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In a vacuum, if electrons are accelerated by a certain voltage, giving the electrons a specific de Broglie wavelength and were incident on a piece of metal, providing the wavelength was roughly the diameter/distance between two of the atoms in the metal, would the electrons interact with the metal as a wave due to diffraction of the elections taking place?

If so, would this mean that the delocalised electrons further in the metal would receive the energy carried by the accelerated electrons in the form of a wave and as a result electrons in the metal would gain energy over a period of time as appose to instantaneous electron emission due to the transfer of energy from electron-electron collisions, as particles?

This is all theoretical, assuming perfect 'conditions'.

I apologise if this makes no sense what so ever. I was just looking at the photoelectric effect and wave particle duality and this question just arose out of curiosity.

Thanks.

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  • $\begingroup$ Wave particle duality is an obsolete concept that doesn't solve any physical problems and hence has little value. What you are referring to is called electron (diffraction) crystallography (en.wikipedia.org/wiki/Electron_crystallography) and it has some applications, albeit in practice x-ray crystallography is used more often. $\endgroup$ – CuriousOne Apr 8 '16 at 2:09
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The diffraction pattern is due to elastic scattering from the "ion core", which is the stationary net charge of the atomic nucleus and it's bound electrons; these elastically scattered electrons don't lose any energy. The electrons which interact with the free electrons are inelastically scattered, and contribute a foggy background to the diffraction pattern.

An electron which scatters more than once contributes to the background, so very thin crystals are used in transmission. For my research I found 100 nm worked well for gold and platinum, and several hundred nm for graphite.

Because of the energies involved this latter process may eject some of the free electrons, or generate x-rays. For an electron microscope the energies may vary from 20 to 200 keV or more; for low energy electron probes the electrons may have only a few eV.

There are many valuable applications for electron probes, including imaging, diffraction, and analysis. Electrons, x-rays, and neutrons each has it's own unique advantages. Electrons require ultrahigh vacuums, while x-rays can be done in air; electrons have much shorter wavelengths than x-rays of the same energy, and interact more strongly. In addition, it is easy to focus electrons, which allows you to image the surface.

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  • $\begingroup$ Or you consider the incident electrons as Bloch waves propagating through the material - much more general when dealing with anything beyond a single crystal. $\endgroup$ – Jon Custer Apr 8 '16 at 17:53
  • $\begingroup$ @Jon Custer: I usually use the wave analysis, but the elastic/inelastic scattering analysis is also useful, and seemed to address the OP better. I used scattering in some polycrystalline scattering models for computer simulations. $\endgroup$ – Peter Diehr Apr 9 '16 at 2:17
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What the scenario here is that you are comparing the photoelectric effect with X-ray diffraction. See a wave is a carrier of energy. IT could transmit energy from one point to another. But it cannot impart it's energy to another particle as a wave, but only by quanta of energy. That's what photoelectric effect tells us. An electron will be excited only if a certain discrete value of energy is given to it. You see that electrons in a metal are in quantized states. So it could not absorb energy continuosly over a period. It absorbs the energy only instantaneously. It absorbs energy only if that energy is the quantized value to get the electron excited. Otherwise it will not absorb that energy. So eventhough you substantiate that there is diffraction happenig inside the metal, the electron energy transfer takes place as a particle collision. You cannot account both particle and wave aspects at the same time. That's Bohr's complementary principle.

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