When a metallic surface is illuminated with monochromatic light of wavelength `lambda`, the stopping potential is `5 V_0`. When the same surface is illuminated with the light of wavelength `3lambda`, the stopping potential is `V_0`. Then, the work function of the metallic surface is

Options

A. `hc//6lambda`
B. `hc//5lambda`
C. `hc//4lambda`
D. `2hc//4lambda`

Explanation

To find the work function of the metallic surface, we can use the photoelectric effect equations based on the stopping potential and the wavelength of the incident light.

### Step-by-Step Solution:

1. **Understanding the Stopping Potential**:
The stopping potential \( V \) is related to the energy of the emitted electrons. The energy of the emitted electrons can be expressed as:
\[ eV = \text{Energy of the photon} - \text{Work function} \]
where \( e \) is the charge of the electron.

2. **Setting Up the Equations**:
For the first condition with wavelength \( \lambda \) and stopping potential \( 5V_0 \):
\[ e \cdot 5V_0 = \frac{hc}{\lambda} - \phi \quad \text{(Equation 1)} \]
where \( \phi \) is the work function of the metal.

For the second condition with wavelength \( 3\lambda \) and stopping potential \( V_0 \):
\[ e \cdot V_0 = \frac{hc}{3\lambda} - \phi \quad \text{(Equation 2)} \]

3. **Expressing the Equations in Terms of Work Function**:
Rearranging both equations to isolate \( \phi \):
- From Equation 1:
\[ \phi = \frac{hc}{\lambda} - 5eV_0 \]
- From Equation 2:
\[ \phi = \frac{hc}{3\lambda} - eV_0 \]

4. **Equating the Two Expressions for Work Function**:
Set the two expressions for \( \phi \) equal to each other:
\[ \frac{hc}{\lambda} - 5eV_0 = \frac{hc}{3\lambda} - eV_0 \]

5. **Solving for \( eV_0 \)**:
Rearranging the equation:
\[ \frac{hc}{\lambda} - \frac{hc}{3\lambda} = 5eV_0 - eV_0 \]
Simplifying the left side:
\[ \frac{hc}{\lambda} - \frac{hc}{3\lambda} = \frac{3hc - hc}{3\lambda} = \frac{2hc}{3\lambda} \]
And simplifying the right side:
\[ 5eV_0 - eV_0 = 4eV_0 \]
Thus, we have:
\[ \frac{2hc}{3\lambda} = 4eV_0 \]

6. **Finding Work Function**:
From the above equation, we can express \( eV_0 \):
\[ eV_0 = \frac{2hc}{12\lambda} = \frac{hc}{6\lambda} \]
Now substituting \( eV_0 \) back into one of the expressions for \( \phi \):
\[ \phi = \frac{hc}{\lambda} - 5\left(\frac{hc}{6\lambda}\right) \]
Simplifying:
\[ \phi = \frac{hc}{\lambda} - \frac{5hc}{6\lambda} = \frac{6hc - 5hc}{6\lambda} = \frac{hc}{6\lambda} \]

### Final Answer:
The work function \( \phi \) of the metallic surface is:
\[ \phi = \frac{hc}{6\lambda} \]