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When a strong light, or even a laser, hits a metal surface or some molecules, and causes lots of electrons to fly away, do the electrons leave together, or do some of them leave first? If it's the latter, then do the valence electrons leave first and then inner layers? If so, how long are the intervals between their leavings?

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – jonsca Mar 26 '18 at 22:27
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In 1905, Einstein demonstrated that the photoelectric effect was a quantum phenomenon. That is, for each photon of light of a minimal specific wavenumber, one electron in a specific orbital is released.

For example, for cesium in its base electron configuration, $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6 6s^1}$, about $\pu{3.43E-19 J}$ is needed to raise that electron to, effectively, an "infinite" orbital (where it is unlikely to be found near the atom). This corresponds to a wavelength of $\pu{\approx 580 nm}$ or wavenumber of $\approx \pu{17200 cm-1}$.

That said, very energetic photons may knock out more than one electron apiece, with some of the photon's surplus energy applied to the kinetic energy of the first electron, and some applied to ionizing the atom further. However, at the minimum photon level, for every photon absorbed, one electron is freed. A brighter light has more photons, so a brighter light releases more electrons (with limitations due to the build-up of free electrons near the $\ce{Cs}$ cathode, the space charge).

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Only electrons that are hit by a photon with the appropriate wavelength will be moved. So that means that they are leaving one by one. This can however ocurr at several places at a time so that means that they can leave simultaneously. In fact, if a current of 1 mA is maintained during 1 second, $\frac{10^{-3}\cdot 1}{1.6\cdot 10^{-19}}=6.25\cdot 10^{15}$ electrons will have left the metal.

Here $I=Q/t$ ($I=$ intensity, $Q=$ charge, $t=$ time) and the charge of one electron is $1.6\cdot 10^{-19} $ Coulomb.

The thing is that the metal remains electroneutral. It is not that a huge number of electrons leaves the metal, stripped off by the incident light. The whole setup works because there is an external electric circuit. Every electron that leaves the metal (or the semiconductor, for that matter) is immediately replaced by another electron returning from the circuit.

The deeper electrons in the atoms are never moved. Once and again the external electrons are pushed to the conductive band and replaced by fresh electrons from the circuit.

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  • $\begingroup$ What if the atoms or molecules are not charged by anything? And the energy of the light is strong enough to make multiple electrons to escape from one single atom or molecule? Then are these electrons flying away simultaneously or one by one or pair by pair? $\endgroup$ – OhLook Jan 24 '18 at 19:00
  • $\begingroup$ Only the electrons from the outer shell are involved in the photoelectric effect. So if there is only one electron, only one can escape. If there were two, after the first one has gone, the atom is charged $+1$ so the second one will need more energy to leave. To leave at the same time means that two photons hit both electrons simultaneously, which is an unlikely event, but there should not be a problem for them to leave both if they also could do it sequentially. "Strong enough" here means that you have a lot of photons, not that their energy is very high. $\endgroup$ – Raoul Kessels Jan 25 '18 at 15:07
  • $\begingroup$ Do you mean that only after the outer shell electrons are gone can the inner ones be gone? If the photons are very dense, then can the inner electrons and outer electrons absorb photons simultaneously and thus fly away simultaneously? Or only after outer ones leave can inner ones absorb photons and leave? $\endgroup$ – OhLook Jan 25 '18 at 18:12
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    $\begingroup$ As far as I know, the inner electrons cannot be moved. Even in the Compton scattering, in which much higher energy photons are used, only free electrons are ejected. See for instance physics.stackexchange.com/questions/31581/… $\endgroup$ – Raoul Kessels Jan 25 '18 at 23:50

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