Maximum kinetic energy of photoelectrons emitted from a metal surface, when light of wavelength `lambda` is incident on it, is 1 eV. When a light of wavelength `lambda/3` is incident on the surfaces, maximum kinetic energy becomes 4 times. The work function of the metal is

Options

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

Answer

To solve the problem, we will use Einstein's photoelectric equation, which relates the energy of the incident photons to the work function of the metal and the kinetic energy of the emitted photoelectrons.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Maximum kinetic energy (K.E.) when light of wavelength \( \lambda \) is incident: \( K.E_1 = 1 \, \text{eV} \)
- Maximum kinetic energy when light of wavelength \( \lambda/3 \) is incident: \( K.E_2 = 4 \times K.E_1 = 4 \, \text{eV} \)

2. **Use Einstein’s Photoelectric Equation:**
The equation is given by:
\[ E = W_0 + K.E \]
where \( E \) is the energy of the incident photon, \( W_0 \) is the work function of the metal, and \( K.E \) is the kinetic energy of the emitted photoelectrons.

3. **Calculate Energy for the First Wavelength:**
For the first case (wavelength \( \lambda \)):
\[ E_1 = \frac{hc}{\lambda} = W_0 + K.E_1 \]
Substituting the known values:
\[ \frac{hc}{\lambda} = W_0 + 1 \, \text{eV} \quad \text{(1)} \]

4. **Calculate Energy for the Second Wavelength:**
For the second case (wavelength \( \lambda/3 \)):
\[ E_2 = \frac{hc}{\lambda/3} = W_0 + K.E_2 \]
This simplifies to:
\[ \frac{3hc}{\lambda} = W_0 + 4 \, \text{eV} \quad \text{(2)} \]

5. **Substitute Equation (1) into Equation (2):**
From equation (1), we can express \( W_0 \):
\[ W_0 = \frac{hc}{\lambda} - 1 \, \text{eV} \]
Substitute this into equation (2):
\[ 3 \left( \frac{hc}{\lambda} - 1 \right) = W_0 + 4 \]
Expanding this gives:
\[ 3 \frac{hc}{\lambda} - 3 = W_0 + 4 \]

6. **Substituting \( W_0 \) Again:**
Substitute \( W_0 \) from equation (1):
\[ 3 \frac{hc}{\lambda} - 3 = \left( \frac{hc}{\lambda} - 1 \right) + 4 \]
Simplifying this:
\[ 3 \frac{hc}{\lambda} - 3 = \frac{hc}{\lambda} + 3 \]

7. **Rearranging the Equation:**
Rearranging gives:
\[ 3 \frac{hc}{\lambda} - \frac{hc}{\lambda} = 3 + 3 \]
\[ 2 \frac{hc}{\lambda} = 6 \]
\[ \frac{hc}{\lambda} = 3 \, \text{eV} \]

8. **Finding Work Function \( W_0 \):**
Now substitute \( \frac{hc}{\lambda} \) back into equation (1):
\[ 3 = W_0 + 1 \]
Thus, we find:
\[ W_0 = 3 - 1 = 2 \, \text{eV} \]

### Final Answer:
The work function of the metal is \( W_0 = 2 \, \text{eV} \).