quz1D_ sol

quz1D_ sol - (c) Is the probability of finding an electron...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
ECE 440: Quiz Section D (11:00 a.m.) Name:_______________________________ 1. (a) How many atoms can be found inside a unit cell of diamond structure, and zinc-blende structure crystal? (2 points) 4 + 6*1/2 + 8*1/8 = 8 Ans: 8 atoms per unit cell (b) How many Ga atoms can be found in a unit cell of GaAs? How many As atoms? (2 points) Ans: Since GaAs is zinc-blende structure => 4 Ga atoms and 4 As atoms per unit cell (c) For Al0.25Ga0.75As, how many Al atoms on average can be found in each unit cell? (1 point) Ans: Al:Ga = 0.25:0.75=1:3, so there will be 4*1/4 = 1 Al atom per unit cell 2. The electron concentration in germanium at T = 300 K is n 0 = 1 x 10 8 cm -3 and the intrinsic carrier concentration is 10 13 cm -3 . (a) Is this material n-type or p-type? (1 point) Ans: ( ) ( ) type p cm n n p o i o = = = 3 18 8 2 2 10 10 10 13 (b) Sketch the band diagram of the sample, labeling the various energy levels ( E C , E V , E F , E i ). (3points) Ans:
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: (c) Is the probability of finding an electron at the energy level E = E C closer to zero one? (1points) Ans: According to Fig. 3-15, the probability is closer to zero. 3. The Fermi-Dirac function is f(E) =1/(1+exp[(E-E F )/kT]). On the energy diagram shown below: (a) Which electron (1) or (3) has the highest occupation probability at equilibrium. (2 points) Ans: Since the energy of electron (1) and (3) is the same, the occupation probabilities are equal at equilibrium. (a) Rank the holes (1), (3), and (4) in descending order of their occupation probability at equilibrium. (3 points) Ans: Probability of finding a hole = 1 f(E). Since f (1) < f(3) < f(4) => 1-f (1) > 1-f(3) > 1-f(4) => hole(1), hole (3), hole(4) in descending order (1) (3) (4) (1') (3') (4') E F x 1 x 2 x 3 x 4 x 5 X E V E C E c E v E i E F E g E g /2 E...
View Full Document

Page1 / 2

quz1D_ sol - (c) Is the probability of finding an electron...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online