材料科学与工程基础作业讲评（课堂PPT）

4-41在20时1m3内含有1021个电荷载体的半导体的电阻率为0.1m，如果在电场为0.15Vmm-1时可传导1A的电流，试计算电导的平均漂移速度=nq=nqv/E,so,v=E/nq=E/(nq)=150/1021*1.6*10-19*0.1=9.375m/s载流子的迁移率，其物理意义为载流子在单位电场中的迁移速度。,第十二次作业中文,1,4-43已知硅和锗在300的电阻率分别为2.3103m和0.46m，试分别计算硅和锗在250的电导率。4-43:ln=C-Eg/(2kT),或=0exp-Eg/(2kT)ln(1/2)=-Eg/(2k)1/T1-1/T2,R=1/硅半导体(Eg=1.2ev,k=8.6210-5ev)：250C,=0.15103(m)-1锗半导体(Eg=0.67ev,k=8.6210-5ev)：250C,=1.14106(m)-1,2,4-49产生介电损耗的原因是什么？主要的影响因素有哪些？定义：电介质在交变电场作用下,电能转变成热能而损耗。或P357。主要原因：漏电电流和极化电流损耗主要影响因素：(1)分子结构：极性大，tg大；基团数目多tg大；(2)小分子及杂质；(3)多相体系；(4)交变电场频率；(5)温度,3,英文书:12.11(a)Calculatethedriftvelocityofelectronsingermaniumatroomtemperatureandwhenthemagnitudeoftheelectricfieldis1000V/m.(b)Underthesecircumstances,howlongdoesittakeanelectrontotraversea25mm(1in.)lengthofcrystal?Solution:(a)v=E=0.38m2.(Vs-1)1000V/m=380m/s(b)25mm/(380m/s)=6.610-5s,4,12.14:(a)Calculatethenumberoffreeelectronspercubicmeterforgoldassumingthatthereare1.5freeelectronspergoldatom.TheelectricalconductivityanddensityforAuare4.3107(-m)-1and19.32g/cm3,respectively.(b)NowcomputetheelectronmobilityforAu.Solution:n=1.5(/M)NA=1.519.32106(g/m3)197(g/mol)6.021023=8.861028m-3=n|e|eSo,e=/n|e|=(4.3107-1m-1)(8.861028m-3)(1.610-19C)=3.0310-3m2/V-s,5,12.38Theintrinsicelectricalconductivitiesofasemiconductorat20and100Care1.0and500(-m)-1,respectively.Determinetheapproximatebandgapenergyforthismaterial.Solution:ln=CEg/(2kT)so,ln1=CEg/(2kT1),ln2=CEg/(2kT2)ln1-ln2=Eg/(2kT2)-Eg/(2kT1)=(Eg/2k)1/T2-1/T1)so,Eg=(ln1-ln2)2k1/T2-1/T1=(ln1-ln500)28.6210-5eV/K(1/373-1/293K)=1.46eV,6,12.51Attemperaturesbetween775Cand1100C,theactivationenergyandpre-exponentialforthediffusioncoefficientofFe2+inFeOare102,000J/moland7.310-8m2/s,respectively.ComputethemobilityforanFe2+ionat1000C.Solution:D=D0exp（-Qd/RT）=7.310-8exp-102000/（8.3141273）=4.7610-11m2/s=neD/(kT)P390=21.610-19C4.7610-11m2/s(1.38110-23J/K1273K)=8.710-10m2/V-s=neD/(kT)P390=24.7610-11m2/s(8.6210-5ev/K1273K)=8.710-10m2/V-s,7,思考题,4-39试述影响离子电导率和电子电导率的主要影响因素离子电导率：温度：温度越高，电导率越大晶体结构：离子大小、电荷影响电导率晶格缺陷：离子性晶格缺陷的生成及浓度大小是决定离子电导的关键电子电导率(金属）：散射温度T-1杂质及缺陷杂质原子（离子）引入新的局部能级塑性形变位错,8,4-47.在下列高聚物材料中，哪些有可能利用高频塑化法加工成型？（1）酚醛树脂；（2）聚乙烯；（3）聚苯乙烯；（4）聚氯乙烯,4-47聚氯乙烯,9,12.8Intermsofelectronenergybandstructure,discussreasonsforthedifferenceinelectricalconductivitybetweenmetals,semiconductors,andinsulators.,10,12.9Ifametallicmaterialiscooledthroughitsmeltingtemperatureatanextremelyrapidrate,itwillformanoncrystallinesolid(i.e.,ametallicglass).Willtheelectricalconductivityofthenoncrystallinemetalbegreaterorlessthanitscrystallinecounterpart?Why?电导率降低。英文书P374，和P375-图12.8,（温度升高，金属电阻率增加，电导率下降，导电性减弱）,11,12.21(a)Computethenumberoffreeelectronsandholesthatexistinintrinsicgermaniumatroomtemperature,usingthedatainTable12.2.(b)Nowcalculatethenumberoffreeelectronsperatomforgermaniumandsilicon(ExampleProblem12.1).(c)Explainthedifference.YouwillneedthedensitiesforGeandSi,whichare5.32and2.33g/cm3,respectively.Solution:(a)n=p=/|e|(e+h)=2.2(-m)-1(1.610-19C)(0.38+0.18m2/V-s)=2.461019m-3,12,(b)forGe:Numberofatoms:N=(/M)NA=5.32106(g/m3)72.59(g/mol)6.021023=4.411028m-3每个原子自由电子数=n/N=(2.461019m-3)/(4.411028m-3)=5.610-10forSi:Numberofatoms:N=(/M)NA=2.23106(g/m3)28.1(g/mol)6.021023=4.781028m-3n=p=/|e|(e+h)=410-4(-m)-1(1.610-19C)(0.14+0.048m2/V-s)=1.331016m-3每个原子自由电子数=n/N=(1.331016m-3)/(4.781028m-3)=2.7810-13(c)每个原子的自由电子数差别大，电导率差别也大。,13,12.34Comparethetemperaturedependenceoftheconductivityformetalsandintrinsicsemiconductors.Brieflyexplainthedifferenceinbehavior.金属减小；半导体为增大。,14,4-52、铁磁性材料和顺磁性材料的磁性来源有何异同？,铁磁性来源于很强的永久磁矩,内部交换场顺磁性来源于永久磁矩,第十三次中文,15,4-55、将2400A.m-1的磁场作用到相对磁导率为5000的材料上，计算磁感强度和磁化强度。,4-55：磁化强度：M=(r-1)H=(5000-1)2400A.m-1=1.2107A.m-1磁感强度：B=0(H+M)=410-7(H.m-1)(2400A.m-1+1.2107A.m-1)=15.1Wb.m-2=15.1teslaB=r0H=5000410-7H/m2400A/m=15.1tesla,16,18.7Themagnetizationwithinabarofsomemetalalloyis3.2105A/matanHfieldof50A/m.Computethefollowing:(a)themagneticsusceptibility,(b)thepermeability,and(c)themagneticfluxdensitywithinthismaterial.(d)Whattype(s)ofmagnetismwouldyousuggestasbeingdisplayedbythismaterial?Why?,（a）m=M/H=（3.2105A/m）/（50A/m）=6400（b）r=m+1=6401=r0=2.5610-3H/m（c）B=H=2.5610-3H/m（50A/m）=0.402tesla（d）铁磁体,17,18.8Compute(a)thesaturationmagnetizationand(b)thesaturationfluxdensityforcobalt,whichhasanetmagneticmomentperatomof1.72Bohrmagnetonsandadensityof8.90g/cm3.,a）Ms=1.72BN=1.72BNA/ACo=1.729.2710-248.91066.0231023/58.9=1.45106A/m（b）Bs=0Ms=（410-7H/m）Ms=1.822tesla,18,18.9Confirmthatthereare2.2Bohrmagnetonsassociatedwitheachironatom,giventhatthesaturationmagnetizationis1.70106A/m,thatironhasaBCCcrystalstructure,andthattheunitcelledgelengthis0.2866nm.,将numberofBohrmagnetons代入公式计算MsMs=nBB/a3=22.29.2710-24/(0.286610-9m)3=1.733106A/m,19,18.25Figure18.25showstheB-versus-Hcurveforasteelalloy(a)Whatisthesaturationfluxdensity?(b)Whatisthesaturationmagnetization?(c)Whatistheremanence?(d)Whatisthecoercivity?(e)OnthebasisofthedatainTables18.5and18.6,wouldyouclassifythismaterialasasoftorhardmagneticmaterial?Why?,20,saturationfluxdensity,remanence,coercivity,软磁性,21,Solution,(a)Thesaturationfluxdensity,(c)Whatistheremanence?,Bs=1.4tesla,Ms=Bs/0=1.4/(410-7H/m)=1.11106A/m,(b)Bs=0(Hs+Ms)0Ms,Br=0.8tesla,(d)Whatisthecoercivity?,(e)OnthebasisofthedatainTables18.5and18.6,wouldyouclassifythismaterialasasoftorhardmagneticmaterial?Why?,Hc=75A/m,22,思考题：,4-50写出下列物理量的量纲：（1）磁感应强度；（2）磁化强度；（3）磁导率和相对磁导率；（4）磁化率。,4-54试述影响金属磁性的因素。正离子的抗磁性和顺磁性；电子的抗磁性和顺磁性。四个主要因素。,23,18.5(a)Explainthetwosourcesofmagneticmomentsforelectrons.电子的轨道磁矩和自旋磁矩。(b)Doallelectronshaveanetmagneticmoment?Whyorwhynot?是。(c)Doallatomshaveanetmagneticmoment?Whyorwhyno？非，具有各层都充满电子的原子结构，其电子磁矩相互抵消，不显磁性。,24,18.12Citethemajorsimilaritiesanddifferencesbetweenferromagneticandferrimagneticmaterials.Ferromagnetic-铁磁性Ferri-magnetic-亚铁磁性-保留了剩余磁矩。磁化率比铁磁性低。,25,18.19Brieflyexplainwhythemagnitudeofthesatura