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PS+5+solution - 3m C2 E7123 = E’z 2mflcz I m a...

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Unformatted text preview: 3m? C2 E7123 = *' E’z 2mflcz I _; m) a , =-.i’_’_"L E365 _ fig ‘ E‘f {éjeflrme Wt %{ :l85é $5 5+3 W— 48136452306AE¢ 8 xz-o-o>3x 3. 3'8 57”. 6 me Farr owtrmfwmgrmd m I '"Y' I Wm. : “‘7 e '5" 7M3 68/, _ w , _ W 49-9 mad [24,6 1 __ ALfIU'O :: e /241 4mm? 3 2/“ 22. e o E"? ‘ 5 < 5* 51. 43'"- WQ 6! : 3112 1a .4 m; 8 4.82. x10. JLV' ss Problem 6.21 For n = 2. E = 0(j:1/2)0r3=1{j=1/2 or 3X2). The eight states are: |1>=I20%%) . --1 }ng [1+(1/2)(3/2)+(3/4)] =1+sx2 _2 —> 2 .2) 2 I2 0 % - 2(1x2)(3x2) 3/3 ‘ ‘ la): 91%) }g = [1+(1/2)(3/2)-(1)(2)+(3,r4)] 21+ __1_/2 = 2/3 ,4) z '2 1 %_ %) “’ 2(1/2}(3/2) 3/2 ‘ In these. four cases. Enj- : —13':cv [1+ 0; — = —3.4 0V(l + %a2)_ 15) = !2 1 g %) f6>=i21%%> _ [1+ (3/2)(5;2)—(1)(2)+(3/4)] _1+ 5/2 _ 4/3 |7>=|21g_%> 9", _ 2(3/2M5/2) _ 15/2 _ ‘ Is) = 21 g— a In these four cases, Enj = —3.4 eV [1 + a: -— - = —3.4 eV(1+ lag). 4 2 4 16 E; = "3.4 eV (1+ 15—432) + prext. 52 = —34 eV (1+ %02) — 1133.3.“ Ba 2 —3 4 CV (1+ 15302) + %,lLBBext 15:1 = -3 4 eV (1 + T591?) — ngBext The energies are: E5 = —3.4 eV (1+ Ila-a2] + 2pBBexL Ea = —3 4 cv (1+ fig?) + gflgBth E7 = —3.4 EV (1+ fie?) — g’uBBext £3 = —3.4 eV (1— 3350:!) — 2;:BBext Problem 7.8 . 1 .1_ . 1 . tI'I-‘uhfljn : fl._-_i. (.——!2,Hl_ut-3J. TNT r1 1 ' . . mm}? = (liv'UIr'J) D = a(¢!u(r1) H 1 3 1 . , 274' x F .1 . = -—.f f."_ a w 4' 3mm” —rl smfidrdfidw = —1 r (I a V r H " R” H 53:16th dr. ' - " “'0‘ . u . u l l"+” ‘ I I r. I, l a a [. ‘. = —— r.-—-’-”-‘“ydy = —i-— [w Emmi“ (r + R ;— 3) — r- "‘“-’" (4- — m + 3) 9 n! . . 1 1 .. n; 1 « A = a_(.filu(r‘) _. {gubvfln = {1—3; C—I'lraf, -r‘_;l‘l__ 1.1,. f'l l'i'fi'. I J"; I I +1: 9 n -a- 1 a - 2W x - 7r '- R 21+ 6" - = —_,~ ri"-"'c""' "" “m -"‘—1"51119d'r'[email protected]= —,; r(."'"-’“ 6“” ‘ ' r “J” 1""51119d6 dr'. "0". 1" n" . u n . a .- .- . : H—R [F'UHHFL'U + H+a} — (“r—"*flttr— RI +u}-l 3' - '2 ,- x g . X = —d [r‘fl’fi/ l"_£r"'0[i'+ R+a)d?' [J _ n , '7“ ., . _{='“-""/- (R—r+rt)dr—ffl‘m/ [If-r'mfi‘r— 3+“)d‘” .0 - R |ni5]:= F[x ] :=_1+ _ Prvblun 2 [(1-(2/3Jx2)Exp[—x]+(1+x)Exp[—2x] 1 + (1+x+ (1 [3) x2) Expt~x1 In16];= PlotIF[x] , (x, 0,. 6]] Out[6}= |n[26}:= Xeq = x /. FindRoot[D[F[x] , {1, 1}] == 0,. {x, 2}] Out[25]= 2 .49283 FEXeq] 0mm]: w 1.12966 |n{30}:= D[F[x], (x, 2}] /. x-o Xeq 0mm]: 0 . 125666 7‘“ a) = j *(0-1'757>- E: 3‘s 0 2,26 431/ R E” = (14+ 35- 0.2zé-e4/ ED .. o :13 43V E3 : (fin—1 7am? [6753”; E7 Mm. m <8 Vf~bmtE~md W C 14:0, n2, ) ...
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PS+5+solution - 3m C2 E7123 = E’z 2mflcz I m a...

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