The work function of aluminium is 4.2 eV. If two photons, each of energy 3.5 eV, strike electrons of aluminium, then the emission of electrons.
A. Will be possible
B. Will not be possible
C. Data is incomplete
D. Depends upon the density of the surface

Hint: The amount of energy needed to remove one electron from a metal surface is known as the work function. Work function is also referred to as surface characteristics, not metal properties overall. We already know that for photoemission to occur, the incident light or photons' energy must be greater than the metal's work function.

Formula Used:
By definition of photoelectric effect,
\[K.E = h\nu - {W_0}\]
Where, h is Planck’s constant, \[\nu \] is frequency of light and \[{W_0}\] is work function.

Complete step by step solution:
Photons are packets of moving energy in the form of light. When photons interact with some metal surfaces it shows a phenomenon called the photoelectric effect where the energy of a photon is absorbed by metal while electrons are emitted. But there are some conditions that are necessary for the photoelectric effect to take place and one of them is a work function. We know that by the photoelectric effect we have,
\[K.E = h\nu - {W_0}\]

In the photoelectric effect if the electron has to come out of a metal surface, then the electron should have energy greater than that of the work function. That is,
\[E > {W_0}\]
\[\Rightarrow h\nu > {W_0}\]
In this case they have given that the two photons, each of energy 3.5 eV are incident on its surface and the work function of aluminium is 4.2eV.
That is here, \[h\nu < {W_0}\]
Therefore, no electron will be emitted.

Hence, option B is the correct answer.

Note: Photons are defined as packets of energy and the energy required to knock out electrons is to be provided all at once. Here the most important thing to keep in mind is that, in order to emit an electron from the surface of the metal, the energy of the photon should be greater than the work function.