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51. The proposed wave function is
ψ
=
1
√
πa
3
/
2
e
−
r/a
where
a
is the Bohr radius. Substituting this into the right side of Schr¨
odinger’s equation, our goal is to
show that the result is zero. The derivative is
dψ
dr
=
−
1
√
πa
5
/
2
e
−
r/a
so
r
2
dψ
dr
=
−
r
2
√
πa
5
/
2
e
−
r/a
and
1
r
2
d
dr
µ
r
2
dψ
dr
¶
=
1
√
πa
5
/
2
·
−
2
r
+
1
a
¸
e
−
r/a
=
1
a
·
−
2
r
+
1
a
¸
ψ.
The energy of the ground state is given by
E
=
−
me
4
/
8
ε
2
0
h
2
, and the Bohr radius is given by
a
=
h
2
ε
0
/πme
2
,so
E
=
−
e
2
/
8
πε
0
a
. The potential energy is given by
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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