A certain metallic surface is illuminated with monochromatic light of wavelength `lamda` . The stopping potential for photoelectric current for this light is `3V_0` . If the same surface is illuminated with light of wavelength `2 lamda` , the stopping potential is `V_0` . The threshold wavelength for this surface for photoelectric effect is

Options

A. `4lamda`
B. `lamda/4`
C. `lamda/6`
D. `6 lamda`

Explanation

To solve the problem, we need to determine the threshold wavelength for the photoelectric effect on a metallic surface illuminated by light of different wavelengths. Here's a step-by-step solution:

### Step 1: Understand the Photoelectric Effect
The photoelectric effect is described by Einstein's equation:
\[ K.E. = E_{\text{photon}} - \phi \]
where \( K.E. \) is the maximum kinetic energy of the emitted electrons, \( E_{\text{photon}} \) is the energy of the incident photons, and \( \phi \) is the work function of the metal.

### Step 2: Write the Energy of the Photons
The energy of a photon can be expressed in terms of its wavelength \( \lambda \):
\[ E_{\text{photon}} = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant and \( c \) is the speed of light.

### Step 3: Set Up the Equations for Each Wavelength
1. For the first wavelength \( \lambda \) with stopping potential \( 3V_0 \):
\[ e \cdot 3V_0 = \frac{hc}{\lambda} - \phi \]
Rearranging gives:
\[ \frac{hc}{\lambda} = 3eV_0 + \phi \quad \text{(Equation 1)} \]

2. For the second wavelength \( 2\lambda \) with stopping potential \( V_0 \):
\[ e \cdot V_0 = \frac{hc}{2\lambda} - \phi \]
Rearranging gives:
\[ \frac{hc}{2\lambda} = eV_0 + \phi \quad \text{(Equation 2)} \]

### Step 4: Solve the Two Equations
Now we have two equations:
1. \( \frac{hc}{\lambda} = 3eV_0 + \phi \)
2. \( \frac{hc}{2\lambda} = eV_0 + \phi \)

We can subtract Equation 2 from Equation 1:
\[ \frac{hc}{\lambda} - \frac{hc}{2\lambda} = (3eV_0 + \phi) - (eV_0 + \phi) \]
This simplifies to:
\[ \frac{hc}{2\lambda} = 2eV_0 \]
Multiplying both sides by \( 2\lambda \):
\[ hc = 4eV_0 \lambda \]
Thus, we can express \( eV_0 \) in terms of \( \lambda \):
\[ eV_0 = \frac{hc}{4\lambda} \quad \text{(Equation 3)} \]

### Step 5: Substitute Back to Find Work Function
Now substitute \( eV_0 \) back into either Equation 1 or Equation 2 to find \( \phi \). Using Equation 1:
\[ \frac{hc}{\lambda} = 3\left(\frac{hc}{4\lambda}\right) + \phi \]
This gives:
\[ \frac{hc}{\lambda} = \frac{3hc}{4\lambda} + \phi \]
Rearranging gives:
\[ \phi = \frac{hc}{\lambda} - \frac{3hc}{4\lambda} = \frac{hc}{\lambda} \left(1 - \frac{3}{4}\right) = \frac{hc}{4\lambda} \]

### Step 6: Relate Work Function to Threshold Wavelength
The work function \( \phi \) is also given by:
\[ \phi = \frac{hc}{\lambda_0} \]
Setting the two expressions for \( \phi \) equal:
\[ \frac{hc}{4\lambda} = \frac{hc}{\lambda_0} \]
Cancelling \( hc \) gives:
\[ \frac{1}{4\lambda} = \frac{1}{\lambda_0} \]
Thus:
\[ \lambda_0 = 4\lambda \]

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