RE-NEET 2026 Physics Mean Free Path MCQ Question

The mean free path $\lambda$ of gas molecules is given by the formula:

$$\lambda = \frac{kT}{\sqrt{2} \pi d^2 n}$$

where $d$ is the diameter of the molecules, $n$ is the number density, $k$ is the Boltzmann constant, and $T$ is the temperature.

Given that the mean free path of gas A is half that of gas B, we can express this relationship as:

$$\lambda_A = \frac{1}{2} \lambda_B$$

Substituting the expressions for mean free path, we have:

$$\frac{kT}{\sqrt{2} \pi (2d_B)^2 n_A} = \frac{1}{2} \frac{kT}{\sqrt{2} \pi d_B^2 n_B}$$

This simplifies to:

\frac{1}{4} \frac{1}{n_A} = \frac{1}{2n_B} \implies n_A = \frac{1}{2} n_B$$ Thus, the correct answer is $n_A = \frac{1}{2} n_B$. The other options fail because they do not satisfy the derived relationship based on the given conditions.