The average translational kinetic energy of `N_2` gas molecules at _________ `""^@C` becomes equal to the K.E. of an electron accelerated from rest through a potential difference of 0.1 volt. (Given `k_B = 1.38 xx 10^(-23) J//K`) (Fill the nearest integer).

Answer

To solve the problem, we need to find the temperature at which the average translational kinetic energy of \( N_2 \) gas molecules equals the kinetic energy of an electron accelerated through a potential difference of 0.1 volts.

### Step-by-Step Solution:

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

2. **Kinetic Energy of the Electron:**
The kinetic energy (K.E.) of an electron accelerated from rest through a potential difference \( V \) is given by:
\[ K.E. = eV \]
where \( e \) is the charge of the electron (\( e = 1.6 \times 10^{-19} \) C) and \( V = 0.1 \) V.

3. **Calculating the Kinetic Energy of the Electron:**
Substituting the values, we find:
\[ K.E. = 1.6 \times 10^{-19} \times 0.1 = 1.6 \times 10^{-20} \text{ Joules} \]

4. **Setting the Two Energies Equal:**
Now, we set the average translational kinetic energy of the \( N_2 \) gas equal to the kinetic energy of the electron:
\[ \frac{3}{2} k_B T = 1.6 \times 10^{-20} \]

5. **Substituting the Value of \( k_B \):**
We know \( k_B = 1.38 \times 10^{-23} \text{ J/K} \). Substituting this into the equation gives:
\[ \frac{3}{2} (1.38 \times 10^{-23}) T = 1.6 \times 10^{-20} \]

6. **Solving for Temperature \( T \):**
Rearranging the equation to solve for \( T \):
\[ T = \frac{1.6 \times 10^{-20} \times 2}{3 \times 1.38 \times 10^{-23}} \]
\[ T = \frac{3.2 \times 10^{-20}}{4.14 \times 10^{-23}} \approx 773.0 \text{ K} \]

7. **Converting Kelvin to Celsius:**
To convert from Kelvin to Celsius, we use the formula:
\[ T(°C) = T(K) - 273 \]
Thus,
\[ T(°C) = 773 - 273 = 500 °C \]

### Final Answer:
The average translational kinetic energy of \( N_2 \) gas molecules at **500 °C** becomes equal to the K.E. of an electron accelerated from rest through a potential difference of 0.1 volt.