Q 11-12-163JEE MainJEE Main 2025 (4 Apr, Shift 1)Easy
The mean free path and the average speed of oxygen molecules at $300\ \text{K}$ and $1\ \text{atm}$ are $3\times10^{-7}\ \text{m}$ and $600\ \text{m/s}$, respectively. Find the frequency of its collisions.
Answer: (C) $2\times10^9\ \text{s}^{-1}$
Between collisions a molecule travels on average one mean free path, so the collision frequency is
$$f = \frac{\bar v}{\lambda} = \frac{600}{3\times10^{-7}} = 2\times10^9\ \text{s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics