The key feature of Bohr's spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid.The rule to be applied is Bohr's quantization condition.
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. `(1)/(n^(2))((h^(2))/(8pi^(2)I))`
B. `(1)/(n)((h^(2))/(8pi^(2)I))`
C. `n((h^(2))/(8pi^(2)I))`
D. `n^(2)((h^(2))/(8pi^(2)I))`

Explanation

To find the quantized rotational energy of a diatomic molecule using Bohr's quantization condition, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Rotational Kinetic Energy**:
The rotational kinetic energy (T) of a diatomic molecule can be expressed as:
\[ T = \frac{1}{2} I \omega^2 \]
where \(I\) is the moment of inertia and \(\omega\) is the angular velocity.

2. **Relate Angular Momentum to Angular Velocity**:
The angular momentum \(L\) of the molecule is given by:
\[ L = I \omega \]
According to Bohr's quantization condition, the angular momentum is quantized and can be expressed as:
\[ L = n \frac{h}{2\pi} \]
where \(n\) is a positive integer (quantum number) and \(h\) is Planck's constant.

3. **Substitute Angular Momentum into the Kinetic Energy Equation**:
From the angular momentum equation, we can express \(\omega\) in terms of \(n\):
\[ \omega = \frac{L}{I} = \frac{n h}{2\pi I} \]
Now, substitute \(\omega\) back into the rotational kinetic energy equation:
\[ T = \frac{1}{2} I \left(\frac{n h}{2\pi I}\right)^2 \]

4. **Simplify the Expression**:
Now, simplify the expression for \(T\):
\[ T = \frac{1}{2} I \cdot \frac{n^2 h^2}{(2\pi I)^2} \]
\[ T = \frac{1}{2} \cdot \frac{n^2 h^2}{4\pi^2 I} \]
\[ T = \frac{n^2 h^2}{8\pi^2 I} \]

5. **Final Result**:
Therefore, the quantized rotational energy of the diatomic molecule in the \(n^{th}\) level is given by:
\[ T_n = \frac{n^2 h^2}{8\pi^2 I} \]