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 need to use the photoelectric effect equation, which relates the maximum kinetic energy (K.E.) of photoelectrons to the wavelength of the incident radiation and the work function (φ) of the metal.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The maximum kinetic energy (K.E.) of photoelectrons emitted from a metal surface when illuminated by light of wavelength λ is given by the equation:
\[ K.E. = \frac{hc}{\lambda} - \phi \]
where:
- \( K.E. \) is the maximum kinetic energy of the emitted electrons,
- \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \( c \) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \( \lambda \) is the wavelength of the incident light,
- \( \phi \) is the work function of the metal.

2. **Setting Up the Equations**:
Let the initial wavelength be \( \lambda_1 = 500 \, \text{nm} \) and the final wavelength be \( \lambda_2 = 200 \, \text{nm} \).
- For \( \lambda_1 = 500 \, \text{nm} \):
\[ K.E_1 = \frac{hc}{500} - \phi \quad \text{(Equation 1)} \]
- For \( \lambda_2 = 200 \, \text{nm} \):
\[ K.E_2 = \frac{hc}{200} - \phi \quad \text{(Equation 2)} \]

3. **Relating the Kinetic Energies**:
According to the problem, when the wavelength changes from 500 nm to 200 nm, the maximum kinetic energy becomes three times larger:
\[ K.E_2 = 3 K.E_1 \]
Substituting the expressions from Equations 1 and 2:
\[ \frac{hc}{200} - \phi = 3 \left( \frac{hc}{500} - \phi \right) \]

4. **Expanding and Rearranging**:
Expanding the equation:
\[ \frac{hc}{200} - \phi = \frac{3hc}{500} - 3\phi \]
Rearranging gives:
\[ \frac{hc}{200} + 2\phi = \frac{3hc}{500} \]

5. **Finding a Common Denominator**:
To simplify, find a common denominator (1000):
\[ \frac{5hc}{1000} + 2\phi = \frac{6hc}{1000} \]
Rearranging gives:
\[ 2\phi = \frac{6hc}{1000} - \frac{5hc}{1000} = \frac{hc}{1000} \]

6. **Solving for the Work Function**:
Thus, we have:
\[ \phi = \frac{hc}{2000} \]

7. **Substituting Values**:
Now, substituting \( h \) and \( c \) in terms of electron volts and nanometers:
\[ hc = 1240 \, \text{eV nm} \]
Therefore:
\[ \phi = \frac{1240}{2000} = 0.62 \, \text{eV} \]

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