The mean free path of a molecule of diameter $5 \times 10^{-10} \mathrm{~m}$ at the temperature $41^{\circ} \mathrm{C}$ and pressure $1.38 \times 10^{5} \mathrm{~Pa}$, is given as ______ m. (Given $\mathrm{k}_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$ ).

(C) $2 \sqrt{2} \times 10^{-8}$

(A) $2 \sqrt{2} \times 10^{-10}$

(B) $10 \sqrt{2} \times 10^{-8}$

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Heat & Thermodynamics

The mean free path of a molecule of diameter

5 \times 1 0^{- 10} m

at the temperature

4 1^{\circ} C

and pressure

1.38 \times 1 0^{5} Pa

, is given as ______ m.
(Given

k_{B} = 1.38 \times 1 0^{- 23} J / K

Answer: (C)

The mean free path of a gas molecule is

λ = \frac{k _{B} T}{2 π d ^{2} P}

, where

d

is the molecular diameter. With

T = 273 + 41 = 314

d = 5 \times 1 0^{- 10}

P = 1.38 \times 1 0^{5}

Pa and

k_{B} = 1.38 \times 1 0^{- 23}

J/K,

λ = \frac{( 1.38 \times 1 0 ^{- 23} ) ( 314 )}{2 π ( 5 \times 1 0 ^{- 10} ) ^{2} ( 1.38 \times 1 0 ^{5} )} .

The factor

1.38 \times 1 0^{- 23}

in the numerator cancels the

1.38 \times 1 0^{5}

in the denominator, leaving

λ = \frac{314 \times 1 0 ^{- 28}}{2 π ( 25 \times 1 0 ^{- 20} )} = \frac{314}{2 π \times 25} \times 1 0^{- 8} .

Evaluating,

\frac{314}{25 π} \approx 4

, so

λ = \frac{4}{2} \times 1 0^{- 8} = 22 \times 1 0^{- 8}

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option