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

Options

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

Answer

To solve the problem, we will use the photoelectric equation and the information provided about the stopping potentials for different wavelengths of light.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The photoelectric effect can be described by the equation:
\[ K.E. = h\nu - \phi \]
where \( K.E. \) is the kinetic energy of the emitted electrons, \( h \) is Planck's constant, \( \nu \) is the frequency of the incident light, and \( \phi \) is the work function of the metal.

2. **Relating Frequency and Wavelength**:
The frequency \( \nu \) is related to the wavelength \( \lambda \) by the equation:
\[ \nu = \frac{c}{\lambda} \]
where \( c \) is the speed of light.

3. **Setting Up the Equations**:
For the first case with wavelength \( \lambda \) and stopping potential \( 3V_0 \):
\[ e \cdot 3V_0 = h \frac{c}{\lambda} - \phi \tag{1} \]

For the second case with wavelength \( 2\lambda \) and stopping potential \( V_0 \):
\[ e \cdot V_0 = h \frac{c}{2\lambda} - \phi \tag{2} \]

4. **Rearranging the Equations**:
Rearranging both equations to express the work function \( \phi \):
From equation (1):
\[ \phi = h \frac{c}{\lambda} - 3eV_0 \tag{3} \]

From equation (2):
\[ \phi = h \frac{c}{2\lambda} - eV_0 \tag{4} \]

5. **Equating the Two Expressions for Work Function**:
Set equations (3) and (4) equal to each other:
\[ h \frac{c}{\lambda} - 3eV_0 = h \frac{c}{2\lambda} - eV_0 \]

6. **Solving for \( \lambda \)**:
Rearranging gives:
\[ h \frac{c}{\lambda} - h \frac{c}{2\lambda} = 3eV_0 - eV_0 \]
Simplifying:
\[ h \frac{c}{\lambda} - \frac{h c}{2\lambda} = 2eV_0 \]
\[ \frac{h c}{2\lambda} = 2eV_0 \]
\[ h c = 4eV_0 \lambda \]
\[ \lambda = \frac{h c}{4eV_0} \tag{5} \]

7. **Finding the Threshold Wavelength**:
The threshold wavelength \( \lambda_0 \) is given by:
\[ \lambda_0 = \frac{h c}{\phi} \]
From equation (4), substituting for \( \phi \):
\[ \phi = h \frac{c}{2\lambda} - eV_0 \]
Thus, substituting back into the threshold wavelength equation gives:
\[ \phi = \frac{hc}{2\lambda_0} \]
Setting \( \phi = \frac{hc}{\lambda_0} \) leads us to:
\[ \lambda_0 = 4\lambda \]

### Final Answer:
The threshold wavelength of this surface for the photoelectric effect is:
\[ \lambda_0 = 4\lambda \]