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H2
HΨ = EΨ (7.1.1)
H Ψ E
H K V
H = K + V (7.1.2)
K = −
2
2m∇ a2 +∇b
2( ) −2
2M∇A2 +∇B
2( )
V = −ke21
rAa+1
rAb+1
rBa+1
rBb−1
rab−1
rAB
⎛
⎝ ⎜
⎞
⎠ ⎟
A B a b
m M rXY X
Y
(7.1.1)
1
L
−
2
2m
∂2
∂x 2+∂2
∂y 2+∂2
∂z2⎛
⎝ ⎜
⎞
⎠ ⎟ +V (x,y,z)
⎡
⎣ ⎢
⎤
⎦ ⎥ Ψ = EΨ (7.1.3)
V (x,y,z) = 0 0 ≤ x,y,z ≤ L
V (x,y,z) =∞
Ψ
Ψ =ϕx (x)ϕy (y)ϕz(z) (7.1.4)
(7.1.3)
−
2
2mϕy (y)ϕz(z)
∂2
∂x 2ϕx (x) +ϕx (x)ϕz(z)
∂2
∂y 2ϕy (y) +ϕx (x)ϕy (y)
∂2
∂z2ϕz(z)
⎛
⎝ ⎜
⎞
⎠ ⎟
+V (x,y,z)ϕx (x)ϕy (y)ϕz (z) = Eϕx (x)ϕy (y)ϕz (z) (7.1.5)
(7.1.4) 0
(7.1.4)
−
2
2m
1
ϕx (x)
∂2
∂x 2ϕx (x) +
1
ϕy (y)
∂2
∂y 2ϕy (y) +
1
ϕz (z)
∂2
∂z2ϕz(z)
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
+V (x,y,z) = E (7.1.6)
V(x,y,z) (7.1.3) α = x y z
vα (α) = 0 0 ≤ α ≤ L (7.1.7)
vα (α) =∞ α < 0 L < α
(7.1.6)
−2
2m
d2
dα 2 + vα (α)⎛
⎝ ⎜
⎞
⎠ ⎟ ϕα (α)
α
∑ = εαϕα (α)α
∑ (7.1.8)
(5.1.3)
−
2
2m
d2
dα 2 + vα (α)⎛
⎝ ⎜
⎞
⎠ ⎟ ϕα (α) = εαϕα (α) (7.1.9)
E
E = εx + εy + εz (7.1.10)
(7.1.9) (7.1.3)
0
−
2
2m
d2
dα 2 ϕ(α) = εϕ(α) (7.1.11)
0 ≤ α ≤ L 0
ϕ(0) =ϕ(L) = 0 (7.1.12)
(7.1.11) 2
sin(kα) cos(kα)
(7.1.12) ϕ(0) = 0 cos(kα)
ϕ(α) = Asin(kα) (7.1.13)
(7.1.12) ϕ(L) = 0 n
kL = nπ
ϕ(α) = AsinnπLα
⎛
⎝ ⎜
⎞
⎠ ⎟ (7.1.14)
(7.1.14) (7.1.11)
ε =
π2 2
2mL2n2 =
h2
8mL2n2 (7.1.15)
n n (7.1.14) (7.1.12)
n = 0 (7.1.14) 0
ϕ(α) = Asin0 • πL
α⎛
⎝ ⎜
⎞
⎠ ⎟ = 0 (7.1.16)
n = 0
n (7.1.14) n A
A
n
(7.1.1) (7.1.14) n
(7.1.15) n
n
n = 1
ε =h2
8mL2 (7.1.17)
0.1 nm L (7.1.17)
3.6•10-20 kJ ≈ 38 eV
(7.1.17)
0
m
L
Ψ(x,y,z) = C sinnxπL
x⎛
⎝ ⎜
⎞
⎠ ⎟ sin
nyπ
Ly
⎛
⎝ ⎜
⎞
⎠ ⎟ sin
nzπ
Lz
⎛
⎝ ⎜
⎞
⎠ ⎟ (7.1.18)
E =h2
8mL2nx2 + ny
2 + nz2( ) (7.1.19)
C nx ny nz
2
1800
r 2
Ekb = EkA + EkB =3h2
8mr2+3h2
8mr2=3h2
4mr2 (7.2.1)
r × r × 2r
2
Eka = 2h2
8mr2(12 +12) +
h2
8m(2r)2(12)
⎡
⎣ ⎢
⎤
⎦ ⎥ =
9h2
16mr2 (7.2.2)
ΔEk = Eka − Ekb =9
16−3
4
⎛
⎝ ⎜
⎞
⎠ ⎟ h2
mr2= −
3h2
16mr2 (7.2.3)
ke2
r (7.2.4)
(7.2.4)
ΔEp = 2ke2
r (7.2.5)
r (7.2.3) (7.2.5)
7.1
ΔE = ΔEk + ΔEp (7.2.6)
highest occupied molecular orbital HOMO
lowest unoccupied molecular orbital LUMO
HOMO LUMO
HOMO LUMO
R.
1981
7.3 10 5
2
7.3
HOMO 3 xy
2pz
HOMO
2pz
HOMO
HOMO 7.4 xy
2 2pz 2pz
xy 2
3
2
p p = |p|
AB
(7.4.1)
pionized = e•rA-B (7.5.1)
e rA-B
(7.5.1) AB
= p
e • rA−B (7.5.2)
HCl
20%
7.6