196 3 Quantum Mechanics – II

or

kR=nπ→k=

nπ

R

(4)

Complete unnormalized solution is

u(r)=Asin

(nπr

R

)

(5)

The normalization constantAis obtained from

∫ R

0

|ψ(r)|^2 · 4 πr^2 dr=1(6)

Using (1) and (5), we get

A=

1

√

2 πR

(7)

The normalized solution is then

u(r)=(2πR)−

(^12)
sin
(nπr
R

)

(8)

From (2) and (4)

En=

π^2 n^2 ^2

2 mR^2

(9)

For ground staten=1. Hence

E 1 =

π^2 ^2

2 mR^2

(10)

The force exerted by the particle on the walls is

F=−

∂V

∂R

=−

∂H

∂R

=−

∂E 1

∂R

=

π^2 ^2

mR^3

The pressure exerted on the walls is

P=

F

4 πR^2

=

π^2

4 mR^5

3.48 The quantityπ

(^2)  2
8 m =
π^2 ^2 c^2
8 mc^2 =
π^2 (197.3)^2
8 × 2 , 200 mcc^2

π^2 (197.3)^2
8 × 2 , 200 × 0. 511
= 42 .719 MeV−fm^2
NowV 0 a^2 = 70 ×(1.42)^2 = 141 .148 MeV−fm^2
It is seen that
π^2 ^2
8 m
<V 0 a^2 <
4 π^2 ^2
8 m
(42. 7 < 141 <169)
From the results of Problem (3.25) there will be two energy levels, one belong-
ing to class I function and the other to class II function.
The particle of mass 2, 200 meor 1,124 Mev/c^2 is probablyΛ-hyperon
(mass 1,116 MeV/c^2 ) which is sometimes trapped in a nucleus, to form a
hypernucleus before it decays(Chap. 10).