When the wavelength of radiation falling on a metal is changed from 500 nm to 200 nm, the maximum kinetic energy of the photo electrons becomes three times larger. The work function of the metal is close to :

Options

A. 1.02 eV
B. 0.61 eV
C. 0.52 eV
D. 0.81 eV

Explanation

To solve the problem, we will use the photoelectric effect equation and the information provided in the question.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The energy of the incident radiation can be expressed as:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the radiation.

2. **Initial and Final Conditions**:
- For the initial wavelength \( \lambda_1 = 500 \, \text{nm} \):
\[ E_1 = \frac{hc}{\lambda_1} \]
- For the final wavelength \( \lambda_2 = 200 \, \text{nm} \):
\[ E_2 = \frac{hc}{\lambda_2} \]

3. **Kinetic Energy Relationships**:
According to the problem, the maximum kinetic energy of the photoelectrons changes from \( K.E. = K \) to \( K.E. = 3K \). The equations for kinetic energy in both cases are:
- Initial:
\[ K = E_1 - \phi \]
- Final:
\[ 3K = E_2 - \phi \]

4. **Setting Up the Equations**:
From the above equations, we can express \( K \) in terms of \( E_1 \) and \( E_2 \):
\[ K = E_1 - \phi \quad \text{(1)} \]
\[ 3K = E_2 - \phi \quad \text{(2)} \]

5. **Substituting Equation (1) into Equation (2)**:
Substitute \( K \) from equation (1) into equation (2):
\[ 3(E_1 - \phi) = E_2 - \phi \]
Expanding this gives:
\[ 3E_1 - 3\phi = E_2 - \phi \]
Rearranging terms:
\[ 3E_1 - E_2 = 2\phi \]

6. **Calculating Energies**:
Now, we need to calculate \( E_1 \) and \( E_2 \):
\[ E_1 = \frac{hc}{500 \times 10^{-9}} \quad \text{and} \quad E_2 = \frac{hc}{200 \times 10^{-9}} \]
Using \( hc \approx 1240 \, \text{eV nm} \):
\[ E_1 = \frac{1240}{500} = 2.48 \, \text{eV} \]
\[ E_2 = \frac{1240}{200} = 6.2 \, \text{eV} \]

7. **Substituting Energies into the Equation**:
Substitute \( E_1 \) and \( E_2 \) back into the rearranged equation:
\[ 3(2.48) - 6.2 = 2\phi \]
\[ 7.44 - 6.2 = 2\phi \]
\[ 1.24 = 2\phi \]
\[ \phi = \frac{1.24}{2} = 0.62 \, \text{eV} \]

### Final Answer:
The work function \( \phi \) of the metal is approximately \( 0.62 \, \text{eV} \).