Question

If \( n \) is the number density and \( d \) is the diameter of the molecule, then the average distance covered by a molecule between two successive collisions (i.e., mean free path) is represented by:

• A. \( \frac{1}{\sqrt{2\pi n d^2}} \)
• B. \( \sqrt{2n \pi d^2} \)
• C. \( \frac{1}{\sqrt{2} n \pi d^2} \)
• D. \( \frac{1}{\sqrt{2n^2\pi^2 d^2}} \)

Hint:

The mean free path decreases with increasing molecular diameter or number density. For smaller \( d \) and \( n \), molecules travel longer distances before colliding.

Answer 1

The Correct Option is A

The mean free path (\( \lambda \)) represents the average distance a gas molecule travels between collisions. Based on the kinetic theory of gases, it is calculated as: \[ \lambda = \frac{1}{\sqrt{2 \pi n d^2}}, \] where: - \( n \) is the number density (molecules per unit volume), - \( d \) is the molecular diameter, - \( \pi \) is the mathematical constant related to circular cross-sections. Derivation Steps:
1. Collision Cross-Section: The effective area for a collision between two molecules is \( \sigma = \pi d^2 \), where \( d \) is the molecular diameter.
2. Collision Frequency: The rate of collisions is determined by \( \sigma \), \( n \), and the average relative velocity of the molecules.
3. Mean Free Path: The mean free path is the inverse of the product of collision cross-section and number density, adjusted by a factor of \( \sqrt{2} \) for molecular motion:
\[ \lambda = \frac{1}{\sqrt{2 \pi n d^2}}. \] Final Answer: The mean free path is: \[ \boxed{\frac{1}{\sqrt{2 \pi n d^2}}} \quad \text{(Option 1)}. \]