Question: The mean free path $\lambda$ for a gas, with molecular diameter $d$ and number density $n$ can be expressed as
[NEET (Sep.) 2020]
A) $\frac{1}{\sqrt{2} n \pi d^2}$
B) $\frac{1}{\sqrt{2} n^2 \pi d^2}$
C) $\frac{1}{\sqrt{2} n^2 \pi^2 d^2}$
D) $\frac{1}{\sqrt{2} n \pi d}$
Answer: $\frac{1}{\sqrt{2} n \pi d^2}$
Solution:
The mean free path $\lambda$ for a gas, with molecular diameterd and number density $n$ is given by the relation $$ \lambda=\frac{1}{\sqrt{2} n \pi d^2} $$
Hence, correct option is (a).
