214
3 Quantum Mechanics – II
= −
∂
∂b
k
b 2 + k 2
=
2kb
(b 2 + k 2 ) 2
Therefore the integral in (4) is evaluated as
2k
a 0
1
a
2
0
+ k
2
(5)
Using the result (5) in (4), putting k = p/, and rearranging, we get
ψ( p) =
2
√
2
π
a 0
5
2
p
2
+
a 0
2
2
or
|ψ( p)|
2
=
8
π 2
(/a 0 )
5
[ p 2 + (/a 0 )
2 ] 4
(6)
3.75 (a) |ψ( p)|
2
=
8
π 2
a 0
5
.4 π p
2
p
2
+
a 0
2
4
(1)
Maximize (1)
d
d p
|ψ( p)|
2
= 0
This gives P most probable = /
√
3 a 0
(b) < p >=
∞
0
ψ
∗
p pψ p .4π p
2 d p
=
32
π
a 0
5 ∞
0
p
3 d p
p
2
+
a 0
2
4
The integral I 1 is easily evaluated by the change of variable
p =
a 0
tan θ. Then
I 1 =
1
8
a 0
4 π/2
0
sin
3 2θdθ =
1
12
a 0
4
Thus < p >=
8
3πa 0
3.76 By Problem 3.71, the probability that
P
r
a 0
= 1 − exp
−
2r
a 0
1 +
2r
a 0
+
2r
2
a
2
0
Put p
r
a 0
= 0.5 and solve the above equation numerically (see Chap. 1). We
get r = 1.337a 0 , with an error of 2 parts in 10
5 .