In a photoelectric effect experiment, photons of energy `5eV` are incident on a metal surface They liberate photoelectron which are just stopped by an electrode at a potential of `-3.5V` with respect to the metal. The work fuction of the metal is

Options

A. `1.5eV`
B. `3.5eV`
C. `5.0eV`
D. `8.5eV`

Explanation

To solve the problem, we need to find the work function (φ) of the metal using the information provided about the photoelectric effect.

### Step-by-Step Solution:

1. **Identify the given values:**
- Energy of the incident photon (E_photon) = 5 eV
- Stopping potential (V_0) = -3.5 V

2. **Understand the relationship in the photoelectric effect:**
The photoelectric effect can be described using the equation:
\[ E_{\text{photon}} = \phi + eV_0 \]
where:
- \(E_{\text{photon}}\) is the energy of the incident photon,
- \(\phi\) is the work function of the metal,
- \(eV_0\) is the energy given to the electrons by the stopping potential.

3. **Calculate the energy due to stopping potential:**
The energy associated with the stopping potential is given by:
\[ eV_0 = e \times (-3.5 \text{ V}) = -3.5 \text{ eV} \]
(Note: The energy is considered as positive when calculating the work function.)

4. **Substitute the values into the photoelectric equation:**
Rearranging the equation gives:
\[ \phi = E_{\text{photon}} - eV_0 \]
Substituting the known values:
\[ \phi = 5 \text{ eV} - (-3.5 \text{ eV}) \]
\[ \phi = 5 \text{ eV} + 3.5 \text{ eV} \]
\[ \phi = 5 \text{ eV} + 3.5 \text{ eV} = 1.5 \text{ eV} \]

5. **Final Result:**
The work function of the metal is:
\[ \phi = 1.5 \text{ eV} \]