JEE Main 2024 Physics question
The average kinetic energy of a monoatomic molecule is 0.414 eV at temperature : (Use J/mol-K)
Answer: (B)
Solution
E = (3/ gives 0.414 × 1.6 × 10⁻¹⁹ = (3/2) × 1.38 × 10⁻²³ × T, so T = 3200 K.
The average kinetic energy of a monoatomic molecule is 0.414 eV at temperature : (Use J/mol-K)
E = (3/ gives 0.414 × 1.6 × 10⁻¹⁹ = (3/2) × 1.38 × 10⁻²³ × T, so T = 3200 K.