When a metallic surface of threshold wavelength `4lamda` is illuminated with light of wavelength `lamda` the stopping potential is V. When the same surface is illuminated by the light of wavelength `2lamda` Stopping potential is `V/3` Threshold wavelength for the metallic surface is

Options

A. `lamda/2`
B. `2lamda`
C. `lamda/3`
D. `4lamda`

Answer

To solve the problem step by step, we will use the concepts of the photoelectric effect and the relationship between stopping potential, wavelength, and work function.

### Step 1: Understand the given information
We have two scenarios:
1. A metallic surface with a threshold wavelength of \(4\lambda\) is illuminated with light of wavelength \(\lambda\), and the stopping potential is \(V\).
2. The same surface is illuminated with light of wavelength \(2\lambda\), and the stopping potential is \(\frac{V}{3}\).

### Step 2: Write the photoelectric effect equation
The maximum kinetic energy of the emitted electrons can be expressed using the photoelectric effect equation:
\[ KE_{max} = h\nu - \phi \]
where \(h\) is Planck's constant, \(\nu\) is the frequency of the incident light, and \(\phi\) is the work function of the material.

### Step 3: Relate stopping potential to kinetic energy
The stopping potential \(V\) is related to the maximum kinetic energy by:
\[ KE_{max} = eV \]
where \(e\) is the charge of the electron.

### Step 4: Express the work function
The work function \(\phi\) can be expressed in terms of the threshold wavelength \(\lambda_0\):
\[ \phi = \frac{hc}{\lambda_0} \]

### Step 5: Set up the equations for the first scenario
For the first scenario (wavelength \(\lambda\)):
\[ eV = h\nu - \phi \]
Substituting \(\nu = \frac{c}{\lambda}\) and \(\phi = \frac{hc}{4\lambda}\):
\[ eV = \frac{hc}{\lambda} - \frac{hc}{4\lambda} \]
This simplifies to:
\[ eV = hc \left(\frac{1}{\lambda} - \frac{1}{4\lambda}\right) = hc \left(\frac{4 - 1}{4\lambda}\right) = \frac{3hc}{4\lambda} \]

### Step 6: Set up the equations for the second scenario
For the second scenario (wavelength \(2\lambda\)):
\[ \frac{eV}{3} = h\nu - \phi \]
Substituting \(\nu = \frac{c}{2\lambda}\):
\[ \frac{eV}{3} = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0} \]
This simplifies to:
\[ \frac{eV}{3} = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0} \]

### Step 7: Substitute the first equation into the second
From the first scenario, we found:
\[ eV = \frac{3hc}{4\lambda} \]
Substituting this into the second equation:
\[ \frac{1}{3} \cdot \frac{3hc}{4\lambda} = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0} \]
This simplifies to:
\[ \frac{hc}{4\lambda} = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0} \]

### Step 8: Eliminate \(hc\) and rearrange
Dividing through by \(hc\):
\[ \frac{1}{4\lambda} = \frac{1}{2\lambda} - \frac{1}{\lambda_0} \]
Rearranging gives:
\[ \frac{1}{\lambda_0} = \frac{1}{2\lambda} - \frac{1}{4\lambda} \]
Finding a common denominator:
\[ \frac{1}{\lambda_0} = \frac{2 - 1}{4\lambda} = \frac{1}{4\lambda} \]

### Step 9: Solve for \(\lambda_0\)
Taking the reciprocal:
\[ \lambda_0 = 4\lambda \]

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