The stopping potential for photoelectrons from a metal surface is `V_(1)` when monochromatic light of frequency `upsilon_(1)` is incident on it. The stopping potential becomes `V_(2)` when monochromatic light of another frequency is incident on the same metal surface. If h be the Planck’s constant and e be the charge of an electron, then the frequency of light in the second case is:

Options

A. (a)`upsilon_(1) - e/h (V_2 + V_1)`
B. (b)`upsilon_(1) + e/h (V_2 + V_1)`
C. (c)`upsilon_(1) - e/h (V_2 - V_1)`
D. (d)`upsilon_(1) + e/h (V_2 - V_1)`

Explanation

To solve the problem, we will use the photoelectric effect equation and the concept of stopping potential. The stopping potential is related to the kinetic energy of the emitted photoelectrons.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The photoelectric effect is described by the equation:
\[ K.E. = h \nu - \phi \]
where \( K.E. \) is the kinetic energy of the emitted electrons, \( h \) is Planck's constant, \( \nu \) is the frequency of the incident light, and \( \phi \) is the work function of the metal.

2. **Relating Stopping Potential to Kinetic Energy**:
The stopping potential \( V \) is related to the kinetic energy by:
\[ K.E. = eV \]
where \( e \) is the charge of an electron. Therefore, we can write:
\[ eV_1 = h \nu_1 - \phi \quad \text{(1)} \]
for the first frequency \( \nu_1 \) and stopping potential \( V_1 \).

3. **Setting Up the Equation for the Second Frequency**:
Similarly, for the second frequency \( \nu_2 \) and stopping potential \( V_2 \):
\[ eV_2 = h \nu_2 - \phi \quad \text{(2)} \]

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

5. **Solving for the Frequency \( \nu_2 \)**:
Rearranging the equation gives:
\[ h \nu_2 = e(V_2 - V_1) + h \nu_1 \]
Dividing by \( h \):
\[ \nu_2 = \frac{e(V_2 - V_1)}{h} + \nu_1 \]

6. **Final Expression**:
Thus, the frequency of light in the second case is:
\[ \nu_2 = \nu_1 + \frac{e(V_2 - V_1)}{h} \]