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# soltn10 - Homework 10 Solutions 9-l M=zpzB=W...

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Homework 10 Solutio ns 9-l M=zpzB=W z(s.n xrc-u /rX0J5 D=(6.53x10-s Is)/ eo | =s.nxtoe tfz (a) 3d sutrehell :r I = 2 =mt=-2, -11, O,7, Z *a r, +l for each nr1 3 3 3 J J 3 fit J, .', -1 0 0 1 1 2 3r-1 3t-l 370 310 3r1 317 I i z a ,, ) ', m, -\n +tn -w +7n -U2 +w -ln +m -7n (b) sz2+1n 3p subehell for a p state, I = 1. Thu6 nrl can take on values -l to l, or -1, O, l. For each I ,n r, rrr can be t:. _tn +ln -1t2 _ln 9-s with s =;, the spin magnitude is lg =[s(r+l1v,r -[tl5]vtJ. The z-component of spin i3 S, = rarl where n, rantef from -s to s in inteter step6 or, in this case, 1112 n, -;, -;, -;, +:. The spin vector S tu indined to the z-a:dsfu an ang)e 0 such that "*e, E= ffiW= #-tr= &, *--r, .#, . & q 0=140.8' , 10F.0" ,75.V , 39.2. .The A- dus obey the pauli Exdurion principle, since the spin 3 of this panide is half-int€gral, a3 it ic for all furmioru. 9-1t For a d electro4 t= Z;t=!; i=z+!,2-! 2' 2 2 s B ? 1135 For i=:; m,=-2. -:.-- - - - ---' 2''I z' Z' z' 2' z' 2 - 3 3 113 ror,=r,nj=-r,- 2,r,,

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g-lg (a) rlnqz-+n=4, t=3, j=; (b) 111=1111 +t))u, tr =l[*Yll*r =f91u" =[ttt]u' 1 L\z^z|J L4J | 2 l (c) I,=m jtt wlrcre m j @nbe -j, - j+1,. ..,j-1, sohere n, canbe 5 3 1135 5 3 r 1 ?
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soltn10 - Homework 10 Solutions 9-l M=zpzB=W...

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