Physics · Kinetic Theory of Gases

JEE Main 2024 — 27 January, Shift 1 — Question 47

The average kinetic energy of a monatomic molecule is 0.414 eV at temperature :

$\left(\right.$ Use $\left.\mathrm{K}_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{mol}-\mathrm{K}\right)$

Answer: B

For monoatomic molecule degree of freedom $=3$ .

$\therefore \mathrm{K}_{\text {avg }}=\frac{3}{2} \mathrm{~K}_{\mathrm{B}} \mathrm{T}$

$\mathrm{T}=\frac{0.414 \times 1.6 \times 10^{-19} \times 2}{3 \times 1.38 \times 10^{-23}}$

$=3200 \mathrm{~K}$

Answer key and solution verified before publishing.

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