 Two-atom neutron diffraction selection

Choose a one-dimensional crystal with cell translation $a$ and two atoms per cell at $0$ and $d=a/2$. Here $d$ is an intracell displacement, distinct from the interplanar spacing $d_{\mathrm{p}}$ above. In a single-scattering approximation, a row of $N$ cells has scattering amplitudeThe sum selects reciprocal lattice positions $Q_h=2\pi h/a$ as $N$ becomes large. At these positions, the two-site crystal basis structure factor and its squared magnitude areConsequently $b_1=b_2=b$ gives even reflections of strength $4b^2$ and extinguishes the odd reflections. Choosing $b_1=b$, $b_2=b/2$ instead gives $9b^2/4$ and $b^2/4$: the odd/even ratio is $1/9$. Equal atomic numbers would not establish equal $b_j$ if the isotopes differ. The figure below plots these chosen coefficients, without assigning them to actual elements. Missing reflections can encode destructive wave interference within a cell, not the absence of matter at a site.