The average kinetic energy of a monatomic molecule is 0.414 eV at temperature: (Use KB = 1.38 × 10⁻²³ J/mol-K) (a) 3000 K (b) 3200 K (c) 1600 K (d) 1500 K

Answer: (b). Average KE of a monatomic molecule = 3/2kBT. 0.414 eV = 0.414 × 1.6 × 10⁻¹⁹ = 6.624 × 10⁻²⁰ J T = 2 × 6.624 × 10⁻²⁰3 × 1.38 × 10⁻²³ = 3200 K So the answer is (b) 3200 K.

The average kinetic energy of a monatomic molecule is 0.414 eV at… | JEE Main 2024 (27 Jan, Shift 1)

Q14JEE Main 2024 · 27 JAN · SHIFT 1Easy

Pressure & Kinetic Energy of a Gas

The average kinetic energy of a monatomic molecule is $0.414\text{ eV}$ at temperature: (Use $K_B = 1.38 \times 10^{-23}\text{ J/mol-K}$ )

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Answer: (b)

Average KE of a monatomic molecule $= \frac{3}{2}k_BT$ .
$0.414\text{ eV} = 0.414 \times 1.6 \times 10^{-19} = 6.624 \times 10^{-20}$ J
$T = \frac{2 \times 6.624 \times 10^{-20}}{3 \times 1.38 \times 10^{-23}} = 3200$ K
So the answer is (b) 3200 K.

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