Explanation:

1. Photocell and Potential:
A photocell is a device that converts light energy into electrical energy. It consists of a photosensitive material, typically a metal, which emits electrons when exposed to light. These emitted electrons can be collected as a current if a potential difference is applied across the photocell.

In this case, the photocell is maintained at a negative potential of 2V. This means that the potential difference across the photocell is 2V, with the negative terminal at a higher potential than the positive terminal.

2. Emission of Photoelectrons:
When light of sufficient energy (greater than the work function of the material) falls on the photocell, it excites the electrons in the material and allows them to overcome the work function barrier. This results in the emission of photoelectrons from the surface of the material.

3. Kinetic Energy of Photoelectrons:
The energy of the most energetic photoelectron can be determined using the equation:

K.E. = eV

Where K.E. is the kinetic energy of the electron, e is the charge of the electron (1.6 x 10^-19 C), and V is the potential difference across the photocell.

In this case, the potential difference is 2V, so the kinetic energy of the most energetic photoelectron is:

K.E. = (1.6 x 10^-19 C) x (2V)

= 3.2 x 10^-19 J

4. Conversion:
To convert the kinetic energy from joules to electron volts (eV), we can use the conversion factor:

1 eV = 1.6 x 10^-19 J

Therefore, the kinetic energy of the most energetic photoelectron is:

3.2 x 10^-19 J / (1.6 x 10^-19 J/eV) = 2 eV

Conclusion:
Thus, the energy of the most energetic photoelectron in this photocell is 2 eV, which is equivalent to option A.

## K.Emax =(Vo)eV=2eV##