Question

# The de Broglie wavelength of a quantum-mechanical particle with momentum p is given by Y=h/p, where...

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.

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 EF = (h2/2m) (3n/8)2/3 ---------------------------(3)

Substituting (2) in (3)

EF = E Y2 (3n/8)2/3

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

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