The mean free path of a molecule of diameter 5 × 10−10 m at the temperature 41°C and pressure 1.38 × 10⁵ Pa, is given as

Q. The mean free path of a molecule of diameter 5 × 10⁻¹⁰ m at the temperature 41°C and pressure 1.38 × 10⁵ Pa, is given as ______ m.

(Given k_B = 1.38 × 10⁻²³ J/K).

1. 2√2 × 10⁻¹⁰

2. 10√2 × 10⁻⁸

3. 2√2 × 10⁻⁸

4. 2 × 10⁻⁸

Correct Answer: 2√2 × 10⁻⁸

Solution

The mean free path of a gas molecule is given by the formula:

λ = k_BT / ( √2 π d² p )

Given diameter of molecule,

d = 5 × 10⁻¹⁰ m

Temperature,

T = 41 + 273 = 314 K

Pressure,

p = 1.38 × 10⁵ Pa

Substituting the values:

λ = (1.38 × 10⁻²³ × 314) / ( √2 π (5 × 10⁻¹⁰)² × 1.38 × 10⁵ )

On simplification:

λ ≈ 2.83 × 10⁻⁸ m

λ = 2√2 × 10⁻⁸ m

Hence, the correct answer is 2√2 × 10⁻⁸.