= Solution
The all-even indices are allowed, but $d_{640}=0.07766\,\mathrm{nm}$ is small. <copper> $K_\alpha$ gives
$$
\sin\theta_B=\frac{0.154\sqrt{52}}{2(0.56)}=0.9917,\qquad \boxed{2\theta_B=165.07^\circ.}
$$
This is near backscattering and outside an ordinary portable instrument whose arm reaches only about $130^\circ$. The numerical angular limit should be checked on the actual model; the mathematical requirement is that its maximum $2\theta$ reach $165^\circ$. It also requires a normal inclined $\tan^{-1}(4/6)=33.69^\circ$ to $[100]$, so the coupled surface-normal scan is insufficient. <atomic scattering factors> are smaller at this larger scattering vector.
\b[A suitable high-angle, independently orientable diffractometer could detect it.] A shorter selected wavelength can instead reduce the angular requirement: with $\lambda=0.071\,\mathrm{nm}$, $2\theta_B\simeq54.4^\circ$, while orientation still needs to be set appropriately. If $\lambda$ exceeded $2d=0.1553\,\mathrm{nm}$ there would be no first-order Bragg solution at all. The tube continuum and other characteristic lines must therefore be distinguished from the specifically calculated <copper> $K_\alpha$ peak.
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