51. The proposed wave function is
1
È = " e-r/a
Ä„a3/2
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
dÈ 1
= - " e-r/a
dr Ä„a5/2
so
dÈ r2
r2 = -" e-r/a
dr Ä„a5/2
and

1 d dÈ 1 2 1 1 2 1
r2 = " - + e-r/a = - + È .
r2 dr dr Ä„a5/2 r a a r a
The energy of the ground state is given by E = -me4/8µ2h2, and the Bohr radius is given by a =
0
h2µ0/Ä„me2, so E = -e2/8Ä„µ0a. The potential energy is given by U = -e2/4Ä„µ0r, so

8Ä„2m 8Ä„2m e2 e2 8Ä„2m e2 1 2
[E - U] È = - + È = - + È
h2 h2 8Ä„µ0a 4Ä„µ0r h2 8Ä„µ0 a r

Ä„me2 1 2 1 1 2
= - + È = - + È .
h2µ0 a r a a r
The two terms in Schrödinger s equation cancel, and the proposed function È satisfies that equation.