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. ` - 1 V`
B. `- 3 V`
C. `+ 3 V`
D. `+ 4 V`

Explanation

To solve the problem step by step, we will follow the principles of photoelectric effect and the relationship between energy, work function, and stopping potential.

### Step 1: Understand the Given Information
- Energy of incident photons (E1) = 5 eV
- Maximum kinetic energy of emitted photoelectrons (Kmax) = 2 eV
- Energy of second incident photons (E2) = 6 eV
- We need to find the stopping potential (V0) relative to the cathode (C).

### Step 2: Calculate the Work Function (Φ)
Using the photoelectric equation:
\[ K_{max} = E_{1} - \Phi \]
Substituting the known values:
\[ 2 \, \text{eV} = 5 \, \text{eV} - \Phi \]
Rearranging gives:
\[ \Phi = 5 \, \text{eV} - 2 \, \text{eV} = 3 \, \text{eV} \]

### Step 3: Analyze the Second Incident Photon Energy
For the second case where photons of energy E2 = 6 eV are incident on the cathode:
Using the photoelectric equation again:
\[ K_{max}' = E_{2} - \Phi \]
Substituting the values:
\[ K_{max}' = 6 \, \text{eV} - 3 \, \text{eV} = 3 \, \text{eV} \]
However, it is given that no photoelectrons reach the anode. This implies that the stopping potential must be equal to or greater than the maximum kinetic energy of the emitted photoelectrons.

### Step 4: Determine the Stopping Potential (V0)
The stopping potential (V0) is defined as the potential required to stop the most energetic photoelectrons from reaching the anode. The relationship is given by:
\[ eV_0 = K_{max}' \]
Since Kmax' is 3 eV (the maximum kinetic energy of the emitted electrons), we have:
\[ V_0 = 3 \, \text{V} \]

### Step 5: Establish the Relation of Stopping Potential to Anode-Cathode Potential
The stopping potential is defined relative to the cathode. Since the anode is at a higher potential than the cathode, the stopping potential relative to the anode (V_ac) is:
\[ V_{ac} = -V_0 \]
Thus:
\[ V_{ac} = -3 \, \text{V} \]

### Final Answer
The stopping potential of the anode relative to the cathode is:
\[ V_{ac} = -3 \, \text{V} \]

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