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 the photoelectric effect equation, which relates the energy of the incident photon to the work function of the metal and the kinetic energy of the emitted electrons. The equation is given by:

\[ E = W + KE_{max} \]

Where:
- \( E \) is the energy of the incident photon,
- \( W \) is the work function (the minimum energy required to remove an electron from the metal),
- \( KE_{max} \) is the maximum kinetic energy of the emitted electrons.

### Step-by-Step Solution:

1. **Identify the given values**:
- Maximum kinetic energy of emitted electrons, \( KE_{max} = 2.6 \, eV \)
- Work function of the metal, \( W = 4.2 \, eV \)

2. **Use the photoelectric effect equation**:
Substitute the known values into the equation:
\[ E = W + KE_{max} \]

3. **Plug in the values**:
\[ E = 4.2 \, eV + 2.6 \, eV \]

4. **Calculate the energy of the incident photon**:
\[ E = 6.8 \, eV \]

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 \( 6.8 \, eV \).