A photoelectric surface is illuminated successively by monochromatic light of wavelength `lambda` and `(lambda)/(2)`. If the maximum kinetic energy of the emitted photoelectrons in the second case is `3` times than in the first case , the work function of the surface of the material is
`(h = Plank's constant , c = speed of light )`

Options

A. `(hc)/( 3 lambda)`
B. `(hc)/( 2 lambda)`
C. `(hc)/(lambda)`
D. `( 2hc)/(lambda)`

Explanation

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

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two cases where a photoelectric surface is illuminated by light of wavelength \( \lambda \) and \( \frac{\lambda}{2} \). We need to find the work function \( \phi \) of the material given that the maximum kinetic energy of emitted photoelectrons in the second case is 3 times that in the first case.

2. **Using Einstein's Photoelectric Equation**:
The maximum kinetic energy \( K \) of the emitted electrons can be expressed as:
\[ K = E - \phi \]
where \( E \) is the energy of the incident photons and \( \phi \) is the work function.

3. **Calculating Energies for Both Cases**:
- For the first case (wavelength \( \lambda \)):
\[ E_1 = \frac{hc}{\lambda} \]
Thus, the maximum kinetic energy \( K_1 \) is:
\[ K_1 = \frac{hc}{\lambda} - \phi \quad \text{(Equation 1)} \]

- For the second case (wavelength \( \frac{\lambda}{2} \)):
\[ E_2 = \frac{hc}{\frac{\lambda}{2}} = \frac{2hc}{\lambda} \]
Thus, the maximum kinetic energy \( K_2 \) is:
\[ K_2 = \frac{2hc}{\lambda} - \phi \quad \text{(Equation 2)} \]

4. **Relating the Kinetic Energies**:
According to the problem, \( K_2 = 3K_1 \). Substituting the expressions for \( K_1 \) and \( K_2 \):
\[ \frac{2hc}{\lambda} - \phi = 3\left(\frac{hc}{\lambda} - \phi\right) \]

5. **Expanding and Rearranging**:
Expanding the right-hand side:
\[ \frac{2hc}{\lambda} - \phi = \frac{3hc}{\lambda} - 3\phi \]
Rearranging gives:
\[ \frac{2hc}{\lambda} - \frac{3hc}{\lambda} = -3\phi + \phi \]
Simplifying:
\[ -\frac{hc}{\lambda} = -2\phi \]

6. **Solving for the Work Function**:
Dividing both sides by -2:
\[ \phi = \frac{hc}{2\lambda} \]

### Final Answer:
The work function \( \phi \) of the surface of the material is:
\[ \phi = \frac{hc}{2\lambda} \]