The average rotational kinetic energy of hydrogen molecule at a temperature `T` is `E`. The average translational kinetic energy of helium at same temperature will be:

Options

A. `(2E)/(3)`
B. `(5E)/(3)`
C. `E`
D. `(3E)/(2)`

Explanation

To find the average translational kinetic energy of helium at the same temperature \( T \) given that the average rotational kinetic energy of a hydrogen molecule at temperature \( T \) is \( E \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Degrees of Freedom**:
- For a diatomic molecule like hydrogen (H₂), the average rotational kinetic energy is associated with its rotational degrees of freedom. A diatomic molecule has 2 rotational degrees of freedom.
- The average rotational kinetic energy \( E \) can be expressed as:
\[ E = 2 \times \left(\frac{1}{2} kT\right) = kT \]
- Here, \( k \) is the Boltzmann constant and \( T \) is the absolute temperature.

2. **Relating \( kT \) to \( E \)**:
- From the above equation, we can conclude that:
\[ kT = E \]

3. **Translational Kinetic Energy of Helium**:
- Helium (He) is a monatomic gas, and for a monatomic gas, the average translational kinetic energy is associated with its translational degrees of freedom. A monatomic gas has 3 translational degrees of freedom.
- The average translational kinetic energy \( K \) can be expressed as:
\[ K = 3 \times \left(\frac{1}{2} kT\right) = \frac{3}{2} kT \]

4. **Substituting \( kT \) with \( E \)**:
- Now, substituting \( kT \) from step 2 into the equation for translational kinetic energy:
\[ K = \frac{3}{2} E \]

5. **Final Answer**:
- Therefore, the average translational kinetic energy of helium at the same temperature \( T \) is:
\[ K = \frac{3E}{2} \]

### Conclusion:
The average translational kinetic energy of helium at temperature \( T \) is \( \frac{3E}{2} \).