Question

The mean free path of a gas molecule of diameter d and number density n (number of molecules per unit volume) is inversely proportional to:

• A. n d
• B. n^2 d
• C. n d^2
• D. n d

Hint:

Think of the collision space geometrically: a molecule sweeps out a target collision cylinder of cross-sectional area = d^2. The higher the density (n) or the wider this target area ( d^2), the more frequently collisions occur, meaning the mean free path () shrinks: 1Area Density = 1 d^2 n.

Answer 1

The Correct Option is C

Step 1: Picture what mean free path means.
A gas molecule keeps moving and bumping into other molecules. The average distance it travels between two bumps is called the mean free path. We want to know how this distance depends on molecule size and crowding.

Step 2: Think about how often collisions happen.
A molecule sweeps out a tube as it moves. Any other molecule whose centre comes within a distance $d$ (the diameter) gets hit. So the tube has a cross-section area that grows with $d^2$.

Step 3: Bring in the number density.
If there are more molecules packed in a given space, the moving molecule meets them sooner. So a higher number density $n$ means a shorter mean free path. The path gets shorter as $n$ goes up.

Step 4: Write the standard formula.
Kinetic theory gives the mean free path as \[ \lambda = \frac{1}{\sqrt{2}\,\pi\, n\, d^2}. \]

Step 5: Read off the proportionality.
The constants $\sqrt{2}$ and $\pi$ do not change. So $\lambda$ is inversely proportional to $n\,d^2$. \[ \lambda \propto \frac{1}{n\,d^2} \]

Step 6: State the answer.
The mean free path is inversely proportional to the product of the number density and the square of the diameter. \[ \boxed{n\,d^2} \]