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I don't understand how electrons and photon interact with each other when a metal surface is illuminated with light. I've read that below a certain threshold frequency or wavelength of light, no matter how much we increase the intensity or time of exposure, no electrons get emitted because the photons don't have sufficient energy and assuming that only one photon interacts with one electron. It is considered to be discrete rather than continuous. But i think it could be continuous.

Let's consider a photon of red light whose wavelength is above the threshold wavelength (i.e insufficient energy to emit electron). Let this red photon hit an electron. The electron just goes to higher energy level but is not freed since energy required is less. Now since the electron remains in higher energy level for some time, it has high chance that it gets hit by another photon of red light since the speed of light is very high. The same electron can be hit by the next photon since the time next photon arrives is very very less than electrons transition time. So can't the electron in higher level get the energy of 2nd photon and travel to next higher level before it goes to ground level and so on it can be freed from the atom?

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    $\begingroup$ Please rerwrite your question in a coherent matter. This is a stream of consciousness and very hard to read. Please erase all superfluous dots and question marks and state clearly what you ask about, what your question is and where you see problems. $\endgroup$
    – Martin
    Oct 9, 2014 at 12:58
  • $\begingroup$ In reality electrons and photons are part of the same field. Electrons have one negative charge and spin 1/2, photons have no charge and spin 1. These quantum numbers and their interaction with each other are the description of one object, which is the quantum field. So what you are really looking at is the quantum field changing its configuration in the only way it can: by changing the momenta and number of its quantized states. $\endgroup$
    – CuriousOne
    Oct 9, 2014 at 14:38

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What you are describing might be possible - that a series of interactions with photons might cause an electron to first be excited, and then to be emitted.

However, the probability of an individual electron interacting with two photons in quick succession without an intermediate interaction with other electrons (and thus an opportunity to shed the excess energy) is effectively tiny. Macroscopically speaking, if the flux of incident photons was really this high, then the material would increase in temperature, the electron motion would increase, and the probability (Boltzmann statistics) of an electron escaping would indeed go up.

But that would no longer be the photoelectric effect. It would be called "heating". As @garyp points out in his answer (and Jon Custer in a comment to this question) a highly focused and intense laser beam can be (and has been) used to do "two step" excitation of electrons, demonstrating that the effect you were searching for does indeed exist in nature.

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  • $\begingroup$ Indeed, that would be analogous to multiphoton ionization of atoms - but you need a whole lot of photons. Interestingly enough, a quick Google resulted in articles like 'High-order multiphoton photoelectric effect at midinfrared laser wavelengths', A.T. Georges, Phys Rev. A66, 063412 (2002). So, yes, it does happen in short intense laser pulses. $\endgroup$
    – Jon Custer
    Oct 9, 2014 at 13:14
  • $\begingroup$ @JonCuster - indeed: there is also something called "two photon microscopy" in which very intense laser pulses excite an electron with two photons to a new level from which an emission is possible at an otherwise inaccessible wavelength: this allows one to observe effects at just the focal point of the laser, eliminating a lot of scatter effects. It's not called the photoelectric effect though. $\endgroup$
    – Floris
    Oct 9, 2014 at 13:19
  • $\begingroup$ Oh...so that would not be a photoelectric effect then...yes i just need to confirm about its possibility and actually I don't have much knowledge about quantum things... $\endgroup$
    – Sagaryal
    Oct 18, 2014 at 6:50
  • $\begingroup$ It would be "photo electric" effect but not "the" photoelectric effect... $\endgroup$
    – Floris
    Oct 18, 2014 at 6:53
  • $\begingroup$ Green laser pointers (used to) work by frequency doubling a 1064 nm diode-pumped solid state laser via an interaction with two photons in quick succession without an intermediate interaction with other electrons, and some of those lasers are not tiny, though it's in a crystal rather than a metal. $\endgroup$
    – uhoh
    May 2, 2020 at 2:59
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This is possible, and it's done. It's used as a form of spectroscopy. It's called two-photon photoemission. See, for example this.

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electron can't escape from atom in your case, why because if electron is at energy level say E1 (got it from photon of light) and another photon colloide it, then electron can have energy to take transition from E0 energy level to E1(if electron is at E0 level), in this case electron is at E1 enegy level so it requires E2-E1 amount of energy to take transition from E1 to E2.

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  • $\begingroup$ Multiphoton, non-resonant ionization has been studied for over 2 decades now. In intense laser fields the electron can absorb multiple (even 10 or more) photons at a time to escape deep atomic orbital levels. $\endgroup$
    – Jon Custer
    Oct 9, 2014 at 16:31

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