 Solution

The all-even indices are allowed, but $d_{640}=0.07766\mathrm{nm}$ is small. copper $K_{\alpha }$ givesThis 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.

A suitable high-angle, independently orientable diffractometer could detect it. A shorter selected wavelength can instead reduce the angular requirement: with $\lambda =0.071\mathrm{nm}$, $2{\theta }_B\simeq 54.4^{\circ }$, 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.