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 the photoelectric effect equation derived from Einstein's theory. The maximum kinetic energy (K.E.) of the emitted electrons can be expressed as:

\[ K.E. = h\nu - \phi \]

Where:
- \( K.E. \) is the maximum kinetic energy of the emitted electrons,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( \nu \) is the frequency of the incident light,
- \( \phi \) is the work function of the material.

The frequency \( \nu \) can be related to the wavelength \( \lambda \) using the equation:

\[ \nu = \frac{c}{\lambda} \]

Where \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)).

### Step-by-Step Solution:

1. **Express Kinetic Energy in Terms of Wavelength**:
The kinetic energy can be rewritten in terms of wavelength:
\[ K.E. = \frac{hc}{\lambda} - \phi \]

2. **Set Up the Initial and Final Conditions**:
Let \( \lambda_1 = 500 \, \text{nm} \) and \( \lambda_2 = 200 \, \text{nm} \).
The initial kinetic energy \( K.E_1 \) when \( \lambda_1 \) is:
\[ K.E_1 = \frac{hc}{\lambda_1} - \phi \]
The final kinetic energy \( K.E_2 \) when \( \lambda_2 \) is:
\[ K.E_2 = \frac{hc}{\lambda_2} - \phi \]

3. **Use the Given Condition**:
According to the problem, the maximum kinetic energy increases to 3 times:
\[ K.E_2 = 3 K.E_1 \]
Substituting the expressions for kinetic energy:
\[ \frac{hc}{\lambda_2} - \phi = 3 \left( \frac{hc}{\lambda_1} - \phi \right) \]

4. **Expand and Rearrange the Equation**:
Expanding the equation gives:
\[ \frac{hc}{\lambda_2} - \phi = \frac{3hc}{\lambda_1} - 3\phi \]
Rearranging this leads to:
\[ \frac{hc}{\lambda_2} - \frac{3hc}{\lambda_1} = -3\phi + \phi \]
Simplifying further:
\[ \frac{hc}{\lambda_2} - \frac{3hc}{\lambda_1} = -2\phi \]

5. **Solve for Work Function \( \phi \)**:
Rearranging gives:
\[ 2\phi = \frac{3hc}{\lambda_1} - \frac{hc}{\lambda_2} \]
Thus,
\[ \phi = \frac{1}{2} \left( \frac{3hc}{\lambda_1} - \frac{hc}{\lambda_2} \right) \]

6. **Substitute Values**:
Convert wavelengths from nm to meters:
\[ \lambda_1 = 500 \times 10^{-9} \, \text{m}, \quad \lambda_2 = 200 \times 10^{-9} \, \text{m} \]
Substitute \( h = 6.626 \times 10^{-34} \, \text{Js} \) and \( c = 3 \times 10^8 \, \text{m/s} \):
\[ \phi = \frac{1}{2} \left( \frac{3 \cdot 6.626 \times 10^{-34} \cdot 3 \times 10^8}{500 \times 10^{-9}} - \frac{6.626 \times 10^{-34} \cdot 3 \times 10^8}{200 \times 10^{-9}} \right) \]

7. **Calculate**:
Performing the calculations:
\[ \phi = \frac{1}{2} \left( \frac{3 \cdot 6.626 \cdot 3 \times 10^{-26}}{500} - \frac{6.626 \cdot 3 \times 10^{-26}}{200} \right) \]
Simplifying gives:
\[ \phi = \frac{1}{2} \left( \frac{5.0 \times 10^{-26}}{500} - \frac{9.9 \times 10^{-26}}{200} \right) \]
After calculating, we find:
\[ \phi \approx 0.62 \, \text{eV} \]

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