Photons with energy `5 eV` are incident on a cathode `C` in a photoelectric cell . The maximum energy of emitted photoelectrons is `2 eV`. When photons of energy `6 eV` are incident on `C` , no photoelectrons will reach the anode `A` , if the stopping potential of `A` relative to `C` is

Options

A. 5
B. 3 V
C. 1 V
D. 4 V

Answer

To solve the problem step by step, we will use the concepts of photoelectric effect and stopping potential.

### Step 1: Determine the Work Function (Φ)
The maximum kinetic energy (K.E.) of the emitted photoelectrons can be found using the equation:
\[ K.E. = E_{photon} - \Phi \]
Where:
- \( K.E. \) is the maximum kinetic energy of the emitted photoelectrons,
- \( E_{photon} \) is the energy of the incident photons,
- \( \Phi \) is the work function of the cathode.

From the problem, we know:
- \( E_{photon} = 5 \, eV \)
- \( K.E. = 2 \, eV \)

Substituting the values into the equation:
\[ 2 \, eV = 5 \, eV - \Phi \]

Rearranging gives:
\[ \Phi = 5 \, eV - 2 \, eV = 3 \, eV \]

### Step 2: Analyze the Situation with 6 eV Photons
Now, when photons of energy \( 6 \, eV \) are incident on the cathode, we need to find the stopping potential \( V_s \) such that no photoelectrons reach the anode.

The maximum kinetic energy of the photoelectrons when \( 6 \, eV \) photons are incident can be calculated as:
\[ K.E. = E_{photon} - \Phi \]
Where:
- \( E_{photon} = 6 \, eV \)

Substituting the values:
\[ K.E. = 6 \, eV - 3 \, eV = 3 \, eV \]

### Step 3: Relate Kinetic Energy to Stopping Potential
The stopping potential \( V_s \) is related to the maximum kinetic energy by the equation:
\[ K.E. = e \cdot V_s \]
Where \( e \) is the charge of an electron (1 eV corresponds to 1 volt of stopping potential).

Substituting the maximum kinetic energy:
\[ 3 \, eV = e \cdot V_s \]
Thus,
\[ V_s = 3 \, V \]

### Conclusion
The stopping potential of the anode \( A \) relative to the cathode \( C \) is:
\[ V_s = 3 \, V \]

### Final Answer
The stopping potential of \( A \) relative to \( C \) is **3 V**.

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