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JEE Main202427 Jan 2024Morning ShiftPhysicsKinetic Theory of GasesActual

The average kinetic energy of a monatomic molecule is 0 . 414 eV at temperature: (Use K B = 1 . 38 × 10 - 23 J mol - 1 K - 1 )

Options

  1. A3000 K
  2. B3200 K
  3. C1600 K
  4. D1500 K

Correct answer

B. 3200 K

Step-by-step solution

For monatomic molecule, degree of freedom = 3 . The formula for the average kinetic energy is given by K avg = 1 2 f K B T . . . 1 From equation (1), it follows that 0 . 414 eV × 1 . 6 × 10 - 19 J 1 eV = 3 2 × 1 . 38 × 10 - 23 J mol - 1 K - 1 × T ⇒ T = 0 . 414 eV × 1 . 6 × 10 - 19 J 1 eV × 2 3 × 1 . 38 × 10 - 23 J mol - 1 K - 1 = 3200 K

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