The mean free path $\lambda$ of molecules of a gas is given by

(A) $\dfrac{2}{\pi n d^2}$
(B) $\dfrac{1}{2\pi n d^2}$
(C) $\dfrac{1}{\sqrt{2}\,\pi n d^2}$
(D) $\dfrac{1}{2\pi n d}$, where n is the number of molecules per unit volume and d is the diameter of the molecules.

Question

The mean free path $\lambda$ of molecules of a gas is given by

(A) $\dfrac{2}{\pi n d^2}$
(B) $\dfrac{1}{2\pi n d^2}$
(C) $\dfrac{1}{\sqrt{2}\,\pi n d^2}$
(D) $\dfrac{1}{2\pi n d}$, where n is the number of molecules per unit volume and d is the diameter of the molecules.

Accepted Answer

The mean free path lambda is the average distance a molecule travels between successive collisions. Two dependences combine to give its formula: (i) lambda is inversely proportional to the number density n = N/V, since a denser gas means more frequent collisions, and (ii) lambda is inversely proportional to d^2 (not d), since it is the molecule's collision cross-section, proportional to the square of its diameter, that determines how likely a collision is. A fuller kinetic-theory derivation (accounting for the fact that both colliding molecules are moving, not just one) introduces an additional factor of 1/sqrt(2). Putting these together gives lambda = 1/(sqrt(2) pi n d^2). [!ANSWER] (C) 1/(√2 π n d²)

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set sz1 markMCQ
$Q. The mean free path of a gas is directly proportional to:$
(a) pressure
(b) temperature
(c) the square of the molecular diameter
(d) number density
›Reveal solutionSolution

The mean free path lambda = kT / (root-2 x pi x d^2 x P), so at constant pressure it is directly proportional to temperature T.

The mean free path is given by lambda = 1 / (root-2 x pi x d^2 x n), where d is the molecular diameter and n is the number density. Using the ideal gas relation n = P/(kT), this becomes lambda = kT / (root-2 x pi x d^2 x P). At constant pressure, lambda is therefore directly proportional to the absolute temperature T. By contrast, lambda is inversely related to pressure, inversely related to the square of the molecular diameter, and inversely related to the number density — so temperature is the only listed quantity to which it is directly proportional.

✓Final answer
$The correct option is (b) temperature.$
CBSE 2023Set ANNUAL1 markMCQ
$Q. If 'n' is the number of molecules per unit volume and 'd' is the diameter of the molecules, the mean free path 'λ' of molecules is$
(a) $\sqrt{\dfrac{2}{\pi n d}}$
(b) $\dfrac{1}{2\pi n d^{2}}$
(c) $\dfrac{1}{\sqrt{2}\pi n d^{2}}$
(d) $\dfrac{1}{\sqrt{2}\pi n d}$
›Reveal solutionSolution

Mean free path follows from the molecular collision cross-section.

A molecule of diameter $d$ moving among others of number density $n$ sweeps out a cylindrical volume of cross-section $\pi d^{2}$ per unit path length. Since every molecule is in motion, the relative speed between two colliding molecules is on average $\sqrt{2}$ times a single molecule's mean speed, so the number of collisions per unit path length is $\sqrt{2}\,\pi d^{2} n$ .

The mean free path is the average distance travelled between collisions:

$\lambda = \frac{1}{\sqrt{2}\,\pi n d^{2}}$

✓Final answer
$\lambda = \dfrac{1}{\sqrt{2}\,\pi n d^{2}}$

Q.The mean free path $\lambda$ of molecules of a gas is given by (A) $\dfrac{2}{\pi n d^2}$
(B) $\dfrac{1}{2\pi n d^2}$
(C) $\dfrac{1}{\sqrt{2}\,\pi n d^2}$
(D) $\dfrac{1}{2\pi n d}$ , where n is the number of molecules per unit volume and d is the diameter of the molecules.

Concept understanding — Mean Free Path

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