When a metalic surface is illuminated with light of wavelength `lambda,` the stopping potential is V. the same surface is illuminated by light of wavelength `2lambda` the stopping potential is `(V)/(3)` The thershold waelength for the surface is

Options

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

Answer

To solve the problem, we will use the principles of the photoelectric effect. The stopping potential is related to the kinetic energy of the emitted electrons, which is determined by the energy of the incident photons and the work function of the material.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The energy of a photon is given by the equation:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the incident light.

2. **Kinetic Energy and Stopping Potential**:
The kinetic energy (KE) of the emitted electrons can be expressed in terms of the stopping potential \( V \):
\[ KE = eV \]
where \( e \) is the charge of the electron.

3. **First Case (Wavelength \( \lambda \))**:
When the metallic surface is illuminated with light of wavelength \( \lambda \), the stopping potential is \( V \):
\[ eV = \frac{hc}{\lambda} - \phi \]
where \( \phi \) is the work function of the material.

4. **Second Case (Wavelength \( 2\lambda \))**:
When the surface is illuminated with light of wavelength \( 2\lambda \), the stopping potential is \( \frac{V}{3} \):
\[ e\left(\frac{V}{3}\right) = \frac{hc}{2\lambda} - \phi \]

5. **Setting Up the Equations**:
From the first case:
\[ eV = \frac{hc}{\lambda} - \phi \quad \text{(1)} \]
From the second case:
\[ \frac{eV}{3} = \frac{hc}{2\lambda} - \phi \quad \text{(2)} \]

6. **Expressing Work Function \( \phi \)**:
Rearranging equation (1) gives:
\[ \phi = \frac{hc}{\lambda} - eV \]
Substituting this expression for \( \phi \) into equation (2):
\[ \frac{eV}{3} = \frac{hc}{2\lambda} - \left(\frac{hc}{\lambda} - eV\right) \]

7. **Simplifying the Equation**:
Rearranging gives:
\[ \frac{eV}{3} = \frac{hc}{2\lambda} - \frac{hc}{\lambda} + eV \]
This simplifies to:
\[ \frac{eV}{3} = \frac{hc}{2\lambda} - \frac{2hc}{2\lambda} + eV \]
\[ \frac{eV}{3} = -\frac{hc}{2\lambda} + eV \]

8. **Combining Terms**:
Bringing all terms involving \( eV \) to one side:
\[ eV - \frac{eV}{3} = \frac{hc}{2\lambda} \]
\[ \frac{2eV}{3} = \frac{hc}{2\lambda} \]

9. **Finding the Threshold Wavelength**:
Rearranging gives:
\[ \lambda = \frac{hc}{\frac{4eV}{3}} = \frac{3hc}{4eV} \]
The threshold wavelength \( \lambda_0 \) is related to the work function \( \phi \):
\[ \phi = \frac{hc}{\lambda_0} \]
Equating the two expressions for \( \phi \):
\[ \frac{hc}{\lambda_0} = \frac{hc}{\lambda} - eV \]

10. **Final Calculation**:
Solving for \( \lambda_0 \):
\[ \lambda_0 = 4\lambda \]

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