When the energy of the incident radiation is increased by `20 %` , kinetic energy of the photoelectrons emitted from a metal surface increased from `0.5 eV to 0.8 eV`. The work function of the metal is

Options

A. `0.65 eV`
B. `1.0 eV`
C. `1.3 eV`
D. `1.5 eV`

Explanation

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

### Step-by-Step Solution:

1. **Understand the Problem**:
- The energy of the incident radiation is increased by 20%.
- Initial kinetic energy of photoelectrons = 0.5 eV.
- Final kinetic energy of photoelectrons after the increase = 0.8 eV.
- We need to find the work function (φ) of the metal.

2. **Define Initial Energy**:
- Let the initial energy of the incident radiation be \( E \).
- The initial kinetic energy of the emitted photoelectrons can be expressed using Einstein's photoelectric equation:
\[ E - \phi = KE_{initial} \]
- Substituting the values:
\[ E - \phi = 0.5 \quad (1) \]

3. **Define Final Energy**:
- After increasing the energy by 20%, the new energy of the incident radiation becomes:
\[ E_{new} = 1.2E \]
- The corresponding kinetic energy of the emitted photoelectrons is now:
\[ E_{new} - \phi = KE_{final} \]
- Substituting the values:
\[ 1.2E - \phi = 0.8 \quad (2) \]

4. **Solve the Equations**:
- From equation (1), we can express \( E \) in terms of \( \phi \):
\[ E = \phi + 0.5 \quad (3) \]
- Substitute equation (3) into equation (2):
\[ 1.2(\phi + 0.5) - \phi = 0.8 \]
- Expanding this gives:
\[ 1.2\phi + 0.6 - \phi = 0.8 \]
- Simplifying further:
\[ 0.2\phi + 0.6 = 0.8 \]
- Subtracting 0.6 from both sides:
\[ 0.2\phi = 0.2 \]
- Dividing by 0.2:
\[ \phi = 1 \text{ eV} \]

5. **Conclusion**:
- The work function of the metal is \( \phi = 1 \text{ eV} \).