In photoelectric emission process from a metal of work function `1.8 eV` , the kinetic energy of most energetic electrons is `0.5 eV`. The corresponding stopping potential is

Options

A. `1.3 V`
B. `0.5 V`
C. `2.3 V`
D. `1.8 V`

Explanation

To find the stopping potential in the photoelectric emission process, we can use the relationship between the work function, the kinetic energy of the emitted electrons, and the stopping potential.

### Step-by-Step Solution:

1. **Understand the Concept**:
- The work function (φ) of a metal is the minimum energy required to remove an electron from the surface of the metal.
- The kinetic energy (KE) of the most energetic emitted electrons is the energy they possess after overcoming the work function.
- The stopping potential (V_s) is the potential difference needed to stop the most energetic electrons.

2. **Given Values**:
- Work function (φ) = 1.8 eV
- Kinetic Energy (KE) = 0.5 eV

3. **Use the Energy Conservation Principle**:
- The energy of the incoming photons (E) can be expressed as:
\[ E = \text{Work Function} + \text{Kinetic Energy} \]
- Therefore, we can write:
\[ E = φ + KE \]
- Substituting the known values:
\[ E = 1.8 \, \text{eV} + 0.5 \, \text{eV} = 2.3 \, \text{eV} \]

4. **Relate Kinetic Energy to Stopping Potential**:
- The kinetic energy of the most energetic electrons is equal to the work done against the stopping potential:
\[ KE = e \cdot V_s \]
- Here, e is the charge of the electron, and V_s is the stopping potential.

5. **Calculate Stopping Potential**:
- Rearranging the equation gives:
\[ V_s = \frac{KE}{e} \]
- Since we are working in electron volts (eV), we can directly use the kinetic energy value:
\[ V_s = KE = 0.5 \, \text{eV} \]

6. **Final Answer**:
- The stopping potential is:
\[ V_s = 0.5 \, \text{V} \]

### Summary:
The corresponding stopping potential is **0.5 V**.