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第五章:多电子原子 : 泡利原理. Atomic Physics 原子物理学. 第一节 氦的光谱和能级. 第二节 两个电子的耦合. 第三节 泡利原理. 第四节 元素周期表. H 原子:. 类 H 离子:. 碱金属原子:. 若核 ( 实 ) 外有两个电子,由两个价电子跃迁而形成的光谱如何?能级如何?原子态如何?. He : Z= 2 Be : Z=4=2 1 2 +2 Mg : Z=12=2 (1 2 +2 2 ) +2 Ca : Z=20=2 (1 2 +2 2 +2 2 ) +2 - PowerPoint PPT Presentation
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: Atomic Physics
H H
()HeZ=2BeZ=4=212+2MgZ=12=2(12+22)+2CaZ=20=2(12+22+22)+2SrZ=38=2(12+22+32+22)+2BaZ=56=2(12+22+32+32+22)+2RaZ=88=2(12+22+32+42+32+22)+2
- +
:
,n,l,j, ,
:
1
:
1. 1s 1s .3. 1s2
,
:
:
:
::S, P, D, F----S, P, D, F----2
:
, 1sn
:11s1s 1S03
41s2s1S01s2s3S1, 21s1s 3S11s1s 1S01s2s3S13 3S1
:
5 Be(4)Mg(12)Ca(20)Sr(38)Ba(56)Ra(88)Zn(30)Cd(48)Hg(80)+2.
:
1. 1s1s1s 1s 2s2p 3s 3d, , n1
:
2. n1l1n2l2 1s1s 1s2s1s2s 1s2p
:
nl; nl n2Lj1,n2Lj2; n1l1n2l2
:
Enlj
:
l1,l2,s1,s2G5,G6 G1,G2G3,G4G1(s1,s2)G2(l1,l2)G3(l1,s1),G4(l2,s2)G5(l1,s2)G6(s2,l1)
:
:1.
:
1): k1=l,k2=s,k=j j=l+s, l-s
:
2): ::::
s=0,1 .s=0j=l;
:s=1
l sjl jl l1 l2 ss=0s=12s+1=1,3;
:s=0,1 .s=0j=l;s=1
3) (L,S,J)::
:
33p4d12
1 1s 1s 1s1s 1S0Review 12 3s 3s
:2: 1s1s: 1s2s: 3s3s: 3s3pn1l1 n2l2 n3l3)
G1(s1,s2)G2(l1,l2)G3(l1,s1),G4(l2,s2)G5(l1,s2)G6(s2,l1)G5,G6
S =0 , 1L J S=0 J=L L J S=1J=L+1LL-1 L J
e1 e2 L S=0 S=1J J 1s 2s 0 0 1 1s 2p 1 1 012 1s 3d 2 2 123 1s 4f 3 3 234 1.
n=2,3
n=2,3n=3,4n=3,4n-4,5 2.
n =3,4n =3,4
2.
:New
:
:
:
jj
:JJ
4j-j3p4d
3
:3p4d 12
l j
:4
1
:1. Laporte
.
:
2) Laporte Laporte
:
l =1, Laporte
:
21 2 Laporte He
:Hg
:
He1s1s
(1s1s)1S0(1s1s)3S1;
1925Pauli
:
Pauli1) 1869 Mendeleev11Na
Wolfgang PauliPauliZeeman1925Pauli1940PauliBohr
1945
1n n2 ll . n ln=1, 2, 3l=0, 1, 2(n-1)
32
:
2Pauli Pauli
:
3Pauli He1s1s
S=1 .(1s1s)1S01He
:
: .
s1 s2 S=1Pauli
:
Pauli
2
:
Z
Pauli
Z
:
Pauli345
:
n l , nl m :1
:
l 1s2s
:
:
l, s
slater
:
Z
,,,18,8
n n l n l ,KLMNO..
1nl n 2 3
: 28183228818 2818Ar3p3d193d4s
4s, 3d3d4s4s 3d4s
1 S 2 S L 3 nl J J |LS|> JL+S1 Hund
pppd LS213
L-S J J J+1JJ-1
ML=0MS=0L=0, S=0, J=0 l=1pnp1np5 np2np4 Hund
(1)S (2)L (3)J (4)2s+1Lj (1) (2)(3)(4)
LSZ6Z7Z17:SLL-SJJ
: The end
npnp nl npnp +10-1 -1/2+1/2
MsMLML Ms221131111MsMLMLMsMLMsMLMsL=2S=01D2L=1S=13P2,1,0L=0S=01S011
npnp1S01D23P210 L=() S=0 L= S=1
17, 6Pauli
:
1 2
:
2 l
:
n nn: 1) n, lml(2l+1)mlmsl nl=2(2l+1).3
:
2 j 2j+1
:
19KK181,KK
:
...... 1
:
1n n Zn (Moseley) 3d32D ZT32D (Moseley)
:P235 27.5
4S 42S (Moseley)(Moseley)32S42DE=-hcTT
:
n=3 n=4 Z=20~2121 1920 4s21 3d .1920
:
1/23/2- 1/23/25/27/21/2