UV light of $$4.13 \text{ eV}$$ is incident on a photosensitive metal surface having work function $$3.13 \text{ eV}$$. The maximum kinetic energy of ejected photoelectrons will be:

UV light of energy 4.13 eV is incident on a metal surface with work function 3.13 eV. We need to find the maximum kinetic energy of ejected photoelectrons.

Recall Einstein's photoelectric equation.

$$ KE_{max} = h\nu - \phi $$

where $$h\nu$$ is the energy of the incident photon and $$\phi$$ is the work function of the metal.

Substitute the given values.

$$ KE_{max} = 4.13 - 3.13 = 1.00 \text{ eV} $$

Explanation: The work function $$\phi = 3.13$$ eV is the minimum energy required to eject an electron from the metal surface. Any photon energy in excess of the work function is converted into kinetic energy of the ejected electron. Since the incident photon has 4.13 eV and 3.13 eV is used to overcome the binding energy, the remaining 1.00 eV becomes the maximum kinetic energy of the photoelectron.

The correct answer is Option (3): 1 eV.