The average translational kinetic energy of `O_(2)` (molar mass 32) molecules at a particular temperature is `0.048 eV`. The translational kinetic energy of `N_(2)` (molar mass 28) molecules in (eV) at the same temperature is (JEE 1997)
(a) 0.0015 (b) 0.003 ( c) 0.048 (d) 0.768

Options

A. `0.0015`
B. `0.003`
C. `0.048`
D. `0.768`

Answer

To solve the problem, we need to determine the translational kinetic energy of nitrogen (N₂) molecules at the same temperature where the translational kinetic energy of oxygen (O₂) molecules is given as 0.048 eV.

### Step-by-Step Solution:

1. **Understanding Kinetic Energy Relation**:
The average translational kinetic energy (KE) of gas molecules is given by the formula:
\[ KE = \frac{3}{2} k T \]
where \( k \) is the Boltzmann constant and \( T \) is the absolute temperature. This formula indicates that the kinetic energy is directly proportional to the temperature.

2. **Given Data**:
- For O₂ (molar mass = 32), the average translational kinetic energy at temperature \( T \) is:
\[ KE_{O_2} = 0.048 \text{ eV} \]

3. **Temperature Dependency**:
Since the average translational kinetic energy is directly proportional to temperature and we are considering the same temperature \( T \) for both gases, we can conclude that the kinetic energy for any gas at a given temperature will be the same regardless of the molar mass of the gas.

4. **Applying the Concept to N₂**:
Therefore, at the same temperature \( T \), the translational kinetic energy for nitrogen (N₂) will also be:
\[ KE_{N_2} = KE_{O_2} = 0.048 \text{ eV} \]

5. **Conclusion**:
Thus, the translational kinetic energy of nitrogen molecules at the same temperature is:
\[ KE_{N_2} = 0.048 \text{ eV} \]

### Final Answer:
The translational kinetic energy of \( N_2 \) at the same temperature is **0.048 eV** (Option C).