The de-Broglie wavelength of a neutron in thermal equilibrium is $\frac{h}{\sqrt{2mk_{B}T}}$ or $\frac{h}{\sqrt{3mk_{B}T}}$

For a single particle moving in one dimension, the average kinetic energy is:

$E=\frac{1}{2}k_{B}T$

In three dimensions, the kinetic energy of the particle would be:

$E=\frac{3}{2}k_{B}T$

But if we only use the expression for the one-dimensional thermal energy, some textbooks express the momentum for the de Broglie wavelength as:

$p=\sqrt{2mk_{B}T}$

Thus, the de Broglie wavelength λ becomes:

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mk_{B}T}}$

So, depending on the context and assumptions about energy distribution, you may see either $\sqrt{3mk_{B}T}$  or $\sqrt{2mk_{B}T}$ in textbooks. The formula with $2mk_{B}T$  typically applies when focusing on the one-dimensional mean energy in thermal equilibrium.