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JEE Mains · Physics · STD 11 - Kinetic theory of gases
  • JEE Mains 2024
  • Hard
  • MCQ

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

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Answer & Solution

Correct Answer

(B)  \(3200 \mathrm{~K}\)

Step-by-step Solution

Detailed explanation

For monoatomic molecule degree of freedom \(=3\).

\( \therefore \mathrm{K}_{v-\mathrm{g}}=\frac{3}{2} \mathrm{~K}_{\mathrm{B}} \mathrm{T} \)

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

\( =3200 \mathrm{~K}\)

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