The de-Broglie wavelength of neutrons in thermal equilibrium at temperature T is

de Broglie wavelength of neutronsin thermalequilibrium at temperature T is

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

where $m$ is the mass of the neutron 

$k_B$ is the Boltzmann constant 

h is the Planck's constant

$$\begin{gathered} \text{ Here, }m=1.67\times {10}^{-27} \mathrm{kg} \\ {\mathrm{k}}_{\mathrm{B}}=1.38\times {10}^{-23} \mathrm{J}{ \mathrm{K}}^{-1} \\ \mathrm{h}=6.63\times {10}^{-34} \mathrm{J} \mathrm{s} \\ \therefore \mathrm{\lambda }=\frac{6.63\times {10}^{-34}}{\sqrt{2\times 1.67\times {10}^{-27}\times 1.38\times {10}^{-23}\times \mathrm{T}}} \\ =\frac{3.08\times {10}^{-34}\times {10}^{25}}{\sqrt{\mathrm{T}}} \mathrm{m} \\ =\frac{30.8\times {10}^{-10}}{\sqrt{\mathrm{T}}}=\frac{30.8}{\sqrt{\mathrm{T}}}\stackrel{\mathrm{o}}{\mathrm{A}} \end{gathered}$$