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I don't know if electrons work as particles or waves or maybe both in photoelectric effect.

How is Photoelectric Effect actually described by Wave-Particle Duality?

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  • $\begingroup$ It's not the electron that we consider as wave, when analysing the photoelectric effect, but that the E/M field consists of particles and is not a wave. So, for the photoelectric effect, the wave-particle duality is about the E/M field and not the electron $\endgroup$ – user3257624 Mar 30 '16 at 20:39
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Depending on the experiment electrons can behave either like a wave or a particle.

In the photoelectric effect is the electron is mainly exhibiting particle behavior. The kinetic energy of the electron is equal to the energy of the photon minus the binding energy of the electron.

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  • $\begingroup$ In the photoelectrit effect, the important thing is the particle nature of the photon (which before that was considered as wave) and not whether the electron manifests as wave or particle. $\endgroup$ – user3257624 Mar 30 '16 at 20:04
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The "particle wave duality" comes because the experimental behavior of particles in the microcosm is mathematically described by quantum mechanics. In quantum mechanics a "particle's" position can only be calculated from a probability distribution, the complex conjugate square of the wave function. The wave function is a solution of the quantum mechanical boundary problem at hand . Thus an electron is not running around an atom in an orbit, but its probable location is described by an orbital, a probability locus. The incoming photon has its associated probability distribution/wave-function.

The photo electric effect comes when a photon of energy equal or larger than the quantized binding energy of the electron , scatters off the atom and the electron is ejected. The mathematical expression for the scattering involves an overall wave function, whose complex conjugate square gives the probability distribution for the problem . Each individual electron ejected will have an angle and energy in the electron detector,a particle property, the cumulative distribution of many such electrons will be governed by the probability function, i.e. the wave nature of the interaction.

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  • $\begingroup$ Side note: Wave functions aren't probability distributions, their squares are. Overall, succinct answer. $\endgroup$ – Tamoghna Chowdhury Mar 30 '16 at 9:01
  • $\begingroup$ @TamoghnaChowdhury I expanded $\endgroup$ – anna v Mar 30 '16 at 10:22
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There are a lot of experimental results that are inconsistent with the assumption that electromagnetic radiation is a wave field.

A. Assuming that radiation is a wave field then it would be absorbed gradualy by the electron and the time needed to trasnfer the necessary for extraction energy to the electon is ~1sec. However, electrons are extracted after ~10^-9 sec after the radiation impacts. This suggests that energy is absorbed instantly, as there exists packets of energy (photons) carrying the required energy which are absorbed by the electron instantly and not gradually.

B. Photoelectring current appears when the frequency of the electromagnetic wave is greater than a lower bound, regardless of the radiation's intensity. So the above hypothesized particles's energy should somehow related to the frequency of the E/M radiation (Planck's quanta!)

These considerations, amongst others, are the qualitative results that suggest the particle nature of the E/M field. After making the assumption that photons do exist and are the quanta of energy that Planck had talked about, then the correct experimental results occur directly.

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