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Why is the effective mass of electron different for metals

  1. Mar 10, 2015 #1
    ... and semiconductors?

    I'm trying to get some basic principles sorted in my head and I cant seem to find a straight answer to this. The lecturer drew a dispersion graph to explain it but I'm still a little confused.

    I understand since ##m^{*}\propto \frac{1}{\frac{\partial^2 E}{\partial k^2}}## then the curvature of the E vs k graph will determine the effective mass. BUT a metal will have curvature, a semi conductor will have curvature, so why should the effective mass be different?

    Is there a physical scenario that I'm missing here?

    Thanks for any ideas on this.
     
  2. jcsd
  3. Mar 10, 2015 #2

    ZapperZ

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    The degree of curvature depends on many different things, including the strength of the many different couplings between the charge carrier and other scattering agents (electron-electron scattering, electron-phonon scattering, electron-impurity scattering, etc.). This is a many-body phenomenon, and we simplify all these interactions and lump them (under the Fermi Liquid theory) into the effective mass.

    Zz.
     
  4. Mar 11, 2015 #3
    Thanks for the reply. I'm not sure how this relates to the differing effective mass for metals and semiconductors, are saying the curvature of the dispersion graph for a semiconductor is different from that of a metal, and that's why the effective mass differs? Please could you tell me how it differs and why?

    thanks again
     
  5. Mar 11, 2015 #4

    ZapperZ

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    First of all, even within metals themselves there can be variation in the effective mass. You only need to look at where the Fermi energy crosses the band. My simply changing the charge career density (adding or subtracting careers), the Fermi level will cross at different part of the band and thus, with different curvature. So there, you are already getting different effective mass, all within the SAME metal.

    Different material will have different band structure, even within metals themselves. There's no reason to expect that the Fermi level will always cross the band at the same curvature for all materials. So with this alone, you can easily see why a semiconductor will have a very different effective mass than metals, considering that the charge carrier in a semiconductor tends to be at the bottom of the conduction band for electrons, and at the top of the valence band for holes. This tends to have very different curvature than metals, which usually tend have Fermi crossing somewhere in the middle of the band (but not always).

    Zz.
     
  6. Mar 14, 2015 #5
    Thats very helpful! thanks
     
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