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| QUANTUM PHYSICS
OF ATOMS, MOLECULES, SOLIDS,
NUCLEI, AND PARTICLES
Second Edition
[Walxyalra) = Wolxa)Wstxa)]
1
A
Robert Eisberg
Robert Resnick
Prepared by Edward Derringh
This supplement contains solutions to most cf the
more-invelved problems in the QUANTUM PHYSICS text;
with one exception, solutions to problems ir the
Appendices are not included,
The supplement is directed toward instructors, and
this has influenced the presentation. Not every
algebraio step is exhibited, The units have not been
displayed explicity in every equation. (SI units are
adopted in the supplement, mainly because they are
briefer than the text notation.) Rules with regard
to significant figures have not been strictly
observed, although there should bo no cutlandish
Tt is a pleasure to thank Prof, Richard Shurtlo£f
Institute cf Tecinology) for preparing
the solutions to the problems in Chapter 18.
Preparation of the supplement, including eholes of
problema, was left to the undersigned, ho vas also
his own typist and illustrator. Be would appreciato
a note, of up to moderate asperity, from those who
detect an error and/or mistake,
December 24, 1984 Edward Derringh
dl Montocmery Drive
Plymouth, MA 02360
nn
2 2
P «férias = Ppfuldu = BA loç) Bi
1
8
vw = cf, = ESBEX IO. 5.4509 x 1014 mz,
5.50 x 107
BZ ip = me,
=c = a
“ 2 s.51x107
Therefore,
Vap = My + v9) = 5.46 x 1034 may
do = vo = 9.9 x 10h ne,
Eince
ent) = Ci ta 2 yr,
mumerically;
entmd/o? = 1.006 x 1933,
hu = 4.37,
EaE 3 = 78.04,
ôriv) = (1.006 x 1072) (78.04) = 2,289 x 1978,
The area of the hole is
Au mm (5 x 10)? = 7.054 x 10º mê.
Hence, finally,
P = k(7.854 x 109) (2.598 x 109) (1.289 x 10 15) (9.9 x 2015,
Pe 7,51 W.
15
0 L= amrêgrÊ = qui? x 208)205.67 x 1078) (570099,
L=3.685 x 102 u,
Le Gm? 2,
26
dn. L 3
de = 5 - dao = 8.094 x 10º kg/s.
(3x 10)
(bl The mass lost in one year is
oem DEE a (4.094 x 10) (3.156 x 107) = 1.252 x 1077 xg.
The desired fraction is, then,
3
£ = 8 2 1.292 x 101
A.
= 6.5 = 10
2.0 x ra
lo
ta) The solar constant S Ls defined by
L
s=“E,
am?
EF = Earth-sun distance, Lan * rate of energy cutput of the
sun. Let R = radius of the earth; the rate P at which energy
impinges on the earth is
L
p= ES aê = mês.
The average rate, per m, CÍ arrival Of energy at the earth's
Surface is
Ed
= - ME rs,
ER amR
238 Wmê É 401352 W/mê) = 338 wjm?.
FE 338 = qu! = (5.67 x 10 Syrº,
T = 280 X.
1-19
=E «mr É
Rpfa) = É opta) SCE
with x = he/AkT. At A = A o x = 4.965, by Problem 18. Thus,
Brad = 42.4030 005 nte).
Now find x such that Rola) = 0,2860400):
mor aê Cr
tr = (0.2)42.403m EE
nto &-a nº
5
E = 4.203,
-l
xm= 1.882, x = 10.136.
Mmericaliy,
- B
ie tEda 2
rm (1.38 x 102) (3) E
1 = 4.88 x 1072,
so that
À = 4.758 x 1073, 1.882 = 2.55 mm,
à; = 4.798 x 102/10.136 = 0.473 mm.
zo
If x= Def ade then, by Problem 18,
=. x
x
x
rim RE:
1-21
Ey Problem DO,
so that the wavelengths souçht must satisfy
5
ma = k-1707 ED,
É a -1 qhey 4
Again let
x = hefakr.
In terms cf x, the preceding equation becomes
É
— = à,
a
Solutions are
x = 2.736; a, = 8.050.
Since, for À or
x = 4.565,
these solutions give
Me 1.815 ut
da = OGIA
ru
Let à" = 200 nm, A” = 400 mm; then,
É A = (3.82) É A
he/r"kT "
Sep = a.saio?.
1
e -
Mumerically,
a
3,
+ = testa poa sso x 2) À asso,
tá x 107) (1.38 x 10 9)
so that
35587/T
[5 = 5
SIA = (3.82) (4)” = 0.1194.
Let x = 0588, enon,
q = 0.194 = 1,
x = 7.375;
SBT 2 9.398,
T = os = 18,000 E.
CRAPTER THO
Ei]
The photoclectric equation is
he = VA + WoA.
With V,
o” 1.85 V for À = 300 rm, and V5 = 0.82 V for à = 400 mm,
he =8.891 x 108 + 3x 10upa
26
pe =s.255 x 102 +4 x 107%.
Hence,
8.891 x 1026 4 3 x 1077u, = 5.255 x 1028 4 4 x 1074,
(bi wo = 3.636 x 109 3 = 2,27 ev.
Therefore,
he = 8.891 x 1028 + (3 x 1077) (3.636 x 101),
he = 19.799 x 10 É Jem,
tab É 26 pe
n= BB xD 6.604 x 10 35.
2.998 x
tel “g" her
3.636 x 102? = 19.799 x 107287,
dp = 5.445 x 107 m = 544.5 1
x
In a magmetio field
E = mu/0B.
2-33
The number of particles stopped/scattered betwom distances x
and x+dx is dl(x) = clixpãx. Hence, for a very thick siab that
ultimately etops/scatters all the incident partícles, the
average distance a particle travels is
x = SE - ea.
E o se ay op o
the limits cn all x integrais being x = 0 tox==
CHAPTER THREE
st
É «pis = Ta Do
ão ele Tux
=31
cá - 03 x 10 0 ka
1.813 x 107! = a ,
a = 1.675 x 107 rg;
wvidentiy, the particle is a nevtron.
mm
ta) E = plo + Ei (6 + E)? = pº 4 ER,
er
poda? + axegt o Da + ot,
Bit E = evandE, = mc, so that
2
2teu) (mc)
1 ES 1 le
UE, j= 2 Ji = ( = y = “Uma,
and
EE, = evfome?,
Thezefore,
1 = = great da
(5) Monrezsesviseio Limit: eu cemc?; set 2 + aU/2mpe? = 1 to
get
x = /(mçen 5 = /tomgo = ar
B
M
8
jo ML aço, te pe = vips, pagão
imp ) ivfe)
Numerical ly,
=34
po = (6.626 x 107º pos) (2,996 x 20º m/0),
(1.602 x 10 JfMev) (O * mir)
he = 1,240 x 10? mem;
bence, sa
3 .
à trm) = ni a Modo,
219
ta)
=34
p = Bo Eifé x ds (2990 x Jg no). 0,2400100,
*oqUme « e
1,602 x 10 3/mev)
Pepe sr,
== (0.120)? + (0.51)? + E = 0.525 mv;
E =E-E,=0.5258- 6.510 = 0.014 MV = 14.8 key,
Cb
p = Silo me Hd. E 126 Ke.
These are garmma-rays, or hard x-rays.
lc) The electron microscope is preferable: the quema-rays are
difficult to focus, and shieldins would be required,
228
ta) set ax = 10
h 6.626 x 1078
p=ip-qlo- Lu
e
p = S228 x 17 pams 2.998 x MÓ m/s ( D.S868 ev
e 1.602 x 101 pay 8
E = (p%e2+ Ed! = (10.566)? + (5112)% = 511.00095 keys
K = E - Eg = 511.00085 - SI] = 0.95 ev.
Atoric binding erergies are cn the order cÉ a few electron volts
so that this result is consistent with finding electrons inside
ators.
(bj 4x = 10 m; hence, p = 9.868 Mev/c, from (a).
pie? + ed) = (9.8682 + 0.512) = 9.B812 Mey;
X=E- E, = 9.812 - 0.510 = 9.37 Me.
This is approximately the average binding energy per nucleon,
so electrons will tend to escape frem nuclei.
(c) For a neutron or proton, p = 9.868 MeV/c, from (b). Using
938 MeV as a rest eneray,
E = (pie end" (5.868? + 9362)
= 0.052 Mev.
= 538.052 Mev;
E =E- E = 938.052 - 338
This last result is much less than the average binding energy
per nucleon; thus the uncertainty principle is consistent with
finding these particles confined inside nuclei.
dz
ta) Since p, > sp, and x 2 dx, for the smallest E use p= Ap,
and x= bx tocbtain
« EiiçãE 2
E FE ltP) + AOlax)
With ”
tp dx = =
the minimum energy becomes
2
1,h,2 2 2
E = etquio + SClém) so + ac(um?.
32mºmlhx)
(b) Set the derivative equal to zero;
2
E A + 2 k
SDS * O Tendo au) CO nO + (0 = qto
Substituting this into the expression for E above gives
Fato TOS = time
3=34
la) Let the crack be of sero width and 4x = horizontal aíming
error (i.e., drop point not exactly above crack). This implies
am inítial horizontal speed v, given by
Bence, the total horizontal distance X from crack to impact
point is
x=mx+xe dx + ÉS L
Mm 9º mw
To minimize, set
Jan = ES BA
0h) ZEH = 10m B/9= 1, m = 0.001 kg, then x = (1023035
= 3 x 1916 m, aperesdimate ly.
Therefore,
17
= 3
Put an electron behind each - Re
siit and cbserve any recoil Ant
due to its collísion with a
d
Ay << qe
Doe to the collísicn, the photon's mementum changes. Tn order
not to destroy the interference pattem,
[o
Ren
e, «Pol,
m the order cÉ the fringe. By conservation cf mmentum, this is
aiso the uncertainty in the electron's momentim. Hence, for the
electron, it is required that, in order not to destroy the
pattern,
(tvi (ap) << tê a =
z
4-18
The periods of revolution of electron and proton are equal:
2rr 2
4.
The motior is about the center cf mass 0F the electror-proton
system, so that
e mt * Ee
Therefore,
“pr ves Ge
vie É sir rod m/s.
“2
(a) Frequency of the fixst Lines w Thy = Rg - pop .
Frequency of the series limit: v = en = est - 0).
m
Therefore,
eu uy" ria
6) 2
My e, /0+1)
= =.
pe agf 51)*
+
ta) Euu2 TA, = TE = 26.6 =;
Ê 5
E = 26.6 - 10.2 = 16.4 ev.
o)
Emi * 13.6 + 16.4 = 30.0 ev.
+
(a) By mementum conservation,
Dam.
c
umbinins this with energy conservation gives
2
AB e gm + MZ ay + AM? a ta 4 DT
Uh) hua)
——.. Atol (AE)
by = ty =hy a
* om Tomé
vo = va + dim,
ve vt Ea.
tb) Since% = c/h vg = C/Agy
nego dia.
1.4
E = (13.65 - =) = 12.089 ev.
Po
Naglacting recoil:
do "Eos = 102.6 mm.
24
With reccil:
-18
at. 2009) 062x108) ox 10?,
do me (a q.e73 x 1027) (2.998 x 109)?
A = 661 mm.
4-4
The kinetic energy of the electron is
K = (0.511 Mov) ds - D.
2 12 x 197
2.998 x 10º
= 0.0400,
2 2
1.622 (3.6
Mag 2 A AR 544 av.
Hence,
En = 54.4 + 409.3 = 463,7 ev,
à m 280 = 2.674 mM,
4-35
(a) Eiydrogen Ba: ah “Rã 3.
Helim, 2 = 2: A) ad “ROS mp
If Ag Agr them Zi .
= . nç=4
3engdo o n=6.
4h) Now take into acoxemt the reduced mass us
2.4 24
1,2 MyfDie 1,2 Mel) e mo
- J i = à = (Rd.
“a “meo amo Me “o amiite Pg
mm m md) F
dy ço O Mel qa Pe "TR" Fell TO
Therefore,
Mge > gr
so that
ds Ee ==
ada A da
Hence, compared to the hydrogen Ho line, the helium 6+4 line
savelength is a littie shorter.
(b) Since à « pb (the factor 2? 45 combined with 1/r7 = L/nê
to alive equal values for E and Be),
AA = (4.084 x 1079) (656.3 ne) = 0.268 mem.
442
The momentum associated with the angle 6 is L = Iy. The total
emergy E is
2
E=Kegul =.
L is indopendent of & for a freely rotating cbject. Hence, by
26
the Eilson-Somerfelá rule,
É LO = nh,
LS d0 = L(23) = (28) (2n) = nbs
UE) = nh,
E = RÉ
a
CHAPTER FIVE
va
ln) The time-deperdent part of the wavefunction is
etitA Cm) À ABA (ritmo
Timrefore,
16% = 2m * vo!
hj Since E = ho = Zehho,
E= ma".
(1) The Limiting x can be fomd frm
ta
According to Example 5-6, the normalizing integral is
4
3 ã
(ZE/C)
2mb dx 22 Atl x
1-2" E 2 o E sin FE
LRE/O - 2 (2E/c) I,
1a sêmS + pêo tcmid,
ta
Problem 5-3tc) provides the limits on x; the wavefunction is
1/8 2
7 = tom SA ot
tri)
27
32
E Cm), | 2?
5 - tm dio E ax = HE) Ee du,
e É
Temo! T= te,
(b) This seme relation, U = T= 4E, is obeyed by the classical
oscillator also.
5-15
Vse the notation
be), = pede te cane E am
Rs da lz ="
Clearly
ta
bp); = meo de + Tdx= (mply — dk,
implying that (xp) and (xp), cannot both be real. Also, by
integrating by e
mo
+.
bel; - ate [o Eae) = ap TE ax.
Thus,
tel, = pdf.
TE (xp), is real, this last relation says that Ge), 15 real
also, which contradicts the first Finding above. Bence Up), às
complex and therefore so is Ge). Moe try
==
=. 2,2 ,
=» sobre dç + Bor eu
—
= = Hb), + teplsh = Sit), + Obg,
*p = Re(xp);
so that this new Xp 1s real, as desired.
al
With V = 0, the energy cf the photon is
E= po.
Replacing the energy E anê momentum p by their operators gives
Mm
ira dio e
Nou set Fix,t) = plbdTIt) and divide the equation by 7 to get
niZ-elã-r
where K is independent of x and t. Write E = kjc and the tuo
equations directly above become
EE = suor + qe qt,
É = my + peelx
Hence, for the photon,
Ta ele te-et) E
22
(a), (b) The curvature of 4 15 proporticnal to |V - E]: where
fu = El is 1arge the function cecillates rapidly in x, and
where |v - E| às small it oscillates less rapidiy (hence, nodes
are close together in the former case, farther apart in the
latter). In the first state, |V - E| is just large encush to
tum 4 over: no nodes. The 10th state will have 10-1 = 9 nodes,
Losing to an odd function since V 1s symmetrical about the
origin. The wavefunction decays exponential ly wherever VE, the
classicalIy forbidden region. For further discussion, see
Eemple 5-12, which treats the símilar símpie harmente
cacillator potential.
35
tc) Classícaliy, the probability density function P is given by
P=BêN,
pê che nomalization constant. Energy conservaticn gives vi
E = iê tc,
the upper sign for 10. Using this,
PPBio).
To determine B, use the nomalization condition
(remo
afeto sê =1,
mat e
veDorE=V= hx. Evalusting the integrais cives
P-dEZEIS:
Particular values are:
PIO) = C/4E; PE/C) = P(-E/C) = =; Soo) = tc2/er?,
td) The graph of the classical density function resembles that
fer the símple harmnic oscillator, the lack of a horizontal
tangent at the origin being the main difference cn a rough
sketch.
+P
due
co ca a
ns e a 2 0 e
ai
36
5-24
See Problem 5-25.
5=25
With no burp, the wavefuncrion will be sinusoldal inside the
elassical region of motion and a decaying exponential outside,
The lowest energy wavefunction will contain no nodes.
In the present situation, in the region of the bp the
curvature of the wavefunction will be less than outsido
bump, since the curvature-is proporticnal to E - V. This will
upset the good behavior of tha mavefunction at larse x
associated with the valve E, corresponding to the first bound
state without the bump. To Tcorpensate for this reduced
curvature in the region of the bamp, a larger curvature (as
compared with the no-bump case) is needed outside the bamp.
Here V = O so that the curvature is proporticnal to E. Hence,
a larger E is required: that is, the first elgenvalue with bump
is greater than the first eigenvalue without the Dump.
5=25
By essumption,
EE + Uma,
Vi = burp potential enerar, Y = wavefunction with no bap in
potential. The integral is the area under & curve cf PT vs E
Now W = O except mhere it 15 equal to Vo/10, Clearly the area
will be larger 1f tho burp is located where Y*Y 1s relatively
laroe (i.e., in the center for Ty) than 1£ the bp is placed
vhere y*y is small, 1.6., at the edge in this case. Evidently
then, E, 1s larger for the centered bump.
..... ado e eme DT,
al
E |
pe + ny x
S=27
Schrodinger's equation is
E Be-vy-o.
In the region in question, V = V5 = constant, E < V,, so that
?=-Bm->0.
Hence,
v= 2%, pe,
e the general solution, However, pls) = 0, requiring E = 0,
vm no,
as the wavefunction.
Em |
Bince | is real, the probability density P is
Pevipo qê = Ae DE,
Recalling that x is measured from the center of the binding
region, the suggested criterion for D gives
neah) 1 12 Ba (ta)
EP Pao
b «ge,
2 aa, - E)
sa
Use the scheme suggested in Problem 5-26:
E= E +JriVIda,
ma elgenvalue and eigenfunction of the lowest
energy of the infinite, flat, square well potential. From
cegiaa 3
(a) Assumina a wave incident from the left:
region Li = AI, poli, 4 = (2mE) Ay
region 2 p= ME, gil = (mtv, - eia:
region 3: qa coltXs po ÍEiX put De O since thero exists
enly a wave moving to the
richt in this region.
Contânvity of the wavefunction at x = 0 and x = a require that:
A+B=P+G, “o
reta + quê = qolhia, (11)
Contiruity cf dy/êx at these seme points yields
IA -IkD= RP +k6, (us)
ka o] ik,a
=kype + cd = 1k Coffia, tv)
tb) From (1), A+B=Gm=F
From (11), ta+B-ge Fi, gola. qolhia,
peted , potua , c(eta = a, = qelhia, tiia)
Prom (141), Aik, - Elk, = «KA +B-G) + kG,
Alik, + k,) + Bl, — dk,) = 26k,. titia)
From (iv),
ge RIA 4 BG) + a a tk colia,
43
-Aetaa - epa 4 Gk teta ses a csxpoltio, (iva)
Now work with (is), (iiia), (iva). From (lida),
=. -
[e Fe 1A Oo + 1kp) + BO, = 4x,)).
Subgtituting this into (Lia) gives
AL + UA + (4 = ax, fe ba)
tab)
+ Bits - dx, /2ç e + (ã + dk 2 Je ME) a colhia,
and (iva) becomes
A(O + Dm jo dy — 4/3 poi (ve)
x
+ Bilk = 1x,/2k,pebeo + ik 42 ,je ia - e ph ela,
Solve for E in (iib) and substitute into (ivb); i£ q, qº are
defined by
q=1+iky Gel,
the result may be written
+
espiao pude qt
+qe
k
+ Siqufza — que ia, = 10 qholkia.
k
How solve for C/A, using the definitions of q, q*, such as
= tê = atos
etc., to obtain
aiik ) grikia
“is aa
Im
“4
Hence, the transmission coefficient is
wc ate asi fee lia
- - ,
VMA quêo as - qressa qua 4 grhsa!
T
T=160k,/%,) Zig rara 2 hem? qq) qu,
T = 160 DI + ESA E? 4 1606, 2,
2,22
(+ K/e)
qa 14 CIDA as taça,
16 (ky/ko)
Finally,
v
2 2 md õ
5 “me Dao
so that
vêrtv, - E?
km qts?
T= LATE Ee Vr
a ksa 2
T=(1+ eis = é er
sda D
o o
E-5
, À
Z£ ka >> 1, then €42 »» e *28 and the transmission coefficien
becomes, under these circumstances,
Za
Now O < E/v, < 1 and therefore
E És
sEn-Bica
Yo
the upper limit cocuring at E/4, = 1/2. Hence, 1€ 248 > 4,
aZra
160 -É)
o “a
> 1.
Since, in fact, it is assumed that ka >>» 1,
Desa
>> 1,
16-23
o o
and therefore, under these conditions,
E E, clksa
T=16qll-ge a
Wo“
s
FRPTE A
E
%
regioes à regioes 2 region 3
e m——| emma mem mol —
E =
Region 1: q = palbiX , po bx
region 2: v = FoloX, co Dux,
region 3: v= celkix
In these equations,
= es - e
k = (EIA, k = (mê Yo) e
(a) Continaity of the wavefunction at x = 0 and x = a qives
A+BRF+G,
polkia , co kia « golkia
Continuity 0£ db/dx at x = 0 and x = a gives
1k,A - 1x,B = IP - 16,
Fix eloa as 1 gertksa ai ix yoeltia,
(b) These are the ssme as the corresponding expressions in
Probjem 6-5, if in the latter ky 15 replaced with -iks. Making
this alteration in the expression for T in Problem 6-5 yíclds
for the nes transmission coefficient,
2
2 no
= ind E
Eno 163) doa qa),
16(k,/k9)
Using the expressions for k,, k, given above reduces this to
A aiksa,2 0
= ade » P
1 EtÉ -
Vo Yo
&s
(a) The opacity of a barrier is proportional to 2w,a2/Hº and
therefore the lower masa particle (proton) has the higher
probability cf getting thzcugh.
Cb) with vg = 10 Mev, E = 3 Mey, a 0a, Lt folic that
E
16 Veil = o) = 3.36.
The required massos are m = 1.67) x 1027 xg, m, « 29. For
the proton ka = 5.603 and, using the approximate forma,
= 3.360 205.800 3,06 x 10º.
Since mg Mp, as noted above, KA = 215.803) = 8.207. Bence,
for the deuteron,
Tg = 3.360 28207) 2 2,5 x 2077,
s
=19,2
(a) we B- (9x 10º) (Usado o
0 ter 2» 107
=13
vo = Sela gls (432 me.
1.6 x 10 3
(b) E=10kT=(10)(1.38x 10200”) = 1.38 x 193 -
8.625 x 1072 Mey = 0.024.
tc) Mumerically, a=2Zr" -r'=2x 135 m; also,
“to En)
16 at = o - 0.032; ka = y a = 0.91.
Te(l+ Sasha 2, cr D.0073.
td) The actual barrier can be V
considered as a series cf
barriers, coach of constant
height but the heights decreasina
with r; hence Vo-E diminishes
with r and the probability of Im
penetration is greater than for E
ah equal width barrier cf o
constant height Vos
eus E
. d—- - .
1
mal oj ngnã DO
52
In terms of C, the transmitted wave amplitude, the solutions are
5
asl=z c,
6
selicão,.
p= da,
5
c=êle.
The desired transmission probability is
VC" yomc
E. o -2E€
VaRPA x DAMA RA
e
ec Eat = RE AA.
10 - 2º! 10 - =! qo - =**) (10 — 29)
Bow,
at; = qikaçãa
qo = =*ºy (10 = 2º) = 100 - Io(e la 4 qSlka, 41,
(10 = 2*É) (10 = 2Ê) = 101 = Z0cosska.
Tr
—— BR
= ZOcostka”
&-20
ta) In the lowest energy state n = 1, y has no nodes, Bence &y
must correspond to n=2, by ton=3. Since E cn dE =
RA
Ery/Er = 3222; Egy = 9 8%
lb) Ey the same analysis,
2/8 = 2/2, m=1ev.
|
E Ee
(o) The energy in question Ls
mn EE,
and therefore the eneroy of the adjscent level is
Erin + nº dE,
E -E 2 2
Sn mtupon $a
sk, Es nm mn
(b) Tn the classical limit n + e; but
un gl = tin B$do o,
es En = n
meaning that the energy leveis get so close together as to be
Andistinguishable. Hence, quantum effects are not arparent.
ta
The eigenfunctiens fer odd n are
W = Bo cos lnme/a) .
Fer nocralization,
fz
“1 = fax = Eljcos (nme/ajáx = tm [a du,
aa o
1 = 282 (a/ne) (nn/6) = 5 E
Es
for all odd n and, therefore, for n = 3.
À
54
cs
By virtue cf Problem 6-24, the normalized eigenfuncticns are
tg = “É costrex/a).
= À x cos? trurxfa) dx =0.
t
E = Êlccs inuox/a) (1H Êrcce (nr /a) Je - Zoom elru du = 0
ja em
(8)
x2(n=2) = 0.07782,
This changes little with n since the size of the box is fixed,
ter
- 2
p= fumo (124º Ecos (rms) Jó,
Lia ax
krm
p = - Pta Sa = 222,
=
Pê) = as.ga0h)2,
55
This incrsases sharpiy with n since E = nº, E « pê, the particle
moving faster at higher energies.
&-26
(a) Usina the results of the previous problem,
2 h
ax = vê = Aa -—+p
nm
== nrtb.
Hence, for n = 3,
exp = sta - Ep ar) = 2.624.
E
(b) The other results are
n=L dep = 05H,
n=2, bépe 6H.
The increase with n is due mainly to the uncertainty in p: ses
Problem 625.
(c) From (a), the limits as n + — are
ax + da hp + e.
[im 1
am ha ta
[no = drmenêgia = fem - cos (ÊB) ja,
— 4a =
(hs = Fel (coszu - cosujdu = 0,
— =
the integrand being an even function ci us
56
2
The Schrodinger equation in three rectanquiar coordinates is
De Bay. Me-weco.
Inside the cubical region sbere V = 0,
yo, mo,
ax? avê así w
wlxyrz) = KlxbYiy)Ztad.
Then, if * denotes the derivative cf a function with respect to
its independent variable,
CT. E o aenê-o. e
This gives
E = aê, X = Asin(k 0 + Boos(k 2d,
ke = real constant. Sinilarly
* = Csin(ky) +Dmos(kyl: & = Esin(k 2) + Foos(k 2).
Also, from (4),
KE + n2 +12 = men,
Since V = « outside the cubical region, p = O at the bondary:
D=X(0) =Y(0) = 2(0) + B=D=P=0;
0 =X) = Ya) =zla) + Kan kann ka mn
with Meg, = 12,3, ee» « Henoe
v= (CAB) sin (nax/2) inn xy/a) sin in vz /a) ,
És
e - Ev? no) Eloi ong erd.
a
(a) Let M = mass of wing. The zero-point eneroy is
By = (0 + Mu = hhu = h/27,
T = pericã of oscillation. The actual energy of oscillation is
E = aê = us dAÊ = De MAPS.
Thus, the value of Mat which E = E, is
ht 16.626 x 1074
Ne SEo =
“ma dm (10)
This is less than the mass cf an electron. Hence E >» E, and
tha cbserved vibration is not the zero-point motion.
(b) Clearly then, n >> 1 and therefore
E=nhv= 2 Mr +» n= rm.
As an exemple, take M = 2000 kg:
= 1.68 x 109 pg.
Therefore,
a - -
E = 501.055 x 168) — ata .6 x 10181,
4.1 x 10
E = 0.051 ev.
7, ao EE od E 1)
“ao Zap (ame) 2),
For the limits on 1,0,4 ses Problem 7-8. Now,
fora =6,
- o?
210 “Tea >
For the states withm, = lr
af
(are EE
1
z" ZE,»
tora = E Paço rata mA,
and therefore
- 2
“ag —— + 5 e T/RacirPpacdado,
Sina, (img) ag
7 am.
ato + seas
This is the same as (*) above, Hence, regardless of the value of
Mr
Va = E
(b) In the case of L = 0,
Sooo artes 3020 - efagpet/do,
giving
= poe À reg soft (2 — /ag)fe "dart sangandaco,
s
º 2 2 2
domo» ee” dx = — Tanlêneas (2) = 2,»
(c) These results are expoctod since, with V « 1), the average
potential energy seen by the electron 1s the same for all n=2
states, regardiess Of à. Thus, the expectation value Of an
energy will be the same for these states.
LI
R(r) must satisfy E9.7-17:
+2E. Br -vr=2a + DÊ.
É Ê
Apr
maR=,
E ada, ak
a EM
pr,
Substituting these into the radial equation gives
ais np? der! -vty Ze + prl2,
Now E is a constant independent of r, and V = k/r; thus the to
tems in () are proportional to 2º, xº2, As r approaches zero,
dE *, rel, hence, () + 0, and the equation is satisfied.
7=11
(a) To avoid infinities, integrate radially to à finite limit R:
a = 41 - cose),
“quad de “ qued Ea
P(Z3.5%) = 4. 1474.
db) Por this state:
as =5/2 =r/2a
Vono = a7tsão o Pre É coss,
Since this is already normalized,
p= sêar te "en,
o
P = Mt = costy),
PM23.5*) = 11.444.
112
(a) See sketch, following page.
() P = (J0s*6-D?
8 = 54.7º, 125.3º.
te)
3 henoe Po = O at 0088 = 11//3, giving
Pa 0 (0 = 0"),
Ca * 1,
(3oost6 =D? = 1,
3c0628 = 1 =b,
6=35.3", 90º, 144.7".
ru
Let (3,2,-1) represent y(n=3,L=2,mp=-1); (2,0,0)* represent
Wetn=2,1=0,m,-0) etc. Then it is required to ahos that
Ult, = 5(13,0,0)*(3,0,0) + (3,1,0)*(3,1,0) + (3,1,=1)*(3,1,=1)
+ (3,1,1)*(3,1,1) + (3,2,0)*(3,2,0) + (3,2,0*(3,2,1)
+ (3,2,-1)8(3,2,-1) + (3,2,2)*(3,2,2) + (3,2,=2)2(3,2,-2))
is independent cÉ 6,4. Now
(3,2,-20)*(3,2,-2) = (3,2,2)*(3,2,2),
(3,20 "(3,2,-D = (3,2,0º(3,2,0,
(3, L=1)*(3,L,=1) = (3,1, 0*(3,1,0,
66
and therefore
Vs = 843,0,0)*(3,0,0) + (3,1,0)4(3,1,0) + 2(3,1,1)*(3,1,1)
+ 13,2,01*(3,2,0) + 203,2, 1*(3,2,1) + 203,2,2)903,2,2)).
Nos substitute the specific expressions for the various
wavefunctions appearing in the above.
L=2 terms:
2((3,2, 0*(3,2,1) + (3,2,2)9(3,2,2)) =
—s ut 28/280 (1 + 2co02e - 30060)
218) nas,
(3,2,0)*(3,2,0) ai pte2r/das a0osdo - 12.
Hence, the sum cf these terma is
Z n8erdr/3as
Za e
3480) mag,
Independent of 6,4j.
E=1tems:
213,1, 0*(3,1,1) + (3,1,0)*(3,1,0) =
2 E2, 1,2. -2r/3as
as No
independent of 6,0. The L = O terms depend on r only. Thus, all
terms in V5b, have been acocunted for and their sum found to be
independent of direction, so that VãU, 1s spherically
syemetric since it depends om Fr only.
J-16
ta) Leop = Mlsing ds + coticoss da,
tores = Te Pre Eai dO,
Ma au
Jo = target; ooo
= hay a-
Therefore,
Lesopbaj- * Nlicotosing + cotêcosd) by...»
Le cpa Hootolcose + isinpdy,,, = Hootõe!ty,
(bl This result cannot be put into the form
Eesoptai- = Mano
with C independent of r,6,4.
Eu
The cperator is question is given by
2
1 2.
sinta aq?
2 me 2. 2,
12, = tod Esteio do +
By Bq.7-13 this may be written
2, = qo - du Do),
But
E E = og E = ouro + DR- êste - Re),
E? Mb =1G + Dy- ue -wrk,
by Bg.7-17.- Echrodinger's equation is
vê = 2lty - ey.
ré
Substituting these last tvo results into the expression for LÊ
gives e
Eat =uL+ no.
Ez
(a) Let the arca of the cllipse be A; then,
mriA-TA
T the pericã cf revolution. The angular mementum is L = nx/d8/0ty
also, da = tras, so that
Lema mi,
since dA/dt equals a constant in classical mechanies 1f the
force is central. Therefore,
L.d le
nsid ds
(b) This result is identical to Bj.8-5, derived assming a
circular orbit. Ba º
Em
The first apparatus produces two bears, one with spin parallel
(in quantum mechanical teres) to the direction of the field (42),
the other with spin antiparallei. This latter besm is blocked by
the first diaphragm. Hence, a “polarized" besm of atoms enter
the second apraratos, field direction +2!. This second ragnet
produces a new space quantization along 7". In analogy with the
passing of polarized light throuch polaroid (except that the
angle for no transmissicn is 90* in the optical case, 180º in
the atomic), the second magnet allows only the projection of the
entering spins along +2' (not -z"') to pass. Thus, 3f I' is the
intensity of the bem entering the second apparatus ani 1 the
intensity of the unpolarized bes entering the first,
1=SI'(1 + com) = KIA(1 + cosa).
EF!
3
25
The deflectina force is
F = u,88,/d2,
here
Ve 7 Sygr
gince £ = O. If D is the deflection and F is constant,
2 2
D = hatí = 40F/m) (Lv)
L = length cf maqet and v = speed of the atoms. Thus,
2
dB
2 z
D= dl fm) (dB, /dz) My: ge” E meg eg
For atoms enitted from the oven, mw? = 2K7 with T = 1233 K.
Herce,
-23
E. (AD ( 8038 x 1) (1203) 19-0005) . 25 jm.
= (0.5)2(2) 19.27 x 102% (4)
gts
Bs
(a) The orbit and spin eneraies are (aumpB and taum.
Hence, with respect to B = O,
E= SpumçE + ER = + Aa -m+ my E.
(bj Fora = 2, E 0,1 qiving the result:
k m Tg
o o
DE (units oÊ mB)
-1
o
+
74
Thus the energy levei dingram appears as follows:
tuo-fold deg.
E=0Q a 1
asi mo à NE tuo-fold deg.
|
-1 twofold deg.
lc) The maximum separation is
rag TMB = 10.2 eU,
a(9.27 x 102% = (10,2) (1.6 x 10715),
B=4.4x 10º,
=
Since 4,) > 0 and & = & the relation
“30 + uhz IML +) - ate +
IU+D ame += A+ DD.
ti) E = 0. In this case (A) reduces to
j6+0 > 3/4.
But for 2 =0,5=5 (the cnly possibility), so that the
relation in question is satisfied,
75
(14) RO. (a) $=2 + 4. Potting this into (A) aives
L > [3(t + 0),
which clearly is satisfied for all 4 > O,
bl j=L-An,n=1, 3,5, Putting this into (a) yielês
-Mm=-3 > (6 +). 18)
o<(m-DWE+D,
which is satisfied for t > 4 (i,e., j> 0), so the relation is
obeyed here also.
n=3 m this event, (B) gives
qe Ii +.
Evidently this is not satisfied for £ > &, E < O, but is for
D £z <&Botthenj=l-3/2<0, whichis impossíble.
a
=5 (B) now reduces to
o<-zmmem-s.
5 satisfied for some £, e.g. L = 1, bot then j = 8 - 5/2
In fact, put 2 =) + 5/2 to que
0 < 2232 - 77) - 64,
a
which does not hold for j > 0.
Results similar to the last apply to n = 7,9, .«.. eto. Bence,
since j > O the inequality is restricted to the values cf 3
given in the problem.
8-19 4
1
(a) Largest j =4+4 = 9/2: 8
largest. = 9/2. The 1
mag! of the vectors are: N
3= 7/0 + DM = (W/2, SE
L= LU + A = (20)K, + =
Pl E
E = vista + 1))JK = 3N/2. o
76
Apply the law of cosines to the L,5,J triangle:
32 = LÊ + 8? - 218008 (180º - 0),
32 = 20 + 24 2/(2008) Joosô,
6 = cos tie) = S8.81º.
ft) Sinos 1, às antiparaliel to À and bg is antiparaliel to 8,
the angie betueen foi, = S8.91º.
lei
3,
op == 6 = 25.24".
Eu
Define the relativistic energy as
Eq KM
een enftri e
É = v/c. The relativistic momento p is
p=zv=mesti - 65,
Bow
To express K in termas of p note that
= “a -s0-89%, n-pãl=1+p'ê,
so that
E= meta +pBi- ne mea + mp. E + ce.) = 1h,
2 4
Ke mo (ap? = dor) E ne
7
Hence,
2 4
Egck+Vve EE su,
rel Es Bmje?
and therefore 2
Z 4 P
ey “Eres Ea o Eta t+.
Po emo =
If p= Er then
4432
Ep =P /êmçe
Now using classical expressions, ín the spirit of the
apprecimation,
ca
Esv=E pl = imite - 2,
so that
E - - Emp,
With E = constant and V = e2/4re,, the above yields the final
quoted remalt directly.
15
(a) The integrals to examine are
tupteBução Seg leBivçãr.
Since both w; dp are single electron eigenfunctions, each has
the form (n,t,m,). Hence each be written
Yntmg em] integral may
eftm Lama) * (E) (nam dr.
Nou thê parity Of É is odd: P(É) = «E; the parity cf (nst,m,)
18 (=1)º so that the parity of tnetamp tn temp) às (092 and
therefore is even regardiess of whether L is odd o even. Thus
the parity cÉ the integrand above is odd, and the integral over
all space vanishes,
(b) Electric dipole moments constant ir time do not exist,
since the governing integral above is zero. Only integrals
Bz
again since the integrand is odd,
(iii) Transitions between an odd and an ever-n levels
Pe [esneesmama = 2)usin (mu) cos (mu) du,
ka :
p « Açfioin(wo) , sinbntnem), pyoosin(rm) , costu(ntm),,
(rem) Crema)
evaluating the final integral. Cleariy the above is not zero
for n even, m odd, Bence, transítions are permitted only betasen
levels such that cre is cf odd n, the other even n; i.e., the
selection rule is
me, 33, À5, ces a
B-20
The eigenfunctions are
0) = age! m=0,1, 2 e.
Pg = ren.
Put Roosg = =,, Ring = =. Dropping the absolute value signs,
L- sf copio = Efe sero tom,
tee Efeito tia M=m, Mp
since
cos = elf + el).
Nom
21 2m Zz
[ee = |costngjdo + ferem = Msinzm - icos2m + 1).
o
Fence, ifn=1, 2, 3 «e
[E =0.
On the other hand, 1f n = O, then
Zu Zn
its «o =
Thus, 1, É O only if either (1) Mm +1=0, cr (ii) m=1=0:
that Ls, only df êm = É.
Since
sinp = que - ty,
ea Aom+do — Albm-l4
,-E AMO ingão = efe e Jão.
The integrand is similar to the one for 1. Therefore, the
selection rule is im = 21.
f-2
By Eq. 8=43,
Bo o
Poa vao
But F,, depends only on âm; since gm =m, - mg = +1 for both
transitions, Po" mg Hence,
E]
The probability densities ape
L I II
vEbs = lug tLvg (2)v vç (2) + VE VÊ vg (DV 62)
AA
III Iv
= va CLDAA (Ze (OD eg (2) + 45 (Dea ya (Do, (2).
Making the switch interchanging 1 and 2, temes 1 ad Iv
interchançe, as do II and III. Thus,
uso = Mr tir É + Tv) + HIV É II É TED) = dio
AA AA
9-4
From Example 9-2:
Va e Arlo DM (4,03) + tg Me (DO) + 4 (DE (209 (0)
= 4 (vg (2403) = Vig (Dy, (Dp 63) — Va DA (2 (DD
As an example, interchange particles 1 and 3 to qet
1
"a = Tt vg (203) + Va UM (DD vg 3) + Va Dag te 0)
- Va (Uva (3) - V Dvd (3) - Vig DD y 024 0,
ne
The same result is achieved if, instead, particles 1 amd 2, cr
2 and 3 are interchanged,
B5
96
The antisymmetric function for three particles is
1 .
tw FaTtta vg (20, 03 + Va (Dy 12D (03 + + lh (2) vg 13)
- 4 (Uva (vç (3) - “a Ca (2) 63) a a LDA (hp 63
Upon formina Jiabçdr there appears the following terms:
(1) Six terms, each the square cf those above; for exmple,
SOU Vg E CI (DM (24 arg,
= UVA CM, (Ddr, HW (2) vg (2) des) UN (DO, (BD),
= QHIMI) = 1,
assumina that each save function 1s normalized. Hence, these
terms add to 6.
(ii) Cross tezms; for exmple,
DG g a Bm 4 Cy tm ta 3 ddr,
= Va OD (Ddr, HUVg (2a (Dr, MUNDO (DD
= (o)tuo) = 0,
doe to the orthogonal ty of the eigenfunctiona. All of these
cross terms vanish.
Tras, the total integral equals 6 and, since 31 = 6 also, “a
is normalized as originally written.
=
Py asmumptico
ta tepacetenin!
Look first at the symetric space functions; since both of the
electrons are In the se (gromd) state,
8
1
“pace = tro Uyog 1) + go! ng th
Ea ao 2a, va 2 33/2Rra/asy,
Y, must be antisysmetric since the space function was chosen
to be symmetrio [electrons in the same level). The coulcrb
energy is
pias,
Ameg Fiz
vhece r;, é rj — 7 is the distance beiweun the electrons. Thus
V = Wgmindr dação,
in which 9,0, are the spin variables. Now V is independent of
the spin of the electrons, so that 1f the spin wave funcrion is
normalized, then
V= WicaceVspace Hp
Putting in the wave function qives
is escoa veda, riar,sing sino, 108,84 )84,+
"5%
in heh E5 = 85000290 0ye8gehyedo)-
Now suppose that the antisyemetric space function had been
chosen. With both electrons in the ground state, this will be
1 -
Vepace = 75!Vo0 Méygg 12) = Pyoo!2 og UU) = 0.
Tt may be concluded then that with both electrons in the ground
state, the electron spin must be ín the antisymmetric (singlet)
etate, coulcmb interaction, being positive, will increase
the ground state energy over that calculated ky imering it.
2
(c) If the electron is bond, vir) < O. Clearly 12/20? >0
(recall that L is a constant for central forces). For small
encugh Fr. 12/om? >> |vitr)| and v' > 0, indicating a repulsive
core in the cne-dimensicnal fommalism, Oy df va r”,n>2
will this core disappear (unless L= 0.
Iê
(a) The potential in question is V'r
vw va L/amê,
How, in electron volta,
E, rp É tem sold,
drag (r/ag)
Clearly, for E = 0, V"' = VW; ses Problem 9-3, For L = 1,
Vo) ay
s3
=ilike = +
=v+ = =(27.2,
tr/a)? ca]
tb) E, = 220 ev.
ic) The classical region ef motion is shown on the sketch: it
falis within the range for which PY tr), Fig.9-10, is large
=
(d) For L = O see Fig.9-13. The classically permitted region
there falls within r=0.Za, a bit smaller than for £ = 1.
This result also corresponds roughly to Fig.9-10, where Po is
large at r « 0.54. There ís qualitative agreement betueer
classical and quantum results.
+ (27.2) (rag)?
E
(a) From Fig.9-15, the lonization energy for the first electror
is ZM ev. In the ground state the energy of the , from
Fig.9-6, is -78 eV, Thus the enerqy after the First electron is
removed is -78 + 24 = -54 ev. The eneroy with both electrons
removed is zero; thus the enercy needed to remove the remaining
electron is 54 ev.
lb) With the first electron gore, the helium atom resembles a
hydroqen atom with Z = 2. For such an atom the qround state
energy will be
= 22(E,p) = 22(-13.6) = -54.4 ev,
& 1H
and therefore 54.4 ev are required to ionize it. Agreement with
ta) is excellent.
os o grelalad, eo
0 SRH qe
L ——— A — E<o
"Ne misioa 1 hola
in hshell
k + ph hai Kel
4
(c) In the x-ray diagram the levels are inverted, the ground
state is taken as Zero energy, and the energy is plotted on a
logaritimic scale.
IE Es
(6) When the hole is in an inner shell the energy differences
for likely transitions are large; thus the x-ray diagram,
plotting logE, is easier to handle than the standard diagree.
(e) Mhen the hole is in an outer shell, the transitions are
more likely to be optical, and tho associatod energy differences
are small. Hence, the standard diagram is adequate,
9-24
The photom energies are
E=HE, Equev) = éto,
Use of the last expression gives the following:
K, Go): A =0.020mm, E = 59.0 key;
Ko ObR): À =0.0184 1, E = 67.4 ke;
K MK): A = 0.0179 mm, E = 69.3 key.
For the absorption edge, E = 1.2400/0.0178 = 69.7 key = energy
needed to ionize the atom by removing an electron from tha K
shell. Hence, E = O (ground state) + 69.7 = 68.7 ke = ensray
cf the atem vith a hole in the E shell. Then E = 69.7 - 59.0
= 10,7 key; similarly, E, = 69.7 - 67.4 = 2.3 key and E =
69.7 - 69,3 = D.d key,
95
617 k
E
(Mu) m7 tr E
e,
a N
G=25
ta) The line is emitted abén a hole jurps from then = 2 to
anm = 3 level. The eneray required for this is approximately
the energy needed to ionize the atom by removins an n= 2
electron. Using the cne-electron fórmula with =: - 10,
z
5 - 203.6) - E O,2(33,6) = -870 ev.
Thus the required voltage is about AJ0 V.
4b) The wavelength is cbtained from
em =E,- E, 2 -E, = 870 ou; À =L4mm.
39-26
(a) The empirical formula is
vi=cm-a?;xt=cêa- a).
Trus a plot of XY us. £ Is a straight line with a Z-intercept
GE 8 = à emAid slope OEM. Poa Bia-ó-1h, the Z-intercept = a
= 1.7. Also
slope = A «1002093; C=8,6x 19é mi,
(6) Fora: a = 1,7 Ca Rm 1x 108 ml,
2
la) The E absorption edge [n = 1) should be aiven by
Erê * 3.6 ev (3, mn),
edoe
cotatts By = (13.6 ELA? = 6.5 key;
zon: Ey= (13.6) EB? 7.83 ev.
For the inner electrons, the wave functions are essentialLly
tydrocente, with an appropriate effective 3. Por the E, line,
domo
tino = Fi te/ag) Reel,
with the selection rule & = £l. The matrix element is
4
De = ietuçoel = 557 à
Since
É = rsingcoegl + rsinesing] + roostk,
the Z-dependence folloss from
nte 32R/ 2a ar = 2º — Ls,
(82/29)
The lifetime becomes
”
T-]= A » de
£ VePEL,Pb
By Exemple 9-8,
mo = By = Cage
Therefore,
7
Fb 1.41 =
Tg larg) =24M x10,,
Tas 10º s + Ta * 2.44 x 10716 5.
10z
10-7
For the configuration 4834, £, = 0, 2, =2,8,=8, =. Hence,
&'=Zonly, s'=LO. With&'=2,s!=1,3!'=3,2,1qivino
2,1 levels. For &' = 2, 8'=0, )' = 2 only, resultina in a
lo, estate, By the Lande interval rule, the separation is
do, -?o3/00, - o) = 3/2.
The eneray shifts themselves are
SE = R(ING! + DD) (a! + 1) — stds* + 1)),
do: 4 =0;
Jo, ME = 4Ks
giving
JD dE = =
*o,: as =.
The latter three shifts chey the Lande rule.
EE uau oasis
jt=ã 3a,
E aÃ
tm! =a -—
doc anl>-—— —— =-—
O =] 3,
ta
“a
a =
Gplitiing seculo Rd 3
"om
ras
+
103
10-8
The magnitudes of the vectors
are:
= IM L'= vd 8! = vd.
Apelyina the cosine Law:
gts! Lê. 29'L'cosã,
2 = 12 +6 -272c088,
6 = 19.47º.
Again resorting tô the cosine
rules
1? es? 4 Lê 4 25'L'coso,
12=2+6 + 2/1204,
p= 5.7.
Turning to the mecnetic momentos
ne = Um
“= a = “eu
vê = uy? + ujó + 2ujuicoss,
RR az + a/Izubcoesa.14" + qu = 6.69035 .
ve = ul +2- Zepucosa,
A = 6 + 21,599 - 2/5(4.6503)c0s0 + a = 29.50º,
so that F0,=5") = 29.50º = 19.67* = 10.03".
104
os
(a) On Fig.9-13, the colums reveal the last shell being filled,
the row the numbers of electrons in that shall. Therefore,
hs o 182, 282, 2º, 382,
at: 182, 282, 20º, 382, 3p!s
Mas. 152, 252, 2º, 382, 32.
(b) Lg; the configuration represents a filled shell, and tus
aii the angular romenta are zero, leading to 7
Di: there is a síncio valence electron (s = 8! = 4); thus
às' +1=2; &º = 1 giving a P-stai 7 3º = 3/2, 1/2 with the
smaller j* 1ying Jouer, leading to .
: here there are tão £t = 1 electrens; s' = 1,0 and t' =2,1,
O. For the lover enerqy pick the larger 5'. This gives as
possibilities:
= 3t= 32 D31
W=1; 3" =2,1,0: Pã
3,
5.
4 =0 5º =1; 1
The ?D, and 25, states are prohibited by the
Exctuslon princípio. DE te Des “states, the smallest )' los
lowest; hence the ground state configuration should be a
10-11
For a single multiplet s* and &* have the ssme valve for each
level, By the interval mile,
is
Ep = Ke E, i
E = Mg E $
E Rs 3
E
A As. '2
195
Therefore
Eme do dd Ut Up 3405
Since j=1, j,=d j=3% d)=2 3)m 1 5)" 0 Bt
jt=Lta!, Est-], ces |L'-s' |,
so that
plrst=4, 4t-at=0 Lt=8'=2,
and hence the results are
Qt=st=2; 3! =4,3,2,1,0.
Jg-1s
tal The g-factor is
ita) + sta!
g=1+ + ”
(1) For q > 2,
tie) est (ste) — Et(uta). te)
I£ jt= 14º +s*, this becomes
Dema s dtst),
which is impossible. But 1f j' = L' - s!' the relation gives
5" > E",
so that j'=s* - L*, a contradíiction. So try )' =8s' = 4" in
(*), which will now reduce to
2 <s',
which is acceptable, For example, s' = 2, A'=1,)j'=2-1=
1, civing q = 5/2 > 2, as required,
(11) For the case q < 1, the fommila for q requires that
3!) + s! fatal) < EMMA. qua)
If 39! =84º +8*, (4) becomes
106
posts + L's!,
which is impossible. But 1£ )' = 2º - =", (*4) reduces to
2º > 9",
For exmple, 1º =2,8'=1,5j'=2-1=1,9=4<l.
(bl Consider this second caser L' = 2, s'=1,j'=1, g=i
Then,
Lee ovM St = 26, Tm /2m, ROD,ÊO = 150º.
E = 2, dra Lj equal in lensth to É', js
ce ireçes É; de ae aoia Ti rs foda
indicatins that q = 6.
AEseemassentaas
za
10-17
(a) Ip, yielês three levels, the JP; five, 2P) three, and
the “Pp qives one; thus the total mwber is 12.
Wb) For the 35 electron £ = 0; 22441) = 2;
For the 3p electron £ = 1; 2(21+1) = 6.
Clesrly, with (6) (2) = 12, the field has removed the deceneracy
completely. E
18
For a sinalet s = (, so there is only orbital angular rementum
to consider, The potential eneray of orlentation is
me- qi.
107
For orbital anular rements q = 1 so that
La O. cut?
Lºm ue
1 É às in the z-direction,
H Mm ue
B=-,LB=-YLA=-mb = um,
giving rise to 22+1 levels. Since s = O, 4 = jj mking this
substitution, and inserting a factor q = 1, leads to
“E = uam,
which agrecs with Eg.10-22 in the case g = 1.
10-18
In the classical model, picture the memetic field buildina vp
to a valve B from zero in time T. Faraday's law requires am
induced electric field E,
pose
This imparts an additional velocity 4v to the electrons, which
circulate either clockwise or counterciockwise, Hence,
E
asa Er E
erB er
Tempra
The new spoeds aro
era
vv
yieldirg frequencies v given by
“2 Ter O MO
E-hy-hy ifÊB
E =
+
corresponding to Eg.10-22 with q = 1, &mj = 1.
CHAPTEP ELEVEN nd
=? The Debye specific heat is, wlth y = 6/T,
ta) Ba. 11-25 is - ma? = . 13 7
E = By 1 E-1
* Wir
Ey definition, For y << 1, the second tem becomes
hu/kT
=. du, 2, Y = 1
Na ( - né 1 Qevegimred) = PENELA nes?
fe) Let x = k7/hv; T+ 0 implies x + O also, Then, in tens oÊ x | implying that
fx ,
e =2 Lim = 1.
vd + . poe -1
Ae x+0, 0!* > 1; therotoro q, + 381/72, tence, foz mali The first tem is y
x, á
- ame ço a |-E-a.
G = E faço, | =
Nem, D
IÉ y 16 small then, over the range of integration, so 15 X«
Pet Pu rl+ sos extdegee. Expandi ng the integrand, ”
x
2ué qué
y
Hence E E E é alzo ia
r 4|-Eas + 4y E
Lim xe * = «, A .2 restos 1 + 3
x+ 0 o
and Therefore
Ema -m Lim RÉ - 1) = 3h
coa nei O "
For small T, the Dulono-Petit result.
op o ne O qr per 2, 11-10
(a) Let the sample be a square of sides a, so that A = a?, and
oriented as shosm, p.ll4. If standing waves are set up in the
112 material, then,
na
udjacent
undrs
Elx,t) = Enpsin(2mx/A )stndrvt,
Elyt) = EgySin(2ny/A, ) sinZave.
There must be nodes at the houndary: hence,
ina/A = na Zme/a, =n Bol, = 012, ss
If the wave makes an anqle a with the x-axis, & with the praxis,
ard has wavelenath À, then
A = A/cosa; = A/cosg,
2,.24
ng rm BB? (costa + cos?B) = 2,
Ea tn) “
E-Ememp!
us
Each frequency is represented by 3
e a Ro ig
z£ (mê + np! = 2av/v, then the
points nen, represent v. It
follows that all points on a
circle of radius 2au/v stand for
frequency v. The density cf the
no is TS Thus with
== (notas 23%, the muber of
o ce ds
ntxjdr = Rar « iordr,
and this is the allowed mmber of frequencies batwomn v and
tdos
z
ptuidy = 38) Paul = 2a = Biivoy,
(b) There are still gy modas par role, £o that
) m
(rias my EB vão,
0
where À 1s chosen so that there are N atoms in the samples
= 22,2); qm ã
3 À tu) Ye viIN mA) a
The average energy of each oscillator is
E E Er
so that the total enerey of each mole will be
E= Entao» fre Ea.
IE x = bu/kT,
é
RAS Sai
o
But
A= mv tmvês
using this gives
2 x
e =-1
“m o
Finally, observe that x = hy /k7 is dimensionlesa; bence,
hy/k has the dimensions of a temperature; let walk = 0h,
giving
Hm
E= See, fe
Pr
corpo os 6 Timm).
(e) The specific heat per mole is
Hr
8 copie tado
If T/o <<, 6/7 >> IT; tut
Lin É p= 0,
être TT 4
7
and
s/T
selo [e ste peter
a -=1-— = a Einite .
B/Des |e* - 1 e-1
o o
Thus, 2
Limic) = T.
im
Al=il
(a) Since the atoms are distinquishahle, use Eoltzmann
statístics:
pp per
dai EMT
4 n=aA; n=a E
since E, = 0, E, = E. With NM = total rumber of atoms,
nm +n;=M n=al ref E) ae
1+e
The total energy is
E=mã + mão = n8 = 68 = MEO
1l+e
(b) As usual, the specific heat is
mc tÊyZe E/S
=
n= = rs
VT gre
(ci Let x = E/kT; then,
=%
nn.
use
Por x << 1, T large, e «1 and c,/Mk = ia?. On the other
hand, with x >> 1 Tamall, CX = 0, amd c/8k = x2e *, A aketch
oÉ the specific heat vs. T is on the next paso.
122
11-24
The eneray of a Bose system 19, with x = E/kT,
2
E- [proa = UE cano) * oem E am.
hn
o
ec -1
0
As in Problem 11-23, put
“ID. + Pica se, RT
et. 1
witha = e? tho energy becomes
E = ELtand Mer 2a (2 (e 4 pa 2 4 le 4 Jam
H
But,
k
e” 2 Kas = verba = nr) = n 2d
Therefore,
3/2 =
E- etglnpenaa + Ta 4 VE 4 nd).
1=25
“The average eneroy per particle in a Fermi gas is,
approximately,
Since the average energy in a classical cas is 3kT/2, quantum
degeneracy occurs when the second tem in the above is not
regligible compared to the first; 1.e., certainly when
193
soa
v 3, rã
Now 1£ KT << E, then
: :
É = Ecomyn2 Do Eat?
Substitutirs this into the inequality above yields
Hmm,
which clearly is satisfied. Thus, when E >> KT, quantum effects
must be considered.
11-26
ae T=0K,
1 0xE<Es
0 E> Ep-
Hence, the total nunber of particles is
25 3,5
r- mea = [ma = frio MEL”.
h
o
Thus, the averace eneroy per fermicn is
o
E- drama -ifpma.
o o
3,5% 3,4
E = 1 fntuçam 1 16 pz
iss RE 3 = 387/5,
using the expression for N found first.
ng) =
124
li-28
The depth of the well às Vo = my + E = 4.8 ev + Bo. Honco,
1t is cnly necessary to compute the Fermi energy. If KT «<<
Epe then,
à = EA.
The electron density is
a 19,3 x 10 ) (6. =s9x 108 w*.
151
Using this and m = 9.11 x 101 kg gives
E =5.53 e; Vou 4.8+5.53= 10.3 ev.
Algo
It 1s aivon that
2E<2E
nm = Eixm)*; nt) =
0 E> E
(ai The mmber of particles is
Fe
n= urna - | inftamm dar = 4(2m)! be,
o
so that
E = tn /amaê,
(bl The average energy is
femme - frotoem - 4 em para,
o
Usirg the expression for N found in (a) gives E = E/3.
11-30
For silver, =5.5eVat Ta
300 E (T << 105 E, above waich
the material is classical in
behavior). Since KT = 1/40 ev,
kT << Ep, and the Ferai
distribution is close to the
T=0FK distribution.
àpprocimate the distribution by
n = 1 for enorcies less than
&-kT, n=0 for enercies
areater than , and in tha
transition reg. the straight
line
kT
dio E ks E<E +.
(Any reasonable approximation will yield the síme final Desult.)
The number $ of partícles with energies greater than the Fermi
encray is
E A qui
PIRES =|NBnindE = 3 (2?) AU sã + E TaE,
3,4
= Am) 4 452,1, K,52. np, EM
E nr EE Mto asia
E = 2ev tam?) Sermb mo,
the total number of particles, the value calculated at T =
OK may be used: see Problem 11-26, Hence
É = dona = IM ev, 5 no = 973,
ev,
8 5.5ev
CHAPIER THEIVE
Ag=1
The potential energy of E and CI” separated by distance r is
=19,
v= 1 Ida 1 (eis ,. x , 6 = 10 2
E meters) "
im, TF ires r
9 -19,
v = (210 041.62 1000) cy o Láá gy tr in mm).
E (in mm) x 10
The required dissociatior energy is the necative cf the energy
reguired to assenble a KC1 molecule from neutral atoms of K and
1, initially am Infinite distance apart. This latter process
irvolves
(1) comoving an electron from the E atom; energy nesded
= 4,34 EV;
(Ly attaching the electron to the Cl atam; eneray
required = -3.82 eV li.e., energy is released);
(444) moving the neviy created jons from infinity tô
their equilibius positions at a separation oÉ
0.279 mm; this requires
Lis (o,
- Tag = 5161 ev
of energy (energy is Liberated).
Hence, the energy needed to form the molecule ls 4.34 - 3.82 -
5.16 = -4.64 ev. Thus, to dissociate a KCl molecule requires
+4,64 EV Of energy.
lEs
A bound EBr molecule must have negative total energy. Hence,
the possible separstion distances are Eounded by that for which
= O. By Problem 12-1, the total energy of a KBr molecule is
E” na 3.5- 4,3) e,
126
127
Rg in mma Hence, Pomax * Pº is given by
ldd - 0.8 -0; Rº= Bm.
2-5
(a) 1£ a = WÊ/20e7, then
np = nptzr + peste),
at the desired level,
Gn fer = nte (2 -atr +03) =0,
«yr ES!
re tit-va Et
(b) Prom p.426, for HCL, 1 = 2.66 x 1079 gem; alo,
-23
xr = Li28l x 10 1 (600). 6,05932 ex;
1.602 x 10
2 q.oss x 102?
E- 955 x 10 = 0.00261 ev.
(2.66 x 1097) (1.602 x 101º)
Therefore, from la),
= (OD 0 395 + red.
lg
The rotational energies are
Ep=rir+ Z
From Table 12-1, p.429, for E,. KÊ/2 = 7.56 x 10? ev. Thus,
F
g
g
g
E
sible light, visible light photons
must be energetic enough to excite the oxygen cr nitrosen
EM aa 4
barely sufficient (but close). All other visible light photons
are less enerçetic.
12-23
For the laser,
À = 694.3 1m; = 8.320 x 1014 ma,
using e = 3 x 10º m/s.
2 a E
ta) Ep = rlz + DH/2I. mo gr = 2; E = 6/20 = 14.88 x 10
ev. Eg = 0, so My! mB, - By = E, giving vº = 3.593 x 10 a
The Raman lines have frequency v=v! = 4.321 x 1019 - 4 x 10
= 0.4317 x 10) Hz, Or Ay = 694.9 nm
IE there are enough molécules ín the r = 1 state, the r = 1 to
r = 3 transition can be chserved. But E; - E) = 1OK2/21 =
24.8 x 104 ev, giving v' = 5.99 x 10:C pr. Again, the line
frequency is v - vº = (0.4321 - 0.0006) x 10! = 0.4315 x 10!
Hz, indicating that A) = 695.2 em.
() The intensity ratio equals the level poculation ratio.
Since the 1, line originates at r = O and 1, with 1 = 1,
s
aging = (2r + 10 Er/kT,
mm * Eno” ana,
With Ey = 2(2.48 x 10 ev), kT = 1/40 eV, this intensity
ratio |s 2,94.
133
fc) Por vibrations, HE = hu, giving v* = vç. Por Ro, u, = "2360
cale yhich transiates into v! = 7.08 x 1013 Hz. Bence, the line
frequencies are
Wa ve
10 1) = 0.4321 x 10! gr * 0.0708 x 101º Hz,
n =vu- v
o1
w = 0.5029 An = 596.5 mm
mi yx 100 pm; 10
“= 0.3613 da = 830.3 m.
(8) As in QD),
Eg/Toy = mig 2 EBD q qtde,
Since vw; = 7.08 x 10 Hz, by, = 0.293 eU, At z00m temperature
KT = 1/40 ev, and these rubers give
= 10.293) (40) -
Log = E =8.1x10 .
12-24
The energy of the rotational levels, above r = O, are given by
E=r(r+ Ka.
For r=:
VÊJZI = 0.0005 ev tr =0)
=p - '
pp = 0.0004 ev qu" = 1).
Thus the energies of all the levels be
tha can now assigqed, and are
&
134
a FRED = PMS
1 —————— |.opas e
arT—— E v.2ms 27 7
1 FE rem 3 n
D TF na ' +
euersg
lo [e] 172 34 SETE
3 n.nnEs 3
2 mao , T
L nom à
1 dn 5
These infrared transitions are not seen in molecules with
identical ruclei. These molecules display only the Raman
spectrum. The Raman spectrum is seen in ron-identical ruclei
moecuies also. For these transiticns the selection rules are
av =t, dr =0, *2. The resulting transitions are shon above.
TE the muclei have identical spin each equal to zero, then
one set of the Raman lines will be missing: i,e., those that
originate either on even-r or odd-r levels.
135
t
E 4a
13-1
(a) Metallie: charge density unifom, like en electron sas.
(b] Tonic: altematins positive and negative distribution.
(c) Molecular: molecules retain identíties, charce zero between
them,
(4) Covalent: electrons shared, highest charge density betaser
molecules.
13-4
From the text, construct the following table.
Transparent
er opaque Melting
Type Of Solid in visibler Point Mallesbility Qonducttvity
molecular low
tonic transparent high
covalent* most cpacue high
metallie reflective high
(4) Properties vary, depending on bond eneray.
soft
hara
hard
e]
Eleet. & Th.
poor
poor
fair
excellent
Also, by Problem 13-8, for a metal the resistivity increases
linsarly with T near roce temperature. From p.497, the
conductivity oÉ a semiconductor increases with T.
Use of these considerations gives the following results:
(a) metal
15) covalent, since the conductivity increases with T;
(c) covalent semiconductor;
(d) ionic (L£ covalent, conductivity may increase wlth T);
te) molecular, by virtue oÉ low melting point.
15
(a) vm SÉ = (98) É. since É.8 = rpcosa,
a bok <
E=- tá - in.
O x
136
r
137
Bence,
Et lol pd T - fposo 1 pt,
dney 1º e] 1
2
2 = «b52n + scos?o - coco:
r
“co
2
v=-—E q + 3c0g) E.
(árco) É
(b) The force is derived from the potential eneray.
bebe fito tir do
2
£--—S Pro + 3ecs2s)P + 30in280,) = E,
timeç) E
since a > O, F-<0, indicating an attractive force.
13-6
From Fig.12-1, E = S.lev, r=0.24 mm. Since
E = err,
5.1 = (LE(0.24 x 10%),
E=2.1x 101º nc.
lrE
(a) By Bq-l3-Ltab,
on En.
DE A
E» analogy with the classical Boltamarm das,
mar = Der Var
14Z
the aversos number cf electrons per atom is
ALOj(Z) + (O) 2; E
10 + 90 Liz Maoy "1
Therefore, the formula for the Fermi energy of the alloy aives
2z a E
Pe atioy = MPE, cy = (1.066) 6.52) = 6.95 ev.
(b) The bend width is 62s2/2ma? and depends solely cn the
Internuclear spacino. This, by assumption, is unchanged, end
therefore there is no difference in band wiéth between copper
and the alloy.
13-17
la) From Table 11-2, p.408, E = 3.1 eV for Na. By definition,
4
Ep = KT: Te” 3.6 x 10 K.
(bj Room temperature is about 300 E, much less than the Fermi
temperature above. Hence, tô à quod ptT=0K
approximation,
ânstesd of 300 K and use F-D statistics (classical
not applicable since T << To).
tc) Ey Bample 13-2,
TE 52 = 0.008.
13-18
(a) By definíticn,
4 É
vp = (287/m) = 1.20 x 10 m/s,
since Ep = 4.72 e.
(b) The de Broglie wavelength is
a=b. = sit 0.564 mm.
ll = 10 1, (2.29 x 109)
143
This is comparable to the interatomic spacirs a, which car be
estimated as follows: from Table 13-1, p.d5l,
(ngTt = 2.70 x 101 npc,
so that
nt = (0.70 x 102º) (1.602 x 1922) = 2.723 x 102 mw,
a=n 3 0.30 mm,
13-20
The Fermi distribution is emudged cover a region = 2kT wide about
the Fermi energy. OF the May electrors (1.€., pl di
one valence ejectron per atom; ses Table 13-1), NayT/T
their energy increased upon heating (by Example 13-2, Eca Te
>> The ipi cem Pic sa
z
Uz Cy Tp OD +K,
where K = the constant energy cf the other electrons. Bence,
a
Ca = qr = 2a
Por the lattice,
Sat Re
and since k = R8, +
c
= .21I.2
e =-S = 5 séooo = 0.005,
the Fermi temperature from Problem 13-17 (a).
L3=22
In mmentm space, tuo electrons cccupy each point, cut to the
surface Of a sphere of radius Pp: exterior points are empty.
Electrons with »-momentum = Py. en a circle, oriented as
shom on the sketch, next page. The number É of electrons with
Léa
-mementam = A, ig proportional
to the area the circle: L.e.,
+ e 2rÊ = 2rtpÉ - pd),
= 2mêo - péph,
E=1- p/o).
13-;
The ratio of the mumber Mof electrons in the conduction band
to the total raember 3 is, by Example 13-€,
aum = o ii da cu
Comparina conditions at the tuo temperatures T, and T;:
Blat
28/88, = (7,/7;) epi =>
With T; = 300 E, the rumber rmtio = 20, E = 0.67 ev, this
0.248 = 15/20" 3884/M4, m=37X
145
la) The Fermi distribution is
nto « (ENA gy,
and the density of states is
3,4%
nem =am-E)% à = SRD
h
Bence, the number of conduction electrons is
dad
A. (ro - 2) He FEM 4 iyrlar,
Er
If the material is a seniconductor or insulator, it is expected
that, over the range of integration, (E = Ep) >> KT and tus
also (E - Ep) >> kT, so that
Ra = fa - Ee EH) x,
n refrão ve — 2h !E Fo ag,
E
ns pet
The value cf the integral is /n/2 and therefore, insertíng the
expression for A,
E 3/2 (Ep) e.
E ES (ami) e
(b) For the valence band,
RG) = AME, — 25.
146
The number n, cf holes is given from
eme [rm af, ente art
0 o
a [em
a
Since
lt follows that
Í
m
Ê
É
5
é
Ê
5
:
:
ô
E
number of electrons in the conduction band; 1,0.,
nv = state) 2a Cr,
To obtain this result from direct integration, note that 1f
(EE) >> KT, then (E-£) >> kT over the range cf integration.
147
Therefore,
[a- ne, ya = [x gig - e FDA,
a o
= [a - ni - compra, nt VD,
0
he first integral is
[x -n'w=-2n/2,
write the second as
[x, - te Da 2 qem Pen,
o
KT
But d SkT >> 1, so it may be asgmed that >» 1; then,
replace lower limit with infinity with Little error, and
this second integral becomes
E]
eos E E per 32,
o
Thus,
non eAÃ GS MMA do,
For E use the T = 0 valve:
E
a-ade- uia = Sn?
Therefore, the same expression as obtained above is derived
here also: i.e., for nv.
By Example 14-1, Te =4.2 E; hence,
2 19,2
«À E = 2.008 x 10))—!L602 x 10
FE reg DE a E (4.2)
341.381 x 10
== 1300 mm.
142
The relevant equations are
ds
fra. Io E= 53.
(a) The first of the equations above implies that
dg
fra --2-a8
If p= 0, then da/6t = 0 civing FP = constant, not necessarily
ofi-a - o.
bl) IfE=0,
Fut a current exists on the surface and outer 1 oE the
superconductor; 1.e., j) É O. Hence p = O unless «di = 0.
152
153
J=3
Consider the total field inside the superconductor as the mm
oÊ an external field and an interral field de to the currerts
set up in the material as the extemal field changes. Py Lenz's
law, if the total field Lg zero and one tries to change the
extemai fieid, the induced currents flow so as to set up
internal fields cppositely directed, so that the total field
remains zero. If the material can do this, the currents must be
able to respond precisely to the changing extemal field, 1.€.,
p = O. Hence, Meissner effect implies p = D.
Lonz's law states that an induced current will Flow in a sense
so as to oppose the change that produced it, but this carries
no implication that the original change will be amsul led
completely even 1f p = 0 (above, it was assumed that complete
armuliment cocurs). For example, if dt = E(t), the
current set up may be independent of t, Ín which case Bp = O
(Meissner effect) carmot be satisfied for all time.
Ld-4
sermel parasegartiom
supercondnciino
ê= à
154
148
The isctope effect is that
nr. = constant.
For naturally occuring varadim, in atomic mess units,
M = 0.5576 (50.9440) + D.0024(49.94727) = 50.9416.
Hence,
k
(50.5416) “(5.300) = asse Ta
E. = 5.352 E.
16-10
Eg.'s 14-2, 14-3 are
B-ufirud R=od.
rap a
ug = -uçfis He Hed x=-l.
From Eq. 14-4,
Ã= Tê): ua + nÊê = é.
The last cf these with y = =1 gives B = 0, consistent with the
above.
él
(a) Since
B=tnç,
=6, =5
po lxI]0 03 x10 0 04x 30% nm.
(4x x 10) (1)
155
(b) For me ka-mole, n = 6.022 x 1026, then,
2 =24
Mat E = = 16.022 x 10 2.8) (9.27 x 10 Eos
22.4
lá-iz
(a) mith x = pB/KI,
x .ck
M = ny É E
€ +e
e +e
x = 13 = 0.549.
(b) By Fig. 14-56, at M = Mo B/T = 0.53 T/K. Therefore,
x= 12,
E
0.549 =— E =s(0,53),
1.381 x 10
v= La x 102377.
14-13
The magnetization is
mem ER
asmming Curie's law (y = C/T]. Hence,
8; - BT, + T = 0.01 X.
156
Já=14
The mametization is
x -K
E.e
M= Au “mt
Es
with x = pB/KT. IE uB/KP >> 1d, thenel > 1, e “= 1, 80 that
ef >» e.
x
e
Mem my,
4
tha saturation valve.
14-15
antiparalLel
After sore antiparallel=spin electrons make the transition to
parallel-gpín, as indicated, the excess in cÉ elactrons with
parallel spin over those with antiparallel spin is represerted
by twice tné shaded area, which may be aporaximated by a
rectangle. Thus,
E]
Bv (am h
tn = 24,8) (nt ONCEÇOE = 248) (1146
Ora, Ss
in which the factor of 4 occurs since the upper curve stands for,
Orly half the electrons. The total rurber cf particles N of a
Fermi cas is
157
"= EPs,
so that the excess per unit volume is
É = ImByze,
where n = mumber cÉ electrons per unit voltme. Now E = kT and
Vu "SM = (2hu E) = w giving ,
Mm CGI = SBa/26 = Bye
2
a
(b) For copper Ep = Kip = 7.1 eU (Table 11-2) and n = 1.1364 x
1029 ="3 (Table 13-1). Substitution of these into fonmia
above for the susceptihility aives y = 1.62 x 105.
J4-16
(a) The rumbor of dipoles per unit volume, aligned parallel and
antiparallel to the field is '
n, = cnesb/XT, n, = ent,
The eneroy associated with parallel aligiment is “se, with
antiparallel alicrment +jB. Hence, the total energy is
U=vin (8) +n,CuB)) = pace MET qB/T,
Now, with x = VB/eT,
=X,
x
n,+n =n= ente +€e
cats Also,
Put this into the expression for U above to get
163
' Using this to eliminate E, cives
=4.454,
rma+ (0.55)in(l + Ze ) = à + O. 0126,
= me . 2 hrose
15-9 D=K + » Ko Eu mg area) cose
Conservation 0f rass-energy and of momentum (classical form is
used) require that 15-11
2
qi Ent) +90 = 22, 089
Ma = PpYpCOSO + Mv, c0s6; myvgsina = mypeinê 0 = (Dum96) = micr2) jo,
Using classical expressions for K also, this last gives O = [2(25.98260) - SL.94051Juc?,
(2x7) "sine O = (0.02069) (031.5 Mey) = 23.0 Mey.
=
(May) 15-12
Terefore, (a Du imo + Mg Mom no
(x n)" = (xa) Posso = (Mr cos, O = (2.0141022 + 3.0160500 - 4.0026033 - 1,0086654) (uc2),
2
O = 00.0188835) (ucê) = (0,0188835) (931.5 Me) = 17.59 Mey,
(pg ooele = 287, + By m coe2s — 2166 Rm m) cos. :
(b) In Bg.15-16, let a =H), b=n,A=H', B= Bel; then,
But, , Ba
2 sin?o 2 ME = q Mag = 4 tmgm/mg)! = 2/4; K, = 0.5 May.
sinCa = =1- cos 4,
With these, Eg.15-16 becomes
2
2 - sino K
cos q = . 35.68 = 5K /2 - Ki; K, = 15.9 Mv.
Equating the tro expressions for cos gives 15-13
= = The b
Ema + Er, = Rpg — 20,68,8,) "0086 = 0. indirg energy dE is
ae: 8E = (Pm, + Ra, - Matom) (02),
2 0-K, +
ar = (6(1.0078252) + 6/1.0086654) - 12.0900000Jucê,
AF = (0.098943) (931,5) = 92.165404 MeV.
The average binding eneray per nucleon becomes
SE - SEIS 68 mov.
15-14
fa) The energy release is
E= (Mg - mae? = (2(2.0141022) - (4.0026033) Juc?,
E = (0,0256011) (931.5) = 23.8 Mev.
(b) With the nuclei just touching, their centers are 3 F apart:
2 9, =19,2
a E:
(3x 107Pp (1,6 x 1072)
15-15
(a) vith rt = LD go É m,
22 Z
3.7 3 e 3/3 a920M/3
vm e
5 due" "E (âneçd (1.1 x 10) f
this has the same fcrm as the Coulomb term of the rass formila.
(É) The energy coefficient in V above is
-15,2
a = Bt8.988 x 10º)— —Mutg2 x 10 Do 2 0.7854 Meu,
tai x 1072) (1,602 x 102)
In mass units,
a = 97854
331.5
compared to a, = 0.0007€63 u cf the mass formula,
= 0.000843 u,
165
15-16
(a) The bindirs enersy of Sgll às, with masses in u,
1Eg = (my + 6m = npc? = (11.091118 - 11.009305) (931.5 Mey),
E, = 76.209 Mey.
Similarly, the binding eneray cÉ E ss
ER = tm + , -moe = (11.090278 — 11.011432) (931.5 Mev),
E- = 73.445 MeV-
The difference is 2.764 MeV.
(b) From Problem 15-15,
V= (D.TAM Mev) E
ror fell 2 =6 and for “Bl ga 5. Assming the ame r', it is
required that
DES s% = 2.764; 1º = 3.13P,
[c] This is someshat larger than the mean radius (= 2.5 F) of
the charge distribution of €cl? aiven in Pig.15-6.
1517
(a) mor pe? 2 = 26, A =56, N = 30. The mese formia is Bg.
15-30 and the terms are the following:
Mass of separate parts: 1.007825(26) + 1.008665(30) =56.4634 u.
Volume term: -ajA = -0.01691(56) = -0.94696 u.
Surface tem: pa (0.01911) (6928 — O.Z1927 ue
Ooulonb term: a,5 = So. 00976) (26) (58) = 0.134816 u.
Asymetry tom: a, (2 = SA/A = (0.10175) (237/56 = 0.007268 u.
Paíring tem: 2 even, N even so this tem 15 aà =
—(0.012)//56 = -0.001604 0.
(b) To convert to eneray, multipiy by 931.5 Mev/u. To fom the
average binding eneray per nuclecr, crit the first term, add
166
and divido by As
= = [=15.7517 + 4.6528 + 2.24252 + 0.1205 - 0.02668].
volre surface coulob asymetry pairing
There is qc0d agreement with Fig.15-12, except thar the paíring
tem is too small to discriránate from the craph,
(e) The atomic mass is the sum cf the terms in lab: to wit,
55,93664 us
(8) From (bl, AE/A = 8,762 Mev, and agrees well with Fig. 15-10.
IS-1E
(a) The binding energy of 2 45
(Gm; + 60, = mp) (931.5 Mev/u) = 92.1660 Mev,
and thot E “Be! is
(mg + 2 = Mo) (931.5 Moju) = 28.2970 May.
The difference, BE(C) - 3BE(He) = 7.275 MeV, and equals the
binding energy of C cn the atpha-partíicie model.
tb) The binding energy of *0!8, calculated as above, is just
127.62463 Mev. Then, ten, BE(0) - 4BE(Be) = 14,44 MeV, amd is the
binding energy of O on the alpha-particle mdel.
(c) Teg muter of bonde de NIN = 19/21 = 3 for clê (1=3) and
6 for *0lÉ q = 4).
2 q iusis Ji Mo!B, E bonis <>
(4) The energy per bond for (12 4a 7.275/3 = 2.43 mew; Pol6
gives 14.44/6 = 2.41 Mev,virtually identical.
15-21
cl, Protons: an even number; 3 = 0, P even.
Neutrons: 5, 3 in lp; 1 = 3/2, P= (tubo
odd; hence, (3/2, cad).
= Pven E, even Z: (0, even).
ta)
157
B461 3 even, 3=0, Peven. N= 33, last single neustron
in 1Es,9' 3 = 5/2, P= (IJ = 1, civing (5/2, od).
Ra73 Z eve, 3=0, Peven. F=ad4l, last síngie nestron
in 165,9) 1 = 9/2, Po (1)! = 4], predicting (9/2,
even) .
28,61
(b) Discrepancy is : observed is (3/2, odd), and predicted
is (5/2, cd). The 1fs,ar 223,2 levels are very close together,
So cne neutron in 2p,,, couples with the single nestron in
1f + leaving a mucleon in 2p, giving l + 3/9 cultas: tias
5/2 far
5/2. Parity is unaffected since (-1)* = (-nl = (p?=a.
15-23
ta) By Example 15-10, spin £ = 5/2.
(b) By Berple 15-11, parity (-Dº = (-1)? às ever.
(e) Urpaired neutron 1s in a 1d, state: Le 2, j=5/20
j= +. By Fig.15-19, the lower Scheidt line gives -1.6y;,
1.e., negative.
td) 3 = 8, which is a magic munber; hence, q = d.
15-24
(ap PSL nas odd Z, even 5; VD res close to the upper
Q=2 +) line, so that on this basis g=)-h=7/2-1/2m
3 is expected.
(b) By the shell model, an even rumber of neutrons couple in
pairs; the last proton is in 1f,,,; the f-level has £ = 3,
which agrees with (a).
a
(8) Prom Pig. 15-20, q/2r'? = 0.09 > 0, The desired ratio is
14 q/rr'2 = 1.09.
172
18
For À atores,
Rem o ma,
The number of E atoms increases due to adéiticnal À atoms that
to B ators, but decrenses as B atoms decay to C atoms;
SE = OS RSA A + Ma,
FÊ
Meltiply by e'Eº and integrate:
[os + besta = [peruca
o
fEngna - (enctennta,
o o
(Essyt
na = Mo = oa ER Rg
Bet No = O by assumption. Therefore,
se em,
1
173
lé-10
tal
B=12,N=15 E= =14 E=18,NEI3
1.007825 Z 12.0939 13.101725 14.10955
1.008665 [A=E) 15.129975 14,12131 13. 112645
o =0.45657 =D. 45657 =), 45657
apê 0.17199 o.17159 0.17199
aguada 0366219 0,0425823 0.0498493
aptz-a/22/A O.O0847916 0.00054213 0.00094213
ElOtaçias O o o
Summing gives: 42,27 = 26,98097 u; Es, = 75135.965 Mev;
M,3,27 = 26.982379 u; Es, = 25134,086 Mev;
Ma 7) * 26-988406 us = 25139,700 Mev,
40
Bum
(b) The smallest M is tha most stable: this belongs to Z = 13.
(c) Electron rest mass m = 0,0005486 u; rest energy = mc” =
0.511 Mey. Masses in (a) are atoric messes, and so the various
decay posstnilitios ares
(1) Electron enission by 2 = 12 E =, »-My =
25125.965 = 25134.086 = 1.88 Mov.
(ii) Electron capture by 2 = 14; E = Mn" Ma, =
25139.7 - 25134.086 = 5.61 May.
(dt) Positron enission by 2 = 14; E = 04, » My po
= 5.614 - 2(0.511) = 4.59 Mev.
16-11,
ta) By the conservation cf momentum, with the initial mementum
equal to zero,
E
Wee
174
with E = neutrino energy = (0.00093) (931.5 Mev) = 0.8663 MeV,
ignoring tha kinetic energy WrW2 cf the rucleus. Evidently then,
-13
v = É - — (OS6ED 0d A 1) 2 3,98 x 10º ja.
Ne cm 0.661 x 107) (2.898 x 10º)
(bj ho process may be monitored by detecting the x-ray eniesicn
as another electron êrcps into the hole created by absorption cf
the E-shell electron, and other transitions.
16-12
(a) For the electron,
E = pi? + mão!,
The kinetic energy E = E - mc? so that
Rs? - 2-0 Ke (ne! + pio! me,
zf p= me, then
E = m/0 +07) - 2
putting in the indicated runbers gives the following:
pino 2.8 4.5 6.9
R 35 500 250
moiR/p)* 6.916 4.563 2.292
jm? L93 4 s.972
(b) From the gragh, p.175,
RT tmcÊL4s) = (4 Mev) (E) = 4 Mev,
when K, the kinetic energy cf the electron, = RJ j the enercy of
the antineutrino Ls zero, so that the decay energy E = 4 MV.
15
metro ME
oo. tomo
z 2 q 5 g Tm?
16-14
(a) From Fig.16-12, K = 0.8 MeyV. Fig.16-13 has log” = 0,3;
Fenes, Soma
"
P=2 T= 10008; FT=Z2000s.
In sctuality, FT = 2340 8.
(b) The decay is a little siover tha that cf ly), for which PT
= 1200 a.
18-15
(a) Use FT = 2000 8 and assume Mº = 1. Example 16-5 with FT =
1200 s gets É = 3.7 x 10-62 Jem), Since
2
= LET,
the value of f for the present case will be 2.9 x 192 genê,
tt] The result for this process is a little less then for
Example 16-5 dus to the larger FT.
176
tc) The shell structure cf the initial mucleus, a neutron, is
identical with that of the final rucleus, a proton
energy 1s absent with only a single proton as the final nucleus)
and thus tha eigenfunctions are identical.
J6-16
o e
cent , EE
Let a = characteristic distance between charges, j = cusrent
density and p = charge per mit volime,
u=sa=sa= (gala? = ja,
poemas (oa = pa!,
and thus “
BeB/JÍE. E.
E iz Sa
where E = cB for a plane electromaqnetic wave in a vacam.
1€17
Cosa asp * art? Err.
1£ E = Epsin2rff - vt) « Ego
E x
E = (ar coszr tê - vt) = Ena,
implying that
Vuad/Uap * ar egfAgrtEy = 1'/3.
lé-18
Foltow the reasoning en p.580.
(2,even) to (O,even): |iy - 4g] = 2 = Ly since parity does not
change
and L is even, radiation is
electric quadrupole (L = 2).
17
(Lida) to (Oeven)s li, - Lg] = 1 = L; L is Odd and the parity
; indicating electric dipole.
(Lodd) to (2,even)s ba=sA == L; Lis cdd aná the parity
changes; as above, this Indicates electric
gdipole radiation.
16-20
The integral in Eg.16-26 is
The parity of x is even; Sf U$ , is Of 064 parity, the partty
of the integrand is odd and the integral vanishes. For tha
integral to be different frem zero, Wj must have tha eme
parity as py.
The other integral being considered is
[uv ra Bu or.
These parities are:
Pl! = —y, odã ptBl) = BEE) 2 dE, siso ods
Thus the parity of the cperator, irvolving products of those
above, is ever, Às a consequence, Yg must have the some parity
as 4, or the integral vanishes.
163
The net reaction is
e” 4 pl, Mg, all, 2a?
That is,
ue? + mel, - mude!) - meme)
3 Orotay * 20.91 May,
welágs3L, 302 = 20.91
182
put
0=0 m, = 12,m, =m =1, 0 = 90º, K = 7 MV,
to get
o=n0+-na-Ta: MK, = SK
tad E, mL MV; E = 0.154 Me.
(b) E, = 0.001 Mevs Hm 154 GU.
(e) E = 3X7/2 = 318.617 x 1072) (500) /2 = 0.06463 ev. :
(8) Since '
"Fa a Kao" os a ss
it follows that
£p = 0.06863 eU = (591 MEN), Ê
-16.5546 = nlnHh = -0.167ny mm 59.
16-34
(a) o = tada? - une) - mood,
O = (0.00351) (932.5) = 3,77 Mev.
(b) One Meçaton yLelds 2.6 x 10? Mey; hence 48 Megatons gives
1.248 x 100 me. The number of required fusions ís
30
4 = EMBALA. 0,38265 x 2030,
Since to hydzosen atoms-are required for each fusion, the
minimum mass cf hydrogen necded is
m = (0.28165 x 1030) (2)((2) (1.66 x 10273)
m= 254 ks.
CEAPTER SEVENTEEN
124
(a) Eq.7=17, with & = O às
É iE Be -uneo.
(b) mith R(z) = ulr)/r,
E uv,iê dem
a r
Sê S-acsetá
Substituting these into tha equation for R(r) given in (a)
EB eve
fc) E3.5-43, the time independent one-dimensional
equation is Schrodinger
BBemes,
identical in form to the equatimm 0 (b).
(d) Since utu = (r*Rº) (8) = r2 (Ren),
E, E
É A [ a a
1 o
= qi ferotability of the neutror-eoton separaticn being between
F md co.
163
184
te) The reduced mass is
ve
Use of the reduced mass reduces the problem to cone of relative
motion; 1.€., one nucleon is at the center or origin cf the
coerdinates
17=2
(a) Assume a bond state, EV, .
Forr<r'V=0ané
I
I
1
I
E a |
Em e RÉ
—E = — md
In the vegion r > E, Vim v n 1
so that 1
= 4 -=—3
E Sem E s
(b) Replace E with V, - 4E; the equations above and their
solutions will ber E
2
rar's: E + ivo = ama = 03 u= Asinkr + Bcosk,r;
r>r's É - Bju = 0» u = Cear + pelir,
te) From Cb)
rer's * = Bio, - 88 r>r's K = Bum.
1-3
(a), (b) In the solutions given in Problem 17-2,
(ij) D=0, othemviseu+=asr+e,
(ii) B= O, otherwise R(r) = u(r)/fr + Ímas E + 0,
For the remaining, the conditions at r = r' are
x
(1414) u continucus at r's Asink,r' = ar,
Civ) E contimscus at r': KyAcosk r* - = gpatar”,
These last two conditions thus imply that
host (k rt) =.
Substituting the expression for k, and k, from Problem 17-2(c)
gives the desired relation.
19=4
Making the substitutions indicated in Problems 17-1 (8) and 17-3
results im
cottfitm vg) et a - 0%) = EA,
with x = SEN. Mmericaliy, with r' = 2F=2x 105 m and
E
Vo = 36 May,
dtmgvoes = 1.86,
Fence, the equation becomes
cot(1.86(1 = md) - ds
from the text values, cr grester precision
calca In any event, 2.2 Mey will be used in the
following problem to maintain conformity with the text.
186
as
(a) With 4E = 2.2 Me, Vo = 36 Mv, 2º = 2 F, the constanta k,
el 1
k, = 0.90 F 1 k = 0.23 P .
u = Asíni0.90r), r<r'=2F;
=0.23r
4
u=Ca r>r' = 27.
These expressions mist be equal at r'; that is,
=0.46
asin(1.60) = Ce 0-46, c/a = 1.54.
Also, it is necessary that
[emana E [ema -1.
o o
Putting ir the expressiora for u, and € in terms cf A yields
eotifunt ooo + fo sotetet =1.
Evaluating the integrais,
2 =
[5220.0000 = 1,123; (ptetca = 0.866.
o 2
-
Therefore,
auto + (1.54)2(0.866)) = 13 A = 0.16.
Thus, the firal results are, with os dio F,
utz) = (0.16)sin(0,90r), = < 2,
utr) = (age UBE sa.
(&l For v(r), see Problem 17-2.
mM posou
"
Tre radial probability density P(r) = utu = uí(r).
Vl)
0.12
D.u2 1
1
0.05 :
Em 4
D.0ms ;
E 1
É >
E 1 E 1 4 =(2)