Given below are two statements:Statement I : | vedclass

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(B) According to the equipartition theorem,the average energy associated with each degree of freedom is $\frac{1}{2} k_B T$.For a diatomic molecule,th

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Both Statement $I$ and Statement $II$ are false.

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(B) According to the equipartition theorem,the average energy associated with each degree of freedom is $\frac{1}{2} k_B T$.For a diatomic molecule,the translational degrees of freedom are $3$,so the average translational kinetic energy is $E_{trans} = 3 	imes (\frac{1}{2} k_B T) = \frac{3}{2} k_B T$.The rotational degrees of freedom for a rigid diatomic molecule are $2$,so the average rotational kinetic energy is $E_{rot} = 2 	imes (\frac{1}{2} k_B T) = k_B T$.Statement $I$ is false because rotational energy levels are quantized and do not follow the continuous Maxwell-Boltzmann distribution in the same way as translational energy.Statement $II$ is false because $E_{rot} = k_B T$ while $E_{trans} = \frac{3}{2} k_B T$,so they are not equal.

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