When a piece of metal is illuminated by monochromatic light of wavelength `lambda,` then stopping potential is `3V_(s)`. When the same surface is illuminated by the light of wavelength `2lambda`, then stopping potential becomes `V_(s)`. The value of threshold wavelength for photoelectric emission will be

Options

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

Answer

To solve the problem, we need to find the threshold wavelength (\( \lambda_0 \)) for photoelectric emission based on the given stopping potentials for two different wavelengths of light.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The stopping potential (\( V_s \)) is related to the energy of the incident photons and the work function of the metal. The equation representing the photoelectric effect is:
\[ eV_s = \frac{hc}{\lambda} - \phi \]
where \( \phi \) is the work function of the metal, \( e \) is the charge of an electron, \( h \) is Planck's constant, and \( c \) is the speed of light.

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

3. **Rearranging the Equations**:
Rearranging equation (1):
\[ 3eV_s + \phi = \frac{hc}{\lambda} \]
Rearranging equation (2):
\[ eV_s + \phi = \frac{hc}{2\lambda} \]

4. **Eliminating the Work Function**:
Now, we can eliminate \( \phi \) by subtracting equation (2) from equation (1):
\[ (3eV_s + \phi) - (eV_s + \phi) = \frac{hc}{\lambda} - \frac{hc}{2\lambda} \]
Simplifying this gives:
\[ 2eV_s = \frac{hc}{\lambda} - \frac{hc}{2\lambda} \]
\[ 2eV_s = \frac{hc}{\lambda} - \frac{hc}{2\lambda} = \frac{hc - \frac{hc}{2}}{\lambda} = \frac{hc/2}{\lambda} \]
Therefore:
\[ 2eV_s = \frac{hc}{2\lambda} \]

5. **Solving for Stopping Potential**:
Rearranging gives us:
\[ eV_s = \frac{hc}{4\lambda} \]

6. **Substituting Back**:
Now, substituting \( eV_s \) back into equation (2):
\[ eV_s + \phi = \frac{hc}{2\lambda} \]
Substituting \( eV_s = \frac{hc}{4\lambda} \):
\[ \frac{hc}{4\lambda} + \phi = \frac{hc}{2\lambda} \]
Rearranging gives:
\[ \phi = \frac{hc}{2\lambda} - \frac{hc}{4\lambda} = \frac{hc}{4\lambda} \]

7. **Finding the Threshold Wavelength**:
The work function \( \phi \) can also be expressed in terms of the threshold wavelength \( \lambda_0 \):
\[ \phi = \frac{hc}{\lambda_0} \]
Setting the two expressions for \( \phi \) equal:
\[ \frac{hc}{\lambda_0} = \frac{hc}{4\lambda} \]
Canceling \( hc \) from both sides:
\[ \frac{1}{\lambda_0} = \frac{1}{4\lambda} \]
Thus:
\[ \lambda_0 = 4\lambda \]

### Final Answer:
The value of the threshold wavelength for photoelectric emission is:
\[ \lambda_0 = 4\lambda \]