Kinetic Theory - Result Question 22
22. The mean free path for a gas, with molecular diameter $d$ and number density $n$ can be expressed as :
[2020]
(a) $\frac{1}{\sqrt{2} n \pi d^{2}}$
(b) $\frac{1}{\sqrt{2} n^{2} \pi d^{2}}$
(c) $\frac{1}{\sqrt{2} n^{2} \pi^{2} d^{2}}$
(d) $\frac{1}{\sqrt{2} n \pi d}$
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Answer:
Correct Answer: 22. (a)
Solution:
- (a) Mean free path for a gas $\lambda_m=\frac{1}{\sqrt{2} n \pi d^{2}}$
Here, $d=$ diameter of a gas molecule and, $n=$ molecular density.
