JEE MAIN - Physics (2025 - 4th April Morning Shift - No. 17)

The mean free path and the average speed of oxygen molecules at 300 K and 1 atm are $3 \times 10^{-7} \mathrm{~m}$ and $600 \mathrm{~m} / \mathrm{s}$, respectively. Find the frequency of its collisions.
$$5 \times 10^8 / \mathrm{s}$$
$9 \times 10^5 / \mathrm{s}$
$2 \times 10^{10} / \mathrm{s}$
$2 \times 10^9 / \mathrm{s}$

Explanation

To determine the frequency of collisions for oxygen molecules, we use the formula for frequency:

$ \text{Frequency} = \frac{1}{T} = \frac{V_{\text{avg}}}{\lambda} $

where $ V_{\text{avg}} $ is the average speed of the molecules and $ \lambda $ is the mean free path.

Given:

Average speed ($ V_{\text{avg}} $) = 600 m/s

Mean free path ($ \lambda $) = $ 3 \times 10^{-7} $ m

Substitute these values into the formula:

$ \text{Frequency} = \frac{600}{3 \times 10^{-7}} = 2 \times 10^9 \, \text{s}^{-1} $

Therefore, the frequency of collisions is $ 2 \times 10^9 \, \text{collisions per second} $.

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