Photoelectric effect – Why does one electron absorb only one photon?

Question

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.

Answer 1

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.

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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.

Answer 2

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.

Answer 3

When light hits a metal, an electron can absorb a photon, but it has to get all the energy it needs to escape from that single photon. It can’t "save up" energy from multiple photons because once it absorbs a photon, if that energy isn't enough, the electron doesn’t just sit there—it enters an excited state. This means it moves to a higher energy level.

However, in a metal, excited electrons don’t stay excited for long. They are surrounded by a dense sea of free electrons and positively charged ion cores, leading to constant collisions and energy exchanges. This causes the electron to lose its extra energy, and it happens within about 10⁻¹⁵ to 10⁻¹² seconds (that’s a femtosecond to a picosecond, unimaginably fast).

So, by the time another photon arrives, the electron has already lost the energy from the first one and is back to its ground state, as if nothing happened. When this new photon comes in, the electron starts fresh—it either gets enough energy in one go, or the cycle repeats.

Now, what if two photons arrive very close together? If they hit at almost exactly the same time, there’s a tiny chance the electron could absorb both before relaxing. But under normal light conditions, this is extremely unlikely because the relaxation happens way too fast. While multi-photon absorption is possible under intense laser light, in everyday scenarios, an electron either gets all the required energy from a single photon or not at all.

That’s why photoemission works only when a single photon has enough energy to free the electron in one shot.