The average kinetic energy of `O_(2)` at a particular temperatures is `0.768` eV. The average kinetic energy of `N_(2)` molecules in eV at the same temperature is

Options

A. `0.0015`
B. `0.0030`
C. `0.048`
D. `0.768`

Explanation

To find the average kinetic energy of \( N_2 \) molecules at the same temperature where the average kinetic energy of \( O_2 \) is given as \( 0.768 \, \text{eV} \), we can follow these steps:

### Step 1: Understand the relationship between average kinetic energy and temperature
The average kinetic energy (\( KE \)) of a gas molecule is given by the formula:
\[ KE = \frac{3}{2} k_B T \]
where \( k_B \) is the Boltzmann constant and \( T \) is the absolute temperature. This formula indicates that the average kinetic energy is directly proportional to the temperature.

### Step 2: Identify the given data
We are given:
- Average kinetic energy of \( O_2 \) at a particular temperature \( = 0.768 \, \text{eV} \)
- The temperature is the same for both \( O_2 \) and \( N_2 \).

### Step 3: Apply the proportionality of kinetic energy to temperature
Since the average kinetic energy is directly proportional to temperature, and both gases are at the same temperature, we can conclude that:
\[ KE_{O_2} = KE_{N_2} \]

### Step 4: Calculate the average kinetic energy of \( N_2 \)
Given that the average kinetic energy of \( O_2 \) is \( 0.768 \, \text{eV} \), we can write:
\[ KE_{N_2} = KE_{O_2} = 0.768 \, \text{eV} \]

### Conclusion
Thus, the average kinetic energy of \( N_2 \) molecules at the same temperature is:
\[ \boxed{0.768 \, \text{eV}} \]