JEE Mains · Physics · STD 11 - 12 . kinetic theory of gases
The average kinetic energy of a monatomic molecule is \(0.414 \mathrm{eV}\) at temperature : (Use \(\mathrm{K}_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{mol}-\mathrm{K}\) )
- A \(3000 \mathrm{~K}\)
- B \(3200 \mathrm{~K}\)
- C \(1600 \mathrm{~K}\)
- D \(1500 \mathrm{~K}\)
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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