物理化学 3 演習問題 No.9 （電子構造とスペクトル、分子） …

1 9 - 1 - 3 No. )(Zeeman 3 n l m K,3,2,1=n L,,,MLK 1,3,2,1,0 =nlK n L,,,dps llllm,1,,2,1,0,1,2,,1, + =KK 12+l m 12+l l m ),,(zyxr ),,(zyxpppp ),,(zyxllll r p )

2 (pr )1(2+=llh z hm z r p l eme2 2 2 s1 l)

3 M 2 No 2 9 - 2 - Stern Gerlach 1 2 Goudsmit Uhlenbeck D 2 2 ) (l l z ) (zl m Goudsmit Uhlenbeck s ) (2s )1(2+ssh s z zs smh sm Goudsmit Uhlenbeck s 21 sm 21 2 2 2 )

4 0( sm ssmsmssss,2,2)1( += h 21=s ssmssmszms,, = h 21 =sm 21=s 21 =sm sms, 2/1,2/1 = 2/1,2/1 = 2 1 ),,(zyx )(),,(),,,( zyxzzyx= )(),,(),,,( zyxzzyx= dxdydzddv= 12= d 12= d 0*= d 0*= d 9 - 3 - 12022022222102211244242 rerZemrZemH + =hh 2=Z Z m)

5 Z sZZi = s )2()1()2,1(BA = 1 A 2 B 1 B 2 A )2()1()2,1(AB = )(),,( r= )(),,( r= 21 )2()2(1)1()1(1 ss I )2()2(1)1()1(1 ss II )2()2(1)1()1(1 ss III )2()2(1)1()1(1 ss IV 4 1 2 II III )2()2(1)1()1(1)2()2(1)1()1(121 sscssc+= c =12dv c 21,2121 ==cc )2()2(1)1()1(1 ss I ())2()2(1)1()1(1)2()2(1)1()1(121 ssss+ II ())2()2(1)1()1(1)2()2(1)1()1(121 ssss III )2()2(1)1()

6 1(1 ss IV 1 2 1 1 9 - 4 - Pauli 1 2 ())2()2(1)1()1(1)2()2(1)1()1(121 ssss Slater )2()2(1)2()2(1)1()1(1)1()1(121)2,1( ssss= 0)2()2(1)2()2(1)1()1(1)1()1(121)2,1(== ssss N )()()()2()2()2()1()1()1(!)

7 1),,3,2,1(212121 NNNNNNNN LMOMMLLL= il is L S

8 =iilL =iisS L S J SLJ+= zil sis ===iiizizMmlL ===issiizizMmsS 9 - 5 - L S J n JSLn}{12+ n }{L L L,4,3,2,1,0 L,,,,,GFDPS S 12+S L,,,DPS L,4,3,2,1 singlet 1 doublet 2 triplet 3 quartet 4 J SLSLSLSLJ + ++=,,2,1,L J }{L n J L S LS LS J LS J 12+J C )2)(2)(1(222pss C )

9 2(2p im sim =iimM =isismM M sM L S LLLLM + =,1,1,0,1,,1,LL SSSSMs + =,1,1,0,1,,1,LL L S L S L S m )(amnp )(anp J 21s 1=n 0=l 0=m 9 - 6 - 0=L 0=S S1 011S 1121ss 2=n s 0=l 0=m 0=M 0=L 0,1=S 1=S 1,0,1 =Ms S3 132S 0=S 0=Ms S1 012S )

10 (Hunt L J J 1 = L 0= S 1,0 = J 00= =JJ +2H M A B m ar br R m = 0 M Ms 0 0 m = 0 0 M Ms 0 1 0 0 0 0 0 -1 9 - 7 - +2H )111(4)(22 0222222 RrreMmHbabae + + = hh +2H )