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NCERT Exemplar · Q11

Q.In a diatomic molecule, the rotational energy at a given temperature (Note: more than one of the given options may be correct.)

(a) obeys Maxwell's distribution.
(b) have the same value for all molecules.
(c) equals the translational kinetic energy for each molecule.
(d) is (2/3)rd the translational kinetic energy for each molecule.
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By the equipartition theorem, a diatomic molecule at temperature TT has average translational KE =32kBT=\tfrac32k_BT (3 degrees of freedom) and average rotational KE =kBT=k_BT (2 degrees of freedom), so rotational energy is 23\tfrac23 of translational energy on average -- option (D). Because individual molecular energies are statistically spread out (the rotational analogue of the Maxwell speed distribution), option (A) is also correct. The correct answers are (A) and (D).

Setting up the physics

A diatomic molecule has two independent kinds of motion at ordinary temperatures: translational motion of its centre of mass (3 degrees of freedom) and rotational motion about the two axes perpendicular to the bond (2 degrees of freedom -- spin about the bond axis itself is negligible, since the moment of inertia there is tiny).

The equipartition theorem states that in thermal equilibrium at temperature TT, each quadratic degree of freedom carries an average energy of 12kBT\tfrac12k_BT.

Computing the average energies

  • Translational: 3 degrees of freedom, so ⟨Etrans⟩=3×12kBT=32kBT\langle E_{\text{trans}}\rangle = 3\times\frac12 k_BT = \frac32 k_BT
  • Rotational: 2 degrees of freedom, so ⟨Erot⟩=2×12kBT=kBT\langle E_{\text{rot}}\rangle = 2\times\frac12 k_BT = k_BT

Taking the ratio:

⟨Erot⟩⟨Etrans⟩=kBT32kBT=23\frac{\langle E_{\text{rot}}\rangle}{\langle E_{\text{trans}}\rangle} = \frac{k_BT}{\tfrac32k_BT} = \frac23

So the average rotational energy is exactly 23\tfrac23 of the average translational energy -- this is option (D).

Why option (A) is also correct …

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