If wavelength of incident radiation changes from `500 nm` to `200 nm` then the maximum kinetic energy increases to 3 times. Find work function ?

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 material and the kinetic energy of the emitted electrons.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two wavelengths of incident radiation:
- \( \lambda_1 = 500 \, \text{nm} \)
- \( \lambda_2 = 200 \, \text{nm} \)
- The maximum kinetic energy of the emitted electrons increases from \( K \) to \( 3K \).

2. **Using Einstein's Photoelectric Equation**:
- The equation is given by:
\[ E = \frac{hc}{\lambda} - \phi = K_{max} \]
- For the first wavelength \( \lambda_1 \):
\[ \frac{hc}{\lambda_1} - \phi = K \]
- For the second wavelength \( \lambda_2 \):
\[ \frac{hc}{\lambda_2} - \phi = 3K \]

3. **Setting Up the Equations**:
- From the first equation:
\[ K = \frac{hc}{\lambda_1} - \phi \quad \text{(1)} \]
- From the second equation:
\[ 3K = \frac{hc}{\lambda_2} - \phi \quad \text{(2)} \]

4. **Eliminating Kinetic Energy (K)**:
- Rearranging equation (1):
\[ \phi = \frac{hc}{\lambda_1} - K \]
- Substituting into equation (2):
\[ 3K = \frac{hc}{\lambda_2} - \left(\frac{hc}{\lambda_1} - K\right) \]
- Simplifying this gives:
\[ 3K = \frac{hc}{\lambda_2} - \frac{hc}{\lambda_1} + K \]
- Rearranging:
\[ 2K = \frac{hc}{\lambda_2} - \frac{hc}{\lambda_1} \]

5. **Expressing K in Terms of Work Function**:
- From the above equation:
\[ K = \frac{1}{2} \left( \frac{hc}{\lambda_2} - \frac{hc}{\lambda_1} \right) \]

6. **Substituting K back to find Work Function**:
- Substitute \( K \) back into equation (1):
\[ \phi = \frac{hc}{\lambda_1} - \frac{1}{2} \left( \frac{hc}{\lambda_2} - \frac{hc}{\lambda_1} \right) \]
- Simplifying gives:
\[ \phi = \frac{hc}{\lambda_1} - \frac{hc}{2} \left( \frac{1}{\lambda_2} - \frac{1}{\lambda_1} \right) \]
- Rearranging:
\[ \phi = \frac{hc}{2} \left( \frac{3}{\lambda_1} - \frac{1}{\lambda_2} \right) \]

7. **Calculating Work Function**:
- Using \( hc = 1240 \, \text{eV nm} \):
\[ \phi = \frac{1240}{2} \left( \frac{3}{500} - \frac{1}{200} \right) \]
- Calculate:
\[ \phi = 620 \left( \frac{3}{500} - \frac{1}{200} \right) \]
- Finding a common denominator (1000):
\[ \phi = 620 \left( \frac{6 - 5}{1000} \right) = 620 \times \frac{1}{1000} = 0.62 \, \text{eV} \]

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