Past exam of the general relativity and cosmology and condensed matter physics paper of Part B of the physics course of the University of Oxford / 2011 / Section VI / 5 / Unlettered 3 / Solution

Use the scattering vector $\mathbf{q}={\mathbf{k}}_{\mathrm{in}}-{\mathbf{k}}_{\mathrm{out}}$; reversing its sign conjugates the following amplitudes without changing the intensities. In the kinematic approximation, the atomic scattering factor of atom $s$ isAtoms at $\mathbf{R}+{\mathbf{d}}_s$ therefore give the scattering amplitudeHere $A_e$ is the single-electron amplitude; $F$ is the unit cell structure factor, and $S_L$ is the geometrical lattice sum. If the crystal has $N_i$ repeats along ${\mathbf{a}}_i$, the latter factor is a product of geometric series, whose magnitudes containThe sharp constructive peaks occur when every $\mathbf{q}\cdot {\mathbf{a}}_i=2\pi h_i$. Thus a permitted reflection has wavevector change equal to a reciprocal lattice vector, with $F(\mathbf{G})\ne 0$ and the additional elastic condition $|{\mathbf{k}}_{\mathrm{out}}|=|{\mathbf{k}}_{\mathrm{in}}|$. A finite crystal broadens these peaks; it does not impose an exact delta-function selection on every scattered photon.