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The photoelectric effect suggests that photons (of frequency more than the threshold value) collide with the electrons and they get ejected.

But knowing the Uncertainty principle which suggests that an electron is more like a wave which is spread out and not accumulated at a fixed point like a particle, how come electrons be ejected?

Can we similarly separate water waves from the water surface by colliding it with a particle? Why/why not?

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  • $\begingroup$ How is it opinion based ? $\endgroup$
    – Ankit
    Jan 18 at 9:13
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    $\begingroup$ Why do you think a spread out particle (the photon) can not interact with another spread out particle (the electron) with a small but finite probability? $\endgroup$
    – A. P.
    Jan 18 at 9:14
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    $\begingroup$ Do you know about quantum mechanics? $\endgroup$
    – my2cts
    Jan 18 at 9:18
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    $\begingroup$ Why on the earth this question is likely to be answered with opinions rather than facts and citations? Whoever was flagging this question should have better to reread the definition of "opinion-based". $\endgroup$
    – GiorgioP
    Jan 18 at 9:21
  • $\begingroup$ @A. P. Sure a wave can interact with a wave but ejection !!!?? $\endgroup$
    – Ankit
    Jan 18 at 9:45
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You are quite correct that the conduction electrons in a metal are delocalised. They are approximately described by a function called a Bloch wave. In principle the electrons are delocalised over the whole piece of metal, though in practice they are delocalised over the distance between scattering events, which I think is generally about $10$ to $100\textrm{nm}$ though I don't have firm figures for this and it will depend on the metal and the temperature. But photons are also delocalised, so what we get is an interaction between two delocalised objects that changes the state of both.

I think where the problem arises is in your use of the word collide, because the interaction is not like a macroscopic collision e.g. between two billiard balls. Neither the electron nor the photon are point particles. Instead what happens is the oscillating electric field of the photon interacts with the electron and the two particles become entangled. In this state we no longer have two distinct particles as the wavefunction describing the two of them cannot be factored into an electron part and a photon part.

The combined wavefunction is time dependent and can evolve in different ways. One possibility is that it will evolve back into the original photon and electron states and there won't be any interaction between the photon and electron. Another possibility is Compton scattering where the state evolves back into a photon and electron but now their energies have changed, and a third possibility is that it evolves into a state describing a free electron and no photon. This third outcome is what produces the photoelectron.

I don't think there is a useful analogy to be made with a water wave as it isn't obvious what the equivalent of the entangled state would be for a water wave.

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  • $\begingroup$ Nice answer. "One possibility is that it will evolve back into the original photon and electron states and there won't be any interaction between the photon and electron.", I think it is worth mentioning that there can be interaction, in the form of elastic scattering, where the energy levels do not change, just that the photon changes angle (momentum vector). $\endgroup$ Jan 18 at 16:33
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Water waves obey classical mechanics. The wave is the change in the motion, and consequently momentum and energy of the water particles, an energy in motion wave. All this with classical mechanics.

The electron is not a wave, it is a point in space, a quantum mechanical particle with a fixed mass. In quantum mechanics there is a solution of a quantum mechanical equation that has the potentials and boundary conditions of the specific problem, in this case the scattering of given energy photons on a metal surface. The complex conjugate square of the wavefunction describing a given electron gives the probability for an electron to be ejected. It is a probability wave that describes the reaction. A photon of an energy higher then the energy binding the electron to the lattice has a probability to scatter with an electron and eject it. That is why the photoelectric effect is plotted on the y axis as the maxmimum photoelectron kinetic energy for that frequency, it is a limit of a more complicated interaction of the reaction photon+ metal-lattice.

photoel

But knowing the uncertainty principle which suggests that an electron is more like a wave

It does not. It just gives an envelope for determining two non commuting variables of the interaction, it says nothing about the nature of the electron.

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One can think of an ejected electron as a plane wave with a certain momentum. Being ejected doesn't mean that it has to be localized, only that it is not bound to the crystal anymore.

If we use water analogy, we could, e.g., think of a wave running away through the entry to a harbor. Dropping a massive object into the harbor would certainly eject some water into the open sea.

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  • $\begingroup$ thanks chuk .. but the electron wave can come out of their respective metal but can it be for water wave ? Can it come out if the sea ? $\endgroup$
    – Ankit
    Jan 18 at 13:25
  • $\begingroup$ I don't think it is a good analogy, but yes, water comes out with a splash, if you drop something inio it. Just like in photoeffect, where the photon energy needs to be sufficiently high to eject an electron. $\endgroup$ Jan 18 at 13:29

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