The occupancy probability function (Eq. 41-6) can be applied to semiconductors a

Question

The occupancy probability function (Eq. 41-6) can be applied to semiconductors as well as to metals. In semiconductors the Fermi energy is close to the midpoint of the gap between the valence band and the conduction band. Consider a semiconductor with an energy gap of 0.68 eV, at T = 280 K. What is the probability that (a) a state at the bottom of the conduction band is occupied and (b) a state at the top of the valence band is not occupied? (Note: In a pure semiconductor, the Fermi energy lies symmetrically between the population of conduction electrons and the population of holes and thus is at the center of the gap. There need not be an available state at the location of the Fermi energy.)

Explanation / Answer

the fermi probabilty function is p=1/([e^(E0-E)/kT]+1) we have E0 ,T,Ef is given to be 0.62/2=0.31 ev substituting the values we get p(0.62)=5.05x10^-6 The probability that the state at the top is not occupied is 1-P(0.62)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.