Given below are two statements:
Statement I In a diatomic molecule, the rotational energy at a given

Q: Why does a diatomic molecule only have 2 rotational degrees of freedom?

The rotation along the internuclear axis has a negligible moment of inertia, so it doesn't contribute to the thermal energy at normal temperatures.

Q: Does vibrational energy also obey Maxwell's distribution?

Yes, at high temperatures when vibrational modes become active, they also follow the equipartition theorem.

Q: What is $k_B$ in the energy formula?

It is the Boltzmann constant, which relates the average relative kinetic energy of particles in a gas with the thermodynamic temperature of the gas.

Q: Why is Statement I true?

Maxwell's distribution applies to the velocity and energy of particles in thermal equilibrium, covering all active degrees of freedom, including rotation.

Q: Would Statement II be true for a monatomic gas?

A monatomic gas has 0 rotational degrees of freedom, so its rotational energy is zero, which is still not equal to its translational energy ($\frac{3}{2} k_B T$).

Question

Given below are two statements:
Statement I In a diatomic molecule, the rotational energy at a given temperature obeys Maxwell's distribution.
Statement II In a diatomic molecule, the rotational energy at a given temperature equals the translational kinetic energy for each molecule.
In the light of the above statements, choose the correct answer from the options given below.

Options:

(A) Both Statement I and Statement II are true.

(B) Both Statement I and Statement II are false.

(C) Statement I is true but Statement II is false.

(D) Statement I is false but Statement II is true.

Sigma Percentile · Accepted Answer

Imagine you are shrinking down to the atomic level and observing a diatomic molecule, like oxygen ($O_2$) or nitrogen ($N_2$). What does it look like? It resembles a tiny dumbbell—two spherical atoms connected by a rigid chemical bond.

This tiny dumbbell isn't just sitting still; it's constantly moving. It can fly through space, which we call **translational motion**, and it can tumble and spin around, which we call **rotational motion**.

### Maxwell's Equipartition Theorem

Let's tackle **Statement I**. In the world of thermodynamics, there is a beautiful and democratic rule known as **Maxwell's Equipartition Theorem**.

This theorem states that in thermal equilibrium, the total energy of a system is shared equally among all its active modes of motion, or **degrees of freedom**.

Because the theorem applies universally to all active modes, the rotational energy absolutely obeys Maxwell's distribution. Therefore, **Statement I is perfectly true**.

### Counting the Degrees of Freedom

Now, let's investigate **Statement II**. To find the energy, we first need to count the ways our molecule can move.

A rigid diatomic molecule can move along the x, y, and z axes. That gives us **3 translational degrees of freedom** ($f_{	ext{trans}} = 3$).

It can also rotate. However, rotation along the axis connecting the two atoms (the internuclear axis) has a negligible moment of inertia, so it doesn't count. It can only meaningfully rotate about the two axes perpendicular to the bond. This gives us **2 rotational degrees of freedom** ($f_{	ext{rot}} = 2$).

In total, the molecule has $f = 3 + 2 = 5$ degrees of freedom.

### The Energy Showdown

According to the equipartition theorem, every single degree of freedom gets an equal slice of the energy pie, exactly $\frac{1}{2} k_B T$, where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.

Let's calculate the **translational kinetic energy**. Since there are 3 translational modes, we multiply:

$KE_{	ext{trans}} = 3 	imes \frac{1}{2} k_B T = \frac{3}{2} k_B T$

Next, let's calculate the **rotational kinetic energy**. With 2 rotational modes, we get:

$KE_{	ext{rot}} = 2 	imes \frac{1}{2} k_B T = k_B T$

### The Final Verdict

Now, we simply compare the two energies. Is the rotational energy equal to the translational energy?

Clearly, $\frac{3}{2} k_B T$ is not equal to $k_B T$. The translational energy is 50% greater than the rotational energy!

This means **Statement II is completely false**.

Our final conclusion is clear: Statement I is true, but Statement II is false, making option (c) the correct choice. Physics is all about keeping track of the details, and here, counting the degrees of freedom leads us straight to the truth!

Select Answer:

Visualized Solution

$D ia t o mi c M o l ec u l e S e t u p$

$M a x w e l l^{'} s E q u i p a r t i t i o n T h eor e m$

$D e g r eeso f F r ee d o m (f)$

$T r an s l a t i o na l K in e t i c E n er g y$

$R o t a t i o na l K in e t i c E n er g y$

$C o m p a r in g E n er g i es$

The Sigma Insight: Kinetic Theory of Gases

Maxwell's Equipartition Theorem

Counting the Degrees of Freedom

The Energy Showdown

The Final Verdict

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } DiatomicMoleculeSetup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Maxwell′sEquipartitionTheorem

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } DegreesofFreedom(f)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TranslationalKineticEnergy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } RotationalKineticEnergy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ComparingEnergies