According to equipartition principle, the energy contributed by each translational degree of freedom and rotational degree of freedom at a temperature T are respectively (\( k_B = \text{Boltzmann constant} \)):
According to equipartition principle, the energy contributed by each translational degree of freedom and rotational degree of freedom at a temperature T are respectively (\( k_B = \text{Boltzmann constant} \)):
Show Hint
- \(\frac{1}{2} k_B T, \frac{1}{2} k_B T\)
- \(k_B T, \frac{1}{2} k_B T\)
- \(k_B T, k_B T\)
- \(\frac{1}{2} k_B T, k_B T\)
- \(\frac{3}{2} k_B T, \frac{1}{2} k_B T\)
The Correct Option is A
Solution and Explanation
According to the equipartition theorem, each degree of freedom contributes \(\frac{1}{2} k_B T\) to the energy of a system at thermal equilibrium.
This applies equally to both translational and rotational degrees of freedom.
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