= Solution
For nonrelativistic <neutrons>, $\lambda\propto E_K^{-1/2}$, so first-order <uncertainty propagation> gives $|\Delta\lambda|/\lambda=\tfrac12|\Delta E_K|/E_K=0.05$. Thus
$$
\boxed{\Delta\lambda_n\simeq3.30\times10^{-12}\,\mathrm m.}
$$
At fixed $d$, differentiating the stipulated first-order relation $d\sin\theta=\lambda$ gives $\Delta\theta=\tan\theta\,\Delta\lambda/\lambda$. Consequently
$$
\boxed{\Delta\theta_n\simeq0.00588\,\mathrm{rad}.}
$$
For <photons>, $\lambda\propto E^{-1}$, so the same fractional energy uncertainty gives $\Delta\lambda_\gamma/\lambda_\gamma=0.10$. Their $18.8\,\mathrm{keV}$ wavelength almost equals the low-energy neutron wavelength, so the conditional central angle is $0.116973\,\mathrm{rad}$ and
$$
\boxed{\Delta\theta_\gamma\simeq0.01175\,\mathrm{rad},}
$$
about twice the neutron angular spread. This difference comes from the different energy-momentum relations, not from substantially different central wavelengths. The small-angle approximation is adequate at $0.117\,\mathrm{rad}$: replacing $\sin\theta$ by $\theta$ changes it by about $\theta^2/6=0.23\%$, and replacing $\tan\theta$ by $\theta$ changes the sensitivity by about $0.46\%$. The numerical angular results retain the preceding geometry qualification; a different physical scattering geometry would require its own angle convention.
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