Differential Scattering Cross Section
Conversion of the differential scattering cross section to an actual
cross section is essential, and can be found in (Lamarsh, 2002) or
really any other neutronics textbook.
We start with the differential elastic scattering cross section. For
\(\ce{^{4}He}\), it’s
\[\sigma_{s} \left(\theta_{c}\right)=\frac{\sigma_{s}}{4\pi}\left( 1 + \cos \theta_{c}\right)\]
Note that the second \(\sigma_{s}\) is a constant. As always, the
neutronics community has some strange notation conventions.
From here, we need to determine the full cross section
\(\sigma_{s}\). We start by integrating over the solid angle.
\[\sigma_{s} =\int_{4\pi}\sigma_{s}\left(\theta_{c}\right)d\Omega\left(\theta_{c}\right)\]
\[\sigma_{s}=\int_{0}^{2\pi}\int_{0}^{\pi}\frac{\sigma_{s}}{4\pi}\left(1+\cos\theta_{c}\right)\sin\theta_{c}d\theta_{c}d\varphi\]
Now, we use \(\mu\) substitution, and from there it is just
integration and algbra.
\[\sigma_{s}=2\pi\int_{-1}^{1}\frac{\sigma_{s}}{4\pi}\left(1+\mu_{c}\right)d\mu_{c}\]
\[\sigma_{s}=2\pi\left[\frac{2\sigma_{s}}{4\pi}\right]=\sigma_{s}\]
So this exercise was a little dumb. Of course
\(\sigma_{s}=\sigma_{s}\), but we can use this to determine more
difficult things, like what fraction of elastically scattered neutrons
appear at angles greater than \(90^{\circ}\) in the center of mass
frame.
To do this, we start with the same scattering cross section, but we only
integrate through angles \(90^{\circ}\) to \(180^{\circ}\). So
we have
\[\sigma_{>90^{\circ}}=\int_{0}^{2\pi}\int_{\frac{\pi}{2}}^{\pi}\frac{\sigma_{s}}{4\pi}\left(1+\cos\theta_{c}\right)\sin\theta_{c}d\theta_{c}d\varphi\]
Then we do our typical \(\mu\) substitution. Note that the
\(\mu\) substitution goes from \(-1\) to \(0\) because those
are the \(\cos\) values at \(\nicefrac{\pi}{2}\) and
\(\pi\), reversed. The reversal comes because
\(d\mu = - \sin \theta d\theta\), so the reversal removes the
negative sign. Then, we can do the algebra and find
\[\sigma_{>90^{\circ}}=2\pi\int_{-1}^{0}\frac{\sigma_{s}}{4\pi}\left(1+\mu_{c}\right)d\mu_{c}\]
\[\sigma_{>90^{\circ}}=\frac{\sigma_{s}}{4}\]
Which shows that \(\nicefrac{1}{4}\) of the neutrons are scattered
through those high angles (backscattered).
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