12.19 A wavefunction can have a node where the value is exactly equal
to zero. For the 2s orbital this occurs at:
12.20Φ(φ+ 2 π) =Φ(φ)
Aeimlφ=Aeiml(φ+^2 π)=Aeimlφeiml^2 π
1 =eiml^2 π=cos(2πml) +isin(2πml)
with ml=0, ±1, ±2, ±3,....
12.21
12.22
−Bsinθ(2sinθcosθ) + 2 Bcosθsin^2 θ= 0
0 = 0
12.23
The second term is zero since:
This leaves the first term and:
E
me
=−
4
2
0
32 πεZ^2
α
πε
=
me
Z^2
2
(^40)
re
mE
e
−−rre
⎛
⎝
⎜⎜
⎞
⎠
αααα⎟⎟+−+
πε
2
2
2
0
2
2
4
2
Z
mm
Z^2
0
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟=
αα
πε
2 αα
2
2
0
2
2
4
re e
me
r
−−rr−+ +Er
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟
Z
ee−αr= 0
d
d
2
2
2
r
Π()rreee=− −αα( −−ααrr+ )−α−αr=αrre−−ααrr− 2 αe
d
dr
Π()rr e e=−( )α −−ααrr+
−+=BBsinθ [sin ] cos sin
θ
θθθ
d
d
(^2220)
sinθ [sin ( sin )] cos sin
θ
θθ θθ
d
d
−+BB 202 =
sinθ sin ( cos ) [ ( )
θ
θ
θ
θ
d
d
d
d
B
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
++11 1 ssin^2 θθ−= 00 ]( cos )B
d
d
d
d
2
2
2
φ
φ
φ
Φ()==( ( )Aim e φφ) Aim e( )
l
im
l
llim =−m ()
l
(^2) Φφ
d
d
d
φ d
φ
φ
Φ()==(Aeimφφ) A im e( )
l
llim
ψ
π
ρ ρ
200
3
0
3
(^14)
224
2
2
=−/ 0
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟ − =
Z
a
e wheenρ= 4
ANSWERS TO PROBLEMS 469
9781405124362_5_end.qxd 4/29/08 9:17 Page 469