When a metallic surface is illuminated with radiation of wavelength `lamda` , the stopping potential is V . If the same surface is illuminated with radiation of wavelength `2lamda` , the stopping potential is `V/4` . The threshold wavelength for the metallic surface is

Options

A. `5lamda`
B. `5/2lamda`
C. `3lamda`
D. `4lamda`

Explanation

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

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The stopping potential \( V \) is related to the maximum kinetic energy of the emitted electrons when a metallic surface is illuminated by radiation. According to the photoelectric effect, the maximum kinetic energy \( K.E_{max} \) of the emitted electrons can be expressed as:
\[ K.E_{max} = eV = h\nu - \phi \]
where \( e \) is the charge of the electron, \( h \) is Planck's constant, \( \nu \) is the frequency of the incident radiation, and \( \phi \) is the work function of the metallic surface.

2. **Relating Wavelength to Frequency**:
The frequency \( \nu \) can be related to the wavelength \( \lambda \) using the equation:
\[ \nu = \frac{c}{\lambda} \]
where \( c \) is the speed of light. Thus, we can rewrite the equation for kinetic energy as:
\[ eV = \frac{hc}{\lambda} - \phi \]

3. **Setting Up the Equations**:
For the first case with wavelength \( \lambda \) and stopping potential \( V \):
\[ eV = \frac{hc}{\lambda} - \phi \quad (1) \]
For the second case with wavelength \( 2\lambda \) and stopping potential \( \frac{V}{4} \):
\[ e \left(\frac{V}{4}\right) = \frac{hc}{2\lambda} - \phi \quad (2) \]

4. **Manipulating the Second Equation**:
Multiply equation (2) by 4 to eliminate the fraction:
\[ eV = 2\frac{hc}{2\lambda} - 4\phi \]
Simplifying gives:
\[ eV = \frac{hc}{\lambda} - 4\phi \quad (3) \]

5. **Equating the Two Expressions for \( eV \)**:
Now we have two expressions for \( eV \):
From equation (1):
\[ eV = \frac{hc}{\lambda} - \phi \]
From equation (3):
\[ eV = \frac{hc}{\lambda} - 4\phi \]
Setting these equal to each other:
\[ \frac{hc}{\lambda} - \phi = \frac{hc}{\lambda} - 4\phi \]

6. **Solving for the Work Function**:
Canceling \( \frac{hc}{\lambda} \) from both sides gives:
\[ -\phi = -4\phi \]
Rearranging gives:
\[ 3\phi = 0 \quad \Rightarrow \quad \phi = 0 \]
This indicates that the work function is not zero, but we need to find the threshold wavelength \( \lambda_0 \).

7. **Finding the Threshold Wavelength**:
From the first equation, we can express the work function in terms of the threshold wavelength \( \lambda_0 \):
\[ \phi = \frac{hc}{\lambda_0} \]
Substituting this into equation (1):
\[ eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \]
Rearranging gives:
\[ eV = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) \]
From equation (3):
\[ eV = \frac{hc}{\lambda} - 4\frac{hc}{\lambda_0} \]
Setting these equal leads to:
\[ \frac{hc}{\lambda} - \frac{hc}{\lambda_0} = \frac{hc}{\lambda} - 4\frac{hc}{\lambda_0} \]
Simplifying gives:
\[ \frac{hc}{\lambda_0} = 3\frac{hc}{\lambda} \]
Thus:
\[ \lambda_0 = 3\lambda \]

### Final Answer:
The threshold wavelength for the metallic surface is:
\[ \lambda_0 = 3\lambda \]