Q 11-12-001NEETNEET 2026Top questionEasy
The mean free path of molecules in an ideal gas A is half that of another ideal gas B. The diameter of the spherical molecules of gas A is twice the diameter of the molecules of B. If number densities of the gases A and B are $n_A$ and $n_B$, respectively, then the correct option is :
Answer: (D) $n_A = \dfrac{1}{2}n_B$
Mean free path: $\lambda = \dfrac{1}{\sqrt{2}\,\pi d^2 n}$, so $\lambda \propto \dfrac{1}{d^2 n}$.
$$\frac{\lambda_A}{\lambda_B} = \frac{d_B^2\, n_B}{d_A^2\, n_A} = \frac{d_B^2\, n_B}{4d_B^2\, n_A} = \frac{n_B}{4n_A}$$
Given $\dfrac{\lambda_A}{\lambda_B} = \dfrac{1}{2}$:
$$\frac{n_B}{4n_A} = \frac{1}{2} \;\Rightarrow\; n_A = \frac{1}{2}n_B$$
Solution by Sreeraj P, M.Sc Physics