The average kinetic energy of a monatomic molecule is 0.414 eV at temperature :
(Use K_B=1.38 × 10⁻²³ J / mol-K)

Solution

Physics sir · 4 steps

A monatomic molecule has 3 degrees of freedom, so its average energy is 3/2k_BT. Solve for T.

1. Count the freedoms. A monatomic molecule can only move along x, y and z, so it has 3 translational degrees of freedom. By equipartition, each one carries 1/2k_BT on average.

   Ē = 3/2k_BT ⇒ T = 2Ē/3k_B

2. Convert to joules. k_B is in joules per kelvin, so the energy must be in joules too: 0.414 eV = 0.414 × 1.6 × 10⁻¹⁹ = 6.624 × 10⁻²⁰ J.
3. Put in numbers. T = (2 × 6.624 × 10⁻²⁰)/(3 × 1.38 × 10⁻²³) = (13.248 × 10⁻²⁰)/(4.14 × 10⁻²³) = 3200 K.
4. So. The temperature is 3200 K, option (b).

The idea to keep. Average energy per molecule is f/2k_BT: f = 3 for monatomic, 5 for rigid diatomic. Always convert eV to joules before dividing by k_B.

Answer from the NTA final answer key. Solution by PrepWiser.