1000 Solved Problems in Modern Physics

8.3 Solutions 461

Multiply (1) byψ∗

ψ∗∇^2 ψ+

2 m

^2

(E+U+iW)ψ∗ψ=0(2)

Form the complex conjugate equation of (1) and multiply byψ

ψ∇^2 ψ∗+

2 m

^2

(E+U−iW)ψ∗ψ=0(3)

Subtract (2) from (3)

ψ∗∇^2 ψ−ψ∇^2 ψ∗=

− 4 imW

^2

ψψ∗ (4)

Now the quantum mechanical expression for the current density is

j=



2 im

(ψ∗∇^2 ψ−ψ∇^2 ψ∗)(5)

so that (4) becomes

divj=−

2



Wψ∗ψ (6)

Sinceψ∗ψis the probability density andW =^1 / 2 vKwhereKis the

absorption coefficient, Eq. (6) is equivalent to the classical continuity equation

∂ρ

∂t

+divj=−

v

λ

ρ (7)

wherevis the particle velocity and the mean free pathλ= 1 /K. When steady

state has reached the first term on the LHS of (7) vanishes. ProvidedW>0,

the imaginary part of the complex potential has the effect of absorbing flux

from the incident channel.

8.44 (i)cp=

√

2 m(E−U)

λ=

h

p

=

2 πc

√

2 mc^2 (E−U)

=

2 π× 197. 3

√

2 × 939. 6 ×(100+25)

= 2 .56 fm

(ii) W=

1

2

vK=

1

2

cβK

β=

√

2 E

mc^2

=

√

2 × 100

939. 6

= 0. 46

K=

2 W

cβ

=

2 × 10

197. 3 × 0. 46

= 0 .22 fm−^1

2 R= 2 roA

(^1) / 3
= 2 × 1 .3(120)
(^1) / 3
= 12. 82
2 KR= 2. 82
Probability that the neutron will be absorbed in passing diametrically
through the nucleus=(1−e−^2 KR)=(1−e−^2.^82 )= 0. 94