Problem:The Be filter

Beryllium crystallizes in the hcp structure with a lattice constant \(a=2.27\) Å. A thick block of Be powder is often used as a neutron attenuator, since it Bragg scatters neutrons away from the beam.

Question 1

Why is the Be powder transparent for long wavelength neutrons?

Solution

There is a maximal distance between the crystal planes in the unitcell of hcp Be powder \(d_{\rm max}\), corresponding to the minimum reflection \(\boldsymbol{\tau}=(1 0 0)\). That this is the reflection corresponding to the minimal scattering vector is seen from the exercise Bragg scattering from Bravais lattices. Since in the Bragg condition \begin{equation} \frac{n\lambda}{2d} = \sin{\theta} \end{equation} and \(|\sin{\theta}|\leq 1\) and there is a higher limit of \(d\) (\(d_{\rm max}\)), there will also be a higher limit of \(\lambda\) (\(\lambda_{\rm max}\)), above which the Be powder cannot scatter and hence is transparent for those neutrons. The neutrons with \(\lambda \leq \lambda_{\rm max}\) are scattered in all directions, away from the direct beam. Hence these wavelengths are 'filtered' from the neutron beam.

Question 2

Calculate the critical wavelength for neutron transmission. What is the corresponding energy?

Hint

Use the results of the Bragg scattering from Bravais lattices problem.

Solution

\begin{equation} n\lambda = 2d\sin{\theta} \Rightarrow 2\theta = 2 \sin^{-1}\left(\frac{n\lambda}{2d}\right) \end{equation} The largest distance between crystal planes is between the planes which are decribed by the vector \(\mathbf{a}\) in hexagonal units \begin{equation} d_{100}=\sqrt{a^2 - (\frac{a}{2})^2} = \frac{\sqrt{3}a}{2}= 1.97 \, \rm{Å} \end{equation} where \(a=2.27\,\rm{Å}\) is the nearest neighbor distance. From the first above we must have \(\left(\frac{n\lambda}{2d}\right)\leq 1\) so \(n\lambda \geq 2\cdot 1.97 =3.93\,\rm{Å}\).\\ I.e if \(\lambda\geq \lambda_c=3.93\,\rm{Å}\) the Bragg condition can never be fulfilled and Be is transparent for \(\lambda > 3.93\,\rm{Å}\). \(E_c = \frac{81.81}{\lambda^2}=5.29\,\)meV hence Be is transparent for \(E<5.29\,\)meV

Question 3

How can you utilize this to eliminate higher order neutrons scattered from a monochromator?

Solution

A monochromator will both scatter neutrons with wavelength \(\lambda_1=2d\sin{\theta}\) and \(\lambda_2=\frac{\lambda_1}{2}\), \(\lambda_3=\frac{\lambda_1}{3}\) etc. i.e. also neutrons with energy \(E_2=\frac{81.81}{(\frac{\lambda_1}{2})^2}=4E_1\) etc. will pass. However if \(\lambda_2\leq \lambda_c <\lambda_1 \) equivalent to \(E_2 \geq E_c > E_1 \) the higher order scattered neutrons are cut off.