Let's say we have a emitter, emitting light that has frequency f, less than the threshold frequency of a metal.

If you leave light shining onto that metal, for long enough, does the energy of the individual photons accumulate, on the electrons, so eventually they will ionize, or does this not happen? What am I missing?


For simplicity let's consider the photoelectric effect in a thin metal foil:

enter image description here

The first step in the photoelectric effect is when a photon strikes an electron in the metal and transfers all its energy to it. The electron energy is now equal to the photon energy $h\nu$. If this energy is greater then the work function $\phi$ the electron can escape the metal and will emerge with a kinetic energy:

$$ \tfrac{1}{2}mv^2 = h\nu- \phi $$

However the $h\nu \lt \phi$ the electron will in effect bounce off the metal-air interface back into the metal:

PE effect

and the electron will start rattling around inside the metal. The trouble is that the metal has some resistance to the motion of electrons and the electron will very quickly lose its energy and come to a halt. By very quickly I mean less than a nanosecond.

So if a second photon strikes the electron before the electron has slowed to a halt, and while the electron is travelling in the right direction then yes the second photon could add enough energy to eject the electron. So in that case we would have photoelectrons ejected by absorbing two photons.

However this process is very unlikely as the two photons would have to be absorbed within a very short time. In practice the rate at which photoelectrons are ejected by two (or more) photon absorption is very slow though it can be observed in special cases. For example the paper Double-Quantum Photoelectric Emission from Sodium Metal by M. C. Teich, J. M. Schroeer, and G. J. Wolga, Phys. Rev. Lett. 13, 611, 1964 reports observation of exactly this effect in sodium.

  • 1
    $\begingroup$ Peter Quicke: thanks for providing the reference. I have edited it into my answer. $\endgroup$ – John Rennie Aug 25 '17 at 10:29
  • $\begingroup$ Another relevant example is photosystem II in plants, which catalyzes photodissociation of a water molecule using four 680nm (red) photons instead of two 340nm (ultraviolet) photons. $\endgroup$ – zwol Aug 25 '17 at 15:24

If you leave light shining onto that metal, for long enough, does the energy of the individual photons accumulate, on the electrons, so eventually they will ionize, or does this not happen? What am I missing?

Here is what happens :

Electrons ejected from a sodium metal surface were measured as an electric current. Finding the opposing voltage it took to stop all the electrons gave a measure of the maximum kinetic energy of the electrons in electron volts.


The minimum energy required to eject an electron from the surface is called the photoelectric work function. The threshold for this element corresponds to a wavelength of 683 nm. Using this wavelength in the Planck relationship gives a photon energy of 1.82 eV.

If you keep hitting with electromagnetic radiation an opaque metalic material below the critical frequency, the temperature of the material will go up because of atomic and molecular and lattice absorption of the energy of a portion of the impinging photons . This energy will raise the temperature of the lattice, and thuswill also raise the energy of the conduction electrons,

The conduction electrons can, therefore, be treated as an ideal gas. However, the concentration of such electrons in a metal far exceeds the concentration of particles in a conventional gas. It is, therefore, not surprising that conduction electrons cannot normally be analyzed using classical statistics: in fact, they are subject to Fermi-Dirac statistics (since electrons are fermions).


It is clear, from the above equation, that the Fermi energy $\mu$ is generally a function of the temperature $T$.

The work function of the photoelectric effect depends on this energy

As this paper shows, the work function will change with temperature


The results for pure silver metal show an evolution of the work function with temperature; a decrease of this parameter of about 0.2 0.3 eV is observed when the cathode temperature reaches 700 K, independently of the nature of the material. The obtained results are rather approximate. This method is not very accurate at higher temperature, because above 700 K the thermionic current dominates and the measurements of the photoelectric current are falsified by large fluctuations

In this sense , if the frequency is below the frequency of the work function by a tiny amount, the rising of the temperature from the absorbed photons will push the fermi level energies high enough so that the photoelectric effect can appear.

So there will be an effect, but it is a collective thermodynamic one, not on individual electrons as explained in John's answer, because of the 1/137 drop of the crossection (due to the coupling constant) for each photon interaction.

  • $\begingroup$ The Fermi level is not a function of temperature. Increasing temperature does not "raise the Fermi level" $\endgroup$ – Matt Aug 25 '17 at 15:03
  • $\begingroup$ @Matt I will correct it. $\endgroup$ – anna v Aug 25 '17 at 15:24
  • $\begingroup$ The Fermi level can be a function of temperature, since things like thermal lattice expansion will perturb the Bloch functions and thus the band structure. Not a lot, perhaps, but there is interaction. $\endgroup$ – Jon Custer Aug 25 '17 at 21:18
  • $\begingroup$ @JonCuster well , since by definition the Fermi level is defined at absolute zero, at least in my favorite general reference hyperphysics.phy-astr.gsu.edu/hbase/Solids/Fermi.html , I worked around it in my clarification. $\endgroup$ – anna v Aug 26 '17 at 3:56
  • $\begingroup$ Anna - From a solid state physics perspective, the simple thought experiment would be to take the crystal configuration at higher temperatures (i.e. with lattice expansion), and cool it to 0K. One step deeper, and the Fermi level is the energy such that the electron distribution is described by the Fermi function (and all the band structure and density of states factors) with a Fermi energy. $\endgroup$ – Jon Custer Aug 26 '17 at 17:49

Energy from photon comes in packets and these do not accumulate. The point to be noted that a single photon transfers energy to the electron.


No in photoelectric emission quantum nature of light is utilised. Photon gives its total energy to one and only one electron. Energy is not accumulated and there is no time lag between incidence of photon and ejection of electron.


This explains it and some things closely linked to it.


Note how frequency is the only key for the energy of photons since they lack mass and their speed is always that of a light for the medium they travel through.


Since at particle physics it's all about probability, multiple fast enough hits on electron would give it enough energy to leave the metal but it's highly unlikely. A few electron could escape given enough time and having precise enough instrument you could measure it.

Besides that rare occasions, rest of the energy absorbed from the light of lower than required frequency for that metal transfers into heating up the metal. Which eventually dissipates into surroundings.


protected by Qmechanic Aug 25 '17 at 14:00

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