= Solution
For a <photon>, $p_\gamma=E/c$ and $\lambda_\gamma=hc/E$. The <neutron> is nonrelativistic here because $18.8\,\mathrm{keV}\ll m_nc^2$, so $p_n=\sqrt{2m_nE}$ and $\lambda_n=h/p_n$. Using the supplied common energy gives
$$
\boxed{p_\gamma=1.005\times10^{-23}\,\mathrm{kg\,m\,s^{-1}},\quad
\lambda_\gamma=6.595\times10^{-11}\,\mathrm m,}
$$
$$
\boxed{p_n=3.176\times10^{-21}\,\mathrm{kg\,m\,s^{-1}},\quad
\lambda_n=2.086\times10^{-13}\,\mathrm m.}
$$
Equal <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. Then
$$
\boxed{v_n\simeq\frac{E}{m_nc}=5.999\times10^3\,\mathrm{m\,s^{-1}}.}
$$
Its <kinetic energy> would then be about $0.188\,\mathrm{eV}$, connecting this comparison with the next calculation.
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