Two gases-argon (atomic radius $0.07 \mathrm{~nm}$, atomic weight 40) and xenon (atomic radius $0.1 \mathrm{~nm}$, atomic weight 140) have the same number density and are at the same temperature. The ratio of their respective mean free times is closest to:
Correct Option: , 2
(2)
Mean free path of a gas molecule is given by
$\lambda=\frac{1}{\sqrt{2} \pi d^{2} n}$
Here, $n=$ number of collisions per unit volume
$d=$ diameter of the molecule
If average speed of molecule is $v$ then
Mean free time, $\tau=\frac{\lambda}{v}$
$\Rightarrow \tau=\frac{1}{\sqrt{2} \pi n d^{2} v}=\frac{1}{\sqrt{2} \pi n d^{2}} \sqrt{\frac{M}{3 R T}}$
$\left(\because v=\sqrt{\frac{3 R T}{M}}\right)$
$\therefore \tau \propto \frac{\sqrt{M}}{d^{2}} \quad \therefore \frac{\tau_{1}}{\tau_{2}}=\frac{\sqrt{M_{1}}}{d_{1}^{2}} \times \frac{d_{2}^{2}}{\sqrt{M_{2}}}$
$=\sqrt{\frac{40}{140}} \times\left(\frac{0.1}{0.07}\right)^{2}=1.09$