1000 Solved Problems in Modern Physics

8.3 Solutions 473

Writing the Laplacian for spherical geometry (1) becomes

d^2 φ

dr^2

+

2

r

dφ

dr

−K^2 φ=0(2)

Equation (2) is easily solved, the solution being

φ=

C 1 eKr

r

+

C 2 e−Kr

r

(3)

AsKis positive, the first term on the right hand side tends to∞asr→∞.

Therefore,C 1 =0 if the flux is required to be finite everywhere including at

∞.

φ=

C 2 e−Kr

r

(4)

We can calculate the constantC 2 by considering the currentJthrough a

small sphere of radiusrwith its centre at the source.

The net current

J=−

λtr

3

∂φ

∂r

=

λtr

3 r^2

C 2 (Kr+1)e−Kr (5)

where we have used (4)

The net number of neutrons leaving the sphere per second is

4 πr^2 J=

4

3

πλtrC 2 (Kr+1)e−Kr (6)

But asr→0, the total number of neutrons leaving the sphere per second

must be equal to the source strengthQ. Thus from (6)

Q=

4

3

πλtrC 2

or C 2 =

3 Q

4 πλtr

(7)

The complete solution is

φ=

3 Qe−Kr

4 πrλtr

(8)

Therefore the neutron density

n(r)=

φ

ν

=

3 Qe−Kr

4 πλtrνr

=

Qe−r/L

4 πDr

(9)

where

λtrv

3

=D, is the diffusion coefficient andK = 1 /L, Lbeing the

diffusion length.