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

Y^{2} =
h^{2}/p^{2}----------------------(1)

Kinetic energy of the particle E = p^{2}/2m

p^{2} =2mE

Therefore Y^{2} = h^{2}/2mE (from (1))

E Y^{2} = h^{2}/2m
---------------------------------------(2)

Fermi energy of free fermi gas is E_{F} =
(h^{2}/2m) (3n/8)^{2/3}
---------------------------(3)

Substituting (2) in (3)

E_{F} = E Y^{2} (3n/8)^{2/3}

Hence, Fermi energy of free fermi gas(E_{F} ) is
numericalli proportinal to K.E of the particle (E)

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