半导体物理与器件第四课后习题答.docx

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1、半导体物理与器件第四课后习题答Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions _ Chapter 3 3.1 If ao 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 ao were to

2、decrease, the bandgap energy would increase and the material would begin to behave more like an insulator. _ 3.2 Schrodingers wave equation is: -h22Y(x,t)2mx2+V(x)Y(x,t) =jhY(x,t)t Assume the solution is of the form: Y(x,t)=u(x)expjEkx-ht Region I: V(x)=0. Substituting the assumed solution into the

3、wave equation, we obtain: -h22mxjku(x)expjkx-Eht +u(x)Exexpjkx-th =jh-jEhu(x)expjkx-Eht which becomes -h2E2m(jk)2u(x)expjkx-ht +2jku(x)xexpjEkx-ht +2u(x)x2expjkx-Eht =+Eu(x)expjkx-Eht This equation may be written as -k2u(x)+2jku(x)2u(x)2x+x2+mEh2u(x)=0 Setting u(x)=u1(x) for region I, the equation b

4、ecomes: d2u1(x)dx2+2jkdu1(x)dx-(k2-a2)u1(x)=0 where a2=2mEh2 Q.E.D. In Region II, V(x)=VO. Assume the same form of the solution: Y(x,t)=u(x)expEjkx-ht Substituting into Schrodingers wave equation, we find: -h22m(jk)2u(x)expjkx-Eth +2jku(x)xexpjkx-Eht +2u(x)x2expjkx-Eth +VEOu(x)expjkx-ht =Eu(x)expjkx

5、-Eht This equation can be written as: -k2u(x)+2jku(x)2u(x)x+x2 -2mVOh2u(x)+2mEh2u(x)=0 Setting u(x)=u2(x) for region II, this equation becomes d2u2(x)dudx2+2jk2(x)dx -2mVk2-a2+Oh2u2(x)=0 where again a2=2mEh2 Q.E.D. _ 1 Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D.

6、A. Neamen Problem Solutions _ 3.3 We have d2u1(x)dx2+2jkdu1(x)dx-(k2-a2)u1(x)=0 Assume the solution is of the form: u1(x)=Aexpj(a-k)x +Bexp-j(a+k)x The first derivative is du1(x)dx=j(a-k)Aexpj(a-k)x -j(a+k)Bexp-j(a+k)x and the second derivative becomes d2 u1(x)2dx2=j(a-k)Aexpj(a-k)x +j(a+k)2Bexp-j(a

7、+k)x Substituting these equations into the differential equation, we find -(a-k)2Aexpj(a-k)x -(a+k)2Bexp-j(a+k)x +2jkj(a-k)Aexpj(a-k)x -j(a+k)Bexp-j(a+k)x() -k2-a2Aexpj(a-k)x +Bexp-j(a+k)x=0 Combining terms, we obtain -(a2-2ak+k2)-2k(a-k)-k2-a2 Aexp(j(a-k)()x) +-a2+2ak+k2+2k(a+k)-(k2-a2) Bexp-j(a+k)

8、x=0 We find that 0=0 Q.E.D. For the differential equation in u2(x) and the proposed solution, the procedure is exactly the same as above. _ 3.4 We have the solutions u1(x)=Aexpj(a-k)x +Bexp-j(a+k)x for 0xa and u2(x)=Cexpj(b-k)x +Dexp-j(b+k)x for -bxd2Edk2(curveB) so that m*(curveA)Edk2(curveB) so th

9、at m*m*p(curveA)p(curveB) _ 3.15 Points A,B: dEdk0velocity in +x direction Points A,D: d2Edk20 positive effective mass _ 5 Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions _ 3.16 For A: E=Cik2 At k=0.0810+10m-1, E=0.05eV Or E=(0.05)(1.610-1

10、9)=810-21J So 810-21=C(0.081010)21 C1=1.2510-38 (1.2 Now m*=h2(05410-34)2C=-38) 121.2510 =4.4410-31kg or m=4.443710-31*9.1110-31mo m*=0.488mo For B: E=C2ik At k=0.0810+10m-1, E=0.5eV Or E=(0.5)(1.610-19)=810-20J So 810-20=C1(0.081010)2 C1=1.2510-37 Now m*=h2(1.05410-34)22C=(37) 121.2510- =4.4410-32k

11、g =4.443710-32 or m*9.1110-31mo m*=0.0488mo _ 3.17 For A: E-Eu=-C2k2 -(0.025)(1.610-19)=-C2(0.081010)2 C392=6.2510- -h2-1.05410-34 m*=(2C=22()26.2510-39) =-8.887310-31kg or m=-8.887310-31*9.1110-31mo m*=-0.976mo For B: E-E2u=-C2k -(0.3)(1.610-19)=-C2(0.081010)2 C2=7.510-38 2 m*=-h-(1.(054-3422C=10) 227.510-38 =-7.40610-32kg *-7.40610-32 or m=9.1110-31mo m*=-0.0813mo _ 3.18 (a) (i) E=hn E(1.42)(1.610-19 or n=)h=6.62510-34 =3.4291014Hz =hcc31010 (ii) lE=n=3.4291014 =8.7510-5cm=875nm (b) (i) n=E(1.12)(1.610-19)