Calculate de-Broglie wavelength (in $\text{Å}$) for an average helium atom in a furnace at 400 K. Given mass of helium = 4.002 amu.

de-Broglie wavelength, $\mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{p}}$

Mean KE of helium atom$=\frac{3\mathrm{kT}}{2}$

or      $\mathrm{E}=\frac{3\times \left(1.38\times {10}^{-23}\right)\times 400}{2}$

In terms of momentum, the energy is given by $\mathrm{E}=\frac{{\mathrm{p}}^2}{2\mathrm{m}}$

$$\begin{gathered} \begin{array}{l}\text{ or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{p}}^2=2\mathrm{mE}\\ \text{ or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{p}=\sqrt{2\mathrm{mE}}\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{p}=\sqrt{2\times \left(4.002\times 1.66\times {10}^{-27}\right)\times 3\times \left(1.38\times {10}^{-23}\right)\times 200}\end{array} \\ =10.49\times {10}^{-24} {\mathrm{kgms}}^{-1} \end{gathered}$$

     (where mass of helium 4.002 amu = $4.002\times 1.66\times {10}^{-27}\mathrm{kg}$) 

Now, $\mathrm{\lambda }=\frac{h}{p}=\frac{6.63\times {10}^{-34}}{10.49\times {10}^{-24}}=0.63\times {10}^{-10} \mathrm{m}=0.63Å$