The energy of emitted photoelectrons from a metal is `0.9ev`, The work function of the metal is `2.2eV`. Then the energy of the incident photon is

Options

A. `0.9eV`
B. `2.2eV`
C. `4.4eV`
D. `3.1eV`

Explanation

To solve the problem, we need to find the energy of the incident photon using the given information about the energy of emitted photoelectrons and the work function of the metal.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Energy of emitted photoelectrons (K.E. of photoelectrons, Kmax) = 0.9 eV
- Work function of the metal (φ) = 2.2 eV

2. **Use Einstein's Photoelectric Equation:**
Einstein's photoelectric equation relates the energy of the incident photon (E) to the work function (φ) and the kinetic energy (K.E.) of the emitted photoelectrons:
\[ E = K.E. + φ \]

3. **Substitute the Known Values:**
Substitute the values of K.E. and φ into the equation:
\[ E = 0.9 \, \text{eV} + 2.2 \, \text{eV} \]

4. **Calculate the Energy of the Incident Photon:**
\[ E = 3.1 \, \text{eV} \]

5. **Conclusion:**
The energy of the incident photon is **3.1 eV**.

### Final Answer:
The energy of the incident photon is **3.1 eV**.

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