半导体物理与器件第四版课后习题答案3.doc

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1、【精品文档】如有侵权，请联系网站删除，仅供学习与交流半导体物理与器件第四版课后习题答案3.精品文档.Chapter 33.1 If were to increase, the bandgap energy would decrease and the material would begin to behave less like a semiconductor and more like a metal. If were to decrease, the bandgap energy would increase and the material would begin to behave

2、more like an insulator._3.2 Schrodingers wave equation is: Assume the solution is of the form: Region I: . Substituting the assumed solution into the wave equation, we obtain: which becomes This equation may be written as Setting for region I, the equation becomes: where Q.E.D. In Region II, . Assum

3、e the same form of the solution: Substituting into Schrodingers wave equation, we find: This equation can be written as: Setting for region II, this equation becomes where again Q.E.D._3.3 We have Assume the solution is of the form: The first derivative is and the second derivative becomes Substitut

4、ing these equations into the differential equation, we find Combining terms, we obtain We find that Q.E.D. For the differential equation in and the proposed solution, the procedure is exactly the same as above._3.4 We have the solutions for and for . The first boundary condition is which yields The

5、second boundary condition is which yields The third boundary condition is which yields and can be written as The fourth boundary condition is which yields and can be written as_3.5 (b) (i) First point: Second point: By trial and error, (ii) First point: Second point: By trial and error,_3.6 (b) (i)

6、First point: Second point: By trial and error, (ii) First point: Second point: By trial and error,_3.7 Let , Then Consider of this function. We find Then For , So that, in general, And So This implies that for _3.8 J From Problem 3.5 J Jor eV J From Problem 3.5, J J or eV_3.9(a) At , J At , By trial

7、 and error, J J or eV(b) At , J At . From Problem 3.5, J J or eV_3.10 J From Problem 3.6, J J or eV J From Problem 3.6, J J or eV_3.11(a) At , J At , By trial and error, J J or eV(b) At , J At , From Problem 3.6, J J or eV_3.12 For K, eV K, eV K, eV K, eV K, eV K, eV_3.13 The effective mass is given

8、 by We have so that _3.14 The effective mass for a hole is given by We have that so that _3.15 Points A,B: velocity in -x direction Points C,D: velocity in +x direction Points A,D: negative effective mass Points B,C: positive effective mass_3.16 For A: At m, eV Or J So Now kg or For B: At m, eV Or J

9、 So Now kg or _3.17 For A: kg or For B: kg or _3.18(a) (i) or Hz (ii) cmnm(b) (i) Hz (ii) cmnm_3.19(c) Curve A: Effective mass is a constantCurve B: Effective mass is positive around , and is negative around ._3.20 Then and Then or_3.21_3.22_3.23 For the 3-dimensional infinite potential well, when ,

10、 , and . In this region, the wave equation is: Use separation of variables technique, so let Substituting into the wave equation, we have Dividing by , we obtain Let The solution is of the form: Since at , then so that . Also, at , so that . Then where Similarly, we have and From the boundary condit

11、ions, we find and where and From the wave equation, we can write The energy can be written as_3.24 The total number of quantum states in the 3-dimensional potential well is given (in k-space) by where We can then write Taking the differential, we obtain Substituting these expressions into the densit

12、y of states function, we have Noting that this density of states function can be simplified and written as Dividing by will yield the density of states so that_3.25 For a one-dimensional infinite potential well, Distance between quantum states Now Now Then Divide by the volume a, so So mJ_3.26 (a) S

13、ilicon, (i) At K, eV J Then m or cm (ii) At K, eV J Then m or cm(b) GaAs, (i) At K, J m or cm(ii) At K, J m cm_3.27(a) Silicon, (i)At K, J m or cm (ii)At K, J m or cm(b) GaAs, (i)At K, J m or cm (ii)At K, J m or cm_3.28For ; eV; mJ eV; mJ eV; mJ eV; mJFor ; eV; mJ eV; mJ eV; mJ eV; mJ_3.29_3.30 Plot

14、_3.31(a) (i) (ii) _3.32_3.33 or_3.34_3.35andSo Then Or _3.36 For , Filled state J or eV For , Empty state J or eV Therefore eV_3.37(a) For a 3-D infinite potential well For 5 electrons, the 5th electron occupies the quantum state ; so J or eV For the next quantum state, which is empty, the quantum s

15、tate is . This quantum state is at the same energy, so eV(b) For 13 electrons, the 13th electron occupies the quantum state ; so J or eVThe 14th electron would occupy the quantum state . This state is at the same energy, so eV_3.38 The probability of a state at being occupied is The probability of a

16、 state at being empty is or so Q.E.D._3.39(a) At energy , we wantThis expression can be written asorThenor(b) At ,which yields_3.40 (a) (b) eV or or which yields K_3.41 (a) or 0.304%(b) At K, eVThenor 14.96%or 99.7%(d) At , for all temperatures_3.42(a) For Then For , eV Then or(b) For eV, eVAt ,orAt

17、 ,or _3.43(a) At orAt , eVSoor(b) For , eVAt ,orAt ,or _3.44 so or(a) At K, For At (b) At K, eV For , For , At , (eV)(c) At K, eV For , For , At , (eV)_3.45(a) At ,Si: eV,orGe: eVorGaAs: eVor(b) Using the results of Problem 3.38, the answers to part (b) are exactly the same as those given in part (a)._3.46 or eV so K or K_3.47(a) At K, eV eV By symmetry, for , eV Then eV(b) K, eV For , from part (a), eV Then eV_