Photons of energy 5 eV and incident on a cathode made of a photosensitive plate. The maximum kinetic energy of emitted photoelectrons is 3 eV . When photons of energy 8 eV fall on the same metal plate , the stopping potential of anode plate will be

Options

A. `-7V`
B. `-6V`
C. 3 V
D. `-3V`

Explanation

To solve the problem step by step, we will use the concept of the photoelectric effect, which relates the energy of incident photons to the kinetic energy of emitted photoelectrons.

### Step-by-Step Solution:

1. **Identify Given Data**:
- Energy of the first incident photons, \( E_1 = 5 \, \text{eV} \)
- Maximum kinetic energy of emitted photoelectrons, \( K_{e, \text{max}} = 3 \, \text{eV} \)
- Energy of the second incident photons, \( E_2 = 8 \, \text{eV} \)

2. **Use the Photoelectric Equation**:
The photoelectric effect can be described by the equation:
\[ K_{e, \text{max}} = E - \Phi \]
where \( \Phi \) is the work function of the material.

3. **Calculate the Work Function**:
For the first incident photon energy:
\[ 3 \, \text{eV} = 5 \, \text{eV} - \Phi \]
Rearranging gives:
\[ \Phi = 5 \, \text{eV} - 3 \, \text{eV} = 2 \, \text{eV} \]

4. **Calculate Maximum Kinetic Energy for Second Photon Energy**:
Now, using the second photon energy:
\[ K_{e, \text{max}}' = E_2 - \Phi \]
Substituting the values:
\[ K_{e, \text{max}}' = 8 \, \text{eV} - 2 \, \text{eV} = 6 \, \text{eV} \]

5. **Relate Maximum Kinetic Energy to Stopping Potential**:
The stopping potential \( V_s \) is related to the maximum kinetic energy of the emitted photoelectrons:
\[ K_{e, \text{max}}' = e V_s \]
where \( e \) is the charge of an electron (which we can consider as 1 eV for simplicity in this context).

6. **Calculate Stopping Potential**:
Therefore, the stopping potential \( V_s \) is:
\[ V_s = K_{e, \text{max}}' = 6 \, \text{V} \]

### Final Answer:
The stopping potential of the anode plate when photons of energy 8 eV fall on the same metal plate will be **6 V**.