Two flasks `A` and `B` have equal volume at `100 K` and `200 K` and have `4 atm` and `1 atm` pressures, respectively. The flasks `A` contains `H_(2)` gas and `B` contains `CH_(4)` gas. The collision diameter of `CH_(4)` is twice that of `H_(2)`.Which of the following is true about the mean free path `(lambda)` of the molecules?
(`a`) `lambda` of `H_(2)` is twice that of `CH_(4)`.
(`b`) `lambda` of `CH_(4)` is twice that of `H_(2)`.
(`c`) `lambda` of `H_(2)` is four times that of `CH_(4)`.
(`d`) `lambda` of `CH_(4)` is four times that `H_(2)`.

Explanation

To solve the problem, we need to calculate the mean free path (λ) of the gas molecules in flasks A and B, which contain H₂ and CH₄ respectively. The mean free path can be determined using the formula:

\[ \lambda = \frac{2 \sqrt{2}}{\pi d^2 n} \]

Where:
- \( \lambda \) = mean free path
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