1000 Solved Problems in Modern Physics

178 3 Quantum Mechanics – II

∫R

0

|ψ 1 |^2 dτ+

∫∞

R

|ψ 2 |^2 dτ= 1

∫R

0

u^21. 4 πr^2 dr/r^2 +

∫∞

R

u^224 πr^2 dr/r^2 = 1

A^2

∫ R

0

sin^2 krdr+C^2

∫∞

R

e−^2 γrdr= 1 / 4 π

Integrating and using (2), we find

A^2 =

γ

2 π(γR+1)

(4)

Using (4) in (3)

P=

1

γR+ 1

(5)

NowγR=

(

MW

^2

) 1 / 2

R=

(

Mc^2 W

^2 c^2

) 1 / 2

R

=

[

940 × 2. 2

(197.3)^2

] 1 / 2

× 2. 1 = 0. 48

where we have insertedMc^2 = 940 MeV/c^2 ,

W= 2. 2 MeVandR= 2. 1 fm

Thereforep= 0. 481 + 1 = 0. 67

Thus neutron and proton stay outside the range of nuclear forces approxi-

mately 70% of time.

3.27 By Problem 3.25, for the finite well, for class I
αtanαa=β

withα=

(2mE)

(^12)

;β=
[2m(V 0 −E)]
(^12)

AsV 0 →∞,β→∞andαa=nπ/2(nodd)
Therefore,α^2 a^2 =^2 mEa
2
^2 =
n^2 π^2
4
OrE=n^2 π^2 ^2 / 8 ma^2 (nodd)
For class II
αcotαa=−β
AsV 0 →∞,β→∞andαa=nπ/2(neven)