半導(dǎo)體物理與器件第四版課后習題答案3

1、頁眉內(nèi)容Chapter 33.1If 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 decrease, the bandgap energy would increase and the material would begin to behave more like an insulator.3.2Schrodinger'

2、;s wave equation is:Assume the solution is of the form:Region I: V x 0. Substituting the assumed solution into the wave equation, we obtain: which becomesThis equation may be written asSetting u x Ui x for region I, the equation becomes: whereIn Region II, V xVO . Assume the sameform of the solution

3、:Substituting into Schrodinger's wave equation, we find:This equation can be written as:Setting u x U2 x for region II, this equation becomes where again3.3We haveAssume the solution is of the form:The first derivative is and the second derivative becomes Substituting these equations into the di

4、fferential equation, we find Combining terms, we obtain We find thatFor the differential equation in u2 x and the proposed solution, the procedure is exactly the same as above.3.4We have the solutionsfor 0 x a andfor b x 0.The first boundary condition is which yieldsThe second boundary condition is

5、which yieldsThe third boundary condition is which yieldsand can be written asThe fourth boundary condition is which yields and can be written as3.5(b) (i) First point:aSecond point: By trial and error, (ii) First point:a2Second point: By trial and error,3.6(b) (i) First point:aSecond point: By trial

6、 and error,(ii) First point:a2Second point: By trial and error,3.7Let ka y, a xThenConsider - of this function.dyWe findThenFor y ka n , n 0,1,2,. sin y0So that, in general,AndSoThis implies thatd八 dE工，n0for k -dkdka3.8_ 193.4114 10 JFrom Problem 3.5181.0198 10 J6.7868 10 19J196.7868 10 19or E194.24

7、 eV1.6 101.3646 10 18JFrom Problem 3.5,182.3364 10 J199.718 10 J199.718 10or E196.07 eV1.6 103.9(a) At ka , 1a193.4114 10 JAtka 0, By trial and error, 2.5172 10 19J208.942 10 Jor8.942 10 20(b)At1.6 ka 2 ,1.364610 193 a 210 18J0.559eVEg1.164 eVT200K, Eg1.147 eVT300 K, Eg g1.125 eVT400K, Eg g1.097 eVT

8、500 K, Eg1.066 eVT600K, Eg g1.032 eVorAtka.From Problem 3.5, 1.0198 10 18J193.4474 10 J193.103.4474 10191.6 10 192.15eV193.4114 10 JFrom Problem 3.6,2 a 1.515or E197.830 10 J194.4186 10 J194.4186 10191.6 102.76eV1.3646 10From Problem 3.6,18 J1.9242 105.597 10or E5.597 104a18J19J192.375191.6 103.50eV

9、3.11(a)AtkaAtka,1a193.4114 10 J0, By trial and error,1.8030 101.6084 10or1.6084 10(b)Atka1.62 ,19103aor1.3646At ka7.8305.816E 5.816103.13The effective mass is given byWe have*so that m curve A m curve B3.14The effective mass for a hole is given byWe have that *-*so that mD curve A mD curve Bpp19 J19

10、J19l 1.005eV218J,From Problem 3.6,10101019J19J19191.6 103.635 eV3.15由0 dkPoints A,B:velocity in -x directionPoints C,D:匹0 dkvelocity in +x directionPoints A,D:d2E dk20d2E dk2negative effective massPoints B,C:0positive effective mass3.16For A:ECik2At k0.08 10 10m 1, E0.05eVOrE0.05一一 191.6 108 10 21 J

11、So 81021 Ci10 20.08 10Nowm21.05434 2102C12 1.25“38104.4410 31 kgorm314.4437 10319.11 10 31moFor B:ECik2At k101_0.08 10 m , E0.5eVOrE0.5 119,6 10208 10 JSo 81020 C1 10.08 101023.12For T 100K,Now m1.054 102Ci21.25 103.17or m.一 324.44 10 kg4.4437 10 323 mo9.11 10For A: E EC2k28.8873 10 31 kgor mFor B:

12、Eor m8.8873 109.11 10 31EC2k2327.406 10 kg7.406 10T739.11 10 33.18momo3.23For the 3-dimensional infinite potential well, V x 0 when 0 x a , 0 y a, and 0 z a. In this region, the wave equation is:Use separation of variables technique, so let Substituting into the wave equation, we have Dividing by XY

13、Z , we obtainLetThe solution is of the form:Since x, y, z 0 at x 0, then X 00so that B 0 .Also, x, y, z 0 at x a , so thatX a 0 . Then kx a nx whereSimilarly, we have12Y212Z 22k y and 2-kzY y2Zz2From the boundary conditions, we find kya ny and kza nz(a) (i) E hE 1.42 1.6 10or -3h6.625 10 3(ii)(b) (i

14、)_143.429 10 Hz10hc c 3 10-T4E 3.429 1058.75 10 5cm 875nm19E 1.12 1.6 10丁一一 _ 34h6.625 10wheren y 1, 2,3,. and nz 1,2,3,.From the wave equation, we can writeThe energy can be written as(ii)_142.705 1014 Hz10c 3 102.705 101441.109 10 cm 1109nm3.19(c) Curve A: Effective mass is a constant Curve B: Eff

15、ective mass is positive around k 0, and is negativearound k .23.24The total number of quantum states in the 3-dimensional potential well is given (in k-space) by whereWe can then writeTaking the differential, we obtainSubstituting these expressions into the density of states function, we haveNoting

16、thatthis density of states function can be simplified and written as3Dividing by a will yield the density of states so that3.253.20Then andThen orFor a one-dimensional infinite potential well, Distance between quantum states NowNowThenDivide by the "volume" Soa, so3.213.221g E 341.054 10 3

17、4181.055 10g E 3312 0.067 9.11 10、E3J 1Then gcor gc(ii) At T 400K, kT0.0259400300183gc1.43 10 cmEE0.1 eV; g5.6344510m3J1EE0.2 eV;7.9684510m3 J1EE0.3eV;9.7581045m3J 1EE0.4 eV;1.1271046m3J 1For E E ; g 03.293.30Plot3.31(a)(i)Wi12!12 11 10!10! 1210!10! 2 1(ii)Wi12!12 11 10 9 8!8! 128!8! 4 3 2 13.32(a)E

18、EfkT , fE11 exp1(b)EEf5kT,f E11 exp 5(c)EEf10kT ,f E11 exp 103.33or(a)EfEkT , 1f E0.269(b)EfE5kT , 13f E 6.69 10(c)EfE10kT ,1 f_5E 4.54 103.343.26(a) Silicon, mn1.08mo(i) At T 300 K, kT 0.0259eV214.144 10 J _ 55_ 213/27.953 102 4.144 102536.0 10 m196.0 10 cm0.034533eV5.5253 10 21 JThen2539.239 10 m1

19、93or gc 9.24 10 cm(b) GaAs, mn 0.067m。21(i) At T 300 K, kT 4.144 10 J2339.272 10 mor gc 9.27 1017 cm 3 _ _ 21(ii) At T 400 K, kT 5.5253 10 J1.427 1024m 33.27(a) Silicon, mp 0.56mo21At T 300K, kT 4.144 10 J4.116 1025 m 3193or g 4.12 10 cm 21(ii)At T 400K, kT 5.5253 10 J6.337 1025 m 3193or g 6.34 10 c

20、m(b) GaAs, mp 0.48m021(i)At T 300 K, kT 4.144 10 J 2533.266 1025 m 3193or g 3.27 10 cm21(ii)At T 400K, kT 55253 10 J5.029 1025 m 3 193or g 5.03 10 cm3.28 For E Ec；gc 0E Ec 0.1 eV; gc 1.509 1046 m 3 J 1EEc0.2 eV;2.1344610m3 JEEc0.3eV;2.6144610m3 JEEc0.4eV;3.0184610m3 JEEc; fF0.306exp9.32 100.0259EckT

21、T; fFexp -0.30 0.0259 20.0259EckT ; fF0.30 0.0259exp0.0259Ec3kT2; fFexp0.30 3 0.0259 20.0259Ec2kT ; fFexp0.30 2 0.02590.02590.255exp 6.43 100.0259kT子1fF0.25 0.0259 2exp 0.0259The 14th electron would occupy the quantum state nx 2, ny 3, nz 3 . This state is at the same energy, soEf 5.746 eVkT ; 1fF0.

22、25 0.0259exp 0.02593kT2kT ;3.35 and3.38The probability of a state at E1EFEbeing occupied isThe probability of a state at E2EfEbeing empty is orso f1 Ei 1 f2 E2So expEc kT EfkTThen EckT EfEf E kTOr EfEc EEmidgap3.36FororE66, Filled state1.5044 10 18 J1.5044 10 183.39(a) At energy E1, we wantThis expr

23、ession can be written as orThenor(b)At E Ef 4.6kT , which yields3.40FororE7_ 191.6 107, Empty state182.048 10 J182.048 109.40 eV1.6 10 1912.8eVTherefore 9.40 Ef 12.8eV(a)700(b) kT0.02593000.060433 eVorexp0.25150kT0.02orwhich yields T740K3.37(a) For a 3-D infinite potential wellFor 5 electrons, the 5

24、th electron occupiesthe quantum statenx2,ny 2,nz 1 ; so3.761103.761 or E51019 J193.41(a)or 0.304%(b) At T 1000 K, kT 0.08633eV行-2.35 eV1.6 10 19For the next quantum state, which is empty, the quantum state is nx 1, ny 2, nz 2 .Thenor 14.96%or 99.7%(d)At E Ef, f E1-for all temperaturesThis quantum state is at the same energy, so Ef 2.35 eV(b) For 13 electrons, the 13 th electronoccupies the quantum state3.42(a) For E E1nxor E133, ny 2,nz :199.194 10 J199.194 103 ; soexpE1 EfkTexpE1 EfkT191.6 10 195.746 eVThenFor E E2 , Ef E2 1.12 0.30 0.82 eV頁眉內(nèi)容so or (a)At TAt E0K, F