材料科学与工程基础习题评讲ppt课件.ppt

习题讲解,第一次作业,英文 2.6 Allowed values for the quantum numbers of electrons are as follows: The relationships between n and the shell designations are noted in Table 2.1. Relative to the subshells,l 0 corresponds to an s subshelll 1 corresponds to a p subshelll 2 corresponds to a d subshelll 3 corresponds to an f subshellFor the K shell, the four quantum numbers for each of the two electrons in the 1s state, in the order of n l ml ms , are 100(1/2) and 100( -1/2 ). Write the four quantum numbers for all of the electrons in the L and M shells, and note which correspond to the s, p, and d subshells,K: s: 100(1/2); 100(-1/2)L: s: 200(1/2); 200(-1/2) p: 210(1/2); 210(-1/2); 21-1(1/2); 21-1(-1/2); 211(1/2); 211(-1/2)M: s: 300(1/2); 300(-1/2) p: 310(1/2); 310(-1/2); 31-1(1/2); 31-1(-1/2); 311(1/2); 311(-1/2) d: 320(1/2); 320(-1/2); 32-1(1/2); 32-1(-1/2); 321(1/2); 321(-1/2); 32-2(1/2); 32-2(-1/2); 322(1/2); 322(-1/2),2.7 Give the electron configurations for the following ions: Fe2+, Fe3+, Cu+, Ba2+, Br-, and S2-.SOLUTION Fe2+ : 1s22s22p63s23p63d6 Fe3+ : 1s22s22p63s23p63d5 Cu+ : 1s22s22p63s23p63d10 Ba2+ : 1s22s22p63s23p63d104s24p64d105s25p6 Br- : 1s22s22p63s23p63d104s24p6 S 2- : 1s22s22p63s23p6,2.17 (a) Briefly cite the main differences between ionic, covalent, and metallic bonding.(b) State the Pauli exclusion principle.SOLUTION(a) 离子键： 无方向性 球形正、负离子堆垛 取决 电荷数电荷平衡 体积（离子半径） 金属键： 无方向性 球形正离子较紧密堆垛 共价键： 有方向性、饱和性，电子云最大重叠(b)原子中的每个电子不可能有完全相同的四个量子数（或运动状态）,2.19 Compute the percents ionic character of the interatomic bonds for the following compounds: TiO2, ZnTe, CsCl, InSb, and MgCl2 . SOLUTION由公式：,已知：TiO2, XTi = 1.5 and XO = 3.5,ZnTe,已知：XZn = 1.6 and XTe = 2.1 ，故，%IC=6.05% CsCl,已知： XCs = 0.7 and XCl = 3.0 ， 故： %IC=73.4% InSb,已知： XIn = 1.7 and XSb = 1.9， 故： %IC=1.0% MgCl2,已知：XMg = 1.2 and XCl = 3.0故： %IC=55.5%,2.24，On the basis of the hydrogen bond, explain the anomalous behavior of water when it freezes. That is, why is there volume expansion upon solidification? 水冻结时结晶，非球形的水分子规整排列时受氢键方向性和饱和性的更强限制，不能更紧密地堆积，故密度变小，体积增大。,2-7影响离子化合物和共价化合物配位数的因素有那些？离子化合物： 体积 电荷共价化合物： 价电子数 电子云最大重叠,第二次作业,2.18 Offer an explanation as to why covalently bonded materials are generally less dense than ionically or metallically bonded ones. 共价键需按键长、键角要求堆垛， 相对离子键和金属键较疏松,2.21Using Table 2.2, determine the number of covalent bonds that are possible for atoms of the following elements: germanium, phosphorus, selenium, and chlorine. SOLUTION Ge : 4 P : 3 Se : 2 Cl : 1,2-6按照杂化轨道理论，说明下列的键合形式：（1）CO2的分子键合 C sp 杂化 （2）甲烷CH4的分子键合 C sp3杂化 （3）乙烯C2H4的分子键合 C sp2杂化 （4）水H2O的分子键合 O sp3杂化 （5）苯环的分子键合C sp2杂化 （6）羰基中C、O间的原子键合 C sp2杂化,2-10 当CN=6时，K+离子的半径为0.133nm (a) 当CN=4时，对应负离子半径是多少？ (b) 当CN=8时，对应负离子半径是多少？若(按K+半径不变) 求负离子半径, 则： CN=6 R = r/0.414=0.133/0.414 = 0.321 nm CN=4 R = r/0.225=0.133/0.225 = 0.591 nm CN=8 R = r+/0.732=0.133/0.732 = 0.182 nm,第三次作业,3.48 Draw an orthorhombic unit cell, and within that cell a 121 direction and a (210) plane.,3.50 Here are unit cells for two hypothetical metals: a. What are the indices for the directions indicated by the two vectors in sketch (a)? b What are the indices for the two planes drawn in sketch (b)?,（a）direction 1, x y zProjections 0a b/2 cProjections in terms of a, b, and c 0 1/2 1Reduction to integers 0 1 2Enclosure 012direction 2, x y zProjections a/2 b/2 -cProjections in terms of a, b, and c 1/2 1/2 -1Reduction to integers 1 1 -2Enclosure 11 2 （b）Plane 1， ：1/2 ： ； 0：2：0 ； (020) Plane 2， 1/2：-1/2 ： 1 ； 2：-2：1； (2 2 1),3.51* Within a cubic unit cell, sketch the following directions:,a b,（c）0 1 2 （d）1 3 3（e）1 1 1 （f）1 2 2,（g）1 2 3 （h）1 0 3,3.53 Determine the indices for the directions shown in the following cubic unit cell:,Direction A: x y z -2/3a b/2 0c -2/3 1/2 0 -4 3 0 4 3 0,Direction A: Direction B: x y z x y z -2/3 a b/2 0c 2/3 a -b 2/3 c -2/3 1/2 0 2/3 -1 2/3 -4 3 0 2 -3 2 4 3 0 2 3 2,Direction C Direction D x y z x y z 1/3a -b -c a/6 b/2 -c 1/3 -1 -1 1/6 1/2 -1 1 -3 -3 1 3 -6 1 3 3 1 3 6,3.57 Determine the Miller indices for the planes shown in the following unit cell:,plane Ax y za /3 b/2 -c/21/3 1/2 -1/23/1 2/1 -2/1(3 2 2),plane B (1 0 1),3.58 Determine the Miller indices for the planes shown in the following unit cell:,lane A 以（0，1，0）为新原点x y z2/3a -b c/22/3 -1 1/23/2 -1/1 2/13/2 -2/2 4/2(3 2 4),plane B ( 2 2 1),3.61* Sketch within a cubic unit cell the following planes:,a,3.61* Sketch within a cubic unit cell the following planes:,ab c,d,e,f,g,h,3.62 Sketch the atomic packing of (a) the (100) plane for the FCC crystal structure, and (b) the (111) plane for the BCC crystal structure (similar to Figures 3.24b and 3.25b).,(a) FCC: (100) plane (b) BCC: (111) plane,3.81The metal iridium has an FCC crystal structure. If the angle of diffraction for the (220) set of planes occurs at 69.22(first-order reflection) when monochromatic x-radiation having a wavelength of 0.1542 nm is used, compute (a) the interplanar spacing for this set of planes, and (b) the atomic radius for an iridium atom.SOLUTION：已知：铱FCC的（220） 晶面,2= 69.22;= 0.1542 nm； n = 1 由公式： n= 2d sin 和 dhkl = a /( ) 故 解：（a）d220 = 0.1542 / (2sin34.61) = 0.1357（nm）（b） a = d220 = 2 d220 ； 又 FCC的 2R = a/ = 2d220 R = d220 = 0.1357（nm）即：其（220）晶面间距为0.1357nm；铱原子半径也为0.1357nm。,2-14计算（a）面心立方金属的原子致密度；（ b）面心立方化合物NaCl的离子致密度（离子半径r(Na+)=0.097，r(Cl-) =0.181）；（C）由计算结果，可以引出什么结论？解：（a）面心立方金属 一个晶胞中有四个金属原子，且a=4R / 2 有：PF = 4(4R3/3) / a3 =16R3(22) / (3)(64R3) = 0.74答：面心立方金属的原子致密度为0.74,（b）解： 面心立方化合物NaCl, 且: r(Na+)=0.097，r(Cl-) =0.181 一个晶胞中有四个Na离子、四个Cl离子有：PF = 4(4r3/3)+4(4R3/3) / (2r+2R)3 =16(0.0973+0.1813) / 3 (8)(0.097+0.181)3 = 0.64答：面心立方化合物NaCl 的离子致密度为0.64（c）结论: (1) 同种原子晶体的致密度只与晶胞类型相关,与原子尺寸无关 (2) 化合物晶体的离子致密度与离子大小相关,2-18在体心立方结构晶胞的（100）面上按比例画出该面上的原子以及八面体和四面体间隙。,八面体四面体,2-24 方向为1 1 1的直线通过1/2, 0, 1/2点,则在此直线上的另外两点的坐标是什么?,解： 该直线过为: u=1/2 v= 0 w= 1/2 u = 0: v = - 1/2 , w = 0 . 即过 (0, -1/2 , 0)点 u = 1: v = 1/2 , w = 1 . 即过 (1, 1/2 , 1)点,(0, -, 0),(1, , 1),(, 0, ),2-27 一平面与三轴的截距为a=1, b=-2/3, c=2/3, 则此平面的米勒指数是什么？,解:该平面与三轴的截距为1，-2/3，2/3 取倒数为： 1/1, -3/2, 3/2，同分后去分母，得： 2/2, -3/2, 3/2 米勒指数为(2 3 3),2-29 氯化钠晶体被用来测量某些X光的波长，对氯离子的d111间距而言，其衍射角2为27.5(a)X光的波长是多少？（NaCl晶格常数为0.563nm）(b)若X光的波长为0.058nm，则其衍射角2是多少？ 解：(a) NaCl， 面心立方，晶格常数为0.563 d111 = a / (h2 + k2 + l2)1/2 = 0.563 /1+1+1 = 0.563/ 3,又 n= 2d sin , 衍射角2=27.5 当 n = 1时， = 2d sin /n = 2(0.563 /3) sin(27.5/2) = 0.154 (nm)答： X光的波长为0.153nm(b) 已知：= 0.058nm 2= 2 arcsin(/ 2d) = 2 arcsin0.058 / (20.325) =10.24答：当X光的波长为0.058nm，则其衍射角2 为10.24 ,2-31请算出能进入FCC银的间隙位置而不拥挤的最大原子半径。解：因为FCC面心立方的四面体和八面体间隙如下所示：,对于FCC来说， 最大间隙为 0.414R又 Ag的晶胞参数 a = 4.0857 A， 而FCC, R= a2/4 最大间隙0.414R = 0.414(a2/4) =0.4144.08572/4 = 0.598 (A) 答：能进入FCC银的间隙位置而不拥挤的最大原子半径为0.598A。,2-32碳原子能溶入fcc 铁的最大填隙位置：(a)每个单元晶胞中有多少个这样的位置？(b)在此位置四周有多少铁原子围绕？,第四次作业,(a) fcc的最大间隙为八面体间隙,每个原子有1个,每个晶胞有4个(b) 有6个原子围绕,2-34请找出能进入bcc铁填隙位置的最大原子的半径（暗示：最大空洞位在1/2，1/4，0位置）。,2-34 解: bcc中八面体间隙为扁球状,两个方向的半径为0.633R和0.154R,故能进入间隙的最大原子半径为0.154R(不变形时) 而四面体间隙半径为0.291R,能容纳此半径的原子,显然此间隙能进入原子半径最大(0.291R).,2-35碳和氮在-Fe中的最大固溶度分别为8.9%和10.3%，已知碳、氮原子均占据八面体间隙，试分别计算八面体间隙被碳原子和氮原子占据的百分数。解：-Fe为面心立方， 则：每个晶胞中4个铁原子，且八面体间隙数 = Fe原子数,将固溶度视为摩尔百分数， 并设八面体间隙被碳原子占据的摩尔百分数为X, 被氮原子占据的摩尔百分数为Y,则有： X/(X+Y+4)=8.9% 和 Y/(X+Y+4)=10.3%由上面两个方程计算得到： X=0.44；Y=0.51答：在八面体间隙占据的摩尔百分数分别为：C：0.44/4=11%； N：0.51/4=12.8%,2-45在钢棒的表面，每20个铁的晶胞中有一个碳原子，在离表面1mm处每30个铁的晶胞中有一个碳原子。温度为1000时扩散系数是310-11m2/s，且结构为面心立方（a=0.365nm）。问每分钟因扩散通过单位晶胞的碳原子数是多少？解：由已知可以计算出碳的浓度： C2=1/30*(0.36510-9m)3=0.681027 /m3 C1=1/20*(0.36510-9m)3=1.031027 /m3 由J= -D dC/dx =-(310-11m2/s)(0.68-1.03)(1027/m3) /(10- 3m) =1.051019/m2s Ju C= (1.051019/m2s) (0.36510-9m)2 (60s/min) =84原子/min 答：每分钟因扩散通过单位晶胞的碳原子数是84。,2-49 计算550时铜在铝中的扩散系数（D0=1.510-5m2/s，Q*=191KJ/mol-1）。解：D = D0 e -Q/RT = 1.510-5 e -1911000 / (8.314823) = 1.1310-17 (m2/s)答：550 时铜在铝中的扩散系数为1.1310-17 (m2/s),英文 5.1 Calculate the fraction of atom sites that are vacant for lead at its melting temperature of 327. Assume an energy for vacancy formation of 0.55 eV/atom解：有题可知，Qv=0.55 eV/atom,答：327 时铅原子的空位分数为2.4110-5,5.8 Below, atomic radius, crystal structure, electronegativity, and the most common valence are tabulated, for several elements; for those that are nonmetals, only atomic radii are indicated.,Which of these elements would you expect to form the following with copper:(a) A substitutional solid solution having complete solubility?(b) A substitutional solid solution of incomplete solubility?(c) An interstitial solid solution?,（a） 完全置换固溶体： Cu-Ni 半径差15%， 同为FCC，电负性相近，电价相同 （b）部分置换固溶体：Cu-Ag；Cu-Al；Cu-Co；Cu-Cr；Cu-Fe；Cu-Pd；Cu-Pt；Cu-Zn； 虽半径差15%，但或晶型不同，或电负性或电价相差较大。（c）间隙固溶体：Cu-C；Cu-H；Cu-O 后者半径小很多,5.10 (a) Suppose that Li2O is added as an impurity to CaO. If the Li+ substitutes for Ca2+, what kind of vacancies would you expect to form? How many of these vacancies are created for every Li+ added? (b) Suppose that CaCl2 is added as an impurity to CaO. If the Cl- substitutes for O2-, what kind of vacancies would you expect to form? How many of the vacancies are created for every Cl- added?,（a）在CaO中Li+ 置换Ca2+ 产生氧空位。一个Li+ 置换一个Ca2+，为保持电中性，每两个Li+ 置换后，减少的正电荷由形成一个氧空位减少的负电荷平衡。 （b）在CaO中Cl- 置换O2- 产生钙空位。一个Cl- 置换一个O2-，为保持电中性，每两个Cl- 置换后，减少的负电荷由形成一个钙空位减少的正电荷平衡。,6.6 Compute the number of kilograms of hydrogen that pass per hour through a 5-mm thick sheet of palladium having an area of 0.20 m2 at 500. Assume a diffusion coefficient of 1.010- 8 m2/s, that the concentrations at the high- and low-pressure sides of the plate are 2.4 and 0.6 kg of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained.,解：由已知5mm厚的板的扩散通量 J = -DdC/dx = -(110-8m2/s) (0.6 - 2.4) (kg/m3) (510-3m) = 3.610-6 kg/m2s 每小时通过的氢原子数 M = J S t = (3.610-6 kg/m2s)(0.20 m2) (3600s/h) =2.59 10 -3 kg/h答：500每小时通过该钯板（厚度：5mm, 面积： 0.20 m2 ）的氢原子数目为2.59 10 -3 kg/h,6.24 Carbon is allowed to diffuse through a steel plate 15mm thick. The concentrations of carbon at the two faces are 0.65 and 0.30 kgC/m3Fe, which are maintained constant. If the pre-exponential and activation energy are 6.210-7 m2/s and 80,000 J/mol, respectively, compute the temperature at which the diffusion flux is 1.4310-9kg/m2-s 解： J= - D dC / dx D = -J (dC / dx ) = - 1.4310-9 kg/m2s (0.3 - 0.65) (kg/m3) / 0.015 m = 6.1310-11 m2/s 又：D = D0 exp(-Q/RT )即： 6.1310-11= 6.210-7 exp-80000/(8.314T) T = 1044K=771答：满足该条件下扩散通量为1.4310-9kg/m2-s时的温度 为771 ,