12.19 A wavefunction can have a node where the value is exactly equal
to zero. For the 2s orbital this occurs at:
12.20 Φ(φ + 2π) = Φ(φ)
Ae
im l φ
= Ae
im l (φ +2π)
= Ae
im l φ e
im l 2π
1 = e
im l 2π
= cos(2πm l ) + i sin(2πm l )
with m l = 0, ±1, ±2, ±3, . . . .
12.21
12.22
− B sin θ(2sin θ cos θ) + 2B cos θ sin
2
θ = 0
0 = 0
12.23
The second term is zero since:
This leaves the first term and:
E
me
= −
4
2
0
2
32π ε Z
α
πε
=
m e
Z
2
2
0
4
re
mE
e
e
r
r
−
−
+
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟ +
− +
α
α
α
α
πε
2
2
2
0
2
2
4
2
Z
m m
Z
2
0
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟ =
α
α
πε
α
α
2
2
2
0
2
2
4
re
e
m e
r
E r
r
r
−
−
−
+
+
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟
Z
e e
r
−
=
α
0
d
d
2
2
2
r
r
r e
e
e
r
r
r
Π( )
(
)
= − −
+
−
=
−
−
−
α α
α
α
α
α
α
r re
e
r
r
−
−
−
α
α
α
2
d
dr
r
r
e
e
r
r
Π( ) ( )
= −
+
−
−
α
α
α
−
+
=
sin
[sin ]
cos sin
B
B
θ θ
θ
θ
θ
d
d
2
2
2
0
sin
[sin ( sin )]
cos sin
θ θ
θ
θ
θ
θ
d
d
B
B
−
+
=
2
0
2
sin
sin
( cos ) [ (
)
θ θ
θ θ
θ
d
d
d
d
B
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
+
+
11 1 s sin
]( cos )
2
0
0
θ
θ
−
=
B
d
d
d
d
2
2
2
φ
φ
φ
φ
φ
Φ( )
( ( )
)
( )
=
=
A im e
A im e
l
im
l
im
l
l
( )
= −m l
2
Φ φ
d
d
d
d
φ
φ
φ
φ
φ
Φ( )
(
)
( )
=
=
Ae
A im e
im
l
im
l
l
ψ
π
ρ ρ
200
3
0
3
4
1
2 2 4
2
2
0
/
=
−
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟
=
−
Z
a
e
whe en ρ = 4
ANSWERS TO PROBLEMS
469
9781405124362_5_end.qxd 4/29/08 9:17 Page 469