 Solution

For cubic lattice planes, $d_{hkl}=a/\sqrt{h^2+k^2+l^2}$. The face-centred cubic selection rule permits indices that are all odd or all even. ThusThe first two spacings exceed the quoted 0.19 nm real-space resolution, whereas $d_{220}$ does not. This restriction is for the resolved image and its transform; a diffraction instrument need not have the same cutoff.

A natural candidate is a beam near a local $⟨110⟩$ zone axis. For example, the $[110]$ zone law $h+k=0$ includes four $\{111\}$ reflections and two $\{200\}$ reflections, all within that real-space cutoff. Their reciprocal-space length ratio is $g_{200}/g_{111}=2/\sqrt{3}=1.1547$. The angle between $(1\bar{1}1)$ and $(1\bar{1}\bar{1})$ reciprocal vectors is ${\cos }^{-1}(1/3)=70.53^{\circ }$. In contrast, a $[001]$ view gives a fourfold arrangement of the resolvable $\{200\}$ family, while the lowest ordinary $[111]$ zone reflections are $\{220\}$ and would not be resolved at the stated cutoff.

In a five-domain decahedral interpretation, neighboring tetrahedral fcc regions meet on $\{111\}$ twin planes around a common near-$⟨110⟩$ axis. Five unstrained wedges supply $5\times 70.53^{\circ }=352.64^{\circ }$; elastic distortion accommodates the remaining $7.36^{\circ }$. Their rotated diffraction contributions need not resemble a single six-spot $[110]$ pattern. Near-$⟨110⟩$ fcc domains in a multiply twinned particle are a defensible orientation proposal, not a unique full indexing. The printed transform has no reciprocal-space scale or indexed spot list, so it cannot establish every domain rotation or exclude an overlapping-particle contribution. Calibrated spacings and transforms from individual regions would settle those distinctions. The five-domain geometry is independently observed in atomic-resolution gold-particle measurements.^[ref]