材料科学与工程基础习题评讲-1_ppt课件

单击此处编辑母版标题样式单击此处编辑母版副标题样式* 1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 subshellsK: 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+ : 1s22s22p63s23p63d6Fe3+ : 1s22s22p63s23p63d5Cu+ : 1s22s22p63s23p63d10Ba2+ : 1s22s22p63s23p63d104s24p64d105s25p6 Br- : 1s22s22p63s23p63d104s24p6S 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 ）2-7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. SOLUTIONGe : 4P : 3Se : 2Cl : 12-6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 nmCN=4 R = r/0.225=0.133/0.225 = 0.591 nmCN=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 bn （ c） 0 1 2 （ d） 1 3 3n （ e） 1 1 1 （ f） 1 2 2n （ g） 1 2 3 （ h） 1 0 33.53 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 0Direction 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 24 3 0 2 3 2Direction C Direction Dx y z x y z1/3a -b -c a/6 b/2 -c1/3 -1 -1 1/6 1/2 -11 -3 -3 1 3 -61 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:plane 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:a3.61* Sketch within a cubic unit cell the following planes:ab cd e fg h3.62 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) plane3.81The 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 n 由公式： n= 2d sin 和 dhkl = a /( ) n 故 解：（ a） d220 = 0.1542 / (2sin34.61) = 0.1357（ nm）n （ b） a = d220 = 2 d220 ；n 又 FCC的 2R = a/ = 2d220n R = d220 = 0.1357（ nm）n 即：其（