Question

The de Broglie wavelength of a quantum-mechanical particle with momentum p is given by Y=h/p, where h is plancks constant. Use this to show that Fermi energy for a free Fermi gas is, with a numerical factor, equal to the kinetic energy of a particle whose de Broglie wavelength is equal to the average interparticle spacing in the gas.

Answer 1

de Broglie wavelength of a quantum-mechanical particle with momentum p is given by

Y= h/p

Y² = h²/p²----------------------(1)

Kinetic energy of the particle E = p²/2m

                          p² =2mE

Therefore Y² = h²/2mE (from (1))

              E Y² = h²/2m ---------------------------------------(2)

Fermi energy of free fermi gas is E_F = (h²/2m) (3n/8)^2/3 ---------------------------(3)

Substituting (2) in (3)

E_F = E Y² (3n/8)^2/3

Hence, Fermi energy of free fermi gas(E_F ) is numericalli proportinal to K.E of the particle (E)