When photons of energy 5 eV fall on a photosensitive surface, the maximum K.E. of the photoelectrons emitted from the surface is 2.5 eV. What is the stopping potential ?

Options

A. 5 V
B. 2.5 V
C. 7.5 V
D. 1 V

Explanation

To find the stopping potential when photons of energy 5 eV fall on a photosensitive surface and the maximum kinetic energy (K.E.) of the emitted photoelectrons is 2.5 eV, we can use the photoelectric effect equation:

1. **Understanding the Photoelectric Effect**:
The energy of the incident photon (E) is used to overcome the work function (Φ) of the material and provide kinetic energy (K.E.) to the emitted electrons. This can be expressed as:
\[ E = \Phi + K.E. \]

2. **Given Values**:
- Energy of the photon (E) = 5 eV
- Maximum kinetic energy of the photoelectrons (K.E.) = 2.5 eV

3. **Finding the Work Function (Φ)**:
Rearranging the equation gives us:
\[ \Phi = E - K.E. \]
Substituting the known values:
\[ \Phi = 5 \, \text{eV} - 2.5 \, \text{eV} = 2.5 \, \text{eV}. \]

4. **Calculating the Stopping Potential (V₀)**:
The stopping potential (V₀) is the potential required to stop the emitted photoelectrons, which is equal to the maximum kinetic energy of the photoelectrons in electron volts. Therefore:
\[ V₀ = K.E. = 2.5 \, \text{V}. \]

5. **Final Answer**:
The stopping potential is:
\[ V₀ = 2.5 \, \text{V}. \]