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Alkali earth metal atoms
碱土金属原子

He: Z=2  He:Z=2
Be: Z = 4 = 2 × 1 2 + 2 Z = 4 = 2 × 1 2 + 2 Z=4=2xx1^(2)+2Z=4=2 \times 1^{2}+2  是: Z = 4 = 2 × 1 2 + 2 Z = 4 = 2 × 1 2 + 2 Z=4=2xx1^(2)+2Z=4=2 \times 1^{2}+2
Mg: Z = 12 = 2 × ( 1 2 + 2 2 ) + 2 Z = 12 = 2 × 1 2 + 2 2 + 2 Z=12=2xx(1^(2)+2^(2))+2Z=12=2 \times\left(1^{2}+2^{2}\right)+2  毫克: Z = 12 = 2 × ( 1 2 + 2 2 ) + 2 Z = 12 = 2 × 1 2 + 2 2 + 2 Z=12=2xx(1^(2)+2^(2))+2Z=12=2 \times\left(1^{2}+2^{2}\right)+2
Ca: Z = 20 = 2 × ( 1 2 + 2 2 + 2 2 ) + 2 Z = 20 = 2 × 1 2 + 2 2 + 2 2 + 2 Z=20=2xx(1^(2)+2^(2)+2^(2))+2Z=20=2 \times\left(1^{2}+2^{2}+2^{2}\right)+2  钙: Z = 20 = 2 × ( 1 2 + 2 2 + 2 2 ) + 2 Z = 20 = 2 × 1 2 + 2 2 + 2 2 + 2 Z=20=2xx(1^(2)+2^(2)+2^(2))+2Z=20=2 \times\left(1^{2}+2^{2}+2^{2}\right)+2
Sr: Z = 38 = 2 × ( 1 2 + 2 2 + 3 2 + 2 2 ) + 2 Z = 38 = 2 × 1 2 + 2 2 + 3 2 + 2 2 + 2 Z=38=2xx(1^(2)+2^(2)+3^(2)+2^(2))+2Z=38=2 \times\left(1^{2}+2^{2}+3^{2}+2^{2}\right)+2  锶: Z = 38 = 2 × ( 1 2 + 2 2 + 3 2 + 2 2 ) + 2 Z = 38 = 2 × 1 2 + 2 2 + 3 2 + 2 2 + 2 Z=38=2xx(1^(2)+2^(2)+3^(2)+2^(2))+2Z=38=2 \times\left(1^{2}+2^{2}+3^{2}+2^{2}\right)+2
Ba: Z = 56 = 2 × ( 1 2 + 2 2 + 3 2 + 3 2 + 2 2 ) + 2 Z = 56 = 2 × 1 2 + 2 2 + 3 2 + 3 2 + 2 2 + 2 Z=56=2xx(1^(2)+2^(2)+3^(2)+3^(2)+2^(2))+2Z=56=2 \times\left(1^{2}+2^{2}+3^{2}+3^{2}+2^{2}\right)+2  三: Z = 56 = 2 × ( 1 2 + 2 2 + 3 2 + 3 2 + 2 2 ) + 2 Z = 56 = 2 × 1 2 + 2 2 + 3 2 + 3 2 + 2 2 + 2 Z=56=2xx(1^(2)+2^(2)+3^(2)+3^(2)+2^(2))+2Z=56=2 \times\left(1^{2}+2^{2}+3^{2}+3^{2}+2^{2}\right)+2
Ra: Z = 88 = 2 × ( 1 2 + 2 2 + 3 2 + 4 2 + 3 2 + 2 2 ) + 2 Z = 88 = 2 × 1 2 + 2 2 + 3 2 + 4 2 + 3 2 + 2 2 + 2 Z=88=2xx(1^(2)+2^(2)+3^(2)+4^(2)+3^(2)+2^(2))+2Z=88=2 \times\left(1^{2}+2^{2}+3^{2}+4^{2}+3^{2}+2^{2}\right)+2  Ra: Z = 88 = 2 × ( 1 2 + 2 2 + 3 2 + 4 2 + 3 2 + 2 2 ) + 2 Z = 88 = 2 × 1 2 + 2 2 + 3 2 + 4 2 + 3 2 + 2 2 + 2 Z=88=2xx(1^(2)+2^(2)+3^(2)+4^(2)+3^(2)+2^(2))+2Z=88=2 \times\left(1^{2}+2^{2}+3^{2}+4^{2}+3^{2}+2^{2}\right)+2
Calculate L S L S quad vec(L)* vec(S)quad\quad \vec{L} \cdot \vec{S} \quad in 4 D 3 / 2 4 D 3 / 2 quad^(4)D_(3//2)\quad{ }^{4} D_{3 / 2}
计算 L S L S quad vec(L)* vec(S)quad\quad \vec{L} \cdot \vec{S} \quad 范围 4 D 3 / 2 4 D 3 / 2 quad^(4)D_(3//2)\quad{ }^{4} D_{3 / 2}
4 D 3 / 2 : L = 2 , S = 3 2 , J = 3 2 ; J ^ 2 = L ^ 2 + S ^ 2 + 2 L ^ S ^ L S = 1 2 ( J ^ 2 L ^ 2 S ^ 2 ) = 1 2 [ J ( J + 1 ) L ( L + 1 ) S ( S + 1 ) ] 2 = 3 2 4 D 3 / 2 : L = 2 , S = 3 2 , J = 3 2 ;  由  J ^ 2 = L ^ 2 + S ^ 2 + 2 L ^ S ^  得  L S = 1 2 J ^ 2 L ^ 2 S ^ 2 = 1 2 [ J ( J + 1 ) L ( L + 1 ) S ( S + 1 ) ] 2 = 3 2 {:[quad^(4)D_(3//2):L=2","S=(3)/(2)","J=(3)/(2);],[" 由 " hat(J)^(2)= hat(L)^(2)+ hat(S)^(2)+2 hat(L)* hat(S)" 得 "],[ vec(L)* vec(S)=(1)/(2)( hat(J)^(2)- hat(L)^(2)- hat(S)^(2))],[=(1)/(2)[J(J+1)-L(L+1)-S(S+1)]ℏ^(2)=-3ℏ^(2)]:}\begin{aligned} & \quad{ }^{4} D_{3 / 2}: L=2, S=\frac{3}{2}, J=\frac{3}{2} ; \\ & \text { 由 } \hat{J}^{2}=\hat{L}^{2}+\hat{S}^{2}+2 \hat{L} \cdot \hat{S} \text { 得 } \\ & \vec{L} \cdot \vec{S}=\frac{1}{2}\left(\hat{J}^{2}-\hat{L}^{2}-\hat{S}^{2}\right) \\ & =\frac{1}{2}[J(J+1)-L(L+1)-S(S+1)] \hbar^{2}=-3 \hbar^{2} \end{aligned}

For 3p4d,find atomic states
对于 3p4d,求原子态

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p把子和 d 起子在 IS 耦合 中 形成的能级

Pauli exclusion principle
泡利排除原则

1、Four quantum number  1、四个量子数
n : n = 1 , 2 , 3 , l : l = 0 , 1 , 2 , , ( n 1 ) m l : m l = 0 , ± 1 , ± 2 , , ± l m s : m s = ± 1 / 2 ( n , l , m l , m s ) n : n = 1 , 2 , 3 , l : l = 0 , 1 , 2 , , ( n 1 ) m l : m l = 0 , ± 1 , ± 2 , , ± l m s : m s = ± 1 / 2 n , l , m l , m s {:[n:n=1","2","3","cdots,l:l=0","1","2","cdots","(n-1)],[m_(l):m_(l)=0","+-1","+-2","cdots","+-l,m_(s):m_(s)=+-1//2],[(n,l,m_(l),m_(s)),]:}\begin{array}{cc} n: n=1,2,3, \cdots & l: l=0,1,2, \cdots,(n-1) \\ m_{l}: m_{l}=0, \pm 1, \pm 2, \cdots, \pm l & m_{s}: m_{s}= \pm 1 / 2 \\ \left(n, l, m_{l}, m_{s}\right) & \end{array}
The state of the electron can be fully determined. For example, physical quantities such as the total energy, angular momentum, spatial orientation of the orbital, and spatial orientation of the spin can all be determined by this set of quantum numbers.
电子的状态可以完全确定。例如,总能量、角动量、轨道的空间方向和自旋的空间方向等物理量都可以由这组量子数来确定。
( n , l , m l , m s ) n , l , m l , m s (n,l,m_(l),m_(s))\left(n, l, m_{l}, m_{s}\right)

2.Pauli exclusion principle
2.泡利排除原则

In an atom,it is impossible for two or more electrons to have completely identical four quantum numbers.In other words,each state in the atom can only accommodate one electron.( n , l , m l , m s n , l , m l , m s n,l,m_(l),m_(s)n, l, m_{l}, m_{s}
在一个原子中,两个或多个电子不可能具有完全相同的四个量子 numbers.In 换句话说,原子中的每个状态只能容纳一个电子。 n , l , m l , m s n , l , m l , m s n,l,m_(l),m_(s)n, l, m_{l}, m_{s}
( 1 s 1 s ) 3 S 1 × ( 1 s 1 s ) 1 S C n 1 = n 2 = 1 l 1 = l 2 = 0 m l 1 = m l 2 = 0 ( 1 s 1 s ) 3 S 1 × ( 1 s 1 s ) 1 S C n 1 = n 2 = 1 l 1 = l 2 = 0 m l 1 = m l 2 = 0 {:[(1s1s)^(3)S_(1)xxquad(1s1s)^(1)S_(C)quadsqrt()],[n_(1)=n_(2)=1quadl_(1)=l_(2)=0quadm_(l_(1))=m_(l_(2))=0]:}\begin{gathered} (1 s 1 s)^{3} S_{1} \times \quad(1 s 1 s)^{1} S_{C} \quad \sqrt{ } \\ n_{1}=n_{2}=1 \quad l_{1}=l_{2}=0 \quad m_{l_{1}}=m_{l_{2}}=0 \end{gathered}
Pauli exclusion principle
泡利排除原则
m s 1 = 1 2 , 1 2 m s 2 = 1 2 , 1 2 m s 1 = 1 2 , 1 2 m s 2 = 1 2 , 1 2 m_(s_(1))=(1)/(2),-(1)/(2)quadm_(s_(2))=(1)/(2),-(1)/(2)m_{s_{1}}=\frac{1}{2},-\frac{1}{2} \quad m_{s_{2}}=\frac{1}{2},-\frac{1}{2}
Electrons with the same n and I quantum numbers are called isoelectronic electrons
具有相同 n 和 I 量子数的电子称为等电子电子
同科电子
2 p 3 p atomic state what about for 2 p 2 2 p 2 2p^(2)2 \mathrm{p}^{2}
2 p 3 p 原子状态呢 for 2 p 2 2 p 2 2p^(2)2 \mathrm{p}^{2}

for 2 p 2 2 p 2 2p^(2)2 \mathrm{p}^{2}
find M L M L M_(L)\mathrm{M}_{\mathrm{L}} and M S M S M_(S)\mathrm{M}_{\mathrm{S}}
2 p 2 2 p 2 2p^(2)2 \mathrm{p}^{2} find M L M L M_(L)\mathrm{M}_{\mathrm{L}} M S M S M_(S)\mathrm{M}_{\mathrm{S}}

l 1 = 1 , m l 1 = 1 , 0 , 1 l 2 = 1 , m l 2 = 1 , 0 , 1 s 1 = 1 2 , m s 1 = 1 2 , 1 2 s 2 = 1 2 , m l 2 = 1 2 , 1 2 m L = m l 1 + m l 2 = 2 , 1 , 0 , 1 , 2 l 1 = 1 , m l 1 = 1 , 0 , 1 l 2 = 1 , m l 2 = 1 , 0 , 1 s 1 = 1 2 , m s 1 = 1 2 , 1 2 s 2 = 1 2 , m l 2 = 1 2 , 1 2 m L = m l 1 + m l 2 = 2 , 1 , 0 , 1 , 2 {:[l_(1)=1","m_(l1)=1","0","-1],[l_(2)=1","m_(l2)=1","0","-1],[s_(1)=(1)/(2)","m_(s1)=(1)/(2)","-(1)/(2)],[s_(2)=(1)/(2)","m_(l2)=(1)/(2)","-(1)/(2)]:}quad=>m_(L)=m_(l1)+m_(l2)=2,1,0,-1,-2\begin{aligned} & l_{1}=1, m_{l 1}=1,0,-1 \\ & l_{2}=1, m_{l 2}=1,0,-1 \\ & s_{1}=\frac{1}{2}, m_{s 1}=\frac{1}{2},-\frac{1}{2} \\ & s_{2}=\frac{1}{2}, m_{l 2}=\frac{1}{2},-\frac{1}{2} \end{aligned} \quad \Rightarrow m_{L}=m_{l 1}+m_{l 2}=2,1,0,-1,-2
m L m L m_(L)\mathrm{m}_{\mathrm{L}} -1 0 1
2
1
0
-1
-2
m_(L) -1 0 1 2 1 0 -1 -2 | $\mathrm{m}_{\mathrm{L}}$ | -1 | 0 | 1 | | :--- | :--- | :--- | :--- | | 2 | | | | | 1 | | | | | 0 | | | | | -1 | | | | | -2 | | | |
For 2 p 2 2 p 2 2p^(2)2 \mathrm{p}^{2} probable m l m s combination m l m s  combination  m_(l)quadm_(s" combination ")\mathrm{m}_{l} \quad \mathrm{~m}_{\mathrm{s} \text { combination }}
对于 2 p 2 2 p 2 2p^(2)2 \mathrm{p}^{2} 可能 m l m s combination m l m s  combination  m_(l)quadm_(s" combination ")\mathrm{m}_{l} \quad \mathrm{~m}_{\mathrm{s} \text { combination }}
\backslashbox m 1 m 1 m_(1)\mathrm{m}_{1} m s m s m_(s)\mathrm{m}_{\mathrm{s}} -1 0 +1
+2 (1,-) (1,-) ( 1 , + ) ( 1 , ) ( 1 , + ) ( 1 , ) (1,+)(1,-)(1,+)(1,-) ( 1 , + ) ( 1 , + ) ( 1 , + ) ( 1 , + ) (1,+)(1,+)(1,+)(1,+)
+1 (1,-) (0,-) ( 0 , ) ( 1 , + ) ( 0 , + ) ( 1 , ) ( 0 , ) ( 1 , + ) ( 0 , + ) ( 1 , ) {:[(0","-)(1","+)],[(0","+)],[(1","-)]:}\begin{aligned} & (0,-)(1,+) \\ & (0,+) \\ & (1,-) \end{aligned} ( 1 , + ) ( 0 , + ) ( 1 , + ) ( 0 , + ) (1,+)(0,+)(1,+)(0,+)
0 (1,-) (-1,-) ( 1 , + ) ( 1 , ) ( 0 , + ) ( 0 , ) ( 1 , ) ( 1 , + ) ( 1 , + ) ( 1 , ) ( 0 , + ) ( 0 , ) ( 1 , ) ( 1 , + ) {:[(1","+)(-1","-)],[(0","+)(0","-)],[(1","-)(-1","+)]:}\begin{aligned} & (1,+)(-1,-) \\ & (0,+)(0,-) \\ & (1,-)(-1,+) \end{aligned} (1,+) (-1,+)  (1,+)(-1,+)
-1 (0,-) (-1,-) ( 0 , + ) ( 1 , ) ( 0 , ) ( 1 , + ) ( 0 , + ) ( 1 , ) ( 0 , ) ( 1 , + ) {:[(0","+)(-1","-)],[(0","-)(-1","+)]:}\begin{aligned} & (0,+)(-1,-) \\ & (0,-)(-1,+) \end{aligned} (0,+) (-1,+)  (0,+)(-1,+)
-2 (-1,-) (-1,-) (-1,+) (-1,-)  (-1,+)(-1,-) (-1,+) (-1,+)  (-1,+)(-1,+)
\backslashboxm_(1)m_(s) -1 0 +1 +2 (1,-) (1,-) (1,+)(1,-) (1,+)(1,+) +1 (1,-) (0,-) "(0,-)(1,+) (0,+) (1,-)" (1,+)(0,+) 0 (1,-) (-1,-) "(1,+)(-1,-) (0,+)(0,-) (1,-)(-1,+)" (1,+) (-1,+) -1 (0,-) (-1,-) "(0,+)(-1,-) (0,-)(-1,+)" (0,+) (-1,+) -2 (-1,-) (-1,-) (-1,+) (-1,-) (-1,+) (-1,+)| \backslashbox$\mathrm{m}_{1}$$\mathrm{m}_{\mathrm{s}}$ | -1 | 0 | +1 | | :--- | :--- | :--- | :--- | | +2 | (1,-) (1,-) | $(1,+)(1,-)$ | $(1,+)(1,+)$ | | +1 | (1,-) (0,-) | $\begin{aligned} & (0,-)(1,+) \\ & (0,+) \\ & (1,-) \end{aligned}$ | $(1,+)(0,+)$ | | 0 | (1,-) (-1,-) | $\begin{aligned} & (1,+)(-1,-) \\ & (0,+)(0,-) \\ & (1,-)(-1,+) \end{aligned}$ | (1,+) (-1,+) | | -1 | (0,-) (-1,-) | $\begin{aligned} & (0,+)(-1,-) \\ & (0,-)(-1,+) \end{aligned}$ | (0,+) (-1,+) | | -2 | (-1,-) (-1,-) | (-1,+) (-1,-) | (-1,+) (-1,+) |
min L m s min L m s min_(L)m_(s)\mathrm{min}_{\mathrm{L}} \mathrm{m}_{\mathrm{s}} -1 0 1
( 1 , + ) ( 1 , ) ( 1 , + ) ( 1 , ) (1,+)(1,-)(1,+)(1,-)
1 ( 1 , ) ( 0 , ) ( 1 , ) ( 0 , ) (1,-)(0,-)(1,-)(0,-) ( 1 , + ) ( 0 , ) ( 1 , ) ( 0 , + ) ( 1 , + ) ( 0 , ) ( 1 , ) ( 0 , + ) {:[(1","+)(0","-)],[(1","-)(0","+)]:}\begin{aligned} & (1,+)(0,-) \\ & (1,-)(0,+) \end{aligned} ( 1 , + ) ( 0 , + ) ( 1 , + ) ( 0 , + ) (1,+)(0,+)(1,+)(0,+)
0 (1,-)(-1,-) ( 1 , + ) ( 1 , ) ( 0 , + ) ( 0 , ) ( 1 , ) ( 1 , + ) ( 1 , + ) ( 1 , ) ( 0 , + ) ( 0 , ) ( 1 , ) ( 1 , + ) {:[(1","+)(-1","-)],[(0","+)(0","-)],[(1","-)(-1","+)]:}\begin{aligned} & (1,+)(-1,-) \\ & (0,+)(0,-) \\ & (1,-)(-1,+) \end{aligned} ( 1 , + ) ( 1 , + ) ( 1 , + ) ( 1 , + ) (1,+)(-1,+)(1,+)(-1,+)
-1 (0,-)(-1,-) ( 0 , + ) ( 1 , ) ( 0 , ) ( 1 , + ) ( 0 , + ) ( 1 , ) ( 0 , ) ( 1 , + ) {:[(0","+)(-1","-)],[(0","-)(-1","+)]:}\begin{aligned} & (0,+)(-1,-) \\ & (0,-)(-1,+) \end{aligned} ( 0 , + ) ( 1 , + ) ( 0 , + ) ( 1 , + ) (0,+)(-1,+)(0,+)(-1,+)
-2 (-1,+)(-1,-)
min_(L)m_(s) -1 0 1 (1,+)(1,-) 1 (1,-)(0,-) "(1,+)(0,-) (1,-)(0,+)" (1,+)(0,+) 0 (1,-)(-1,-) "(1,+)(-1,-) (0,+)(0,-) (1,-)(-1,+)" (1,+)(-1,+) -1 (0,-)(-1,-) "(0,+)(-1,-) (0,-)(-1,+)" (0,+)(-1,+) -2 (-1,+)(-1,-) | $\mathrm{min}_{\mathrm{L}} \mathrm{m}_{\mathrm{s}}$ | -1 | 0 | 1 | | :--- | :--- | :--- | :--- | | | | $(1,+)(1,-)$ | | | 1 | $(1,-)(0,-)$ | $\begin{aligned} & (1,+)(0,-) \\ & (1,-)(0,+) \end{aligned}$ | $(1,+)(0,+)$ | | 0 | (1,-)(-1,-) | $\begin{aligned} & (1,+)(-1,-) \\ & (0,+)(0,-) \\ & (1,-)(-1,+) \end{aligned}$ | $(1,+)(-1,+)$ | | -1 | (0,-)(-1,-) | $\begin{aligned} & (0,+)(-1,-) \\ & (0,-)(-1,+) \end{aligned}$ | $(0,+)(-1,+)$ | | -2 | | (-1,+)(-1,-) | |

Pauli exclusion principle
泡利排除原则

For 2 p 2 2 p 2 2p^(2)2 \mathrm{p}^{2} probable atomic states?
对于 2 p 2 2 p 2 2p^(2)2 \mathrm{p}^{2} 可能的原子状态?
1 D 2 3 P 2 , 1 , 0 1 S 0 1 D 2 3 P 2 , 1 , 0 1 S 0 ^(1)D_(2)^(3)P_(2,1,0)^(1)S_(0){ }^{1} \mathrm{D}_{2}{ }^{3} \mathrm{P}_{2,1,0}{ }^{1} \mathrm{~S}_{0}
For 2 p 3 2 p 3 2p^(3)2 \mathrm{p}^{3} probable atomic states?
对于 2 p 3 2 p 3 2p^(3)2 \mathrm{p}^{3} 可能的原子状态?
2 D 5 / 2 , 3 / 2 2 P 3 / 2 , 1 / 2 4 S 3 / 2 2 D 5 / 2 , 3 / 2 2 P 3 / 2 , 1 / 2 4 S 3 / 2 ^(2)D_(5//2,3//2)^(2)P_(3//2,1//2)^(4)S_(3//2){ }^{2} \mathrm{D}_{5 / 2,3 / 2}{ }^{2} \mathrm{P}_{3 / 2,1 / 2}{ }^{4} \mathrm{~S}_{3 / 2}
For 3 d 2 3 d 2 3d^(2)3 \mathrm{~d}^{2} probable atomic states?
对于 3 d 2 3 d 2 3d^(2)3 \mathrm{~d}^{2} 可能的原子状态?
3 F 4 , 3 , 2 3 P 2 , 1 , 0 1 G 4 1 D 2 1 S 0 3 F 4 , 3 , 2 3 P 2 , 1 , 0 1 G 4 1 D 2 1 S 0 ^(3)F_(4,3,2)^(3)P_(2,1,0)^(1)G_(4)^(1)D_(2)^(1)S_(0){ }^{3} \mathrm{~F}_{4,3,2}{ }^{3} \mathrm{P}_{2,1,0}{ }^{1} \mathrm{G}_{4}{ }^{1} \mathrm{D}_{2}{ }^{1} \mathrm{~S}_{0}
If only 2 isoelectronic electrons, L + S = L + S = L+S=L+S= even(偶数)exist L+S=Odd(奇数)not fit Pauli exclusion principle
如果只有 2 个等电子电子, L + S = L + S = L+S=L+S= 偶数存在 L+S=奇数(奇数)不符合泡利排除原则

Only for 2 isoelectronic electrons,can not judge N>2
仅对 2 个等电子电子,不能判断 N>2
例: initial 3 P 2 , 1 , 0 3 P 2 , 1 , 0 ^(3)P_(2,1,0){ }^{3} \mathbf{P}_{2,1,0} add 1d electron, find possible atomic state
例: initial 3 P 2 , 1 , 0 3 P 2 , 1 , 0 ^(3)P_(2,1,0){ }^{3} \mathbf{P}_{2,1,0} add 1d electron, find possible atomic state

initial: l = 1 s = 1 l = 1 s = 1 quadl^(')=1quads^(')=1\quad l^{\prime}=1 \quad s^{\prime}=1  初: l = 1 s = 1 l = 1 s = 1 quadl^(')=1quads^(')=1\quad l^{\prime}=1 \quad s^{\prime}=1
d electron: l = 2 s = 1 / 2 l = 2 s = 1 / 2 quad l=2quad s=1//2\quad l=2 \quad s=1 / 2  d 电子: l = 2 s = 1 / 2 l = 2 s = 1 / 2 quad l=2quad s=1//2\quad l=2 \quad s=1 / 2
L-S:  L-S:
S = 3 / 2 , 1 / 2 L = 3 , 2 , 1 S = 3 / 2 , 1 / 2 L = 3 , 2 , 1 {:[S=3//2","1//2],[L=3","2","1]:}\begin{aligned} & S=3 / 2,1 / 2 \\ & L=3,2,1 \end{aligned}
L-S:  L-S:
S = 3 / 2 S = 1 / 2 L = 3 4 F 9 / 2 , 7 / 2 , 5 / 2 , 3 / 2 2 F 7 / 2 , 5 / 2 L = 2 4 D 7 / 2 , 5 / 2 , 3 / 2 , 1 / 2 2 D 5 / 2 , 3 / 2 L = 1 4 P 5 / 2 , 3 / 2 , 1 / 2 2 P 3 / 2 , 1 / 2 S = 3 / 2 S = 1 / 2 L = 3 4 F 9 / 2 , 7 / 2 , 5 / 2 , 3 / 2 2 F 7 / 2 , 5 / 2 L = 2 4 D 7 / 2 , 5 / 2 , 3 / 2 , 1 / 2 2 D 5 / 2 , 3 / 2 L = 1 4 P 5 / 2 , 3 / 2 , 1 / 2 2 P 3 / 2 , 1 / 2 {:[,S=3//2,S=1//2],[L=3,^(4)F_(9//2)","7//2","5//2","3//2,^(2)F_(7//2,5//2)],[L=2,^(4)D_(7//2,5//2,3//2,1//2),^(2)D_(5//2,3//2)],[L=1,^(4)P_(5//2,3//2,1//2),^(2)P_(3//2,1//2)]:}\begin{array}{ccc} & S=3 / 2 & S=1 / 2 \\ L=3 & { }^{4} F_{9 / 2}, 7 / 2,5 / 2,3 / 2 & { }^{2} F_{7 / 2,5 / 2} \\ L=2 & { }^{4} D_{7 / 2,5 / 2,3 / 2,1 / 2} & { }^{2} D_{5 / 2,3 / 2} \\ L=1 & { }^{4} P_{5 / 2,3 / 2,1 / 2} & { }^{2} P_{3 / 2,1 / 2} \end{array}
23 V  23 伏
: 3 d 3 ( n 1 2 N l = 1 2 × 10 = 5 ) , S = m s i = 1 2 + 1 2 + 1 2 = 3 2 2 s + 1 = 4 , L = 1 2 n ( 2 l + 1 n ) = 1 2 × 3 ( 2 × 2 + 1 3 ) = 3 J = | L S | = | 3 3 2 | = 3 2 jmin 4 F 3 / 2 : 3 d 3 n 1 2 N l = 1 2 × 10 = 5 , S = m s i = 1 2 + 1 2 + 1 2 = 3 2 2 s + 1 = 4 , L = 1 2 n ( 2 l + 1 n ) = 1 2 × 3 ( 2 × 2 + 1 3 ) = 3 J = | L S | = 3 3 2 = 3 2  jmin  4 F 3 / 2 {:[:3d^(3)(n(:(1)/(2)N_(l)=(1)/(2)xx10=5),quad S=summ_(si)=(1)/(2)+(1)/(2)+(1)/(2)=(3)/(2):}],[=>2s+1=4","quad L=(1)/(2)n(2l+1-n)=(1)/(2)xx3(2xx2+1-3)=3],[=>J=|L-S|=|3-(3)/(2)|=(3)/(2)],[" jmin "{:^(4)F_(3//2):}]:}\begin{gathered} : 3 d^{3}\left(n\left\langle\frac{1}{2} N_{l}=\frac{1}{2} \times 10=5\right), \quad S=\sum m_{s i}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{3}{2}\right. \\ \Rightarrow 2 s+1=4, \quad L=\frac{1}{2} n(2 l+1-n)=\frac{1}{2} \times 3(2 \times 2+1-3)=3 \\ \Rightarrow J=|L-S|=\left|3-\frac{3}{2}\right|=\frac{3}{2} \\ \text { jmin } \begin{array}{c} { }^{4} F_{3 / 2} \end{array} \end{gathered}
25 Mn (锰)
3 d 5 ( n = 1 2 N l = 5 ) , S = m s i = 1 2 + 1 2 + 1 2 + 1 2 + 1 2 = 5 2 2 s + 1 = 6 , L = 1 2 n ( 2 l + 1 n ) = 1 2 × 5 ( 2 × 2 + 1 5 ) = 0 J = S = 5 2 3 d 5 n = 1 2 N l = 5 , S = m s i = 1 2 + 1 2 + 1 2 + 1 2 + 1 2 = 5 2 2 s + 1 = 6 , L = 1 2 n ( 2 l + 1 n ) = 1 2 × 5 ( 2 × 2 + 1 5 ) = 0 J = S = 5 2 {:[quad3d^(5)(n=(1)/(2)N_(l)=5)","quad S=summ_(si)=(1)/(2)+(1)/(2)+(1)/(2)+(1)/(2)+(1)/(2)=(5)/(2)],[=>2s+1=6","quad L=(1)/(2)n(2l+1-n)=(1)/(2)xx5(2xx2+1-5)=0],[=>J=S=(5)/(2)]:}\begin{gathered} \quad 3 d^{5}\left(n=\frac{1}{2} N_{l}=5\right), \quad S=\sum m_{s i}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{5}{2} \\ \Rightarrow 2 s+1=6, \quad L=\frac{1}{2} n(2 l+1-n)=\frac{1}{2} \times 5(2 \times 2+1-5)=0 \\ \Rightarrow J=S=\frac{5}{2} \end{gathered}

24 Cr ground state
24 Cr 基态

Cr : 3 d 5 4 s 3 d 5 4 s 3d^(5)4s3 d^{5} 4 s,  铬 : 3 d 5 4 s 3 d 5 4 s 3d^(5)4s3 d^{5} 4 s
3 d 5 4 s = 3 d 5 + 4 s 3 d 5 : s 1 = 5 2 , l 1 = 0 4 s : s 2 = 1 2 , l 2 = 0 S = s 1 + s 2 , , | s 1 s 2 | = 3 , 2 { S = 3 L = 0 , { J = 3 2 s + 1 = 7 7 S 3 3 d 5 4 s = 3 d 5 + 4 s 3 d 5 : s 1 = 5 2 , l 1 = 0 4 s : s 2 = 1 2 , l 2 = 0 S = s 1 + s 2 , , s 1 s 2 = 3 , 2 S = 3 L = 0 , J = 3 2 s + 1 = 7 7 S 3 {:[3d^(5)4s=3d^(5)+4s],[3d^(5)quad:s_(1)=(5)/(2)","l_(1)=0],[4s quad:s_(2)=(1)/(2)","l_(2)=0],[S=s_(1)+s_(2)","cdots cdots","|s_(1)-s_(2)|=3","2],[{[S=3],[L=0],{[J=3],[2s+1=7]:}],[^(7)S_(3)]:}\begin{gathered} 3 d^{5} 4 s=3 d^{5}+4 s \\ 3 d^{5} \quad: s_{1}=\frac{5}{2}, l_{1}=0 \\ 4 s \quad: s_{2}=\frac{1}{2}, l_{2}=0 \\ S=s_{1}+s_{2}, \cdots \cdots,\left|s_{1}-s_{2}\right|=3,2 \\ \left\{\begin{array}{l} S=3 \\ L=0 \end{array},\left\{\begin{array}{l} J=3 \\ 2 s+1=7 \end{array}\right.\right. \\ { }^{7} S_{3} \end{gathered}
{ 26 Fe 3 d 6 n = 6 , l = 2 S = 1 2 [ 2 ( 2 l + 1 ) n ] = 2 { J = L + S = 4 2 S + 1 = 5 L = 1 2 [ 2 ( 2 l + 1 ) n ] [ n ( 2 l + 1 ) ] = 2 5 D 4 25 Mn 3 d 5 n = 5 , l = 2 { S = 1 2 [ 2 ( 2 l + 1 ) n ] = 5 2 L = 1 2 [ 2 ( 2 l + 1 ) n ] [ n ( 2 l + 1 ) ] = 0 { J = L + S = 5 2 2 S + 1 = 6 26 Fe 3 d 6 n = 6 , l = 2 S = 1 2 [ 2 ( 2 l + 1 ) n ] = 2 J = L + S = 4 2 S + 1 = 5 L = 1 2 [ 2 ( 2 l + 1 ) n ] [ n ( 2 l + 1 ) ] = 2 5 D 4 25 Mn 3 d 5 n = 5 , l = 2 S = 1 2 [ 2 ( 2 l + 1 ) n ] = 5 2 L = 1 2 [ 2 ( 2 l + 1 ) n ] [ n ( 2 l + 1 ) ] = 0 J = L + S = 5 2 2 S + 1 = 6 {:[{[26Fe,3d^(6),n=6","l=2],[S=(1)/(2)[2(2l+1)-n]=2,,{[J=L+S=4],[2S+1=5]:}],[L=(1)/(2)[2(2l+1)-n]*[n-(2l+1)]=2,],[^(5)D_(4),]:}],[25Mn],[3d^(5)quad n=5","l=2],[{[S=(1)/(2)[2(2l+1)-n]=(5)/(2)],[L=(1)/(2)[2(2l+1)-n]*[n-(2l+1)]=0]{[J=L+S=(5)/(2)],[2S+1=6]:}]:}\begin{aligned} & \left\{\begin{array}{ccl} 26 \mathrm{Fe} & 3 d^{6} & n=6, l=2 \\ S=\frac{1}{2}[2(2 l+1)-n]=2 & & \left\{\begin{array}{l} J=L+S=4 \\ 2 S+1=5 \end{array}\right. \\ L=\frac{1}{2}[2(2 l+1)-n] \cdot[n-(2 l+1)]=2 & \\ { }^{5} D_{4} & \end{array}\right. \\ & 25 \mathrm{Mn} \\ & 3 d^{5} \quad n=5, l=2 \\ & \left\{\begin{array} { l } { S = \frac { 1 } { 2 } [ 2 ( 2 l + 1 ) - n ] = \frac { 5 } { 2 } } \\ { L = \frac { 1 } { 2 } [ 2 ( 2 l + 1 ) - n ] \cdot [ n - ( 2 l + 1 ) ] = 0 } \end{array} \left\{\begin{array}{l} J=L+S=\frac{5}{2} \\ 2 S+1=6 \end{array}\right.\right. \end{aligned}
45 Rh: 4 d 8 5 s 1 4 d 8 n = 8 , l = 2 { S 1 = 1 2 [ 2 ( 2 l + 1 ) n ] = 1 L 1 = 1 2 [ 2 ( 2 l + 1 ) n ] [ n ( 2 l + 1 ) ] = 3 5 s 1 n = 5 , l = 0 { S = S 1 + S 2 = 3 / 2 L = L 1 + L 2 = 3 J = L + S = 9 / 2 2 S + 1 = 4 { S 2 = 1 2 L 2 = 0 4 F 9 / 2 45  Rh:  4 d 8 5 s 1 4 d 8 n = 8 , l = 2 S 1 = 1 2 [ 2 ( 2 l + 1 ) n ] = 1 L 1 = 1 2 [ 2 ( 2 l + 1 ) n ] [ n ( 2 l + 1 ) ] = 3 5 s 1 n = 5 , l = 0 S = S 1 + S 2 = 3 / 2 L = L 1 + L 2 = 3 J = L + S = 9 / 2 2 S + 1 = 4 S 2 = 1 2 L 2 = 0 4 F 9 / 2 {:[45" Rh: "quad4d^(8)5s^(1)],[4d^(8)quad n=8","l=2],[{[S_(1)=(1)/(2)[2(2l+1)-n]=1],[L_(1)=(1)/(2)[2(2l+1)-n]*[n-(2l+1)]=3]:},5s^(1)quad n=5","l=0],[{[S=S_(1)+S_(2)=3//2],[L=L_(1)+L_(2)=3],[J=L+S=9//2],[2S+1=4]:},{[S_(2)=(1)/(2)],[L_(2)=0]:}],[,^(4)F_(9//2)],[,]:}\begin{array}{cc} 45 \text { Rh: } \quad 4 d^{8} 5 s^{1} \\ 4 d^{8} \quad n=8, l=2 \\ \begin{cases}S_{1}=\frac{1}{2}[2(2 l+1)-n]=1 \\ L_{1}=\frac{1}{2}[2(2 l+1)-n] \cdot[n-(2 l+1)]=3\end{cases} & 5 s^{1} \quad n=5, l=0 \\ \left\{\begin{array}{l} S=S_{1}+S_{2}=3 / 2 \\ L=L_{1}+L_{2}=3 \\ J=L+S=9 / 2 \\ 2 S+1=4 \end{array}\right. & \left\{\begin{array}{l} S_{2}=\frac{1}{2} \\ L_{2}=0 \end{array}\right. \\ & { }^{4} F_{9 / 2} \\ & \end{array}

电子在每一闭
壳层的总数


------ 36 (Kr)  ------ 36 (KR)
3 s 3 p = 6 2 } 8 3 s 3 p = 6 2 8 {:_(3s)^(3p)=[6],[2]}8quad\left.{ }_{3 s}^{3 p}=\begin{array}{l}6 \\ 2\end{array}\right\} 8 \quad------ 18 ( Ar ) 18 ( Ar ) 18(Ar)18(\mathrm{Ar})
3 s 3 p = 6 2 } 8 3 s 3 p = 6 2 8 {:_(3s)^(3p)=[6],[2]}8quad\left.{ }_{3 s}^{3 p}=\begin{array}{l}6 \\ 2\end{array}\right\} 8 \quad 18 ( Ar ) 18 ( Ar ) 18(Ar)18(\mathrm{Ar}) ------

2 p = 6 2 s } 8 2 p = 6 2 s 8 {:2p=[6],[2s]}8\left.2 p=\begin{array}{r}6 \\ 2 s\end{array}\right\} 8
1 s 22 ( He ) 1 s 22 ( He ) 1s longrightarrow22(He)1 s \longrightarrow 22(\mathrm{He})
图5.7 碳族元素在激发态 ps 的能级比较