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 (\( \lambda \)) quantifies the average distance a gas molecule traverses between successive collisions. According to the kinetic theory of gases, this is expressed by the formula: \[ \lambda = \frac{1}{\sqrt{2 \pi n d^2}}, \] where \( n \) denotes the number density (molecules per unit volume), \( d \) represents the molecular diameter, and \( \pi \) is the constant associated with circular cross-sections.

 

Derivation Outline: 
1. Collision Cross-Section: The effective collision area between two molecules is given by \( \sigma = \pi d^2 \). 
2. Collision Frequency: This rate depends on \( \sigma \), \( n \), and the average relative molecular velocity. 
3. Mean Free Path: It is the reciprocal of the collision cross-section multiplied by the number density, with a \( \sqrt{2} \) factor accounting for molecular motion: 
\[ \lambda = \frac{1}{\sqrt{2 \pi n d^2}}. \]

 Final Result: The mean free path is calculated as: \[ \boxed{\frac{1}{\sqrt{2 \pi n d^2}}} \quad \text{(Option 1)}. \]