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

Solution

Physics sir · 3 steps

The photon pays the exit fee first; whatever energy is left over is the electron's kinetic energy.

1. Energy budget. One photon gives all its energy hν to one electron. The electron spends the work function φ to escape the surface, and the rest stays with it as kinetic energy.

   Kₘₐₓ = hν - φ

2. Put in numbers. Kₘₐₓ = 4.13 eV - 3.13 eV = 1 eV. The photon energy is larger than the work function, so electrons are ejected.
3. So. The fastest photoelectrons come out with 1 eV of kinetic energy, which is option (b).

The idea to keep. Einstein's equation is an energy subtraction: photon energy minus work function gives the maximum kinetic energy. The work function is a cost, so it is never added.

Answer from the NTA final answer key. Solution by PrepWiser.