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Fermi-Dirac distribution for metals

  1. Jan 18, 2014 #1
    Hello everyone!
    I'm a little confused. The Fermi-Dirac distribution is about every electron in a metal or only about the valence electrons?
     

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  3. Jan 18, 2014 #2

    hilbert2

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    As far as I understand it, we think of the valence (conduction) electrons as forming a free electron gas, whose states are occupied as the FD distribution describes. The core electrons are thought to be bound to the atoms they belong to, and don't even have a continuous spectrum of possible states (bound states form a discrete spectrum), so they can't be described with the FD distribution. Well, in principle even the electron gas has a discrete set of states as it is confined to a finite volume, but the spacing between energy levels is very small.
     
  4. Jan 18, 2014 #3
    Thanks a alot! I'll give you an example to see if i got it right.
    The core electrons of lithium occupy the energy level E1s. All the other electrons occupy the levels E2s, E2p, E3s.
    The FD distribution is about all the electrons apart from the ones in E1s? Or only about the ones in E3s?
     
  5. Jan 18, 2014 #4

    hilbert2

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    A lithium atom has three electrons, and in the simple orbital model two of them occupy the 1s orbital and the remaining one occupies a 2s orbital. In the context of the electron gas model of electrical conductivity, we think of the 2s electrons getting delocalized over the entire volume of the metal object.

    In deriving the FD distribution, an assumption is made that the electrons interact very weakly with each other. This is not the case for the core electrons which are rather close to each other and feel their mutual Coulomb repulsion.
     
  6. Jan 18, 2014 #5
    I'm very sorry, I meant Sodium (Na), which has an atomic number 11. So the same questions but for sodium...
     
  7. Jan 18, 2014 #6

    hilbert2

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    All alkali metals have only one conduction electron per atom. In the case of sodium the 3s electron becomes the conduction electron.
     
  8. Jan 18, 2014 #7
    And in deriving the FD distribution we take in consideration only the 3s electrons right, not the 2s, 2p?
     
  9. Jan 18, 2014 #8

    hilbert2

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    Yes, the FDD only applies to conduction electrons. Just remember that the electron gas model is just that, only a model. Even the conduction electrons are actually interacting with their environment, and we usually take this into account by introducing an effective electron mass - the conduction electrons in the lattice behave as they would be heavier than genuinely free electrons.
     
  10. Jan 18, 2014 #9
    Thanks a lot! You 've been very helpful!
     
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