NEET Physics Question - Thermal Physics

The mean free path for a gas, with molecular diameter $d$ and number density $n$ can be expressed as: (2020)

$\frac{1}{\sqrt{2} n \pi d^2}$

$\frac{1}{\sqrt{2} n^2 \pi d^2}$

$\frac{1}{\sqrt{2} n^2 \pi^2 d^2}$

$\frac{1}{\sqrt{2} n \pi d}$

Solution:

From the kinetic theory of gases, the mean free path $\lambda$ (the average distance a molecule travels between collisions) is derived as $\lambda = \frac{1}{\sqrt{2} \pi n d^2}$.

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Solution:

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