= Solution
For a face-centered cubic lattice, allowed reflections have indices all odd or all even. The first two are \b[$\{111\}$ and $\{200\}$], with
$$
d_{111}=\frac{a}{\sqrt3}=0.23556\,\mathrm{nm},\qquad
d_{200}=\frac a2=0.20400\,\mathrm{nm}.
$$
For first-order <Bragg diffraction>, $2d\sin\theta=\lambda$. The detector intercepts the beam at the full scattering angle $2\theta$, giving
$$
r=L\tan(2\theta)\simeq L\frac{\lambda}{d}.
$$
The inner detector radius supplies the minimum length for each ring:
$$
L_{\min,111}\simeq\frac{2.7\,\mathrm{mm}\times0.23556}{0.007053}
=90.2\,\mathrm{mm},
$$
$$
L_{\min,200}\simeq\frac{2.7\,\mathrm{mm}\times0.20400}{0.007053}
=78.1\,\mathrm{mm}.
$$
Thus \b[both rings reach the annulus when $L\gtrsim90.2\,\mathrm{mm}$]. At this minimum the 200 ring is at roughly $3.12\,\mathrm{mm}$, well inside the $10\,\mathrm{mm}$ outer radius. Both remain within the detector for approximately
$$
\boxed{90.2\,\mathrm{mm}\lesssim L\lesssim289\,\mathrm{mm}}.
$$
Equality at the inner edge gives vanishing acceptance for an ideal infinitely narrow ring, so practical operation requires a small margin. The calculation concerns ring positions; individual nanoparticle orientation and the finite probe convergence also affect observed diffraction contrast. Using the Bragg angle instead of the full scattering angle would introduce an erroneous factor of about two.
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