材料科学与工程基础作业讲评-9.pptx

1、4-62 已知碲化锌的Eg=2.26eV，它对哪一部分可见光透明？,第十四次作业中文,=hc/Eg=0.55m h=6.62*10-34J.s/1.6*10-19=4.13*10-15eV.s c=3*108m.s-1 对波长0.550.7 m的可见光透明,4-63、当可见光垂直入射并透过20mm厚的某种透明材料时，透射率为0.85，当该种透明材料的厚度增加到40mm时，透射率为多少？已知折射率为1.6。,R=（n1-1）2/（n1+1）2=0.053 T=（1-R）2e-l；0.85=（1-0.053）2e-*0.02；=2.6 T=（1-R）2e-l=0.808,19.9 Compute

2、the velocity of light in calcium fluoride(CaF2),which has a dielectric constant r of 2.056(at frequencies within the visible range)and a magnetic susceptibility of-1.4310-5.,r=1+m=1-0.0000143=0.9999757n=（rr）1/2=1.43v=c/n=2.10108m/s另解：r=/0,=0 r r=/0,=0 r,19.19 The fraction of nonreflected radiation t

3、hat is transmitted through a 10-mm thickness of a transparent material is 0.90.If the thickness is increased to 20 mm,what fraction of light will be transmitted?,IT=I0（1-R）2e-l=I0e-l0.9=e-l=e（-0.01）=10.54IT=I0e（-0.02）=0.81I0,19.22 Briefly explain what determines the characteristic color of(a)a metal

4、 and(b)a transparent nonmetal.,金属：吸收和再发射透明非金属：透射、部分吸收（选择性吸 收）。,19.24 The index of refraction of quartz is anisotropic.Suppose that visible light is passing from one grain to another of different crystallographic orientation and at normal incidence to the grain boundary.Calculate the reflectivity at

5、the boundary if the indices of refraction for the two grains are 1.544 and 1.553.R=（n2-n1）2/（n2+n1）2=（1.553-1.544）2/（1.553+1.544）2=8.42710-6,19.25 Briefly explain why amorphous polymers are transparent,while predominantly crystalline polymers appear opaque or,at best,translucent.,半晶聚合物，结晶与非晶区密度差异造成折

6、射指数差异，当晶区尺寸大于可见光波长，造成光散射，造成不透明或半透明。,4-59叙述下列概念：（1）反射、透射和吸收：（2）旋光性：（3）非线性光学性质：（4）光泽：（5）发光：（6）光敏性,吸收：光穿过介质时，引起介质的价电子跃迁，或使原子振动而消耗能量，或因电子能量转变成分子的动能即热能，从而构成光能衰减。反射：光在两相交界面上返回入射相中透射：光在两相交界面上进入透射相中旋光性：偏振光的偏振面转过一定角度非线性光学性质：分子在强光场的作用会产生极化光泽：材料表面某一方向反射光较强发光：材料吸收高能辐射，然后发射可见光,4-66为什么有些透明材料是带色的，有些是无色的？,选择性吸收；内部折

7、光指数不均匀,4-70有哪些方法可以改善材料的透明性。,增加聚合物材料透明性的方法 加速成核或由熔体急剧冷却减少球晶大小；拉伸球晶转变为取向微丝,19.4 Distinguish between materials that are opaque,translucent,and transparent in terms of their appearance and light transmittance.,不透明，透过率接近0半透明透明，透过率接近1,19.7(a)Briefly explain why metals are opaque to electromagnetic radiati

8、on having photon energies within the visible region of the spectrum.(b)Why are metals transparent to high-frequency x-ray and-ray radiation?,高能级(价带）只部分填充电子，具有可见光频率的辐射，会激发电子到费米能级以上的空能级（吸收再发射光子）。对高频辐射无吸收(不会产生电子跃迁，因为没有合适的空轨道),19.13 It is desired that the reflectivity of light at normal incidence to the

9、 surface of a transparent medium be less than 5.0%.Which of the following materials in Table 19.1 are likely candidates:sodalime（石灰钠）glass,Pyrex glass,periclase（方镁石）,spinel（尖晶石）,polystyrene,and polypropylene?Justify your selections.,R=（ns-1）2/（ns+1）2soda-lime glass：R=（ns-1）2/（ns+1）2=（1.458-1）2/（1.45

10、8+1）2=0.0347Pyrex glass：R=（ns-1）2/（ns+1）2=(1.47-1)2/(1.47+1)2=0.03615periclase：R=（ns-1）2/（ns+1）2=(1.74-1)2/(1.74+1)2=0.0729spinol：R=（ns-1）2/（ns+1）2=(1.72-1)2/(1.72+1)2=0.07PS：R=（ns-1）2/（ns+1）2=（1.60-1）2/（1.60+1）2=0.05325PP：R=（ns-1）2/（ns+1）2=(1.49-1)2/(1.49+1)2=0.03873soda-lime glass、Pyrex glass、PP a

11、re likely candidates,19.26(a)In your own words describe briefly the phenomenon of luminescence.(b)What is the distinction between fluorescence and phosphorescence?,（a）吸收能量后发出可见光叫发光（b）延迟发射 10-8 s，荧光 延迟发射 10-2 s 10s，磷光,4-71 化学腐蚀、电化学腐蚀和物理腐蚀有哪些区别？试举例 说明。化学腐蚀：材料（包括金属材料、无机非金属材料和高分子材料）在接触酸、碱、盐、大气、有机溶剂等环境介质

12、时，受到环境介质的化学侵蚀作用，引起材料不可逆的化学变化，导致材料的破坏。例如某些金属在酸性条件下，置换酸中的氢离子而发生腐蚀。电化学腐蚀：金属材料在发生化学腐蚀的同时，伴随有电子的迁移，同时金属以离子的形式溶解而发生的腐蚀破坏类型。例如工业纯的锌放入稀硫酸中,在杂质和锌晶粒之间有电流流动，而锌以离子形式溶解的腐蚀。物理腐蚀：是材料在环境介质作用下，没有化学变化，而是以物理变化发生破坏的腐蚀类型。主要是高分子材料。例如溶剂分子渗入到体型高聚物发生溶胀、软化，是强度降低。,4-74 在一块单向玻纤增强不饱和聚酯单面板中，若纤维直径是20m，纤维体积分数是0.45，Ef=4Gp，Em=80MPa。

13、求（a）单位体积界面的面积；（b）当该复合材料沿纤维方向受到200MPa拉应力时，玻纤和基体各承受的拉应力是多少？,解（a）设该单向板为 1cm1cm1cm=1cm3正立方体 单根纤维的体积V1=（0.002cm/2）21cm=3.1410-6cm3 单根纤维的界面面积S1=0.002cm1cm=0.00628 cm2 则该1cm3正立方体单向板复合材料界面面积：Sc=(1 cm3 0.45S1)/V1=900cm2,b）Ff/Fm=VfEf/(1-Vf)Em=0.454000MPa/(0.5580MPa)=40.90=F/A（假定Ac=1m2)F=200 106N Fc=Ff+Fm=40.9

14、0 Fm+Fm=200 106N Fm=4.77 106N Ff=200-4.77=195.23 106N m=4.77 106N/0.55m2=8.68MPa f=195.23 106N/0.45m2=433MPa,4-78 一边长为25mm的立方体用薄铝板和硫化橡胶（厚度分别为0.5mm和0.75mm）交替层压制成。问该复合板的热导率为多少？（a）平行板方向（b）垂直板的方向（Al=0.22 R=0.00012 单位Js-1m-1K-1）,解（a）model:铝板和橡胶板分别都为25mm/（1.25mm）=20层,VA1=10/25=0.4 VR=15/25=0.6 l=V A1 A1+V

15、RR=0.4*0.22+0.6*0.00012=0.088+0.000072=0.088(Js-1m-1K-1),(b)Solution:因为：1/t=V A1/A1+Vr/r=0.4/0.22+0.6/0.0012=1.82+5000=5001.82t=1/5001.82=2.0*10-4(Js-1m-1K-1),15.11 A continuous and aligned fiber-reinforced composite is to be produced consisting of 30 vol%aramid fibers and 70 vol%of a polycarbonate

16、matrix;mechanical characteristics of these two materials are as follows:Also,the stress on the polycarbonate matrix when the aramid fibers fail is 45 Mpa.For this composite,compute(a)the longitudinal tensile strength,and(b)the longitudinal modulus of elasticity.,(a）the longitudinal tensile strength

17、cl=45*0.7+3600*0.3=31.5+1080=1111.5(Mpa)(b)in terms of Equation:Ecl=Em(1-Vf)+Ef*Vf Ecl=2.4*0.7+131*0.3=1.68+39.3=40.98 GPa,(15.17),?,15.15(a)Verify that Equation 15.11,the expression for the fiber loadmatrix load ratio(Ff/Fm),is valid.,(a)m=f m/Em=f/Ef Fm/Am/Em=Fm/(Am*Em)=Ff/Af/Ef)=Ff/(Af*Ef)Ff/Fm=(

18、Af*Ef)/(Am*Em)=(Ef*Vf)/(Em*Vm)=(Ef*Vf)/Em(1-Vf),Vf=Af*l,Vm=Am*l,15.20 It is desired to produce an aligned carbon fiber-epoxy matrix composite having a longitudinal tensile strength of 750 MPa(109,000psi).Calculate the volume fraction of fibers necessary if(1)the average fiber diameter and length are

19、 1.210-2 mm(4.7 104 in.)and 1 mm(0.04 in.),respectively;(2)the fiber fracture strength is 5000 MPa(725,000 psi);(3)the fiber-matrix bond strength is 25 MPa(3625 psi);and(4)the matrix stress at fiber failure is 10 MPa(1450 psi).,4-76 简述纳米材料的基本物理效应。,思考题,思考题：CMC中陶瓷基体增韧机理？为什么纤维增强复合材料的抗疲劳性能明显优于基体本身的抗疲劳性能？为何复合材料的实测力学强度与理论计算强度存在明显差异且分散性较大？界面层对增强相和基体相的结合强度是否越强越好？为什么？4-75，4-77，15.4，15.5，15.8,15.16思考题：为何纳米微粒易团聚，难以分散？为何金属微粒的电阻随其粒径的减小而增大甚至变成绝缘体？为何纳米复合材料的力学性能远优于相同组分的常规复合材料？,