Question

The atomic masses of He and Ne are 4 and 20 a.m.u. respectively. The value of the de Broglie wavelength of He gas at -73${}^o$C is M times that of the de Broglie wavelength of Ne at 727${}^o$C. M is:

Answer 1

Since, $\lambda =\frac{h}{mV}=\frac{h}{\sqrt{2MK.E}}$ (since $K.E.\propto T$)

$\Rightarrow$$\lambda \propto \frac{1}{\sqrt{MT}}$

For two gases,

$\frac{{\lambda }_{He}}{{\lambda }_{Ne}}=\sqrt{\frac{M_{Ne}{T}_{Ne}}{M_{He}{T}_{He}}}=\sqrt{\frac{20}{4}\times \frac{1000}{200}}=5$