How Does the Fermi Wave Vector of Sodium Compare to Its Brillouin Zone Boundary?

[Rory9]

Homework Statement

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.

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.

 

[Jhero]

Thank you for your post. In order to calculate the Fermi wave vector for sodium, we first need to understand the concept of Fermi energy and Fermi level. Fermi energy is the maximum energy that an electron in a solid can have at absolute zero temperature, while Fermi level is the energy level at which the probability of finding an electron is equal to 1/2. In a free electron metal, the Fermi level lies at the top of the energy band.

To calculate the Fermi wave vector, we can use the formula you have mentioned:

(4/3)πkF^3 = (N/L)(2π/L)^3

where N is the number of electrons, and L is the length of the metal. For sodium, we have 1 valence electron per atom, so N = 1. The length of the metal can be calculated using the lattice parameter a = 4.23 Angstrom. Therefore, L = Na = 4.23 Angstrom.

Substituting these values in the formula, we get:

(4/3)πkF^3 = (1/4.23)(2π/4.23)^3

Solving for kF, we get:

kF = (3/4)×(1/4.23)^(1/3)π^(2/3) = 0.92 Å^-1

Now, for the Brillouin zone boundary, we need to consider the reciprocal lattice of the bcc structure. The reciprocal lattice of bcc is also bcc, with lattice parameter a* = 4π/a = 4π/4.23 = 0.94 Å. This means that the Brillouin zone boundary is located at a distance of 0.94 Å from the origin.

Therefore, we can see that the Fermi wave vector (0.92 Å^-1) is slightly smaller than the distance to the Brillouin zone boundary (0.94 Å). This makes sense because the Fermi level lies at the top of the energy band, which means that the Fermi wave vector will be slightly smaller than the distance to the Brillouin zone boundary.

I hope this helps clarify your doubts. Let me know if you have any further questions. Keep up the good work!Scientist