196
3 Quantum Mechanics – II
or
k R = nπ → k =
nπ
R
(4)
Complete unnormalized solution is
u(r ) = A sin
nπr
R
(5)
The normalization constant A is obtained from
R
0
|ψ(r )|
2
· 4πr
2 dr = 1
( 6 )
Using (1) and (5), we get
A =
1
√
2π R
(7)
The normalized solution is then
u(r ) = (2π R)
−
1
2 sin
nπr
R
(8)
From (2) and (4)
E n =
π
2 n
2
2
2m R 2
(9)
For ground state n = 1. Hence
E 1 =
π
2
2
2m R 2
(10)
The force exerted by the particle on the walls is
F = −
∂ V
∂ R
= −
∂ H
∂ R
= −
∂ E 1
∂ R
=
π
2
2
m R 3
The pressure exerted on the walls is
P =
F
4π R 2 =
π
2
4m R 5
3.48 The quantity
π
2
2
8m
=
π
2
2 c
2
8mc 2 =
π
2 (197.3)
2
8×2,200 m c c 2
=
π
2 (197.3)
2
8 × 2, 200 × 0.511
= 42.719 MeV − fm
2
Now V 0 a
2
= 70 × (1.42)
2
= 141.148 MeV − fm
2
It is seen that
π
2
2
8m
< V 0 a
2
<
4π
2
2
8m
(42.7 < 141 < 169)
From the results of Problem (3.25) there will be two energy levels, one belonging to class I function and the other to class II function.
The particle of mass 2,200 m e or 1,124 Mev/c
2 is probably Λ-hyperon
(mass 1,116 MeV/c
2 ) which is sometimes trapped in a nucleus, to form a
hypernucleus before it decays(Chap. 10).