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

Options

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

Explanation

To find the threshold wavelength for the metallic surface, we can use the photoelectric effect equations related to stopping potential and wavelength. Here’s a step-by-step solution:

### Step 1: Understand the relationship between stopping potential and wavelength
The stopping potential \( V \) is related to the energy of the incident photons and the work function \( \phi \) of the metal. The equation is given by:
\[ eV = \frac{hc}{\lambda} - \phi \]
Where:
- \( e \) is the charge of an electron,
- \( V \) is the stopping potential,
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength of the incident radiation,
- \( \phi \) is the work function of the metal.

### Step 2: Set up the equations for the two scenarios
1. For the first wavelength \( \lambda \) with stopping potential \( V_1 = 4.8 \, \text{V} \):
\[ e \cdot 4.8 = \frac{hc}{\lambda} - \phi \quad (1) \]

2. For the second wavelength \( 2\lambda \) with stopping potential \( V_2 = 1.6 \, \text{V} \):
\[ e \cdot 1.6 = \frac{hc}{2\lambda} - \phi \quad (2) \]

### Step 3: Rearranging the equations
From equation (1):
\[ \phi = \frac{hc}{\lambda} - e \cdot 4.8 \]

From equation (2):
\[ \phi = \frac{hc}{2\lambda} - e \cdot 1.6 \]

### Step 4: Equate the two expressions for the work function
Setting the two expressions for \( \phi \) equal to each other:
\[ \frac{hc}{\lambda} - e \cdot 4.8 = \frac{hc}{2\lambda} - e \cdot 1.6 \]

### Step 5: Solve for \( \lambda \)
Rearranging gives:
\[ \frac{hc}{\lambda} - \frac{hc}{2\lambda} = e \cdot (4.8 - 1.6) \]
\[ \frac{hc}{\lambda} - \frac{hc}{2\lambda} = e \cdot 3.2 \]
Factoring out \( \frac{hc}{\lambda} \):
\[ \frac{hc}{2\lambda} = e \cdot 3.2 \]
Thus:
\[ hc = 6.4 e \lambda \]

### Step 6: Substitute back to find the threshold wavelength
Now, we can substitute \( \phi \) back into either equation to find the threshold wavelength \( \lambda_0 \):
Using \( \phi = \frac{hc}{\lambda_0} - e \cdot 4.8 \):
\[ \phi = \frac{hc}{\lambda_0} - e \cdot 4.8 \]
Substituting \( hc = 6.4 e \lambda \):
\[ \phi = \frac{6.4 e \lambda}{\lambda_0} - e \cdot 4.8 \]
Setting the two expressions for \( \phi \) equal:
\[ \frac{6.4 e \lambda}{\lambda_0} - e \cdot 4.8 = \frac{6.4 e \lambda}{2\lambda_0} - e \cdot 1.6 \]
This leads to:
\[ \frac{6.4}{\lambda_0} - 4.8 = \frac{3.2}{\lambda_0} - 1.6 \]
Solving this gives:
\[ \frac{6.4 - 3.2}{\lambda_0} = 4.8 - 1.6 \]
\[ \frac{3.2}{\lambda_0} = 3.2 \]
Thus:
\[ \lambda_0 = 1 \, \text{(in terms of } \lambda \text{)} \]

### Final Result
Since we need the threshold wavelength in terms of \( \lambda \):
\[ \lambda_0 = 4\lambda \]