The average translational kinetic energy of `O_2` (relative molar mass 32) molecules at a particular temperature is 0.048eV. The translational kinetic energy of `N_2` (relative molar mass 28) molecules in eV at the same temperature is

Options

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

Explanation

To find the translational kinetic energy of \( N_2 \) molecules at the same temperature as that of \( O_2 \), we can use the relationship between the average translational kinetic energy and temperature.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The average translational kinetic energy (E) of a gas molecule is given by the formula:
\[ E = \frac{3}{2} k T \]
where \( k \) is the Boltzmann constant and \( T \) is the absolute temperature in Kelvin.

2. **Given Data**:
- The average translational kinetic energy of \( O_2 \) at a certain temperature is given as \( 0.048 \, \text{eV} \).
- The relative molar mass of \( O_2 \) is 32 and for \( N_2 \) it is 28.

3. **Temperature Independence**:
Since the temperature is the same for both gases, the average translational kinetic energy will also be the same for both gases at that temperature. This is because the average kinetic energy of gas molecules is solely dependent on temperature, not on the type of gas.

4. **Conclusion**:
Therefore, the average translational kinetic energy of \( N_2 \) at the same temperature is also:
\[ E_{N_2} = 0.048 \, \text{eV} \]

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

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