Fermi Energy and Conduction Electrons

[tommowg]

The problem:
A simple cubic metal has an electron density such that the Fermi energy just touches the edge of the first Brillouin zone. Calculate the number of conduction electrons per atom for this condition to be fulfilled.

The attempt at a solution:
I know that the electron density for a simple cubic (SC) is n=1/a³
It says that the Fermi Energy just touches the surface of the 1st Brillouin Zone, well the 1st BZ occurs at k=+- π/a
I know that the Fermi Energy is given by: Ef = ħ²/2m * (3π²n)^2/3 where n is the electron density again.

Now I can sub in the value of n for a simple cubic into this Fermi Energy equation, however I do not know how I can get the total number of conduction electrons from this information.

Any pointers in the right direction would be greatly appreciated.

Thanks,

Tom

 

[BigJDubs]

Solution:The total number of conduction electrons per atom is given by the equation: N = (2/3)VF, where VF is the Fermi volume. The Fermi volume is the volume of a sphere of radius equal to the Fermi wavevector. Using the above equation for the Fermi energy, we can calculate the Fermi wavevector as follows:kF = (2mEf/ħ2)1/2.Substituting in the values for the electron density, the Fermi energy and the Planck's constant, we have:kF = (2m(ħ2/2m * (3π2/a3)2/3)/ħ2)1/2 = (3π2/a3)1/3The Fermi volume is then just 4π/3 * (3π2/a3)3/3 = 4π/a3.Finally, substituting in the Fermi volume into the equation for the number of conduction electrons per atom, we have:N = (2/3)VF = (2/3)(4π/a3) = 8π/3a3 Hence, the total number of conduction electrons per atom is 8π/3a3.