Calculate the typical wavelength of neutrons after reaching thermal equilibrium with their surroundings at $373 \mathrm{~K}$. For simplicity, assume that the particles are travelling in one dimension.
From the equipartition principle, we know that the mean translational kinetic energy of a neutron at a temperature $T$ travelling in the $x$-direction is $E_{\mathrm{k}}=\frac{1}{2} k T$. The kinetic energy is also equal to $p^2 / 2 m$, where $p$ is the momentum of the neutron and $m$ is its mass. Hence, $p=(m k T)^{1 / 2}$. It follows from the de Broglie relation $\lambda=h / p$ that the neutron's wavelength is
$$
\lambda=\frac{h}{(m k T)^{1 / 2}}
$$
Therefore, at $373 \mathrm{~K}$,
$$
\begin{aligned}
\lambda & =\frac{6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{\left\{\left(1.675 \times 10^{-27} \mathrm{~kg}\right) \times\left(1.381 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}\right) \times(373 \mathrm{~K})\right\}^{1 / 2}} \\
& =\frac{6.626 \times 10^{-34}}{\left(1.675 \times 10^{-27} \times 1.381 \times 10^{-23} \times 373\right)^{1 / 2}} \frac{\mathrm{kg} \mathrm{m}^2 \mathrm{~s}^{-1}}{\left(\mathrm{~kg}^2 \mathrm{~m}^2 \mathrm{~s}^{-2}\right)^{1 / 2}} \\
& =2.26 \times 10^{-10} \mathrm{~m}=226 \mathrm{pm}
\end{aligned}
$$
where we have used $1 \mathrm{~J}=1 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2}$.