The de-Broglie wavelength associated with thermal neutrons at room temperature $\left(27{}^{\circ }C\right)$ is

Kinetic energy of most of the neutrons i.e. the most probable energy of a neutron at temperature $T$ is

$\frac{1}{2}m{v}^2=k_{B}T$

$\Rightarrow \frac{p^2}{2m}=k_{B}T$

$\Rightarrow p=\sqrt{2m{k}_{B}T}$

de-Broglie wavelength possessed by most of the neutrons is given by

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2m{k}_{B}T}}$

Given $m_n=1.675\times {10}^{-27} \mathrm{kg}$ , $k_B=1.38\times {10}^{-23} {\mathrm{Jmol}}^{-1}{\mathrm{K}}^{-1}$, $T=27+273=300 \mathrm{K}, h=6.63\times {10}^{-34} \mathrm{Js}$

$\Rightarrow \lambda =\frac{6.63\times {10}^{-34}}{\sqrt{2\times 1.67\times {10}^{-27}\times 1.38\times {10}^{-23}\times 300}}$$=1.78\times {10}^{-10} \mathrm{m}$$\approx 1.8 \stackrel{\circ }{\mathrm{A}}$