Past exam of the condensed matter physics course of the University of Oxford / 2019 / 1 / e / Solution

In cubic coordinates write $\mathbf{G}=(2\pi /a)(h,k,l)$. The face-centred cubic selection rule allows integer triples only when $h,k,l$ are all even or all odd. One way to derive it is to sum the four conventional-cell lattice positions:This vanishes for mixed parity and equals four otherwise. The two-atom diamond atomic basis adds the factorFor all-odd indices this is nonzero. For all-even indices it vanishes when $h+k+l=2(\mathrm{mod}4)$ and survives when the sum is divisible by four: this is diamond-basis diffraction extinction.

Exclude the forward beam $000$ and order the remaining families by $h^2+k^2+l^2$. The first is $111$, with squared length three. The candidate $200$, with squared length four, is extinguished by the diamond atomic basis. The next allowed family is $220$, with squared length eight. ThusFor fixed incident wavelength, $|\mathbf{G}|=2k\sin \theta$ increases with scattering angle, so these are the two lowest-angle powder rings. The two atoms are counted per primitive cell; a conventional silicon cube contains four lattice points and eight atoms.