F.E.G. fermi sphere radius problem

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SUMMARY

The discussion focuses on calculating the electron concentrations required for the Fermi sphere to contact the zone faces in Body-Centered Cubic (BCC) and Face-Centered Cubic (FCC) structures. The key formula used is kf = (3*pi^2*n)^(1/3), where n represents the number of electrons per atom. For BCC, the Brillouin zone edge is at the {110} planes, leading to the equation (2^(3/2)*pi/a)^3/(3*pi^2) = n. The FCC structure involves the {110} and {200} planes, which also requires similar calculations for accurate electron concentration determination.

PREREQUISITES
  • Understanding of Fermi spheres and electron concentrations
  • Familiarity with BCC and FCC crystal structures
  • Knowledge of Brillouin zones and reciprocal space
  • Proficiency in using the formula kf = (3*pi^2*n)^(1/3)
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  • Research the implications of electron concentration on material properties in BCC and FCC structures
  • Study the derivation and applications of the Brillouin zone in solid-state physics
  • Explore advanced topics in reciprocal lattice vectors and their significance
  • Learn about the role of lattice parameters in determining electronic properties
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Homework Statement



Calculate the electron concentrations (# electrons/atom) needed for the fermi sphere to contact the zone faces (first bril. zone edges) in BCC and FCC structures.

Homework Equations



kf = (3*pi^2*n)^(1/3) where n is # electrons per atom.

For cubic structures, dhkl=a*(h^2+k^2+l^2)^(-1/2), and the reciprocal magnitude of d is 2*pi/dhkl

The Attempt at a Solution



In BCC structure, the brill. zone edge is at the {110} planes, so that dhkl = a*2^(-1/2) and this distance in reciprocal space is 2^(3/2)*pi/a. (where a is the lattice parameter). Setting this reciprocal distance equal to the fermi vector,
(2^(3/2)*pi/a)^3/(3*pi^2) = n

I'm just unsure about this, maybe someone could give me a shove in the right direction. For the FCC structure, the edges of the FBZ are at the {110} and {200} planes.
 
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