214 3 Quantum Mechanics – II
=−
∂
∂b(
k
b^2 +k^2)
=
2 kb
(b^2 +k^2 )^2
Therefore the integral in (4) is evaluated as
(
2 k
a 0)/[(
1
a 02)
+k^2]
(5)
Using the result (5) in (4), puttingk=p/, and rearranging, we getψ(p)=(
2
√
2
π)(
a 0)^52 /[
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 πp^2
/[
p^2 +(
a 0) 2 ]^4
(1)
Maximize (1)
d
dp|ψ(p)|^2 = 0This givesPmost probable=/√
3 a 0(b)<p>=∫∞
0ψ∗ppψp. 4 πp^2 dp=
(
32
π)(
a 0) 5 ∫∞
0p^3 dp/[
p^2 +(
a 0) 2 ]^4
The integralI 1 is easily evaluated by the change of variable
p=(
a 0)
tanθ. ThenI 1 =
(
1
8
)(
a 0
) 4 ∫π/ 20sin^32 θdθ=(
1
12
)(
a 0
) 4
Thus<p>= 3 π^8 a 03.76 By Problem 3.71, the probability that
P(
r
a 0)
= 1 −exp(
−
2 r
a 0)(
1 +
2 r
a 0+
2 r^2
a 02)
Putp(
r
a 0)
= 0 .5 and solve the above equation numerically (see Chap. 1). We
getr= 1. 337 a 0 , with an error of 2 parts in 10^5.