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43. The proposed wave function is
ψ
=
−
1
32
π
a
e
ra
where
a
is the Bohr radius. Substituting this into the right side of Schrödinger’s equation,
our goal is to show that the result is zero. The derivative is
d
dr
a
e
=−
−
1
52
π
so
r
d
dr
r
a
e
2
2
−
π
and
11
2
1
1
2
1
2
2
r
d
dr
r
d
dr
a
e
ar
a
F
H
G
I
K
J
+
L
N
M
O
Q
P
+
L
N
M
O
Q
P
−
π
.
The energy of the ground state is given by
Em
e
h
4
0
22
8
ε
and the Bohr radius is given
by
ah
m
e
E e
a
==
−
2
0
8
ππ
ε
0
,.
so
The potential energy is given by
Ue
r
2
4
πε
0
,
so
88
84
8
8
12
1 12
2
2
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 Spring '08
 Any
 Physics

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