A metal of work function 2.5 eV is irradiated by light. If the emitted electrons have maximum kinetic energy 1.8 eV, then the energy of irradiating light is

Options

A. 0.7 eV
B. 3.8 eV
C. 4.3 eV
D. 4.8 eV

Explanation

To solve the problem, we will use the photoelectric equation, which relates the maximum kinetic energy of emitted electrons to the energy of the incident light and the work function of the metal. The equation is given by:

\[ KE_{max} = E_{incident} - \phi \]

Where:
- \( KE_{max} \) is the maximum kinetic energy of the emitted electrons.
- \( E_{incident} \) is the energy of the incident light.
- \( \phi \) is the work function of the metal.

### Step-by-Step Solution:

1. **Identify the given values**:
- Work function (\( \phi \)) = 2.5 eV
- Maximum kinetic energy (\( KE_{max} \)) = 1.8 eV

2. **Write the photoelectric equation**:
\[ KE_{max} = E_{incident} - \phi \]

3. **Rearrange the equation to find the energy of the incident light**:
\[ E_{incident} = KE_{max} + \phi \]

4. **Substitute the known values into the equation**:
\[ E_{incident} = 1.8 \, \text{eV} + 2.5 \, \text{eV} \]

5. **Perform the addition**:
\[ E_{incident} = 4.3 \, \text{eV} \]

6. **Conclusion**:
The energy of the irradiating light is **4.3 eV**.