JEE Main202628 January 2026Evening ShiftPhysicsKinetic Theory of GasesActual
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 ).
Options
- A10 2 10⁻⁸
- B2 2 10⁻⁸
- C2 2 10⁻¹⁰
- D2 10⁻⁸
Correct answer
B. 2 2 10⁻⁸
Step-by-step solution
Mean free path is calculated using = k_B T 2 d^2 P Given: d = 5 10⁻¹⁰ m, T = 41°C = 314 K, P = 1.38 10^5 Pa, k_B = 1.38 10⁻²³ J/K = 1.38 10⁻²³ 314 2 (5 10⁻¹⁰)^2 1.38 10^5 = 1.38 10⁻²³ 314 2 25 10⁻²⁰ 1.38 10^5 Numerator: 433.32 10⁻²³ Denominator: 1.414 3.14159 34.5 10⁻¹⁵ = 153.18 10⁻¹⁵ = 433.32 10⁻²³ 153.18 10⁻¹⁵ = 2.83 10⁻⁸ m = 2 2 10⁻⁸ m