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Width of a Wigner-Seitz cell

  1. Aug 25, 2009 #1
    1. The problem statement, all variables and given/known data

    Treating Sodium (Na, bcc, a= 4.23 Angstrom) as a free electron metal with a single valence electron per atom, calc. the Fermi wave vector and compare with the distance to the Brillouine zone boundary closest to the origin.

    3. The attempt at a solution

    I have obtained the Fermi wave vector via

    [tex](4/3)\pi k_{F}^{3} = (N/L)(2\pi/L)^{3}[/tex]

    and obtain 0.92 inverse Angstrom.

    For the Brill. zone, I understand that this must be constructed from the reciprocal lattice, and I have found (using [tex]a_{1}^{*} = (2\pi/V)(a_{2} \times a_{3})[/tex] that this gives an fcc set of lattice vectors (but perhaps this move was unnecessary?).

    I have been given the hint that [tex]|a^{*}| = 4\pi / a[/tex], but when I determine the magnitude of my [tex]a_{1}^{*}[/tex] I find it to be a factor of root 2 less than this.

    Is there some simple formula that is being used here to decide what the lattice width must be for this Wigner-Seitz cell? Should I think of the 'origin' as being at the centre of this cell. I'm not sure what it is I should be doing in order to 'compare with the distance to the Brillouine zone boundary closest to the origin'.

    Cheers.
     
  2. jcsd
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