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

Explanation

To solve the problem, we need to understand the relationship between the average translational kinetic energy of gas molecules and temperature. The average translational kinetic energy of a molecule in an ideal gas is given by the formula:

\[ KE = \frac{3}{2} k T \]

where:
- \( KE \) is the average translational kinetic energy,
- \( k \) is the Boltzmann constant,
- \( T \) is the absolute temperature.

### Step-by-Step Solution:

1. **Identify Given Information**:
- The average translational kinetic energy of \( O_2 \) at a certain temperature is given as \( 0.048 \, \text{eV} \).
- We need to find the average translational kinetic energy of \( N_2 \) at the same temperature.

2. **Understand the Dependence on Temperature**:
- The average translational kinetic energy is dependent only on the temperature and not on the type of gas or its molar mass. This means that at the same temperature, all gases will have the same average kinetic energy.

3. **Apply the Concept**:
- Since the temperature is the same for both gases, we can conclude that the average translational kinetic energy of \( N_2 \) will be equal to that of \( O_2 \).

4. **Conclusion**:
- Therefore, the average translational kinetic energy of \( N_2 \) is also \( 0.048 \, \text{eV} \).

5. **Select the Correct Option**:
- The correct answer is option (c) \( 0.048 \, \text{eV} \).

### Final Answer:
The translational kinetic energy of \( N_2 \) at the same temperature is \( 0.048 \, \text{eV} \).

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