In photoelectric experiment energy of $$2.48 \text{ eV}$$ irradiates a photo sensitive material. The stopping potential was measured to be $$0.5 \text{ V}$$. Work function of the photo sensitive material is :

In a photoelectric experiment, light of energy 2.48 eV is incident on a photosensitive material, and the stopping potential is 0.5 V. We need to find the work function.

Recall Einstein's photoelectric equation.

$$ E_{photon} = \phi + KE_{max} $$

where $$\phi$$ is the work function and $$KE_{max} = eV_0$$ (stopping potential times electron charge).

Express in electron volts.

Since the stopping potential is 0.5 V, the maximum kinetic energy is $$eV_0 = 0.5$$ eV.

Solve for the work function.

$$ \phi = E_{photon} - KE_{max} = 2.48 - 0.5 = 1.98 \text{ eV} $$

The correct answer is Option (3): 1.98 eV.