When a centimeter thick surface is illuminated with light of wavelength `lamda`, the stopping potential is V. When the same surface is illuminated by light of wavelength `2lamda`, the stopping potential is `(V)/(3)`. Threshold wavelength for the metallic surface is

Options

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

Explanation

To solve the problem, we need to use the photoelectric effect equations and the information provided about the stopping potentials for two different wavelengths of light.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The kinetic energy (KE) of the emitted photoelectrons can be expressed as:
\[ KE = E_{\text{incident}} - \phi \]
where \(E_{\text{incident}}\) is the energy of the incident photons, and \(\phi\) is the work function of the material.

2. **Energy of Incident Photons**:
The energy of a photon can be calculated using the formula:
\[ E_{\text{incident}} = \frac{hc}{\lambda} \]
where \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength of the light.

3. **First Case (Wavelength \(\lambda\))**:
When the surface is illuminated with light of wavelength \(\lambda\), the stopping potential is \(V\). Thus, we can write:
\[ eV = \frac{hc}{\lambda} - \phi \]
Rearranging gives:
\[ \phi = \frac{hc}{\lambda} - eV \quad \text{(Equation 1)} \]

4. **Second Case (Wavelength \(2\lambda\))**:
When the surface is illuminated with light of wavelength \(2\lambda\), the stopping potential is \(\frac{V}{3}\). Thus, we can write:
\[ e\left(\frac{V}{3}\right) = \frac{hc}{2\lambda} - \phi \]
Rearranging gives:
\[ \phi = \frac{hc}{2\lambda} - \frac{eV}{3} \quad \text{(Equation 2)} \]

5. **Setting Equations Equal**:
Since both equations represent the work function \(\phi\), we can set them equal to each other:
\[ \frac{hc}{\lambda} - eV = \frac{hc}{2\lambda} - \frac{eV}{3} \]

6. **Clearing the Equation**:
To eliminate the fractions, we can multiply through by \(6\lambda\):
\[ 6hc - 6eV\lambda = 3hc - 2eV \]

7. **Rearranging Terms**:
Rearranging gives:
\[ 6hc - 3hc = 6eV\lambda - 2eV \]
Simplifying leads to:
\[ 3hc = (6\lambda - 2)eV \]

8. **Solving for \(\lambda_0\)**:
We know that the work function can also be expressed in terms of the threshold wavelength \(\lambda_0\):
\[ \phi = \frac{hc}{\lambda_0} \]
Setting this equal to either expression for \(\phi\) allows us to find \(\lambda_0\). Using Equation 1:
\[ \frac{hc}{\lambda_0} = \frac{hc}{\lambda} - eV \]
Rearranging gives:
\[ \frac{hc}{\lambda_0} + eV = \frac{hc}{\lambda} \]

9. **Finding the Threshold Wavelength**:
From the previous steps, we can derive:
\[ \frac{1}{\lambda_0} = \frac{1}{\lambda} - \frac{eV}{hc} \]
Using the relationship from the stopping potential equations, we can find:
\[ \lambda_0 = 4\lambda \]

### Final Answer:
The threshold wavelength for the metallic surface is:
\[ \lambda_0 = 4\lambda \]