Solutions Quantum Physics - R. Eisberg & R. 2nd Resnick

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

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(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.


37/209

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 =


38/209

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) *>,


39/209

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


53/209

(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


56/209

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


57/209

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)


65/209

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),


71/209


72/209

*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


74/209

(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,


75/209

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,


77/209

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.


78/209

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) .


79/209

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


80/209

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.


82/209

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:


86/209

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


88/209

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 oneel 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


105/209

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


107/209

(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


108/209

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 ^


109/209

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*.


110/209

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


112/209

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 -

Solutions Quantum Physics - R. Eisberg & R. 2nd Resnick

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Transcript of Solutions Quantum Physics - R. Eisberg & R. 2nd Resnick