= Two-atom neutron diffraction selection
{title2=$I\propto|b_1+b_2e^{i\mathbf Q\cdot\mathbf d}|^2$}
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 amplitude>
$$
A_N(Q)=\left(b_1+b_2e^{iQd}\right)\sum_{r=0}^{N-1}e^{iQra}.
$$
The 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 are
$$
F_h=b_1+(-1)^h b_2,\qquad
I_h\propto|F_h|^2=
\begin{cases}
(b_1+b_2)^2,&h\text{ even},\\
(b_1-b_2)^2,&h\text{ odd}.
\end{cases}
$$
Consequently $b_1=b_2=b$ 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.
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