# Width of a Wigner-Seitz cell

1. Aug 25, 2009

### Rory9

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

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

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 $$a_{1}^{*} = (2\pi/V)(a_{2} \times a_{3})$$ that this gives an fcc set of lattice vectors (but perhaps this move was unnecessary?).

I have been given the hint that $$|a^{*}| = 4\pi / a$$, but when I determine the magnitude of my $$a_{1}^{*}$$ 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.