9 Neutron attenuation
If a flux of neutrons ([latex]\phi_0[/latex]) is impinged on a target, some neutrons will interact with the medium, but many will pass through with a resultant flux ([latex]\phi[/latex]) on the other side.
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The probability that the neutrons interact with the medium is the cross section ([latex]\sigma[/latex]).
If the macroscopic cross section is defined as –
[latex]\Sigma = \frac{\rho N_A}{A}[/latex], the units of the quantity are in [latex]\frac{1}{cm}[/latex].
Then, the mean free path can be defined as –
[latex]\lambda \equiv \frac{1}{\Sigma}[/latex], which are in units of [latex]cm[/latex].
The mean free path then can be considered as the expected value of the distance a neutron travels between interactions.
To solve for flux, it should be intuitive that the resultant flux will be less than the initial flux –
[latex]-d\phi = \Sigma_T \phi dx[/latex],
with the solution –
[latex]\phi(x) = \phi_0 e^{-\Sigma_T x}[/latex] if [latex]\phi(0) = \phi_0[/latex].
Additional notes
