Q 11-12-048JEE MainJEE Main 2024 (27 Jan, Shift 1)Easy
The average kinetic energy of a monatomic molecule is $0.414\ \text{eV}$ at temperature:
(Use $k_B = 1.38\times10^{-23}\ \text{J K}^{-1}$)
Answer: (B) $3200\ \text{K}$
For a monatomic molecule, average kinetic energy $= \frac32 k_B T$:
$$T = \frac{2E}{3k_B} = \frac{2\times0.414\times1.6\times10^{-19}}{3\times1.38\times10^{-23}} = \frac{1.325\times10^{-19}}{4.14\times10^{-23}} \approx 3200\ \text{K}$$
Solution by Sreeraj P, M.Sc Physics