Photon of 5.5 eV energy fall on the surface of the metal emitting photoelectrons of maximum kinetic energy 4.0 eV . The stopping voltage required for these electrons are

Options

A. 5.5 V
B. 1.5 V
C. 9.5 V
D. 4.0 V

Answer

To solve the problem, we need to determine the stopping voltage required for the emitted photoelectrons. Here are the steps to arrive at the solution:

### Step-by-Step Solution:

1. **Understanding the Given Data:**
- Energy of the incident photon (E_photon) = 5.5 eV
- Maximum kinetic energy of emitted photoelectrons (K_max) = 4.0 eV

2. **Using the Photoelectric Equation:**
The photoelectric effect can be described by the equation:
\[ E_{\text{photon}} = \text{Work Function} (\phi) + K_{\text{max}} \]
Rearranging this gives us:
\[ \phi = E_{\text{photon}} - K_{\text{max}} \]

3. **Calculating the Work Function:**
Substituting the given values into the equation:
\[ \phi = 5.5 \, \text{eV} - 4.0 \, \text{eV} = 1.5 \, \text{eV} \]

4. **Understanding Stopping Voltage:**
The stopping voltage (V_0) is the potential difference needed to stop the most energetic photoelectron. It can be related to the maximum kinetic energy of the emitted electrons:
\[ K_{\text{max}} = e \cdot V_0 \]
where \( e \) is the charge of the electron (approximately 1 eV when considering energy in electron volts).

5. **Calculating the Stopping Voltage:**
Rearranging the equation gives:
\[ V_0 = \frac{K_{\text{max}}}{e} \]
Since \( K_{\text{max}} = 4.0 \, \text{eV} \):
\[ V_0 = 4.0 \, \text{V} \]

6. **Final Answer:**
The stopping voltage required for the emitted photoelectrons is:
\[ V_0 = 4.0 \, \text{V} \]

### Summary:
The stopping voltage required for the photoelectrons emitted from the metal surface when a photon of 5.5 eV energy strikes it is **4.0 V**.