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Metals and free electron capture

  1. Jan 6, 2012 #1
    I am not a condensed matter physicist. Can someone explain the situation for free electrons in a metal to me. It seems like they would head straight for the nearest positively charged lattice nuclei. As it got very close <3 fermi it would radiate a lot because it is accelerating. So after a few back and forths it would be stuck at the nuclei without enough energy to leave. But clearly this is not what happens. The question is why?
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  3. Jan 7, 2012 #2

    Simon Bridge

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    Because the conduction-band electrons have enough energy to leave the nuclei.

    The proximity of the nuclei in the lattice lowers the potential barrier between them and causes the discrete atomic energy levels to split and broaden into bands. If the element has enough electrons to spill over into the conduction band it is a conductor and the conduction electrons are not tethered to their nuclei. They cannot lose energy to drop to a lower band because all those energy levels are already occupied (Pauli Exclusion Principle).

    The actual situation can get quite complex - but the bottom line is that about 2 electrons per atom (metals) don't feel a strong-enough localized +ve charge to hold them.
    (Google for "band theory of solids".)
  4. Jan 7, 2012 #3


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    As Simon Bridge has stated, this CAN become a complicated issue, and in some case, apparently contradictory.

    When atoms form a solid, they are in close proximity to each other. What this means is that their wavefunction will now overlap, especially the outermost shells. The overlapping can be quite complicated. It might overlap with its nearest neighbor, its next nearest neighbor, its next next nearest neighbor, etc. When one does a calculation on such a thing, one get "bands" of energies.

    Now, if you look at the Schrodinger equation/Hamiltonian, you have the kinetic part, and the potential part. At some point, the kinetic part can be significantly larger than the potential part, and this means that for all practical purposes, the particle is "free" and is not confined to a particular potential. This is what happens in a conduction band for a conduction electron.

    But there is a caveat in this, as often is the case. In many solids, the "free electron" model works quite well. The Ohm's law that we know and love can be derived using the Drude model that is built upon the presence of such free electrons. However, there are also many cases in which such free electron model breaks down, and breaks down spectacularly. In such a case, one can no longer ignore the presence of a potential field. One such example is to model a periodic potential or Bloch potential. This allows for one to still produce bands, but also allows for the ability to introduce weak periodic potential to the system.

    Why would one need to do that? One good reason is the photoelectric effect. A photoelectric effect cannot occur on a free-electron. We can see this in the reflection mode photoelectric effect, where a photon hits the surface of a metal, gets absorbed, and an electron comes out of the surface of the metal, with a momentum component in the opposite direction as the incoming photon. There is a clear violation of conservation of momentum here if all we have is the photon, and a free electron that comes out of the metal.

    What actually happens is that there is a very weak "coupling" of the conduction electrons to the lattice ions. So one can imagine this periodic weak potential that, while still allowing the electrons to move about the crystal, allows for this coupling between the electrons and the lattice. So in the photoelectric effect, the lattice ions are the ones absorbing the recoil momentum of the electron and the momentum of the photon, and all is well with the universe with no violation of conservation of momentum. Without this weak potential, there's no coupling between the lattice and the electron, and we would be in deep doo doo.

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