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(D) For a diatomic molecule,the translational degrees of freedom are $f_t = 3$ and the rotational degrees of freedom are $f_r = 2$.According to the eq

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$3/2$

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(D) For a diatomic molecule,the translational degrees of freedom are $f_t = 3$ and the rotational degrees of freedom are $f_r = 2$.According to the equipartition theorem,the average energy associated with each degree of freedom is $\frac{1}{2} k_B T$.Average translational kinetic energy $E_t = f_t 	imes \frac{1}{2} k_B T = 3 	imes \frac{1}{2} k_B T = \frac{3}{2} k_B T$.Average rotational kinetic energy $E_r = f_r 	imes \frac{1}{2} k_B T = 2 	imes \frac{1}{2} k_B T = 1 k_B T$.The ratio of average translational kinetic energy to rotational kinetic energy is $\frac{E_t}{E_r} = \frac{\frac{3}{2} k_B T}{k_B T} = \frac{3}{2}$.

The ratio of average translational kinetic energy to rotational kinetic energy of a diatomic molecule at temperature $T$ is

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