A diatomic molecule is made of two masses m1 and m2 which are separated by a distance r. If we calcu

Q: Why do we use reduced mass instead of total mass?

Using reduced mass allows us to mathematically simplify a complex two-body rotation around a center of mass into a much simpler equivalent one-body rotation around a fixed axis.

Q: Why is the moment of inertia $I = \mu r^2$?

If you calculate the moment of inertia about the center of mass using $I = m_1 r_1^2 + m_2 r_2^2$ and substitute the distances $r_1$ and $r_2$ in terms of the total bond length $r$, it perfectly simplifies to $\mu r^2$.

Q: Can Bohr's rule be applied to molecules?

Yes! While Bohr originally proposed it for electrons, the quantization of angular momentum is a fundamental quantum principle that applies to any rotating system, including entire molecules.

Q: What is the difference between $h$ and $\hbar$?

$\hbar$ (pronounced h-bar) is simply the reduced Planck's constant, defined as $h / 2\pi$. It is used to make angular momentum equations cleaner.

Q: Why is the energy inversely proportional to $r^2$?

A larger bond length $r$ means a larger moment of inertia (the molecule is harder to spin). For a fixed quantized angular momentum, a larger inertia results in lower rotational kinetic energy.

Question

A diatomic molecule is made of two masses $m_1$ and $m_2$ which are separated by a distance $r$. If we calculate its rotational energy by applying Bohr's rule of angular momentum quantisation, its energy will be given by ($n$ is an integer)

Options:

(A) $\frac{(m_1 + m_2)^2 n^2 \hbar^2}{2 m_1^2 m_2^2 r^2}$

(B) $\frac{n^2 \hbar^2}{2(m_1 + m_2) r^2}$

(C) $\frac{2 n^2 \hbar^2}{(m_1 + m_2) r^2}$

(D) $\frac{(m_1 + m_2) n^2 \hbar^2}{2 m_1 m_2 r^2}$

Sigma Percentile · Accepted Answer

### The Rigid Rotor: A Microscopic Dumbbell

Imagine you are looking at a diatomic molecule—perhaps a molecule of carbon monoxide or hydrogen chloride.

In the realm of classical mechanics, we can visualize this molecule as a tiny, rigid dumbbell. It consists of two atoms with masses $m_1$ and $m_2$, separated by a fixed bond length $r$.

When this molecule rotates in space, it doesn't just spin haphazardly. It rotates around a very specific point: its **center of mass**.

Understanding this rotation is the first crucial step in solving our problem. We need to bridge the gap between classical rotational mechanics and the quantum world.

### The Magic of Reduced Mass

To find the rotational kinetic energy, we first need to determine the **moment of inertia** ($I$) of this two-body system.

Calculating the moment of inertia directly using the distances of each mass from the center of mass ($r_1$ and $r_2$) can be algebraically tedious.

Instead, physicists use a brilliant mathematical trick: the concept of **reduced mass** ($\mu$).

By using the reduced mass, we can mathematically transform our two rotating atoms into a single "effective" particle that rotates at a distance $r$ from a fixed axis.

The reduced mass is defined as:
$$ \mu = \frac{m_1 m_2}{m_1 + m_2} $$

With this elegant substitution, the moment of inertia of the entire molecule simplifies beautifully to:
$$ I = \mu r^2 $$

### Enter Niels Bohr: Quantizing the Rotation

Now, we introduce the quantum twist. The problem asks us to apply **Bohr's rule of angular momentum quantization**.

Niels Bohr originally proposed this rule for electrons orbiting a nucleus, but its fundamental principle applies to any quantum rotation.

Bohr stated that the angular momentum ($L$) cannot take just any continuous value. It must be an integral multiple of the reduced Planck's constant ($\hbar$).

Mathematically, this is expressed as:
$$ L = \frac{nh}{2\pi} = n\hbar $$

Here, $n$ is an integer ($1, 2, 3, \dots$) representing the quantum state of the rotation. This is a profound realization: the molecule can only spin at specific, allowed "gears" or energy levels!

### Synthesizing Mechanics and Quantum Theory

With both our classical moment of inertia and our quantum angular momentum defined, we can now find the **rotational kinetic energy** ($E$).

From classical mechanics, the relationship between kinetic energy, angular momentum, and moment of inertia is given by the master equation:
$$ E = \frac{L^2}{2I} $$

This equation is the rotational analog of the linear kinetic energy formula ($E = p^2 / 2m$).

Now, let's substitute our quantum and classical expressions into this master equation.

Replacing $L$ with $n\hbar$ and $I$ with $\mu r^2$, we get:
$$ E = \frac{(n\hbar)^2}{2(\mu r^2)} $$

Squaring the numerator gives us the raw quantized energy state:
$$ E = \frac{n^2 \hbar^2}{2 \mu r^2} $$

### The Final Algebraic Flourish

We are almost at the finish line. Our current expression is correct, but it is written in terms of the reduced mass $\mu$.

The options provided in the question are written explicitly in terms of the individual masses $m_1$ and $m_2$.

Therefore, we must expand the reduced mass back into its full fractional form. Let's substitute $\mu = \frac{m_1 m_2}{m_1 + m_2}$ back into our energy equation:
$$ E = \frac{n^2 \hbar^2}{2 \left( \frac{m_1 m_2}{m_1 + m_2} \right) r^2} $$

When we divide by a fraction, we multiply by its reciprocal. The term $(m_1 + m_2)$ elegantly flips up into the numerator.

This algebraic rearrangement yields our final, beautiful expression for the quantized rotational energy:
$$ E = \frac{(m_1 + m_2) n^2 \hbar^2}{2 m_1 m_2 r^2} $$

Looking at our options, this matches **Option (d)** perfectly!

### Beyond the Problem: The Birth of Spectroscopy

Why does this result matter? This isn't just an abstract textbook exercise; it is the mathematical foundation of **microwave spectroscopy**.

Because the rotational energy is quantized, a diatomic molecule can only absorb or emit photons of very specific energies when it transitions between these states.

By measuring the exact frequencies of microwave radiation absorbed by a gas, scientists can determine the energy difference $\Delta E$ between rotational levels.

Since Planck's constant ($\hbar$) and the atomic masses ($m_1, m_2$) are known, they can use this exact formula to calculate $r$—the precise bond length between the atoms!

It is a stunning example of how a simple quantum rule allows us to measure the invisible architecture of molecules.

Animated Solution for Physics - Atoms and Nuclei: A diatomic molecule is made of two masses $m_{1}$ and $m_{2}$ which are separated by a distance $r$ . If we calculate its rotational energy by applying Bohr's rule of angular momentum quantisation, its energy will be given by ( $n$ is an integer)

Select Answer:

Visualized Solution

Diatomic Molecule Setup

Moment of Inertia & Reduced Mass $μ$

Bohr's Quantisation Rule

Rotational Kinetic Energy

Substituting $L$ and $I$

Expanding Reduced Mass $μ$

Final Expression

The Way Forward

The Sigma Insight: Bohr's Atomic Model and Energy Levels

The Rigid Rotor

The Magic of Reduced Mass

Enter Niels Bohr

Synthesizing Mechanics and Quantum Theory

The Final Algebraic Flourish

Beyond the Problem

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