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Delocalization of states in valence band du to doping

  1. Apr 1, 2006 #1
    How delocalization of the states in the valence band occurs. Can somebody explain how many kinds of localization of states are there in semiconductors.
     
  2. jcsd
  3. Apr 1, 2006 #2
    If you use the envelope wave function, you get an H-atom like energy spectrum (e.g. for n-doping):
    [tex]
    E_D(n) = E_C+\frac{m*}{c^2 m}E_H = E_C - 13.6 eV \frac{m*}{j^2c^2m}[/tex]
    where Ec is the energy of the conduction band, m* the effective mass of an electron, and j an integer.
    From that, you get an effective Bohr radius of about 5nm, which is two orders of magnitude larger than a0.
     
  4. Apr 2, 2006 #3
    i could not understand it much though i get the idea of spread in band due to 5nm
     
  5. Apr 3, 2006 #4
    The idea behind it is:
    Before it gets doped, the semiconductor is perfectly happy with its electron distribution. Then it gets, e.g., p-doped. Nothing holds the extra hole anywhere except for the potential of the doping atom, so you can calculate its effective Bohr radius as said above, taking all the potentials into account.
     
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