A photon of energy `4 eV` is incident on a metal surface whose work function is `2 eV`. The minimum reverse potential to be applied for stopping the emission of electrons is

Options

A. `6 V`
B. `4 V`
C. `2 V`
D. `8 V`

Explanation

To solve the problem, we need to determine the minimum reverse potential (stopping potential) required to stop the emission of electrons when a photon of energy 4 eV is incident on a metal surface with a work function of 2 eV.

### Step-by-Step Solution:

1. **Identify the Energy of the Photon**:
The energy of the incident photon is given as \( E_{\text{photon}} = 4 \, \text{eV} \).

2. **Identify the Work Function of the Metal**:
The work function of the metal surface is given as \( \phi = 2 \, \text{eV} \).

3. **Calculate the Maximum Kinetic Energy of the Emitted Electrons**:
According to the photoelectric effect, the maximum kinetic energy (\( KE_{\text{max}} \)) of the emitted electrons can be calculated using the formula:
\[ KE_{\text{max}} = E_{\text{photon}} - \phi \]
Substituting the values:
\[ KE_{\text{max}} = 4 \, \text{eV} - 2 \, \text{eV} = 2 \, \text{eV} \]

4. **Determine the Stopping Potential**:
The stopping potential (\( V_0 \)) is the potential required to stop the most energetic emitted electrons. This potential is equal to the maximum kinetic energy of the electrons in terms of volts. Therefore:
\[ V_0 = KE_{\text{max}} = 2 \, \text{V} \]

5. **Conclusion**:
The minimum reverse potential required to stop the emission of electrons is \( V_0 = 2 \, \text{V} \).

### Final Answer:
The minimum reverse potential to be applied for stopping the emission of electrons is **2 V**.

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