The mean rotational kinetic energy of a diatomic molecule at temperature T is :

Options

A. `(1)/(2) kT`
B. `(5)/(2) kT`
C. `k T`
D. `(3)/(4) kT`

Answer

To find the mean rotational kinetic energy of a diatomic molecule at temperature T, we can follow these steps:

### Step 1: Understand the Degrees of Freedom
A diatomic molecule has translational and rotational degrees of freedom. For a diatomic molecule:
- Translational degrees of freedom: 3 (movement in x, y, and z directions)
- Rotational degrees of freedom: 2 (rotation about two axes perpendicular to the bond axis)

### Step 2: Total Degrees of Freedom
The total degrees of freedom (F) for a diatomic molecule is:
\[ F = 3 \text{ (translational)} + 2 \text{ (rotational)} = 5 \]

### Step 3: Average Kinetic Energy Formula
The average kinetic energy per degree of freedom is given by the formula:
\[ \text{Average Kinetic Energy} = \frac{F}{2} k T \]
where \( k \) is the Boltzmann constant and \( T \) is the temperature.

### Step 4: Calculate Mean Rotational Kinetic Energy
Since we are interested in the mean rotational kinetic energy, we only consider the rotational degrees of freedom (F = 2):
\[ \text{Mean Rotational Kinetic Energy} = \frac{2}{2} k T \]
This simplifies to:
\[ \text{Mean Rotational Kinetic Energy} = k T \]

### Conclusion
Thus, the mean rotational kinetic energy of a diatomic molecule at temperature T is:
\[ \text{Mean Rotational Kinetic Energy} = k T \]

### Final Answer
The correct answer is \( k T \).

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