I have a doubt about photoelectric effect and the nature of light in general. From what I understood, in order to ionize a piece of some material, I need an electromagnetic wave with a frequence greater or equal than a certain frequence, meanwhile the intensity of the incident wave doesn’t matter. But does the intensity of the wave transfer some energy to the material (like the one calculated with the poynting vector?)? If I increase the intensity of the wave (i.e. its amplitude) will this energy be able to ionize the material at a certain point? So the final question might be: does an electromagnetic wave possess two energy? One associated with the frequency and one associated with the intensity?
If you want to understand the photoelectric effect and energy of light you have to be very very careful to keep the terminology correct.
The energy of a electromagnetic wave (1 photon) is completely determined by its frequency.
The intensity of the electromagnetic wave(s) is how many photons per second are hitting on some area of surface. You can't combine photons to get a "bigger" photoelectric effect.
The gist of the photoelectric effect is this. The most loosely bound electrons in the material still have some specific binding energy. Until the photon has at least this much energy then it can't knock an electron free. But if it has more than that energy, then the electron gets the excess energy.
So if the binding energy is 4 eV:
- A 3 eV photon doesn't knock out an electron
- A 10 eV photon knocks out an electron with 6 eV of kinetic energy
The other "kind of energy transfer" (from the intensity of the photons) creates heat. So if a huge number of photons were absorbed by the material then it gets hot. If the material gets hot enough then it will emit electrons. (But this is not the photoelectric effect.) This was the purpose of the heater in old glass vacuum tubes. It heated the metal cathode so that it would release electrons.
The energy of a electromagnetic wave is completely independent of its amplitude. The amplitude merely describes how many quanta of energy are propagated. In essence the higher the intensity the more photons with the certain energy.
A good analogy for the photoelctric effect is a big room with an infinite amount of kids in it. They need to pay 5€ to get out of the room. If someone throws in 10 cent pieces no child will ever have enough money to escape the room because there are an infinite amount of children. If I throw in 5€ bills however a child can take it and instantly has enough money to leave the room.
In the analogy the value of the money is the energy and the amount of bills or coins is the intensity.
When the photoelectric effect was first examined the common understanding was much like yours that if the light behaves like a wave the metal should be able to accumulate energy over time until it emitted an electron but this was not the case because there is usually only single photon absorption by an electron. This photon therefore needs the right amount of energy.