= Solution
{c}
In cubic coordinates write $\mathbf G=(2\pi/a)(h,k,l)$. The <face-centred cubic selection rule> allows integer triples only when $h,k,l$ are all even or all odd. One way to derive it is to sum the four conventional-cell lattice positions:
$$
F_{\mathrm{fcc}}=1+(-1)^{h+k}+(-1)^{h+l}+(-1)^{k+l}.
$$
This vanishes for mixed parity and equals four otherwise. The two-atom diamond <atomic basis> adds the factor
$$
F_{\mathrm{basis}}=b(\mathbf G)\left[1+e^{i\pi(h+k+l)/2}\right].
$$
For all-odd indices this is nonzero. For all-even indices it vanishes when $h+k+l=2\pmod4$ and survives when the sum is divisible by four: this is <diamond-basis diffraction extinction>.
Exclude the forward beam $000$ and order the remaining families by $h^2+k^2+l^2$. The first is $111$, with squared length three. The candidate $200$, with squared length four, is extinguished by the diamond <atomic basis>. The next allowed family is $220$, with squared length eight. Thus
$$
\boxed{|\mathbf G_1|=\frac{2\pi\sqrt3}{a}\quad(111),\qquad|\mathbf G_2|=\frac{2\pi\sqrt8}{a}=\frac{4\pi\sqrt2}{a}\quad(220).}
$$
For fixed incident <wavelength>, $|\mathbf G|=2k\sin\theta$ increases with scattering angle, so these are the two lowest-angle powder rings. The two atoms are counted per primitive cell; a conventional silicon cube contains four lattice points and eight atoms.
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