My sir told me that Energy required to remove an electron from one orbit to another depends on hf * n. Where n means the no of photons that will strike on a metal surface and hf is energy of one photon.(Means energy of each photon is going to be 2eV)[Like the example below I have written ].

My question is that if I continuously hit many photons on a metal surface.Why doesn’t the electron come out then since the energy also increases or hf’s.

If work function(energy required of an electron ) is 3eV. My hf value is 2eV and n = 6.Now wouldn’t the total energy value increase and electron should come out (Like from metal surface).?This as I know means that there are 6 photons of energy 2eV.

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    $\begingroup$ You have it all backwards. Roughly speaking, the number of photons is always 1. If your electron needs 3 eV, then no number of photons having 2eV will do the job. $\endgroup$ Commented Dec 20, 2020 at 14:40
  • $\begingroup$ Please explain more clearly. Your first sentence mentions the energy necessary to "remove" an electron. Remove ? Are you speaking of absorption of light, or emission of light ? In both cases, the absorbed or the emitted energy does not depend on the number of photons. Later on, you say that "you hit" many photons... What do you mean ? Photons cannot be hit ! They can hit a target. But they cannot be hit. $\endgroup$
    – Maurice
    Commented Dec 20, 2020 at 14:46
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    $\begingroup$ Have you heard about the photoelectric effect, explained by A. Einstein in 1905 ? In a shortcut. if photons have lower then the threshold energy, it does not matter how many of them is colliding. They will not cause the photoemission. ( Well, then can cause thermal electron emission, but that is other case. ). $\endgroup$
    – Poutnik
    Commented Dec 20, 2020 at 15:06
  • $\begingroup$ @Maurice photoelectric effect $\endgroup$
    – Alchimista
    Commented Dec 21, 2020 at 10:38
  • $\begingroup$ @IvanNeretin - multiphoton ionization is definitely a thing... $\endgroup$
    – Jon Custer
    Commented Jan 5, 2021 at 17:09

2 Answers 2


Though the explanation of the photoelectric effect is clear, and you should see how Albert Einstein won his 1921 Nobel prize for this, to some extent, it is possible for multiple photons to be absorbed, either simultaneously (or in sequence, if an intermediate metastable state allows it), to cause phenomena that would seem to be beyond their energy.

In "multiphoton fluorescence process, two or more photons are simultaneously absorbed by a molecule, which then emits a single photon at a different wavelength." That is, the energy of the photon emitted in fluorescence may be greater than that of either incident photon.

The Max-Planck-Institut für Kernphysik has an excellent description of the photoelectric effect and multiphonton effects.

One can consider these effects through classical physics, too: as more photons are "squeezed" in time and space, the electric field increases. In the extreme, high intensity fields produced by femto- and attosecond lasers not only tear off inner electrons, they can even cause nuclear particles to be ejected!

So your idea that perhaps the energy of photons could be summed is not entirely wrong... but it requires the photons to interact simultaneously with matter. This only takes place where the radiant flux is immense. Unless you're working with multiphoton microscopy or laser-induced nuclear fusion, the photoelectric effect is the accepted answer.

  • $\begingroup$ What does it mean by photon to interact simultaneously with matter. Like 2 photons join each other ? $\endgroup$
    – Srijan
    Commented Dec 23, 2020 at 17:03
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    $\begingroup$ To some extent, photons might interact with each other: sciencenews.org/article/lhc-atlas-photons-interact-physics . However, "simultaneously," here, means that within quantum limits, the wave functions supplement each other. This is the Planck on which quantum mechanic stands. $\endgroup$ Commented Dec 23, 2020 at 19:47

The photoelectric effect is explained in such a way that some people get it right away, but others (including physicists before Einstein) don't/didn't.

Try this: take a photon gun that shoots one photon at a time, with a selectable energy. Fire it at the metal target. Here's what will happen: At low energies (reddish light), the photon is absorbed (let's ignore reflection!) and shakes up the sea of electrons - makes it warmer, but does not make a splash. Turn up the energy of the photon gun (toward the blue end of the spectrum) and shoot again. Now the photon is energetic enough so that when it hits the metal - the sea of electrons in the metal - it actually bangs an electron so hard that it pops out.

The photoelectric effect simply says that electrons are bound into a metal and low energy photons are absorbed by the whole sea of electrons because no one electron is excited enough to leave the metal. But a high energy photon can knock an electron so hard - so fast - that it gets knocked out of the metal before it can distribute its energy to other electrons in the sea of electrons.

Consider the interaction of a photon with the metal as having 3 possible outcomes: a) reflection (which we have ignored), b) absorption with temperature increase (very familiar to people who go out into sunlight with black clothing), and c) electron ejection because the photon energy is enough to excite an electron so much that it is ejected. This third possibility is measured by physicists with specialized equipment that captures the ejected electron.

But we normal human beings can't observe that. Or can we? Sunlight with its considerable blue and UV components can bleach fabric and cause sunburn by generating free radicals (the residue of an ejected electron!). But an infrared lamp will not bleach fabric, and won't cause sunburn (although it can cause thermal burns - which are somewhat different, but I suppose equally painful, so don't experiment on yourself).

  • $\begingroup$ Amazing answer.Thank you very much.Just some more clarity form what you wrote.What happens if we do not ignore reflection ? And By black clothing , does it mean that the work function of black is so high that it absorbs a lot of photons and heat up. $\endgroup$
    – Srijan
    Commented Dec 20, 2020 at 16:53
  • $\begingroup$ While the answer is explanatory and more visual, we should avoid making macroworld analogies which are not correct, such as a photon banging an electron hard that it pops out or a high energy photon knocking the electron hard and fast. Modern physicists will perhaps frown upon these macroworld similarities. $\endgroup$
    – ACR
    Commented Dec 20, 2020 at 17:15
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    $\begingroup$ @srijan nahar: I don't have numbers on work function of black material, but since anything will eject electrons if you hit it hard enough, it would seem that you have to hit black stuff harder. $\endgroup$ Commented Dec 22, 2020 at 16:06
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    $\begingroup$ @M. Farooq: You are quite right in insisting on the best analogies for introducing people to an understanding of complex phenomena. Modern physicists (and mathematicians) have a unique mental ability to condense complicated processes into simple symbols which are unfamiliar to beginners. Sometimes a picture (worth 1000 words) is more explanatory than a symbol worth 1e^6 words. $\endgroup$ Commented Dec 22, 2020 at 16:16

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