Question

Sodium crystallizes in the bcc structure with the lattice constant a=4.23 Å. a. Calculate the carrier concentration of Na and give the result in units of cm-3 . b. Calculate the Fermi wave length and give the Fermi energy in eV.

Answer 1

1. Since sodium has BCC crystal structure, there are 2 atoms in the unit cell.

Lattice constant of sodium is a = 4.23 A^o

Here 1 A^o = 10^-8 cm

Their fore, a= 4.23×10^-8 cm

The Atomic concentration is 2 atoms in a cube of volume a³,

i.e.         n_at = 2/a³

                                = 2/ (4.23×10^-8)³

                                =2.642×10²² cm^-3

For a BCC cell, the lattice parameter a and the radius of the atom R is in the following relation

R = a√3/4

                                =4.23×10^-8×√3 /4

                                =1.831×10²² cm^-3

b)Fermi enengry E_f =h²/2m (3N/8∏V)^2/3

Dependences only on electron density n=N/V =2.642×10²² cm^-3

                h=1.05×10^-27, m=9.1×10^-28

                E_f= 36.46×10^-2 (N/V)^2/3

                = 3.24 eV

wave lenght

h = planks constant

c= velocity

Ef = fermi energy

= 5.76 ×10-²⁶ cm^-3