The work function of a metal is \[6eV\] . If two photons each having energy \[4eV\] strike with the metal surface
(i). Will the emission be possible?
(ii). Why?

Hint: You can start by explaining the photoelectric effect and work function. Then write down that the photoelectron will not be ejected. Then for the second part, explain how the photoelectrons are not ejected when the energy of the photon is less than the work function.

Complete step-by-step answer:
Photoelectric effect - When electromagnetic like light electrostatic radiation strikes the surface material electrons are ejected out of the surface of the material. The electrons ejected out of the surface in this manner are called photoelectrons.
Work function – Work function is the minimum energy that the photon of the incident light should have for the photoelectron to be ejected out of the surface of the material.
(i). For the photoelectron to be ejected out, the energy of the photons hitting the surface of the metal should be equal or more than the work function. So in this case the photoelectron will not be ejected out of the surface of the metal even when two photons of energy \[4eV\] hit the surface of the metal.
(ii). This happens because the energy of the photons striking the surface of the metal is lower than the work function.
Changing the intensity of the incident light will not lead to the ejection of photoelectrons but increasing the energy of the photons will lead to the ejection of photoelectrons.

Note: The phenomenon of photoelectric emission is very important. Because this phenomenon can only be explained if we consider that light has a particle nature. Because if the light had only wave nature, then the photoelectrons would have emitted even if the energy of the photons striking the surface of the metal was less than the work function.