3.3 Solutions 229
3.3.7 Approximate Methods .........................
3.97ΔE=<ψ|δU|ψ>
δU=U(interaction energy of electron with point charge nucleus)δU=e^2 /r−3 e^2
2 R(
R^2 −
r^2
3)
forr≤R=0forr≥R(a) First we considern=1 stateΔE=(
e^2
πa 03)∫ R
0exp(
−
2r
a 0)[
1
r−
3
2 R
+^1 / 2
r^2
R^3]
4 πr^2 dr=
2 e^2
a^30∫ R
0e−2r/a^0[
2 r−3 r^2
R+
r^4
R^3]
drNowR= 10 −^13 cm 10 −^8 cm=a 0 , the factor exp(
−^2 ar 0)
≈ 1
ΔE=
(
2 e^2
a 03)(
R^2 −R^2 +
R^2
5
)
=
4
5
(
e^2
2 a 0)(
R
a 0) 2
=(0.8)(13.6)
(
10 −^13
0. 53 × 10 −^8
) 2
= 3. 87 × 10 −^9 eV
(b)n= 2ψ 200 =(
1
8 πa 03)^12 (
2 −
r
a 0)
exp(
−
r
2 a 0)
ΔE=
(
e^2
8 πa 03)∫R
0exp(
−
r
a 0)(
2 −
r
a 0) 2 [
1
r−
3
2 R
+^1 / 2
r^2
R^3]
4 πr^2 drHere also exp(
−ar 0)
∼1, for reasons indicated in (a)
When the remaining factors are integrated we getΔE=(
e^2
2 a 0)(
R^2
a^20)[
2
5
−
1
6
R
a 0+
3
140
R^2
a 02]
≈
2
5
.
(
e^2
2 a 0)(
R
a 0) 2 (
asR
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
2 m)
d^2
dx^2+^1 / 2 mω^2 x^2 +qEx]
ψn=Enx (1)Now,^1 / 2 mω^2 x^2 +qEx=^1 / 2 mω^2[
x^2 +^2 mqExω 2