When a metallic surface is illuminated with monochromatic light of wavelength `lambda`, the stopping potential is `5 V_0`. When the same surface is illuminated with the light of wavelength `3lambda`, the stopping potential is `V_0`. Then, the work function of the metallic surface is

Options

A. `hc//6lambda`
B. `hc//5lambda`
C. `hc//4lambda`
D. `2hc//4lambda`

Explanation

To find the work function of the metallic surface, we can follow these steps:

### Step 1: Understand the photoelectric equation
The photoelectric effect can be described by the equation:
\[ K_{\text{max}} = E - \phi \]
where:
- \( K_{\text{max}} \) is the maximum kinetic energy of the emitted electrons,
- \( E \) is the energy of the incident photons,
- \( \phi \) is the work function of the metal.

The energy of the incident photons can be expressed as:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength of the incident light.

### Step 2: Set up the equations for the two scenarios
1. For the first case with wavelength \( \lambda \) and stopping potential \( 5V_0 \):
\[ K_{\text{max}} = e \cdot 5V_0 \]
Thus, we have:
\[ e \cdot 5V_0 = \frac{hc}{\lambda} - \phi \]
Rearranging gives:
\[ \phi = \frac{hc}{\lambda} - e \cdot 5V_0 \quad \text{(Equation 1)} \]

2. For the second case with wavelength \( 3\lambda \) and stopping potential \( V_0 \):
\[ K_{\text{max}} = e \cdot V_0 \]
Thus, we have:
\[ e \cdot V_0 = \frac{hc}{3\lambda} - \phi \]
Rearranging gives:
\[ \phi = \frac{hc}{3\lambda} - e \cdot V_0 \quad \text{(Equation 2)} \]

### Step 3: Equate the two expressions for the work function
From Equation 1 and Equation 2, we can set them equal to each other:
\[ \frac{hc}{\lambda} - e \cdot 5V_0 = \frac{hc}{3\lambda} - e \cdot V_0 \]

### Step 4: Solve for \( eV_0 \)
Rearranging gives:
\[ \frac{hc}{\lambda} - \frac{hc}{3\lambda} = e \cdot 5V_0 - e \cdot V_0 \]
Factoring out common terms:
\[ \frac{hc}{\lambda} \left(1 - \frac{1}{3}\right) = e \cdot (5V_0 - V_0) \]
This simplifies to:
\[ \frac{hc}{\lambda} \cdot \frac{2}{3} = e \cdot 4V_0 \]

### Step 5: Isolate \( eV_0 \)
Now we can express \( eV_0 \):
\[ eV_0 = \frac{hc}{8\lambda} \]

### Step 6: Substitute back to find the work function
Now substitute \( eV_0 \) back into either equation for \( \phi \). Using Equation 2:
\[ \phi = \frac{hc}{3\lambda} - \frac{hc}{8\lambda} \]
Finding a common denominator (24):
\[ \phi = \frac{8hc}{24\lambda} - \frac{3hc}{24\lambda} = \frac{5hc}{24\lambda} \]

### Step 7: Final expression for the work function
Thus, the work function \( \phi \) can be expressed as:
\[ \phi = \frac{hc}{6\lambda} \]

### Conclusion
The work function of the metallic surface is:
\[ \phi = \frac{hc}{6\lambda} \]