Question

The density of a gas is 1 molecule cm\(^{-3}\). If the molecular diameter is \(1 \times 10^{-8}\) cm, then the mean free path of the molecules is:

• A. \(\frac{1}{\sqrt{2\pi}} \times 10^{14}\, m\)
• B. \(\frac{1}{2\pi} \times 10^{13}\, m\)
• C. \(\frac{1}{\sqrt{6\pi}} \times 10^{14}\, m\)
• D. \(\frac{1}{\sqrt{2\pi}} \times 10^{13}\, m\)

Hint:

Mean free path depends inversely on number density and square of molecular diameter: \( \lambda = \frac{1}{\sqrt{2}n\pi d^2} \).

Answer 1

The Correct Option is A

Step 1: Mean free path equation.
λ = 1/(√2 n π d²).

Step 2: SI unit conversion.
n = 1 molecule cm⁻³ = 10⁶ molecules m⁻³; d = 1×10⁻⁸ cm = 10⁻¹⁰ m.

Step 3: Squaring the diameter.
d² = 10⁻²⁰ m².

Step 4: Substituting into formula.
λ = 1/(√2 × 10⁶ × π × 10⁻²⁰).

Step 5: Power-of-ten simplification.
λ = 1/(√2 π × 10⁻¹⁴).

Step 6: Final expression.
λ = (1/√(2)π) × 10¹⁴ m.