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(B) The mean free path $\lambda$ of a gas molecule is given by the formula $\lambda = \frac{1}{\sqrt{2} \pi d^2 n}$,where $d$ is the diameter of the m

Vedclass · Accepted Answer

$25$

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(B) The mean free path $\lambda$ of a gas molecule 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 that the number density $n$ is the same for both molecules $A$ and $B$,the ratio of their mean free paths is $\frac{\lambda_A}{\lambda_B} = \frac{d_B^2}{d_A^2}$.Substituting the given values,$d_A = 10 \, \mathring{A}$ and $d_B = 5 \, \mathring{A}$,we get:$\frac{\lambda_A}{\lambda_B} = \left( \frac{5}{10} ight)^2 = \left( \frac{1}{2} ight)^2 = \frac{1}{4} = 0.25$.Expressing this in the required format,$0.25 = 25 	imes 10^{-2}$.Thus,the ratio is $25 	imes 10^{-2}$.

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