We all familiar with the famous PE observation and Einstein's explanation. If a photon has energy less than the electron's binding energy, then there will be no PE observed. However, I am wondering if a low energy photon that has energy equal to M-K shell energy difference and able to excite the K-shell electron to M shell can then subsequently interact with the second photon through photo-electric effect to escape. Possible, even if the probability is near 0, or impossible?
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Perhaps you can start your exploration from searching on-line "two photon microscopy" or even "three photon microscopy". There are some interesting material to read. – verdelite Jan 08 '20 at 17:52
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This is a second-order process, so it has lots probability unless the field is very strong, but yes, it's possible. – Emilio Pisanty Jan 08 '20 at 17:52
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i think we need to distinguish between the low energy visible/uv photoelectric effect and the X-ray photoelectric effect. The second harmonic generation mentioned by Emilio works in low energy PE as dscussed in Photoelectric effect, low frequency light. Trying to do the same with X-rays would be a very different matter. – John Rennie Jan 08 '20 at 17:54
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1I believe there's also something called tunnelling ionisation, where the electric fields of the photons in a laser ''add up'' because they are coherent. The resulting field can be comparable to the field near the atoms, and could free the electrons. Locally, the valance electron of the atom effectively finds itself in a superposition of the potential of the atom and a potential which comes from the electric field of the laser. This gives rise to an ''effective'' potential that allows for the electron escape through quantum tunnelling. It can be used to create plasmas with a laser. – Philip Jan 08 '20 at 19:28
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Yes, two photon photoelectric effect is possible.
The first experimental observation of the two photon photoelectric effect was in 1964.
They used Cs3Sb1 and laser irradiation, with 1.17 eV, which as you describe is far below the threshold (the work function of the electron), but the process is still satisfactory. You can see that this is a higher order process and the linear function (one photon process) will dominate, but at high intensities the higher order process is possible, as you write, but the probability will be little.
https://aip.scitation.org/doi/10.1063/1.1754079
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 1000W/m2=10−1W/cm2. 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 107W/cm2) the effect might be measurable. They also note, that thermal heating by the laser field may make the pure second order effect unobservable.
Photoelectric effect – Why does one electron absorb one photon?
Árpád Szendrei
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There is two-photon photoemission, for example on a silver surface where the electron is first excited to a long-lived image state on the surface. The second photon then excited this bound state to a state with sufficient energy to overcome the work function.