210
3 Quantum Mechanics – II
∇
2
=
d
2
dr 2 +
2
r
d
dr
(2)
Inserting (2) in (1) and performing the integration we get
< T >=
2
2ma
2
0
But a 0 =
2
me 2
Or
2
m
= a 0 e
2
Therefore, < T >=
e
2
2a0
Also < E >=< T > + < V >=
e
2
2a0
−
e
2
a0
= −e
2
/2a 0
3.65 The normalized eigen function for the ground state of hydrogen atom is
ψ 0 = 1/
πa
3
0
1
2 e
−r/a 0
where a 0 is the Bohr’s radius
(a) The probability that the electron will be formed in the volume element dτ
is
p(r )dr = |ψ 0 |
2 dτ =
e
−
2r
a 0
πa
3
0
.4πr
2 dr
=
4
a
3
0
r
2 e
−2r/a 0 dr
Maximum probability is found by setting
d p
dr
= 0
d
dr
4r
2 e
−
2r
a 0
a
3
0
=
8r
a
3
0
e
−
2r
a 0
1 −
r
a 0
= 0
Therefore r = a 0
(b) < r >=
∞
0
ψ
∗ r ψdτ =
1
π
a
3
0
∞
0
r exp
−
2r
a 0
.4πr
2 dr
=
4
a
3
0
∞
0
r
3 exp
−
2r
a 0
dr
Let
2r
a 0
= x; dr =
a 0 dx
2
< r >=
a 0
4
∞
0
x
3 e
−x dx =
a 0
4
3! =
3a 0
2
3.66
u
2
210 dτ = A 2
2
e
−2x x
2 cos
2
θr
2 sin θ dθ dr dϕ
=
1
π
1
2a 0
3 ∞
0
exp
−
r
a 0
r
2
4a
2
0
r
2 dr
+1
−1
cos
2 θ d(cos θ)
2π
0
dϕ
where we have put x = r/2a 0 . Put r/a 0 = y, dr = a 0 dy