Problem: Attenuation of the neutron beam

Consider a neutron beam penetrating a sample. For simplicity we can assume the sample to be shaped like a slab with the large area perpendicular to the neutron beam.

Question 1

Derive the exponential decay of the beam inside the sample, this equation from the Basics of neutron scattering page, \(\Psi(z) = \Psi(0) \exp(-\mu z)\).

Solution

Consider a thin slab of material with thickness \(dz\) and area \(A(z)\). The number of neutrons scattered by this piece of material is \(\Psi(z) \Sigma A(z) dz\). The number of neutrons after passage of the slab will have decreased by this amount, i.e.

\( dN = - \Psi(z) \Sigma A(z) dz \, . \)

Flux is neutrons pr. second hitting an area divided by the area whence

\( d\Psi = -\Psi \Sigma dz \quad\Rightarrow\quad \dfrac{d\Psi}{dz} = -\Psi \Sigma , \)

and the solution to this first order differential equation is

\( \Psi(z) = \Psi_0 e^{-\Sigma z} . \,\)

The constant \(\Sigma = \mu\) is also called the attenuation coefficient.

Question 2

Prove that when there is both absorption and scattering, the total attenuation coefficient is the sum of the individual coefficients.

Solution

Similarly the decrease of neutrons in the beam after \(dz\) due to absorption is

\( d N_a = -\Psi(z)\mu_a A(z)dz .\, \)

The total decrease is \(d N_\text{tot} = dN + d N_a\) giving decrease in flux

\( d \Psi_{\rm tot} = -\Psi_{\rm tot}(z)(\mu+\mu_a) dz \,\)

with solution

\( \Psi_{\rm tot}(z) = \Psi_0 e^{-(\mu+\mu_a) z} . \)