At temperature T, the average kinetic energy of any particle is (3/2) KT. The de Broglie wavelength follows the order:

Kinetic energy of any particle=$\frac{3}{2}\mathrm{KT}$

Also K.E.=$\frac{1}{2}{\mathrm{mv}}^2$

$$\begin{gathered} \frac{1}{2}{\mathrm{mv}}^2=\frac{3}{2}\mathrm{KT} \\ \mathrm{v}=\sqrt{\frac{3\mathrm{KT}}{\mathrm{m}}} \end{gathered}$$

De-broglie wavelength

=$$\begin{gathered} \mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{mv}} \\ \mathrm{\lambda }=\frac{\mathrm{h}}{\sqrt{3\mathrm{KTm}}} \\ \mathrm{\lambda } \mathrm{\alpha } \frac{1}{\sqrt{\mathrm{m}}} \end{gathered}$$

mass of thermal neutron > mass of thermal electron > mass of visible photon

$\therefore$order of debroglie wavelengths is Visible photon > Thermal electron > Thermal neutron