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JEE MainPhysicsKinetic Theory of Gases

A particular gas molecule is observed to undergo 10^5 collisions in a time interval of 10 s . If the mean free path of the gas molecules is 40 nm , what is the average speed of the molecule?

Options

  1. A4 10⁻¹⁸ m/s
  2. B400 m/s
  3. C4 10⁻³ m/s
  4. D4 10⁻⁴ m/s

Correct answer

B. 400 m/s

Step-by-step solution

The collision frequency f is the number of collisions per unit time. Given: Number of collisions, N = 10^5 Time interval, t = 10 s = 10 10⁻⁶ s = 10⁻⁵ s Mean free path, = 40 nm = 40 10⁻⁹ m The collision frequency is: f = N t = 10^5 10⁻⁵ = 10¹⁰ Hz The average speed v_ avg is related to the collision frequency and mean free path by: v_ avg = f v_ avg = 10¹⁰ 40 10⁻⁹ = 400 m/s Answer: 400 m/s

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