bei48482_FM

Example 12.8

A neutron passing through a body of matter and not absorbed in a nuclear reaction undergoes

frequent elastic collisions in which some of its kinetic energy is given up to nuclei in its path.

Very soon the neutron reaches thermal equilibrium, which means that it is equally likely to gain

or to lose energy in further collisions. At room temperature such a thermal neutronhas an

average energy of ^32 kT0.04 eV and a most probable energy of kT0.025 eV; the latter figure

is usually quoted as the energy of such neutrons.

The cross section of^113 Cd for capturing thermal neutrons is 2  104 b, the mean atomic

mass of natural cadmium is 112 u, and its density is 8.64 g /cm^3 8.64  103 kg/m^3. (a) What

fraction of an incident beam of thermal neutrons is absorbed by a cadmium sheet 0.1 mm thick?

(b) What thickness of cadmium is needed to absorb 99 percent of an incident beam of thermal

neutrons?

Solution

(a) Since^113 Cd constitutes 12 percent of natural cadmium, the number of^113 Cd atoms per cubic

meter is

n(0.12)

5.58 1027 atomsm^3

The capture cross section is  2  104 b  2  10 ^24 m^2 , so

n(5.58 1027 m^3 )(2 10 ^24 m^2 )1.12 104 m^1

From Eq. (12.20), NN 0 enx, so the fraction of incident neutrons that is absorbed is

  1 enx

Since x0.1 mm  10 ^4 m here,

 1 e(1.12^10

(^4) m (^1) )(10 (^4) m)
0.67
Two-thirds of the incident neutrons are absorbed.
(b) Since we are given that 1 percent of the incident neutrons pass through the cadmium sheet,
N0.01N 0 and
0.01enx
ln 0.01nx
x4.1 10 ^4 m0.41 mm
Cadmium is evidently a very efficient absorber of thermal neutrons.
The mean free path of a particle in a material is the average distance it can travel
in the material before interacting there. Since en^ xdxis the probability that a particle
interact in the interval dxat the distance x, we have, by the same reasoning as that
ln 0.01

1.12 104 m^1
ln 0.01

n
N

N 0
N 0 N

N 0
N 0 N 0 enx

N 0
N 0 N

N 0
8.64 103 kg/m^3

(112 u/atom)(1.66 10 ^27 kgu)
444 Chapter Twelve
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