Question

Find the ratio of de Broglie wavelength of molecules of hydrogen and helium which are at temperatures $27^{\circ} C$ and $327^{\circ} C$, respectively.

Answer 1

Average K.E. per molecule of a gas at the given temperature $(T)$ is given as,
$\frac{1}{2} m v^{2}=\frac{3}{2} k T$
where, $m=$ molecular mass of the gas
$\therefore mv =\sqrt{3 mkT }$
But, $\lambda=\frac{ h }{ mv }=\frac{ h }{\sqrt{3 mkT }}$
$\therefore \frac{\lambda_{ H }}{\lambda_{ He }} =\sqrt{\frac{ m _{ He } T _{ He }}{ m _{ H } T _{ H }}}$
$=\sqrt{\frac{(4)(273+327)}{(2)(273+27)}}=\sqrt{4}=2$