When a metallic surface is illuminated with monochromatic light of wavelength `lambda` stopping potential for photoelectric current is `3 V_(0)`. When the same metal surface Illuminated with a light of wavelength `2lambda`, the stopping potential is `V_(0)` The threshold wavelength for the surface is

Options

A. `6lambda`
B. `4lambda`
C. `(4lambda)/(3)`
D. `8lambda`

Explanation

To solve the problem, we need to determine the threshold wavelength of the metallic surface based on the given stopping potentials for two different wavelengths of light.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The photoelectric effect states that when light of sufficient energy (frequency) shines on a metal surface, it can eject electrons. The energy of the incident photons can be expressed as:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the incident light.

2. **Setting Up the Equations**:
For a given stopping potential \( V_s \), the energy of the incident photons minus the work function \( \phi \) of the metal gives the kinetic energy of the emitted electrons:
\[ E = \phi + eV_s \]
where \( e \) is the charge of an electron.

3. **Case 1**: For the first wavelength \( \lambda \) with stopping potential \( 3V_0 \):
\[ \frac{hc}{\lambda} = \phi + 3eV_0 \quad \text{(1)} \]

4. **Case 2**: For the second wavelength \( 2\lambda \) with stopping potential \( V_0 \):
\[ \frac{hc}{2\lambda} = \phi + eV_0 \quad \text{(2)} \]

5. **Expressing Work Function**:
From equation (1):
\[ \phi = \frac{hc}{\lambda} - 3eV_0 \]
From equation (2):
\[ \phi = \frac{hc}{2\lambda} - eV_0 \]

6. **Equating the Two Expressions for Work Function**:
Set the two expressions for \( \phi \) equal to each other:
\[ \frac{hc}{\lambda} - 3eV_0 = \frac{hc}{2\lambda} - eV_0 \]

7. **Solving for \( V_0 \)**:
Rearranging gives:
\[ \frac{hc}{\lambda} - \frac{hc}{2\lambda} = 3eV_0 - eV_0 \]
Simplifying:
\[ \frac{hc}{2\lambda} = 2eV_0 \]
Therefore:
\[ eV_0 = \frac{hc}{4\lambda} \]

8. **Substituting Back to Find \( \phi \)**:
Substitute \( eV_0 \) back into either equation to find \( \phi \):
\[ \phi = \frac{hc}{\lambda} - 3\left(\frac{hc}{4\lambda}\right) \]
Simplifying gives:
\[ \phi = \frac{hc}{\lambda} - \frac{3hc}{4\lambda} = \frac{4hc}{4\lambda} - \frac{3hc}{4\lambda} = \frac{hc}{4\lambda} \]

9. **Finding the Threshold Wavelength**:
The threshold wavelength \( \lambda_0 \) is defined by:
\[ \phi = \frac{hc}{\lambda_0} \]
Setting \( \phi = \frac{hc}{4\lambda} \):
\[ \frac{hc}{4\lambda} = \frac{hc}{\lambda_0} \]
Cancelling \( hc \) gives:
\[ \lambda_0 = 4\lambda \]

### Final Answer:
The threshold wavelength for the surface is \( \lambda_0 = 4\lambda \).