A diatomic molecule has moment of inertia I. By Bohr's quantization condition its rotational energy

Q: Why can we apply Bohr's model to a molecule instead of just an atom?

Bohr's quantization of angular momentum ($L = \frac{nh}{2\pi}$) is a fundamental quantum rule that applies to any rotating system, not just electrons in atoms. While modern quantum mechanics uses the Schrödinger equation, Bohr's condition gives the correct energy dependence for a simple rigid rotor.

Q: Why is $n=0$ not allowed in this context?

If $n=0$, the molecule would have zero rotational energy and zero angular momentum. In a strict quantum mechanical treatment, a confined system cannot have exactly zero energy due to the Heisenberg uncertainty principle. Even in the simple Bohr theory, the lowest allowed state is typically taken as $n=1$ to represent the ground state of rotation.

Q: What is the difference between this rotating molecule and an electron in a hydrogen atom?

In a hydrogen atom, a single electron orbits a central, massive proton. Here, two masses rotate around their common center of mass. The moment of inertia $I$ effectively replaces the $m_e r^2$ term used for the electron.

Q: Does the moment of inertia $I$ change with the quantum number $n$?

In this specific problem, we assume the molecule acts as a 'rigid rotor'. This means the bond length between the atoms is perfectly constant, regardless of how fast it spins. Therefore, the moment of inertia $I$ is treated as a constant.

Q: How does this quantized energy relate to real-world observations?

Transitions between these specific rotational energy levels give rise to the rotational spectra of molecules. When a molecule jumps from a higher $n$ to a lower $n$, it emits a photon. Because the energy gaps are small, these photons are typically in the microwave region of the electromagnetic spectrum.

Question

A diatomic molecule has moment of inertia $I$. By Bohr's quantization condition its rotational energy in the $n$th level ($n = 0$ is not allowed) is

Options:

(A) $\frac{1}{n^2}\left(\frac{h^2}{8\pi^2 I}\right)$

(B) $\frac{1}{n}\left(\frac{h^2}{8\pi^2 I}\right)$

(C) $n\left(\frac{h^2}{8\pi^2 I}\right)$

(D) $n^2\left(\frac{h^2}{8\pi^2 I}\right)$

Sigma Percentile · Accepted Answer

### The Universality of Bohr's Quantization

When we first learn about Bohr's atomic model, it is usually in the strict context of a hydrogen atom—an electron orbiting a central proton. However, the true beauty of Bohr's postulates lies in their universality. The quantization of angular momentum is not just a quirk of electrons; it is a fundamental rule of the quantum realm that applies to any rotating system.

In this problem, we are asked to apply Bohr's quantization condition to a macroscopic-like system: a rotating diatomic molecule. Imagine a tiny dumbbell spinning end-over-end. Classically, this molecule could spin at any speed, possessing any arbitrary amount of rotational energy. But quantum mechanics tells a different story.

### The Master Equation: Angular Momentum

Let's start by defining the angular momentum of our rotating molecule. From classical mechanics, we know that the angular momentum $L$ of a rigid rotor is the product of its moment of inertia $I$ and its angular velocity $\omega$:

$$L = I\omega$$

Now, we inject the quantum rule. Bohr's second postulate states that the angular momentum must be an integral multiple of $\frac{h}{2\pi}$ (often written as $\hbar$). Therefore, we can write:

$$L = \frac{nh}{2\pi}$$

where $n$ is the principal quantum number ($n = 1, 2, 3, \dots$).

### Bridging the Classical and the Quantum

By equating our classical expression with the quantum condition, we can find the allowed, discrete values for the angular velocity $\omega$:

$$I\omega = \frac{nh}{2\pi}$$

$$\omega = \frac{nh}{2\pi I}$$

This equation is profound. It tells us that the molecule cannot spin at just any rate. It can only spin at specific, quantized angular velocities dictated by the integer $n$.

### Calculating the Rotational Energy

Our ultimate goal is to find the rotational kinetic energy $K$ of the molecule. The classical formula for rotational kinetic energy is:

$$K = \frac{1}{2}I\omega^2$$

Now, we simply substitute our quantized expression for $\omega$ into this energy equation:

$$K = \frac{1}{2}I\left(\frac{nh}{2\pi I}ight)^2$$

Let's carefully expand the squared term:

$$K = \frac{1}{2}I\left(\frac{n^2h^2}{4\pi^2 I^2}ight)$$

Notice that one factor of $I$ in the numerator cancels with one in the denominator. Multiplying the constants in the denominator ($2 	imes 4 = 8$), we arrive at our final, elegant expression for the quantized rotational energy:

$$K = \frac{n^2h^2}{8\pi^2 I}$$

Rewriting this to match the format of the options:

$$K = n^2\left(\frac{h^2}{8\pi^2 I}ight)$$

This perfectly matches option (d). The rotational energy levels of a diatomic molecule are proportional to the square of the quantum number $n$. This quadratic spacing is a hallmark of the rigid rotor model in quantum mechanics and forms the basis for understanding microwave rotational spectroscopy!

Animated Solution for Physics - Atoms and Nuclei: A diatomic molecule has moment of inertia $I$ . By Bohr's quantization condition its rotational energy in the $n$ th level ( $n = 0$ is not allowed) is

Select Answer:

Visualized Solution

Visualizing the Rotating Molecule

Bohr's Quantization Condition

Classical Angular Momentum

Finding Angular Velocity

Rotational Kinetic Energy

Substituting $ω$

Final Simplification

Conclusion

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

The Universality of Bohr's Quantization

The Master Equation

Bridging the Classical and the Quantum

Calculating the Rotational Energy

Similar Questions

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)

Comprehension Passage

A diatomic molecule has moment of inertia $I$ . By Bohr's quantization condition its rotational energy in the $n^{th}$ level ( $n = 0$ is not allowed) is

It is found that the excitation frequency from ground to the first excited state of rotation for the CO molecule is close to $\frac{4}{π} \times 1 0^{11} Hz$ . Then the moment of inertia of CO molecule about its centre of mass is close to (Take $h = 2 π \times 1 0^{- 34} J-s$ )

In a CO molecule, the distance between C (mass $= 12 amu$ ) and O (mass $= 16 amu$ ), where $1 amu = \frac{5}{3} \times 1 0^{- 27} kg$ , is close to

A particle of mass $m$ moves in a circular orbit in a central potential field $U (r) = \frac{1}{2} k r^{2}$ . If Bohr's quantization conditions are applied, radii of possible orbitals and energy levels vary with quantum number $n$ as

A particle of mass $m$ moves in a circular orbit in a central potential field $U (r) = U_{0} r^{4}$ . If Bohr's quantisation conditions are applied, radii of possible orbitals $r_{n}$ vary with $n^{1/ α}$ , where $α$ is .................