Monochromatic light incident on a metal surface emits electrons with kinetic energies from zero to 2.6 eV. What is the least energy of the incident photon if the tightly bound electron needs 4.2eV to remove?

Options

A. `1.6eV`
B. From `1.6eV to 6.8 eV`
C. `6.8 eV`
D. More than `6.8eV`

Explanation

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

### Step-by-Step Solution:

1. **Understand the Given Information:**
- The maximum kinetic energy (KE) of the emitted electrons is given as 2.6 eV.
- The energy required to remove the tightly bound electron (work function, W) is given as 4.2 eV.

2. **Recall Einstein's Photoelectric Equation:**
\[ E = W + KE_{\text{max}} \]
where:
- \( E \) is the energy of the incident photon,
- \( W \) is the work function of the metal,
- \( KE_{\text{max}} \) is the maximum kinetic energy of the emitted electrons.

3. **Substitute the Known Values:**
- Substitute \( W = 4.2 \, \text{eV} \) and \( KE_{\text{max}} = 2.6 \, \text{eV} \) into the equation:
\[ E = 4.2 \, \text{eV} + 2.6 \, \text{eV} \]

4. **Calculate the Energy of the Incident Photon:**
\[ E = 4.2 \, \text{eV} + 2.6 \, \text{eV} = 6.8 \, \text{eV} \]

5. **Conclusion:**
- The least energy of the incident photon required to emit electrons from the metal surface is **6.8 eV**.

### Final Answer:
The least energy of the incident photon is **6.8 eV**.