1、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 more like an insulator._3.2 Schrodingers wave equa
2、tion 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, . Assume the same form of the solution: Substituting into
3、 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 Substituting these equations into the differential equation
4、, 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 second boundary condition is which yields The thir
5、d 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) First point: Second point: By trial and error, (
6、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(a) J From Problem 3.5 J Jor eV(b) J From Problem 3.5, J J or eV_3.9(a) At , J At , By trial and error, J J or eV(b) At , J At . From
7、 Problem 3.5, J J or eV_3.10(a) J From Problem 3.6, J J or eV(b) 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 by We have so that _3.14 The effec
8、tive 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 So Now kg or _3.17 For A: kg or Fo
9、r 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(a) (b) _3.22(a) (b) _3.23 For the 3-dimensional infinite potential well, when , , and . In this r
10、egion, 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 conditions, we find and
11、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 density of states funct
12、ion, 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) Silicon, (i) At K
13、, 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_3.28(a) For ; eV; mJ eV; mJ eV; mJ eV; mJ(b) For ; eV; mJ eV; mJ eV; mJ eV; mJ_3.29(a)(b)_3.30 Plot_3
14、.31(a) (b) (i) (ii) _3.32 (a) , (b) , (c) , _3.33 or (a) , (b) , (c) , _3.34(a); ; ; ; ; (b) ; ; ; ; ; _3.35and So 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 ;
15、so J or eV For the next quantum state, which is empty, the quantum state 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
16、The probability of a state at being occupied is The probability of a state at being empty is or so Q.E.D._3.39(a) At energy , we want This expression can be written as or Then or (b) At , which yields _3.40 (a) (b) eV (c) or or which yields K_3.41 (a) or 0.304%(b) At K, eVThen or 14.96%(c)or 99.7%(d
17、) At , for all temperatures_3.42(a) For Then For , eV Then or (b) For eV, eVAt , or At , or _3.43(a) At or At , eVSo or (b) For , eVAt , or At , 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, or Ge: eV or GaAs: eV or (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(a) or eV so K(b) 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_
