The mean free path $\lambda$ of a gas molecule of diameter $d$ at number density $n$ is approximately:
A$\lambda = \dfrac{1}{\sqrt{2}\pi d^2 n}$
B$\lambda = \sqrt{2}\pi d^2 n$
C$\lambda = \dfrac{1}{n d}$
D$\lambda = \pi d n$
Answer & Solution
Correct answer: A. $\lambda = \dfrac{1}{\sqrt{2}\pi d^2 n}$
The mean free path is the average distance a molecule travels between collisions. Standard kinetic-theory derivation gives:
$\lambda = \dfrac{1}{\sqrt{2}\pi d^2 n}$
where $d$ is the molecular diameter and $n$ is the number density.
Note the $d^2$ (cross-sectional area), $\sqrt{2}$ factor from accounting for the relative motion of two molecules colliding (both move, not just one), and the inverse-density dependence (more molecules around means more collisions, shorter path).
At STP, $\lambda$ for air is about 70 nm — small but huge compared to molecular size (a few Å).
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