 Solution

For nonrelativistic neutrons, $\lambda \propto E_K^{-1/2}$, so first-order uncertainty propagation gives $|\Delta \lambda |/\lambda =\frac{1}{2}|\Delta E_K|/E_K=0.05$. ThusAt fixed $d$, differentiating the stipulated first-order relation $d\sin \theta =\lambda$ gives $\Delta \theta =\tan \theta \Delta \lambda /\lambda$. ConsequentlyFor 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 $0.116973\mathrm{rad}$ andabout 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.