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de Broglie wavelength of thermal neutrons
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The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T (Kelvin) and mass m, is
Thermal KE = $\frac{3}{2}kT$, so $\lambda = h/\sqrt{3mkT}$.
de-Broglie wavelength $\lambda = \frac{h}{mv} = \frac{h}{\sqrt{2m(KE)}}$ $= \frac{h}{\sqrt{2m\left(\frac{3}{2}kT\right)}}$ $\lambda = \frac{h}{\sqrt{3mkT}}$