0. For the given concentration, if the ratio of the diameters of the molecules of two gases is $1: 2$, then…
0. For the given concentration, if the ratio of the diameters of the molecules of two gases is $1: 2$, then the ratio of their mean free paths is
$4: 1$
$2: 1$
$1: 1$
$1: 2$
Solution
Ratio of the diameter of the molecules
$\frac{d_1}{d_2}=\frac{1}{2}$
Mean free path is given by
$\begin{aligned} & \lambda=\frac{1}{\sqrt{2} \pi \mathrm{d}^2 \mathrm{~h}} \\ & \frac{\lambda_1}{\lambda_2}=\frac{\mathrm{d}_2^2}{\mathrm{~d}_1^2}=\left(\frac{2}{1}\right)^2=\frac{4}{1}\end{aligned}$