Physics - Modern Physics Question with Solution | TestHub

The de-Broglie wavelength of a gas molecule is given by $\lambda =\frac{h}{\sqrt{3mkT}}$.

Thus, $\lambda \propto \frac{1}{\sqrt{mT}}$.

For hydrogen and helium molecules, the ratio of their de-Broglie wavelengths is:

$\frac{{\lambda }_{H_2}}{{\lambda }_{He}}=\sqrt{\frac{m_{He}{T}_{He}}{m_{H_2}{T}_{H_2}}}$

Given $m_{He}=2m_{H_2}$, $T_{H_2}=27^{\circ }C=300$ K, and $T_{He}=127^{\circ }C=400$ K.

Substituting these values:

$\frac{{\lambda }_{H_2}}{{\lambda }_{He}}=\sqrt{\frac{2m_{H_2}\times 400}{m_{H_2}\times 300}}=\sqrt{\frac{8}{3}}$

Answer: $\sqrt{\frac{8}{3}}$