Problem:Simple Bragg scattering, the monochromator

Consider the Bragg law for scattering of waves by a crystal. A reciprocal lattice vector, \({\boldsymbol\tau}\), is always perpendicular to a set of lattice planes, and has the length

\begin{equation}\label{eq:taubragg} |{\boldsymbol\tau}| = n \dfrac{2 \pi}{d} , \end{equation}

where \(n\) is an integer.

Question 1

Argue why \(n\) is needed in equation \eqref{eq:taubragg}.

Solution

The \(n\) is needed in equation \eqref{eq:taubragg} since Bragg-scattering from the same set of crystal planes can occur also for neutrons of double (triple etc) wavevector. I.e. the crystal can scatter wavelengths
\(\lambda, \lambda/2, \lambda/3, \ldots, \lambda/n\) if they are present in the incoming neutron spectrum.

Question 2

Show that the Bragg law can be written as \(\tau = 2 k \sin \theta\), following the diffraction in Bragg geometry figure on the Diffraction from crystals page.

Solution

Figure 1: A sketch of the SANS instrument in terms of \(q\)-range and resolution.

On Figure xx--CrossReference--fig:Bragg--xx it is seen that \(\sin{\theta}=\dfrac{l}{d}\). The condition for constructive interference between rays of wavelength \(\lambda=2\pi/k\) scattered from the upper and lower crystalplane is \(2l=n\lambda\). Combining the two equations and rewriting \(d= n \frac{2\pi}{\tau}\) we reach at

\( n\lambda = 2l = 2d\sin{\theta} \Rightarrow \tau = 2k\sin{\theta} \)

Question 3

Show that the Bragg law can be derived from the (crystal) momentum conservation law \(\hbar {\bf k}_{\rm i} = \hbar {\bf k}_{\rm f} + \hbar {\boldsymbol\tau}\).

Solution

For elastic scattering \(|\mathbf{k}_i| =|\mathbf{k}_f| = k \), hence by trigonometry \(q/2=k\sin{\theta}\)

\( {\boldsymbol\tau} = \mathbf{k}_i - \mathbf{k}_f \Rightarrow |{\boldsymbol\tau}| = |\mathbf{k}_i - \mathbf{k}_f| = q= 2k \sin{\theta} \)

Question 4

Determine the scattering angle, \(2 \theta\), for 5 meV neutrons scattering off a pyrolytic graphite monochromator crystal with \(\tau_{(002)} = 1.8734\) Å\(^{-1}\).

Solution

It is useful to have the relation \(E\)[meV]\(=\frac{\hbar^2 k^2}{2m_n}= 2.072 k^2\) where \(k\) is in reciprocal Ångstrøm. Hence \(k=\sqrt{\frac{5.00}{2.072}}= 1.553\) Å\(^{-1}\). For \(\tau_{(002)}=1.8734\) Å\(^{-1}\) we get

\( 2\theta = 2 \sin^{-1}\left(\dfrac{\tau}{2k}\right) = 74.19^\circ \)