THEORY AND PROBLEMS OF

MODERN PHYSICSSECOND EDmONRONALD GAUTREAU, Ph.D.Pmfessor of PhysiC-'s New Jersey Instilule of Technology

WILLIAM SAVIN, Ph.D.Professor of Physics New Jersey Institute of Technology

SCHAUM'S OUTLINE SERIESMcCRAW-HILLNew York San francisco Washington. D.C. Auckland Bogota Caracas Lisbon London Madrid ~exico City Mihm MOnlreal New Delhi San Juan Singapore Sydney Tokyo Toronto

ABOUT THE AUTHORS

Ron.ad Cautrnu. Ph_D.. aoo WiUiarn Sa~';n. Ph.D.. are Profcssor'! (If Physics at New Jersey Institute ofT;It

- 1'

iJ' '" = '" '" il.vl f'hfl '" '" = "' az2 ilz'l

CHAP: 6]

RELATIVISTIC SPACE- TIME MEASUREMENTS

35

Substituting these in the wave equation. we obtain ; 1>

~ + ;y- +

[t21>

if! rP;):2 -

I ; rP [t2 rP ;12 rP c 2 iIP = ;h~ + i~)~!

Ii! q, I if! q, + ii::;i - -;.! all

so that the equation is invariant under Lorentz trdnsfonnations. Recall that the wave equation is not invariant under Galilean IJansfonnations (Problem LHI).

Supplementary Problems6.24.An unstable particle with a mean lifl-'time of 4 JIS is forme speed ofthe beam aCIOgl; the screen'! Ans. 7.85 x lOs mils (note: since c = I.R6 x 1O~ mils. tile sweep speed is large r than c.) Show that tile expressions x 2 +.v2 +:? transformations.

6.36.

6.37.

.h1 and rtx2 + d);2 + J z 2 -

~dll arc not invariant LInder Galilean

Relativistic Velocity Transformations~~;~cive ~I~O< ~i~ty ~~tran;sfonnatiOl1S.~

(Fig. i - l). One obseIver, 0 ', moves along the common axis at a constant respect to a second observer. O. Each observer measures the velocity of a single particle, ~!~t~~~~ (II~. " .)" u:r) .and Q ' recording (~ . ~. U:) for the components of the partic1e's velocity. coordinate transfonnations one finds the fOllowing Lcrenu ve/ocily lronsjim nations (see

....'e consider an arrangement identical to that for the Lorentz

x-x

r

t

IIx

= 7 ld" )u -, 1----'(;-v--;

Ux -

V

u'. ~ u,.,/1Y

I

(O'{d, ) (vlil )llx

u'. _ u,,/I (0'1T.)=' 1

( v/c2)11~ =

(0.'''')./'

(0.6)'

(-O.6c) , - -, - (O.693c)

,

.22&

TIle speed measured by observer 0' is

II' =

Ju? + u,/ =

J(0.9I3c)1

+ (O .226c)2 =

O.941c

and the angle 4/ the velOCity makes with the x' -axis is

, II, O.226c tan tft = -'- = - - = 0.248

II,.

O .913c

0'

'~

13.9"

7.6.

Consider a radioactive nucleus that moves with a constant speed ofO.5c relative to the laboratory. The nucleus decays and emit is the rest

8.36. 8.37.

At what velocity must a panicle move such that its kinetic ent:rgy equals its rest energy?

Ails.

0.86&IS

Suppose that the relativistic mails of a particle is 5% larger than its rest mass. What An.~. 0.]05e velocity?

its

8.38.

What is the ratio ofthc relativistic mass to the rest mass for (0) an electron. (b) a proton. when it acceleratt:S Ails. (0) ]0.]5; (b) 1.015 ffom rest throogh a potential difference of 15 megavolts'! What is the m ails of an electron if it moves through a potential difference that would, according to classical physics, accelerate the electron to the speed of light? Ails. ~ mo Refer to Problem 8.20. What are the velocity and momentum of each nO?

8.39.

8.40.8.41.

Ans.

0.84'; 209 MeV Ie

Suppose that electrons in a uniform magnetic field of nux density 0.03 T move in a circle of radius 0.2 m. What arc the velocity and kinetic energy of the electrons? AilS. 0.962c; 1.]6 MeV What is the minimum energy required to accelerate a rocket ship to a speed of 0.& if its final payload rest mass is 5000 kg'! AIlS. 3 x 102(1 J A 0.8 MeV electron moves in a magnetic field in a cilCular path with a radius of 5 cm. What is the magnetic induction'! An.\. 8.07 x 10- 2 T Compute the radius of II 20 MeV electron moving at right angles to a uniform magnetic field of nux density 5T. AIlS. l.37cm A partiele of rest mails mil moving with a speed ofO.6c collides with and sticks to a similar particle initially at rest. What are the rest mass and velocity of the composite particle'! Ans. 2.12m(): O.]]]e A particle with a rest mass Inn and kinetic energy ]m02 makes a completely inela'!tic collision with a stationary particle of rest mass 2m n. What are the velocity and rest mass of the composite particle'! AII.~. O.64~:: 4.58mo A n'" -meson whose rest energy is 140 MeV is crealed 100 km above sea level in the earth'S atmoSphere. The n'" -meson has II total energy of 1.5 )( lOS MeV and is. moving vertically downward. If it dis inlegrates 2 )( IO- ~ s after its creation. as determined in its own frame of reference. at what a ltitude above sea level does the disintegrdtinn occur? Am:. 9].6km

8.42.

8.43.

8.44.

8.45.

8.46.

8.47.

The Ouantum Theory of Electromagnetic Radiation and Matter

Electromagnetic Radiation PhotonsTIl~11rHl'0I1\, OF PHOTONSto quantum interpretation, electromagnetic radiation conslsts of particle-like discrete called photons or quanta. Each photon has an energy E that depends only on the the radiation and is given by

E = hv =

h~

,.

c

x 10- )4 J . s is Planck's cunstant. S:::t:~~:.:.:tr4Vcl at the speed of light. they must. according to relativity theory, have zero rest mass; ~ is entirely ki netic. If a photon exists. then it moves at the sp eed o f light, c; if it ceases to c, it ceases to exist. For mo = 0, the relativistic momentum-energy relation (Section 8.5)

E='"c. Thus, each photon has a momentum ofcc

E hl" h p= - = - = -

1

F,..., the .",nn,m pc,i.' of view, a beam of electromagnetic energy is composed of photons traveling at the speed c. intcilbity o f tht: beam will be proportional to the number of photons crossing a unit area UJdt lime. if the beam is monochromatic (of one frequency), the intensity J will be given by

J = (energy of one phOlon) x

number of photons . arca )( tnne

For ..""0IIi~,,:e in calculations, the following expressions in nonstandard units can be used:h = 4. LJ6 x lO- ls eV _s

hc = 12.4 kcV -'All!!!~.u:!'.il1l 0- ] keV = 1.602)( 10- 19 J and I A = lO- lU m_

59

60

ELEcrROMAGNETIC RADIATION- PHOTONS

[CHAP. 9

9.2

THE PHOTOELECTRIC EFFECT

In a photoelectric experiment, light shines on a melal surface in an evacuated tube and electrons are emitted from this surface, as shown in Fig. 9-1. The frequency l' and intensity I of the light, the retarding voltage V. and the material of the emitter can be varied. If the electrons are sufficiently energetic they will be able to overcome the retarding potential V and will reach the collector and be recorded as current j in the a mmeter A. In order to be able to reach the collector. the electrons must have a kinetic energy equal to or greater than the electrical potential energy that they must gain in going between emitter and collector, i.e.!m v ,

.

2

> - eV

Emitter

v

.Fig. 9-1

If their energy is less than this value, they will be turned back before reaching the collector and will not be recorded as current. The experimental results are:

].

The current begins almost instantaneously. even for light of very low intensity. The delay between when the incident light strikes the surface and when the electrons are observed is of the order of 10- 9 s and is independent of the intensity. When the frequency and retarding potential are held fixed, the current is directly proportional to the intensity of the incident light. When the frequency and light intensity are held fixed, the current decreases as the retarding voltage is increased, reaching zero for a certain stopping voltage. Vs' This stopping voltage is independent ofthe intensity. For a given emitter material. the stopping voltage varies linearly with the frequency according to the relation

2. 3.

4.

eV. =hv - eWoThe value of the constant term, eWo varies from material to material. but the slope h remains the same for all materials. being nwnerically equal to Planck's constant (see Problem 9.10). For a given material there exists a threshold frequenq. V,h ' below which no electrons will be emitted, no matter how great the light intensity.

5.

CHAP. 9)

ELECTROMAGNETIC RADlATION- PHmDNS

6'

A wave picture oflight can explain only result (2), the increase of current with intensity, since the more intense the light, the more energy transmitted by the wave, and the more electrons that should be emitted. The other results, however, are completely inexplicable in terms of a \\'ave picture (see Problem 9.9). The quantum interpretation of light consisting of photons explains all the experimental results. In the quantum picture the energy carried by a photon is absorbed by a single electron. If the electron is ejected from the material, the difference between the energy absorbed by the electron and the energy with which the electron wa" bound to the surface appears as kinetic energy of the electron. The electrons are bound to the surface with varying energies, but the binding energy of the lco.st tightly bound electrons depends on the material of the emitter. The energy required to remove these least tightly bound electrons is called the workjunclion,l/J, of the material. Hence, the electrons will be ejccted with various kinetic energies ranging from zero to a maximum value given by Maximwn kinetic energy of emitted electron

= (energy carried by photon) - (binding energy of the least tightly bound electron)thereby explaining experimental result (3). Since KIfIIIIJ. =e~p

the maximum-encrgy relation becomes

where l/J = eWo . Hence the linear relation of result (4) is explained, along with the existence ofa threshold frequency (result (5 given by

Below this threshold frequency the incident photons will not have sufficient energy to release even the least tightly bound electrons, no matter how intense the light. The short delay time of experimental result (1) is also explained, because the absorption of a photon occurs almost instantaneously. Finally, thc morc intcnse the light. the larger the photon density. and hence the more electrons that will be ejected, thereby explaining result (2).

9.3

THE COMPTON EFFECT

The wave interpretation predicts thai when electromagnetic radiation is scattered from a charged particle the scattered radiation will have the same frequency as the incident radiation in all directions. Arthur H. Compton, in 1922. demonstrated that if the quantum interpretation of electromagnetic radiation is accepted, then the scattered radiation will have a frequency that is smaller than that of the incident radiations' and also depends on the angle of scattering. Cnmpton's analysis involved, in effect, viewing the scattering of electromagnetic radiation from a charged particle as a perfectly elastic, billiard baJJ type of collision between a photon and the effectively free charged particle. as shown in Fig. 92. Even though the details of the interaction are not known, conservation of energy and momentum can be applied. h is found that the scattered photon undergoes a shift in wavelength, /l}., given by .. .. uJ'.=J'. - . J'. = h - (, -cos 0)moc

(see Problem 9.27). The quantity II/moe is usually called the Compton wavelength; its value for an electron is 0.0243 A. Note thaI the shift in the wavelength depends only on the scattering angle () and is independent of the incident photon's energy. Compton verified his theoretical relationship experimentally by scattering Xrays (i. = 0.7 A) from graphite. The energy of the X-rays (1.8 x lif eY) is several orders of magnitude larger than the binding energy of the outer carbon electrons. so treating these electrons as free particles is a good approximation.

62

ELECfRQMAGNETIC RADlAnQN- PHffiONS

(CHAP. 9

y

~ ---QI-----~

. -'" ,-h,.

I

._ .. Jp- "I).'(b)AI~

a

...

x

lCfoueriaa

Fig. 9-2

9.4

PAIR PRODUCfION AND ANNIHILATION

In the process of pair production the energy carried by a photon is completely converted into maUer, resulting in the creation of an electron--positron pair, as indicated in Fig. 9-3. (Except for its charge, a positron is identical in all ways to an electron.) Since the charge of the system was initially zero, two oppositely charged particles must be produced in order to conserve charge. In order to produce a pair. the incident photon must have an energy at least equal to the rest energy of the pair; any excess energy of the photon appears as kinetic energy of the particles.

Heavy nucleus, 11.10(0)

o

Befort pair production

(b)

Aller pair productioD.

Fig. 9-3

Pair production cannot occur in empty space (see Problem 9.38). Hence, in Fig. 9-3 the presence of a heavy nucleus is indicated. The nucleus carries away an appreciable amount of the incident photon 's momentum, but because of its large mass, its recoil kinetic energy. K :::::: rll2Mo. is usually negligible compared to the kineric energies of the electron-positron pair. Thus. energy (but not momentum) conservation may be applied with the heavy nucleus ignored, yieldinghv = m+c 2 + m~ c? = K ...

+ K_ + 2moc?

since the positron and the electron have the same rest mass. mo = 9.11 x 10- 3 1 kg. The inverse of pair production can also occur. In pair annihilation a positron-electron pair IS annihilated, resulting in the creation of two (or more) photons. as shown in Fig. 9-4. At least two photons

CHAP: 9]

ELECTROMAGNETIC RADIATION- PHOTONS

63

must be produced in order to conserve energy and momentum. In contrast to pair production, pair annihilation can take place in empty space and both energy and momentum principles are applicable, so thatEinirial

= =

E rlJllll

0'

Pinir ial

Pfinal

0'

where k is the photon's propagaJion vector, Ikl = 2n/ A..

E. _m.cJ_mgeJ+ K ..

0--~

(a)

Before pair annihilation

(b)

Afw pair annihilallon

Fig. 9-4

Both pair production and pair annihilation can occur with other particles and antiparticles, such as a proton and an antiproton (see Problem 9.92).

9.S ABSORPTION OF PHOTONS The intensity of a beam ofmdiation will be reduced as it passes through material because photons will be removed or scauered from the forward direction by some combination of the photoelectric effect, the Compton effect, and pair production. The reduction in intensity obeys the exponential attenuation law(9. /)

Here 1 0 is the intensity of the radiation incident on the absorber and Jl (the linear absorption coefficient) is. for a given photon energy, a constant that depends on the particular absorbing material. For any given material, Jl will vary with the energy (or wavelength) of the radiation because different interactions predominate at different energies.

Solved ProblemsTHE THEORY OF PHOTONS9.1 Find the wavelength and frequency ofa I.OkeV photon.). ==- he = 12.4keV . j.. = 12.4;" E I.OkeV c 3 x l08 m/s 17 v=' = 12.4x lO IO m =2.42 x lO Hz

Am.

64

ELECTROMAGNETIC RADIATION- PHOTONS

[CHAP. 9

9.2

Find the momentum ofa 12.0MeV photon.Ans.

p =- = 12 McV /c

,

9.3

Calculate the fiequency of the photon produced when an electron of 20 keV is brought to rest in one collision with a heavy nucleus.Ans.Assuming all the kinetic energy of the electron is m;cd 1 0 produce the photon, we haveEi~" ...l = Erinal

K+4 = hl'+420 )(

Ill' cV =V

(4. 136

X

10- 15 cV . s)l'

= 4.84 )( lO u Pr

h,

e

which contradicts the momentum expression. Essenlially the same problem has been solved in a somewhat difTen.>nt manner in connection with the phOioelectric effect (Problem 9. 17).

9.26

Detcnnine the maximum scattering angle in a Compton experiment for which the scattered photon can produce a positron-elcctmn pair.

70

ELECTROMAGNETIC RA DIATION- PHOTONS

[CHAP. 9

Ans.

The threshold wavelength for positron- deenon pair production is (see Problem 9.3 1)0'

Substituting rilis resull in the Compton equation, v.e find

;: = ;. + 2i-t~( I ~ cos (})The righi-hand side of this expression is the sum of twO positive-definite terms. Hence, if2.!th (1 - cosO) 2: i" h

then J.' > i.'h and pair production cannot occur. Taking the equality.0'COSOth

\o\e

find for O'h:

=!

0'

Note thai Ihis resuh is indepcndem of the energy of the incident photon.

9.27

Derive the Complon equation, ;.' - A = (h /m4lc)( I - cos U).Ans.p

Refer In Fig. 9-7. The photon is treated as a particle of energy E = hll = he/i. and momentum = hI i.. From conservation of energy:he ;J ill" 2 -:- +mot =--;;- +mcI.

J.

Squaring and rearranging v.e obtain(mc-)J 2

=

112(.1 -2 -2 .,2 (.01,. A

+ ... )--. - ., ,., .

,,2

2h~ c?

211"'1)2 ., 2 2 +~ .. ,- (I. - ) ) +(mor)

t.,.

(I)

From conservation of momentum we obtain the vector diagnlln shown in Fig. 9-7. Since p" = p _ p',

Pc P.. = P, 2

= {) 2

h ,2{1. '2 +p' 2 - 2p . p' = . 2,

,

..2

2 - 2, '.. , ' cos 0 ) +,.

(1)

Substituting (I) and (1) in the relation (m~ )2 = (Pc, l + (mu~ )2 , we obtainhlt? 2 '2 2h 2cl 2hmoC'1'J' 27iI (J. +A. ) - ---;-y-+ - . - .-,- V. - ;) ). , . A: . . . . 1 .+{1II0() -

.2 2 _ h ,2 ,'2 2 . 1 .'1 (A + J.I. I.

2

- 2,.... cos )+(moc-)

. ,

0

"

Solving, we obtain the Compton relationship

., - ,. = 2 Sin . ern; " . =. Sill, 'I' , 2we obtainDsinq, = 111. Taking n = I. we then have (2.15A)sin50.0 = (IV,I. = L65A

10.17. In the experiment described in Problem 10.1 6. Davisson and Gemlcr used 54.0eV e lectrons. Determine the effective accelerating potential of the nickel cryswl.Ans.

The de Broglie wavelength of 54eV electrons ish h ;( = - - - = - - =., 1110 1'

he

JimoK

J2 (/I/,i 1)K

_ ..

=

124x IOJeV.A .. .. = 1.67A J2(0.5 11 x IOI'cV)(54eV)

This is diflerent from the observoo wavelength of 1.65 A. The kinetic energy corresponding to t. = 1.65 A is found from. 1 .= he/2 (mnc1 )K'Ji. _ - ' -- -, _~' _

(hd- _ ( 12.4 x IlreVA)' - 5 J 'V 2 - 5. t' 2(1110("2);,- 2(0.51 I x llY' cV)(1.65A)

,

",

Therefore, the eflcctive accelerating potential of the nickcl cryst:,jl isV,. = 55.3 V - 54.0 V = 1.3 V

THE PROBABILITY INTERPRETATION OF DE BROGLIE WAVES10.18. Determine the photon flu x associated with a beam o f monochromatic light of wavelength 3000 A and inte nsity 3 x 10- 14 W/ml.AI/s.

E = I/1' =-=

he

(6 .63x lO- J.l J s}t3xlO~ m/s)

3x l0 7 m H I 3x lO- J/sm1 III = - = - .. ------ - - = 4.5 hI' 6.63 x 10 .'> J/phOlon

i.

= 663 . x

I - " Jh 0 /p oton

X

~ photons photons 10 - - , - = 4.5 - , s msemL

On the average, 4.5 photons will strike a I em! lIrea tof photogra.phic fil m, say) during a period of I s. Of course. only integral numbers of photons can be observed. Thus. for a givcn I em! area.....-e might ObSt-'TVe 3 photons or 5 photons in a one-st.'Cond interval, but never 4.5 photons. Only if an average is taken ovt'r many intervals will the average number approach 4.5 photons. Also, for a givcn one-second interval, the arriving photons may cluster within a fixed I em! area. Only after a long period of time will the photon positions approach a uniform distribution. 10.19. Suppose h = 6.625 x 10- ) J. s instead o f 6 .625 x 10- ]4 J ..\'. Balls of mass 66.25 gmms are thrown with a speed of 5 m/s into a house through two tall, narrow, parallel windows spaced O.6m apart. the choice o f window as target being rando m at each toss. Detennine the spacing between the fringes that would be formed on a wall 12 m behind the windows.

CHAP. 101

MATfER WAVES

91

An.)".

The

6.63 x 10- 14 J . s

- 4nlO- 1 (5 x 10- 3 kg)(2 m/s)

5.28 x lO- lU m = 5.28 x 10- 20 A a value thai is clearly unmeasurable.

The minimum lIIlcertainty is 5.28 x 1O- 2-

(6.63

4nl0 )mov

=

x 10- )4 J . s)1 _(0.6)2

4n lO 3(9. 11 x 10- 31 kg)(1.8 x l ()IIm/s)

=257x 1O- lO m = 2.57 A.

The minimum lln(;ertainty is 2.57 A.

CHAP. 10)

MATTER WAVES

93

10.22. What is the uncertainty in the location of a photon of wavelength 3000A. if this wavelength known to an accuracy of one part in a million?Ans.

IS

The momentum of the photon is given byp= - =i.l.:

m.:

12.40 x J03 e V.A eV = 4.13 (3 x WI A)c c

The ul\l;cnainty in the photon momentum is (y."Orking wilh rnagnirudes only);IIp == - ~

I

1

hi' nA. =

AJ. PT

=P )( 10- , = 4.13 x

10- -

"V,

from which"'> --~--=

h - 417.llp

he 4ne!J.p

12.4x lO'eVA . = 239x 106 A = 239 mm 4ne(4.13 x 1O- ~cV/c) .

10.23. What is the minimum uncertainty in the energy state of an alom if an electron remains in this state for 10- 1l s?Ans.

The lime available for measuring the energy is 10 -~s. l1aereforc. from {I.6.1 ~ hl4n,~E

h > -- 417. 6.t

~

he 12.4xIOI eV.A 7 - = . ==0.329 x 10- eV 417.c 6.t 411'(3 x 1()8 m/s)( I 0 ~ s){ 1010 Aim)

The minimum uncenainty in the energy of a slate, r = h/(417.r), where t is the mean lifetime of the ~ci ted state, is 1:h by fJ~ h/ 411

= s/ R; therefore

I'lps 6s = (6 L/ R){R6.0) = ALl10

For a state of fixed angular momentum (e.g. an electron in a 130hr orbit, which will be discussed in Chapter II), the uncertainty in the angular momentum, AL, is 7..ero. Therefore, the uncertainty in the angular position, 6.8. is infinite, so that the position of the particle in the orbit is indetenninale.

10.27. Ifwe assume that E = !1t = !lX/v.Ans.

! mv for a particle moving in a straighlline, show that!1E !1t :::. h/4rr, where - 1 J _ (mv)2 _ p2 m - 2m -2m -2

Taking differentials of both sides of this expression. we obtainI1E =-~-- =v l1p m m

p6p

mv6p

Then . from 11p/h .:: h/ 4n.

M h - /h > v - 4n

0'

h 66.( > - 4.

10.28. A particle of mass m is confined to a one-dimensionalline of length L. From arguments based upon the uncertainty principle, estimate the value of the smallest energy that the body can have.Ans.Since the particle must be somewherc in the given segment. the uncertainty in its position. 11x. cannot be greater than L. If L\x is set equal to L. the uncertainty relation L\x6p,, ':: h/4n in tum implies that the momentwn must be uncertain by the amount 6.P. .:: h/ 4nL. We are looking for the smallest pOssible value of the energy and hence. s ince K = p" 2/2m. the smallest possible 11',,1. We identify the uncertainty in 11',,1 with that in p". and assume that the uncertainty interval is symmetrical about 11',,1. Then (see Fig. 10-8).0'

I ( -- > , 6.P > (P"I } I "I - 2 4nL - 8nl-

h) h

01

14

4\,.1

1,.1Fig. 10-8

1

.

CHAP. IOJ

MATTER WAVES

95

Thus, the minimum magnitude of p~ is hl8nL, and

This value compares reasonably well, considering the crudeness of our argument, with the value

h' EI =8mUfrom Problem 10.20. The rcsuh further illustrates that, under the uncertainty principle, bound systems cannot have zero energy.

10.29, Calculate the minimum kinetic energy of a neutron in a nucleus of diameter 10Ans.

14

m

The s ituation is that of Problem 10.28, with L equal 10 the nuclear diameter. Thus,

Krold

= 2m

I (

8nL

h)'= 2(mc2) I (he )' I [12.4 8n(10-'" 10- ' M,V . A]' 8nL = 2(940 MeV) A) = 0.013 MeVx

10.30. If an electron were in the nucleus of Problem 10.29, what would be its minimum kinetic energy?Ans.

For an e1ed:ron. a relativistic calculatio n is nece~ry. As in Problem 10.28, the minimum magnitude of the momentum is

(Km., + Eoi = (lplmind +E02(Km ;,,+0.5 11 MeV)2 = [ 12.4 x IO- J MeV 4' 810(10- A)

ft. ]' + (0.5 11 MeVl

Solving, KrN~ = 4.45 MeV. When the emission of electrons (#-rays) from nuclei was first observed, it was believed that the eled:rons must res ide inside the nucleus. The energies of the emitted d edrons, however, \\.'erc often a few hundred keVand not the minimum 4 MeV predicted by the foregoing calculation. It was therefore concluded that elIXlrons are not nuclear building blocks. ( See also Problem 17.1.)

10.31. The position ofa particle is measured by passing it through a slit of width d. Find the corresponding uncertainty induced in the particle's momentum.AilS.When monochromatic waves of ....'3velength J. pa~ through a slit of width d, a diffra97 x 10- A - ) 2 - ' II, f1;;

1 0

7700 A).

I.

,

, '(' ')

In Problem 11 . 1 it was found thaI when II/ = Ithcwavelcngths r:mgc from 9 12A lo 12 1S A, so mal none of these lie in the optical region. For tit = 2 the longcst wavelength corrc:;ponds to nil = 3, givmg

;. = 6563 Aand the shonest wavelength corresponds 1 0tI"

=

0:::,

giving

--:- = (1.097 x 10 - A- ) - - I.

,

,

'(' ')22 :)(:2

i. = 3646 A

Hence, some oflne wavelengths in the Ualmcr series (tit = 2) lie in the optical tegion. To detennine these wavelengths set i. = 3800 A and solve for II".- - - - = (1097 x 10 3.R x lOl A '

,

-, A

' ('

) ---4 II~

,)1/"

0'

flu

= Q.9

Therefore. the lines in the optical region arc given by'7

I.

,

= (1.097 x 10- A ) - - 2 4 tlu

,

' (' ')(II{

= 3.4 . 5..... Q

S ince the

shone.~1

wavelength of the Paschen series

= 3) is0'

j = (I Ji97x IO -, A

,

-,(3 1-

, ')00 2

;.= 8200A

all other series will give rise to lines lying out'> idc the optical region.

11.5. Evalute the io nization potential of hydrogen. Fi. in units of eY.Am.2n!k 2e4m

Ej11.6.

=

h2

=

21! ~ (.I.. ~l {mc2 )(IId 2

=

2n: 2 ( 14.40cV Al 2(O.51 1 x Ifl' eV) (12.40 X If)J eV. A)2

1J.6eV

Find the wave!enb>th o f the photon that is emiut.xI w hen a hydrogen atom undergot.-s a transition from nIl = 5 to nl = 2.Ans.

From the Bohr model, the energy k:\"cls arc EN = ( - I J.6eV }/n~ . Hence,

E2 =

IJ.lleV -----v--= - JAOeV

5 =

--,-

IJ.6eV = - O.544eV 5-

From the Bohr posrulates the cnergy of the emitted photon is E;

= - O.544eY -

(- 3AOeY)

= 2.86eV

The wavelength of this photon is given by. he 12A x lo-~ eV A /. = _. = - - _._-_ ..- = 4340 A E.. 2.86cY

CHAP. IIJ

TH E BOHR ATOM

'09

This problem can also be solved using Rydberg's fonnula.

, = R(' ') = 1.097 x 10 _ ,(I ') ;. II; - liZ 22 - 52Solvi ng.';' = 4340 A.

11.7. Dclt:nnine the ionization energy of hydrogen if the shortest WeT. the highest state that {;an be reached corresponds to n = 3. Hence (Problem 11 .8) there arc three possible wavelengths that will be emitted as the atom returns to the ground stale. conesponding 1 0 the transitions 3 -+ 2,2 -+ I and 3 -+ I. These wavelengths are found from

- = (1.097 x 10- A- ) - - - 2 A 22 3

1

, '( 1 1)

0'

i.=6563A;= 121 5.4.

0'

0'

). = 1026A

1I.11. According to the Bohr Iht:ory, how many revolutions \\/i ll an electron make in the first excited stale of nydrogcn if the lifetime in Ihal slate is 10- 11 5?AilS.

By (lIA) and (11_0, the radius and orbital velocity for the state n = 2 are given byr2v2

= 4{0.529.4.) = 2_ 12 A = 2. 12 X 10- 10 m II. e 3 x I()'!mjs if ="'2 = 2(137) 2( 137) = LID x I m/s

= 4ft

The angular velocity is IhenIf)

= - ='2

V2

LlO x IWm/s 10 2. 12 x 10- m

= 0.52

X

10 rad/s

16

and the total number of revolutions isN =(I)/

= (0.52 x lOll>rad /s)( I 0- 11 s) = 8.3 x Iif rev6.2R rad j rcv

2n

11.12. Dt.1ermine the com . "Ction to the wavelength of an t.-mitted pho ton when the R:coil kinetic energy of the hydrogen nucleus is taken into account.Ans.

Assuming that the atom is initially al rest. wnservation of energy gives

E,, = E,+E, +K

0'

E,, -E1_ R: = Khe he he

where K is Ihe nuclear kinelic energy. The first term on the left is 1/.1.0, and the second is 1/ A, where j-{l and t. are the uncorrttted and actual wavelengths. Thus1;'0

- J:= he

K

-- = }0(1

l-Ao

lK he

The recoil momentum of the nucleus is p

= J2MK. Then, by conservation of momentum,0'

K = -2M ).2

.'

CHAP. II]

THE BOHR ATOM

III

from which12.40 x IO l eV A 2(939 x l()f>eV)i. 6.60 x 10 - 6 A

Since the wavelengths are of the order). "" iO l therefore negligible.

A.

the fractional change is of the order 10- 9 and is

11.13. For hydrogen, show that whL'I111 I the frequency of the t.mitted photon in a transition from II - I equals the rotational frequency.Am. TIle rotational frequency in state n is

II

to

w v" 2nh? 11111 4n 2k 2me4 2n;:' 2nrn ;:. 2nn 2 h2 j4n Zkmfil;:' nl li lThe frequency of the emitted photon is

I'=C~=CRoo [l

(n _ 1)2

I

- .n!l. ]=CR '>i; nl(n 2n - 1 - Il

Forn I,

which is the same as the rotational frequency given above. This problem illustrates Bohr's corresponding princil,Ie, which states that for large II a quantum equation should go O\Ier into the wrTesponding c1a.~ica' equation. According to classical theory, radiation emitled from a rotating charge will have a frequem;y equal to the rOialional frequency.

11.14. An electron rotates in a circle around a nucleus with positive charge Ze. How is the electron's velocity related to the radius of its orbit?Ans.Equaling the Coulomb force to the (electron mass) x (centripetal ac{;clern.tion),

11.15 How is the total energy of the e1t:clton in Problem 11.14 related to the radius of its orbit?Am.

The e lectrical potential energy of the electron isI.)

=qY = (- e)Y = -e - - - = - --

k(Ze)

,

kz.e2

,

The kinetic energy of the electron is foond by using the resuU of Problem 11 . 14:

2 kzil K =1 mv2 = :m kZe __ = __ 2 " mr 2rThe total energy is then E=K+U = - - - - - = ._-- =1 1.) 2r r 2r 2

kZe>

kZth of the H" line (3 -+ 2 An.\. (tI) -6.8eV. -1.7eV. -0. 76eV; (b) 131 3 A transition) of posirronium.

Electron Orbital MotionA~(;ULAR ~IO)IENTUM

FROM A CLASSICAL VIEWPOINT

(""";(,)

,,,- 1

.-,

.________________ -

OE-------------OE ______________

Ed""

, ,!.

-

OE - - -OE -

- - - - OE OE(')

, ,

.-,

$

.-,

Withoui mapelic field

Willi magnetic: field

Fig. 14-4 Ans.

After the external magnetic field is applied, the new value (E;) of each particlc's energy equals the original value (E,,) plus the interaction energy:

the new levels wi ll be displaced from the old hy an amount (sec Problem 13.71. Since m. = 6. = c"8j2m, with a spin - ~ particle occupying the lower sublevel and a spin +~ particle

CHAP. 14]

THE PAULI EXCLUSION PRINCiPLE

\39

occupying the upPer sublevel [Fig. 14-4(h)J. In CQntrast to the sitlJlltion withoulthe ITIlIgtletic field prest:nt, the particle in the n = 2 level will havt: its spin - ~ if tht: system is in the ground statt:.

14.4.

In a one-dimensional infinite square \vell of length a there are 5 x 109 electrons per meter. If all the lov.'CSt energy levels are filled detennine the energy of the most energetic electron.Am'.

Since there are 1'....0 electrons in each energy level. tht: total number of electrons up to and including the last or- nth level is N = 2n. The number of eI(.'Ctrons per unit length is tht:rcfore-=-= S)(IOm~

N a

2n a

'J

1

so that

~ = (2.S

x

I09 m~ I)(IO- l(Ji) = O.25A~ 1

The energy of the nth t:nergy level is thus

En = 8 - rna

n2h2

(2 "~ 1 2 (1 2.4x I WeV . Al 3 ,= (n)2(hd - 8( m{-' )= O. SA ) " (0511 a . x I~ \F' e VI =2. SeV

14.5. If a nucleus is approximated by a one-dimensional square well, there will be about I neutron per 10- 15 m. In this approximation, detennine the energy of the most energetic neutron. The neutron rcst mass is 938 MeV.Am.

Refer to Prob1t:m 14.4.

Supplementary Problems14.6. What is tht: energy oftht: photon that would be emitted in a transition from the n infinite square wdl of Problem 14. 1? An.... 5.20eV

= 3 to the II = 2 level in the

14.7.

4 m. What are the values of the first three energy levels A one-dimensional inlinitt: square well has length 10- 1 An.f. 2.05 MeV: 8.20 MeV; 18.45 MeV for a neutron (m n = 938 MeV) in the '.vell? Rept:at Problt:m 14.7 for an electron in a I A well. Repeat Problem 14.2 for neutrons.

14.8. 14.9.

Ans.

37.6eV; 150t:V; 338eV

Ans.

6ISMeV; 1024MeV; 1634Mt:V"flip~

14.10. Determint: Ihe ent:rgy n.'"quired 10 cause the uppermost electron in ProhJt:m 14.3 to Ans. 2.32 x 10- 4 t:V magnt:lic field is 2T.

its spin if the

any-Electron Atoms d the Periodic TableCTIRO:SODPIIC NOTATION FOR ELECTRON CONFIGURATIONS IN ATOMS

deal of infonnation about the character of many-electron a10mic sta1cs can be found by a furs! "pp,-oxi" ma,;"n thai each electron moves independently in the field of the nucleus and the produced by the other e1ectrons. The other existing intenctioos are treated separatefy, as will In such an independent particle nwdel the quantum numbers n.l. m,. and m, (lIe usedelectron's state. n, the integer wlues that the quantum number I can take on are

1=0,1.2..... -1win be designated by a Iomrrcase letter according to the following scheme.

Valueof/: Letter symbol :

o

1 2 3 4 5 spdfgh

foi' specifying the number of eJeccrons in a particular orbit, defined by the quanlW11 to give n, followed by the letter symbol for I, with the number of electrons as 8 post.. orbits are sequentially written one after the other, thereby defining an electron an example. the configuration for the five electron h2 2s2 2pt. 3$2 3pt. 1.\.2 2s2 2p/' 3$2 3pt. 4s 2 3d lO4p b I.r 2s2 2pt. 3$2 3!1' 4.r 3dlO 4,1' S.r 4d IO Spb

2 10 18

3654

15.12. Alkali metals have one electron more than a noble gas. Give the electron configurations of the firstfour alkali metals.Am.

When a panicular lip subshell (that is, the p subsht!ll of a particular shell II) is completely filled, corresponding 10 onl! of the noble gases, the next-higher-Z a tom will have an (II + l )s I!ll!Ctmn added 1 0 the noble Sa!) core shown in Table 15.4. Thus we have the ground-state electron configurations given in Tabll! 15-5.

CHAI). 15J

MANY-ElECfRON ATOMS AND THE PERIODIC TABLE

149

Table I5-SNoble Gas Core

Alkali Metal

Electron Configuration

U(Z = 3)Na(Z = II ) K(Z = 19)

Rb(Z = 37)

Ir2s ' I;: 2f2 2p6 3s 1 Ii' 2~ 'l/I' 3.r 31' 4S1 I.~ 2~ 21' 3.r2 31'4s 2 3d 10 4,1' ~. I

H, N,A' 3"" " 3d' )r z.t12,t 3s 2 31' 4 r 3d}

15. 15. Knowing that and

ILs

=

-2G:)S

show in a ve(.1or diagram that IL and J are not parallel.

150

MANY-ELECTRON ATOMS AND THE PERIODIC TABLE

(C H AP. 15

IJI-VJ(J + I) A

Fig. 15-5Ans.

The vector relations J = L

+ Sand f.l =

POL + f.l.~ are shown in Fig. 15-5. Because

IfLsl = 2' ....1 lSI ILlthe two triangles are not similar. and .... and J are not parallel.

15.16. Refer to Problem 15. 15. Calculate the projection of the tolal magnetic moment vector fL on thevector J.An,f.

From Problem 15. 15.p.

=

ILl + J.I. ~' =

.

.

-2m - (L + 2S) = - 2m - (J +5)

,

,

The projcction of .... o n J is

fLJ = (_.!'..)JJ + J SIJINow,

2m

IJIoj

.,Therefore.,

L L = (J - S) . (J - S) = J

+5 , S - 2J S

J . S = ! (J . J

+ S S -

L L)

.... . J =

(_.!'..) J o j + i (J' J +5 5 - L L )2m

lJI~

IJI

(_ .!'..) J(J + 1),,2 + HJ(] + 1)111 + S(S + nil! - L(L + I)h!) 2m ,/J(1 + I }I! ~ ( _ ") ./1(1+ + 1(1+ I) + S(5 + I) - L(L+ I)] 2m 2/(} + I)

1)[1

;;i ( -

;:')./1\1 +, _ , 7 -

1),.

The quantity

+

J(1+ I)+S(S+ I) - L(L+ I) 21(J+ 1)

CHAP. 15)

MANY-ELECTRON ATOMS AND THE PER IODIC TABLE

15 1

is called the Lundl~ y-jact(Jr. As will be seen in the following problems. the y-factor is needed to calculate the relative spliuing of different energy levels in weak magnetic fields.

IS.17. Detennine the value of the energy splining of an atom in a magnetic fie ld 8 if il is assumed that the splitting depends only on the component of ... along J .Ans.

From Problem 15. 16, the componcnt of I' along J isJI) = ( - eh)JJ(J

2m

+ I)

y

or. in vector nmation,1'.1 = JI).:!. =

IJI

(-2"m ) JJ(J + I) y JJ(J + I) h : ; - -2ma 1~2 3p6 4.s2 3d lo 4pf> 5s 2 4d l0 5p 6 6s 1 ( s~ Cs)

15.29. The alkaline ('orth have two electrons more than a noblc gas. Give the electron configuration ofthc first four.Ans.

Bc(Z 4): Mg(Z = 12) : CalZ = 20): Sr(Z = 38) :

=

1.~2rI s2 2~2 2/l3s 2 b 2 2~ 2p 63.5 2 3p6 4;: IS2 2s2 2/l3s2 3p6 4i 3d lO 4/16 5.s2

15.30. Give the electron configuration for the halogen with one ek-clron less than the noblc gas Xc.Ans.

I.; 2~2 21'(' 3.\'2 31'6 4...2 3dllJ 41'6 5.~2 4dJlI 51'S(H I)

15.31. What transi tion clement has thc electron configuration I s2 2s1 2p{' 3.~2 31'6 3cf' 4s1?

Ans.

26Fe

15.32. Givc in SpeCtral notation the pOssiblc states of an atom whichAns.

ha~

a closed corc plus onc d elcctron.

!D~/2: l D ,l !2

15.33. Listthc possiblc states (M,., Ms) of l\\IO equivalent nd electrons,Ans.

The 45 allowable combinations of quantum numbers (10 things taken 2 at a lime) yield 23 stales: ( 4.01. ( 3 .11. ( 3.0). ( 3. - 1), ( 2. I). ( 2.0). ( 2, - 1). ( I. I ). I I.O). ( I , - I), (0. 1). 10,0), CO, - I I.2S 1/2 ~1atCS

15.34. For a field of2T. calculate the Zeeman encrgy splitting ofthcAn,\',

2 P I/l

and

in Na.

3.86

)1ates for the e lectron transition, arn.l Z*(' is the net ~ilive charge acting on the e lectron. For K2 and 1-2 Irdnsition."+"_1

(2)

the electric quadrupole moment reduces to

.2 = 2Z (I>' _ il) 5

(3)

wheR: Ze is the lotaJ nuclear charge. If the average nuclear radius is taken to be ~ = i?b (the volume of the ellipsoid is jruz2b), with Ro + ~ Ro = h, show that the electric quadrupole moment is

II =AM.

~ Rle~o)Ill,

If b = II, + 611, omd Ill, =

a'b, !hen, fo, 611, R,. a' =R., + ~Ro

---.!'L ~ R!(' _DRo) I + bRo RoR,

b' - h

1lI, [, +2e~') + e~'Y] -R!(' - ~)

=R![3e~) + (~)'] ~ 3R!e~)Hence,

17.5. For I~Gd the quadrupole moment is 130fin2. If Ro is given byAm.

Ro =

(1 .4fm)AIf), find

lJRoIRo.

The average radius is

Ito =

(1.4fin)A11l

= (1 .4 fm)(15S)IJ} = (1.4 fmX5.J7) = 7.52 fin

178

PROPERTIES OF NUCLEI

[CHAP. 17

From Problem 17.4.

:i

=

6Z Ii:. (MO) 5 Ro0

130fm2 = 6(64) (752fm)15

(bR

O)

Ro

OR _ 0 = 2.99 X 10- 2 = 2.99% N,This shows that for 'llGd the nucleus is almost spherical, deviating from sphericity by only 2.99"1u of the average radius .

17.6. Delcnnine the possible stales of a deuteron if its total angular moment has quantum number J = 1.An.f.

The total angular momentum (J) of the d euteron is the vCClouum orthc orbital angular momentum for the neutron- proton bound system (L) and the total intri nsic spin o flhe neutron-prolon system (S). Since both neulron and proton have spin S die lolal intrinsic spin is 0 (singlet statc) or 1 (triplct stale). Since J = L + Sand J = I and S = O. 1, the o nly possible values for L arc 0, I and 2. In the spectroscopic notation of Section 15.3, the possible deuteron states are lSI' J P1 ' I P I and 3 1, The ground slate of the deuteron is a mixture of . 1SI a nd ]D I .

=!.

17.7. Determine the radii of a 1"'0 and a 208 Pb nucleus.Ans.

From R = roAl j l = (J.4fm)Al fl ,Ro = ( 1.4fm)(16)1 /l = 3.53fm

Rpt,

= (1.4 fm)(208 )l l = 8.29fmj

17.8. Determine the approximate dens ity of a n ucleu s .If the nucleus is Irealed as a uniform sphere,

. Dc m; lly

= ---~

mass

volume

A x (mass ofanucleon) A(!.7 x 1O- 27 kg) = = 15 x j nR3 j n( 1.4 x 1O- ISAlfl m)l .

A cubic inch of nuclear matcrial would weigh about I billion IonS!

17.9. Determine the stable nucleus that has a rddius 1/3 that of 189 05.Ans.

Since R ex: A I f3 .

A ~-~ 7

18927

corresponding to lU.

17.10. Calculate the binding energy of ' ~~Te.Ans.

The binding energy is given by BE = (Zmr )c2 + (Nm~)Cl - Mnur:c'-

=

(52 x I J)07 825 u + 74 x 1.008665 u - 125.903322 u) x 93 1.5 MeV / u = 1.066

X

103 MeV

o r i.066GeY.

CHAP. 17J

PROPERTIES OF NUCLEI

179

17.11. What is the energy required to remove the least tightly bound ncutron from ~Ca?Ans.

From conservation of mass--energy,

M~:.l? + E = (M" Ca + m,,)C (39.%2 589u)(931.5MeV lu ) + E = (38.97069 1 u + 1.008665 u)(931.5 MeV l u)

E = 15.6M eV

17.12. Dctennine the electrical potential energy of the protons in a nucleus if it is assumed that the charge is unifonnly spherically distributed.Am.

Con. remove a neu tron (it ha~ two neu tron pairs).

CHAI~

181

NUCLEAR MODELS

187

IS.S.

" Mirror" nuclei have the same odd value of A, but the values of Nand Z are interchanged. Determine the mass difference between two mirror nuclei which have Nand Z differing by one unit.An.f.

The term A - 2Z in the scmicmpirical mass formula can be written as A - 22 =N +Z - 2Z =N-Z so that if Nand Z differ by one unit, A - 2Z = J. If we now subtract the tv.'O masses M z rrom each other. the (A - 2Zlterm will cancel, Icaving. for constant A =- 2Z + I,M z ,.,

+, and M z

-

MJ'.

= (fII,l - m")[(Z + I) -

Zj

+ blA ~ I J3 [(Z + 1)2 -

Z2 )

= mp -

m"

+ b]A2/1

IS.6. The masses on1Na and H Mg are 22.989771 u and 22.994125u, respectively. From these data determine the consmnt h) in the semiempirica l mass formulaAns.

The two nuclei arc mirror nuclci. From Problcm 18.5,My. .. 1h] = 6.42-

My' = mp - m" + h1A2/3

22.994125 u - 22.989771 u = 1.007825 u - 1.008665 u + b3 (23 )2/3X

10 ~ 4 U

= 0-598 MeV

18.7. From Problem 17. 12, the Coulomb energy of a nucleus is. for large Z,Ec = - -. 5 R

3 kZ'e'

CalculateAns.

"1 in the semiempirical mass fonnula, taking ro = 1.5 fm.Et =3 kz2 il Z2 5" roA' j) = b1 A1 0

For a nucleus. R = roA ' i3 , and the Coulomb energy is

Therefore.

h]- '5-;:;;- -

_ 3kt? _ 3(1.44MeV.fm) _ O 5(1 .5 fm) - .58MeV

If ru is taken as 1.4 fm, the value of h3 becomes 0.62 MeV. These answers agree rea...anably well with the value of hl found in Problem 18.6.

18.8. Using the liquid drop mode. find the most stable isobar for a given odd A.An.~.

For odd A. h5 = 0 in the scmiempirical mass fonnula and the binding energy is found to be BE = h ,A = b 2A 2/3 - b]Z2A- l fl _ b4(A _ 2Z)2A- 1The most stable isobar (A = constant) is the one with the maximum binding energy. This is found by setting d(RE)/dZ = O.

U~ing

b1 = 0.58 MeV and b4 -= 19.3 MeV gives

z ,IS.9. For A = 25.43,77, find the most stable nuclei.

0.015A2J3

A

+2

188

NUCLEAR MODELS

[C H A P. 18

Ans.

The result of Problem 18$ gives, for A = 25,~

A

Z = 0.01 5 A2/l

+ 2 = iO~O :C'::5'X::-25':)" 'I"'+-::: 2=

25

11.7 "'" 12

and U Mg is in fact l>1ablc. It is li)unJ experimentally lhal nAI and RNa arc not stab le. For A = 43,

z=

_

43(0.0 15 )(43 )21. 1+ 2-19.7 ::::::20

and it is found experimentall y that ~~Ca is stahle. while for A = 77,77

1lKand ;~Sc arc unstable.

Z= and it is found experimentally that

(O:: ~ _O :: ' :C 5J::: (77 )2/.l + 2 == 33.9 "'" 34

USc is stable , while H As and j~Br are unstable.

18.10. Show that in an o rbit of givcnj there may be al most 2) + I nucleons. Demon::.1rate that for p stales (J = I) this is consistent with the fact that the Pauli pri nciple allows 2(21 + I) = 6 nucleons.Ans.

For given ),1111

= jJ - I .... , - (j - I). - j

a tota l of 2) + 1 values. Therefore the Pauli principle allcw,'s 2) + 1 nucleons in the orbit. A p state is split into a Pll2 orbit. which may contain 2j + I = 2(3/2 ) + I = 4 nucleons. and a PI /2 orbit. which may contain 2j + I = 2( 1/2 ) + I = 2 nucleons. The 1 0 Iai is 6 nucleons.

IS.11 . For A = 50 the known masses arc: ~~Sc. 49.951 730 u; ~Ti . 49.944 786 u; ~~ V . 49.947 164 u; ~~C r. 49.946055 u; ~~ Mn, 49.954 215 u. From these data estimate the constant h 5 the strength of the pairing leon. in the semiempirical mass fonnula .Ans.

For fixed even A and for Z odd (whence Z + I is even. elc. ) the semiempirical written asM(Z) = tllZ lM(l

ffia n.:ul nm ~hclls II.":. (tlc C ll C I-g~ ~ap

(~'rr~spnndil l~ to IWUl ron lllil!;l!.: Ilulllhl'r ::!UJ.

CII .- \\'.

uq

:-\1 n.J .A\{ :o.lol>n.s

Iljl

..til.'

~,~{ ' OJ i!l.kl:; a nl'UlfIR., _ 1.\1." . + nlr, -

\I_(" . ff

1 3t' . '):'fI("~ln

1 1.0IJSM.S uk~

:i'J .IJt.. .:':"i~luI19.115T\1c\' f u) _

1"i 6.:':\k"

\\hllc the hm.lmg en .. ~y Hflhe 1/: ._ ncUlmn III "l 'a, ~

IU

(.,!....

. III ~

.\1.a

- L'n. ,i. tlfthc 1/.: an,\ 101 , : shells:1 5h~\ k\ '

,) .- BE .

IU:: .

KJrd ..k\ _ 7 2hf\.J,:V

HUb. C'lIlsldcr OJ .. hell Illudel ill ",,hid) the nudellns ar.. in IlOli r.- III e'-l.ually .. pan:d ..:nerg)" le\"ds. If one starh w,th Ihl' sallie tlllmocr ,.1" neutmns and pmhms. call-ulah: the encr!!y needed to ..:hange n proton pilirs 1 0 nt:llln.ns and InIln.' ,h"lIl t l) Ileutron orhits .. 11/1' Ih,' rrnt>km I .. tlht,mnll'd In I ;g. 1 ~-' . Ifl\1l' lilldlllue iLns is Itl ha\c X J~U ln'l l'i and X prolons. wt' .,klin~ 1/ _ ;\' 1, whICh I. . 'n jl (" Ih~' l1umhcr "I' IlUcknl1 Ilall'S ' 0 b )( 11~

1)-2.)[1 - J - 5 I -

'lll -- Il]

.., I.\' - Xr - ..,t .-l

,

~

~n

,

Thi.. Icnn i... dinxt l ~- r.:!;II.:,lwlhe Io~ t.-I .- 2/ 1~.-I

It.'rm in th.: v.:nllemrmelll Jl1:t';;S formula. which , ...

;m lX pI\.'loo!>lOoO liof I:'.: ncliUnl! ... r prl,ltlll e.'llt.~' ln.:tg}

~

j,

~

-4

';' '" '"

;;

~

~ TT

j,

~~

';'

'-;rolon: *'

,.,..

T';'

",,"1rOIll>

'"

:E '*'

j,

'*' ! Tj,

'*' !Tj,

:'.2'

:cmro!FiJ;:. 18-7

:roto!

~eu~

Supplementar~'

Prohlems:-;:;:IIf!_

IH.1 7. l alnllale th..' ... ",.Iint: en~'rg~ ' ..... r Jllldl'''11 flOr III J ~ I It.-. Ih l ~ l . I,) ::{ ' ;t..,1",.ta) : IE\!.:\, :

(,I)

t/1I : h t' lI.k\': 1" o f nucleons berore ,lilt! oncr a decay mlL..t be equal.RADlQ-K1lH

apply. In nuclear decay. however. it is found that a law of conservation of nucleons also

m:cw

LAW

In a

radioactive decay an initially WlstOible nucleus. C'.dlcd the parent, emits a particle and cOilled the da /lgh/t'r:. cfTt..'Ctivc1y. the birth of the daughter arises from the death of the may be either Ihe same nucleus in a lower energy $tatc. as in thc case of i'-dccay. or an as arises from ~- and /l-tlecays. No mailer what types of particles are cmincd all the ~me mdirmcli\'t' 11t'I'ay 11l1l~ If there are initially No unstable parent nuclei ",,,,,!>,, N of parents that will be left after a lime t is (Problem 19. 1)( /9. /)

io/. " , IICd thc

c1f-'tY1Y ('tJlUIIlIll

or d b;;llIt'gl'flliOll ( 'Ol/Slalll and depends on dlC particular decay

Equalion ) is 01 slalistical. nol a dClcnnin istic. law: it gi\'cS the (xpccted number N of parents thai """" eafter 8 I. Howcver. if Nil is vcry largc (as it always is in applications), the actual number and CJq>Il O. so thai cm:rgy is released is called an e.'WtlJermic or exoergic reaction; lhe reaction can occur even if bolh initial part icles arc at rest. If Q < O. energy is absorbed or consumet.l and the reaclion is calk'll endothermic or ent/(Jergic; the reaction cannol occur unless the bombard ing particle has a certain threshold kinelic energy (sec Problem 20.8 ). If Q = 0 and iflhe particles are the same before and after the reaction. we have an ('las/i(- ct)lh~ion.

20.5

NUCLEAR CROSS SECfIONS

When a target material is bombarded wilh incident part icles to produce a nuci... -ar reaction, Ihere is no guamntcc that a particu lar bombanling projectile will interact with a target nucleus to bring about the reaction. A cmss section,". is a quantity that measures the probability that a nuclear reaction will occur in a given region of large1 mah..-rial. II is defined as

" ="i-""-';,c..:.= 'C;'-= "7== = "7= =---C number of projectiles incident per second per areaThe larger the value of ", the more probable will it be for a particular reaction to occur. A cross section has the dimensions of area and is usually measured in tenns of a unit called the barn. where I bam = 10' 2)( m 2 so that one barn is of the order of the squ~rc of a nucle-dr radIUS. If the number of target nudei per unit volume in a material is n. the number N"" of particles scanered when a beam of Nu project iles is incident on a thickness T of the material is (sec Problem 20. 12)

number o f reactions per second per nucleus

Nsc = Nu( I - e- n(}T)Cross sectionS will be diftcrent for diflcrent reactions. and for a given reaction will val)' with the energy of the bombaru ing particle. If the reaction is endothermic. the cross section will be Lero if the energy is below the threshold value.

20.6

NUCLEAR FISSION

One of the moot pmctical nuclear reactions is the formation of a compound nucleus when a nucleus with A > 230 absorbs an incident neutron. Many of these comp()Und nuclei will then split into two medium-ma"s nuclear fragments and additional neutrons. This type of reaction is called lIudearfissj(Jn. In a nuclear reactor. the number of fiss ions per unit time is controlled by the absorption of excess neutrons so that, on the average. onc neutron from each fission produces a m .:v. fis."ion. The libemted heat is used to make steam to drive turbines and genemtc electrical power. If the real,1ion is uncontrolled. SO that each fission results in more than one neutron capablc of producing further fissions, the number of fissions will increase geometrically. resulting in all the ene rgy of the souree being released over a short time interval. producing a nuclear bomb. A typical fission reaction isB5lJ ' 92 + on

,'M'UJ' '12

-

"X l,

+" 'll z, Y + f;n

with ZI + Zz = 92. A I + A 2 r E = 236. and /: an integer. The mtio of the masses ofthe fission fmgments. MI 1M2, is found experimentally to be roughly 3/2. The number /; of neutrons released in the fission ing of a particular dement will depend upon the final fragment s that are produced. For the above reaction the avemge number of neutrons released in a fission is found experimentally to be about 2.44, the &detional number resulting from an average taken over all reaction products. The two decay fragments usually have a neutron- proton ratio approx imately equal to that of the original nucleus. Therefore. in Fig. 20-3. they lie above the stabil ity curve, in a region where nuclei are

CHAP. 201

NUCLEAR REACfIONS

211

N

t'issioo

Stable nuclei

zFig. 20-3

neutron-rich and undergo bela decay. Usually it will require a chain of several beta decays, each decay reducing thc N /2 ratio, bero!\: a ~table nucleus is reached (Problem 20.23). A fission reaction liberates about 200 MeV of energy for each fission (Problem 20.24). This is much greater than the few MeV relea'>ed in a typical exothermic reaction where the final products include only one particle comparable in mass to the originallarget nucleus. This 200 MeV is distributed approximately as follows:(a)

170 MeV is kinetic energy of the fi ssion fragmenls 5 MeVi~

(b)

the combined kinetic encrgy of fis~ion

neutron~

(el(d)

15 MeV i ~ and V -ray energy 10 MeV is neutrino energy liberated in the

tr

Ir

decays of Ihe fission

fragment~

In many fission reactions the formation of the compound nucleus occurs most readily with thermal of energy ::::: 0.04 eV From the above it is ~een that the neulrons released in a typical fission reaction have large kinelic encrgies of about 2 MeV How these fast nculrons are slowed down to 'ilcilitate further fission~ is demonstrall'd in Problems 20.18 and 20AO.neulron~

20.7

NUCLEAR .USION

As implied by ils name. thefilsion reaction is one in which "",0 nucleons or relatively light (A < 20) nuclei combine 10 form a heavier nucleus, with a resulting release of energy. An example of a fusion reaclion is the formation of a deuteron from a proton and a neutron:

'H' I +0 11 -+ ' IH

Q = 2.23 MeV

Another fusion reaclion is the formation of an a-particle by Ihe fusion of two deuterons:

Q= 23.8 MeVAllhough these eOl.'J"gics are much smaller than the energy released in a typical fission reaction (:::::200 MeV), the energy per unit mas~ is larger because of the ~maller ma sse~ of Ihe participating particles. The release of energy in fusion can be underslood from Fig. 18 1, which shows that fo r light nuclei the binding energy per nucleon generally increases with in,,:reasing mass number A. Consl.",uently, the heavier

212

NUCLEAR REACTIONS

(C HAP. 20

nucleus fonned from the fusion of two lighter nuclei will have a larger binding energy per nucleon than either of the two original nuclei. But higher binding energy mean ~ lower rest mass (Section 18.1). and the lost Test mas~ appears as released energy. The reactions that seem most promising for use in the first practical fusion reactor are the D - D reactions

~H(d. ")~He~H(d,p)~ 1-1and the O-T reaction

Q = 3.27McYQ=4.OJMeV

Q= 17.59MeVReaction series 1cn0\Vl1 as the clJ.rbon or Bethe cycle and the proton- proton or Critchfield cycle are believed to occur in stars. These cycles are illustrated in Problem~ 20.27 through 20.29 and Problem 20.41.

Solved ProblemsNote: Where required, atomic masses have been taken from the Appendix.

20.1. When ~Li is bombarded with 4 MeV deuterons, one reaction thai is observed is the fonnalion of two a-particles, each wjlh 13.2 MeV of energy. Find Ihe Q-value for this rea(..1ion.Ans. Q=(K~I

+ Kd -

KJ = (l3.2MeV

+ 13.2MeV) -

4MeV = 22.4 MeV

20.2. Detennine thc unknown particle in Ihc rollowing nuclear reactions: (a) ' ~O(d,p)X, (b) X(p, a)~~Y, (e) ' ~~Te(X, d)';jl. .Ami. (0)

In the process I~O(d,p)X a neUlron is added 10 I~O to fonn X, which is I~o. In the process X(p, (X)~Y a prolon and two neUirons have been removed from X to form ~~Y. so X is :SZr. In the process InTe(X, d)'HI a deuteron (fH) and 1~11 have been fonned rrom 1~~Te and X. Therefore X must have two protons and a total of four nucleons, and is ;He.

(b)(e)

20.3. Detennine the compound nucleus and some or the possible rea(..1ion prodUcts when a-particles are incidenl on I;F.Ans.

The compound nucleus has 2 =21 +22 = 2 +9 = II and A = AI +A2 = 4 we have~ He +I~ F-+

+ 19= 23. Therefore

[RNa]

This nucleus can th(..'11 decay 10 many products, such as

HNa +,'nNe+p nNe+d

20.4. Calculate the Q-valucs for the reactions (a) I~O(},. p)I~N. (h) I~Sm(p. a)I:YPm.

CHAP. 20]

NUCLEAR REACflONS

213

Am.

For the reaction M;(mi'

m ,)M

f the Q-value is

Q = [M; +mj - (Mf(a)

+ mf)JC

Q = (15.994915 u + Ou - (15.000 108 u + 1.007 825 u)J(931.5 MeV l u) = - 12.13 MeV Q = (149.917276u + 1.007825 u - (146.915 108u + 4.002 603 u)](931.5 MeV l u) = 6.88McV

(b)

20.5. Calculate the mass excess (Problem 18.23) for (a) ~~Ca, (b) I~Te.

Am.

(a) (b)

0

o=4 1.958625u - 42u= - 0.041375u=-38.540MeV = 129.906238u - 130u = - 0.093762 u = -87.337MeV

Nuclear data are often given in tenns of mass excess rather than atomic weight.

20.6.

Using the data Nucleus1920S 1910S

Mass Excess- 0.038SS0u

" "dI

-0.039030 +0.014102 +0.0 16 050

find the Q-value for the reaction 1~~Os(d, r)'~!0s.Am.

Since total A is conser....ed in any reaction. we can replace rest maSses by mass exL"e$SeS in cakulating

Q. Thus

Q= (M;+m, - (M, +mfl]C= (-0.D38 SSOu + 0.014 102 u - (-0.039030u

+ 0.0I60S0u)K931.5 McV/ u) =

-1.37 MeV

20.7. As observ... xt in the laboratory system, a 6 MeV proton is incident on a stationary 12C target. Find the velocity of the center-of-mass syslem. Take the mass of the proton to be I u.Am.Using a nonrclativistic treatmenl, the proron velocity is foond from K,

= tmit?:

V=

~

Kj H f,;2(6MCV) 7 - - =1' - -:2= (3 x 10 m/s ) (I X93 5M) = 3.4 1 x 10 mls m, m,t u I. cV/um V",, =~ v=

j2K;

By 120.2),

M,+mi

Iu 7 do 2 (3.41 x 10 m/s) = 2.62 x hrm/s I u+lu

in the direction of the proton.

20.8.

For the endothermic reaction Mj(m;, mr)Mr detcnnine how the Q-value is related to the threshold energy of thc incoming particle. Use a nonrelativistic treatment.Am.The desired an.'iwer is obtained most easily by fir:.1 doing the calculation in the center-of-mass system. where the total momentum is 7.erO, and then transfonning the results to the laborarory system. Using the notation of Fig. 20-2. \\1,! have for the total initial kinetic energy in the center-of-mass system

K,nn = tm;tf2

+ ~M; V'2

214

NUCLEAR REACfIONS

[CHAP. 20

Transfonning 10 the laboratory system via (20.3), and recalling that the target particle is at rest in the laboratory system,

M, K;tab ( -) mj+Mi

(I)

Equation ( 1) is the general relation between the initial kinetic energies measured in the laboratory and center-of-mass systems. The Q-value of the reaction. which depends only on resl masses, is the same in both systems:

(1)The Ihreshold energy, K'hc = _Q(Mi~mi)

(4)

A more revealing expression for Kthlah can be obtained by considering the kinetic energy, K, of the center of mas~' (in the laboratory system) when the incident particle has the threshold energy. We have, using (20.2) and (4),

(5)Equation (5) siaies that the incident particle must have sufficient energy to start the endothermic reaction (- Q) and to account for the gross motion of the system (K*, which remains unchanged in the reaction).

20.9. Find the Coulomb barriers of t ~o, ::Nb and ~B i as seen by a proton.Ant.The Coulomb barrier is the energy needed to bring the proton to the edge of the nucleus (Fig. 20-4). If we define !J. :=; R + r -:= ro(A 1/3 + '). thenEc = k(Ze)e

a

= k ro(AI!3+ I) =

zi'

(I.44M'fin V.fin)( Z )1.4 Ai/3+1

-:= (I.03MeV) A i/3+ 1

( Z)

For I~O,EC =

( 1.03MeV)C61/~ + ,) =

2.34MeV

CHAP. 20)

NUCLEAR REACTIONS

215

Fig. 20-4

For 1~Bi,Ec =

(1.03MeV)(209~; + I) =

12.33 MeV

20.10. Refer to Problem 20.9. Compare the Coulomb barrier, E(", with the threshold energy for the reactions.Ans.

"O(p ,f\I~o 8 ,'" 8 The Q.value fur a reaction is

~tNb(p, d):iNb

~Bi(P, d)~Bi

and. from (4) of Problem 20.8,

For t~O(p,d) t ~O:

Q = (15.994915 u + 1.007 825u - 15 .003070u - 2.014 102 u)(931.5 MeV /u) = -13.44 MeVKtt. = (13.44MeV) (For :tNb(p, d)~Nb:

16U+ lU)16u

= 14.28 MeV

Q

= (92.906382u + 1.007825 u - 91.907 211 u - 2.014 l02u)(931 .5 MeV /u) =: -6.62 MeVKit. = (6.62

MeV)C3~3: IU)

= 6.69 MeV

For ~Bi(p. d)~Bi:

Q = (208.980394 u + 1.007 825u - 207.979731 u - 2.014 102 u)(931.5MeV /u) = -5.23 MeVKill

+ 1U) = 5.26 MeV = (5.23 MeV) ( 209U 209u

For t:O(p, d)'~O, Kth Ec and the reaction will occur with a large probability at the threshold energy. For ~Bi(p, d)~Bi, Kth Ec and the reaction will hardly l.'ver occur at the threshold energy. because the proton never gets close to the ~Bi nucleus. In the ~tNb(p, d):~Nb reaction the threshold energy (6.69 MeV) is slightly less than the Coulomb barrier (7.61 MeV), so one might expect no reaction, httause the proton just doesn't reach the :~Nb nucleus. Bul in fact the reaction :~Nb(p, d):~Nb is seen at the threshold energy. This is an example of Coulomb barrier tunneling, where the proton, even though below the Coulomb barrier, does manage 10 reach the :~Nb nucleus.

216

NUCLEAR REACTIONS

[CHAP. 20

20.11. If there are n scattering cen(crs (nuclei) per unit volume, each of area If, in a thin target ofthiclrness dT. find the ralio R oflhe area coveret.l by ~cattering cenlcrs to the total area of the target.Ans.

In a Ihin target no nucleus hides another nucleus. Hence,

R "'" IOlalarcaof scatteringcenlers = volume of target x n x area of target area of large!

(J

= (AdD x n x A

(J

= nqdT

20.12. Obtain an expression for the number of particles scattered from a beam of area A comaining No particles, after it traverses a thickness T of target material containing n scattering centers per unit volume, each of cross-sectional area (J.Art~.

Consider a thin slice. of thickness dT, of the target material. Any lime an incident particle encounters onc of the scattering centers in Ihis thin slice, the incident particle will be scattered. Therefi:>rc, the ratio oftlle number of scattered particles to the number of particles N incident on the thin slice will be the same as the ratio or the total area or the scattering centers to the area of the beam, so that from Problem 20. 11 \hoe have Number of scanl.>red partides Numberor incident partides~

total area of scattering centers area of target

0'- - = - - = nadT

dN""

dN

N

N

(The minus sign is used because an increase of scattered particles. dNsc corresponds to a decrease of incident particles. - dN .) Integrating this expression we obtain dN = - [ ,No

N

J'0

nadT

0'

- In - =naT No

Nj

0'

with No and N, the initial and final numbers of particles in the beam. The number of scanercd particles is then given by

20.13. For a hypothetical scattering target 10- 3% or an incoming neutron beam is scattered. Ir the target has a density of I .06 x I Q4 kg/m 3 A = 200 and the total neutron cross section per nucleus. (1, is 1.1 barns, find the target thickness.AtL~.

The number or scanering centers per unit volume isn= (

6.02 x 1020 nUciei/kmol) 00 I (1.06xI04 kg/m1 ) = 3.19xI02 nudei/m 3 2 kg/kIno

,ndI/G

= (3. 19 x lO:n-: m - 3){1.I x 1O- 2l!m 2) = 3.51 m- I

From rroblcm 20. 12, the number or scanercd partides is given by

N"" = No(l - e - ... T)and with Noc/No = IO-~ we haveIO-~

=

I _

e -u.~1

m" )T

0'

For small x, e- r

~

1 - x. and we haveIO-~

T=

3.51 m

. = 2.85xI0- 6 m

CHAP. 20J

NUCLEAR REACTIONS

217

20.14. When 5.30 MeV tl-particles from a 2~PO source are incident on a :Be target it is found that uncharged but otherwise unknown radiation is produced. Assuming that the unknown radiation is ,-rays, calculate the energy the j'-rays have as they leave the !Be target in the forward direction. [This problem, together with Problems 20.15 and 20.16, illustmtes the reasoning that led Chadwick in 1932 to the discovery of the neutron.1Ans.

The as.~umed reaction is 1Se(IX, i)I~C. Taking the nucleus to be at rest and the a-partide kinetic energy to be 5.30MeV, wc have from conservation of mass-energy(MIk

:ae

+ M~}t! + K"Kr

= Mot?

+ Ko + K l"

(9.0 12 186u + 4.002 603u)(931.5 MeV l u) + 5.30MeV = (13.003354 u)(931.5 MeV / u)

+ Kc + Ky(1)

+ Kc =

16.0 MeV

When the y my and l~C nucleus move in the same direClion as the incident a-particle, we have from conservation of momentum

0'For the matenal particles, in a noorclativistic treatment,0'

(2)

IX ~ J2(Mc')K

Thus.P2l' = J2(4 u x 931.5 MeV/u)(5.30MeV)

I99MeV

Pc'" = J2 (1 3 u x 931.5 MeV l u)Kc = 156K~/2

and fer the "l'-ray photon. , = Ky = Prc. Substituting in (2): 199 MeV = K7 + 156K~/Z Solving (I) and (3) simultaneously, we obtainK; = 14.6MeVKc = IAMeV

(3)

20.IS. In separate cxperimenL" the unknown radiation of Problem 20.14 is incident on a proton-rich paraffin target and a I~N target. Still assuming this radiation to be photons. detennine the minimum photon energies to produce the observed 5.7 MeV recoil protons and the 1.4 MeV recoill~N nuclei, and compare these energies with the result of Problem 20.14.Ans.

The photons will interact with the target nuclei by Compton scanering. The minimum , will cOlTcspond 10 a head-on collision. In analyzing this collision ,~ may use nonre1ativistic expressions for the particles since the observed kinetic eneJgies arc much less than the rest energies ofthe target particles.. Thus (primes refer to conditions after collision).hl'mu> = hr'

+ K'

(energy conservation)

and, since all momenta are along the x-axis, (momentum conservation) - =- - +mov' c c MUltiPIng the second equation by c and adding it to the first equation, we obtain. after using mov' = 2moK',2hl'm,n = ';2moClK' +K' = ..j"j(i(J2m oClhi'hll

+& )

Since 2mo2

K', we may neglect the

&

in the parentheses to obtain

h . = .jK'(moCl) v.... n 2

218

NUCLEAR REACTIONS

[CHAP. 20

For the proton target.

For the I;N target.I/I'mln ~

#4 \1cvii 14 II ) = mass of electron)

Type of Intcrm.1ioo

Masslcs. quantized accon:l.ing to

m, = / . / -!,

1... - 1

Each particle in a multiplet corresponds to a value of with the values arran!,red in order of decreasing charge. Thus. for the pions, = + 1.0, - I for 71 + , nO, n - . respectively, and, for the nucleons, ni, = +~, for the proton and neutron. respectively. Antiparticle multiplets have the same isotopic spin as the corresponding particle multiplet, but nil for an antiparticle is the negative of ni, for the corresponding particle. Table 21-3 gives I and m, for the mesons and baryons.

m,

m,

Table 21-3

Baryons

~,1

1

,1

0

- jn

- I

940 MeV IllOMeV IllXIMeV

I'

01

.+"E.0n+

1320 M eV

I01

1670 MeVMesons 138 MeV 496 MeV 549 MeV

0-

.'A'n'

"E.-

.n-

j0

K'

K'

"

It is found thai in all strong interactions the total isotopic spin (added as a vector) is conserved, and that in all strong and electromagnetic processes the total mi' is conserved. Another quantity that obeys a conservation law is parity. which is conserved in a reaction if the mirror image of the reaction (involving the antiparticles) also occurs. It is found that parity is conserved in strong and electromagnetic interactions but is not conserved in weak interactions.

21.8 SHORr-LIVED PARTICLES AND THE RESONANCES Because of their extremely short lifetimes, particles such as the nO and 11 (tm < 10- 16 s) and the resonances (1m < 10- 2 1 s) do not leave observable tracks in instruments like bubble chambers. Their existence is inferred by measuring the energies and momenta of the tinal decay products and working backw.ud through the conservation laws to see if the measured results are coosistent with the assumption

l'11:W 111

PARTICI t- I'IIYSI("S

uf th ... existence of the ul\obser1.';,bk tnlt'nnctliillt' particle. For cxampk, whcn tlw K tll'cays . what ohS('rwd is a ;'1:- panick ,II\U two ;-rays, so th", it might tx: hclic\t:u that the uecay scheme is

IS

Hllwc\"C"r, it I)' found cxpcnmentally th,,1 in Ihe c~llI~r-(lf,mass system the i. t p;tnkk i... mom)('netl!~tic (St ... rrobkm 21.210:). Thi s fact rulrs out =

11/ 2

is at rest,(I)

(m,r + K[) + m2(.2

In the center-of-ma.~s system the total momcnlllm is zem, and at the 'hreshold energy all the final particles arc created at rest Therefore.(2)

For a system of particles the quantity E 2 - (pc)2 is invariant, where is the sum of'he energies of the particles andp is the magnitude ofthe vector sum of the particle momenta. TIx:refore, s ince the total momentum in the laboralOry system is just the momentum of the projectile

m"

E~b - (Plcr == E~n[( 11112Also. for the particle

+ K,) + m2c2f - (p,d

= [(Af,

+ M2 + ... + M~Yf(m l c)2

(3)

m"(p, C)2 = E: - (ml c )l = (K,

+ m,?i _

(4)

Elimlflation of(p[c)2 between (3) and (4) gives a linear equation for K I with solUlionKl~

= K. =

- -

I

2m2

L(m .

+ m2 -

MI - MI - ... - M,,)e](m,

+ ml + MI + Ml + ... +M,,)(5)

~-- Q(ml + ml+MI+M2+ + M,, )

I

2m,

in tenns of the (negative) Q-valuc of the reaction. Note that, in a low-energy approximation, we could usc the classical mass relationship

MI +M2

+ ._ . +M~

=

ml

+m2

We \O,Quld then obtain from (5)

in agreement with the nonre1ativistic resuh of Problem 20.8, cq (4)

21.10. Find the threshold energyAIlS.

fOT

the reaction p

+ P _ P + p + TfJ.==- 135 MeV

For the reaction,

Q = [m"

+ mp -

(mp + mp + mw)JC = - mne?

so, by Problem 2 1.9,Klh =- ,Q (m,, +mp+m,,+mp+m,,) = mp 2mp( -

Q _)(4m,, + mn~

135 MeV = 2(938 MeV) [4(938 MeV) + 135 MeV] = 280MeV This is the minimum energy that an acccleratOf" must give to a proton to produce a rro meson by the above reaction_

21-11. From the reaction rrAns.

+p

_

n

+ y delennine the possible values of the

spin of a

7( -

meson.

From conservation of intrinsic angular momentum we have

., IS 2"

where 151 = Js(s + I )Ii . On the right side we have s~ 'The prOlon has lp _ I so Sn 0 or r - Z' IS or I .

= ! and $ .. = I. So the total angular momenrum

238

PARTICLE PHYSICS

{CHAP. 21

21.12. Evaluate the quantityAns.

fO

= h/mnCZ.

If One views the strong interaction as the exchange of asuch as

1(

meson. then inside a nucleus processes

can occur, even though mas..'l-Cncrgy consen:alion forbids the process for free nucleons. According to quantum mechanics, conservation of energy can be violated in the amoun t m~c2 if the lime for the process is of thc order givcn by the Heisenberg uncertainty principle:0' 0'

,o

~ --

m~c2

Therefore. strong interaction processes occur on a lime scale of aboUi 10- 24 s.

21.13. Estimate the mass of aAm.

11:

meson.

If the range a of the 1[-rncson field is abou t the size of a nucleus and i f it is assumed that the 11: meson travcis at nearly the speed of light. then a = CTo. when: TO is the lime for thc 'It meson to navel the extcnt of the nucleus. Selling a ::: 1.4 fm (the approximate nuclear size) and TO = fI/m. r?-, the time consistent with the Heisenberg uncertainty principle (Problem 21.12), we obtain

This comparcs well with the observed masses of 1T mesons (Table 2 1-1).

21.14. Construct a stmngcness number versus charge number plot similar to Fig. 21-1 for the nine spin-O mesons listed in Table 21-1. Discuss the features of the resulting symmetry pattern.Ans.The charge number and strangeness number of the nine spin-O mesons in Table 2 1-1, counting both panic les and antiparticles. arc as follows. MesonIT' IT" IT

" +10- I

.'I'

0 0

0

K' K"

+10 0 - I 0

ie"K'/ ,(

+1 +1 +1- I

0

0

0

When the strangeness numbers arc ploued versus charge numbers. using a sloping axis lor Ihe charge number. the pattern shown in Fig. 21-3 is obtained. Figure 21-3 shows a hexagonal paltern where the six particles on the perimeter of the hexagon lie opposite their antiparticles. The remaining three particles. which arc their own antiparticles, lie al Ihe center of the hexagon.

PARTH ' IJ .

I"IIY~ I ( 'S

2W

Fi,::.2 1J

21.IS, &1(lfl! 1 ~64 . rhl! Ib lluwtrlg. nine "Pin. ~ h;tty.,n ranKks had been ulY.tati!>tical mechanic!> it is as!>umcd that any given particle has a n "jntrinsic probability" gi of occupying the fth cell (Lgi = I). For example, if the cells are energy intervals and jfthere are twice as many states in 6.Ej as in 6.Ej , then it should be twice as probable fo r a particle to have an energy in the interval 6.Ei as in dEj . Under this assumption, obtain an expres!>ion for the probability of finding a particular distribution of particles (11,.112 ... , II,'.Ans.

Consider one particular way of filling the cells. in which. say. particles (X"PI"" go to cell 1; P> ... gO 10 cell 2; etc. 'The probability of (x, going to cel l I is g,; the probability of (x , and PI GII(I . .. going to cell I is'll'

8, x8, x xg, = g,n,and the probability of the entire filling is

Each way of realizing the distribution has this probability, and, by Problem 25. 14. there are

N!such ways. )'Ienee the probability we seck isP -

N'

0 - 1I,!'12!"'II,!

g,~'81~1

.. '8r M,

(I)

25.16. Refer to Problem 25. I 5. Statistical mechanics assumes that among all distributions of particles. the onc of highest probability corresponds to equilibrium of the system. Find this most probable distribution subject to the conditions that the number of particles and total energy of the ~)'stcm are constant.Ans.

The problem is to maximize(I)

(the numocr r of cells is not important, provided it

Slays

fixed) subject to rwo constraints:LII;=N(2)

Fixed numocr of particles : Fixed tolal energy:

L E;tI, =

Elol

(3)

where N and E,~ are given. The energy constraint assumes that each particle in the ith cell has the same energy ,. It will be convenient to replace Po by the monotonically increasing function of Po_ ", 1 - "' I , 1 n p. ,-L,.II,n8, Lnll,.N.

(remember: N is fixed so far as the maximization is concerned). Further we assume that N is very large ( ~ IOH) and, more important. that all the III are large enough to allow the usc of Stirling'S

284

CLASSICAL STATISTICS: THE MAXWELL- BOLTZMANN DISTRIBUTION

[CHAP: 25

fonnula (Problem 25.13). Thus we writeIn

~~ = Ln, Ing; - Ln, Inn, +)211':

(4)

dropping the last sum since. by (2). il is constant and will have a zero derivative upon ma;llimiling. Treating the IIi as continuous variables, .....'C maximize the function (4), subject to constraints (2) and (3), by the method of Lagmnge multipliers. That is, we form the function

F(tt,. til" .. ; A,. ).2) =

L "i lng, - L II,

Inn; - i"

(L: fI,

-

N) - ;'2(:[ Ern, - EIOt )).2- Taking the panial

and find its maximum, subject 10 no constraint on the variables derivatives of F, we obtain the conditions

"I- '12, ... ; i."

::;:- = Ingi -Inn; - 1 - A, - " 1' = 0

OF "" i

.

(i = 1,2 .... )'Ii:

plus (2) and (3) from the i.-dcrivalivcs. Solving for the maximizingti,

=

KiC- l - i" C- ili:,

=

Ag;c- fI

(5)

where A = e -

l - ;.,

and

P=

).2 are unknown constants.

25.17. Refer to Problem 25.16. Show that the Maxwell- Boltzmann distribution follows from (5) under the following assumptions (a) the "cells" are infinitesimal volumes dv.Td vy dL'z in velocity spaee (sec Problem 24.6) and it is equiprobable for a particle to be found in each cell (i.e., all the g; are equal); (h) the avernge of the particle energies has the value given by kinetic theory, Ea' "8 = ~kT.An.~.

Under the assumption of equiprobability, the intrinsic probability lhat a particle will be found in the (h .dl': is jU!'1 infinitesimal cell dL'._gdvxd~~dv:

where g is a eOO!'1ant. Each particle in the cell has the same energy,E =: !mv. ,2

+ !nw,? + !mv,2(I)

Thus (5) of Problem 25.16 gives for the number of particles in the cell(/n. = Ae- /lE: dvx d v.,_dv,

W 'd S

where .....'C have absorbed the conStant g into A. Nonnalizing as in Problem 25.6---except that, there, already knO\vn- we obtainA = N (m/2)1/2 fi l/2

p

nl/2

(2)

Finally, proceeding as in Problem 25.7- where, again.

II was

already known- we find(3)

3 ~,.~ ::: 2/tEquating this to 3kT12 gives

p=

IlkT and from (21.

A =N ( - m- ) '" 2nkT

With these values of p and A. (I) is the Maxwell- Boltzmann distribution.

25.18. Find the most probable distribution of N intrinsic probabilities for the cells are

= 5 distinguishable particles among r = 3 cells if the

CHAP. 25)

CLASSICAL STATI STICS: THE MAXWELL- BOLTZMANN DISTRIBUTION

285

Ans.

By (I) of Problem 25. 15. the probabiliry of a distributio n ('I" '12' '13) is

since ' 11 + '12 + '11 = 5. Therefore, the most probable distribution is that for which the number of ways of realizing it,

is greatest. Because the interchanging of the particles in any'12

rwo ccl ls. e.g.,'1 2 = 3,'1).

== 2.

resuhs in the same value of X, we calculate X in Table 25-2 only tor the ca'lC 'I, ;:: '1 2 ;::

Table 25-2

".5 43 32

", ",0 I 2 I 2

X

0 0 0I

I

510

I

20 30

From the table it is seen that there will be th ree(2,2, I),

mO~1

probable distributions, each with X = 30: and(1.2,2)

(2, I, 2),

25,19.

FOT Problem 25.18, if the energy ofa particle in cell 1 is zero, in eell 2 is f, and in eel1 3 is 2l.. find the most probable distribution if the total energy is fixed at EI04 = 3l..

Ans.

The

con~1rnint

0'

eliminates all distributions except (3, 1. 1) and (2.3.0). From Table 25-2 it is secn that the fonner is the most probable. For equal intrinsic probabilities, the Maxv."C II-Boltzmann distribution has the fOnT)'I,

=

Ae- /J,

The distribution (3, I, I), which puts most oflhe particles in the lowest energy state, is much closer to the Maxwell- Boltzmann distribution than is (2,3,0).

Supplementary Problems25.20. Find the ratio of the root-mean-square speedMaxweil- Bolt7Jl1ann distribution.t'"",",

to the average speed,

L'a"i'

for a gas described by the

An.\"...;r;rrg :: 1.08

25.21 . For the cireuil of Problem 25.10, L = 1 mH, C = I ~F, R = 0.1 n and T = 300K. Find the rms vohageproduced in the inductance.Ans.

6.44 x 10- 8 V

286

CLASSICAL STATISTICS: THE MAXWEU- BOLlZMANN DISTRIBUTION

[CHAP.2S

2S.22. Six di~1iog uishablc particles are to be distributed into 3 cells. Find the number of differen t combinations of particles thai can produce (0 ) the d istribution (b, O,O). (h ) the distribUlion (4. 1, I). AIlS. (a ) I; (b) 30 25.23. In Problem 25-22 suppose thai the three cells have respective panicle energies oro, t. and 2(. What is the most probable distribution with lotal energy 6( ir lhe intrinsic probabilities for the cells arc equal? Ans. 2 particles in each cell 25.24. Refer 10 Problem 25. 19. What is the most probable distribution for a 1 00ai energy 4f? 25.25. Consider 50 clislinguishablc particles distribu ted into eight cells as follows:Cell1An.~.

(2.2. 1)

2

J

4

5

6

7

8

Number of particles

6

8

9

0

7

2

10

8

Ans_

'-low many combinatio ns can be made o f the 50 particles which yield this distribution? 1.% x 16M'

25.26. For a two-state system with energies ( and 2( find fo r the most probablc di).1ributio n for the averagc encrgy, if'"

", =". = >.'.

An.\".

E~,-~ = - c - 2/1.

-='='0,,,;,:-:;.+ c - fJ 0, 0: ::: 0, and = I, we have

(26.4)where the gamma fUllctioll, r(x), obeys r(x + I) = xr(x). Particular values are

r(l) = ,JiiIn the !>pecial case 1%=-0,( =

r(II+I) =- II!

(II an integer)

+1, (26.4) becomes

r' & ~dq = rcp+IK(p+l) Jo - I26.S BLACKBODY RADIATION

(26.5)

The walls of a cavity maintained at a temperature T continuou!>ly emit and ab!>orb electromagnetic radiation (photons), and in equilibrium the amounts of energy emitted and absorbed by the wall!> arc equal. The radiation inside the cavity can be analyzed by opening a small hole in one of the walls of the cavity; the escaping photon!> constitute what is called blackbody radiatioll. Quantum physics was born when Max Planck discovered around 1900 the correct expression for the experimentally observed spectral distribution of blackbody radiation, i.e . the fraction of the total radiated energy with frequency between v a nd v + dv.

Solved Problems26.1. Consider, fo r simplicity. a cubic cavity of side I whose edge!> define a set of axe!> and whooe wall!> are maintained at a temperature T. Maxwell 's equations for electromagnetic wavcs !>how that the rectangular components of the wave vector

290

QUANTUM STATISTICS: FERMI- DIRAC AND BOSE- EfNSTEfN DISTRIBUTIONS [CHAP 26

mlL'>t satisfY the boundary conditions

k.,l k,1 -n = fi._ --n = fI,. . n where n .... fly , and liz are positive integers. (These boundary conditions ensure an integral number of half waves in each edge of the cube.) Each triplet (n.. , n,., II,) represents, in classical tenns. an electromagnetic mode ofoscillatioll for the cavity; we shali consider these modes as photon .\tat.\". Find the number of modes in the frequency interval between I' and \' + d\', given that there are two independent polarization directions for each mode.- = fl x

k,1

AIlS.

Each allowed frequency v corresponds to a ccnain mode (n.. , n.l .,

":)

and can be written as

v ="j = 211: =21 "1'1 OK \

,,

E

Fig, 26-2

26.9. Obtain an expression for E fn the Fenni energy at T = 0 K. for an electron gas in a metal. Am,', In Problem 26.7 we found that the number of :states ('lpn". n,) in the energy interval between E am.! + dE was given by dS = g(E) d =2 V(2m)Jl lJr

Sinee for each sci of quantum nu mbers In,. n" ne) there are two possible eleclron spin Ori(.'Otallons, we must multiply g eE ) by a factor of 2 to get the actual density of states for the clecrron gas. Thus, the total numhcr N of fermions in the system is given byN

"3

l

E I12 dE

=

.2[

(\

fmX(E)tIE

=

4 n V(2m) JIl

h

"", IOdE J(\ eIF.' t:,I/H + I

At T = 0 K the Fcnni- Dirae distribution function is (sec Prohlem 26.8)

FH )

=

1

lor

E < [0

Fm = ()

fM

t: > E,..,

so the limits on the integral can be changed. to giveN =

4n V(2m)~{lh-1

J I).

( ," l f2 E = 4 n V(2m)l /2 d

(~E!./2 )3 to

k'

0'

f?

flo = 8m

(3N)UJ nV

As the temperature increases, it is found (sec Problem 26. 13) that the Fenni energy remains about equal to E'II'

26.10. Metallic potassium has a density of 0.86 x 10J kg/mJ and an atomic weight of39. Find the Fenni energy for the electrons in the metal if each potassium atom donates one elcctron 10 the elcctron gas.An~.

First

\Jo"C

calculate the number or clct:trons per unit volume. N j V:

N (6.02 x 102(' aloms/kmol)(0.86 x 103 kgjm 3 ) 2~ atoms 0'8 elct:troll$ - ~ k k = 1.33 x 10 - ,- = 1.33 x I J V 39g/mol Ill" mThco. from Pmhlem 26.9.

E = ~ ( 3N ) 1/J =(he)2l (3N )!/J = (1 2.4 x 10- 7cV. rn)2 ( 3 x 1.33 x IOlll m_1) 213 = 2.05eVto

8m nV

8mc

nV

~(O.5 11 x IlY'eV)~

It

At 300K. kT = 0.026eV. so it is secn thal !:.i

flO is much greater than kT at room temperature.

CHAP. 26} QUANTUM STATISTICS: FERMI-DIRAC AND BOSE-EINSTEIN DISTRIBUTIONS

295

26.11. Find the average kinetic energy per particle for a Fcnni gas o f N particles at T = 0 K.Am.

Usi ng

th~ d~n s ity

of states 2g() fouod in Problem 26.9, we have I

E~'''=N dJ/ =N= 4nJl(2m)1I2Nh1

J

2

r(l

EFr08(E)JE =E

4nJl(2m)1/2 Nhl

r(l

ff'. .- f.i ".r+ 1

El/2 (IE

Jo

rE ,. Elf2

d.

= 4nJl(2",lf! (~e/2 ) Nh} 5 (0

Substituting the value of N fou nd in Problem 26.9. "-'C find that

,vi = i roThus, even at 0 K, the ekctrons.. on the average, have a Si7.able kinetic ~nergy. This ocrurs because the Pauli exclusion principle will not al low a ll the ela.-untls to occupy the IOWf..'St ent.Tgy levels.. so that ~Icctmn.o; wi ll be found with all energtcs up to EjO.

26.12. Using the nonnalizalion condition thai the total number of particles is a fixed number. obtain, for of an electron gas in tenns of /0' the low lempt.'f'aturcs, an expression for the Fermi energy fermi energy at T = 0 K.

Ei

Am.

lbe number of panidt.'S is given byN = [

F,..,)8(E)d = C

r

E1flFFO(IE

whtTC. from Problem 26.9. th~ dcn.o;ity of Slates is geE) = CE 1 /2 d with C = 4nf/(2m)li2/hl . Equation (26 .J) gives. for p and keeping only the fu'Sltcnn ofthc sc.'ties,N

=

C I; EI/2FFf)d ~ 2~ [;/2 +

=!

2(kTi( 1 -

D { (2)(iEj /2 )]I II)

= 2C e12 [1 ~ lkT)'] 3 I +8:.;l~tting

T _ OK in (I ). we obtain

2C 'I' N_ -T E10

'"

E =(3N ),n=(8nJl(2m)1t1 3Nh' )>1'= "' - (3N)'" 10 2C Rm nV

whic h agrees with the resu lt of Pmblcm 26.9 . Sinee kT/EJ is small, we sec from (I) that 1 does not change rapidl y with wmpcraturc. 1llc~fore w~ can sc..'t EI = Ero in the second krm on thc right of (1), to obtain. af\(.'T a lso substituting N = 2Ce:-~2 /3,

2C 'I' = 2C EI' [I +8 n' 3 Elo 3'jfrom which

(kT)']Ero

r = EJnFinally.~call ing

[

1+

~2 ( : :) ']->/l

that for small x

296

QUANT UM STATISTICS: FERMI- DIRAC AND BOSE- EINSTEIN DISTRIBUTIONS

[CH AP. 26

we can write

26.13_ Silver has a Fenni energy o f 5.5 eV al T = 0 K . Using the results of Problem 26.12, estimate the size of the first-order correction at T = 300 K.Am'.

TIlc first-order correction is _ n 2 (krl 12 Eru

",. _ n2 [(8.6\712

x 10- 5 cV/K)(300 K)] l. 5.SeV

= _ IO- 4 cV

an almOM insignificant change.

26.14. At 0 K, silver has a Fenni energy of 5.5 eV and a work function of 4 .6 eV What is the averageelectrostatic potential energy seen by the free electrons in silver'?

Ans.

The \.10m fu nc tion. 1'. is the minimwn energy rt."quin' E,., and everywhere negative if E, < E.... In either case the integrand in (I) \muld be one-signed, and the integnll YtOUld be nonzero. Hence, E, = E,.,. 1\ more refined analysis shows that the density of Slates is not exactly symmetrical about E,.,. so the Fermi energy will not be precisely at the midpoint oflhe gap. However. the error introduced by our assumption is quite small.

~t:.. -E/ )I It.T

27.6. In Problem 26.7 the density of states in the free electron theory of metals W'dS found to be_ S.J2ttm!{2 V 1{2 g() h) Assume that this same expression holds for electrons in the conduction band, but with E on the right side replaced by E - E(" where Ec is the energy at the bottom of the conduction band . Taking the Fenni energy at the center of the gap, show that the number of valence electrons per unit volume in the conduction band at a temperature T varies like

nt = Ce -~f2kTwhere C

= 2(21tm..kn3/ 2 / If and g = Ec - Er is the size of the energy gap.

An.f.

I\t ordinary temperatures kT :::::: O.026eV. so that in the conduction band E - Ef kT. and the approximafe exprcs.

151.96

152 154 15'

152 153

155-

'54'

64

Gadolmium

Gd

157.25

148-