11-2（泡利不相容）

Alkali earth metal atoms
碱土金属原子

He: Z=2 He：Z=2
Be:

Z = 4 = 2 \times 1^{2} + 2

是：

Z = 4 = 2 \times 1^{2} + 2

Mg:

Z = 12 = 2 \times (1^{2} + 2^{2}) + 2

毫克：

Z = 12 = 2 \times (1^{2} + 2^{2}) + 2

Ca:

Z = 20 = 2 \times (1^{2} + 2^{2} + 2^{2}) + 2

钙：

Z = 20 = 2 \times (1^{2} + 2^{2} + 2^{2}) + 2

Sr:

Z = 38 = 2 \times (1^{2} + 2^{2} + 3^{2} + 2^{2}) + 2

锶：

Z = 38 = 2 \times (1^{2} + 2^{2} + 3^{2} + 2^{2}) + 2

Ba:

Z = 56 = 2 \times (1^{2} + 2^{2} + 3^{2} + 3^{2} + 2^{2}) + 2

三：

Z = 56 = 2 \times (1^{2} + 2^{2} + 3^{2} + 3^{2} + 2^{2}) + 2

Ra:

Z = 88 = 2 \times (1^{2} + 2^{2} + 3^{2} + 4^{2} + 3^{2} + 2^{2}) + 2

Ra：

Z = 88 = 2 \times (1^{2} + 2^{2} + 3^{2} + 4^{2} + 3^{2} + 2^{2}) + 2

Calculate

\vec{L} \cdot \vec{S}

^{4} D_{3 / 2}

计算

\vec{L} \cdot \vec{S}

范围

^{4} D_{3 / 2}

\begin{aligned} ^{4} D_{3 / 2} : L = 2, S = \frac{3}{2}, J = \frac{3}{2}; \\ 由 {\hat{J}}^{2} = {\hat{L}}^{2} + {\hat{S}}^{2} + 2 \hat{L} \cdot \hat{S} 得 \\ \vec{L} \cdot \vec{S} = \frac{1}{2} ({\hat{J}}^{2} - {\hat{L}}^{2} - {\hat{S}}^{2}) \\ = \frac{1}{2} [J (J + 1) - L (L + 1) - S (S + 1)] ℏ^{2} = - 3 ℏ^{2} \end{aligned}

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

[

]

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

\begin{array}{cc} n : n = 1, 2, 3, \dots & l : l = 0, 1, 2, \dots, (n - 1) \\ m_{l} : m_{l} = 0, \pm 1, \pm 2, \dots, \pm l & m_{s} : m_{s} = \pm 1 / 2 \\ (n, l, m_{l}, m_{s}) \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})

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}

）
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}

\begin{matrix} (1 s 1 s)^{3} S_{1} \times (1 s 1 s)^{1} S_{C} \sqrt{} \\ n_{1} = n_{2} = 1 l_{1} = l_{2} = 0 m_{l_{1}} = m_{l_{2}} = 0 \end{matrix}

Pauli exclusion principle
泡利排除原则

m_{s_{1}} = \frac{1}{2}, - \frac{1}{2} 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 3 p 原子状态呢 for

2 p^{2}

for $2 p^{2}$
find $M_{L}$ and $M_{S}$
为 $2 p^{2}$ find $M_{L}$ 和 $M_{S}$

\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} \Rightarrow m_{L} = m_{l 1} + m_{l 2} = 2, 1, 0, - 1, - 2

$m_{L}$	-1	0	1
2
1
0
-1
-2

For

2 p^{2}

probable

m_{l} m_{s combination}

对于

2 p^{2}

可能

m_{l} m_{s combination}

\backslashbox $m_{1}$ $m_{s}$	-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,+)(-1,+)
-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,+)

$\min_{L} m_{s}$	-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}

probable atomic states？
对于

2 p^{2}

可能的原子状态？

^{1} D_{2}^{3} P_{2, 1, 0}^{1} S_{0}

For

2 p^{3}

probable atomic states？
对于

2 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}

For

3 d^{2}

probable atomic states？
对于

3 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}

If only 2 isoelectronic electrons，

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 =

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}

add 1d electron, find possible atomic state
例： initial

^{3} P_{2, 1, 0}

add 1d electron， find possible atomic state
initial:

l^{'} = 1 s^{'} = 1

初：

l^{'} = 1 s^{'} = 1

d electron:

l = 2 s = 1 / 2

d 电子：

l = 2 s = 1 / 2

L-S: L-S：

\begin{aligned} S = 3 / 2, 1 / 2 \\ L = 3, 2, 1 \end{aligned}

L-S: L-S：

\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

\begin{matrix} : 3 d^{3} (n ⟨ \frac{1}{2} N_{l} = \frac{1}{2} \times 10 = 5), S = \sum m_{s i} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2} \\ \Rightarrow 2 s + 1 = 4, L = \frac{1}{2} n (2 l + 1 - n) = \frac{1}{2} \times 3 (2 \times 2 + 1 - 3) = 3 \\ \Rightarrow J = | L - S | = | 3 - \frac{3}{2} | = \frac{3}{2} \\ jmin \begin{matrix} ^{4} F_{3 / 2} \end{matrix} \end{matrix}

25 Mn (锰)

\begin{matrix} 3 d^{5} (n = \frac{1}{2} N_{l} = 5), 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, 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{matrix}

24 Cr ground state
24 Cr 基态

Cr :

3 d^{5} 4 s

, 铬：

3 d^{5} 4 s

，

\begin{matrix} 3 d^{5} 4 s = 3 d^{5} + 4 s \\ 3 d^{5} : s_{1} = \frac{5}{2}, l_{1} = 0 \\ 4 s : s_{2} = \frac{1}{2}, l_{2} = 0 \\ S = s_{1} + s_{2}, \dots \dots, | s_{1} - s_{2} | = 3, 2 \\ {\begin{cases} S = 3 \\ L = 0 \end{cases}, {\begin{cases} J = 3 \\ 2 s + 1 = 7 \end{cases} \\ ^{7} S_{3} \end{matrix}

\begin{aligned} {\begin{array}{ccl} 26 Fe & 3 d^{6} & n = 6, l = 2 \\ S = \frac{1}{2} [2 (2 l + 1) - n] = 2 & {\begin{matrix} J = L + S = 4 \\ 2 S + 1 = 5 \end{matrix} \\ L = \frac{1}{2} [2 (2 l + 1) - n] \cdot [n - (2 l + 1)] = 2 \\ ^{5} D_{4} \end{array} \\ 25 Mn \\ 3 d^{5} n = 5, l = 2 \\ {\begin{matrix} 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{matrix} {\begin{matrix} J = L + S = \frac{5}{2} \\ 2 S + 1 = 6 \end{matrix} \end{aligned}

\begin{array}{cc} 45 Rh: 4 d^{8} 5 s^{1} \\ 4 d^{8} 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} n = 5, l = 0 \\ {\begin{cases} S = S_{1} + S_{2} = 3 / 2 \\ L = L_{1} + L_{2} = 3 \\ J = L + S = 9 / 2 \\ 2 S + 1 = 4 \end{cases} & {\begin{cases} S_{2} = \frac{1}{2} \\ L_{2} = 0 \end{cases} \\ ^{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} = \begin{array}{l} 6 \\ 2 \end{array}} 8

－－－－－－

18 (Ar)

_{3 s}^{3 p} = \begin{matrix} 6 \\ 2 \end{matrix}} 8

18 (Ar)

------

2 p = \begin{array}{r} 6 \\ 2 s \end{array}} 8

1 s ⟶ 22 (He)

图5．7 碳族元素在激发态 ps 的能级比较

Alkali earth metal atoms碱土金属原子

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

Pauli exclusion principle泡利排除原则

2．Pauli exclusion principle2.泡利排除原则

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}}为 2 p 2 2 p 2 2p^(2)2 \mathrm{p}^{2} find M L M L M_(L)\mathrm{M}_{\mathrm{L}} 和 M S M S M_(S)\mathrm{M}_{\mathrm{S}}

Pauli exclusion principle泡利排除原则

24 Cr ground state24 Cr 基态

电子在每一闭 壳层的总数

Alkali earth metal atoms
碱土金属原子

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

Pauli exclusion principle
泡利排除原则

2．Pauli exclusion principle
2.泡利排除原则

for $2 p^{2}$
find $M_{L}$ and $M_{S}$
为 $2 p^{2}$ find $M_{L}$ 和 $M_{S}$

Pauli exclusion principle
泡利排除原则

24 Cr ground state
24 Cr 基态

电子在每一闭
壳层的总数