JEE Physics 2025 Question with Solution

Given: Mean free path $\lambda = 3 \times 10^{-7} \, \text{m}$ and average speed $\bar{c} = 600 \, \text{m/s}$.

Find: Frequency of collisions $f$.

The frequency of molecular collisions is related to the mean free path and average speed by

$$f = \frac{\bar{c}}{\lambda}$$

Substituting the given values,

$$f = \frac{600}{3 \times 10^{-7}}$$

Therefore,

$$f = 2 \times 10^{9} \, \text{s}^{-1}$$

Therefore, the frequency of collisions is $2 \times 10^{9} \, \text{s}^{-1}$. The correct option is C.

Given: $\lambda = 3 \times 10^{-7} \, \text{m}$ and $\bar{c} = 600 \, \text{m/s}$.

Find: The collision frequency $f$.

Concept Used: The frequency of molecular collisions is given by

$$f = \frac{\bar{c}}{\lambda}$$

where $f$ is in $\text{s}^{-1}$.

Step 1: Write the formula.

$$f = \frac{\bar{c}}{\lambda}$$

Step 2: Substitute the values.

$$f = \frac{600}{3 \times 10^{-7}}$$

Step 3: Simplify.

$$f = \frac{600}{3} \times 10^{7}$$ $$f = 200 \times 10^{7}$$ $$f = 2 \times 10^{9} \, \text{s}^{-1}$$

Therefore, the frequency of collisions is $2 \times 10^{9} \, \text{s}^{-1}$.