At a given temperature, rotational kinetic energy of diatomic gas is `K_(0)`. Find its translational and total kinetic energy.

Explanation

To solve the problem, we need to find the translational kinetic energy and the total kinetic energy of a diatomic gas given its rotational kinetic energy \( K_0 \).

### Step-by-Step Solution:

1. **Understanding Degrees of Freedom**:
- A diatomic gas has 5 degrees of freedom: 3 for translational motion (movement in x, y, z directions) and 2 for rotational motion (rotation about two axes).

2. **Relating Translational and Rotational Kinetic Energy**:
- The translational kinetic energy \( K_t \) and the rotational kinetic energy \( K_r \) are related by the ratio of their degrees of freedom. For a diatomic gas:
\[ \frac{K_t}{K_r} = \frac{3}{2} \]
- Given that the rotational kinetic energy \( K_r = K_0 \), we can express the translational kinetic energy as:
\[ K_t = \frac{3}{2} K_r = \frac{3}{2} K_0 \]

3. **Calculating Total Kinetic Energy**:
- The total kinetic energy \( K_{total} \) is the sum of the translational and rotational kinetic energies:
\[ K_{total} = K_t + K_r \]
- Substituting the values we have:
\[ K_{total} = \frac{3}{2} K_0 + K_0 \]
- To combine the terms, we can express \( K_0 \) as \( \frac{2}{2} K_0 \):
\[ K_{total} = \frac{3}{2} K_0 + \frac{2}{2} K_0 = \frac{5}{2} K_0 \]

### Final Answers:
- **Translational Kinetic Energy**: \( K_t = \frac{3}{2} K_0 \)
- **Total Kinetic Energy**: \( K_{total} = \frac{5}{2} K_0 \)