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MCQEasyJEE 2026Mean Free Path & Avogadro's Number

JEE Physics 2026 Question with Solution

The mean free path of a molecule of diameter 5×10−10 m5\times10^{-10}\,m at temperature 41∘C41^\circ C and pressure 1.38×105 Pa1.38\times10^5\,Pa is given as _____ mm. (Given kB=1.38×10−23 J/Kk_B=1.38\times10^{-23}\,J/K)

  • A

    22×10−82\sqrt{2}\times10^{-8}

  • B

    102×10−810\sqrt{2}\times10^{-8}

  • C

    2×10−82\times10^{-8}

  • D

    22×10−102\sqrt{2}\times10^{-10}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: molecular diameter d=5×10−10 md = 5\times10^{-10}\,\text{m}, temperature T=41∘CT = 41^\circ C, pressure p=1.38×105 Pap = 1.38\times10^5\,\text{Pa}, and Boltzmann constant kB=1.38×10−23 J/Kk_B = 1.38\times10^{-23}\,\text{J/K}.

Find: the mean free path λ\lambda.

Concept: The mean free path of a gas molecule is

λ=kBT2πd2p\lambda=\frac{k_BT}{\sqrt{2}\pi d^2 p}

Step-by-step Calculation

First convert temperature into kelvin:

T=41+273=314 KT=41+273=314\,\text{K}

Now substitute the values into the formula:

λ=(1.38×10−23)(314)2π(5×10−10)2(1.38×105)\lambda=\frac{(1.38\times10^{-23})(314)}{\sqrt{2}\pi(5\times10^{-10})^2(1.38\times10^5)}

Simplifying as shown:

=314×10−232π×25×10−20×1.38×105=\frac{314\times10^{-23}}{\sqrt{2}\pi\times25\times10^{-20}\times1.38\times10^5}

On evaluation,

λ≈2.8×10−8 m\lambda\approx2.8\times10^{-8}\,\text{m}

which is

λ≈22×10−8 m\lambda\approx2\sqrt{2}\times10^{-8}\,\text{m}

Therefore, the correct option is A.

Common mistakes

  • Using temperature in degree Celsius directly is incorrect because the mean free path formula requires absolute temperature. Convert 41∘C41^\circ C to 314 K314\,\text{K} before substitution.

  • Forgetting to square the molecular diameter is incorrect because the formula contains d2d^2. Use (5×10−10)2(5\times10^{-10})^2, not 5×10−105\times10^{-10}.

  • Omitting the factor 2π\sqrt{2}\pi in the denominator gives a larger incorrect value. Use the complete expression for mean free path exactly as given.

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