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When I read about the photoelectric effect, I came across this:

"The electrons could not absorb more than one photon to escape from the surface, they could not therefore absorb one quanta and then another to make up the required amount – it was as if they could only embrace one quantum at a time. If the quantum absorbed was not of sufficient energy the electron could not break free. So 'escape energy' could only be transferred by a photon of energy equal or greater than that minimum threshold energy (i.e. the wavelength of the light had to be a sufficiently short). Each photon of blue light released an electron. But all red photons were too weak. The result is no matter how much red light was shown on the metal plate, there was no current."

So what is the physical explanation of "electrons could not absorb more than one photon"? How do we know its exactly one? For example, how do we know that changing the frequency doesn't change how much photons will get absorbed by one electron? One could argue that all photons have the same energy at whichever frequency but when you change the frequency, an electron could simply absorb more photons, thus gaining more energy.

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    They may absorb more than one photon, but the probability for that process scales with $E^2$, so for weak fields the absorption of one photon (scaling with $E$) dominates distinctly. So with enough intensity, electrons will be released from the surface, even if illuminated with radiation whose frequency is below the surface binding energy of the electrons. – Sebastian Riese Jun 03 '15 at 14:40
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    So the statement "one electron absorbs one photon" is only true under relatively weak fields(low intensity)? – TheQuantumMan Jun 03 '15 at 14:42
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    Exactly, but "relatively weak fields" are pretty strong by practical standards. – Sebastian Riese Jun 03 '15 at 15:30
  • Could you give me a source which describes in detail the things that you say please? – TheQuantumMan Jun 03 '15 at 15:41
  • Here is one, specifically on this topic: http://iopscience.iop.org/0022-3727/30/22/003/pdf/0022-3727_30_22_003.pdf – Sebastian Riese Jun 03 '15 at 15:53
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    Look at their "rather low laser power" in the range of $10^{10},\mathrm{W/cm^2}$ (that is pretty high intesity, sunlight has about $10^{-1},\mathrm{W/cm^2}$). – Sebastian Riese Jun 03 '15 at 15:56
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    @SebastianRiese: I believe that is an answer, not a comment. ;) – ACuriousMind Jun 03 '15 at 18:30
  • I believe that too! – TheQuantumMan Jun 03 '15 at 18:45

2 Answers2

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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.

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My view is different so don't downvote without considering it enough. The photoelectric effect theory was to laid down foundation of quantum mechanics, how matter interact with energy in quantum fashion.

There are many objection on explanation of the effect, but here take first. If electrons in an atom have quantized energy and they interact with energy in quantized manner, then how an electron absorbs energy more than potential function.

Also how could it be explained that higher frequency's photon emits electrons with higher kinetic energy. Is fraction of energy utilised for atomic structure. This shows that this is classical way of interaction, not quantum.

Also, intensity of light depends upon square of frequency, so double the frequency intensify light four times, $\varepsilon=h\nu$ and power is rate of energy, $p=\varepsilon\nu=h\nu^2$. Intensity can't remain constant on varying frequency. Classical relation of radiation power involves square of frequency time speed of wave.

Also current is saturate sooner after threshold frequency for given intensity of light, and graph shows current is linearly proportional to intensity of incident light. So it doesn't prove quantum nature of light.

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    It is certainly true, that the the photoelectric effect does not prove the quantum nature of light. But that doesn't fundamentally change the discussion (especially, since there is other evidence, that the em field is quantized). You get the same results for the higher order effects if you consider an atom in a classic time-varying external field.

    I also don't get what is "classical" about emission from a bound state into the continuous spectrum, that's also a normal quantum mechanical process.

    – Sebastian Riese Oct 01 '22 at 21:03
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    Sebastian Reese: an atom can absorb or emit photon because it has system of potential to store energy. A free electron can't necessarily completely absorb photon as per present understanding, it collide with photon as particle. So energy is not quantized for free electron. This description can be given as classical field, as in cyclotron where fields are used to change the velocity of charged particles. I discussed the experimental result, that intensity is not independent of frequency in classical case also, so a small increment in frequency is many times increase in intensity. – Neil Libertine Oct 02 '22 at 02:19
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    Sebastian Riese: Intensity depends upon square or cube of frequency and not independent of it. So a intensity related photo current can be related as frequency dependent. In your answer laser showed that intensity is also dependent upon amplitude otherwise no photo current below cutoff. – Neil Libertine Oct 02 '22 at 02:38