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. zero
B. 1 eV
C. 2 eV
D. 10 eV

Answer

To solve the problem, we need to apply the concept of the photoelectric effect, which can be summarized by the equation:

\[ K.E. = E_i - W_0 \]

Where:
- \( K.E. \) is the kinetic energy of the emitted photoelectrons,
- \( E_i \) is the energy of the incident photons,
- \( W_0 \) is the work function of the metal.

### Step-by-Step Solution:

1. **Identify the Energy of Incident Photons (\( E_i \))**:
The energy of the incident photons is given as \( 6 \, \text{eV} \).

2. **Identify the Work Function (\( W_0 \))**:
The work function of the metal surface is given as \( 4 \, \text{eV} \).

3. **Apply the Photoelectric Effect Equation**:
Substitute the values of \( E_i \) and \( W_0 \) into the equation for kinetic energy:
\[ K.E. = E_i - W_0 = 6 \, \text{eV} - 4 \, \text{eV} \]

4. **Calculate the Kinetic Energy**:
\[ K.E. = 2 \, \text{eV} \]

5. **Conclusion**:
The minimum kinetic energy of the emitted photoelectrons is \( 2 \, \text{eV} \).

### Final Answer:
The minimum kinetic energy of the emitted photoelectrons will be \( 2 \, \text{eV} \).

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