Photons of energy `6 eV` are incident on a metal surface whose work function is `4 eV`. The minimum kinetic energy of the emitted photo - electrons will be

Options

A. `0 eV`
B. `1 eV`
C. `2 eV`
D. `10 eV`

Explanation

To solve the problem, we need to determine the minimum kinetic energy of the emitted photoelectrons when photons of energy \(6 \, \text{eV}\) are incident on a metal surface with a work function of \(4 \, \text{eV}\).

### Step-by-Step Solution:

1. **Identify the given values:**
- Energy of the incident photon, \(E = 6 \, \text{eV}\)
- Work function of the metal, \(\phi = 4 \, \text{eV}\)

2. **Understand the photoelectric effect:**
- According to the photoelectric effect, when a photon strikes a metal surface, it can transfer its energy to an electron. If the energy of the photon is greater than the work function of the metal, the electron can be emitted.

3. **Calculate the kinetic energy of the emitted electrons:**
- The kinetic energy (KE) of the emitted electrons can be calculated using the formula:
\[ KE = E - \phi \]
- Substituting the values:
\[ KE = 6 \, \text{eV} - 4 \, \text{eV} = 2 \, \text{eV} \]

4. **Determine the minimum kinetic energy:**
- The minimum kinetic energy of the emitted photoelectrons occurs when all the excess energy goes into kinetic energy. However, it is important to note that in an ideal scenario, the minimum kinetic energy can be zero if the energy is just enough to overcome the work function.
- Since the question asks for the minimum kinetic energy, we consider the possibility that all excess energy could be lost in other forms (like heat or work done against external forces). Therefore, the minimum kinetic energy of the emitted photoelectrons can be:
\[ KE_{\text{min}} = 0 \, \text{eV} \]

### Final Answer:
The minimum kinetic energy of the emitted photoelectrons is \(0 \, \text{eV}\).

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