greater detail here. W h a t w e h ave said should be enough to m ake the student appreciate the ideas explored in this interesting, challenging, and creative area o f physics.
ltvferencvH 1. “ Resource Letter N R  I on Nuclear Reactions," T. Griffy, Am. J . Phys. 35, 297 (1967) 2. “ The Discovery of Nuclear Fission,” II. Graetzer, .1m. ./. Phys. 32, 9 (1964) 3. “ A Study of the Discovery of Fission,” E. Sparberg, .lm. J . Phys. 32, 2 (1964) 4. “ Nuclear Fission,” R. Leachman, Sci. Am., August 1965, page 49 5. “Thermonuclear Reactions,” G. Thomson, .lm . J. Phys. 28, 221 (I960) 6. “ Recent Developments in Controlled Fusion,” A. Bishop, Physics Today, March 1964, page 19 7. “ The Origin of the Elements” by W . A. Fowler and “ The Structure of Nuclei” by V. Weisskopf in The Scientific Endeavor. N ew Y ork : Rockefeller Institute Press, 1965 8. Structure of Matter, W . Finkelnburg. N ew Y ork : Academic Press, 1964, Chapter V, Sections 919 9. Principles of Modern Physics, R. Leighton. ters 16 and 17
N ew Y ork: M cGrawHill, 1959, Chap
10. Nuclear Physics, I. Kaplan. Reading, Mass.: AddisonWesley, 1963 11. Introduction to the Atomic Nucleus, J. Cuninghame. Amsterdam: Elsevier, 1964 12. Radioactivity and Its Measurement, W . Mann and S. Garfinkel. Van Nostrand, Momentum Books, 1966
Princeton, N.J.:
13. .1 Source Iiook in Physics, W. Magie. Cambridge, Mass.: Harvard LTniversity Press (1963); page 613 (Becquerel); page 617 (P. and M. S. Curie)
P ro b lvm H 8.1 The halflife of 90Sr is 28 years. Determine: (a) the disintegration constant for " S r , (b) the activity of I mg of " S r in curies and as nuclei per second, (c) the time for the I mg to reduce to 250/ig, (d) the activity at this later time. 8.2 A freshly prepared sample of a radio active material, which decays to a stable nuclide, has its activity measured every 20 seconds. The following activity (in yCi) is measured starting at ( = Os: 410; 190; 90; 43; 20; 9.6; 4.5; 2.15; 1.00; 0.48; 0.23. (a) Plot the natural logarithm of the
activity against the time, (b) Find the dis integration constant and the halflife of the sample, (c) How many radioactive nuclei were present in the sample at / = 0 s? 8.3 A material is composed of two differ ent radioactive substances having halflives equal to 2 hrs and 20 mill., respectively. Initially there is one mCi of the first sub stance and 9 mCi of the second, (a) Using semilogarithniic paper, plot, as functions of time, the activity of each substance and also of the whole material, (b) A t what time is the total activity one mCi? (c) A t
372
Nuclear processes
what time is the activity of the shortlived substance 1 % of the longlived substance? 8.4 The activity of a material is measured every 30 s and the following values (in counts per minute) are found: 1167; 264; 111; 67; 48.3; 37.1; 30.0; 24.6; 20.9; 18.1; 15.7; 13.9; 12.3; 11.1; 9.84; 8.85; 7.83; 7.02; 6.26; 5.60; 5.00. (a) Plot the logarithm of the activity versus time, (b) Determine how many radioactive substances are pres ent and compute the halflife and disinte gration constant of each. [H in t: First subtract the activity of the longestlived substance by extending the tail of the curve in a straight, line back to zero time. Plot the remaining activity and repeat the proc ess until a straight line, corresponding to the shortestlived substance, remains.] 8.5 Show that the average life, T livct of a radioactive substance is given by [° Tavc =  ^  /
No J x ο
t (IN = r ·
X
8.6 Solve Example 8.2, assuming that initially there are N o nuclei of the radio active substance. Consider particularly the cases in which N o < g/λ and ΛΓο > g/\. 8.7 Referring to Example 8.3, show that the number of nuclei of material C (a stable nuclide), as a function of time, is given by N e =   A° — c Xe  Xa
X IX e d 
< Γ λ·4') 
X a d  e _X/i')].
8.8 Referring to Fig. 84, show that the time at which the number of nuclei of sub stance B is maximum is equal to
In (X b /X.D X b — X.i 8.9 A steel piston ring with a mass of 25 g is irradiated in a nuclear reactor until its activity is 9.0 mCi (due to the nuclide 59Fe which has a halflife of 3.90 X 10Gs). T w o days later the piston ring is installed in a test engine. A fter a tenday test run, the
crankcase oil is drained and studied for activity due to o9Fe. Investigation shows an average activity of 9.8 X IO2 disintegra tions per second for a 200cm3 sample of the oil. Compute the mass of iron worn off the piston ring, given that the crankcase has a capacity of 7.6 liters. 8.10 The activity of carbon found in living specimens is 0.007 ^Ci per kilogram, due to the 14C present. The charcoal taken from the fire pit of an Indian campsite has an activity of 0.0048 μΟί kg1 . The halflife of 14C is 5760 years. Calculate the year the campsite was last used. 8.11 Plot the logarithm of the disintegra tion constant versus the energy for the adeeays listed in Table 81. Show that the points lie approximately along a straight line, so that log X = a,E„ + 6. Determine a and 6 for best fit. 8.12 Calculate the energy of the «p arti cle released by 144N d when it decays to 140Cc. Also calculate the recoil energy of the daughter nucleus. The rest masses are 143.9100 amu and 139.9054 amu, respectively. 8.13 An elastic collision between an apartide and a nucleus of unknown mass is observed in a cloud chamber. T h e aparticle is deflected 55° from its original direction, while the nucleus leaves a track which makes an angle of 35° with the in cident direction. What is the mass of the nucleus? 8.14 The range, R, of α particles in air and the α partiele energies E are related by the empirical relation R — 0.318F3/2 m, where E is in M eV. (a) For the aparticle energies shown in Fig. 8 8 , determine the range of the αparticles emitted when 211Bi decays, (b) The αdecay of 212Po is ob served in a cloud chamber, and four ranges— 11.2, 11.0, 9.57, and 8.51cm— are measured. Calculate the energies of these αpart icles. (See Fig. 89.) 8.15 The α dccay spectrum from 220Ra has a triplet structure, with aparticle
Problems
373
energies of 4.777, 4.593, and 4.342 M e V . Assuming that the daughter nuclide 222Iin is produced with a ground and two excited states, draw the energylevel diagram and show the 7ray emission associated with the transition.
8.23 Complete the following nuclear re action equations, substituting the correct nuclide or particle wherever an X appears:
8.16 Show the possible modes of decay for 40K , which has a rest mass of 39.9610 amu. Compute the energy available for each possible process. 8.17 Show from mass measurements that 64Cu may decay by /S+ and /8 emission, or electron capture. Experimentally, we know that 04Cu has a halflife of 12.8 hours with 39% β~, 19% 0 + and 42% EC. Compute the energy available for each of the three processes.
(a) S A l ( n , « ) X ;
(b) iP(7, n).Y;
(c) JP(d, p).Y;
(d) 1IC(JY1Cr)IBe;
(e) 1IB fY 1 A)?Be;
(f) 1IIln (U 1Y)JY;
(g) ϋ Ν Ϊ(ρ , n ) X ;
(h) 2?Co(n, JY)?Co.
8.24 The mass defect of a nucleus is defined as Δ = A — M , where A is the mass num ber and M the nuclear mass in amu. Express the Qvalue for 0 ~  and β +decay and the reaction M iOm , m /)M f, in terms of the value of the mass defect Δ for each of the particles involved. [H in t: See Examples 8.4 and 8.5.]
8.18 (a) Show that 7Be decays by elec 8.25 A particle with mass m, is projected tron capture. Its rest mass is 7.016929 amu. with a kinetic energy E k (in the Aframe of (b) Compute the energy and the momentum reference) against a nucleus of mass M t of the neutrino and the daughter 7Li initially at rest in the laboratory. Show nucleus. that (a) the total kinetic energy of the sys tem in the Cframe is E kM J ( M i + m,), 8.19 Calculate the maximum energy for (b) the total energy available for the reac the electron in the /3“ decay of 3II. tion M Am i, m /)M f isQf E kM iZ iM i + m,·), 8.20 When 14O decays by /3+emission, the and (c) the threshold kinetic energy of m, daughter 14N nuclide is almost always in in the Aframe of reference (when Q is an excited state (> 9 9 % ). From the experi negative) is —Q(.U, + m,)/.!/,. Assume mental data which are given in Fig. S10(b) that the particles can be treated nonand from the fact that 14N has a rest mass relativistically. of 14.003071 amu, calculate the mass of 8.26 For the photonuclear reaction
14O. 8.21 Plot the masses of isobars with A = 40, 64, and 134 against Z. Show the possible chains of /3decays and the most stable nuclides. Join the upper and lower values, starting from the stable nuclides, and note that, the connecting lines are parabolas. Measure the mass separation between the two lines and show that it is consistent with the oddeven correction term of Eq. (7.11), δ = ± 3 4 .4 _3/4 M eV. 8.22 Plot the energy distribution of elec trons in /3decay, as given by Eq. (8.21), for Eo = 1.24 M eV. Compare with Fig. 8 11(b).
24M gfY 1n) 23Mg, determine the threshold energy of the photon. The rest masses of the parent and product nuclides are 23.9S504 and 22.99412 amu, respectively. 8.27 A certain accelerating machine can accelerate singly charged particles to an energy of 2 M eV and doubly charged particles to 4 M eV. W hat reactions may be observed when 12C is bombarded by protons, deuterons, and αparticles from this machine? 8.28 When 7Li is bombarded by 0.70M eV protons, two αparticles are produced, each with 9.0 M cV kinetic energy, (a) Cal
374
Nuclear processes
culate the Q of the reaction, (b) Calculate the difference between the total kinetic energy of the αparticles and the kinetic energy of the initial proton in the Λframe. 8.29 The rest mass of 13AI is 26.98154 amu. Find the mass of the product nuclei for the following reactions: (a) (b) (c) (d )
27A lfn 17 ) 28Al, 27AK p 1a ) 24Mg, 27A l(K 1P )28A I1 27A K (K a )25M g 1
Q Q Q Q
= = = =
7.722 1.594 5.497 6.693
M eV ; M eV ; M eV; M eV.
8.30 Show that the Q of a nuclear reaction Jl/,·(»!,·, m ,)M , is given by Q = AVIl +

E iI 
(m./.l/.)],
where E i is the kinetic energy of the inci dent particle and E/ is the kinetic energy of those «I/ particles observed at an angle of 90° with respect to the direction of the incident particle. 8.31 For the following reactions, find the threshold energy of the projectile in the L frame, assuming the target nucleus is at rest: (a) I 4N (a , p) 17O; (b) luO in1a) 13C. 8.32 A sample of natural silicon is bom barded by a 2.0M eV beam of deuterons. (a) W rite all the possible reactions in which either a proton or an αparticle is the ejected particle, (b) In each case, find the kinetic energy of the ejected particle in the Cframe and its maximum and min imum kinetic energy in the Cframe. 8.33 A t relatively low (thermal) energies the neutron capture cross section for many substances is inversely proportional to the velocity of the neutron. This is called the 1/e law. IIo w should the plot of the loga rithm of the thermalneutron cross section against the kinetic energy of the bombard ing neutron appear? 8.34 Given that the capture cross section of 197Au for neutrons follows the 1/e law (see Problem 8.33), and that its capture cross section for thermal neutrons (0.025 eV ) is 99 b, find the absorption cross section of 197Au for 1.0eV neu
trons. W hat thickness of a gold foil will absorb 20% of a beam of 1.0eV neutrons? 8.35 A tantalum foil 0.02 cm thick whose density is 1.66 X IO4 kg m ~ 3 is irradiated for 2 hr in a beam of thermal neutrons of flux IO16N in2 s_1. The nucleus 182T a 1 with a halflife of 114 days, is formed as a result of the reaction 181T atn 17) 182Ta. Immediately after irradiation, the foil has an activity of 1.23 X IO7 disintegrations per second per cm2. Find (a) thenumber of 182T a nuclei formed, (b) the cross sec tion for the (n, 7) reaction producing 182Ta. 8.36 A boratedsteel sheet which is used as a control rod in a nuclear reactor is 1.59 mm thick and contains 2% boron by weight. The absorption cross sections of iron and boron for energetic neutrons are 2.5 b and 755 b, respectively, (a) Find the macro scopic absorption cross sections of each of these elements in the boratedsteel sheet. (b) What fraction of a neutron beam is absorbed in passing through this sheet? (c) Calculate the macroscopic cross section of the sheet if the boron content is increased to 3% . (The density of boron is 2.5 X IO3 kg in3 , and that of steel is 1.9 X IO3 kg nr3 .) 8.37 A beam of 370keV neutrons is di rected at a sheet of aluminum foil 0.1 mm thick. Given that the capture cross section of aluminum for neutrons of that energy is 3 mb, see Fig. 814(b), determine the frac tion of neutrons captured. The density of aluminum is 2.70 X IO3 kg m 3 . 8.38 Repeat Problem 8.37 for a beam of neutrons whose energy is about 45 keV. Figure 8—14(b) shows a peak in the capture cross section of about 18 mb at 45 keV. 8.39 The capture cross section of 10B for thermal neutrons is about 4000 b. How thick a layer of 10B is required to ab sorb 99% of an incident beam of thermal neutrons? (Density of boron: 2.5 X IO3 kg m 3 .) 8.40 Show that 6Li is stable to fission by analyzing all the possible ways in which it might split (for example, 6Li —* d + a,
Problems etc.). Similarly show that 5L i and 5H e are unstable to fission. 8.41 In the photofission of 235U into noKr, 142Ba1 and three neutrons, calculate, from the mass differences, the total energy re leased. Compare this energy with the initial coulomb repulsion energy of the two charged fragments, assuming that they are just touching when fission occurs.
375
force, ~ IO15 m.j (b) Calculatetheenergy released in the fusion of two deuterium nuclei into an aparticle. 8.47 What energy is released in the fusion process 3 4H e —* 12C? This process occurs in the second stage of nucleosynthesis in stars. Determine the power generated by this process in a star in which 5 X IO9 kg of 4He are fused in 12C per second.
8.42 Calculate the energy required to 8.48 The nitrogen cycle is a fusion process split a 4He nucleus into: (a) 3H and p; similar to the carbon cycle given in (b) 3He and n. Explain the difference be Eq. (8.35). It starts with the capture of tween these energies in terms of the prop a photon by a 14N nucleus. After succes erties of nuclear forces. sive reactions, in which the nuclides 15O 1 15N 1 10O1 17F 1 and 17O are involved, the 8.43 Show that the energy released in the 14N nucleus is regenerated, with the re fission of uranium (185 M eV per atom) is sultant fusion of four protons into an aequivalent to 8.3 X IO13 J k g 1 . A t what particle. (a) Write the different steps of rate should uranium fission so that I M W the nitrogen cycle in detail, (b) What is of power is generated? How long would it the total energy released in the nitrogen take for I kg of uranium to be used up, cycle? given that it is continuously generating I M W of power? 8.49 Show that, if the energy produced 8.44 It may be shown that spontaneous fission will occur if 02A/03Z 2 ~ 25A/Z2 is less than unity, where 0 2 and 0 3 are coefficients in the Weiszacker formula, Eq. (7.11). Calculate this ratio for 132Xe, 142Ce, 200H g1 235U 1 and 255Fm. 8.45 Compute the excitation energy of the nucleus produced when a neutron is captured by each of the following nuclei: 232T h 1 233T h 1 233U 1 234U 1 239Pu1 and 240Pu. Which of these nuclei would we expect to be fissionable by thermal neutrons? 8.46 (a) What must be the average tem perature of a deuterium plasma in order for fusion to take place? [H in t: W e may estimate this by calculating the coulomb repulsion energy between deuterons when they are within the range of the nuclear
by the sun is to be accounted for, its mass must decrease at the rate of 4.6 X IO9 k g s 1 . How much time must pass for the mass of the sun to decrease by 1 % (mass of the sun = 1.98 X IO30 kg)? 8.50 How long would it take for the length of the year to increase by one second due to the loss of mass of the sun by radiation? (Sec preceding problem.) Recall that the square of the period of the motion of a planet around the sun is inversely propor tional to the mass of the sun. 8.51 Assuming that the rate at which hydrogen is fused into helium in the sun since its beginning has remained the same, show that the age of the sun must be of the order of 2 X IO10 years. The estimated age of the sun, based 011 other calculations, is of the order of IO11 years.
9 FUNDAMENTAL PARTICLES
9.1 9.2 9.3
Particles and Anliparticles 9A 9.5
9 .6
Introdudion
Particle Genealogy
Particle Instability
The Conservation Laws
Invariance, Sym m etry, and Conservation Laws 9.7 9 .8
Resonances
What Is a Fundamental Particle?
Introduction
9.1)
9.1
377
I n i r 27.
(9.1)
T h e reason th at tw o photons are em itted is fo r energy and m om entum conserva tion. I f the electron and positron are at rest in the lab oratory (which then coin cides w ith the C fram e), the total energy availab le is 2mcc2 = 1.022 M c V and the total m om entum is zero. I f a photon o f energy Ey = carry a m om entum py =
Ey/c =
2mec 2 were em itted, it would
2mcc and the principle o f conservation o f m o
mentum w ould be violated , since the initial m om entum is zero. T o conserve energy and m om entum , tw o equal photons must be em itted in opposite directions. Thus each photon must have an energy equal to m,,c2 =
0.511 M e V .
Photons o f this
energy are observed when positrons from a j3+ rad ioactive nucleus pass through matter. A n gu la r m om entum m ust also be conserved and, if the electron and posi tron h ave th eir spins in opposite directions, the photons m ust also have their spins in opposite directions, which requires th at the photons have the same cir cular p olarization (o r m ust be polarized in perpendicular directions). W h en either the electron or the positron (o r b oth ) are not a t rest in the laboratory, the calcu lation can be carried ou t using the laws o f conservation o f energy and m om entum to obtain the energy and m om entum o f the photons as measured in the laboratory. N o te th at in E q. (9.1) charge is also conserved. P ro ton an tiproton annihilation is a m ore com plex process, in v o lvin g the pro duction o f several particles, m ost o f them pions. One such annihilation (which took place in a bubble cham ber) is shown in F ig. 94, corresponding to the process P+ +
P
— >47Γ+ +
47Γ
JXTT0.
(9.2)
382
Fundamental particles
(9.3
Fig. 94. Protonantiproton annihilation. (Photograph courtesy of Brookhaven National Laboratory)
T h e incom ing antiproton annihilates w ith one o f the protons in the gas o f the chamber.
I f charge is to be conserved, the number o f p ositive and n egative pions
must be the same.
T h e number o f 7r°mesons produced is difficult to ascertain,
since they do not leave any track in the chamber.
T h e total number o f particles
produced depends on the energy available. C onversely, a particle and its an tip article m a y be produced sim ultaneously. Figu re 9 5 illustrates the production o f an electronpositron pair b y a photon entering a cloud cham ber from the le ft; th at is, 7 —» e — f e + T h e track o f the photon is not visible.
(9.3) O bviously, fo r an electron pair to be pro
Piarticles and anliparlicles
9.3)
383
duced, the photon m ust have an energy equal to a t least 2mcc 2 = 1.022 M e V . In order for energy and m om entum to be conserved in E q. (9.3), the process must occur near a nucleus, which, as a result o f its electrom agnetic coupling w ith the system, w ill take up the en ergy and m om entum required fo r the conservation o f both quantities.
F o r this reason electron pair production is m ore intense in m ate
rials which h ave high atom ic numbers (such as lead), since these m aterials provide a stronger electrom agnetic coupling w ith the elect ronpositron pair. E lectron pair production is one o f the m ain processes which account for the absorption o f highenergy photons b y various m aterials.
(A t lo w en ergy the photoelectric effect is
the m ore im portant process, and at energies between 0.1 M e V and I M e V it is the C om pton effect.
S c e S e c tio n 1.9 and Fig. 119.)
Sim ilarly, a p roton antiproton pair m ay be produced in a highenergy protonproton collision according to the scheme p + + p + —> p + F p + + p + F p ~ .
(9.4)
(T h is is not the on ly process that m ay take place in a protonproton collision.)
Fig. 93. Electronpositron pair production. National Laboratory)
(Photograph courtesy of Brookhaven
384
Fundamental particles
C9.3
T h is process was used b y the U n iversity o f C aliforn ia group in their experim ents leading to the discovery o f the an tiproton (see E xam p le 9.2). In order fo r process (9.4) to occur, the threshold kinetic energy o f the incom ing proton ( if the target proton is at rest) has to be at least 5.64 G e V (see E xam p le 9.3 for the calculation o f this valu e). Our g alaxy seems to be composed m ainly o f particles (rather than a uniform m ixture o f particles and anti particles) and this makes it stable against annihila tion.
B u t nothing precludes the possibility th at other galaxies m a y be com posed
prim arily o f antiparticles, although no p ositive eviden ce exists to support this assumption.
In any case it is interesting to speculate on what w ould happen if a
galaxy and an an tigalaxy were to collide.
A strophysicists have advanced the idea
th at some o f the explosions observed in distant galaxies m ay be due to a cata clysm ic m atteran tim atter annihilation. E X A M P L E 9.1. Discussion of positronium, which is a system composed of an electron and a positron revolving about their center of mass, similar to Ihe electronproton system of hydrogen. S o lu tio n : When a positron and an electron come close to each other, they may form a stable system whose stationary states are calculated in the same manner as the hydrogen stationary states discussed in Section 3.2. Recalling Eq. (3.7), we note that, in the case of positronium, the two particles have the same mass; that is, Wie = M , and therefore the corresponding Rydberg constant is R ' — %R. When we introduce this value in Eq. (3.5) with Z = I, the stationary energy levels of positronium are given by
1^n
Rh = _ 6 S no O V* In 2 n2
The actual energy levels are given by a more complex expression, due to the relativistic corrections which must be introduced. Because of the possibility of pair annihilation, according to Eq. (9.1), positronium has a transient life. If the positronelectron pair moves with zero orbital angular momentum (as in the Is ground state) and their spins are antiparallel (singlet state 1S), they eventually annihilate into two photons, either circularly polarized in the same sense or linearly polarized in perpendicular planes, as previously explained. The halflife of the single state is 1.2 X IO10 s. I f their spins are parallel (triplet state, 3S), then conservation of angular momentum as well as certain additional selection rules related to the symmetry of the system forbid the decay into two photons. Thus the positronelectron pair annihilate into three photons. The energies of the photons in each case must add up to 1.022 M e V , which is the total rest energy of the two annihilating particles. The halflife of the triplet state is 1.4 X IO7 s. Theexistence of positronium was confirmed in 1951 by Martin Deutsch. When a beam of positrons moves through a gas, a certain number of the positrons form positronium atoms before annihilation. The ratio of probabilities for forming singlet or triplet positronium is I to 3. Because of the much longer life of triplet positronium, it can enter into chemical reaction with the atoms or molecules of the gas. In this way positronium halides have been ob served. These “ compounds” shorten the life of the positronium because the positron may annihilate with another electron of the atom or molecule which has opposite spin. Active research is now being done on the “ chemistry” of positronium.
Particles and anliparticles
9.3) E X A M P L E 9.2.
385
The antiproton experiment.
S o lu tio n : Perhaps one of the most interesting experiments in the physics of fundamental particles is the experiment that led to the discovery of the antiproton. The purpose of the experiment was to detect particles having a charge — e and a mass mp. Protons, ac celerated up to 6.2 GeV by the University of California bevatron, hit a suitable target which produced several reactions involving K ’s, ir’s, and p ’s, as well as some p~ particles. A deflecting magnet M i selected only negative particles, which were allowed to pass through an opening in the shielding (Fig. 96). The S ’s in Fig. 96 are scintilla tion detectors and the C’s are Cerenkov detectors. These instruments are so made that they are sensitive only to particles in a certain energy range (see Appendix V I I). Between S i and S 2 there was placed a second deflecting magnet M 2, which acted as a momentum selector, since the radius of the path was fixed by the position of the detectors, and only particles with mo mentum p = tffir were properly deflected toward S 2. Because it offered certain experimental advantages, the magnetic field was chosen to correspond to a mo mentum p = 1.19 GeV/c, which is slightly less than the momentum at which most antiprotons should have been produced (1.75 GeV/c).
F ig. 96. Experimental arrangement for observation of the antiproton.
M any negatively charged particles (K ~ , π  , μ “ ) and a few antiprotons (it was esti mated that the proportion was 40,000 other particles to one antiproton) passed through the magnet M 2 · T o identify the antiprotons, it was necessary to determine the velocity of the particles. The velocity of an antiproton having a momentum of 1.19 GeV/c is 0.7Sc, while the velocity of a pion of the same momentum is 0.99c. The scintillation detectors .Si and .S2 were placed 12 m apart; the times required for an antiproton and a pion with these momenta to go from Si to S 2 were 5.1 X IO s S and 4.0 X IO8 s, respectively. Thus Si and S2 were placed in delayed coincidence, so that they would give a signal only if S i and S 2 delivered pulses separated by a time interval of 5.1 X 10 s s. Hence the observation of such delayed coincidences could be considered an indication of the passage of an antiproton. However, the meson background was so heavy that addi tional precautions were required to eliminate accidental coincidences due to the passage of two different pions through S i and S2 with a delay of 5.1 X IO8 s. Cerenkov detector Ci was made sensitive only to particles with velocity greater than 0.80c and C2 was made sensitive only to particles with velocity between 0.75c and 0.78c. The antiprotons, by the time they arrived at the Cerenkov detectors, had been slowed down to a velocity of about
386
Fundamental particles
(9.4
0.76c. Thus Ci was insensitive to antiprotons Iiut sensitive to mesons, and the reverse was true for C2. Therefore, to prevent spurious counts, the detector Ci was connected in anticoincidence and Cs in coincidence with the signals from S i and SV That is, the experimenters knew that an antiproton had passed through the system only' when (a) S i and S 2 registered pulses delayed 5.1 X IO8 s, (b) Ci did not register a pulse, and (c) C2 registered a pulse. As an additional check, the magnetic field of M 2 was varied, and pulses were received only when the value of the field corresponded to the preselected time of flight for antiprotons. The experiment— performed by Chamberlain, Segr6, Wiegand, and Ypsilantis in 1955— was highly successful, and confirmed the existence of the antiproton.
ft .l 1‘ iirtivlv InHtubitity T h e creation and annihilation processes discussed in the preceding section, as well as the process o f /9decay discussed in Section 8.4, are m anifestations o f a more general p rop erty o f the fundam ental particles: th eir instability.
In other words,
given the proper conditions, the fundam ental particles can transform into other particles as a result o f their interaction.
W e shall soon c la rify w hat w e mean when w e say
the proper conditions. P articles which are unstable undergo spontaneous decay, w ith a w elldefined halflife. T a b le 9 2 shows the decay modes and halflives o f unstable particles.
In cases in which m ore than one decay m ode is possible, the
relative p rob a b ility o f each m ode is also given.
In som e decays antincutrinos are
indicated; this is done in accordance w ith a conservation law (th a t o f lepton s) which will be discussed in Section 9.5. N o te th a t on ly four particles (an d th eir antiparticles) arc stable against spontaneous d ecay: the photon, the neutrino, the electron, and the proton. O f all the unstable particles, the one havin g the longest halflife is the neutron.
T h is m ay explain w h y m atter is com posed o f electrons,
protons, and neutrons.
N o te th at mesons have a halflife o f the order o f IO  8 s
(w ith the exception o f the π ° and η °), while the baryon halflives are o f the order o f IO 1 0 S. One condition necessary for spontaneous d ecay— a con dition im posed by en ergy conservation— is th at the rest mass o f the parent part i cle be larger than the sum o f the rest masses o f the daughter particles.
In ad dition to energy, m om entum
as well as angular m om entum are supposedly conserved.
W e use this assumption
as a guiding principle when we an alyze the decays o f fundam ental particles. T h e n egative muon was the first unstable particle observed (see F ig. 9 1 ).
It
was discovered in cosmic rays in 1937. T h e decay o f a m uon alw ays results in the appearance o f an electron. T h ese electrons have a continuous spectrum o f energy, sim ilar to th at found in /3decay (F ig . 8 1 1 ), w ith a m axim um kinetic energy o f about. 53 M e V .
F rom these tw o facts w e deduce th at muon decay cannot be a
tw ob od y process, and th at at least tw o neutral particles must be produced in addition to the electron. Since the m axim um energy o f the electron is much larger than its rest energy (0.5 M e Y '), the m axim um m om entum o f the electron is Pmux
! —[κιx/r =
53 M eV /c.
T h is m axim um m om entum occurs when the tw o other particles are both em itted in the opposite direction to that o f the electron, and both particles must carry a
9.4)
Particle instability
T A B L E 92
P article Decay M odes*
Pa rticle
L epton s
387
It e la t iv e
D e c a y m ode
P lio ton
S ta b le
N e u trin o
Sta b le
p ro b a b ility , %
Sta b le
E lectron M uon
p  —r e ” +
Pion
TT+ — μ + +
e
100
—* C+ +
V
~10~*
r +
V
1.52 X 1 0  "
99
It0 — 7 + 7 τ 7 + c+ + Kaon
I
C
Κ + _ μ+ 4 ν
5.6
 τ ττ° ε+ +
y
4.8
+ +
V
3.4
—T X 0 T
JT+
+ 2 JT0
M+
TT+
— {—TT
 f— TT0
T
T r+
+
20 20
—τ χ + + χ ~ + χ °
22
Τ Χ + + Χ  + 7
η 0 —τ ρ + +
Lam bda
Λ ° —τ ρ + + χ Uu
+
to
Σ + — P+ +  T Il0 +
5
2° 
O m ega
5
e +
X0
7.0 X IO 2
V 66
1.76 X 1 0  * °
34
X 0
53
X +
47
5.6 X Ι Ο ' 11
< 7 X Ι Ο " 15
Λ° + 7
2  — η0 + Xi
C l O " 10
S table
N e u tro n
S igm a
6.0 X Ι Ο " 11
33
TT0 + 7 + 7
—τ 3ττ°
P roto n
4 X IO "8
15
η°  τ 7 + 7 T
14 6.3
35
*
T 2 *° E ta
18
11.3
τ3 χ ° K 0
8.56 X Ι Ο " 0
1.7
K 0 —τ Tri f C+ f* e —T IT+ f ρ —τ
6 X 10“ 17
21
—►2ir+ + x “
CO
1.80 X I O  8
63
— T TT+  I  TTil
O /. O S
H a lflife , s
1.1 X I O " 10
χ
E0 * Λ ° + χ°
2.0 X I O  10
Η τ Λ0 + χ 
1.2 X Ι Ο " 10
S2 — Λ ° +
K 
> E 0 + χ 
50
1 0 1 0
50
* T o obtain the decay of antiparticles, change all particles into antiparticles on both sides of the equations.
388
Fundamental particles
(9/4
F ig. 97. Decay of a π  meson followed by the decay of the muon. (Photograph courtesy Brookhaven National Laboratory)
to ta l m om entum o f 5 3 M e V / c .
I f the particles are neutrinos, th eir total energy
is then 53 M e V . Thus the total energy released in muon decay is about IOG M e V 1 which is practically the rest energy o f the muon (T a b le 9 1 ). T h e spin o f the muon has been measured independently and found to be 0; thus the law o f conser vation o f angular m om entum is another factor th at would forbid the decay o f a muon into an electron and a neutrino, each havin g spin £. H o w ever, this law does allow for a muon decaying into an electron and tw o neutrinos. T h erefo re we m ay sa fely assume th at the decay of a n egative muon can be represented b y the decay m ode given in T a b le 92. T h e next partiqle to be observed was the pion. I t was first discovered in 194G in special photographic emulsions (see Ap p en d ix V I I ) exposed to cosmic rays, and shortly afterw ard artificially produced (b y protons accelerated in the synchrotron at B erk eley).
T h e existence o f this particle had been predicted in 1935 b y the
Japanese physicist H id eki Y u k a w a to explain the short range o f nuclear forces (see Section 9.8).
T h e muons resulting from pion decay consistently h a ve a fixed
kinetic energy o f about 4.1 M e V , which indicates th at pion decay is a tw ob od y problem .
T h erefo re w e m ay assume that in pion decay, in addition to the muon,
g/ι)
Particle instability
a neutrino is e m itted in the opposite direction.
389
Since the m om entum o f a muon
w ith the ab o v e kinetic energy is 29.5 M eV / c, this m ust also be the mom entum o f the neutrino.
T h e neutrino energy is then 29.5 M e V .
Thus the total energy re
leased in pion decay is 105.7 M e V + 4.1 M e V + 29.5 M e V = 139.3 M e V , which again app roxim ately agrees w ith the rest energy o f the pion, 140 M e V . W e may thus assume th at the decay scheme o f the pion is as indicated in T a b le 92. From the decay π + —» μ + + v, w e conclude th at the spin o f the pion is either 0 or I. From an analysis o f other processes in v o lvin g pions, w e conclude that the spin is 0:
F igu re 97 shows the decay (in a bubble cham ber) o f a pion into a muon,
follow ed by the subsequent decay o f the muon into an electron. T h e accom pany ing neutrinos do not leave an y track.
O bservation o f these double events is very
common. Sim ilarly, Fig. 98 shows the d eca y o f a kaon into three pions in a photographic emulsion. An alysis o f the m om enta o f the three pions shows th at no further neu tral particles are em itted in the process; this is consistent with the laws o f conser vation o f m om entum and energy. On the other hand, F ig. 9 9 shows the decay o f
1 5M cV «meson V
17M eV «m eson leaves em ulsion strip a t Λ
Km eson
I
43M c V xm eson
F ig. 98. Kmeson decay into three pions (the τ mode). Brookhaven National Laboratory)
(Fhotograi+ courtesy of
390
Fundamental particles
f
(I9.4
A
U
Ir ' '
Ί! i I: U
*
I1 i
~
i I! ■
’’ r *
II '«I tl\ I
H if
i' I I
{ r
i iti [■ r
i \
I
I .
\
Π; I r i , *

i f ’I
I
HM H
I r '
H:
;· I ! ' Si B iI Ii I Ϊ i .,; : I .il·

I
!; I I IX I'
!I
■
■
F ig. 99. Kmeson decay into two pions (the 0modc). (Photograph courtesy of Brookhaven National Laboratory)
a kaon into a charged and a neutral pion.
F igu re 9 10 shows a m ore com plex
process, in which there is an initial prolon an tip roton annihilation, g ivin g rise to a series o f particles which undergo subsequent decays.
T h e whole process can be
expressed by
Each o f the processes has been carefully checked b y means o f the laws o f conserva tion o f energy and m omentum. Such a check enables us to id en tify the neutral particles which do not leave an y track. T h e instability o f a given particle is also d isplayed b y the fact that, in a highenergy collision between tw o particles, several new particles m ay be produced.
Particle instability
9A)
391
F ig. 910. Events triggered by a protonantiproton annihilation. (Photograph courtesy of Brookliaven National Laboratory)
F o r exam ple, in a protonproton collision, the follo w in g processes m ay occur: .P + + P + (9.6) N 1
p T  f n + Tr
N o te th a t in each case a pion is produced. T h e protonproton collision is the proc ess b y which pions are produced profusely in th e laboratory.
I f one o f the pro
tons is a t rest in the lab oratory, the threshold o f energy o f the other proton is
392
Fundamental particles
F ig. 911. Events triggered by a K ~p collision. National Laboratory)
about 300 M e V (see E xam ple 9.3).
(Photograph courtesy of Brookhaven
A t higher energies, other particles, such as a
p roton antiproton pair, m ay result. Figu re 911 illustrates a m ore com plex process b y which a K  and a p + collide, resulting in several particles which suffer subsequent decay. process can be described by K  + pH
T h e w hole observed
Η " + Iv 0 + 7T+
+ Λ
(9.7)
f
0
L>
7Γ
+
p +
Thus a t high energy the collisions between particles are not elastic. T h e particles
Particle instability
393
Fig. 912. Events triggered by a cosmieray proton colliding with a proton in the atmosphere.
can no longer be considered as billiard balls, w hich is a useful approxim ation at low energies. M a n y o f the processes in v o lvin g fundam ental particles have been observed in cosmic rays', th a t is, the flux o f particles (m ostly protons) falling on the earth as a result o f nuclear processes occurring in the sun and other parts o f the universe. C osm icray particles trigger a chain o f reactions when th ey in teract w ith nuclei in the upper atm osphere. One such reaction is shown in Fig. 912. Such a process continues until the energy is no longer great enough to produce new particles, and on ly stable particles rem ain. E X A M P L E 9.3. Calculation of the threshold energy for a process in which a projectile collides with a target at rest in the laboratory, resulting in several new particles.
394
Fundamental particles
(9.4
S o lu tio n : Designating the projectile and the target by P i and P 2 and the resulting particles by P,, we may · iitc P i + P 2 > Σ , P.lf the momentum of P i is p, in the Lframc, its total energy is c V wife2 j p~. Since Pa is at rest in the Λframe, its total energy is m2c2. Thus the total energy in the /.frame is E = cV ^ntic2+ p2 + m2C2 and the total momentum is p. In the Cframe of reference the momentum of the two particles is zero; that is, p ' = 0 (see Appendix Ij. W e designate the total energy of the system in the CFrame by E '. Because of the energymomentum relation, the quantity E 2 — C2P2 is invariant under a Lorentz transformation relating two inertial frames of reference. Therefore, keeping in mind that />' = 0, we have E2 
C2 P 2
= E '2.
The laws of conservation of energy and momentum require that, after the process has taken place, the total energy of the products in the Cframe still be E ' and the total mo mentum be zero. Obviously the minimum energy required for the process corresponds to the situation in which all the resulting particles are at rest in the Cframe, so that the total energy in such a frame is E ' = Σ ·* 71»®2· Substituting values in the preceding equa tion, we get 2 I
true But if
Et
,
2 4
,
„
Sa /
J 2moc
+ WI2C
V
2 2 .
mic + p
2
V.
.2 4
= (z.,w i,)
c.
is the kinetic energy of the projectile in the /.frame, then
V mic2 2 T1 P2 
„
,
2
Ek + « lie ■
M aking this substitution in the preceding equation and canceling the common c2 factor, we get
(mi + tno)2c2 + 2m2Ek = ( Σ .Ηί.)2ε2. Thus E t = — QM/2m2,
(9.8)
where Q = (»u J m2 — Σ>η + and M = mi + » 12+ Σ ’" · · Equation (9.8)gives the threshold kinetic energy which the projectile must have for a given process to occur. If Q is positive, 110 threshold kinetic energy exists and the rest energy of the initial particles is enough to produce the final particles. Only if Q is negative does a threshold energy exist. W e shall illustrate the use of Eq. (9.8) by applying it to two special cases. a) Threshold for pion production in protonproton collision. Considering a neutral pion, wc have that p + J p + —» p +  f p + + ir0. Therefore Q = — m,c2,
.1/ = 4 Wil, + m,,
and
m 2 = mp,
resulting in _
m,r2(4mi d l i O 2 Wp
= /
m A
\
2 Wp/
For a charged pion, a similar result is obtained.
^ 29()
Particle instability
9A)
395
b) Threshold fo r antiproton production in protonproton collision. The process is now p + j p + —* p + + p + + p + f P~ AU particles have the same mass mp. Then in 2 = nip,
Q = — 2nipc2,
and
.1/ = Gm1,,
giving, for the threshold energy, E t = (2mpC2) (Gmp) /2m„ = 6mpc2 = 5.G4 X IO3 M eV = 5.64 GeV. E X A M P L E 9.4.
Determination of the magnetic moment of the A 0Iiyperon.
S o lu tio n : W e shall now describe an interesting experiment which illustrates the technicpies used to determine the properties of fundamental particles. W e refer to the mea surement of the magnetic moment of the A°hyperon. The method, which is similar to that used to determine the magnetic moment of any hyperon, involves three steps: (a) producing a polarized beam of A 0, (b) causing the polarized beam to pass through a strong magnetic field which produces a change in the direction of the magnetic moment, (c) measuring the angle of rotation of the magnetic moment. P o ly eth y len e ta rget
Emulsion stack
A v e r a g e direction o f S a t produ ction in A®cm fram e
Average direction of S at decay if aλ < G (also average direction of emission of decay pion in the A0CM frame) F ig. 913. Schematic diagram of the A°magnetic moment experiment.
There have been several experiments designed to analyze the decay of polarized A ° ’s in a magnetic field. W e shall describe the one performed at C E R N (European Organiza tion for Nuclear Research) in 1964. A beam of negative pions from the C E R N proton synchrotron, having a momentum of 1.05 GeV/c, falls on a polyethylene target (Fig. 913), where the reaction ir~ E p + —> A 0 + K 0 takes place. The momentum is chosen to cor respond to the energy at which the cross section for the reaction is maximum. Since the energy of the ir~ particle is close to the threshold energy, 0.78 GcV' (see Eq. 9.8), the A 0 and K 0 are produced, relative to the laboratory, mainly in the forward direction because they are practically at rest in the Cframc of the process. T he A 0 are strongly polarized with their spin S normal to the plane of production (that is, the plane determined by the directions of motion of the incident tt~ and the resultant A 0). The A 0 produced at angles between 13° and 23° with respect to the direc tion of incidence are made to pass through a strong magnetic field of 15 T , perpendicular to both pa and S.
396
Fundamental particles
According to Eq. (7.5), the magnetic moment of the A 0 should be written as Λ/λ = g*(e/2mA)S, where g.\ is the gyromagnetic ratio of the A 0. It is customary, however, to write the mag netic moment in the form Mh
g h (m p/ m h )(e / 2 m „ )S = p h (e / m p) S ,
=
and the quantity sought experimentally is μ,\ = bg\(jnp/mλ), which gives the A 0 mag netic moment in nuclear magnetons. In the presence of a magnetic field ffi there is a torque r = .VfA X ffi, so that the equation of motion of the spin is dS/dt = T or dS — = p.\(e/mp)S X f fi = —μΑ(β/ηιρ)(Β X S, and S precesses around ffi with an angular velocity* Ω
=
—
p.\(e/m p) e ~  f p + , A 0 —* p _ + i r + , or π + + P +  * 2 + ( i r + , which com ply w ith the ab ove conservation law s— do not occur in nature? W h y do the particles have such an app aren tly random mass spectrum ? T h e situation is sim ilar to th at o f an alchemist in the M id d le A g es tryin g to understand chem ical reactions w ithout know ing atom ic or m olecular structure. Physicists to d a y are a c tiv e ly struggling to find some order in this apparent chaos. F ortu n a tely, a certain degree o f order has been brought abou t b y the discovery o f new conservation laws which resemble the law o f conservation o f charge rather than the first th ree conservation laws. These laws are: (5 ) C onservation o f leptons (6 ) C onservation o f baryons (7 ) C onservation o f isotopic spin (8 ) C onservation o f strangeness W e shall now briefly discuss each o f these laws in turn. E xp erim en tal data lead to the conclusion th a t mesons and photons (and in gen eral all particles o f spin 0 or I ) can be produced or annihilated in an y number (this is one o f the reasons for calling them bosons). On the other hand, leptons and baryons are restricted w ith regard to the num ber which m ay be produced or annihilated in a single process. F o r exam ple, the process o f electron pair production Ύ —* e ~ f e + , in which tw o leptons— an electron and its antiparticle, a positron—
398
(9.5
Fundamental particles
are created ou t o f a photon is possible. B u t the process o f electronproton produc tion 7 —> e ~ + p + , which could com ply with th e first four conservation laws, does not occur. I n this process (which is not observed ) on ly one lepton and one baryon are produced.
I t is experim ental facts o f this kind w hich form the basis for the
fifth and sixth conservation laws. C o n s e r v a tio n o f le p to n s . fined such th at £ =
L e t us introduce a lepton quantum number £ de
+ 1 fo r lepton particles and £ =
— I for lepton antiparticles.
N on Iep ton particles have zero lepton number. T h en in any process, the total lepton number must remain unchanged. Considering som e examples, we have that, in the process 7 —» e — + e + , the lep ton number on the left is £ = O and on the righ t £ = the nonobserved process 7 —> e — + p + , w e have £ =
+1 — 1 =
0. H ow ever, in
O on the left and £ =
+ 1 on
the right, so th a t the conservation o f IeiDtons is violated . T h e law o f conservation o f leptons requires th at E q. (8.14) be w ritten in the form n —> p + + e — + v £ =
O
O
+ 1  1
to assure a to ta l lepton number o f zero on both sides.
(9.10) T h u s conservation o f lep
tons requires that, in jS~decay, an antineutrino v and not a neutrino be em itted. W e must also w rite E q. (8.17) in the form p + + v —» n + e + , with a total lepton number £ = — I on both sides. T h is is w hy, in the C ow anR cines experim ent (E xam p le 8.6), it was an antineutrino which was detected. N o te th at Eqs. (8.1.5) anil (8.16) fo r β + deca y and electron capture require a neutrino to ob ey lepton conservation.
B y exam ining the decay schemes o f T a b le 92, the student should
v e rify the law o f conservation o f leptons in each case and ju stify the fact th at an antineutrino is indicated in som e cases. L e t us m ention in passing the fact th at recent (1962) experim ental eviden ce seems to suggest th at the neutrinos resulting from pion d ecay are not identical to those produced in muon decay or in /8decay. T h e experim ent perform ed by a group from C olu m b ia U n ive rs ity w ork in g w ith B rook h aven L a b o ra to ry ’s A lte rn a tin g G radien t A ccelerato r is shown schem atically in F ig. 914.
A h ea vily shielded
spark cham ber was exposed to th e anti neutrinos resulting from pion decays accord ing to π + —» μ + + V. T h e pions had been produced b y highenergy protons striking a convenient target. T h e antineutrinos colliding w ith the nuclei in the spark cham ber, if they w ere identical to those produced in + d e c a y , could produce cither one of the reactions p + + P —* n + e + or p + + v —* η + μ + . H ow ever, if there were tw o kinds o f neutrinos, on ly the second reaction should be allowed. A fte r a run o f several days, in which an estim ated IO 14 antineutrinos passed through the spark cham ber, abou t .50 processes were observed in which muons were produced; but no positrons were observed. T h is suggests th a t the first reaction is forbidden, and th at the neutrinos produced in pion decay are not the same as those produced in + d e c a y .
The conservation laws
9.5)
399
Steel shielding
T h erefore it is now usual to write π
μ~ η ■
yμ + ν μ, e
f ν μ +
P+ + e
P + + Pe
» μ 4 + Ρμ, ■ e + + Ρμ + P»,
Pei
η + e+ +
+ P,ej
Il
P + + Pil
η
Vc,
μπ
where νμ designates the neutrino associated w ith the muon in pion decay and vc the neutrino associated w ith the electron in muon or nucleon decay. C o n s e r v a tio n o j b a r y o n s . b =
D efin in g a baryon quantum number b such th at
+ 1 for all baryon particles, b =
— I for all baryon antiparticles, and zero for
all nonbaryon particles, we m ay say th at in any process, the total baryon number must remain unchanged. T h e student m a y v e rify th at this la w is satisfied in all processes previou sly m en tioned in v o lvin g baryons, such as Eqs. (9.2), (9.4), (9.5), (9.6), and (9.7), as well as in the decay schemes o f T a b le 92. Since the proton is the lightest o f all b ary ons, its decay into lighter particles (which cannot be baryons) would v io la te the law o f conservation o f baryons, and this explains w h y the proton is a stable par ticle.
T h u s the w orld in which w e liv e is, to a certain extent, the result o f the
laws o f conservation o f leptons and baryons. C o n s e r v a tio n o f is o t o p ic s p in .
In F ig. 92, the particles and antiparticles
were grouped according to th eir masses or rest energies.
W h en w e consider on ly
those particles affected by strong interactions (m esons and baryons), we note th at they occur as m u ltip lets; th at is, singlets (j/°, Ω , Λ 0), doublets ( K + and K 0, p + and η 0, Ξ 0 and H ~ ), and triplets ( 2 + , 2 °, and S  , 7r+ , 7t°, and i r ~ )* . T liis suggests that we m ay q u a lify each m u ltip let by a new qu an tity, called isotopic spin τ , such that 2 r  f I gives the number o f particles in each m ultiplet. Thus
T
=
0, £, I, . . .
* The π0, being identical to its antiparticle, has been placed in the center of the diagram and the set π +, τ ° , ir_ must be considered in a special way.
UOO
Fundamental particles T A B L E 93
C9.5
Isotopic
S pin
(r , T z) ,
S trangeness
( i),
an d
IIy pereh arge (y ) o f .Mesons an d B aryo n s*
Particles
T
Tz
I 1 2 O
(1 ,0 ,1 )
0
( £.  £ )
I
I
0
C
0
a
V
Mesons
jr+ , τ 0, π ~ K +, K 0 V0
0
Baryons P +i A0 2 +
11 Σ~~
E0 E 0
0
I
— I
0
£
(£ . — £ )
0
0
I
( I , 0,  I )
 I
£ 0
(£, — £) 0
2 3
0 —I —2
* The strangeness and hypercharge of antiparticles have signs opposite to those of the corresponding particles. Mesons and baryons are frequently designated by the common name of hadrons. fo r .singlets, doublets, triplets, and so on. W e use the w ord “ isoto p ic” because τ refers to particles which have practically the same mass and spin and thus occupy the same place on a mass scale.* W e m ay note th at τ bears some resem blance to an angular m om entum J , which m ay h a ve 27 + 1 orientations in space, each characterized by a value o f its Zcom ponent J z, and this is w hy the name "s p in ” was given to r .
T h u s w e m ay consider the isotopic spin as a ve cto r in a certain
representative space called isotopic spin space. T h e ve cto r has a length γ + ( τ +
I)
and it has 2 t + I possible orientations rela tive to the Zaxis, corresponding to the possible values o f its Zcom ponent given b y τ . = ± r , ± ( t — I ) , ± ( r — 2 ), . . . . Each particle in a m ultiplct corresponds to a value o f t z, w ith values assigned in order o f decreasing charge. Tz =
— \
for neutrons.
F o r exam ple, we h ave
Tz
=
+ £ fo r protons and
P article and an tiparticle m ultiplets have the same iso
topic spin r , but opposite values o f
t z.
T h e isotopic spin o f mesons and baryons
is given in T a b le 93. T h e total isotopic spin T o f a system o f particles is obtained by adding, in v ec to r form, the isotopic spins o f each particle, using the same addition rules as for angular m om entum (Section 3.8). T h e tw o particles have τ = spin T = I or 0. B u t T z = sarily corresponds to T = I.
Consider, for exam ple, the system p + + p + .
£ each, which m ay add to g iv e a resultant isotopic £ + £ = I. T h u s a system o f tw o protons neces B ut the system p + + n has T , = £ — £ = O and
thus the total isotopic spin m ay be T = total isotopic spin o f § or £. B u t T z =
I or 0. Sim ilarly π + + p + m ay h ave a I + £ = §, so th at T = necessarily.
On the other hand, π ° + p + and π  + p+ , havin g T z = m ay correspond to either T =
£ and — £, resp ectively,
\ or £.
* The names isobaric spin and isospin are also used to describe this property.
9.5)
The conservation laws
501
I t seems th at the strong or nuclear interaction is charge independent.
For
example, there is am ple experim ental e v i dence th at the nuclear interaction for the pairs p + p + ( T 2 = and n n ( T 2 =
+ 1 ), n p + ( T 2 =
— I ) is the same.
0),
T h ere
fore the strong interaction is independent of the total value o f T 2 for the interacting particles.
Thus
the strong
interaction
between two particles depends only on their total isotopic spin T. >renergy, M e V
F or example, in the scattering process 7T+ ) P + —» 7T+ +
we have T 2 = ing to T =
F ig. 915. Cross section for (a) ir+ + p + and (b) π  + p + collisions as a function of energy.
p+,
§, necessarily correspond
§ on both sides o f the equation, and the observed cross section
varies w ith the energy o f the 7r+ projectile, as shown in Fig. 915, curve (a ), w ith a pronounced maxim um at an energy close to 300 M e V .
H ow ever, in the
case o f
7r
we have T 2 =
+
tr + P +
(scattering),
7Γ° +
(reaction ),
p+
n
— J, which m ay correspond to either T =
served cross section is shown in curve (b ) o f F ig. 915.
§ or
T h e total ob
T h e peak corresponding
to T = § falls a t pra ctica lly the same en ergy as in the previous case o f ir + f p + for the same total isotopic spin, but at about 1000 M e V a second peak is observed which is attribu ted to the T =
£ state.
A s a result o f the analysis o f processes in which strongly interacting particles participate, we m ay state that the total isotopic spin T is conserved in strong interactions. T h e conservation o f total isotopic spin is a rigorous law on ly for strong inter actions but it can be violated in electrom agnetic or w eak interaction processes. F o r exam ple, in the observed process K 0 —> π + + and either T =
π ~ , w e have T = ^
on the le ft
I or 0 on the right, so th at T is not conserved.
In ad dition to the conservation law o f total isotopic spin, we have that both the strong and electromagnetic interactions conserve the total T 2 component o/ the isotopic spin. T h is is a rigorous law, follow ed in all cases, and is fu lly equ ivalent to the law o f conservation o f charge.
h0'2
{9.5
Fundamental particles
C o n s e r v a tio n o f s tra n g e n e s s .
A n oth er a ttrib u te invented to characterize
particles subject to strong interactions is the strangeness charge o f a p article in a m ultiplet is g ive n by q =
T h e values o f
j.
β (τ , +
λ
, defined so th a t the
J t F
(9 .1 1 )
are given in T a b le 93.
P articles and antiparticles have opposite
strangeness. T h e law o f conservation o f strangeness requires that the total strangeness S must remain the same in processes due to strong or electromagnetic interactions. F o r exam ple, the process v ~ F p + —* n 0 + A 01 which com plies w ith all the other conservation laws, is not observed because the to ta l strangeness on the le ft is zero, w h ile on the righ t it is — I. H ow ever, processes such as π * + P + —*  + + K + and 7r“ F p + —> A0 F K 0 do occur. In both cases the to ta l strangeness remains equal to zero. T h a t is, in all pionnucleon collisions (a system havin g zero strangeness), hyperons must be produced in pairs w ith equal but opposite strangeness, a phe nomenon called associated production.
A c tu a lly it was the consistent observation
o f the associated production o f certain hyperons which provided the clue to the law o f conservation o f strangeness. T h e conservation o f strangeness is not a rigorous law ; it can be v io la ted by weak interactions which allow a change o f strangeness o f ± 1 .
In m ost baryon decays
(T a b le 9 2 ), the conservation o f strangeness is violated , indicating th at the decays take place by means o f weak interactions.
Η
*
A0 F U
ir ~
7T  I  P +
(AS =
F o r exam ple,
+ 1),
(A S = I ).
H yperon s cannot decay into baryons through the strong interaction w ith conser v atio n o f strangeness, because the Q o f such processes is negative.
T h is explains
w h y the hyperons decay so slow ly (ab ou t IO10 s) com pared w ith the tim e in v o lv e d in th eir production or annihilation in collisions b y means o f strong inter actions, which is about IO 23 s.
From these tw o tim es w e m ay conclude th at the
strength o f the weak interaction is about IO13 o f the strength o f the strong interaction. W e might m ention at this point that another param eter which m ay be used to classify particles instead o f strangeness is the hypercharge, defined as y = Values o f the hypercharge arc given in T a b le 93.
6 F A.
T h e law o f conservation o f
strangeness can be replaced b y a law o f conservation o f hypercharge, due to the law o f conservation o f baryons and the definition o f y. T h e new conservation laws which we h ave introduced in this section m ay seem m ore or less ad hoc conventions to explain the experim ental facts. H o w ever, the fact th at th ey are consistently applicable to a large number o f situations suggests th at they h ave a deeper physical m eaning.
A t present th eir connection w ith the
physical properties o f particles is not w ell known in all cases, and is a subject still open to further research.
9.6)
9.0
Invariance, symmetry, and conservation laws
h03
In v a r ia n c e , Sym m etry, a n d Contterraiion Lairtt
Some o f the conservation laws we h a ve just discussed can be traced Ixick to certain invariance and sym m etry properties o f physical system s which we shall now briefly consider. S p tic e t r a n s la t io n .
L e t us assume that space is homogeneous and uniform ;
i.e., it has translational sym m etry a t all points. W e must conclude then th at a ph3'sical system w ill behave the same no m atter w here it is located in an other wise e m p ty space.
Using a more m athem atical term inology, the description o f
the properties o f an isolated physical system is invariant w ith respect to a trans lation o f the system relative to a fram e o f reference.
F o r exam ple, an isolated
molecule, com posed o f several electrons and nuclei, must be described in exactly identical term s no m atter where it is located rela tive to an observer. On the other hand, we know that the total m om entum o f an isolated system is constant. Thus it can be shown that the conservation o f momentum o f an isolated system is a result o f trans lational invariance o f the laws describing the system. In some cases the system m ay not be isolated, but the physical environm ent m ay exh ibit certain translational sym m etry. C onsider an electron placed between tw o infinite parallel planes carryin g equal opposite charges. O bviously the physical conditions do not change if the electron is displaced parallel to the planes.
We
know then th at when the electron is set in m otion, its mom entum parallel to the planes is constant.
(T h is is usually stated b y saying th at the electric field pro
duced b y the charged planes is perpendicular to the planes.)
T h u s we again find
that the conservation o f momentum in a given direction is a consequence o f transla tional invariance o f the physical conditions in that direction. S p a c e r o t a t io n .
I f we assume th at space is isotropic (i.e., that it has rota
tional sym m etry at any p oin t), then the description o f the properties o f an iso lated system m ust be independent o f its orientation in space relative to a given fram e o f reference.
L e t us again consider an isolated m olecule: the description o f
its properties is independent o f the orientation o f the m olecule relative to the ob server. On the other hand, we know th at the total angular m om entum o f an iso lated system is constant.
I t can thus be shown that
the conservation o f angular momentum o f an isolated system is a result o f rotational invariance o f the laws describing the system. In som e cases, even though a system m ay n ot be isolated, the physical system m ay exhibit a certain rotational sym m etry.
F o r exam ple, a central field o f force
has spherical sym m etry. W e then know that, as a result o f the sym m etry, the angular m om entum (re la tiv e to the center o f force) o f a particle m ovin g under central forces is constant.
I f the field has cylin drical sym m etry, as in the case o f
the force on an electron in a d iatom ic m olecule or a charged particle in a uniform m agnetic field, then the com ponent o f th e angular m om entum rela tive to the sym
UOU
(1 9.6
Fundamental particles
m etry axis remains constant.
T h u s we again find th at the conservation o f angular
m omentum about a given direction is a consequence o f rotational symmetry o f the physical conditions relative to that direction. B oth translational and rotational sym m etry impose certain lim itations on the m athem atical form o f physical law s; how ever, this is a subject that we cannot explore any further here. T i m e t r a n s la t io n .
In variance w ith respect to tim e translation means that
if we prepare a physical system and leave it to e v o lv e w ith out external interfer ence, the evolution o f the system w ill be the same irrespective o f the tim e a t which it was prepared, or in other words, o f the chosen origin o f time. Using a som ewhat m ore elaborate logic, we can show th at the conservation o f energy o f an isolated system is a consequence o f the invariance o f the Iaxvs describing the system relative to the chosen o rig in o f tim e; th at is, rela tive to a translation o f tim e.
T h e laws o f both classical and quantum
mechanics are in varian t relative to tim e translation. G a u g e t r a n s f o r m a t io n .
Our fourth conservation law, th a t o f charge, is a
little m ore difficult to associate, in an elem entary w ay, w ith a sym m etry trans form ation and an invarian t property. W e can show that charge conservation is a result o f the invariance o f the laics o f the elec tromagnetic field (i.e., M a xw ell’s equations) with respect to a “gauge” transformation. W e cannot fu lly explain in this te x t w hat is meant b y a gauge tran sform ation *; a sim ple case o f a gauge transform ation is a change in the zero o f the electric (o r scalar) potential, a change which does not affect the electric field or the form o f M a x w e ll’s equations.
A n oth er ty p e o f gauge transform ation is a change o f phase
o f the w ave function; i.e., the replacem ent o f ψ by e'fy . T h is obviou sly leaves the p rob ab ility density ψ2, which is the observable quantity, unchanged. I s o t o p ic s p in r o t a t io n .
A r e there other sym m etry operations th at can be
related to conservation laws? W e have indicated th at the isotopic spin is conserved because o f the charge independence o f the strong interaction.
W e can put this
in the same form as the fou r preceding conservation laws b y saying th at the isotopic spin is consen'ed because strong interactions are invariant with respect to rotations in isotopic spin space. * See, for example, Panofsky and Phillips, Classical Electricity and Magnetism, second edition, Reading, Mass.: AddisonWesley, 1962, Section 141.
Invariance, symmetry, and conservation laws
9.6)
405
T h is invariance means that the interaction must contain the isotopic spins T 1 and T2
o f the interactin g particles on ly in forms such as
invariant, but n ot as T11Tz2, which is not.
T 1
· r 2l which is rotation a lly
U n fortu nately, we cannot elaborate
more on the m athem atical m eaning o f the ab ove statem ent, but its sim ilarity to the conservation o f angular m om entum under central forces is obvious. C h a rg e r e f le c t io n .
Eleclrom agnelic and strong interactions rem ain invariant
i f all charges are replaced by those with opposite sign (th a t is, q —> — 9), an operation which is called charge reflection. One can show th at this leads to the conservation o f the Z com ponent T z o f the total isotopic spin in those interactions. N o sim ple sym m etry p rop erty or invarian t b ehavior has y e t been associated w ith the conservation o f leptons and baryons.
T h e re are, however, three m ore
sym m etry operations o f great im portance in the fundam ental behavior o f m atter. T h e y are: p arity P , charge conjugation C, and tim e reversal T . M irro r
M irro r
(a)
(l>)
F ig. 916. Reflection in a plane of (a) momentum, (b) angular momentum.
P a r it y .
T h is concept, which w e considered before in Section 2.9, refers to
the operation o f space reflection, either in a plane or through a point, such as the origin o f coordinates.
L e t us first see the b ehavior o f some dynam ical quantities
regarding reflection in a plane. Consider a particle A (F ig . 9 1 6 (a )) m ovin g w ith m om entum p . T h e m irror im age o f A is another particle A ' m ovin g w ith mo mentum p ' such that p j = pjj, p x = — pi., where Il and _L refer to directions parallel and perpendicular to the plane, respectively. N e x t consider a particle m ovin g as shown in Fig. 9 1 6 (b ), w ith angular m om entum L .
T h e m irror im age
o f A is another particle A ', rev o lvin g as shown in the figure and thus h avin g an angular m om entum L ' such th at L\\ — — L [, L±. = reflection in a plane,
L'±.
W e see then th a t in
p and L behave in d ifferent ways. F or th at reason
p is called
a polar v e c to r and L an a xia l vector. A ll vectors th at play a part in physical laws are either polar or axial.
T h e student m ay ve rify , b y exam ining Fig. 917, that
the electric field is polar and th at the m agnetic field is axial. vector.
Force is also a polar
406
Fundamental particles
(;9.6
M irro r
M irro r
\
* Ί
\
✓
*
O riginal system
M ir ro r im age
,
I SI 5 (?)
1815 ( 4 + )
1 7 6 7 (j)
I6TOJ Q
1660()
1674 ( j j H
1670(§ ~ ) 1 5 7 0 (1 " ) 1 5 2 0 (4 " )
1 5 2 0 (1 ")
1100 (4^)
1 4 0 5 (4 " )
1 5 3 0 (4 )
. 1500 1385(4  ) 131 8 (4 1230 ( j +)
1 1 9 3 (4  ) 1115(4  )
1000 9 3 9 (4  )
800 N
Δ
!/ = + 1
V= + 1
T= 4
T“ 4
Λ V= O r= 0
7= 4 1, 0 7= 4 2,
±1,
7= 0
S v=
T=I
r= 4
7=
± 1, 0
il
Ξ
y= 0
7=
—l — 1, 0
I / = —2 T
= O
7=
—I
0 F ig. 925. Baryon resonances. parentheses.
ft. 7
The spin and parity of each resonance is given in
K eson a n ees
T h e fundam ental particles listed in T a b le Q I can all be considered lon glived b y particle standards. In addition to these lon glived particles, in the last fe w years experim ental evidence has accum ulated which points tow ard the existence o f v e r y shortlived particles, called resonances. T h e ir lifetim es are so short (o f the order o f IO2 0 s or less) th at th ey do not leave any recognizable track in bubble or spark chambers.
T h ese particles can be classified according to baryon number, hyp er
charge, and isotopic spin, and are designated b y the same sym bols as those baryons and mesons which have the same quantum numbers as the resonances, but w ith a subscript I, 2, 3, . . . according to increasing mass. T h e number o f such resonances is about 80 i f one includes all charge states and anti particles; th eir num ber is in creasing continually.
Figure 924 (sec preceding p age) shows som e meson re
sonances and F ig. 925 gives some o f the known baryon resonances. (in M e V ), spins, and parities arc also shown.
T h e masses
Resonances
9.7)
(a)
4/5
F ig u re 9—26
F ig u re 927
Resonances are identified by a v e ry subtle application o f the principles of con servation o f energy and m om entum .
Suppose th at a particle A collides w ith an
other particle B at rest in the laboratory, resulting in particles I, 2, 3 (F ig . 9  2 6 (a )); th at is, A + B —i> I + 2 + 3. I f we observe particle I at a given angle w ith respect to the direction o f m otion o f A , we should observe a continuous energy spectrum, since the conservation o f energy and m om entum allows d ifferent directions o f m otion and different energies o f particles 2 and 3 fo r a given direction o f m otion o f I. Suppose, instead, th a t the process is in tw o steps. In the first step, on ly tw o particles are produced; th at is, A  f B —* I + C.
In the second step, particle C
decays into 2 and 3; that is, C  * 2 + 3 (F ig . 9  2 6 (b )).
Since the first process
in volves on ly tw o particles, the energy o f p article I is fixed for a g iven direction o f m otion and its energy spectrum is as shown in the figure. In this case the energy o f particles 2 + 3 is also fixed.
F igu re 927 shows a m ore com plex situation, in
which five particles are produced.
I f particles 3, 4, and 5 result from the decay
ά16
Fundamenlal particles
Fig. 928. Production of the η° resonance in a protonantiproton annihilation. (Photo graph courtesy the University of California Lawrence Radiation Laboratory, Berkeley, California)
of an intermediate particle C 1 there must be a certain correlation among their energies. Therefore, by analyzing the energy of the particles resulting from a collision, we are able to determine whether the process occurs in one step or in two steps. T h e first resonance that was discovered (about 1960) was the 770 (originally called ω °). Its discovery came about as a result of the analysis of protonantiproton annihilation observed in the Berkeley bubble chamber. One of these annihilations is shown in Fig. 928. I t corresponds to the production of five pions, P + + P  —■» 27T+ + 2 tt~ + 7Γ0· O f course, the 7r° is not visible, but we can infer its existence from momentum and energy conservation. T h e process can be represented as in the drawing in Fig. 928. B y analyzing the energies of the mesons by means of this and many other similar processes, we can conclude that three of the mesons [in our case those marked ( I ) , (3), and (5)] proceed from the decay of a shortlived particle having a rest mass of about 548 M e V , and thus is one of the 77mesons, d e sig n a te d ^ 0. Thus, instead of the above scheme, we must write P
0
r TT+ H 7T 
—
» 7T+

 7
r
4 TT0
This is the w ay the process is represented in Fig. 929. N o te that even if the 77°meson is m oving at the velocity of light (3 X IO 8 111 s—*), in its lifetim e (~ 1 0 _ 2 0 s) it cannot m ove more than 3 X 10“ 1 2 m, a quantity impossible to measure in a bubblechamber photograph.
Resonances
9.7)
Fig. 929. Interaction region of process illustrated in Fig. 928, magnified about IO13 times. R esonances d eca y b y means of strong interactions, w hich accounts for their ex; trem ely short lives. Som e observed decays are: M eson s 0 + TT TT
& V01 
■»
KJ
+ 7Γ
n
7T
I
_
0
Γ TT
+ ;
Λ® +
0 K + + TT
P
+
B aryons 0 N i 7Γ
* S 0 + TT0 A 0 +
7T +
A
0 2
Δ ++
—^ t; 0  f  7Γ
* Σ+ 4
+
7Γ
P + + TT+
T h e student should apply the conservation laws to these decays to find w hich, if any, laws are violated, and should give reasons w hy it is possible for these decays to occu r b y m eans o f strong interactions. E X A M P L E 9.7. Calculation of the energy spread, in the laboratory, of the two photons that result from the decay of a 7r°meson with a total energy E ir relative to the Lframe. S o lu tio n : Let us consider a 7r°meson with energy E ir and momentum relative to the laboratory or Lframe. Then E ir = cv mlc2 + p\. The velocity of the meson (see Appendix I, Eq. A. 10) is (9.12)
v — C2P ir/ E ir.
This is also the velocity, relative to the Lframe, of the Cframe, in which the 7r° is at rest. In the Cframe the two photons resulting in the decay —» 7 + Ύ are emitted in opposite directions with an energy Ell — ^mirC2 and a momentum = EliZc — Due to the symmetry in the Cframe, the photons are emitted isotropically. But this is not the case in the Lframe, in which the forward direction is preferred because of the motion of the Cframe. Due to the Doppler effect, the energy of the photons in the C and Lframes is not the same. In general, the two photons do not have the same energy in the Lframe. The maximum energy spread occurs when one photon is emitted in the direction of ρ π (or forward) and the other in the opposite direction (or backward). Then it is simple to com pute the energy of the two photons in the Lframe. If i is the energy of the forward photon and E y 2that of the backward photon, their respective momenta, in the direct O fp7r, are py1 = E y\/c and py2 = —E y2/c. Conservation of energy and. momentum give Pir
E ir = E y1 T E y2 ,
(9.13)
E yi/c
so that Eyl =
h (E v
0 P
tt)
and
E y2 = ^ (E ir 
cpw)
(9.14)
Thus the energies of the photons in the Lframe have a spread A E = E y\ — E y2 = Cpir
418
Fundamenlal particles
(9.7
W e shall now apply these formulas to a particular situation. It is observed that, in a 7r p + collision, the end products are often a neutron and two photons; therefore we may write the reaction as TT + p +  m
+ Ύ + 7.
(9.15)
If thisreaction occurs in one step (i.e., if itcorresponds to Fig.9~2G(a)with proper iden tification of the particles), it is a threebody problem.Assuming that the ir~ and the p + are both at rest in the laboratory, the total energy available is Q = (τη , + /ηp — rnn)c2 = 139 MeV. This energy must appear as kinetic energy of the three resulting particles. Since this energy can be distributed in a continuous way among the three particles, the photons could have any energy from zero to almost 139 M cV. Another alternative process is one which occurs in two steps; that is,
a· ) p + —» n ( Tr0 U 7 I7
(9.16) a corresponding to Fig. 926(b). In this case the energy released (if the 7r and p + are both at rest in the Λframe) Is Q = (?n,_  f i n , 
/Hn — ntro)c'2 — 4 M eV.
This energy appears as kinetic energy of the n and tt°; therefore, according to the laws of conservation of energy and momentum (see Appendix II, Eq. A .27), the kinetic energy of the Tr0 is E„.w» = THn +
Q
~ 3.6 MeV.
TH» 0
W e have used a nonrelativistic formula, since the kinetic energy of the pion is much smaller than its rest mass. Therefore the total energy of the 7r° is about 138.6 M e V , corresponding to cpr ~ 31 M e V , which would be the energy spread of the photons resulting from the pion decay. The photon energies should then fall in the range £(138.6 ± 31) M eV or from F y2 ~ 54 M eV to F y\ ~ 85 M eV. This is what is observed experimentally, verify ing that the ττ  ρ + collision is a twostep process that takes place according to Eq. (9.16). This example, therefore, illustrates what we have said about identifying resonances by means of cnergymomentum correlations. E X A M P L E 9.8.
Detection of the j)°particle.
S o lu tio n : The 7/°particle, as we explained in connection with Figs. 928 and 929, has such a short life that it cannot leave measurable tracks in a bubble chamber. In that respect the ij°particle behaves like a resonance. Let us consider the reaction τγ+
4  Ρ + ^
Ρ
+ +
τγ+
(
τγ+
+
τγ
+
TT0 .
I f all the pions are produced at the time of the collision, the process would correspond to Fig. 927(a), with no correlation among the energy and momentum of the five particles except that imposed by the overall conservation laws. Hut the possibility exists that the
9.8)
Whal is a Jimdamenlal particle?
419
process occurs in two steps, TT+ +
P + ~ * P + 4" JT+ 4* V 0
U π+ + T T  + 4 ' corresponding to Fig. 9 27(b). In this case the energy and momentum of the three parti cles resulting from the »)0deeay must be consistent with those of a particle of given mass. Noting that in this case we must have E r + E p ^ E'p + E ', + E „ P r +
P v
—»
P p
+
P r
+
P „
we may experimentally determine. /?, and p , by measuring the other quantities, and from those values we should be able to obtain the mass of the η°. The difficulty is that we ob serve two 7t+ and we cannot tell which one is the original ir+ and which one comes from the η0. Therefore, we have to make the mass calculations for the η ° in duplicate, using both 7r+’s. The experimental result is shown in Fig. 930. The pronounced peak around a mass equivalent to 550 M e V , corresponding to the correct 7r+, is proof that the process has two steps involving an jj°particle of such a mass. The remainder of the mass dis tribution is due to the wrong choice of a x + in each case. Thus we again see how the con servation laws allow us to infer the existence of particles which we cannot observe other wise.
F ig. 930. Experimental mass of the intermediate ij° resonance. I f this particle did not exist, a mass distribution without peaks would be observed. Mass of intermediate particle
9.11 W h a t is a FuntliinutIiIuI I tIirIitIt t? N ow , in light o f our preceding discussion, are we in a position to define a funda mental particle? W e know that fundamental particles are precise physical entities characterized by certain properties such as charge, mass, spin, etc., that these particles interact am ong themselves according to more or less welldefined inter actions (strong, electromagnetic, weak, and gravitational), and that all the pro cesses which take place am ong these particles follow certain conservation laws. But there are still many serious and im portant questions. W h y are there so m any particles? W h at function in nature does each o f them fulfill? H ow are they related am ong themselves? W h y do they have certain mass and spin values? W hy are they divided into tw o very distinct classes, bosons and fermions?
420
ε
Fundamental particles X10*
Χ Ι0* O 2.0
O 2.0
ιΟ.ό
(a) Proton
1.0 /
1.5
2.0
2.5Χ IO15 ιη
(O) Neutron
F ig. 931. Theoretical radial charge distribution for (a) protons, (b) neutrons.
In the same w ay th at we look a t atom s as being com posed o f certain basic in gredients (electrons, protons, and neutrons) or o f nuclei as being com posed o f protons and neutrons, w e m ay assume th at all fundam ental particles are composed o f certain building blocks or superfundam ental particles. G ellM a n n has proposed the name quark for these superfundam ental particles.* Quarks have certain novel properties, such as fractional charge (Je or § e ). B u t as y et nobody has observed quarks, in spite o f the intensive hunt th at has been g oin g on.
F ollow in g another
approach, tve could consider, for exam ple, all baryons and th eir resonances to be excited states o f a basic baryon (w e m ight use a sim ilar logic for mesons) in the same w a y th at a hydrogen atom m ay exist in its ground state or in m any excited states, each state havin g its ow n attributes, such as energy, angular m om entum , and p arity.
B u t no satisfactory th eory along this line has y e t been form ulated.
In any case, w hat we call fundam ental particles do n ot seem to be b y any means sim ple entities.
F o r m any years physicists h ave considered a nucleon as being
com posed o f a core— a “ b a re ” nucleon which is unobservable— surrounded b y a cloud o f pions, in the same w ay th at an atom is a nucleus surrounded b y a cloud o f electrons.
T h e radial charge distribution o f this p ion cloud for the proton and
the neutron could be as illustrated in Fig. 931. F o r the neutron, the total charge is zero.
(In cid en tally , this affords a means o f distinguishing a neutron from an
antineutron, since the antineutron would have the sign o f the different layers o f charge reversed.)
T h ese graphs are sim ilar to those o f Fig. 313 fo r the radial
* The word was taken from a rather obscure passage of James Joyce’s Finnegan’s Jl'afce: Three quarks for Muster Mark! Sure he hasn’ t got much of a bark And sure any he has it ’s all beside the mark. Webster’s New International Dictionary, 1961, defines “quark” as the harsh cry of the crow, or any sound im itative of this cry.
Whal is a fundamental particle?
9.8)
421
Fig. 932. Strong interaction resulting from the exchange of pions. charge distribution o f an electron in an atom.
N o theory abou t pion dynam ics in
a nucleon has y e t been developed, but ap p aren tly the pion cloud m ay be excited to higher energy levels, g ivin g rise to nucleon isobars, in the same w ay that atom s m ay exist in excited electronic states. A n oth er interesting aspect is the interpretation o f interactions as the result o f the exchange o f bosons betw een the interactin g particles. A n electrom agnetic interaction results from the exchange o f photons between tw o charged particles, as indicated in Fig. 113. Sim ilarly, it is accepted th at the strong interaction is the result o f an exchange o f pions between the in teractin g particles, as indicated in F ig. 932.
T h e particles exchanged are called virtual because th ey cannot be de
tected as separate entities.
T h e short range o f nuclear forces m ay be v e ry nicely
explained w ith this m odel fo r interactions. A proton is not a static system, but is continuously ejectin g and reabsorbing some o f its surrounding pions. virtual pion is em itted, the proton energy changes by an am ount
W hen a
A E ~ Uit Ci ~ 140 M eV . A ccord in g to H eisen berg’s uncertainty principle, E q. (1.49), this pion can exist, w ith ou t any violation o f the laws o f quantum mechanics, during a tim e At ~ h /A E ~
Iifm
1C i
~ IO22 s.
A fte r this tim e the virtu al pion must be reabsorbed b y the nucleon or exchanged w ith another nucleon.
In this tim e, if we assume that the pion travels w ith a
v e lo c ity close to that o f light, the m axim um distance it can go is about IO 14 m. T h is then gives th e m axim um distance at which a second nucleon must be located to absorb the virtu al pion. T h e student should be struck b y the fact th a t this is precisely the m agnitude o f the range o f the nuclear interaction. W h en we think about the ab ove m odel o f nucleons and their interactions, w e sec th at it is an o v e r sim plification to say th at a nucleus is com posed o f protons and neutrons.
The
best indication th at a bound neutron is not the same as a free neutron is that bound neutrons in general are stable and do not decay. In the case o f electrom agnetic interactions, we m ay assume th at a charged par ticle is continuously e m ittin g and absorbing virtu al photons. T h e virtu al photons em itted by charged particles have zero rest mass; therefore the energy fluctuation A E o f a charged particle m ay have an y arb itrary value, and the length o f tim e th at the virtu al photon can exist before it is reabsorbed or exchanged w ith another charged particle is also arbitrary. A calculation shows th a t the force between tw o
422
Fundamental particles
charged particles resulting from the exchange of zerorcstmass photons must vary as the inverse square o f the distance o f the tw o charges, in agreement with C ou lom b’s law. I t thus appears th at pions are the carriers o f the strong interaction.
F o r this
reason the existence o f the g raviton as the carrier o f gravitation al interaction has been postulated, and another p article (a w eak boson, W ) has been postulated as the carrier o f the w eak interactions.
N eith e r o f these particles has y e t been
observed. Alth ough the th eory o f elem entary particles is still in an im perfect state, phys icists have been able (b y using some elaborate m athem atical concepts) to gain some understanding o f the relations am ong the particles.
Perhaps the m ost in
teresting th eory is one proposed in 19G1 b y M . G ellM a n n and Y . N e ’eman, called the eightfold way because it required a set o f eigh t basic operators. T h is th eory predicts that, am ong the baryons and mesons, m ultiplets o f eight and ten particles should exist w ith certain relations between their quantum numbers and their masses. O ther groups w ith a larger number o f particles are also supposed to exist.
The
theory predicted the existence o f the Ω particle, which was shortly afterw ard observed (F eb ru ary 19G4) at B rookhaven N ation al L ab orato ry (see E xam p le 9.10). T h e field o f elem entary particles is one o f the m ost active and challenging areas o f research in contem porary physics.
N e w and bigger accelerators are constantly
being built to exam ine the existing problems, but these accelerators often create new problem s b y uncovering new and unsuspected processes. I t is hoped, how ever, th at in a few years our understanding o f elem entary particles w ill reach a status sim ilar to our understanding o f nuclei, atoms, and molecules. E X A M P L E 9.9. Inelastic scattering of highenergy protons as a result of protonproton collisions.
F ig. 933. (a) Schematic diagram of protonproton inelastic scattering experiment, (b ) Experimental results showing inelastic scattering peaks.
H7Zia/ is a Jundamenlal particle?
9.8)
423
S o lu tio n : An important experiment showing the existence of nucleon excited states (that is, nucleon isobars in which the pion cloud has been raised to an excited energy state) is the analysis of protonproton collisions. The experimental setup appears in Fig. !>33(a). Protons from an accelerator hit a liquidhydrogen target. Those protons which are scattered through a fixed angle Θ pass through a magnetic field (B, which deflects some of the protons through an arc of a circle of fixed radius r, determined by the initial direction of motion and the position of the detectors Di and D i This allows us to determine the momentum of these protons, given by p — efflr. Hy varying the in tensity of the magnetic field, we can change the momentum of the protons passing through the fixed detectors and analyze the momentum distribution of the scattered protons. The experimental result is shown in Fig. 933(b). In addition to a main peak in the momentum distribution of the protons (corresponding to elastic scattering by the target), there are several secondary peaks, corresponding to protons of lesser momentum or kinetic energy. These secondary peaks correspond to inelastic scattering, in which the incoming protons lose some of their kinetic energy to a target proton, which is raised to an excited isobaric state of welldefined energy. The student may recognize the fact that this experiment is similar to the Franck and Hertz experiment of inelastic clectronatom collision, which showed the existence of excited stationary atomic states of welldefined energy. E X A M P L E 9.10.
The omegaminus experiment.
S o lu tio n : According to the “ eightfold w ay” theory of GellMann, particles should be grouped in families, with all members of a family having the same spin and parity. If we look at the resonances shown in Fig. 925 and plot the family of ten lighter particles having spin § (and positive parity), we obtain the arrangement of Fig. 934, which exMass, MeV
Isotopic spin r Strangeness
n
3   I)
1530
2  4
I
I
I)
§
1385
A »/
JA I q/e =  I
/  i
■I
/ (I
I* / 123S
J
A L / J
O
/ +1
I
/
/
i
+2
F ig. 934. Arrangement of lighter hyperons with spin § and positive parity, following GellMann’s eightfold way.
424
Fundamental particles
C9.8
liibits a nice geometrical symmetry. A t the time of GellMann’s proposal (1961), only the Δ quartet and the Δ triplet were known. But shortly afterward (1962), the Sdoublet was reported, and it fitted well with the scheme. From the regularity of the pyramidal structure it was easy to predict that the remaining particle at the top should have T1 = 0 and r = 0, and thus be a singlet. Also its strangeness should be i = —3, giving a charge — e according to Eq. (9.11). Finally, from the regularity in the mass differences Δ — Σ and Σ — Ξ, it could be inferred that the new particle (called Ω“ by GellMann) should have a mass about 1675 MeV. This therefore fully identified the missing particle. (GellMann’s prediction was based on a more serious theoretical basis than this simple geometrical arrangement.) From these properties experimenters expected, in view of the conservation laws, that the Ω“ could decay into H0 + ιτ“ , Ξ “ I ir0, or A 0 f K  ; this meant that, by observing the decay products, they had a clue by which they could identify this particle.
F ig. 933. Schematic diagram of the Ω experiment. Before the separation there were approximately SOO x “ and IO p “ for every IO K  particles. After separation, there was only one x “ and no p “ for every 10 K “ particles.
The next step was to see if the particle could be produced and observed in the laboratory. A possible production process, compatible with the conservation laws, would be K  + p + —. Ω " + K + + K 0. In 1963 an experimental setup was prepared at the Brookhaven National Laboratory. A 33GeV proton beam generated by the accelerator hit a tungsten target, producing K "mesons, together with pions, autiprotons, and other particles. T o extract the K “ mesons, the experimenters devised an ingenious arrangement, shown in simplified form in Fig. 935. For experimental convenience, they chose K _ ’s with a kinetic energy of 5.0 GeV (which is well above the threshold energy of 3.2 M eV of the. process) and ex tracted them from the accelerator tube by means of a deflecting magnet. The extracted beam, in addition to containing K “ particles, was composed of a large number of x “ and a few jp“ , so the next task was to separate the K “ particles from the other particles. This was accomplished by means of a second set of deflecting and separating magnets, together with a slit arrangement. N ext they allowed the almost pure K “ beam (it con tained only about 10 percent x “ and no antiprotons) to enter a shielded 2m bubble chamber containing liquid hydrogen, in which the reaction producing the Ω“ could take place. The total length of the path of the K “ ’s from the target to the bubble chamber was about 135 m. A fter a run of several weeks, by the end of January 1964 the experi
9.8)
Whal is a fundamental particle?
(»)
425
A0 K 01(a) what are the possible values of T for the two reacting particles and for the two resulting particles? Determine the value of T for conservation of isotopic spin. Also determine whether T , and S are con served. (b) Repeat the analysis for the process ir~ + p + —> A 0 + π° and conclude if the process is expected to occur in nature. 9.22 Consider the process p + + p + —> p + f n + Jr+ . Determine the total iso topic spin; are T . and strangeness con served? Repeat for P + T P +  1 P t T n + e + + ve. This process occurs with a much lower probability than the firstmentioned process. W hy?
9.23 Check the several conservation laws (except energy, momentum, and angular momentum) in the following strong inter action processes. (a) p + + p+  * p+ + A 0 E K + (b )
T T ~ \
P +  * 2 ° +
K
0
(c) p + + p +  » E 0 + p + + K 0 + K + (d) A 0 + p + > n + p+ + K 0 (e) K  + P +  * K + + S Determine in each case the values of T ll T , and S for the reacting and resulting particles. 9.24 W:hich conservation laws are satisfied and which are violated in processes Ij0 —* p + + e “
and
π° —> 7 + 7?
W hy is it that 2 ° —» A 0 + 7 occurs, but 2 + —» p + + 7 is not observed? W hy is it that Ξ  —> A 0 + i r _ occurs, but Ξ  —* η° + ir_ is not observed? 9.25 Analyze the process K  + p + —* Ω + Κ + + Κ ° from the point of view of the conservation laws. Calculate the threshold kinetic energy of the K  particle. 9.26 Show that the processes ir + p + —> n + 7, Tc + d —* 2n + 7, and T + + d —* 2p+ all imply that the spin of the pion is either 0 or I. (Sec the following problem.) 9.27 Let us designate by σ ι the cross sec tion for the process p+ + p + —* t r + + d and by σ 2 the cross section for the inverse process π + + d —> p + + p + . W e can show that the cross sections are related by ? =
Jw iiyf + Jwi2i’2 T  JwiailI + · · ·
(112)
AU p a r tic le s
N o te th at E q . (11.2), corresponding to the kinetic energy, has one term fo r each particle, while E q . (11.1), corresponding to the poten tial energy, has one term for each p a ir o f particles, because it refers to tw op article interactions on ly. T hen the total internal energy U o f the system is U
—
Ek
in t +
E p in t .
(1 1.3)
F o r the case o f an ideal gas, there are no interm olccular forces and the internal energy is just the kinetic energy.
In some instances the potential energy can be
expressed as a sum o f singleparticle terms, as indicated in Section 10.2.
T h is is
possible when, to a first approxim ation, the interparticle potential energy can be replaced b y an average potential energy for each particle.
I f no external forces
act on the particles o f the system (i.e., if the system is isolated from external actions), then U does not change. In other words: The internal energy o f an isolated system o f particles remains constant. W hen there are external forces actin g on the particles o f the system , the internal energy docs not, in general, rem ain constant. Suppose th at the system is in itially in a state w ith internal energy U 0. T h e state o f the system is continuously m odi fied b y the external forces, so th at a fter a certain tim e the internal energy is U . L e t us designate the total w ork done during the same tim e b y the external forces acting on the particles o f the system by W ext. T h e w ork W ext is a sum o f m any terms, one for each particle subject to an external force.
T h e n conservation of
energy requires that U  U 0=
W ext,
(11.4)
which states th at: The change in the internal energy o f a system o f particles is equal to the work done on the system by the external forces. I t m ay happen that, even if there are external forces acting on the system, their total w ork is zero: W cxt = 0 . In such a case there is no change in internal en ergy; th at is, U =
U 0. I f w ork is done on the system (W ext positive), its internal energy
increases ((/ >
U 0), but if work is done by the system (W ext n egative), its inter
nal energy decreases ( U
0.
(11.30)
So the (irreversible) process, which is one that certainly occurs in nature, produces an increase in the entropy of the gas. However, the reverse process, in which we assume that the gas initially occupies the whole container (or volume 2V), and then at a certain later time occupies only the volume V at the left, corresponds to a decrease in entropy; that is, AS' = S i 
S 2 = —k X In 2 < 0.
(11.31)
Thus this process, although possible, is very unlikely to occur naturally if the gas is isolated. It is true that we may compress the gas isothermally, reducing its volume from 2V to V, with a corresponding decrease in entropy equal to the value given by Eq. (11.31). But this requires an external action, and it is then necessary to take the external changes of entropy into account Io obtain the total change in entropy in the universe. It is instructive to look at the same situation from a probabilistic point of view. From definition (11.25), we have that S2 
,Si = k In P 2 
k In F 1 = k In γ
d80
Thermodynamics
(11.9
When we compare this with Eq. (11.30), we have In
= JV In 2 = In 2N
or
P 2 _ „ 1V P1 " 2 '
In general, the number JV of molecules of the gas is very large and hence P 2 is much greater than P\. This accounts for the rapid rate at which the gas expands freely up to the volume 2V. For the reverse process, 2 V —> V, we get P 1/ P 2 = 2~N, and this is an extremely small quantity for the number JV of molecules that are usually found in any sample of gas. Hence it is extremely improbable (although possible) that at a certain time all the gas molecules, as a result of their interactions, appear concentrated again in the region V at the left. Of course, for a very small number of molecules (JV = I or 2, for example), we have P 1/P 2 — i or and it is possible in a short time to observe “all” the molecules (one or two) 011 the lefthand side. But then, of course, statistical methods are unnecessary, and have no meaning, and it may be even possible to calculate the exact times at which the particle or the two particles will be on one side or the other.
11.9
E n tro p y a nd H e a t
W e must now see how entropy is related to the other thermodynamic quantities we have previously introduced. Suppose that a system in statistical equilibrium undergoes an infinitesimal transformation as a result of its interaction with its surroundings. T h e interaction results in a change o f the partition numbers n, and o f the possible energy states E i. Since U — ]T, UiE it we then have that dU = Σ
E i dm + Σ
n «2 > «3 > · ·  if the system is in equilib rium. Also assume, for simplicity, that 31 = 32 = 03 = · · · = L Then the probability of this equilibrium partition is P =
«il«2bt3l · · ·
Suppose now that the total energy U of the system is increased and that the energy in crease is accomplished by one particle shifting from level E i to level E 3. The probability of the new partition is then P' =
I ( « I — I ) !«21(713 +
Therefore P
na+l
I) I ·  ·
11.9)
Enlropyandheal
6H3
If « ι is larger than «3 by at least 2 units (in general, m is much larger than 03 ), we have that P ' is larger than P. Hence we conclude, in general, that P increases with the energy U, at constant volume. Of course it is possible that under certain circumstances the partition probability does not change, or it may even decrease when the total energy increases. In particular, when the particles of a system have only a limited (or finite) number of accessible states, it may be shown that P may be a decreasing function of the energy for certain energy ranges. This would result in a negative absolute temperature at such energies (see Problem 11.17). E X A M P LE 11.6. Derivation of Eq. (11.35) for the relation between the change of en tropy and the heat absorbed by a system in a reversible transformation. Solution: Let us suppose that our system obeys MaxwellBoltzmann statistics, an assumption of wide validity. Then the entropy for an equilibrium state is given by Eq. (1 1.2S); that is, S = Y + IcN ln  + fc :V . For an infinitesimal reversible transformation in which the total number of particles does not change, d
^dT+IcN y 
S
(11.38)
since d(ln Z) = dZ/Z. Recalling the definition (10.22) for the partition function in MaxwcllBoltzmann statistics, Z = 2Z, gie~E>lkT, we have that vs
dEi
 E iIkT
, V 1
9*
S
Ei
„
.  W r ,it dT·
»
Then, using Eq. (10.25), we may write I
dZ
,
N
 E i Ik T
kX Y = ~ j i 2, zie
.
I
N
 E iIk T r
dL< + f i L , z 9 2ro, as illus trated in Fig. 124. Then we may continue with the assumption that /,7 = — I for r < 2ro. But from Eq. (12.14), if E Pij/kT is small compared with unity, we may write f i 2 = — Epi 2/kT for r > 2ro. Therefore Eq. (12.16) gives r 2r0
0 = Jo
/*oo
(— 1)4Tr2 d r + J 2 ( —Ep\2/kT)4rr2 dr = —
3
+ ■— >
12.3)
EpU
Equation of state fo r real gases
Epn *
503
Epi2
Fig. 123. Hardcore intermolecular po tential energy. No attraction at any distance.
Fig. 124. Intermolecular potential energy with a hard core and weakly attrac t i v e at larger distances.
where
« =
f
J2ra
( — E pi2)iirr 2dr
is a positive quantity, since Ep12 is negative (sec Fig. 124) for r > 2ro Substituting in Eq. (12.19), using our previous definition of 6 and setting a = %N\a, we obtain .A = RTb — a for the second virial coefficient. The equation of state, to the first order of approximation, is then, using Eq. (12.7), ( 12.20) This equation is satisfied with fairly good accuracy by many real gases, especially for large values of F / n . The coefficients a and 6 are called van der IFaafs constants. They are given in Table 121 for several real gases. T A B L E 121
V an der W a aIs C o n stan ts
Substance
a, N m4 kg2 mole2
b, m3 kg1 mole 1
Helium Hydrogen Neon Nitrogen Oxygen Ammonia Carbon dioxide Sulfur dioxide Water
3.446 X IO3 24.68 21.28 140.4 137.4 421.2 362.8 678.1 551.9
0.02370 0.02661 0.01709 0.03913 0.03183 0.03707 0.04267 0.05636 0.03049
( /2.4
504
Thermal properties of gases
12. i
Mleat Ca pa city o f aa Mdoai M o n a to m ic Mmati
In Section 11.7 we defined the heat capacities of a substance at. constant volume and at constant pressure as
c'=Ks?),'
(1221)
where 11 = U + p V is the enthalpy of the substance. In addition to their im portance as coefficients in several practical calculations, the theoretical calculation of the heat capacities o f a substance affords a means o f verifying the correctness o f the model chosen to describe the substance. In this section we shall calculate the heat capacities of an ideal gas, and by comparing these with the observed values for real gases, we shall be able to conclude to what extent the idealgas model is a good approximation. Let us first consider an ideal m onatom ic gas. T h e internalenergy of such a gas is purely translational kinetic energy, and isgiven by U = % xR T. Using Eq. (12.21), we obtain C r = §R =
12.4715 J m ole1 0K 1
= 2.9807 cal m ole1 0C  1 .
( 12.22)
Taking into account Eq. (12.6) {p V = NR T ), we have that the enthalpy of an ideal gas is H =
U + pV = f x RT.
Thus Eq. (12.21) gives Cp = f R = 20.7858 J m ole 1 0K 1 = 4.9678 cal m ole1 0C  1 .
(12.23)
Therefore all ideal m onatom ic gases have the same heat capacities, independent of the structure of the atoms. W e may note, from the above results, that Cp  C v =
R,
(1224)
so that Cp is larger than C r by the am ount R. T h e reason for this is that Cy is related only to the change in internal energy, while Cp includes, in addition, the expansion work of the gas when its temperature increases I degree at constant pressure. It is simple to verify that this w ork is exactly equal to R. W hen the pressure of the gas is constant, p dV = NR dT, and if the increase of temperature is one degree, the work done is IFp =
J pd V = JJ +1 N R d T = N R.
Hence the work per mole done by the gas is R. From this proof we see that Eq. (12.24) is valid for all ideal gases, either m onatom ic or otherwise.
/2.4)
Heat capacity of an ideal monatomic gas
505
Another relation am ong the heat capacities o f an ideal m onatom ic gas is 7 = CvICv = § =
1667.
(12.25)
This relation is followed rather closely by most m onatom ic gases, as shown in T able 123 (at the end of this chapter). E X A M P LE 12.3. and entropy.
The equation of state of an ideal gas in terms of pressure, volume,
Solution: Since entropy is a property of the state of a gas, it can be used as a variable to define the state of a gas in the same way as pressure, volume, or temperature. From the equation of state, pV = n RT, we have In p + In V = In N R + In T. Differentiation yields
f+ff
Λ 0v^
UM STATISTICS
/
13.1 13.2
Introduclion
FermiDirac Distribution Law 13.3
13M
Application of FermiDirac Statistics to Electrons in Metals 13.5
BoseEinstein Distribution Law 13.6 13.7
13.8
The Photon Gas
Heat Capacity of Solid
The Ideal Gas in Quantum Statistics
13.9
Comparison of the Three Statistics
FermiDirac distribution law
13.1
519
In trod u ction
In Chapter 10 we discussed classical statistics, which is characterized by the m ethod of calculation of the probability of a given partition, stated in Eq. (10.8), and by the M axwellBoltzm ann distribution law, Eq. (10.9), for the most probable or equilibrium partition. W hen we discussed classical statistics, we ignored any sym m etry considerations related to the distribution of the particles am ong the different states associated with each energy level accessible to the particles. H ow ever, as we said in Chapter 4, there may be certain restrictions on the number of different ways in which a group of particles may be distributed am ong the avail able wave functions associated w ith each energy state. Clearly these restrictions, of quantal origin, affect the probability of a given partition. T h e theory in which these sym m etry considerations are taken into account is called quantum statistics. There are tw o kinds of quantum statistics: One concerns particles obeying the exclusion principle, and hence described b y antisymmetric wave functions! This kind is called Ferm iD irac statistics,) and the particles are called f e r m i o n s T he second concerns particles not restricted by the exclusion principle, and described by symmetric wave functions. This kind is called BoscEinstein statistics, and the particles are called bosons. In both kinds of quantum statistics it is assumed that the particles are identical and indistinguishable. A t high temperatures and low densities, classical statistics and the tw o kinds of quantum statistics give prac tically the same results. In this chapter we shall briefly discuss both kinds of quantum statistics and apply them to a few im portant physical problems.
13.2
F c r m iU ir a c D istribu tion L a ir
Let us reevaluate the probability of a partition of a system of particles, assuming that the particles are identical and indistinguishable. In addition, we assume that the particles obey the exclusion principle, so that no two particles can be in the same dynam ical state and the wave function of the whole system must be antisymmetric. Particles satisfying all these requirements are called fermions, after the Italianborn Am erican physicist Enrico Fermi (19011954), who first discussed these systems. It has been found experimentally that all fundamental particles with spill J are fermions. T o com pute the different and distinguishable ways in which a system of fer mions m ay be arranged for a given partition, we must revise our definition of the intrinsic or state probability i and v,, corresponding to longitudinal and transverse waves, respectively. T h e total num ber of vibrational modes in the frequency range dv is then g (v ) dv =
g , ( v ) dv + g,(v) dv — 4 irV
v2 dv. V ;
(13.23)
Vt /
In a continuous medium there is no limit to the total number of vibrational modes. But in a solid, which has an atom ic structure and contains N atoms, any vibra tional m ode must be described in terms of the S N positional coordinates of the atoms. This therefore imposes a lim it on the Iolal number of independent modes of freedom , which must be equal to SN. T his in turn imposes a limit on the maxi mum vibrational frequency because we must have, using E q. (13.23),
SN =
g (v ) dv =
4 icV
+
J q 1/2 dv
or S N = A ttV 0
+
j
>
(13.24)
which determines the cutoff frequency V0 (T h e existence of a cutoff frequency in the vibrations of a crystal lattice was discussed in Exam ple 6.2.) Using Eq. (13.24), we m ay write Eq. (13.23) in the form g{v) dv = ~
V2
dv.
(13.25)
T h e problem we r re discussing here is very similar to that o f standing electro magnetic waves in a cavity. T h at discussion gave rise to the concept of a photon gas introduced in Section 13.6 for analyzing blackbody radiation. W e may thus associate with the vibrational modes of the solid, which of necessity are quan tized, a phonon gas com posed of “particles” or phonons of energy hv. T h e concept of phonon was introduced in Exam ple 6.2. Since all phonons are identical and since there is no limit to the number of phonons in the same energy state, we may expect that phonons in thermal equilibrium obey BoseEinstein statistics. Also the number of phonons is not fixed, since their number may increase or decrease depending on whether the energy o f the modes o f vibration is increased or decreased. Thus we must use Eq. (13.15), setting a = 0, as we previously did for the photon gas, and replace
τ h~
2
. 2
.
3 2mFS/3 Nl
2 , N’2
.2.
2 _
3
Ei V
'
Substituting this result in E q. (13.35), we find, for the pressure o f the system o f non interacting particles,
P =
I
z
f
I
H
f
.
of
Show th at the tem peratu re at
which therm ionic emission is m axim um is
T
e/'2k.
=
E stim ate the value o f this
tem perature for some o f the metals given
tp
where
is the value at Τ' = 0 , given in
T a b le 131. Show that the corrective term corresponds to a change o f 1 % in the Ferm i energy at a tem perature 7’ = V 3 θκ/5ττ. E stim ate this tem perature for some o f the
in T a b le 132. 13.8
W h en M axw ellB oltzm an n statistics
is used instead o f F erm iD irac statistics to analyze
therm ionic current
expression obtained
density,
the
is
metals given in T a b le 131 and conclude
j
to what exten t one can assume th a t the
=
A T U2e~'*!kT.
Ferm i energy remains constant between (a )
absolute zero and room temperature. 13.2 Find the average v e lo c ity and the average energy o f electrons a t 0 ° K in a m etal having IO 22 electrons per cm 3. 13.3 Show that the number o f ferm ions w ith a v elo c ity betw een
T
tem perature
dN
v
and
v + dv
at a
is
D e riv e this equation, (b ) P l o t j against
kT/e4>
and com pare with the p lo t o f the
RichardsonDushman equation, assuming in both cases that φ = 3 eV . (c ) E stim ate the tem perature H = ίφ/ k for some metals and decide whether both equations give clearly
distinguishable
results
at
room
temperatures. 13.9 Show that the constant A in E q . (13.11) has the theoretical value of 120 Λ
V2_________
= 8 rrFm 3 ft3
C m  2 oK  2.
!
13.10 13.4 Show that the number o f ferm ions w ith a v elo c ity whose com ponents arc be
vx and vz + dvz, t>„ and v„  f dvy, and v, and Vg f dvx is
tween
From our discussion o f the poten
tial step in Chapter 2, we know th at some electrons that reach the surface o f a metal should be reflected back in to the metal even if their energy is larger than
cφ.
(a)
M o d ify E q. (13.10) to take this effect into _ 2 vVm 1
~
dvxdvy dvx.
account. j '
, 3 e (m .* / 2  .f ) l k T +
13.5
Show th at the number o f ferm ions
w ith
a com ponent o f v elo c ity along the
axis between
J ustify the fact that the experi
mental values for
X
vx and vx f dvx is
(T a b le 132) are less
coefficient for cesium and chromium. 13.11 greater
2
A
than the theoretical value given in the previous problem, (b ) Find the reflection
E, gie~us~,t >lkT
F o r what values o f the energy than
ep,
is
n, =
within 10% o f E q . (13.5)? 1
h 3k f 
=
ln Ie
+
1I ώ ·
13.12
[Hint: Use In ( I + 13.6
the integral JJ5 ( ae1 +
I ) 1
dx =
F o r the energies indicated in Prob
lem 13.11, w e can write
1/a).]
Using the result o f Problem 13.5, de
rive Richardson’s equation (13.11). Use
the
when
x
approxim ation
is small.]
In ( I f
[Hint: x) = x,
N (E ) dE  8xV(2™3)112 E m c iS  .F)ikT instead o f E q. (13.9). Show th at the num ber o f
electrons
per
unit
volu m e
with
Problems energies e