第 1 章 作业 答案.doc

Chapter one home work1. (P80 3-3) Calculate the atomic radius in cm for the following:(a) BCC metal with a0=0.3294nm and one atom per lattice point; and(b) FCC metal with a0=4.0862 and one atom per lattice point.Solution:(a) In BCC structures, atoms touch along the body diagonal, which isa0 in length. There are two atomic radii from the center atom and one atomic radius from each of the corner atoms on the body diagonal, so: =0.14263nm=1.4263cm(b) In FCC structures, atoms touch along the face diagonal of the cube, which is in length. There are four atomic radii along this lengthtwo radii from the face-centered atom and one radius from each corner, so , =1.44447 =1.44447cm2. (P80 3-4) determine the crystal structure for the following:(a) a metal with a0=4.9489, r=1.75, and one atom per lattice point; and(b) a metal with a0=0.42906nm, r=0.1858nm, and one atom per lattice point.Solution:We know the relationships between atomic radii and lattice parameters are in BCC and in FCC. (a) 1.75 so its crystal structure is FCC;(b) =0.186nm so its crystal structure is BCC.3. (P80 3-5) the density of potassium, which has the BCC structure and one atom per lattice point, is 0.855g/cm3. the atomic weight of potassium is 39.09g/mol. Calculate(a) the lattice parameter; and(b) the atomic radius of potassium.Solution(a) For a BCC unit cell, there are two atoms in per unit cell, atomic mass is 39.09g/mol, density =0.855g/cm3Avogadros number NA=6.02atoms/mol0.855g/cm3=So a=0.53=5.3(b)then r=0.229cm=2.294. (P81 3-20) determine the indices for the directions in the cubic unit cell shown in Figure 3-32.The procedure for finding the Miller indices for directions is as follows:1. Using a right-handed coordinate system, determine the coordinates of two points, which lie on the direction.2. Subtract the coordinates of the “tail” point from the coordinates of the “head” point to obtain the number of lattice parameters traveled in the direction of each axis of the coordinate system.3. Clear fractions and/or reduce the results obtained from the subtraction to lowest integers.4. Enclose the number in square brackets . If a negative sign is produced, represent the negative sign with a bar over the number. SolutionDirection A1. Two points are 0,0,1 and 1,0,02. 0,0,1-1,0,0=-1,0,13. no fraction to clear or integers to reduce4.Direction B1. Two points are 1,0,1 and ,1,02. 1,0,1-,1,0=,-1,13. 2(,-1,1)=1,-2,24.Direction C1. Two points are 1,0,0 and 0,12. 1,0,0-0,1=1, -,-13. 4(1, -,-1)=4, -3, -44.Direction D1. Two points are 0,1, and 1,0,02. 0,1, -1,0,0=-1,1, 3. 2(-1,1, )=-2,2,14.5. (P82 3-22) Determine the indices for the planes in the cubic unit cell shown in Figure 3-34.The procedure for finding the Miller indices for planes is as follows:1. Identify the points at which the plane intercepts the x, y, and z coordinates in terms of the number of lattice parameters. If the plane passes through the origin, the origin of the coordinate system must be moved!2. Take reciprocals of these intercepts.3. Clear fractions but not reduce to lowest integers.4. Enclose the resulting numbers in parentheses (). Again, negative numbers should be written with a bar over the number.SolutionPlane A1. x=-1, y=, z=2. 3. Clear fractions: -3, 6, 44. ()Plane B1. x=1, y=-, z=2. 3. Clear fractions: 3, -4, 04. Plane C1. x=2, y=, z=12. 3. Clear fractions: 3, 4, 64. (346) 6. (P82 3-23) Sketch the following planes and directions within a cubic unit cell:(a) 101 (b) 00 (c) 12 (d) 301 (e) 01 (f) 23(g) (0) (h) (102) (i) (002) (j) (10) (k) (12) (l) (3) 7. Calculate the angle between 100 and 111 in Al.Solution: The crystal structure of Al is Fcc. We can calculate the angle between 100 and 111 as8. Use a calculation to verify that the atomic packing factor for the FCC structure is 0.74.Solution:In an FCC, there are four lattice points per cell: if there is one atom per lattice point, there are also four atoms per cell. The volume of one atom is 4r3/3 and the volume of the unit cell is a3: Packing factor = 44r3/3 a3Since for FCC unit cell, a=4r/, packing factor =o.749. 写出溶解在-