I was reading photoelectric effect , which was completely explained by Einstein. And a bit difficult question , atleast seems to me is whether photoelectric effect is inelastic collision or elastic collision . Every where it is written that photoelectric effect is example of inelastic collision . But as far as I know in inelastic collision some energy become internal energy of the system , there maybe some energy loss as heat or something . But here in photoelectric effect photon is being absorbed by electron but where is the energy loss? No energy is being lost here the energy $h\nu $ is absorbed by electron and its kinetic energy increases so no energy loss occurs . So in my opinion it is not an inelastic collision . So what is the reality , is it inelastic or elastic collision ?
[Physics] Is photoelectric effect example of inelastic collision
photoelectric-effectquantum mechanics
Related Solutions
More than one photon can be absorbed, but the probability is minute for usual intensities. As a scale for "usual intensities" note that sunlight on earth has an intensity of about $1000\,\mathrm{W/m^2} = 10^{-1}\mathrm{W/cm^2}$.
The intuitive reason is, that the linear process (an electron absorbs one photon) is more or less "unlikely" (as the coupling between the em. field and electrons is rather weak), so a process where two photons interact is "unlikely"$^2$ and thus strongly suppressed. So for small intensities the linear process will dominate distinctly. The question is only, at what intensities the second order effects will become visible.
In the paper by Richard L. Smith, "Two-Photon Photoelectric Effect", Phys. Rev. 128, 2225 (1962) the photocurrent for radiation above half of the cutoff frequency but below the cutoff frequency is discussed. They note that for usual intensities the photocurrent will be minute, but that given strong enough fields such as those observed in a focus spot of a laser (on the order of $10^7\,\mathrm{W/cm^2}$) the effect might be measurable. They also note, that thermal heating by the laser field may make the pure second order effect unobservable.
The more recent paper S. Varró, E. Elotzky, "The multiphoton photo-effect and harmonic generation at metal surfaces", J. Phys. D: Appl. Phys. 30, 3071 (1997) dicusses the case where high intensities (on the scale of $10^{10}\,\mathrm{W/cm^2}$) produce even higher order effects (and unexpectedly high, coherent non-linear effects, that is absorption of more than two photons by one electron). Their calculations explain the experimental observations of sharp features in the emission spectra of metal surfaces.
Historical fun fact: The 1962 paper is so old, that it talks about an "optical ruby maser"; lasers where so new back then, they did not even have their name yet.
The term photoelectric effect is used to describe two different phenomena. The first is the one you are thinking of, i.e. the ejection of photoelectrons from a metal surface by visible/uv light, and the second is the ejection of electrons from atoms by X-rays (as used in X-ray photoelectron spectroscopy).
Ejection of photoelectrons by X-rays is basically ionisation. I'm not sure I'd restrict the term ionisation to mean only removal of the outermost electrons as your lecturer seems to.
Gases don't have a work function like a solid or liquid, because their atoms/molecules don't overlap strongly enough to form an electronic band structure. So there is nothing analogous to the visible/uv light photoelectric effect in metals. However the X-ray photoelectic effect is the same in gases as in solids and liquids. The energy of X-rays is so much higher than the bonding energy in solids and liquids that the photoelectron ejection is the same for all three phases.
Best Answer
When a photon interacts with an atom, three things can happen:
elastic scattering, when the photon keeps its energy, and changes angle
inelastic scattering, when the photon gives part of its energy to the atomic system and changes angle
absorption, when the photon gives all its energy to the atom, and the valence electron will move to a higher energy level, and then the electron will eventually move back to a lower energy level and emit a photon
At low energies, Rayleigh scattering happens, 1. elastic.
In your case, photo electricity is when 3. happens. That is at higher energies.
At even higher energies, 2. happens, Compton scattering.