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 quantization it energy will be ( n is an integer )

Options

A. `((m_(1)+m_(2))^(2)n^(2)h^(2))/(2m_(1)^(2)m_(2)^(2)r^(2))`
B. `(n^(2)h^(2))/(2(m_(1)+m_(2))r^(2))`
C. `(2n^(2)h^(2))/((m_(1)+m_(2))r^(2))`
D. `((m_(1)+m_(2))n^(2)h^(2))/(2m_(1)m_(2)r^(2))`

Answer

To find the rotational energy of a diatomic molecule made of two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \), we will apply Bohr's rule of angular momentum quantization. Here’s a step-by-step solution:

### Step 1: Understanding Angular Momentum Quantization
According to Bohr's rule, the angular momentum \( L \) of a rotating system is quantized and given by:
\[ L = n \hbar \]
where \( n \) is an integer and \( \hbar = \frac{h}{2\pi} \) (with \( h \) being Planck's constant).

### Step 2: Relating Angular Momentum to Rotational Energy
The rotational kinetic energy \( E \) of a system can be expressed in terms of its angular momentum:
\[ E = \frac{L^2}{2I} \]
where \( I \) is the moment of inertia of the system.

### Step 3: Calculating the Moment of Inertia
For a diatomic molecule, the moment of inertia \( I \) about the center of mass can be calculated using:
\[ I = m_1 r_1^2 + m_2 r_2^2 \]
where \( r_1 \) and \( r_2 \) are the distances from the center of mass to each mass. The distances can be defined as:
\[ r_1 = \frac{m_2 r}{m_1 + m_2}, \quad r_2 = \frac{m_1 r}{m_1 + m_2} \]

Substituting these values into the moment of inertia formula:
\[ I = m_1 \left(\frac{m_2 r}{m_1 + m_2}\right)^2 + m_2 \left(\frac{m_1 r}{m_1 + m_2}\right)^2 \]

### Step 4: Simplifying the Moment of Inertia
After simplification, we find:
\[ I = \frac{m_1 m_2 r^2}{m_1 + m_2} \]
This is also known as the reduced mass \( \mu \):
\[ \mu = \frac{m_1 m_2}{m_1 + m_2} \]
Thus, we can express the moment of inertia as:
\[ I = \mu r^2 \]

### Step 5: Substituting into the Energy Equation
Now, substituting \( L \) and \( I \) into the energy equation:
\[ E = \frac{(n \hbar)^2}{2I} = \frac{(n \hbar)^2}{2 \mu r^2} \]

### Final Expression for Rotational Energy
Thus, the rotational energy of the diatomic molecule is given by:
\[ E = \frac{n^2 \hbar^2}{2 \mu r^2} \]