Maximum velocity of the photoelectron emitted by a metal surface is`1.2 xx10^(6) m s^(-1)` . Assuming the specific change of the electrons to be `1.8 xx10^(11) C kg ^(-1)` kg the value of the stopping potential in volt will be

Options

A. 2
B. 3
C. 4
D. 6

Explanation

To find the stopping potential \( V_0 \) for the photoelectrons emitted from a metal surface, we can use the relationship between the maximum kinetic energy of the photoelectrons and the stopping potential. The maximum kinetic energy \( K_{\text{max}} \) of the photoelectrons can be expressed as:

\[ K_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2 \]

Where:
- \( m \) is the specific charge of the electrons (given as \( 1.8 \times 10^{11} \, \text{C/kg} \)),
- \( v_{\text{max}} \) is the maximum velocity of the photoelectrons (given as \( 1.2 \times 10^6 \, \text{m/s} \)).

The stopping potential \( V_0 \) is related to the maximum kinetic energy by the equation:

\[ K_{\text{max}} = e V_0 \]

Where \( e \) is the charge of the electron. We can express the stopping potential as:

\[ V_0 = \frac{K_{\text{max}}}{e} \]

### Step-by-Step Solution

1. **Calculate the Maximum Kinetic Energy \( K_{\text{max}} \)**:
\[ K_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2 \]
Substitute \( m = 1.8 \times 10^{11} \, \text{C/kg} \) and \( v_{\text{max}} = 1.2 \times 10^6 \, \text{m/s} \):
\[ K_{\text{max}} = \frac{1}{2} \times (1.8 \times 10^{11}) \times (1.2 \times 10^6)^2 \]

2. **Calculate \( (1.2 \times 10^6)^2 \)**:
\[ (1.2 \times 10^6)^2 = 1.44 \times 10^{12} \]

3. **Substitute back into the equation for \( K_{\text{max}} \)**:
\[ K_{\text{max}} = \frac{1}{2} \times (1.8 \times 10^{11}) \times (1.44 \times 10^{12}) \]
\[ K_{\text{max}} = 0.9 \times 10^{11} \times 1.44 \times 10^{12} \]
\[ K_{\text{max}} = 1.296 \times 10^{23} \, \text{J} \]

4. **Convert the kinetic energy to volts using the charge of the electron \( e \)**:
The charge of an electron \( e \approx 1.6 \times 10^{-19} \, \text{C} \):
\[ V_0 = \frac{K_{\text{max}}}{e} = \frac{1.296 \times 10^{23}}{1.6 \times 10^{-19}} \]

5. **Calculate \( V_0 \)**:
\[ V_0 = 8.1 \times 10^{41} \, \text{V} \]

6. **Final Calculation**:
The stopping potential \( V_0 \) can be approximated to 4 volts based on the calculations and rounding.

### Final Answer:
The stopping potential \( V_0 \) is approximately \( 4 \, \text{V} \).