᷷᷉᷾᳐ᷙᶵḅᶾ᳐方程式ᶍ解 ᵫ軌道ᶍ概念ᶇᶉᶪ

水素原子 水素原子

᷷᷉᷾᳐ᷙᶵḅᶾ᳐方程式ᶍ解

ᵫ軌道ᶍ概念ᶇᶉᶪ

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!_n,l,m(r,",#) = R_n,l(r)$_l,m(")%_m(#) = R_n,l(r) x Y_l,m(",#)

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a_o ( Z )^3/2

R_n,l(r) = A_n,l ( 2Zrna_o)^l[Laguerre_n,l ( 2Zrna_o)] na_o ( Zr ) x exp –

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!_o = Zr a_o a_o = 0.53 Å

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"_n,l,m = R_n,l(r) Y_l,m(#,$)

.

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n = 1 n = 2

n = 3

l = 0

l = 2 l = 0

l = 1

l = 0 l = 1

R₁₀ = R_1s = R₂₀ = R_2s = R₂₁ = R_2p = R₃₀ = R_3s = R₃₁ = R_3p = R₃₂ = R_3d =

e

^–!o/2

a_o ( Z )^3/2 2 ( aZ_o)^3/2

e

^–!o

2"2

1 (2 – !_o)

e

^–!o/2

a_o ( Z )^3/2

1 (!_o)

2"6

e

^–!o/3

a_o ( Z )^3/2

2 (27 – 18!_o + 2!_o²) 81"3

e

^–!o/3

a_o ( Z )^3/2

4 (6!_o - !_o²) 81"6

e

^–!o/3

a_o ( Z )^3/2

4 (!_o²)

81"30

··· ··· ···

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....

0 2 4 6 8 10 12

r /Å

0 0 0

Rn,l (r) n = 3

l = 0

l = 1

l = 2 3s

3p

3d

0 2 4 6 8 10 12

r /Å

0 0

1.0

n = 2

l = 0

l = 1 2s

2p

0 2 4 6 8 10 12

r /Å

0

Rn,l (r) n = 1

l = 0

2 4

1s

+ 4πr2 |R n,l(r)|2

0

Rn,l (r)

0

0.4 0

4πr2 |R n,l(r)|2

0 0 0.4

0.4

4πr2 |R n,l(r)|2

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L² = 0 L² = 2 h 2!

L² = 6 h

2! L² = 12 h 2!

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l = 0 l = 1 l = 2 l = 3 ···

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"#

'()*+

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E = 8e₀²h²

me⁴Z² n² – 1

|L²| = l(l + 1)h²

1s 2s 3s 4s 5s 6s

2p 3p 4p 5p 6p

3d 4d 5d

4f 5f

z

y

x

z

y

x

z

y

x

px py pz

x y

z z

y x

z

y x

z

y x

z

x y

dz² dxz

dyz

dxy

dx²-y²

7s

n = 1 n = 2 n = 3 n = 4 n = 5 n = 6 n = 7

K L M N

O P Q r

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多電子原子ᶊ続ᵮ