3.3 Solutions
229
3.3.7 Approximate Methods
3.97 ΔE =< ψ|δU |ψ >
δU = U (interaction energy of electron with point charge nucleus)
δU = e
2
/r −
3e
2
2R
R
2
−
r
2
3
for r ≤ R
= 0
f o r r ≥ R
(a) First we consider n = 1 state
ΔE =
e
2
πa
3
0
R
0
exp
−
2r
a 0
1
r
−
3
2R
+
1
/ 2
r
2
R 3
4πr
2 dr
=
2e
2
a
3
0
R
0
e
−2r/a 0
2r −
3r
2
R
+
r
4
R 3
dr
Now R = 10
−13 cm 10
−8 cm = a 0 , the factor exp
−
2r
a 0
≈ 1
ΔE =
2e
2
a
3
0
R
2
− R
2
+
R
2
5
=
4
5
e
2
2a 0
R
a 0
2
= (0.8)(13.6)
10
−13
0.53 × 10 −8
2
= 3.87 × 10
−9 eV
(b) n = 2
ψ 200 =
1
8πa
3
0
1
2
2 −
r
a 0
exp
−
r
2a 0
ΔE =
e
2
8πa
3
0
R
0
exp
−
r
a 0
2 −
r
a 0
2
1
r
−
3
2R
+
1
/ 2
r
2
R 3
4πr
2 dr
Here also exp
−
r
a0
∼ 1, for reasons indicated in (a)
When the remaining factors are integrated we get
ΔE =
e
2
2a 0
R
2
a
2
0
2
5
−
1
6
R
a 0
+
3
140
R
2
a
2
0
≈
2
5
.
e
2
2a 0
R
a 0
2
as
R
a 0
<< 1
=
1
/ 2 × 3.87 × 10
−9
= 1.93 × 10
−9 eV
where we used the result of (a)
3.98 Schrodinger’s equation in the presence of electric field is
−
2
2m
d
2
dx 2 +
1
/ 2 m ω
2 x
2
+ qEx
ψ n = E n x
(1)
Now,
1
/ 2 mω
2 x
2
+ qEx =
1
/ 2 mω
2
x
2
+
2qEx
mω 2