 Solution

For a photon, $p_{\gamma }=E/c$ and ${\lambda }_{\gamma }=hc/E$. The neutron is nonrelativistic here because $18.8\mathrm{keV}\ll m_n{c}^2$, so $p_n=\sqrt{2m_{n}E}$ and ${\lambda }_n=h/p_n$. Using the supplied common energy givesEqual kinetic energies do not imply equal momenta: the massive particle obeys $E=p^2/(2m_n)$, while the massless photon obeys $E=pc$. The neutron momentum is about $316$ times larger, so its de Broglie wavelength is about $316$ times smaller.

To give the neutron the photon wavelength, set its momentum to $E/c$, not its kinetic energy to $E$ again. ThenIts kinetic energy would then be about $0.188\mathrm{eV}$, connecting this comparison with the next calculation.