Fermi Energy, number of electrons and holes

  • Thread starter bcjack
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Fermi Energy, Quantum mechanics

Electron levels and degeneracies thereof in a system are:
W1 = 0 eV, 10^23 /cm3 Valence band
W2 = 0.9 eV, 5x10^20 /cm3 Donor level
W3 = 1 eV, 2x10^23 /cm3 Conduction level

Total number, n, of electrons in the system is (10^23 + 5 x 10^23) /cm3.

Determine Fermi energies, number of electrons in level W3, and number of vacancies (holes) in levels W1 and W2 at i) 290 K and ii) 174 K.

My first thoughts of doing this problem is by adding the number of electrons in the 3 states together and solve for Ef with the fermi distribution.

f(e)=1 / [1 +e^ (E-Ef)/kT ]
n = g(e) * f(e) density of states times the fermi distribution

n_total = n1 + n2 + n3 = 10^23 / [1 +e^ (0-Ef)/kT ] + 5x10^20/ [1 +e^ (0.9-Ef)/kT ] + 2x10^23 / [1 +e^ (1-Ef)/kT ] = (10^23 + 5 x 10^23) /cm3

and solve for Ef.

Is this a reasonable approach?

To find the number of electrons and holes, i would then plug the Ef back into the equation for each energy level.

any thoughts would be appreciated. thanks!
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