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 will use the principles of the photoelectric effect and Einstein's photoelectric equation. The stopping potential is related to the energy of the incident photons and the work function of the metal.

### Step-by-Step Solution:

1. **Understanding Stopping Potential**:
The stopping potential \( V_s \) is related to the energy of the incident photons and the work function \( \phi \) of the metal. According to Einstein's photoelectric equation:
\[ eV_s = E - \phi \]
where \( E \) is the energy of the incident photons given by \( E = \frac{hc}{\lambda} \).

2. **First Case (Wavelength \( \lambda \))**:
For the first case, when the wavelength is \( \lambda \) and the stopping potential is \( 3V_s \):
\[ e(3V_s) = \frac{hc}{\lambda} - \phi \quad \text{(1)} \]

3. **Second Case (Wavelength \( 2\lambda \))**:
For the second case, when the wavelength is \( 2\lambda \) and the stopping potential is \( V_s \):
\[ eV_s = \frac{hc}{2\lambda} - \phi \quad \text{(2)} \]

4. **Setting Up the Equations**:
From equation (1):
\[ 3eV_s = \frac{hc}{\lambda} - \phi \quad \text{(1)} \]
From equation (2):
\[ eV_s = \frac{hc}{2\lambda} - \phi \quad \text{(2)} \]

5. **Subtracting the Equations**:
Now, we can subtract equation (2) from equation (1):
\[ 3eV_s - eV_s = \left(\frac{hc}{\lambda} - \phi\right) - \left(\frac{hc}{2\lambda} - \phi\right) \]
Simplifying this gives:
\[ 2eV_s = \frac{hc}{\lambda} - \frac{hc}{2\lambda} \]
\[ 2eV_s = \frac{hc}{\lambda} - \frac{hc}{2\lambda} = \frac{hc}{2\lambda} \]

6. **Finding \( V_s \)**:
Rearranging gives:
\[ eV_s = \frac{hc}{4\lambda} \quad \text{(3)} \]

7. **Substituting \( V_s \) into Equation (2)**:
Now, substitute \( V_s \) from equation (3) into equation (2):
\[ e\left(\frac{hc}{4\lambda}\right) = \frac{hc}{2\lambda} - \phi \]
This simplifies to:
\[ \frac{ehc}{4\lambda} = \frac{hc}{2\lambda} - \phi \]
Rearranging gives:
\[ \phi = \frac{hc}{2\lambda} - \frac{ehc}{4\lambda} \]
\[ \phi = \frac{hc}{2\lambda} - \frac{hc}{4\lambda} = \frac{2hc}{4\lambda} - \frac{hc}{4\lambda} = \frac{hc}{4\lambda} \]

8. **Finding Threshold Wavelength**:
The work function \( \phi \) is also related to the threshold wavelength \( \lambda_0 \) by:
\[ \phi = \frac{hc}{\lambda_0} \]
Setting the two expressions for \( \phi \) equal gives:
\[ \frac{hc}{4\lambda} = \frac{hc}{\lambda_0} \]
Cancelling \( hc \) from both sides results in:
\[ \frac{1}{4\lambda} = \frac{1}{\lambda_0} \]
Thus, we find:
\[ \lambda_0 = 4\lambda \]

### Final Answer:
The threshold wavelength for photoelectric emission is \( \lambda_0 = 4\lambda \).