1、1.讨论温度(包括熔点)对短程序及长程序的影响。,As temperature increases atomic vibrations increase and atoms can be regarded as moving about their “cages(盒, 箱)” more and more. At the melting temperature the LRO, which establishes the crystallinity of many materials, vanishes; however, SRO being associated with primary bo
2、nd distances, is largely preserved. Materials that are not crystalline have no long range order. At the softening temperature or glass transition temperature there is little change in SRO. Some polymers neither melt nor show a glass transition; rather, they degrade by primary bond scission, which ch
3、anges their composition.,引 言,材料科学基础,第3章答案,2.找一个印花的纺织品或墙纸，画出它们的花样。它们具有LRO吗？壁画有LRO吗？,引 言,The wallpaper in my room has a pattern similar to that above. The lattice is a parallelogram(平行四边形). There is one repeat motif per unit cell. The wallpaper has both SROand LRO: It fills the entire wall!,3.当晶体熔化时短程
4、序如何变化？,SRO changes very little with melting. In fluids the atomic separations are about the same as the bond distances in crystalline solids.,引 言,4.画出四方和菱方单胞,布拉维点阵和单胞,四方单胞 abc；90,菱方单胞 a b c ； 90,14 种 布 拉 菲 点 阵,6.画出一个底心立方结构，这一晶系包含在图3-3中吗？,布拉维点阵和单胞,简单四方,8.BCC的角上的原子彼此接触吗？FCC结构的呢？,SOLUTION: BCC - corner
5、 atoms are separated by Thus, corner atoms do not touch. They are 2.31r apart. FCC - Corner atoms are separated by ao, but atoms touch along the cube face diagonal. Hence, Thus, corner atoms do not touch. They are 2.83r apart. COMMENTS: Corner atoms are closer in the BCC than in the FCC.,布拉维点阵和单胞,SO
6、LUTION: A hexagonal lattice. COMMENTS: One crystalline form of SiO2 is “high temperature quartz”, which has a hexagonal lattice.,布拉维点阵和单胞,9.矿物学家通常用光学显微镜来鉴别成分。晶体的宏观形状经常(但不总是)反映单胞的对称性的，石英晶体被认为是六角的，哪一种胞可表征SiO2这种形式的晶态？,10.计算BCC金属Cr的面对角线长度,GIVEN: Cr is BCC. DATA: According to Appendix C, the atomic radiu
7、s of Cr is 1.25 A. SOLUTION: The length of the face diagonal is ao2. Atoms touch along the body diagonal, which has length ao 3 = 4r. Solving for ao in the latter equation:ao = 4r / 3. Thus, ao 2 = 4r 2 / 3= 4(1.25 A) 2 / 3 = 4.08 A,布拉维点阵和单胞,11.只用对原子测量的资料计算FCC铜的密度。它的测量值是8.96g/cm3。,DATA: Cu is FCC; a
8、tomic weight 63.55 g/mole; atomic radius = 1.278A. ASSUMPTIONS: Atoms are perfect spheres. SOLUTION: FCC 4 atoms/unit cell COMMENTS: This compares favorably with the measured value of 8.96 g/cm3 计算时未考虑到实际原子间距与钢球模型间的差异,布拉维点阵和单胞,12.计算HCP金属镁的密度，把结果与测量值1.74g/cm3比较。解释这一差别的原因。,GIVEN: HCP; measured value o
9、f atomic radius = 1.74A DATA: Atomic weight = 24.3 g/mole ASSUMPTIONS: Atoms are spherical SOLUTION: = Mass/Volume = ( x 24.3 g/mole) / (6.023 x 1023 atoms/mole)(1/11.67 x 10-24 cm3) = 1.73 g/cm3 COMMENTS: The value in tables is a similar 1.74 g/cm3,布拉维点阵和单胞,晶胞中的原子数：121/621/2+3=6,a02r, c/a0=1.633单胞体
10、积Vuc3 31/2 a02c/2,13.说明HCP材料的c/a比率为什么应该接近1.63。为什么有些金属的c/a不同于1.63？,ASSUMPTIONS: Spherical Atoms SOLUTION:COMMENTS: c/a usually is not 1.633 because atoms are not perfectly spherical and bonds may be partly covalent.,布拉维点阵和单胞,14. 钆加热到1260由HCP转变为BCC。HCP的点阵常数是a0.36745nm，c1.18525nm。BCC结构的点阵常数是a0.4060nm。计
11、算因晶体结构转变引起的体积变化。,GIVEN: HCP: ao = 3.6745A, c = 11.8525A. BCC: ao = 4.060A. ASSUMPTIONS: Atoms are spherical SOLUTION:,布拉维点阵和单胞,There are, however, a different number of atoms in each unit cell. Thus, we need to calculate the volume per atom:,(V/atomBCC - V/atomHCP) /VatomHCP = (33.46-23.10)/23.10 =
12、45% increaseCOMMENTS: This is a huge increase, 10.36A3/per atom, which is much larger than anticipated.,66.92,Gd的特性,15.列出5种最低密度的金属。这些材料是要求高的温度或刚度与重量(密度)比的首选。注意，这些金属中哪一个是你熟悉的。,SOLUTION: We begin by examining the metals that have the lowest atomic weights. Using Appendix A, we note that Li (0.53), Be
13、(1.85), B (2.34), Na (0.97), Mg (1.74), and Al (2.70), K (0.86), Ca (1.55), Sc (3.0), Ti (4.54) are the first entries, where the numbers in ( ) are densities in g/cm3. COMMENTS: Be is toxic, but not so in alloy form. It is a very high performance metal. B is used in fiber form for composite reinforc
14、ement. Na spontaneously bursts into flames upon exposure to air. Li, K, and Ca are also not stable as metals. Mg is commonly used (e.g. “mag“ wheels). Al is very common. Ti is used in bicycle construction and aerospace applications.,布拉维点阵和单胞,16. 在立方系中如果原子处在所有的1，0，0和1/2，1/2，0类型位置，它是什么布拉维点阵？,GIVEN: At
15、oms are in 1, 0, 0 - and , , 0-type positions; cubic structure SKETCH:SOLUTION: All the type positions are shown, which defined a FCC cell.,密勒指数,17.给出在FCC金属中由截距1，-1，-2构成的面，这个面的法线是什么？,GIVEN: Cubic structure SOLUTION: The plane is the reciprocal(倒数) of the intercepts = (1, bar -1, bar -1/2) with fract
16、ions cleared = (2, bar -2, bar- 1). The normal to a plane has the same indices in direction format 2- 2- 1.,密勒指数,18.计算Al中的100与111间的夹角。,A=ui+vj+wk B=ui+vj+wk AB=|A|B|cos,密勒指数,cos-11/3,19.在立方单胞中心构造一个坐标系，它的轴平行于方向的坐标系，确定四面体角即由原点到任两个面对角末端方向的角的大小。,SOLUTION:Thus, cos = (-1 + 1 - 1) / = -1/3 = 109.5, the “t
17、etrahedral(四面体) angle”. COMMENTS: Remember this angle. It is important in crystallography and in covalent bonding.,密勒指数,111,111,20. 立方晶体中110与1-11间，(110)与(111)间以及(110)与111间夹角是多大？,GIVEN: Cubic system SOLUTION: For the (110) + (111) set we will use the fact that the angle between the plane normals equa
18、ls that of the plane intersection. For the third set we use the fact that 001 lies in (110). We solve all using the vector dot product:,密勒指数,21.给出图3-38的立方单胞中的点、方向和面的指数。,SOLUTION: For A-F, the origin is the bottom-back-left cube corner unless indicated otherwise. A is 0, 0, ; B is 0, 1, ; C is 0-11 w
19、ith respect to the front-bottom-right corner; D is -20-1 with respect to the front-top-right corner; E is (001); and F is (112). For G-L, the origin is the bottom-front-left corner unless indicated otherwise. G is 0, 0, 1/3; H is 0, , 0; I is 10-1 with respect to the center of the right face; J is -
20、11-1 with respect to the front-top-left corner; K is -110; and L is -223.,密勒指数,25.在一个FCC单胞中计算在(100)，(110)和 (111)面上的原子的面密度。,SOLUTION: In FCC COMMENTS: Thus, the planar density of (111) (100) (110) in FCC,密勒指数,26.计算BCC结构的(100)和(110)的面的密度。,SOLUTION: In BCC, COMMENTS: Thus, the planar density of (110) (
21、100)in BCC,晶体结构的密度和堆垛因子,晶体结构的密度和堆垛因子,29.计算FCC和HCP结构密排面的面间距。,FIND: Separation of close-packed planes in FCC and HCP GIVEN: Close packed planes in FCC are (111) and the c-plane in HCP ASSUMPTIONS: Atoms are spherical SKETCH: Proceed using the same sketch as in problem 12: Atoms 1, 2, 3 define a close
22、packed plane. Number 4 rests on the hole formed by the other three. The distance from plane 1, 2, 3 to atom 4 is 2/3 the distance from x to 4, since 1, 2, 3 is a (111)- type plane.,30.比较FCC晶体中100和111的线密度。,SOLUTION: In FCC, body diagonal =COMMENTS: The linear density of 100 is greater than that of 11
23、1 in FCC.,晶体结构的密度和堆垛因子,31. 由纤维和树脂组成的纤维增强复合材料，设纤维直径的尺寸是相同的。请由计算最密堆棒的堆垛因子来确定能放入复合材料的纤维的最大体积分数。,晶体结构的密度和堆垛因子,SOLUTION: Consider one equilateral triangle, as shown. The packing in this triangle represents that of the entire structure. From stereology the area fraction of this structure equals the volume
24、 fraction. Shown is a unit cell.COMMENTS: This represents the maximum volume fraction of fiber in a composite or fiber in a yarn(纱线), assuming fiber cross-section remains circular.,33.FCC比BCC密排和具有较高的配位数，为什么有些金属结晶成BCC结构？,SOLUTION: Not all metals can be represented by ideally packed spheres that negle
25、ct interactions between the atoms. Atoms may be not spherical, bonds may be partially covalent, or both.,晶体结构的密度和堆垛因子,34. 假设你发现一种材料，它们密排面以ABAC重复堆垛。这种发现有意义吗？你能否计算这种新材料的原子堆垛因子？,DATA: The packing factor of HCP and FCC is 0.74. SOLUTION: The packing factor of the new material is 0.74. This is true regar
26、dless of the sequence, so long as (1) each plane is close packed and (2) there are no AA, BB or CC blocks. COMMENTS: The material would be unique in its stacking sequence. There are presently no known materials that stack ABAC.,晶体结构的密度和堆垛因子,36. 在FCC、HCP和BCC中最高密度面是哪些面？在这些面上哪些方向是最高密度方向？,SOLUTION: Thes
27、e planes and directionsare identified in the text. FCC 111 HCP 001 BCC 110 ,晶体结构的密度和堆垛因子,39.用能装入的最大的球填入FCC结构的八面体空洞中。对BCC结构能否如此？每一种情况下有多少个与球接触的邻居？,间隙原子和尺寸,SOLUTION: Examine the 6 atoms surrounding the central octahedral (八面体) site in Fig. a. Each of the 6 neighbors is ao / 2 away. Hence, the largest
28、sphere in the hole will just touch each of the 6 neighbors. Now examine Fig. b. The 2 atoms above and below the site indicated are ao / 2 away. Those at the corners of the unit cell plane in which the hole lies are further away (ao / 2). Hence, the largest atoms in the octahedral site touches only t
29、wo neighbors - one above and one below.,40.如果在FCC结构中的八面体和四面体位置全部填入原子，它的APF是多少？,GIVEN: In the text are given: 4 octahedral (八面体)sites/cell r/R = 0.4148 tetrahedral (四面体)sites/cell r/R = 0.225 ASSUMPTIONS: Spherical atoms - all sites filled with “perfectly”fitting spherical atoms. SOLUTION: In FCC The
30、 volume of the cell atoms is (4/3)r3 The volume of the octahedral atoms is 4 x (4/3) (0.414r)3 The volume of the tetrahedral sites is 8 x (4/3) (0.225r)3 Thus, the APF with all tetrahedral and octahedral sites filled = (cell atom volume + tetrahedral atom volume + octahedral atom volume) / (cell vol
31、ume) =,间隙原子和尺寸,42.比较BCC结构中的四面体位置和FCC结构八面体位置的数目和尺寸。,SOLUTION: These values are given in the text:r/R or k/R BCC 6 Oct. 0.15524 Tet. 0.291 FCC 4 Oct. 0.4148 Tet. 0.225,间隙原子和尺寸,44. 在铁中加入碳形成钢。BCC结构的铁称铁素体，在912以下是稳定的，在这温度以上变成FCC结构，称之为奥氏体。你预测哪一种结构能溶解更多碳？对你的答案作出解释。,DATA: Atomic radius of Fe depends on the e
32、xtent of covalent bonding. It is 1.241A in BCC and 1.269A in FCC. The atomic radius of carbon is 0.77A. ASSUMPTIONS: Spherical atoms; C goes into holes SOLUTION: The radius of the octahedral sites, the larger ones in FCC, is 0.414 x 1.269 = 0.525AThe radius of the tetrahedral sites, the larger ones
33、in BCC, is 0.291 x 1.241A = 0.361ATherefore, carbon fits easier into FCC iron; however, there are more tetrahedral sites in BCC (24) than octahedral sites in FCC (4). COMMENTS: We expect the FCC crystal structure to have greater C solubility than the low temperature BCC crystal structure. The maximu
34、m solubility of C is 0.02% in BCC iron and 2.1% in FCC iron.,间隙原子和尺寸,45.计算硅的点阵常数。,DATA: Si assumes the diamond cubic structure, which has atoms in the FCC positions and the tetrahedral sites filled. The atomic radius of Si is 1.176A. ASSUMPTIONS: Spherical atoms. SKETCH: Showing only a portion of th
35、e FCC atoms: SOLUTION: 2 atoms touch along 1/4 body diagonal,在每个点阵位置有多个原子的晶体,47.计算组成硅酸盐的基本积木块(SiO4)4-四面体的密度。,DATA: The ionic radius of Si4+ is 0.39 A and that of O2- is 1.32 A, according to Appendix C. The atomic weight of Si is 28.09 g/mole and that of O is 16.00 g/mole. SKETCH: See Fig. 3.7-6a and
36、 the following. Remember that anion and cation touch. Anions do not touch.,SOLUTION: Observation of the figure, after focusing on the body diagonal, leads to the conclusion: (ao3) / 2 = r + R ao = 2(r + R) / 3 = 2(1.32 A + 0.39 A) / 3 = 1.97 A. To calculate mass, M, we need to determine the number o
37、f anions(阴离子) and cations (阳离子)in the tetrahedra. Number of anions = 1/8 x 4 = 1/2. Number of cations = 1.Density = M/V = = 7.84 g/cm3.,在每个点阵位置有多个原子的晶体,51.预测NiO的结构。,DATA: rNi2+ = 0.69A; rO2- = 1.32A. The electronegativity of Ni is 1.91 and that of O is 3.44. ASSUMPTIONS: Ions are spherical. SOLUTION
38、: rNi2+/rO2- = 0.69/1.32 = 0.532 CN = 6, from Table 2-2. The difference in electronegativity is 1.53 bond is 44% ionic. Thus, we predict a NaCl structure, the only MX ionic structure with CN = 6. COMMENTS: In this case, 44% is sufficiently ionic that the NaCl structure is the correct structure.,在每个点
39、阵位置有多个原子的晶体,55.说明为什么像ThO2、TeO2和UO2这类材料结晶为荧石结构。,DATA: Electronegativities: Th = 1.3 and O = 3.44, etc. ionic radii: rTh4+ = 1.20, rO2- = 1.32, etc. ASSUMPTIONS: Ions are spherical SOLUTION: Three criteria need to be satisfied for fluorite structure: (1) stoichiometry(化学计量) of metal to oxide = 1:2 (2) Electronegativity difference large ionic structures,e.g. ThO2 2.14 68% (3) radius ratio needs to be 0.732-1.0,e.g. for ThO2 1.02/1.32 = 0.77 CN = 8 for Th and CN = 4 for O,在每个点阵位置有多个原子的晶体,
