Past exam of the condensed matter physics course of the University of Oxford / 2015 / 2 / b / Solution

For the four conventional face-centered cubic lattice sites $000$, $0\frac{1}{2}\frac{1}{2}$, $\frac{1}{2}0\frac{1}{2}$ and $\frac{1}{2}\frac{1}{2}0$, their centering factor isThis is the face-centred cubic selection rule. Enumerating the nonzero representatives with $h^2+k^2+l^2\le 16$ givesSymmetry-related permutations and signs belong to the same powder ring, not additional rings. For Al, the basis contains one atom, $S=F_{\mathrm{fcc}}{f}_{\mathrm{Al}}$. ThusFor the stated KCl basis,The all-even reflections add the two atomic factors; all-odd reflections subtract them. Ionic $K^{+}$ and $C{l}^{-}$ each have 18 electrons, giving similar X-ray amplitudes. Their difference is much smaller than their sum, so the 111 and 311 rings are weak. Consequently the four strong rings areThis is an accidental diffraction extinction from equal atomic scattering factors in the equal-factor limit, not an extra exact fcc selection rule. Isoelectronic ions can have different radial electron distributions, so real odd reflections need not vanish exactly. Even the cruder neutral-atom estimate $f_K\simeq 19$, $f_{\mathrm{Cl}}\simeq 17$ makes the odd/even basis-intensity ratio only $(2/36{)}^2\simeq 0.0031$, before other ring-intensity factors.

For Si's displaced identical basis atoms,For all-odd indices, the bracket is $1\pm i$, with squared magnitude 2. For all-even indices it vanishes when $h+k+l=2(\mathrm{mod}4)$, and has squared magnitude 4 when the sum is a multiple of four. Thus 200 and 222 vanish, whereas 111, 220, 311 and 400 survive:These are the basis extinctions of the ideal diamond cubic structure. Intensity strengths within each surviving set can still differ because of atomic factors, multiplicity and geometry.