chat: 3 Ef = 5/2 + 2/kTln (mm) me now that the concentration of electrons and holes

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Ef = fermi level

(a) Prove that:
Eg
3
E₁ = 5/2 + ²/(kTin (mm)
Ef
4
me
(b) Now show that the concentration of electrons and holes for intrinsic
semiconductors is:
n = p = √√NcN₂e-Eg/2kT, where Eg = Ec - Ev
(c) Calculate the concentration of electrons and holes in an intrinsic
semiconductor InSb at room temperature (T= 300 K, E, = 0.2 eV, me = 0.01m and
mh = 0.018 m).
(d) Determine the position of the Fermi energy.
Transcribed Image Text:(a) Prove that: Eg 3 E₁ = 5/2 + ²/(kTin (mm) Ef 4 me (b) Now show that the concentration of electrons and holes for intrinsic semiconductors is: n = p = √√NcN₂e-Eg/2kT, where Eg = Ec - Ev (c) Calculate the concentration of electrons and holes in an intrinsic semiconductor InSb at room temperature (T= 300 K, E, = 0.2 eV, me = 0.01m and mh = 0.018 m). (d) Determine the position of the Fermi energy.
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