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5/21/2018 Solutions Quantum Physics - R. Eisberg & R. 2nd Resnick
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Solutions Supplement to Accompany
QUANTUM PHYSICSOF ATOMS, MOLECULES, SOLIDS,NUCLEI, AND PARTICLES
Second Edition
Robert Eisberg
Robert Resnick
Prepared by Edward Derringh
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Thi s suppl ement contai ns sol ut i ons t o mast of themore- i nvol ved probl ems i n the CUANI UMPHYSI CS text ;wi t h one except i on, sol ut i ons t o probl ems i n theAppendi ces are not i ncl uded.
The suppl ement i s di rected toward i nstructors, andthi s has i nf l uenced the presentat i on. Mot every
al gebrai c st ep i s exhi bi ted. The uni t s have not beendi spl ayed expl i ci t y i n every equat i on. (SI uni t s areadopt ed i n the suppl ement , mai nl y because t hey arebr i ef er than the text notat i on. ) Rul es wi t h regardto si gni f i cant f i gures have not been str i ctl yobserved, al t hough there shoul d be no out l andi shvi ol at i ons. Use of symbol s and choi ce of notat i oni s general l y obvi ous and theref ore r ot exhaust i vel ydef i ned f or each probl em.
I t I s a pl easure t o thank Prof . Ri char d Shurt l ef f(Wentworth I nst i tut e of Technol ogy) f or prepar i ngt he sol ut i ons t o t he probl ems i n Chapt er 18.
Preparat i on of t he suppl ement , i ncl udi ng choi ce ofprobl ems, was l ef t t o the undersi gned, whs was al sohi s own typi st and i l l ustrator. He woul d appreci atea note, of up t o moderat e asper i ty, f romt hose whodet ect an er ror and/ or mi stake.
Decenfcer 24, 1984 Edward Derr i ngh
41 Montgomery Dr i vePl ymouth, MA 02360
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Chapt er O n e . . . . . . . . . . . . . . . . . . . . . . . . . 1
Chapt er TMd . . . . . . . . . . . . . . . . . . . . . . . . . 7
Chapt er Thr ee . . . . . . . . . . . . . . . . . . . . . . . . . 13
Chapt er Four . . . . . . . . . . . . . . . . . . . . . . . . . 18
Chapt er F i v e . . . . . . . . . . . . . . . . . . . . . . . . . 27
Chapter S i x . . . . . . . . . . . . . . . . . . . . . . . . . 41
Chapter Seven . . . . . . . . . . . . . . . . . . . . . . . . . 59
Chapt er Ei ght . . . . . . . . . . . . . . . . . . . . . . . . . 72
Chapt er Ni n e . . . . . . . . . . . . . . . . . . . . . . . . . 85
Chapt er T e n . . . . . . . . . . . . . . . . . . . . . . . . . 98
Chapt er El even . . . . . . . . . . . . . . . . . . . . . . 112
Chapt er T we l v e . . . . . . . . . . . . . . . . . . . . . . 126
Chapt er Thi r t een. . . . . . . . . . . . . . . . . . . . . . 136
Chapter Four teen . . . . . . . . . . . . . . . . . . . . . . 152
Chapter Fi f teen . . . . . . . . . . . . . . . . . . . . . . 160
Chapt er S i x t e e n. . . . . . . . . . . . . . . . . . . . . . 169
Chapt er Seventeen . . . . . . . 183
Chapter Ei ghteen . . . . . . . . . . . . . . . . . . . . . . 194
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CHAPTER ONE
The radi ant enerqy cont ai ned i nvol ume dv t hat i s i rovi nq towardA at any t i me, i n t he f requencyi nterval v, v+dv i s
dEp( v)dv = ppl vl dv ^ dV,
where fi i s t he sol i d anql esubtended at dv by A. Wi t h
= Acose/ r2
dv = r2si n6drd0d.
the enerqy beccroes
dE , (v)dv = pT (v)dv Asi n0cos6drd0d4i .
The energy i n thi s f requency i nterval that crosses A i n t i me tf rcmthe ent i re upper hemi sphere i s
rv/2 r2 T\ i-ct
E , (\>)dv = Apf vl dv ^ I I I si necosSdedi j i dr
J 0=ol
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i d
P=ABT
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(a) L = 4 t iP2oT 4 = 41,(7 X 108 ) 2 (5 .6 7 x 10~8) (570 0)4 ,
L = 3 .6 8 5 x 1026 W.
r _
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(b) 338 = oT4 = (5. 67 X 10_8)T4,
T = 280 K.
1-19
c . 2*k5T5 x5v x > W x> T t r X
h e e - 1
wi t h x = he/XkT. At X = max, x = 4. 965, by Probl em
VNnax' = 42 403,1 ( W) 5/ h4c3.
Now f i nd x such t hat ( X) = 0-2RT (Xmax>!
2nk5T5 x5 k5T^", i -rr-- - (0 . 2)42. 40371 7 ,h e ex - 1 h V
If
4. 2403,
xx = 1. 882, x2 = 10. 136.
Numer i cal l y,
1 = he 1 (6. 626 x 10~34) (2. 998 x 108) 1
x (1- 38 x 10- 23) (3) x'
X = 4. 798 x 10' 3/ x ,
so that
Xx = 4. 798 x 10- 3/ 1. 882 - 2. 55 mn,
X2 = 4. 798 x 10- 3/ 10. 136 =0. 473 nm.
1-20
I f x = hc/XmaxkT, then, by Probl em18,
. Thus,
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s
5 - x
Hence,
t (wmax
But,
x = 4.965;\nax
Upon subst i tut i on, these gi ve
x = 4.965; -f-= (4.965 j^) 4.
r W = 170"(he)
1 - 2 1
By Probl an 20,
pT = 170"T mBX (he)4
so that the wavel engt hs sought must sat i sf y
^ hcTxkT - - 170, - 4 .Xs e ' - 1 (he)4
Agai n l etx = hc/ XkT.
I n t erms of x, t he precedi ng equat i on beoanee
x5 _ 170
ex - l " 16 '
Sol ut i ons are
Xx = 2. 736; = 8. 090.
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Si nce, f or X ^,
these sol ut i ons gi ve
X1 - 1-B15W -
X2 = *614W
1-24
Let X' = 200 nm, X" = 400 nm; then.
1 ______1_____ ,, 1 _______1x, 5 ehc/ X' kT _ ~ (3-82) x5 ehc/ x"kT _ ^
hc/ X- kT _ .
ehc/ XkT _ 382
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CHAPTEK TOO
The photoel ect r i c equat i on i s
he = eVQX + wQX.
Wi th VQ = 1. 85 V f or X = 300 nm, and VQ = 0. 82 \
he = 8. 891 x K f 26 + 3 x 10' 7wQ,
he = 5. 255 x 10- 26 + 4 x 10_7wQ.
Hence,
8. 891 x 10"26 + 3 x 10~7w0 = 5. 255 x 10- 26 +
(b) wQ = 3. 636 x 10_19 J = 2. 27 eV.
Theref ore,
he = 8. 891 X 10- 26 + (3 X 10- 7) (3. 636 X ]
he = 19. 799 x 10- 26 J - m,
(fl) h . 19 ,?9 9_x. . 101 6 = 6-6Q4 x 10-34 ,2.998 X 10
(c) wQ = hc/ X0,
3. 636 X 10-19 = 19. 799 x 10_26/ X0,
Xq = 5. 445 x 10 7 m= 544. 5 nm.
2-8
I n a magnet i c f i el d
r = mv/ eB.
4 x 10
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p = mv = erB = (1. 602 x 10- 19) (1. 88 x 10- 4),
p = 3. 012 x 10- 23 k - m/ s,
(3. 012 x 10~23) (2. 998 x 108) _ 0. 05637 MeV
Al so,
Hence,
(a)
c (1.602 x 10- 13)
E2 =P 2C2 + E2,
E2 = (0. 05637)2 + (0. 511)2,
E = 0. 5141 MeV.
K = E - E 0. 5141 - 0. 5110 = 0. 0031 M6V.
(b) The pbDt cmenergy i s
1240Eph(eV>= S = 0 7 1 = -0175 M6V!
wo - Eph - K - 17-5 - 3. 1 = 14. 4 keV.
2- 9
(a) Assumi ng t he process canoperate, appl y conservat i on ofmass- energy and of momentum:
hv +E0
K + Eq hv Ks
*1!
These equat i ons t aken t oget her i npl y that
P = K/ c.
But , t or an el ectron,
E2 = P2c 2 + E ,
(K + Eq)2 - p2c2 + E*
(*)
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p = / ( K2 + 2EQK) / c. (**)
(*) and (**) can be sat i sf i ed together onl y i f Ep ^ 0, whi ch i snot t rue f or an el ectron.
(b) I n the Compton ef f ect , a photon i s present af ter thecol l i si on; thi s al l ows t he conservat i on l aws t o hol d wi thoutcontradi cti on.
2-14
Let n = number of phot ons per uni t vol ume. I n t i me t, al l
photons i ni t i al l y a di stance < ct wi l l cross area A normal t othe beamdi rect i on. Thus,
= Energy = n(hv)A(ct}_ = nhcv = nh_At At X
For two beams of wavel engths X and X2 wi t h 1 =
f i =, = ni A i + Hi ,I 2 n2/ X2 n2 X2'
and theref ore
n / At X
n2/ At ~ X2
The energy densi t y i s p = nhv = nhc/ X. Si nce thi s di f f ers f rom
onl y by the factor c (whi ch i s the same f or bot h beams) , then :Px = p2, t he equat i on above hol ds agai n.
2-26
Set = 20 keV, Kf = 0; K1 = el ect ron ki net i c energy af t er the
f i rst decel erat i on; then
l J = Ki - Ki : r2 = K1 - Kf - Kr *2 = xi + fi x'
wi t h AX = 0. 13 nm. Si nce
he = 1. 2400 keV-nm,
hh
W
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these equat i ons beccme
1. 240020 - K, ;
Sol vi ng yi el ds,
(a) Kx - 5. 720 keV;
( W 0. 0868 rm; X2 - 0. 2168 i m.
2- 28
Appl y the l aws of conservat i on of total energy and of i ranentam.Cent er of Mass Frame
r-VUVUUU-*
Theref ore,
i t r o t
E1 + me2 - i i c2,
- mv = 0.
mv + me 3mgC,
m( l + B) 3n ,
^ 2 I "
8 - 4/ 5.
Hence, m - 5m/ 3 and E' - mc2B - 4UC2/ 3. By the Docpl er shi f t ,wi t h B = 4/ 5, ~
E' = E{( 1 - B) / ( 1 + B)}%= E/ 3,
E - 3E' = Sl OgC2) - 4b1qC2.
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Laboratory Frame
' VWl Ar
- -
-
X
%
Theref ore,
bo that
2-29
(a)
(b)
E + m c2 = 3 0 + 3K,
E/ c = 3(K2 + 2mQc2K) Vc .
2it1qC2 + 3K = 3( K2 + 2mQc2K)
K - |V 2'
E = 2mQc2 + 3K = 4mQc2.
E + Mqc 2 = Mqc2 + 2mQc2 + K,
E - 2(0. 511) + 1 = 2. 022 MeV.
p = E/ c = 2. 022 MeV/ c; p+ = 0;
p_ - i (K2 + 2m()c2K) %= i {l 2 + 2(0. 511) (1)}%= J
P = (2. 022 - 1. 422) = 0. 600 MeV/ c;
%transf erred =
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Use t he Doppl er shi f t t o conver t t he gi ven wavel engths t o
wavel engths as seen i n the rest f rame of the pai r:
2XJ = \'2,
4 1 - P , 1 +1 + B 1 - B'
3B2 - 10B + 3 = 0,
B =i ; v = f .
2-33
The ramber of par t i cl es st opped/ scattered between di st ances xand x-tdx i s dl (x) = oI (x)pdx. Hence, f or a ver y thi ck sl ab thatul t i mat el y stops/ scat ters al l t he i nci dent part i cl es, t heaverage di st ance a par t i cl e travel s i s
. . f l SS . gp/ xl dx _ x e ^_ dx
av / dl op/ I dx fe~xdx ap
t he l i mi t s on al l x i ntegral s bei ng x = 0 to x = .
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CHAPTE2? THREE
f c iX h/ m .
x _ e e* x h/mxv - m - v -
Xe mx vx'
1. 813 x l O- ^- 9- 9- ^10-"31^ ),mx 3
mx = 1. 675 x l ( f 27 kg;
vl dentl y, the par t i cl e i s a neutron.
3-7
(a) E2 =p 2c2 + E ; (K + Eq)2 = p2c2 + E ,
Ri t K = eV and Eg = m c2, so that
. 2KE- , 2(eV) ( mc2) ,
C (2KEq) = ( )* = {- - - - - 2^-- ) = / ( 2meV) ,c c
and
K/ 2EQ = eV/ 2m0c2.
Theref ore,
X =$ = 77 47r ( l + - ^ r ^ .p = 7( i v r l i + c2
(b) Nonrel at i vi st l c l i mi t : eVmgC2; set 1 + eV/2rruc2get J
X =h/ ( 2m0eV)' ! = h/ ( 2m0K)' s =
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Numeri cal l y,
hence.
bj. = (6. 626 x 1Q- 34 J s) (2. 998 x 108 m/s)
(1. 602 x 10- 13 J / MeV) ( 10- 9 m/ m)
he = 1. 2400 x 10"3 MeV- nm;
. . . . 1. 2400 x 10~3 MeV- r mj l - f,2)'5*' ' Eq (MeV) B
3-19
(a)
p = h _ 6. 626 x 10~34 J - s (2. 998 x 108 m/ s) = 0. 12400 MeVX (10- 11 n0 c (1. 602 x 10- 13 J / MeV) c
E2 = (0. 1240)2 + (0. 511)2 E = 0. 5258 MeVs
K * E - E0 0. 5258 - 0. 5110 - 0. 0148 MeV - 14. 8 keV.
( t>) _
P - V 124kev-These are qartma-rays, or hard x- rays.
(c) The el ect ron mi croscope i s pref erabl e: t he gamna- rays aredi f f i cul t t o focus, and shi el di ng wcul d be requi red.
3-28
(a) Set Ax = 10- 10 m.
_ h _ _ 6. 626 x 10~34 J - s
P_4P **
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= 5. 2728 x 10~25 kg- m/ s 2. 998 x 10R m/ s = 0. 9868 keV_
c 1. 602 x 10- 16 J / keV c
E = Ax, f or the anal l est E use px = Apx
and x = Ax to obt ai n
E =
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E = to(4ik >2 +!>C(to>2 =-~ A ,2 +32n m(Ax)
(b) Set t he der i vat i ve equal t o zero:
= _ +C( f i x) =0 *
Subst i t ut i ng thi s i nto t he expressi on f or E above gi ves
Emi n * " "
3-34
(a) Let the crack be of zero wi dt h and Ax * hor i zontal ai mi nger ror (i . e. , drop poi nt not exact l y above crack) . I bi s i npl i esan i ni t i al hor i zontal speed v^ gi ven by
v = Avx x mAx
As a resul t of thi s, t he bal l l ands a hor i zontal di stance x f rant he rel ease poi nt , gi ven by
Hence, t he total hor i zontal di st ance X f ran crack t o i i rpactpoi nt i s
X = AX + x - A X +
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Put an el ect ron behi nd each
sl i t and observe any recoi ldue t o i ts col l i si on wi t h aphoton emergi ng f ran thesl i t , l b det ermi ne whi chel ectron recoi l ed, i t s observeddi spl acement ay must sati sf y
Ay W,
at l east, or even
W
X
Ay 4J .
Due t o the col l i si on, t he ptoton' s manent umchanges. I n ordernot t o dest roy the i nter f erence pat tern,
mX _ mh _ j nh.px 6 - d - pd - pxd-
Py T d'
m the order of the f r i nge. By conservat i on of manentun, thi s i sal so the uncer t ai nt y i n t he el ect ron' s rranentum. Hence, f or theel ect ron, i t i s r equi red that , i n order not t o dest roy thepattern,
(Ay) = W.
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CHAPTER FtXI R
4=1
Consi der an el ect ron osci l l at i ng al ong a di aneter . When at adi stance r f ran t he cent er of t he atom, t he f or ce on t he el ectroni s
F =
where p = e/ ( 4nR3/3) > 0 si nce the net charge on atai t-el ectroni s +e. Theref ore,
- - - r.4tte0
Thi s f orce i s at t ract i ve: i . e. , di rect ed t cward t he cent er of theatcm. Hence,
F - ma.
I f t he el ect ron revol ves i n a ci r cul ar orbi t of radi us R,
F ma.
The t wo f requenci es are seen t o be equal . The equal i t y appl i esal so to osci l l at i ons of arpl i t ude l ess t han R and ci r cul arorbi t s of radi us l ess t han R, si nce t he charge ext er i or t o theanpl i t ude or radi us exer t s zero f orce on t he el ectron f orspher i cal l y symmet r i c charge di st r i but i ons.
4-4
(a) Manent amconservat i on:
Mv = MucosS + mwcos*,
Musi nS = i twsi n*.
I B
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Ki net i c energy conservat i on:
IjMv 2 = Mu2 + !smw2-
The r rcmentumequat i ons gi ve:
mwcosi j ) = M( v - ucos8)mwsini ti = Musi n8.
Hence, 2 2 2 2 2mw = tr(v - 2uvcos6 + u ).
The energy equat i on yi el ds
m2w2 = Mn( v2 - u2) .
Equati ng the t wo expressi ons f or m2w2:
M2 (v2 - 2uvcos6 + u2) = Mn( v2 - u2),
cos0 = &U-B> + +
(Si nce m 0, a mi ni numf or cos0, a maxi i mmf or 0.
Subst i t ute thi s val ue of u i nto the equat i on f or cos0 to get
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The maxi nun f orce f el t by the al pha- par t i cl e i n passage throughthe at omoccurs at t he aton' s surf ace:
p = _ 1 _ (Ze]. L2eIit. 4ne0 r 2
For maxi mumdef l ect i on, suppose thi s f orce i i rparts trcmentun Apperpendi cul ar t o t he or i gi nal di rect i on of mot i on:
Ap = J Tdt = Fm(At) .
_ 1 4Ze2^ " 4ne0 Rv -
Then, ant i ci pat i ng a smal l def l ect i on 8,
Aj>= 1 2Ze2 _ Usurf
rav 4e0 R( %Mv2) Ka
For gol d, Z = 79; suppose Kq = 5 MeV; then
6 = (8 .9 88 X 109)
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. 1 . 2 ,zZe2,2. 1dN = (-7 ) (-- ?) I n-- j - - - -
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The per i ods of revol ut i on of el ect ron and prot cn are equal :
2 n e2nrp
v_ = v ve P
v = (-Ej v .p e e
The not i on i s abcut t he cent er of mass of t he el ect ron- proton
syst em, so thatr_ m
V pr = m r -E = .ee re
Theref ore,m
v.
4- 22
(a) Frequency of t he f i rst l i ne: = c/ X1 = cH{- - - - -m (nr+1)
Frequency of t he ser i es l i mi t : = c/\a = - 0).m
Theref ore, ^
Av " v - v. - - - - - - r.1 (m+1)
(b) 2Avl v < y ( ^ > 2
i vPf cKj j / f S+X)2
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(l>)
t - i 2
K = 26. 6 - 10. 2 = 16. 4 eV.
Eph, l = 13- 6 + 16- 4 = 30- eV-
(n) By nonentumconservat i on,
i!H = Mv.c
(' i i rbi ni ng t hi s wi t h energy conservat i on gi ves
fiE = hvn = hv + >jMv2 = hv + lsM( ) 2 = hv + - r - .0 ^ 2Mc
0 2Mc 2Mc
v0 = v , l + ^ , ,
(b) Si nce \) = c/ X, vQ = c/ XQ,
&E = (13. 6) - A) = 12. 089 ev.1 3
Negl ect i ng recoi l s
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Wi t h recoi l :
f _ J S , . --------(1 2.0 89 ) (1 .6 02 . x. 1 0 -191 _ _ 6 . 440 x lt f 9 ,
A0 2Mc^ (2) (1. 673 x 10 ) (2. 998 x l o V
AX = 661 am.
4- 34
Hi e ki net i c enerqy of t he el ectron i s
K = (0. 511 MeV) (/ i -" p 11
g = 0.0400,2. 998 x 10
t hi s gi ves K = 409. 3 eV. For hel i i m, t he second i oni zat i onpotent i al f r an the gr ound state i s
, a 4 , W.ftffl.2. . 54.4eV.i on n2 x2
Hence,
4-38
E ^ = 54. 4 + 409. 3 = 463. 7 eV,
, _ 1240463. 7
2. 674 nm.
(a) Hydrogen Hq : Xj j1 =
Hel i um, Z = 2: X^1 = 4 1 ^ - ^> =
I f XH = XHe' 2 = nf / 2 - nf = 4,
3 = nj / 2 -* n = 6.
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(Ii) Nowt ake i nto account t he reduced mass y:
_ i _ 2 = (_ L , 2 = hf e 4ne0 4*K3c 4"0 4n3c h ^
mm m m (4m) m
^ ~me-Hnp =rae(1 " nip1' pHe ~ ( 4 ^) ^ =me (1 ~ Ini 1'
Theref ore,
He V
no that
X = >4rh4 - t|}-
l l once, compared to the hydrogen Ha l i ne, the hel i um&4 l i newavel ength i s a l i t t l e shorter.
(b) Si nce X ji * ( the f actor Z2 i s combi ned wi t h 1/ n? - 1/ n?
to gi ve equal val ues f or H and He) , 1
XH ~ He _ He ~ h . _ _H_
*H He He*
M _1 - me/mp _ 3 ^ _ 1 0, 511 = -4
xH 1 - ^ 7 % 4 p 4 938-3 '
AX = (4. 084 x 10- 4) (656. 3 nm) = 0. 268 nm.
4-42
Hi e i ranentun associ at ed wi t h t he angl e 8 i s L = I u. The totalenergy E i s
E = K = %Ioi2 = jr.
L i s i ndependent of 6 f or a f reel y rot at i ng obj ect. Hence, by
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t he Wi 1 son- Scnmer f el d rul e,
i Ld0 = nh,
LSI de = L(2t i) = / (2IE) (2n) = nh,
A/ ( 2I E) = nj i ,
E =
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CHAPTER FI VE
'J J
In) The t i me- dependent part of t he wavef unct i on i s
e- l j i t/ (C/m) = e- i Et / K = e-i2irv)t_
'l1Tef ore,
ft'1 * 2 *
(l>) Si nce E = hv = 2u)lv,
E =
(-) The l i mi t i ng x can be f ound f ran
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Hence, the desi red probabi l i t y i s gi ven by
Prob. = L V l O n l x 2^ .(nH) ln
I f u = - ^ x .
"fc
Prob. *=2l 7 ^ - e"2/ 2du = 2(0. 42) = 0. 84.
The l ast i ntegral i s t he normal probabi l i t y i ntegral .
5-7
(a) Si nce
V =
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Put t i ng these i nto Schrodi nger ' s equat i on qi ves
i , thi sI wxmes
1 = 2A2| si n2(2i Tx/a)dx = - A2 I sJ o " J o
si n2udu = A2 ?,
A = / 0 .a
(b) Thi s equal s the val ue of A f or the ground st at e wavef unct i onnd, i n f act , t he normal i zat i on const ant of al l t he exci t ed
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states equal s t hi s al so. Si nce al l of t he space wave f unct i onsare si npl e si nes or cosi nes, t hi s equal i t y i s understandabl e.
5-11
The wavef unct i on i s
= (f | lisi n(27i x/ a)e"i Et/ >i ,a
and theref orer+a/2
x = ~l x si n2(2ti x/a)dx = 0.
J-a/ 2
As f or x2: r+a/2 2
x2 = - l x2 si n2(2nx/ a)dx = - -=\u2si n2udu = -t(-|-- >)a2,
J -a/2 24 >
x2 = 0. 07067a2.
5-12
The l i near moment umoperat or i s -i tf and theref ore
r+a/ 2 r
= - \ s i r 3^ {- i H ( s i n ~) }dx = - - As i nu cosu du = 0.
J - a/ 28 J oP " a\elrrT = a
Si mi l ar l y, f +a/ 2 2
3x2
_ r+p2 = Ms i n f i 2*2 ( s i n ^J d x ,
a] a rw* aJ' a/2
r712 = - 8ni 2 ( ) 2i si n2udu = 4it2 $ 2 = 2 .
J o
P ---------- . . .
5-13
Let Ax = (x2)1*; tsp = (p2) 1.
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Hence,
Ax4p = (na) (|) = 4mi (*) = ( | n2 - 2) 1* | = 3. 34 | .
(b) I n the ground state,
AxAp = (0. 18a) ( ) = 1. 13
I n t he f i r st exci t ed stat e t he uncer tai nt i es i n posi t i on andnement umboth i ncrease over t he ground st at e val ues, due to thel i i gher energy of t he part i cl e.
5-14
Hi e normal i zed wavef unct i on i s
_ (On)1/ 8 V( Cm) x2/ 2K - i Et / KT /A K e
(irW 7
wi t h E =
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5 . i a 4 ^V ^^d x ,
Jo
0 = & * 4 =* - w -(b) Thi s same rel at i on, U = T = *jE, i s obeyed by the cl assi calosci l l at or al so.
5-15
Use t he notat i on
dx.( xp>i = - i ^ x f dx, , xp) 2 = - i n j f ^
Cl ear l ypH
(xp) 2= - i nW*( x + V) dx = ( xp^ - 1#,
i mpl yi ng that (i gi ^ and (xp>2 cannot bot h be real . Al so, byi ntegrat i ng by
f +CD+ r f
(xp) 2 = - i H{xy*y| - \ x-f ^d x } = my
Thus,(xp)2 = (xp)*.
I f (xp)1 i s real , thi s l ast rel at i on says t hat (xp)2 i s real
al so, whi ch cont radi cts the f i r st f i ndi ng above. Hence (xp)^ i s
carpl ex and t heref ore so i s (xp)2- I fowt ry
x) V dx;
xp = fc{(xp>1 + (xp)2) = >s{(xp)1 + (xp) *>,
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xp = Re(xp) 1#
so that thi s new xp i s real , as desi red.
5-21
Wi t h V = 0, t he energy of t he phot on i s
E = pc.
Repl aci ng t he energy E and moment ump by t hei r operators gi ves
i t * = - i *cf *.
Now set (x, t ) = i |i (x)T(t) and di vi de the equat i on by T to get
--afeii =K'where K i s i ndependent of x and t. Wr i t e K = k>tc and the t woequat i ons di rect l y above become
= - i kcT + T e_i kct ,
| * =i k* - * = ei kx.
Hence, f or t he photon,
y a ei k( x- ct ) _
5-22
(a), (b) The curvature of ip i s proport i onal t o | v - E| : where|v - E| i s l arge the f unct i on osci l l ates rapi dl y i n x, andwhere | V - E| i s smal l i t osci l l at es l ess rapi dl y (hence, nodesare cl ose t oget her i n the f ormer case, f ar t her apar t i n thel at ter) . I n t he f i r st state, | v - E| i s j ust l arge enough toturn over: no nodes. The 10t h st at e wi l l have 10- 1 = 9 nodes,l eadi ng t o an odd f unct i on si nce V i s synmet r i cal about theori gi n. The wavef unct i on decays exponent i al l y wherever V>E, the
cl assi cal l y f orbi dden regi on. For f ur t her di scussi on, seeExampl e 5- 12, whi ch t reat s the si mi l ar si mpl e harmoni coaci l l at or potent i al .
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'E/C
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(c) Cl assi cal l y, the probabi l i t y densi t y f unct i on P i s gi ven by
P = B2/v
,B2 the normal i zat i on constant . Energy conservat i on gi ves v:
E = %mv2 Cx,
the upper si gn f or x>0. Usi ng thi s,
P = B2(| ),s(E + Cx)"*.
Tto determi ne B, use the normal i zat i on condi t i onf C f + E / C
9 v
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5- 25
Wi t h no bump, t he wavef unct i on wi l l be si nusoi dal i nsi de thecl assi cal regi on of not i on and a decayi ng exponent i al outsi de.The l owest energy wavef unct i on wi l l contai n no nodes.I n t he present si tuat i on, i n t he regi on of the turrp thecurvat ure of the wavef unct i on wi l l be l ess t han cut si de thebump, si nce t he cur vat ur ei s propor t i onal t o E - V. Thi s wi l l
upset t he good behavi or of the wavef unct i on at l arge xassoci at ed wi t h the val ue E, cor respondi ng t o t he f i r st boundstate wi t hDUt the bump. Tto compensate f or t hi s reducedcurvature i n t he regi on of t he bump, a l arger curvature (ascanpared wi t h t he no- bump case) i s needed out si de t he bump.Here V = 0 so t hat the curvat ure i s propor t i onal t o E. Hence,a l arger E i s requi red: t hat i s, the f i r st ei genval ue wi t h t umpi s great er than t he f i r st ei genval ue wi t hout t he bunp.
5- 25
By assumpti on.Eh = E1 + / V*VbVdx,
Vb = bump potent i al energy, - wavefunct i on wi t h no t xnp i n thi
potent i al . The i ntegral i s t he area under a curve of Y*Vb vs x
Now vb = 0 except where i t i s equal t o VQ/ 10. Cl ear l y t he area
wi l l be l arger i f t he bump i s l ocated wher e * i s rel at i vel yl arge (i . e. , i n the cent er f or f j ) t han i f t he I xmp i s pl aced
wher e * i s smal l , i . e. , at t he edge i n thi s case. Evi dent l ythen. Eh i s l arger f or t he cent ered bunp.
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Schrodi nger ' s equat i on i s
+ % E - V)4. = 0.dx K
I n the regi on i n quest i on, V = VQ = constant , E > o-y2 0
Hence,
>(i = Ae~qx+ BeqX,
i s t he general sol ut i on. I fcwever, i l i(x=) = 0, requi r i ng B = 0,l eavi ng
* = Aeqx,
as t he wavef unct i on.
5-28
Si nce * i s real , the probabi l i t y densi t y P i s
P = = i 2 = A2e- 2qX.
Recal l i ng t hat x i s measured f ran the cent er of the bi ndi ngregi on, the suggested cr i t er i on f or D gi ves
A2e- 2q(%a) _ e- l A2e- 2q( W(
e- qa - 2qD = p- aq - 1
D 1 - - - - - - - ts- - - - -
** 2{2m(V0 - E)}*
w iUse t he scheme suggest ed i n Probl em5-26:
E = Ex + ' / Wydx,
i n whi ch E. and are ei genval ue and ei genf unct i on of the l owestenergy state of the i nf i ni te, f l at , square wel l potent i al . Fran
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\
Exampl e 5- 9, 10 the t i me- i ndependent part of i s
Hence,
^ = (-) cos (i tx/a).a
/i *Vi|ax = jc o s 2(irx/a ) VQcos (irx/a ) dx = - ^jc o s 3 (nx/a)dx.
3 %/ Wpf l x = - Mcos udu = -j j -.
E _ 2 2 a**
5- 32
The wavef unct i ons i n quest i on are
Vx = (| )%oos(i rx/a)e- i E, t/ >1s *2 = (|)' ssl n( 2/ a) e_1E!t/ >',
wi t h E2 = 4E1> The l i near combi nat i on i s
V =
Normal i zi ng thi s l ast gi ves
v = o1v1 + c2y2.
1 = / *dx,
c J / TJ dx + c2c*/ Y*Y2dx + c ' c ^^d x + cc 1/ | 1
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(a) The total energy i s E = p2/ 2m+ V. But V = 0 i n t he regi onof not i on, so that
Hence,
* = - + C2y2 , ' ^ (Ci y 1 + C2Y2 ,d x -
a ^/ a x 2 = - ( f ) \ ; a2v2 /a x2 = -< ? f ) 2vr
Al so, by Probl em5-32, / Y*Y2dx = / Y^dx = 0 and theref ore
E = ( ( f ) 2c1c*/*' )' i dx + ( ) 2/ *V2dx},
E - c c , L t c c * ^' H 9__2 + 2 2 2 2ma ma
I = c1c*E1 + c2c *2E2.
But
(b) Si nce Cj C* + c2c* = 1,
E = (1 - c2c*) El + c2c*E2 = Ex + c2c*(E2 - E1) .
Wi th 0 < c2c* < 1, thi s means that
E l < E < E2.
Hence, i f the part i cl e can be f ound ei t her i n l evel 1 or 2,maki ng t ransi t i ons between than, i t s average energy, as woul d beexpected, l i es between the energi es of t he t o l evel s.
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(a)' Hi e probabi l i t y densi t y V*Y has a t i me dependence of
e - i ( E2 - E, ) t / H
and theref ore the f requency i s
v = (E2 - E / h.
E = L td = J jL , ____________(6. 626 x 10~3V
But,
1 2ma2 8ma2 8(1. 67 x 10~27) (10_1V (1. 602 x 10- 13) '
Ex = 2. 051 MeV; E2 = 4E;L= 8. 204 MeV.
Hence,
V C 8 J 04 - 2. 051 = 1 4 8 8 X i 021
4. 136 x 10 X
(b) The f requency of the phot on i s the same as i n ( a) . Hi e
photon' s energy i s
hv = 8. 204 - 2. 051 = 6. 153 MeV.
(c) Photons wi t h thi s energy l i e i n the ganma- ray regi on of t hespect rum.
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CHAFTED SI X
Amnme that
: Ce- l k , x
bA*
-----------
*2 =Ae' 1* 2* + Bei J
wtmre A = ampl i t ude of i nci dent - - - - - - - - -ynw, B = ampl i t ude of ref l ected J
wnvr, C = anpl i t ude of theI i nnmi t t ed wave. There i s no wave movi ng i n t he +x- di rect i onI n reqi on I . Al so, i
k .(M il k - i ^ v rki - >i * 2 - - - - - - - if
nmt i nui t y of wavef unct i on and der i vat i ve at x = 0 i mpl y
A + B = C, - k ^ + k2B = - kj C.
' Hi ene equat i ons may be sol ved t o gi ve the ref l ect i on and theI rnnsni ssi on ampl i t udes i n terms of t he i nci dent ampl i tude, thesui ts bei ng:
k2 ~ kl
k2 + kl
2kA, C
k2 +k lA.
ref l ect i on coef f i ci ent F and t ransmi ssi on coef f i ci ent T nowl cane
R . S I . S * - . ( ?LB2. ,k l k2. 2A*A
'kl + k2
vxC*C
V A*A
Hk 2k ,f__k\ I- - - - \ *
W kl + k2
4kl k2_
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6- 5r a t r B B
*r egi on 1 r r gi no 2 r cgi ns 3
- - - - - - - - - - - - - - - - - - -
(a) Assumi ng a wave i nci dent f r an the l ef t:
regi on I s * Ae1* 1* + Be' i kx, = (2mE)
regi on 2: * = Fe- kj X + Ge*2*, k2 = {2m(VQ - E) }V#;
regi on 3: i(i = Ce*k, x + De ^ l X, t ut D = 0 si nce there exi st sonl y a wave movi ng t o ther i qht i n thi s regi on.
Cont i nui t y of t he wavef unct i on at x = 0 and x = a requi re that:
A + B = F + G, (i )
Fe k2a + cek2a - ce* ' 3, (i i )
Cont i nui t y of d i /dx at t hese same poi nt s yi el ds
i kj A - i kj B = - kj F + k2G, (i i i )
- k2Fe"k?a + k2Gek2a = i k1Cei k, a. (i v)
(b) Frcm (i ) , A + B - G = F;
Fran ( i i ) , (A + B - G) e_kza + Gekza = Ce ^3,
Ae' k2a + Bek2a + G( ek2a - e' k2a) = Ce1*5'3. (i i *)
Fr om (i i i ), Ai kx - Bi kj = - k2(A + B - G) + kj G,
A( i k^+ k2) + B( k2 - i k^ - 2(3c2. (i l i a)
- k2e' k2a(A + B - G) + Gk2ek2a = i kJCei ki a,
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-Rk2e~kia - k2Be~k,a+ Gk2
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Hence, the t ransni ssi on coef f i ci ent i s
Vj C*C - 4i ( k1A 2)ei ki a 4i (k]/ k2)el k, a
T = = q* V ksa - q2ekj a - q*2ekj a + q2*** '
T = 16(k1A 2)2( q V 2(ek2a - e k2a)2 - (q2 - q*2)2}"1,
T = 16(k1A 2) 2{( l + k2/ k2)2 (ek2a - e"k2a)2 + 16 ( k ^) 2}"1.
T = {1 + - - - - - i V ( e k2a - e k2a)2} 1.
Fi nal l y,
so t hat
16 (k1/ k2)
,kl ) 2 E . 2 2 V0(k2 " VQ - E' 1 + 1 2 " VQ - E'
V?/ ( V - E) 2 . . , ,
T = (I + -2___P ____ (p 2a - e 2 1 1T U + 16E/ (VQ - E) , e e > '
(pk;a _ p_k2ai 2 _iT = {1 + E e ' >
16 f-a - f-)vo vo
6-6
I f k2a 1, then ekza e k2a and the t ransmi ssi on coef f i ci entbeccmes, under t hese ci rcumst ances,
T {1 +------jT - } _ 1 .
16
Now 0 < EA' o < 1 and theref ore
16 | - 4,
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Si nce, i n f act, i t i s assumed t hat kj a 1,
2k2a
16%11- f 1,
and theref ore, under these condi t i ons,
T = 16 #- ( 1 - r f j e"2*3.
6-7
Regi on 1:
regi on 2:
regi on 3:
tatrgB-f-
* = Ae1* 1* + Be- i k>x,
* = Fei k>x + Ge- l k, x,ik ,x
* = Ce
I n these equat i ons,
kx = (2mE) '>/)(, k3 = {2m( E - VQ) }*/ #.
(a) Cont i nui t y of t he wavef unct i on at x = 0 and x = a gi ves
A + B = F + G,
Fe',1k, . + gg- i kj a Ce'l k , a
Cont i nui t y of di(i /dx at x = 0 and x = a gi ves
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l k A - l k B = l kj F - I Xj G,
Fi k3ei ksa - i k3Ge"i l Csa = U Ce1* 13.
(b) These are t he same as t he correspondi ng expressi ons I nProbl em6- 5, i f i n t he l at t er k2 i s r epl aced wi t h - i kj . Maki ngthi s al t erat i on i n t he expressi on f or T i n Probl em6- 5 yi el dsf or t he new t ransmi ssi on coef f i ci ent ,
T ={1- - e- Vr1.i 6
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(a)vo = 4i r 0 ? = ( 9 x 1q9) ' 6 J -U M 1 - 6 x 102 x 10- 15
. ...6, 912 X 10~]3 J . 4 32
1. 6 x 10 J / MeV
(b) E = l OkT = (10)(1. 38 x 10~23)(107) = 1. 38 X 10' 15 J =
8. 625 X 10~3 MeV = 0. 002VQ.
(c) Numeri cal l y, a = 2r' - r* = 2 x 10- 15 m; al so,
P E / {2m( V- E) }
16 ^ (1 " = -032' k23 = - - - - > ^ ~a = 0. 91.
T = {1 + &i 84..- ..0- 40.3?2}- 1 . 0 . 0 0 7 3 .
(d) Hi e actual bar r i er can be
consi dered as a ser i es of
barri ers, each of const ant
hei qht t ut the hei qht s decreasi ng
wi th r; hence V - E di mi ni shes
wi th r and t he probabi l i t y of
penet rat i on i s areater than f oran equal wi dt h bar r i er of
const ant hei ght V,O'
fc l5/K
1-l
rtj ' o* 3
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(a) Assumi ng a wave i nci dent f r an t he l ef t , t he wavef unct i on i nthe i ndi cat ed regi ons wi l l be
regi on I s \|>= Ael kx + Be- l kl X,
regi on 2 s * = Fe~i kj X + Ge1* 2*,
regi cn 3 s ip= Cei kl X.
The expressi ons f or t he k' s are
/ {2m( E - V0)}/)4, k2 = / (2mE)/7i .
The equat i ons f or the wavef unct i on are i dent i cal wi t h t hose i nProbl em6- 5 i f i n t he l at t er k2 i s r epl aced wi t h i kj (note t he
di f f erent expressi ons f or t he k' s i n t he tvro probl ems, hcwever)Usi ng Tl k kj ) f ran Probl em6- 5 and maki ng t he change gi ves
T . {1 . . (. ** . e- 1* 3)2) ' 1.16( kj / k2>
But,
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These are f our equat i ons f or t he f i ve ampl i t udes A, B, F, G, C.Sol vi ng rel at i ve t o C gi ves
k kA/ C = ei k!a{cosk2a - %i + ) s i nk2a},
kO 1 I V aB/ C Wf c * - r-=) e 1 si nk. a,
kl 2 l
F/ C = >sei ( k>+k ) a,2
G/ C = lsei ( ki_l c2) a(l + i ) .2
The t ransmi ssi on coef f i ci ent i s T = C*C/ A*A. Subst i t ut i on ofthe appropr i ate ampl i t udes gi ven yi el ds the same expressi on for
T cis obt ai ned above. ,
(b) I n or der that T = 1 i t i s requi red t hat si n k a = 0 whi chI n turn requi res
k2a = nn, n = 1, 2, 3, . . . .
I n terms of t he par t i cl e energy E, thi s i s
a = nn.
2 2,.:- n * *E = ^ 2.
2ma
(c) I n the regi on of t he wel l , t hat i s, i n regi on 2, theprobabi l i t y densi t y i s
= (F*ei ksX + G- e' * 2*) (Fe- i k!X + Gei kj X) ,
k k*il/2 = C*C{~| + (1 - - | ) cos2k2(a - x)},
k 2 k 2
eval uated by usi ng t he F, G ampl i t udes f ound i n (a) . Theosci l l atory par t of t hi s probabi l i t y densi t y has a maxi mi mat
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x = a. I f kj a = nir, i t al so has a maxi i mmat x = 0. I hi s
i mpl i es t hat an i ntegral number of hal f rwavel engths f i t abovet he wel l ; i . e. , that
a = n X2/ 2;
but thi s i s equi val ent t o
k-,a = nn,
as obt ai ned i n (b).(d) One exampl e of an opt i cal anal ogue i s t hi n f i l mi nter f erenceas i n the opt i cal coat i ng of l enses.
6- 17
Numeri cal l y a = 2( 4 x 10- 10 m) and K = 0. 7 eV. E = K + VQ where
nf l n2___________(6. 626 x 10~3V ___________#
8ma2 8( 9. 11 x 10- 31) (8 X 10~10) 2(1. 6 x 10- 19)'
E = n2 (0. 588 eV) .
Set n = 1; then
Ex = 0. 588 ev < K,
whi ch i s not possi bl e. Usi ng n = 2 gi ves
E2 = 22E1 = 2. 352 ev, |
VQ = E - K = 1. 65 eV.
The el ect ron i s too energet i c f or onl y hal f i t s wavel engt h t of i t i nt o the wel l ; t hi s may be ver i f i ed by cal cul at i ng thedeBrogl i e wavel engt h of an el ect ron wi t h a ki net i c ener gy overthe wel l of 2. 35 eV.
6- 18
tfkx = (2rr(V0)%, Xk2 = {2m(9V0)}%; J i kj = {2m(4V())}!l;
These rel at i ons can be summari zed as
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regi on 1 r r gi nn 3
- - - - - - - -oc.
Theref ore,
kx = k, k2 = 3k, k3 = 2k.
= Aei kx + Be_i kx,
*2 = Fe- 3i kx + Ge3i kx,
= Ce'2i kx
Matchi ng ^ and d*/ dx at x = 0 gi ves
A + B = F + G,
A - B = 3( G - F) .
At x = a t he same ccti di t i ons yi el d
Fe- 31te + Ge3i ka = Ce2i ka,
Wr i t i ng
- 3Fe- 3i t e + 3Ge3i ka = 2Ce2i ka.
i ka
these l ast equat i ons bcccme
- 1 1 ?Fz + Gz = Cz ,
- 3Fz- 3 + 3Gz3 = 2Cz2.
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I n terms of C, the transni t t ed wave ampl i tude, the sol ut i ons are
. _ 10 - z6 cA 6z C'
B - 2g6 ~ 5 CB 6z C'
F = I Z5C,
The desi red t ransmi ssi on probabi l i t y i s
V A*A (Kk1)A*A A*A"
C*C = ( ~ A * ) ( SS- g. A) - - - - - - - - 2 ^2 - - - - ^ A*A.10 - z* 10 - z (10 - z*6) (10 - z )
^ z*z = e - ^ = I ;
(10 - z*6) (10 - z6) = 100 - 10(e_6i ka + e6i ka) + 1,
(10 - z*6) (10 - z6) = 101 - 20cos6ka.
Hence,
72T = 101 - 20cos6ka
6-20
(a) I n t he l owest energy st ate n = 1, >|i has no nodes. Hence
must cor respond t o n = 2, ^ t o n = 3. Si nce Efi n2 and Ej =
4 ev,
Ej j / Ej = 32/ 22; EI I = 9 eV.
(b) By the same anal ysi s,
Eo/Kj = l 2/22, Eq - 1 ev.
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(a) The energy i n quest i on i s
n2 * ^.2'
2maEn = n
and theref ore t he energy of t he adj acent l evel i s
bo that
n _ En+1 ~ En _ (n + l )2 - n2 _ 2n + 1
En " En n2 n2 *
(b) I n t he cl assi cal l i mi t n * ; t ut
Ur n ^ = Li m * 1 = 0,rt* n n
meani ng that the energy l evel s get so cl ose toget her as to bei ndi st i ngui shabl e. Hence, quant umef f ect s ar e not apparent .
6-24
The ei genf unct i ons f or odd n are
i|in = Bncos( nnx/ a).
For normal i zat i on,
r%a j -nn/2
1 = J>2dx = B2 \c o s 2 (nnx/a)dx = 2B2 (a/ ntt )lcos 2u du,
J - %a J O
1 = 2B2(a/ nn)(nn/ 4) = B2,
Bn = 4
f or al l odd n and, theref ore, f or n = 3.
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By vi r t ue of Probl em6-24, t he normal i zed ei genf unct i ons are
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2 2Thi s i ncreases sharpl y wi t h n si nce E 5-ija
t he i nt egrand bei ng an even f unct i on of u.
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The Schrodi nger equat i on i n t hree rect angul ar coordi nates i s
-2. . 2. .2.
ax 3y 3z K
I nsi de the cubi cal regi on where V = 0,
a! | + + | + ^ = 0 .
3x 3y 3z K
Assume that
(*yz) = X(x)Y(y)Z(z) .
Then, i f ' denot es the der i vat i ve of a f unct i on wi t h respect t oi t s i ndependent var i abl e,
+ T + + = ' (*>
Thi s gi ves
= - k2; X = Asi n (kxx) + Bcos (k x),
kx = real constant . Si mi l ar l y
Y = Csi nl kyY) + Dcos (k y) ; Z = Esi n( kzz) + Fcos( kzz).
Al so, f r an (*),
kx + *y + kz = W -
Si nce V = out si de the cubi cal regi on, >Ji = 0 at t he boundary:
0 = X(0) = Y (0) = Z (0) + B = D = F = 0;
0 - X(a) = Y(a) = Z(a) -* kxa - nxn, k a = i yr k^a =
wi t h = 1, 2, 3, . . . . Hence
^ = (CAE) si n (nxi rx/a) si n (nyi ry/a) si n (nzi i z/a),
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(a) Let M = mass of wi ng. Hi e zero-FOi nt energy i s
EQ = (0 + %)Ko) = y w = h/ 2T,
T = per i od of osci l l at i on. The actual energy of osci l l at i on i s
E = yt A2 = Ij mA 2 = 2h2MA2/ T2.
Thus, the val ue of M at whi ch E = EQ i s
M = M . (6. 626 x i O- Ul I . l j 68 x 10- 33 kg.A-A-K*-
I Ms i s l ess than the mass of an el ectron. Hence E E. and
the observed vi brat i on i s not the zero- poi nt mot i on.(b) Cl ear l y then, n 1 and theref ore
E = nhv - 2tt2MA2/ T2 * n = 2n2M&2/hT.
As an exampl e, take M = 2000 kg:
n 2, 2(2000).(.C i l l _ 6 x 1035>
(6. 626 x 10 ) (1)
6-30
The zero- poi nt energy i s
Eq = Wu = W(C/ ta)%.
Theref ore,
E_ = |i (l .055 x 10- 34) (--- 1C|3 o'fi) t l . 6 x 10- 19) "1,0 4. 1 x 10 1:6
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(a) Usi ng Efl = 0. 051 eV, t he l evel spaci ng wi l l be
AE = A(n + (rtHu = Ho = 0. 102 eV = 2EQ.
(b) The energy 1of the photon = AE = 0. 102 eV.
(c) For t he photon,
1 - V
But
= AE = Ho,
ph ID,
wher e u = cl assi cal osci l l at i on f requency. Unas,
v = 1 . .(0.-.102) .(1. 6 X 10- 19) _ 2- 5 x 1013Hz-
(6. 626 x 10~39)
(d) Photons of thi s f requency ar e i n t he i nf rared spect nm,X = 12, 000 nm.
6- 32
(a)
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CHAPI TK SEVEN
Tl i e t i me- dependent equat i on i s
Let
y(x, y, z, t ) = \ p(x, y, z)T(t ) .
Put t i ng thi s i nt o the f i r st equat i on gi ves
- T( t) ( | + 2! | + a! | ) + v(x, y, z)*(x, y, z)T( t ) = i # f,3x 3y 3z
assumi ng that V does not depend on t expl i ci t l y. Di vi di ng theabove by t he wave f unct i on yi el ds
- i * v2* + v - * t f = = E-
TTi ere are two equat i ons:
V2* + %( E - V)4* = 0,P
f or t he space dependent part of t he wave f uncti on, and
i M
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The reduced mass i s
mnucme m-nuc e /, e x
si nce each nucl eus t o be consi dered i s sur rounded by oneel ect ron. The charges and masses of t he nucl ei are
mH = mp; "b = anp; "fee = * V h m = l ! ZHe =2'
mp = pr oton mass. The mass rel at i ons are approxi mate. Si nce
me 11836'
i t f ol l cws that
*H(1)2 = V 1 - IS6>(1)2 = - " 95V -
*D(1)2 = me{1 - 2( l i 36r}(1)2 = - 9997me;
Ee WHe( 2)2 " 3-9995n,e-
These gi ve t he rat i os:
Ej j/ Ej, = 1. 0002; E / E^ = 4. 0015.
7- 7
(a) Si nce R21 = re~r/,2a, P( r) = 4aQ.
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(a) The potent i al energy and the ground st ate wave f unct i on are
V = - e2/ 4*e0r, ^ = t f ^3/ V r / a\
Theref ore,
rT\ f2ir
liT *
e- r / a o r 2si nedr ded$ ;
V _____ - 2-- fxe~xdx ______
4i *oao\0 4lieoao'
(b) I n the ground state.
E = - ue4/ (4nc0)22X2.
Si nce aQ = 4ne0H2/ ue2,
V = -2- - - ye' = 2E; E = IjV.
(4nG0) 2H
(c) As f or the ki net i c energy,
E = K + V; E = K + V; = K +V; K = -%V.
7- 9
(a) For t he st ate wi t h = 0,
*210 4 ^o 3/ 2( r / aO)e"r / 2a0cos6
4
so that
v?i n - - - - - - - - - - - - - \ ( r3/ a2) e' r / aocos2esi nedrde
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V2 1 0 - - - - - - 3g-2 - a2[x3e~xdx- (*)
210 24a0 J o
For t he l i mi t s on r,0,i ti see Probl em7- 8. New,
fJ ox3e' xdx = 6,so that
e2 1 e4- - - - - - - - - - - - - - - - - - - 2E- .210 4"E0a0 22 (4ne0) W 2
For t he states wi t h m = t l :
and theref ore
V2i +i - - - - - - - f"---14e"r / asi n30dr d6d*,- 64, a3(4, Eo) J a2
V2l l l = - 24a0We 0)]X e dx
Thi s i s t he same as (*) above. Hence, regardl ess of t he val ue of
VV2i = 2E2.
(b) I n t he case of I = 0,
*200 = 4 7 i W a- 3/ 2
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V20 = - 8 5 ^ o i (2 - *> dx - - 8 3 ^ T (2>= 2E2-
(c) These resul t s ar e expected si nce, wi t h V r- \ t he averaqepotent i al energy seen by t he el ect ron i s t he same f or al l n = 2states, regardl ess of I. Thus, the expect at i on val ue of anenergy wi l l be the same f or these states.
7-10
R(r) must sat i sf y Eq. 7- 17:
- + + [ ( E - V) R = III + 1)- .dr M r
I f R = r1,
dR ___d r " " 1 ' dr 2
Subst i t ut i ng t hese i nt o the radi al equat i on gi ves
III + Dr 1- 2 + {Er 8, - Vr*} I III + l l r1' 2.H2
Now E i s a const ant i ndependent of r, and V = k/ r; thus the tvo
terms i n {} ar e proport i onal t o r*, r*- 1. As r approaches zero,
r*- 2 r', r*- 1; hence, {} * 0, and t he equat i on i s sat i sf i ed.
7-11
(a) Tt avoi d i nf i ni t i es, i ntegrat e radi al l y t o a f i ni t e l i mi t R:
rR /-6
P =4ttR
r ri [dv = -I I 2nr2si n0drde = Ml - cose) ,
R-V3* 4"R J r=cJ 6=0
P(23. 5) = 4. 147%.
(b) For t hi s state:
*210 = 4 7 ^ 0 5/ 2re' r / 2a0cOs6-
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Si nce thi s i s al ready normal i zed.
P = J >2dV = -]6Y2-T,)a~5| [r2e' r / acos20r2si n0drd8.
' 0J0
P = %(1 - cos30),
-P(23.5) = 11. 44%.
2=12
(a) See sketch, f ol l owi ng page.
(b) P (3cos20- l ) 2; hence P j ^= 0 at cos0 = i l / / 3, gi vi ng
9 = 54. 7 , 125. 3.
(c)Pnax = 4 (6 = 0- ),
i Pmax = ^
(3c o s 26 - 1 ) 2 = 1 ,
3cos26 - 1 = l l ,
e = 35. 3", 90, 144. 7.
7-14
Let (3,2, -1) represent ip(n=3,11=2,m l ) ; (2, 0, 0)* represent
(n=2 , a=0 ,ma=0 ) etc. I ti en i t i s requi red to stowthat
= f { ( 3 , 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, 1)* (3,2,1)
+ (3, 2, 1)*(3, 2, -1) + (3, 2, 2)*(3, 2, 2) + (3, 2, - 2)*(3, 2, - 2)}
i s i ndependent of 0, $. Now
(3, 2, -2)*(3, 2, -2) = (3, 2, 2)*(3, 2, 2),
(3, 2, - l )*(3, 2, - l ) - (3, 2, 1)*(3, 2, 1),
(3,1, -1) *(3, 1, -1) - (3, 1, 1) *(3, 1, 1),
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*5*3 = | {3, 0, 0)*(3, 0, 0) + (3, 1, 0)*(3, 1, 0) + 2(3, 1>1)*(3, 1, 1)
+ (3, 2, 0)(3, 2, 0) + 2(3, 2, 1) *(3, 2, 1) + 2(3, 2, 2)*(3, 2, 2) }.
Mow subst i t ute t he speci f i c expressi ons f or t he var i ouswavef unct i ons appear i ng i n the above.
I = 2 t erms:
i ndependent of e,i j>. The i = 0 terms depend on r onl y. Thus, al l
terms i n have been account ed f or and thei r sumf ound to be
i ndependent of di recti on, so t hat 41* ^ i s spheri cal l y
syi met r i c si nce i t depends on r onl y.
7- 16
2{(3, 2, 1)*(3, 2, 1) + (3, 2, 2)*(3, 2, 2)} =
2 (8 1) 2ira
(3, 2, 0)*(3, 2, 0) - - - - - r4e' 2r/3a
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Theref ore,
Lx, opt 21- l = HR - % E - v>Rr 2}.V1
37(r2 9r' + U* - 2
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(a) Si nce
Pop - - ! ? * = ei kx'
Pop' 11= f r ~ (' i >') UWe 11" = #**.
and thus t he ei genval ue' i s Kk.1kx
(b) Usi ng * = e , r epl ace k i n (a) wi t h - k t o get theei genval ue ->0c.
(c) These resul t s i ndi cate that measurement s of rrcmentun wi l lyi el d Jflc preci sel y.
(d) \p = si nkx, coskx ar e not ei genf unct i ons si nce 3/ 3x convert ssi ne t o cosi ne and cosi ne t o si ne.
(e) These states, not bei ng ei genf unct i ons of t he moment umoperator , wi l l not yi el d preci se val ues of t he remant un uponmeasurement .
7-20
(a) Wi t h R a const ant and V = 0, t he total energy E i s
E = K + V = K.
But t he ki net i c energy K i s si mpl y
K = !jlu)2 - I j l (L/ I )2 - L2/ 2I ,
where I = rotat i onal i ner t i a akcut t he z-axi s. Hence
E = K = L2/ 2I .
( b ) 2
Lop = Lz, op- - i 1l*'-
Eop = f t
Al so, = y (r,e,
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Let *)T(t).
Subst i tute thi s i nto t he energy equat i on of Probl em7- 20(b) anddi vi de by to obtai n
l i! T * . _ l *i i df * _ i Mi _ E
21 V * 21 * d *2 T d t
where E i s the separat i on const ant . Thus two equati ons emerge:
d2t
d$2and
(b) f = T = -fT.
7-22
(a) Fr an the precedi ng probl em,
5 E - _ i E _ _ . i E dt . T _ e- i Et / Kdt H T; T " n dt, T - e
The normal i zat i on const ant wi l l be i ncorporated i nto $.
(b) The sol ut i on above represent s an osci l l at i on of f requencyoi gi ven by
E - fa.
But thi s i s the de Brogl i e- El nstei n rel at i on. Hence E i s t hetotal energy.
7-23
The equat i on f or i s, f ran Probl em7- 21(a) :
d2 -
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Thi s i s anal ogous t o the cl assi cal si npl e harmoni c osci l l at orequat i on:
d2x 2 ~ + (i) x = 0 .dt
The canpl ex f orms of t he sol ut i on wi l l t heref ore be
* = ce1*1+ c*e_l m>, m = / (2I E) / H.
7- 24
(a) The par t i cul ar sol ut i on bei ng consi dered i s
t =e1*1= cos (m)>) + i 8i n(m)i ).
Si ngl e- val uedness requi res that
4i(0) = 4(2n) 1 = cos(2i roi ) + i si n( 2nm).
Hence,
si n(2i tm) = 0 ; cos( 2nm) = 1 ; m = 0, 1, 2, . . .
(b) By Probl em7-23
* - * , . - 0 . 11 . ! ! . . . .
(c) Sol vi ng t he probl emvi a the ol d quant umt heory gi ves
E _ L2 _ (nH)2 n h 221 21 rT' n l * 2' 3'
Evi dent l y t he new versi on, i n cont rast t o t he ol d theory,i nt roduces an m = 0 (E = 0) state; al so, t he exci t ed l evel s arenow t wo- f ol d degenerate.
(d) Apparent l y there i s no roomf or zero- poi nt energy si nce Ri s assumed constant . I n actual i t y, t he masses, on a mi croscopi cscal e, wDul d be aterns whi ch osci l l at e sl i ght l y, so that I)cannot be assumed t o be f i xed.
7- 25
The compl ete wave f unct i on i s, f ran t he precedi ng probl ems,
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The wavef unct i on must be normal i zed at any t i me t: thus,
= 1,
1 = j (N*e- i m* ) ( Ne* ) ^ = N*N(2n) ,
N * N =i
7- 23
Use * = M ' V * 1; Lz = - i *
(a)2 t t f 2 iT f 2 iT
(b)
f= U*( - :Jo JO
L = m>(.
Al so, f ran (a)
L = mV .
Si nce L2 = L2, measurement s of L wi l l yi el d mtf exactl y.
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CHAPTER EI GHT
(a) I t t he area of t he el l i pse be A; then,
U* = i A = | A,
T t he per i od of revol ut i on. The angul ar moment umi s L = mr 2de/ dt>
al so, dA = >ir2de, so that
t _ dA _ . Adt " t '
si nce dA/ dt equal s a const ant i n cl assi cal mechani cs i f thef orce i s cent ral . Theref ore,
_ e Ur = aL. &_ _ e_HI T 2m 2m' L 2m
(b) Thi s r esul t i s i dent i cal to Bq. 8- 5, der i ved assumi ng a
ci r cul ar orbi t .
8- 4
The f i r st apparatus produces t wo beams, one wi t h spi n paral l el(i n quant ummechani cal terms) t o t he di rect i on of t he f i el d ( +z),t he ot her wi t h spi n ant i paral l el . Thi s l at t er beami s bl ocked byt he f i r st di aphra n. Hence, a "pol ar i zed beamof at oms enterthe second apparatus, f i el d di rect i on +z . Thi s second magnet
produces a new space quant i zat i on al ong z . I n anal ogy wi t h thepassi ng of pol ar i zed l i ght through P olaro id (except t hat theangl e f or no t ransmi ssi on i s 90 i n t he opt i cal case, 180* i nthe at omi c) , t he second magnet al l ows onl y the proj ect i on of theent er i ng spi ns al ong +z (not - z ) t o pass. Thus, i f I 1 i s t hei nt ensi t y of t he beament er i ng t he second apparatus and I t hei nt ensi t y of t he unpol ar i zed beament er i ng the f i rst ,
I = %!(! + cosa) = W0(l + cosa) .
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8 -5
The def l ect i ng f orce i s
F = uzdBz/ dz,
where
W s '
si nce = 0. I f D i s the def l ect i on and F i s constant ,
D = tet2 = !s(F/m) (L/v)2,
L = l ength of maqnet and v = speed of t he atoms. Thus,
D = * 5 (dBz/dz) (L/ v)2;
gs b s
For atoms emi t ted f romt he oven, Ijniv2 = 2kT wi t h T = 1233 K.
Hence,
f z _ 8KTO = 8( 1. 381 x 10~23) (1233) (0. 0005) = ^ T/ m_
62 'LW s (0 5)2(2) (9. 27 X 10- 24) (Is)
8 -6
(a) The orbi t and spi n enerai es are
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Thus t he onecay l evel di agr amappear s as f ol l ows:
m m EX s
- - - - - I
^ 0
\ l
- - 1
\ l
_- 0
1 +% +2
0 +%-+1 tvi o- fol d deg.
B = 0 ^ r 0 two- f ol d deg.
0
* } - i-h J
- 1 t wo- f ol d deg.0
- 1 ~h - 2
(c) The naxi j numseparat i on i s
max " 4bB = 10*2 ev-
4(9. 27 x 10- 24)B = (10. 2) (1. 6 x 10- 19),
B = 4. 4 X 104 T.
8-8
Si nce I,j > 0 and s = *s t he rel at i on
/ { j t j + i d > \Aia+ d ) - /{s(s + i)> |,
becomes
j ( j + 1) > H I + 1) + | - / {3UI + 1)}.
(i) I = 0. I n thi s cas^ (A) reduces t o
j (j + 1) > 3/ 4.
But f or *. = 0, j = % (the onl y possi bi l i t y) , so t hat t herel at i on i n quest i on i s sat i sf i ed.
(A)
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(i i ) *. * 0. (a) j = 8, + %. Put t i ng thi s i nt o (A) gi ves
whi ch cl ear l y i s sat i sf i ed f or al l I > 0.
(b) j = I - Vn, n = 1, 3, 5, . . . Put t i ng thi s i nto (a) yi el ds
n = 1 (B) becomes
0 < (2 - 1) ( + 1),
whi ch i s sat i sf i ed f or I > % (i . e. , j > 0) , so the rel at i on i sobeyed here al so.
n = 3 I n thi s event , (B) gi ves
Evi dent l y thi s i s not sat i sf i ed f or I > V, < 0, but i s for0 0.
Resul t s si mi l ar t o the l ast appl y t o n = 7, 9, . . . etc. Hence,si nce j > 0 the i nequal i t y i s r est r i cted t o the val ues of jgi ven i n the probl em.
> V{3( + 1)},
-n + k(n2 - 2n - 3) > -{3. (. + 1)}. (B)
0 < 38. (-28, + 1)
0 < - 222 + 3 3 1 - 9
0 < - 22j 2 - 77j - 64,
8-10 /
(a) Largest j = 4 + %= 9/ 2;l argest m. = j = 9/ 2. The
L = / {( + 1)}K = / (20)tf,
S = / {s( s + 1) }H = /3tf/2.
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Appl y t he l aw of cosi nes t o the L, S, J t r i angl e:
J 2 = L2 + S2 - 2LScos( 180* - 9),
^ = 20 + | + 2/ {20( | )}cos6 ,
6 = cos_1 = 58. 91*.
(b) Si nce u. i s ant i paral l el t o L and i s ant i paral l el t o 2, +
t he anql e between u. , y = 58. 91".X s
(c)
COS* = f = i 9 ; * = 25. 24*.
8-12
Def i ne the rel at i vi st i c energy as
Er ei = K + V.
Now
K = me2 - nv.c2 = nuc2{--- L- y- - 1},v 0 / ( I - 8 )
B = v/ c. The rel at i vi st i c i ranentump i s
2 -*sp = mv = m0ce -
K V 2 ( p , 2 - ^ 4) 4 ' ^
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and theref ore 2
1 =*W - Ecl = ( ' ^ +V) - ( + VK
I f p =pcl * then
r el - - P4/ 8 c 2.
Mow usi ng cl assi cal expressi ons, i n t he spi r i t of theapproxi mati on,
| ^ + V = E; p4 = 4m2(E - V) 2,
so thatFT - 2EV + v
AE- 1 ---------------t o ? '
Wi th E = const ant and V = - e2/ 4neg, t he above yi el ds the f i nal
quoted resul t di rectl y.
8-15
(a) The i ntegral s t o exami ne are
/ (erj i pj dr; / ( er i dx.
Si nce both are si ngl e el ect ron ei genf uncti ons, each has
the form>l'nj ln = (n. l . m) . Hence each i ntegral may be wr i t ten
e / ( n . f c . m^J d i .
Now the par i t y of r i s odd: P(r) = - r; t he par i t y of (n.H. m)
i s (l )a so that the par i t y of (n. f c. mMn. H. m) i s ( - I )2* and
theref ore i s even regardl ess of whether SL i s odd or even. Thusthe par i ty of the i ntegrand above i s odJ , and the i ntegral over
al l space vani shes.
(b) El ect r i c di pol e i reroents const ant i n t i me do not exi st,si nce t he governi ng i ntegral above i s zero. Onl y i ntegral s
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preceded by e" ** Ef^ , dependi ng on t i me, may be non- zero.
8- 16
For n = 2: SL= 0, m = 0; S. = 1, m = 1, 0, -1. Hence the
i ntegral s t o be consi dered are
(i ) / ( l , 0, 0, )*(er) (2, l , l )dr; (i i ) / (1, 0, 0)*(2, l , 0)drs
( i i i ) / ( l , 0, 0) *(er) (2, 0, 0)dT,
where = (n, , ma). Al so,
r = ( rsi nBsi n*) ! + (rsi n6cosi (i )] + rcosei c.
Subst i t ut i ng t he expl i ci t expressi ons f or the wavef unct i onsgi ves the fol l owi ng f or t he i nt egral s above.
(i )
e/ e- r / a (r) e_r / 2a0ei 4'r2sin6drd6d
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p . O , ^ V ^ d r * 0; j lcos28si n0d6 = j ^ 0 .
Thus here al so / ( l , 0, 0) *( er ) ( 2, l , 0)dx ^ 0 and a t ransi t i on i sal l owed, agai n i n accord wi t h the sel ect i on r ul e si nce here41- 1.
(i l l ) Fi nal l y, t hi s i ntegral gi ves
e/ e~r / a (r) (2 - r / a0)er/'2ar2si nedrded41.
The I , ] t erms vani sh f or t he same reason as i n ( i i ) . But thi st i me the 0 i ntegral i n the k termi s
csi necosede = 0
Thi s t i me, then, / ( l , 0, 0) * (er) ( 2, 0, 0) dT = 0 and the t ransi t i oni s f orbi dden. TTi e sel ect i on rul e i s obeyed si nce M. = 2 between
t he t wo st at es and the sel ect i on rul e i s Ail = 1.
8- 17
I t i s desi red to check t he sel ect i on rul e An = i l by eval uat i ng
X " el **f u*ni du * H nf u*ni du,
si nce t[>(u) i s real and u x.
(i ) nt = 3, nf = 0. I n t hi s case,
I f (3u - 2u3)e_l !U e_l iU udu = 2 3u2 - du.
I = 2{3( i m%) - 2( | nS} = 0.
But An = 3, so t he sel ect i on rul e i s not vi ol ated,
(i i ) nt = 2, nf = 0:
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si nce t he I ntegrand i s of cxi J par i ty) i n - 2 i n thi s i nstance,
(i i i ) n = 1, nf = 0.
2I l uV " du = 2 (k ) ? 0,.p
J0
and An = 1.Thus t he sel ect i on rul e An i l i s obeyed i n these three cases.
8-18
(a) Fran EQ. 8- 43, t he t ransi t i on rate R i s
R =i ! * y
30hc
wi t h
3l Uoex( 1ax|2'J -CO
V - ( C/ m) *5.
The i nt egral i n the expressi on f or R i s, wi t h u = (On) x / ,
..2 _1_.2 u r , -.2
P = 2eA0Ai yr ' i U"xue- ' i U dx = 2eAQA1
The normal i zat i on const ant s Ap and A1 are det ermi ned f ran:
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Sol vi ng these l ast f or AQ, t he expressi on f or P becomes
Put t i ng thi s and the expressi on f or v i nt o R gi ves f i nal l y, f or
8-19
The i nf i ni t e square wel l ei genf unct i ons, apar t f romthe not -
needed normal i zat i on const ant , are
* (x) = si n
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agai n si nce the i nt egrand i s odd.
(i i i ) Transi t i ons bet ween an odd and an even- n l evel s
rHn> r Hmp
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New,
r2ir r 2n r2nr r
le in9de = lalo 'o
coe(n6)d6 + i | si n(n0)d0 - ( si n2mi - i cos27i n + i ) .
Hence, i f n = 1, 2, 3,
r2"
le n6de = 0.Jo
On the ot her hand, i f n = 0, then
r2n r2n
j ei n6d0 = 30 - 2n.
Thus, I x ^ 0 ral l y i f ei t her (i ) tm + 1 = 0, or (i i ) t o - 1 = 0;
that i s, onl y i f Am= i l .
Si nce
s i n ^i f e ^- e ' 1*),
I y = = { I i ( te+1)* - ei (ta- 1, *>d.
The i ntegrand i s si mi l ar t o t he one f or I x - Theref ore, the
sel ect i on rul e i s Am= i l .
B-21
By Bq. 8-43, 3 2
R12 ' J12P12
*01 uoi poi But Pf i depends onl y on Am; si nce Am= - mf = +1 f or bot h
t ransi t i ons, P12 - PQ1- Hence,
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* 01
si nce v =* ( AE/ W. But ,
Em
and theref ore
bim*01 o2
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CHAPTER NI NE
9-3
nie probabi l i t y densi t i es are
, 1 11*S*S = t 4>*(l)*|(2)A)B( l)*a(2)
A AI I I I V
1 #*,
*k = V
The same resul t i s achi eved i f , i nstead, par t i cl es 1 and 2, or2 and 3 are i nterchanged.
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The ant i syi met r i c f unct i on f or t hree part i cl es i s
a = TJ Tt Pa(1) (2) (3) + *B(l )*Y (2)*a (3) + *Y (l )+a (2)*e (3)
- *Y (l)H-B(2)+a - *B( l )*a(2)*Y(3) - * a(l )*Y (2)*B(3)) .
Upon f ormi na fVppp&t t here appears t he f ol l owi ng terms:
(i ) Si x terms, each the square of those above; f or exarpl e,
/ * U) *g (2) ** (3)*a (1) *B (2) *Y (3) dt j ai j dTj
= {J \| (1) *a (1) dTl ) {/ ** (2) (2) dr2) {/ ** (3) (3) * 3},
= UM1H1} = 1,
assumi nq t hat each wave f unct i on i s normal i zed. Hence, theseterms add t o 6.
(i i ) Cr oss terms; f or exampl e,
/
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''space = 7 2 ^ 100(1) *100121 + l OO '*' l00(1)
=M (t )3/2e'2r,/a>(t ,3/2e'2r2/a0)-\(jgpi n must be ant i symmetr i c si nce the space f unct i on was chosen
t o be symmet r i c (el ectrons i n the same l evel ) . The coul crrbenergy i s
v4lt0 r12'
where r 2 f ^ - r2 i s t he di st ance between the el ectrons. Thus
V = /*JVi|'A'3T1dT2do1do2,
i n whi ch Oya2 are t he spi n var i abl es. J fcwV i s i ndependent of
the spi n of t he el ectrons, so t hat i f t he spi n wave f uncti on i snormal i zed, then
V = Space space l a-
Put t i ng i n t he wave f unct i on gi ves
V = - ^- f e - 4( r>+ / a^ r 2dr 1r 2dr 2si nel si ne2d81de2d*1d$2,
* a003 12
i n whi ch r 12 = ( r ^r ^e ^, ^, ^) .
ttowsuppose that t he ant i synmet r i c space f unct i on had beenchosen. Wi t h bot h el ectrons i n t he ground state, thi s wi l l be
' space = > 100(1I W 2) * *100(2)*100(1)) "
I t may be concl uded then that wi t h both el ect rons i n the groundstate, t he el ectron spi n must be i n t he ant i synmet r i c (si ngl et)
state. The coul onb i nteract i on, bei ng posi t i ve, wi l l i ncreasethe ground st at e energy over t hat cal cul ated by i gnor i ng i t.
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Consi der tD ei genf unct i ons sol ut i ons of
u2 A- , u2 d2*,
- h - ^ 1 ( 1 > - | ; ^ r + v * i = E i * i -
Take the canpl ex conj ugat e of (2) t o get
vi2
- f e ^ r + v* I - Ei
(i ) I f Ej j* E^ and the systemi s bound, the wavef unct i cns
approach zero at bot h i nf i ni t i es. Thus the i ntegral vani shes and
Uj l dx = 0.
(11) I n the ccnt i nui mregi on (unbound syst em) , t he wave f unct i onremai ns f i ni t e at l arge posi t i ve and negat i ve x. I n pract i ce,towever, box normal i zat i on i s i nvoked and t he wavef unct i onvani shes at the sur f ace of t he box, so t hat the resul t abovei s achi eved here al so.
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(i i i ) I f Ej = (degenerate case) , const ruct
a = V i + V j '
al so a sol ut i on f or t hi s potent i al . I f a a. ar e chosen proper l y
can be made orthogonal to
J tyj l i j dx = 0 = a*/\pj \(i jdx + a*/ i ), * dx,
a*f >f tpj dx + a< = 0,
and so choose
i j = " / *J *. j dx-
9- 13
- MWr ) |
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where = ground st ate energy of t he hydrogen atan. Val ues of
Z(r) may be taken (not easi l y) f r cmFi g. 9-11.
(b) The resul t s are shown on t he precedi ng page. The f i rstthree l evel s of argon are
I n t he ground stat e (n = 1) of hel i i m. Fi g. 9- 6 gi ves E. o -80 eVand theref ore 1
(b) Wi t h so f ew el ectrons, i t i s not cl ear whet her an i j i ner or
cut er shel l i s bei ng descr i bed. I f = n = 1 an out er shel l i s
i ndi cated; f or an i nner shel l , Z1 =Z - 2 =2 - 2 =0 .
(c) The f act t hat Z1 equal s (or r cughl y equal s) nei t her n = 1
(outer shel l ) nor Z - 2 = 0 ( i nner shel l ) i mpl i es that the
Hart ree met hod i s not appl i cabl e t o hel i un. Thi s i s not ver y
surpr i si ng si nce a stat i st i cal met hod cannot be expected t owork wel l wi t h so f ew par t i cl es (two el ect rons) .
E1= - 3500 eV; E2 = - 220 eV; = - 16 eV.
9- 14
(a) Bq. 9-27 i s, wi t h E^ ref er r i ng t o hydrogen.
- 8 - Z ^ = - Z*(13. 6),
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(a) Fr an Fi g. 9-6, Em , = +30 eV;coul
E = 1 coul 4np r '
30 = (9 x i oV 1-6 X - y ,r ( 1 .6 x 10
coul
r = 0. 048 nm.
(b) E__ _ = +9 ev - r = 0. 16 nm.
9- 17
For the el ect ron r x p and t i s perpendi cul ar to the pl aneof t he orbi t . Now v / 0 anywhere i n the orbi t and theref orep # 0. I f L ^ 0, t hen r ^ 0everywhere and the el ect ron
avoi ds the nucl eus (r = 0) . I fL = 0 , t he el ect ron voul d moveon a st rai ght l i ne t hrough thenucl eus.
(a)
E = K + V = - + V,
E = + p!> + v -
But L = rp (si nce r x p, = 0);hence.
2'
(b) I n one di mensi on ^ = 0, L= 0 and E - Pj / 2m+ V1. Thi s and
the precedi ng equat i on are f ormal l y i dent i cal i f
V = L2/ 2mr2 + V.
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(c) I f t he el ect ron i s bound, V( r) < 0. Cl ear l y L2/ 2nr2 > 0
( recal l t hat L i s a constant f or cent ral forces) . For smal l
enough r, L2/ 2mr2 | v(r)| and V' > 0, i ndi cat i ng a repul si vecore i n the one- di mensi onal f ormal i sm. Oi l y i f V r -*1, n > 2
wi l l t hi s core di sappear (unl ess L = 0).
9-18
(a) The potent i al i n quest i on i s V' i
V' = V + L2/ 2mr2.
Now, i n el ect ron vol ts,
L2 = H I + 1)H2 _ J S L + 1) = (13. 6)ML +_ U.
2mr 2 2mr2 2ma2 (r/ aQ)2 (r/ a0)2
Cl ear l y, f or 1 = 0 , v' = V; see Pr obl em9- 13. For 1=1 ,
v'(r) v
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(b) E2 = - 220 eV.
(c) The cl assi cal regi on of mot i on i s shewn on t he sketch: i t
f al l s wi t hi n the range f or whi ch P2i *r ' Fi 9- 9- 10, i s l arge
(1 = 1).
(d) For !. = 0 see Fi g. 9- 13. The cl assi cal l y permi t ted regi on
there f al l s wi thi n r = 0 . 2aQ, a bi t smal l er than f or i - 1 .Thi s resul t al so cor responds roughl y to Fi g. 9-10, where P->q i s
l arge at r = 0. 5a . There i s gual i t at i ve agreement between
cl assi cal and quant umresul ts.
9-22
(a) Fran Fi g. 9-15, the i oni zat i on energy f or the f i rst el ectron
i s 24 eV. I n t he ground st ate the energy of t he atern, f ranFi g. 9-6, i s -78 eV. Thus t he energy af t er the f i r st el ectron i sraroved i s - 78 + 24 = - 54 eV. The energy wi t h both el ect ronsranoved i s zero; t hus t he energy needed to remove the remai ni ngel ect ron i s 54 eV.
(b) Wi t h the f i r st el ect ron gone, t he hel i umat an resembl es ahydrogen atan wi t h Z = 2. For such an at cmthe ground stateenergy wi l l he
and theref ore 54. 4 eV are requi red to i oni ze i t . Agreement wi th(a) i s excel l ent .
E1 = Z2(E1h) = 22( -13. 6) = - 54. 4 eV,
9-23B = 0mmitmt
L
" kiM\
6
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(c) I n the x- ray di agramt he l evel s are i nverted, t he groundst ate i s t aken as zero energy, and the energy i s pl ot t ed cn al ogar i thmi c scal e.
' *86
eo
*#
nn t i f -1 T 1 T 1 T I T l H f
(d) When the hol e i s i n an i nner shel l t he energy di f f erencesf or l i kel y t ransi t i ons are l arge; t hus the x- ray di agram,pl ot t i ng l ogE, i s easi er to handl e than the st andard di agram.
(e) When the hal e i s i n an out er shel l , t he t ransi t i ons aremor e l i kel y to be opt i cal , and the associ at ed energy di f f erencesare smal l . Hence, the st andard di agrami s adequate.
9- 24
The phot on energi es are
EheX
E(keV)1. 2400X (rm)
Use of t he l ast expressi on gi ves t he fol l owi ng:
K ( M) : Xa
Kg (M*K): X
Ky (N*K): X
0. 0210 rm, E = 59. 0 keV;
0. 0184 rm, E = 67. 4 keV;
0. 0179 nm, . E - 69. 3 keV.
For t he absorpt i on edge, E = 1. 2400/ 0. 0178 = 69. 7 keV = energy
needed t o i oni ze the atcmby r emovi ng an el ect ron f ran t he K
shel l . Hence, ^ = 0 ( gnxmd state) + 69. 7 = 69. 7 keV = energyof the at cmwi t h a hol e i n the K shel l . Then E^ - 69. 7 - 59. 0
= 10. 7 keV; si mi l ar l y, = 69. 7 - 67. 4 = 2. 3 keV and =
69. 7 - 69. 3 = 0. 4 keV.
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() LMN
9- 25
(a) Hi e I*, l i ne i s emi t t ed when a hol e j unps f ran the n = 2 toan n = 3 l evel . The energy requi red f or thi s i s approxi mat el ythe energy needed t o i oni ze the atan by rerrcvi ng an n = 2el ect ron. Usi ng the one- el ect ron f ormul a wi t h = Z - 10,
E2 = - ( ) 2(13. 6) = - I 2'6 )2(13. 6) = - 870 eV.
Thus t he requi red vol t age i s about 870 V.
(b) The wavel engt h i s obt ai ned f ran
_ F - fA VE3 - E2 " -E2 870 eV; A = 1. 4 nm.
9- 26
(a) TTi e empi r i cal f ormul a i s
A- 1 = C( Z - a)2; A- ^ = C%(Z - a) ._!
Thus a pl ot of A vs. Z i s a st rai ght l i ne wi t h a Z- i nterceptof Z = a and a sl ope of / C. Fran Fi g. 9- 18, t he Z- i ntercept = a= 1. 7. Al so
sl ope =0 s = J t - :~17~ (105)! C = 8, 65 x l t)6 m_1-
(b) For a: a = 1. 7; C = = 11 x 106 nf 1.
9- 27
(a) The K absorpt i on edge (n = 1) shoul d be gi ven by
Eedge =
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H t u s ,
Cohal t : E* = (13. 6) P - f 2 -)2= 8. 5 keV,
i ron: ^ = (13. 6) & - f - h 2 = 7. 83 keV.
(b) For phot on energi es great er t han 7;83J W* J ^t hat i ron wi l l absorb the photon di mi ni shes shar pl y. Wi t h 8. 5greater t han 7. 83, and si nce 8. 5 keV f al l s near t he K edge f orcobal t , where t he probabi l i t y f or absorpt i on i s hi gh, a Photon
energy near 8. 5 keV woul d be best .
9- 28
For the i nner el ectrons, t he wave f unct i ons are essent i al l yhydroqeni c, wi t h an appropr i ate ef f ect i ve Z. For t he Kq l i ne,
use
*100 = (z/ a0)3/ 2e_Zr / a,
*210 = 47^0 ^ V 572- 2172300086'
wi t h the sel ect i on rul e t&= i l - Hi e mat r i x el ement i s
Pf l = l / ^ d r l - - & 2 ^ r V 3Zr ^ ^s i n e d 6 d * d r | .
Si ncer = rsi n6cos$I + rsi n6si n*3 + rcosSk,
t he Z- dependence fol l cws f rctn
J o
le- 3Zr / 2a0dr = z
(3Z/ 2aQ)
The l i f et i me becones
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i 3Eoh 3 ^Pb
T = B = 16h 3v 3Pj 1 ^ TH
By Exarrpl e 9-8,
hv - F* = Z*f f .
Theref ore,
TPb ZH. ef f ZPb. ef fT|1 ZPb, ef f ZH, ef f
(___i___) 4 = 2 .44T 82 - 2
H
3 2Vf i . H
3
Wf i . P b
< ^ > 4,Pb, ef f
x 10- 8,
Th = 10~8 s * Tj = 2. 44 x 10"16 s.
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CHAPI TK TFH
J f cl
E = Ej p - E2g = - 3. 50 - (-5. 35) = 1. 85 eV.
For photons.
so t hat
. . , 1240X(nm) - Ef i vT'
X = = 670. 3 nm= 6703 A.
(b) By Exampl e 10-1,
,, hcdE he dE , dEd x =- 2_ = T T x T -
Fr an Tabl e 10-1, dE = 0. 42 x 10~4 eV, and theref ore-4
dX = (670. 3)- 421_Xp510 = 0. 0152 r m= 0. 152 L
10- 3
y 7 fi i(a) The ground st at e conf i aurat i on i s I s 2s 2p 3s , t he f i rstthree shel l s bei ng cl osed; t he possi bl e exci t ed stat es wi l l be
t hose wi t h the opt i cal el ectron i n t he 3p, 3d, 4s, 4p, 4d, 4f , 5s etc.l evel s, ai t 4d, 4f l i e above 5s and 4s l i es bel cw 3d. Al so, f orany n, the l evel wi t h maxi mumpermi t t ed 1 ( =n - 1) correspondst o the hydrogen at an l evel of t hat n. That i s, f or 3d (n = 3,1 = 2),
E = E3H = - 1. 5 ev.
(b) Each l evel , except s-states, i s spl i t i nto t wo l evel s, wi t ht he stat Q of smal l er j bei ng more neoat i ve i n enerqy. The encr oy
spl i t t i ng i s snai l compared t o t he ener gy of t he degeneratestates. Thi s spi n- orbi t spl i t t i ng i s gi ven by
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AE = 7 5 : (j + D - + X) - s( s + 1)}, c
AE =K{j (j + X) - Ml + X) - s(s + X)}.
Assume K = K(r=0) = constant . I n al l states s = , 2s + X = 2,j = I *s, except 1 = 0 wher e j = s. Put t i ng al l t hi s tooethergi ves t he resul t s bel ow.
3s: 1 = 0 , j = X/2; AE = 0; S/ 2-
3p: . = X,r 3 = 3/2; AE = +K; S / 2 .
Ol j = X/2; AE = - 2K;
X/ 2
4s: = 0; j = X/2; AE = 0; S / 2
3d: I = 2 ;r j = 5/2; AE - 2K; S / 2 -
l j = 3/ 2; AE = - 3K; S/ 2 .
4p: I = I ;j j = 3/2; AE = K; S/ 2 '
O*j = 1 / 2 ; AE = -2K; S / 2
5s: I = 0; j = 1 / 2; AE = 0; CNsH
sel ect i on rul es are: At = i x, Aj = 0 , 1 . Uiai ves f or the permi t t ed t ransi t i ons:
2Sj y2 2P1/2 ' S / 2 " 69011of ^ 3s, 4s, 5s l evel s can make atransi t i on t o each of t he t wD l evel s oft he 3p, 4p states.
* 2D, , - . But not t o si nce t hen Aj = 2.111 Al so t o the S- st ate above.
2P, /, 2D, . _, 2D, . These i n addi t i on to the S,3/ 2 / t ransi t i on. L/*
S/ 2 ' S / 2 These are i ncl uded above.
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Pitbnut S5pio-*>rbit Splitting
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(a) Ij, = 1, P.2 = 2; s1 = s2 = %. Thus t he possi bl e val ues of J.'
ar e It = 3, 2, 1; possi bl e s' = 1, 0. The snal l est 4 , s' i s thest ate of maxi i rameneray: i . e. , 4' = 1, s' = 0 . Wi t h 4' = 1 , s' =
0, there i s onl y one possi bl e j ' : t o wi t , j * = 1 .
(b) Si nce 5' = 0 , ' l i es al cnq 3'.
10-5
= 2, H2 = 3; = s2 = lj. The possi bl e 4' = 5, 4, 3, 2, 1; t he
possi bl e s' = 1, 0. For j 1:
j ' = V + s , V + s - 1 , . . . |SL' - s' |
Theref ore, t he possi bl e conf i gurat i ons are:
i l si 1 1
5 1 6, 5, 45 0 54 1 5, 4, 34 0 43 1 4, 3, 23 0 32 1 3, 2, 12 0 2
1 1 2, 1,01 0 1
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For the conf i gurat i on 4s3d, ^ = 0, l2 = 2, = s2 = >s. Hence,
V = 2 onl y, s' = 1, 0. Wi t h V = 2, s' = 1, j ' = 3, 2, 1 gi vi ng
3D, - . l evel s. For i ' = 2, s' = 0, j ' = 2 onl y, resul t i ng i n a
VJ , 12 state. By t he Lande i nterval rul e, the separat i on i s(3D3 - 3D2) / (3D2 - 3Dj ) = 3/ 2.
The energy shi f t s themsel ves are
AE = K{j 1(j 1 + 1) - t ' d' + 1) - s 1( s1 + 1)},
gi vi ng
*D2: AE = 0;
3D3: AE = 4K;
3D2: AE = - 2K;
AE = -6K.
The l at ter t hree shi f t s obey the Lande rul e.
= i . r = 2
- - r = 2 . 3 2
Splitting cb1e 1 - 1 ^
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The magni t udes of the vectorsare:
J' = / 12H; L' = / 6K; S' = / 2K.Appl yi ng the cosi ne l aw:
S' 2 = J ' 2 + L' 2 - 2J ' L' cos0,
2 = 12 + 6 -2/ 72COS0,
0 = 19. 47*.
Agai n resort i ng t o the cosi nerul e:
j ' 2 = s' 2 + L' 2 + 2S' L' cos,
12 = 2 + 6 + 2/ l 2cos*,
= 54. 74.
Turni ng t o the magnet i c moments:
ws - T 5' = 2/2V
' i = F l ' = / 6V
V = v's2 + uj 2 + 2p cos * ,
U2 = 8y + 6p + 4/ l 2ucos54. 74 u = 4. 6903 .
Fi nal l y,
Mg2 = U 2 + U2 2Pj J ucosa,
P = 6 21. 999 - 2/ 6(4. 6903)cosa * a - 29. 50,
so that Mu, - 3 > = 29. 50" - 19. 47 = 10. 03*.
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10-9
(a) On Fi g. 9- 13, t he col umns reveal t he l ast shel l bei nq f i l l ed,the row the numbers of el ect rons i n that shel l . Theref ore,
12Mq: I s2, 2s2, 2p6, 3s2;
13A1: I s 2, 2s 2, 2p6, 3s2, 3pX,
14Si : I s2, 2s2, 2p6, 3s2, 3p2.
(b) 12Mg; t he conf i gurat i on represents a f i l l ed shel l , and thusal l t he angul ar mcment a are zero, l eadi ng t o i Sg.
Al t there i s a si ngl e val ence el ect ron (s = s' = %); thus2s' + 1 = 2; ' = 1 gi vi ng a P- state; j ' = 3/ 2, 1/ 2 wi t h thesmal l er j * l yi ng l ower, l eadi ng to 2p .
14Si : here there are tw> 1 =1 el ectrons; s' = 1, 0 and 8.' = 2, 1,0. For t he l ower energy pi ck the l arger s' . Thi s gi ves aspossi bi l i ti es:
V = 2; j ' = 3, 2, 1;
* =1; j * . 2, 1. 0, \ 1>0
V = 0; j ' = 1; 3Sr
The 3d , and 3S1 st ates are, however , prohi bi t ed by theExcl usi on pr i nci pl e. Of the 3p2 x q states, t he snal l est j ' l i esl owest ; hence t he ground st at e conf i gurat i on shoul d be 3pQ.
1 0 - 1 1
For a si ngl e nul t i pl et s' and I ' have t he same val ue f or eachl evel . By t he i nterval rul e.
*4 = *4
r .s
*3 = 2Kj 4,
S2 = 2Kj 3, - - - - - 3
E! = *12- 7
" ' l
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V * 3 = 3 = j 5/ j 4 = (j 4 + 1) / j 4; j 4 = 3-
Si nce j = 1, j 5 = 4, j 4 = 3, j 3 = 2, j 2 = 1, j x = 0. But
j ' = i ' +s' , t ' +s' - l , . . . U - s' l ,
so that
+ s' = 4, I' - s' = 0; S.' = s' = 2,
and hence the resul t s are
r = s' = 2; j ' = 4, 3, 2, 1, 0.
10- 14
(a) The g- f actor i s
j d' +l ) + s Ms +l ) - t ' ( t ' +l )9 - 1 2j " (j ' +l )
(i ) For q > 2,
j ' ( j ' +l ) < s' (s' +l ) - I ' t t ' +l ) . (*)
I f j ' =8. ' + s' , t hi s becomes
0 < -(V2 + V + H' s' ) ,
whi ch i s i mpossi bl e. But i f j ' =V
- s' t he rel at i on gi vess > I,
so that j ' = s' - V, a cont radi ct i on. So t r y j ' = s' - I ' i n(*), whi ch wi l l now reduce to
I' 2, as requi red.(i i ) For the case g < 1, t he f ormul a f or g requi res t hat
j ' ( j ' +l ) + s' (s' +l ) < t ' t t ' +l ) . (**>
I f j ' = V + s' , (**) becomes
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0 > s 2 + s' + 4' s' ,
whi ch i s i mpossi bl e. But i f j ' = 4' - s' , (**) reduces t o
4' > s' .
For exampl e, 4' = 2, s' = 1, j ' = 2 - 1 = 1, g = %< 1.
(b) Consi der t hi s second case: 4' = 2, s' = 1, j ' = 1, q = %Then,
L' = / 6X, S' = /2tf, J ' = / 2H, *( t ' , 2' ) = 150*.
Si nce g = 1, os = 2, dr aw egual i n l ength to ' , twi ceas l ong as S' ; i t i s seen that i s about hal f as l ong as J ' ,i ndi cat i ng t hat g = %.
10- 17
(a) The yi el ds three l evel s, t he 3P2 f i ve, ^ three, andt he j Pq gi ves one; thus t he total ni tnber i s 12.
(b) For t he 3s el ect ron 1=0; 2(24+1) = 2;For t he 3p el ect ron 4 =1; 2(24+1) = 6.
Cl ear l y, wi t h (6)(2) = 12, the f i el d has removed t he deoeneracycompl etel y.
10-Ift
For a si ngl et s = 0, so there i s onl y orbi t al angul ar mcment umt o consi der . The pot ent i al energy of or i ent at i on i s
AE = - J -fc.
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f or or bi t al ancj ui ar moment um g = 1 so that
U_ . e
L 2m' 11 " HI f $ i s i n t he z- di rect i on,
AE = - . 5 = - LzB = -
gi vi ng r i se t o 2S.+1 l evel s. Si nce s = 0, I = j ; maki ng thi ssubsti tut i on, and i nser t i ng a f actor g = 1 , l eads to
AE = - i gai y
whi ch agrees wi t h Eq. 10- 22 i n t he case g = 1.
10-19
I n the cl assi cal model , pi cture t he magnet i c f i el d bui l di ng upto a val ue B f romzero i n ti ne T. Faraday' s l aw requi res ani nduced el ectr i c f i el d E,
r, i _ BK = dt = *r T'
Thi s i mparts an addi t i onal vel oci t y Av t o the el ectrons, whi chci rcul ate ei t her cl ockwi se or countercl ockwi se. Hence,
Av = aT = T = T = | ( | S) T =m m m 2 T 2m
The new speeds are
yi el di ng f requenci es v gi ven by
v j . gBv 2nr v0 " 4nm'
E = hu = hv0 1 B'
AE = \ilB,
correspondi ng t o Eq. 10- 22 wi t h g = 1, Arnj = 1.
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10-20
(a) For 1P1, j ' = 1 so there ar e t hree l evel s; t he h>2 has j ' =
2 gi vi ng r i se to f i ve l evel s wi t h t he f i el d. For bot h these
st at es s' = 0 so that g = 1 and AE = j . Hence, t he l evel
spaci ngs are the same f or each state.
(b) The sel ect i on rul es are Amj = 0, l l , al l owi ng zero to zero
si nce Aj ' ^ 0 between t hese states. Tti e group I of t ransi t i ons
gi ve t he same wavel engt h as when B = 0; f or I I , AE < A E ^ so
that Xj j > XQ; i n I I I , AE > A E ^ and theref ore Xm < XQ. Al so,
wi t h al l the l evel spaci ngs the same, al l wavel engt hs i n I I are
equal , as are al l i n I I I . Group I has wavel engt hs equal t o t hose
wi t h B = 0 and so are al l equal . Hence, three l i nes appear.
(c) The wavel engt h of a l i ne i s gi ven by
X = *.A AE'
Consi der i ng two t ransi t i ons whose energi es di f f er by A E, thewavel engt hs of these l i nes di f f er by
AX = I W a (AE) = - ^A ^.(AE)2
New cl ear l y.
XI I X0 ~ X0 XI I I '
and consequent l y AXI X = = AX XXI - XQ. The energy AE,
cor respondi ng t o 0, gi ves wavel engt h XQ, whi ch i s
i dent i cal t o that f or ^ when B = 0. Fran Fi g. 10-8, thi s
i s
AE ~ 10 - 3. 6 = 6. 4 ev = 10. 24 x 10- 19 J .
Si nce Am! = 1, A*E = Bf Arn ) =1^6= (9. 27 x XO- 24) (0.1) =
9. 27 x 10- 25 J . Theref ore
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n - n b* a
\
/
33: : 3
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J! = D
(*)iir-orht *plittinBignnrtA)
/
/
s j
n
-- +5 +1
---------------- ------
. +% 0
+1, - 1 tmd- f ni a"-ij +1 dr gcmmcB
--l i -1
- +% 0
So litti hue tljs raif manf-
lcngtl) * B = II lint. .
X3 X33 *333
(!) s' =%, * ; = f t . *' = ) i -i = Usi ng the
rel at i on f or fiE, these gi ve:
1j L
AE (uni ts
1 +% +2
1 ->} 0
0 +% +1
0 -*s - 1
- 1 +>J 0
- 1 ->s -