1000 Solved Problems in Modern Physics

480 8 Nuclear Physics – II

At the surface the neutron density corresponding tor= 9 .6 cm and mean

neutron velocity 2. 2 × 105 cm s−^1

n(r)=

φ

v

=

3 Q

4 πλtrv

e−^9.^6 /L

9. 6

= 2. 93 × 10 −^9 Qcm−^3

8.88 Consider the diffusion equation

∂n

∂t

=S+

λtr

3

∇^2 φ−φΣa (1)

where nis the neutron density,S is the rate of production of neutrons/

cm^3 /s,φΣais the absorption rate/ cm^3 /s and

λtr∇^2 φ

3

represents the leakage

of neutrons.Σais the macroscopic cross-section,φis the neutron flux andλtr

is the transport mean free path.

Since it is a steady state,

∂n

∂t

=0. FurtherS=0 because neutrons are not

produced in the region of interest. As we are interested only in thex-direction

the Laplacian reduces to d

2

dx^2. Thus (1) becomes

λtr

3

d^2 φ

dx^2

−φΣa=0(2)

or

d^2 φ

dx^2

−K^2 φ=0(3)

where

K^2 =

3 Σa

λtr

=

3

λaλtr

(4)

λabeing the absorption mean free path.The solution of (3) is

φ=C 1 eKx+C 2 e−Kx (5)

whereC 1 andC 2 are constants of integration. The condition that the flux

should be finite at any point including at infinity means thatC 1 =0. Therefore,

(5) becomes

φ=C 2 e−Kx (6)

We can now determineC 2. Consider a unit area located in the YZ plane

at a distance x from the plane source as in Fig. 8.14. On an average half

of the neutrons will be travelling along the positive x-direction. Asx →0,

the net current flowing in the positive x-direction would be equal to^12 Q; the

diffusion of neutrons through unit area atx =0 would have a cancelling

effect because from symmetry equal number of neutrons would diffuse in the

opposite direction.