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

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NEET 2020 Official Paper (Held On: 13 September, 2020)
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  1. \(\frac{1}{\sqrt{2} n^{2} \pi d^{2}}\)
  2. \(\frac{1}{\sqrt{2} n^{2} \pi^{2} d^{2}}\)
  3. \(\frac{1}{\sqrt{2} n \pi d} \)
  4. \(\frac{1}{\sqrt{2} n \pi d^{2}}\)

Answer (Detailed Solution Below)

Option 4 : \(\frac{1}{\sqrt{2} n \pi d^{2}}\)
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Explanation:

Mean free path- The mean free path for gas is defined as the average distance of an object that will move in between collisions and it is directly proportional to the temperature and inversely proportional to the pressure and the diameter of the molecule. It is written as;

\(\lambda = \frac{kT}{{\sqrt 2 n\pi {d^2}}}\)

⇒ \(\lambda \propto \frac{1}{{\sqrt 2 n\pi {d^2}}}\)

Where d is the diameter and n is the molecular density of the gas.

Hence, option 4) is the correct answer.

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