The mean free path ($ \lambda $) of gas molecules is given by the formula: $$ \lambda = \frac{1}{\sqrt{2} \pi d^2 n} $$ 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 = 2 d_B $. Using the formula: $$ \frac{\lambda_A}{\lambda_B} = \frac{d_B^2 n_B}{d_A^2 n_A} $$
Substitute $ \frac{\lambda_A}{\lambda_B} = \frac{1}{2} $ and $ d_A = 2 d_B $: $$ \frac{1}{2} = \frac{d_B^2 n_B}{(2 d_B)^2 n_A} $$ $$ \frac{1}{2} = \frac{d_B^2 n_B}{4 d_B^2 n_A} $$ $$ \frac{1}{2} = \frac{n_B}{4 n_A} $$
Rearranging the equation: $$ 4 n_A = 2 n_B $$ $$ n_A = \frac{2}{4} n_B = \frac{1}{2} n_B $$
Final Answer: n_{A}=\frac{1}{2}n_{B}
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