A metallic surface is irradiated by a monochromatic light of frequency `v_(1)` and stopping potential is found to be `V_(1)`. If the light of frequency `v_(2)` irradiates the surface, the stopping potential will be

Options

A. `V_(1)+(h)/(e)(v_(1)+v_(2))`
B. `V_(1)+(h)/(e)(v_(2)-v_(1))`
C. `V_(1)+(e)/(h)(v_(2)-v_(1))`
D. `V_(1)-(h)/(e)(v_(1)+v_(2))`

Explanation

To solve the problem, we need to derive the relationship between the stopping potentials \( V_1 \) and \( V_2 \) for two different frequencies of light \( \nu_1 \) and \( \nu_2 \) incident on a metallic surface. We will use Einstein's photoelectric equation and the concept of stopping potential.

### Step-by-Step Solution:

1. **Understanding Stopping Potential**:
The stopping potential \( V \) is related to the maximum kinetic energy of the emitted photoelectrons. The kinetic energy \( K \) can be expressed as:
\[ K = eV \]
where \( e \) is the charge of the electron and \( V \) is the stopping potential.

2. **Applying Einstein's Photoelectric Equation**:
According to Einstein's photoelectric equation, the energy of the incident photons is given by:
\[ E = h\nu = \phi + K \]
where \( \phi \) is the work function of the metal.

3. **Setting Up Equations for Two Frequencies**:
For the first frequency \( \nu_1 \):
\[ h\nu_1 = \phi + eV_1 \quad \text{(1)} \]
For the second frequency \( \nu_2 \):
\[ h\nu_2 = \phi + eV_2 \quad \text{(2)} \]

4. **Subtracting the Two Equations**:
To eliminate the work function \( \phi \), we subtract equation (2) from equation (1):
\[ h\nu_1 - h\nu_2 = (eV_1 - eV_2) \]
This simplifies to:
\[ h(\nu_1 - \nu_2) = e(V_1 - V_2) \]

5. **Rearranging for \( V_2 \)**:
Now, we can rearrange the equation to solve for \( V_2 \):
\[ V_2 = V_1 + \frac{h}{e}(\nu_1 - \nu_2) \]

6. **Final Expression**:
The final expression for the stopping potential \( V_2 \) when light of frequency \( \nu_2 \) irradiates the surface is:
\[ V_2 = V_1 + \frac{h}{e}(\nu_1 - \nu_2) \]