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) $n_A = \dfrac{1}{2}n_B$
Concept. Mean free path: $\lambda = \dfrac{1}{\sqrt2\pi d^2n}$, so $\lambda \propto \dfrac{1}{d^2n}$.
Step 1. Ratio. With $d_A = 2d_B$:
$$\frac{\lambda_A}{\lambda_B} = \frac{d_B^2n_B}{4d_B^2n_A} = \frac{n_B}{4n_A}$$
Step 2. Solve. $\dfrac{n_B}{4n_A} = \dfrac12 \Rightarrow n_A = \dfrac12n_B$.
Answer. $n_A = \dfrac12n_B$ (Option D).
Solution by Sreeraj P, M.Sc Physics