When radiation of the wavelength `lamda`is incident on a metallic surface ,the stopping potential is 4.8 V. If the same surface is illuminated with radiation of double the wavelength ,then the stopping potential becomes 1.6 V.Then the threshold wavelength for the surface is

Options

A. `2lamda`
B. `4lamda`
C. `6lamda`
D. `8lamda`

Answer

To solve the problem, we will use the concepts from the photoelectric effect and Einstein's photoelectric equation.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two scenarios:
- When radiation of wavelength \( \lambda \) is incident, the stopping potential \( V_0 = 4.8 \, \text{V} \).
- When radiation of wavelength \( 2\lambda \) is incident, the stopping potential \( V_0 = 1.6 \, \text{V} \).

2. **Using Einstein's Photoelectric Equation**:
The maximum kinetic energy (K.E.) of the emitted electrons can be expressed using the stopping potential:
\[ K.E. = e V_0 \]
where \( e \) is the charge of the electron.

3. **Setting Up the Equations**:
From Einstein's photoelectric equation:
\[ K.E. = \frac{hc}{\lambda} - \phi \]
where \( \phi \) is the work function of the metal.

For the first case (wavelength \( \lambda \)):
\[ e \cdot 4.8 = \frac{hc}{\lambda} - \phi \quad \text{(1)} \]

For the second case (wavelength \( 2\lambda \)):
\[ e \cdot 1.6 = \frac{hc}{2\lambda} - \phi \quad \text{(2)} \]

4. **Dividing the Two Equations**:
Dividing equation (1) by equation (2):
\[ \frac{e \cdot 4.8}{e \cdot 1.6} = \frac{\frac{hc}{\lambda} - \phi}{\frac{hc}{2\lambda} - \phi} \]
Simplifying the left side:
\[ 3 = \frac{\frac{hc}{\lambda} - \phi}{\frac{hc}{2\lambda} - \phi} \]

5. **Cross-Multiplying**:
Cross-multiplying gives:
\[ 3\left(\frac{hc}{2\lambda} - \phi\right) = \frac{hc}{\lambda} - \phi \]
Expanding this:
\[ \frac{3hc}{2\lambda} - 3\phi = \frac{hc}{\lambda} - \phi \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{3hc}{2\lambda} + 2\phi = \frac{hc}{\lambda} \]
Subtracting \( \frac{3hc}{2\lambda} \) from both sides:
\[ 2\phi = \frac{hc}{\lambda} - \frac{3hc}{2\lambda} \]
This simplifies to:
\[ 2\phi = \frac{hc}{2\lambda} \]

7. **Finding the Work Function**:
Thus, we have:
\[ \phi = \frac{hc}{4\lambda} \]

8. **Relating Work Function to Threshold Wavelength**:
The work function is also given by:
\[ \phi = \frac{hc}{\lambda_t} \]
Setting the two expressions for \( \phi \) equal to each other:
\[ \frac{hc}{4\lambda} = \frac{hc}{\lambda_t} \]

9. **Solving for Threshold Wavelength**:
Canceling \( hc \) from both sides:
\[ \frac{1}{4\lambda} = \frac{1}{\lambda_t} \]
Thus, we find:
\[ \lambda_t = 4\lambda \]

### Final Answer:
The threshold wavelength \( \lambda_t \) for the surface is \( 4\lambda \).