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.1习题讲解习题讲解.2第一次作业第一次作业英文英文 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.3K: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).4.5.6已知：已知：TiO2,XTi=1.5 and XO=3.5 ZnTe,已知：已知：XZn=1.6 and XTe=2.1，故，%IC=6.05%CsCl,已知：已知：XCs=0.7 and XCl=3.0，故：故：%IC=73.4%InSb,已知：已知：XIn=1.7 and XSb=1.9，故：故：%IC=1.0%MgCl2,已知：已知：XMg=1.2 and XCl=3.0故：故：%IC=55.5%.7.8.9第二次作业第二次作业.10.11.12.13第三次作业第三次作业.143.50 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)?.15.163.51*Within a cubic unit cell,sketch the following directions:.17.18.193.53 Determine the indices for the directions shown in the following cubic unit cell:.20Direction 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 24.21Direction 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 6.223.57 Determine the Miller indices for the planes shown in the following unit cell:plane B(1 0 1).233.58 Determine the Miller indices for the planes shown in the following unit cell:plane B (2 2 1).243.61*Sketch within a cubic unit cell the following planes:.253.61*Sketch within a cubic unit cell the following planes:.26defgh.273.62 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).28.29222lkh22202222 22.30.31.32八面体八面体四面体四面体.33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,).34解解:该平面与三轴该平面与三轴的截距为的截距为1 1，-2/3-2/3，2/32/3 取倒数为：取倒数为：1/1,-3/2,3/2，同分后去分母，得：同分后去分母，得：2/2,-3/2,3/2 米勒指数为米勒指数为(2 3 3).35.36.37.38.39.40对于对于FCC来说，来说，最大间隙为最大间隙为 0.414R又又 Ag的晶胞参数的晶胞参数 a=4.0857 A，而而FCC,R=a2/4 最大间隙最大间隙0.414R=0.414(a2/4)=0.4144.08572/4 =0.598(A)答：答：能进入能进入FCC银的间隙位置而不拥挤的最大银的间隙位置而不拥挤的最大原子半径为原子半径为0.598A。.412-32碳原子能溶入碳原子能溶入fcc 铁的最大填隙位置：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？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？在此位置四周有多少铁原子围绕？第四次作业第四次作业(a)fcc(a)fcc的最大间隙的最大间隙为八面体间隙为八面体间隙,每个每个原子有原子有1 1个个,每个晶每个晶胞有胞有4 4个个(b)(b)有有6 6个原子围绕个原子围绕.42.432-342-34请找出能进入请找出能进入bccbcc铁填隙位置的最大铁填隙位置的最大原子的半径（暗示：最大空洞位在原子的半径（暗示：最大空洞位在1/21/2，1/41/4，0 0位置）。位置）。.44 体心立方体心立方（bcc）八面体间隙数：八面体间隙数：四面体间隙数：四面体间隙数：每晶胞：每晶胞：每晶胞：每晶胞：1/2 6(面面)+1/4 12(棱棱)=6 4 1/2 6(面面)=12 每原子：每原子：6/2=3 每原子：每原子：12/2=6间隙大小（半径）间隙大小（半径）间隙大小（半径）间隙大小（半径）=a/2 R=0.633R =5 a/4 R =0.291R =a/2 R=0.154R .452-34 解解:bcc中八面体间隙为扁球状中八面体间隙为扁球状,两两个方向的半径为个方向的半径为0.633R和和0.154R,故能进入间隙的最大原子半径为故能进入间隙的最大原子半径为0.154R(不变形时不变形时)而四面体间隙半径为而四面体间隙半径为0.291R,能能容纳此半径的原子容纳此半径的原子,显然此间隙能显然此间隙能进入原子半径最大进入原子半径最大(0.291R).46.472-35碳和氮在碳和氮在-Fe中的最大固溶度分别为中的最大固溶度分别为8.9%和和10.3%，已知碳、氮原子均占据八，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。子和氮原子占据的百分数。解：解：-Fe为面心立方，为面心立方，则：每个晶胞中则：每个晶胞中4个铁原子，且八面体个铁原子，且八面体间隙数间隙数=Fe原子数原子数 .48 面心立方面心立方四面体间隙数：四面体间隙数：每晶胞：每晶胞：8 每原子：每原子：8/4=2 间隙大小（半径）间隙大小（半径）=3 a/4 R=0.225R八面体间隙数：八面体间隙数：每晶胞：每晶胞：1(体体)+1/4 12(棱棱)=4 每原子：每原子：4/4=1 间隙大小（半径）间隙大小（半径）=a/2 R=0.414R.49.502-45在钢棒的表面，每在钢棒的表面，每20个铁的晶胞中有一个铁的晶胞中有一个碳原子，在离表面个碳原子，在离表面1mm处每处每30个铁的晶胞个铁的晶胞中有一个碳原子。温度为中有一个碳原子。温度为1000时扩散系数时扩散系数是是310-11m2/s，且结构为面心立方，且结构为面心立方（a=0.365nm）。问每分钟因扩散通过单位）。问每分钟因扩散通过单位晶胞的碳原子数是多少？晶胞的碳原子数是多少？解：由已知可以计算出碳的浓度：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原子原子/min 答：每分钟因扩散通过单位晶胞的碳原子数是答：每分钟因扩散通过单位晶胞的碳原子数是84。.512-49 计算计算550时铜在铝中的扩散系数时铜在铝中的扩散系数（D0=1.510-5m2/s，Q*=191KJ/mol-1）。）。解：解：D=D0 e-Q/RT =1.510-5 e-1911000/(8.314823)=1.1310-17(m2/s)答：答：550 时铜在铝中的扩散系数为时铜在铝中的扩散系数为1.1310-17(m2/s).52英文英文 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，解：有题可知，Qv=0.55 eV/atom答：答：327 时铅原子的空位分数为时铅原子的空位分数为2.411010-5-5.535.8 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.54.55.565.10(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?.576.6 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.58解：由已知解：由已知5mm厚的板的扩散通量厚的板的扩散通量 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：答：500每小时通过该钯板（厚度：每小时通过该钯板（厚度：5mm,面面积：积：0.20 m2）的氢原子数目为）的氢原子数目为2.59 10-3 kg/h.596.24 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：解：J=-D dC/dx D D=-J -J (dC/dx)=-1.4310-9 kg/m2s (0.3-0.65)(kg/m3)/0.015 m =6.1310-11 m2/s 又：又：D=D0 exp(-Q/RT)即：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时的温度时的温度 为为771 Te314.880000Te314.880000
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