JEE Mains · Physics · STD 11 - Kinetic theory of gases
- JEE Mains 2026
- Medium
- MCQ
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 \(k_{B}=1.38\times10^{-23}J/K\)).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(2\sqrt{2}\times10^{-8}\)
Step-by-step Solution
Detailed explanation
\(\lambda=\frac{k_{B}T}{\sqrt{2}\pi\sigma^{2}P}\)
\(=\frac{1.38\times10^{-23}\times(273+41)\times100}{\sqrt{2}\times3.14\times(5\times10^{-10})^{2}\times1.38\times10^{5}}=2\sqrt{2}\times10^{-8}\)
\(=\frac{1.38\times10^{-23}\times(273+41)\times100}{\sqrt{2}\times3.14\times(5\times10^{-10})^{2}\times1.38\times10^{5}}=2\sqrt{2}\times10^{-8}\)
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