Q 11-12-033JEE MainJEE Main 2026 (28 Jan, Shift 2)Easy
The mean free path of a molecule of diameter $5\times10^{-10}$ m at the temperature $41^\circ$C and pressure $1.38\times10^{5}$ Pa, is given as ______ m. (Given $k_B = 1.38\times10^{-23}$ J/K).
Answer: (C) $2\sqrt2\times10^{-8}$
Mean free path:
$$\lambda = \frac{k_B T}{\sqrt2\,\pi d^2 P}$$
With $T = 41 + 273 = 314$ K, $d = 5\times10^{-10}$ m, $P = 1.38\times10^5$ Pa:
$$\lambda = \frac{1.38\times10^{-23}\times 314}{\sqrt2\times 3.14\times 25\times10^{-20}\times 1.38\times10^{5}} = \frac{100\times10^{-23}}{\sqrt2\times 25\times10^{-15}}$$
$$\lambda = \frac{4\times10^{-8}}{\sqrt2} = 2\sqrt2\times10^{-8}\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics