I was watching this experiment (https://www.youtube.com/watch?v=v-1zjdUTu0o) which demonstrates the photoelectric effect, but it does not make any sense to me how it proves light as a particle instead a wave. Can you please explain me? And I also want to know where exactly the electrons that are released by beam of UV light go.
[Physics] How does photoelectric effect prove that light is also a particle
photoelectric-effectwave-particle-duality
Related Solutions
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.
The plate would have to have initially been negatively charged. When the UV light is shined on the plate, negative photoelectrons leave it so that the plate overall becomes neutral (negative charge leaving a surface induces a positive charge resulting in electrical neutrality if the surface was already negatively charged).
By saying the plate was first positively charged would result in it being more positive after photoelectric emission. This means it is not possible for the electroscope to become neutral. If the person in the video states that it has positive charge to begin with, then this person misspoke.
There are no electrons coming from the air and if you did the same experiment in a vacuum, you'd get the same result.
Best Answer
According to the classical theory, light is a wave and this effect would be explained to be due to transfer of light energy on to the Zinc plate. Increasing the intensity should therefore increase the kinetic energy of the ejected electrons.
However, this did not meet up with the experimental observations which showed that light below a particular frequency did not cause any photoelectric effect. Also, increasing the intensity increased the photoelectric current instead.
Einstein then proposed that light behaved as a stream of discrete wave packets called photons which had an energy value associated with them which depended on its frequency. Both observations could be exhaustively explained with this theory.
As shown earlier in the video, the plate is negatively charged which means it has excess of electrons. When hit with UV light, the electrons leave the metal and become dissipated in the surrounding air.