Question

The density of sodium is 0.971 g/cm3 and it has an atomic weight of 22.9897 g/mol. Each sodium atom contributes one valence electron to the metal, so the number of electrons per unit volume, N/V, is equal to the number of sodium atoms per unit volume.

   a) Calculate the number density of atoms per unit volume, N/V.    b) Calculate the Fermi energy of sodium.    c) Calculate the Fermi temperature.   d) At room temperature (298 K), calculate the chemical potential.   e) Calculate the degeneracy pressure for the electron gas.

Answer 1

Density of sodium = 0.971 gm/cm3 = 0.971*106 gm/m3

1 m3 contains 0.971*106 gm of sodium

Atomic wight = 22.9897 g/mol

22.9897 gm sodium contains 6.032*1023 atoms

1 gm contains = 6.023*1023/(22.9897) atoms

1 m3 (0.971*106 gm) will contains 6.023*0.971*1029/(22.987) = 0.2544*1029 atoms= 2.544*1028 atoms

each atom contains 1 free electron.

hence number density is n = 2.544*1028 electrons/m3

(a) number density of atoms = n =N/V = 2.544*1028 atoms/m3

(b) Fermi energy = J

  

m = mass of the electron = 9.1*10-31 kg

= 1.054*10-34 J-s

putting all the values,

Fermi Energy = Ef = 3.154 eV

(c) Fermi temperature =

where k = Boltzman's constant = 1.380*10-23 m2-kg s-2 K-1

=36,602 K

(d) At temperature T> 0,

the chemical potetnial is equal to Fermi energy.

Hence Chemical potential = 3.154 eV

(e) pressure =

= = 0.4*2.544*1028*5.0464*10-19 J = 5.13*109 Pa