The mean free path of molecules in an ideal g | vedclass

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(D) The formula for the mean free path $\lambda$ of gas molecules is given by $\lambda = \frac{1}{\sqrt{2} \pi d^2 n}$, where $d$ is the diameter of t

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$n_A = \frac{1}{2}n_B$

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(D) The formula for the mean free path $\lambda$ of gas molecules is given by $\lambda = \frac{1}{\sqrt{2} \pi d^2 n}$, where $d$ is the diameter of the molecule and $n$ is the number density.Given: $\lambda_A = \frac{1}{2} \lambda_B$ and $d_A = 2d_B$.Using the formula for both gases:$\lambda_A = \frac{1}{\sqrt{2} \pi d_A^2 n_A}$ and $\lambda_B = \frac{1}{\sqrt{2} \pi d_B^2 n_B}$.Taking the ratio: $\frac{\lambda_A}{\lambda_B} = \frac{d_B^2 n_B}{d_A^2 n_A}$.Substituting the given values: $\frac{1}{2} = \frac{d_B^2 n_B}{(2d_B)^2 n_A} = \frac{d_B^2 n_B}{4d_B^2 n_A} = \frac{n_B}{4n_A}$.Rearranging for $n_A$: $4n_A = 2n_B$, which simplifies to $n_A = \frac{1}{2} n_B$.

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