材料科学基础第一章习题答案

材料科学基础第一章习题答案1. (P80 3-3) Calculate the atomic radius in cm for the following:(a) BCC metal with a 0=0.3294nm and one atom per lattice point; and(b) FCC metal with a 0=4.0862Å and one atom per lattice point.Solution:(a) In BCC structures, atoms touch along the body diagonal, which is 3a 0 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 340ra =430a r ==0.14263nm=1.4263810-⨯cm (b) In FCC structures, atoms touch along the face diagonal of the cube, which is02a in length. There are four atomic radii along this length —two radii from the face-centered atom and one radius from each corner, so240r a =, 420a r ==1.44447 Å=1.44447810-⨯cm2.(P80 3-4) determine the crystal structure for the following:(a) a metal with a0=4.9489Å, r=1.75Å, 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 430a r =in BCC and 420a r =in FCC. (a) 420a r ==⨯=49489.42 1.75 so its crystal structure is FCC;(b) 430a r ==44296.03⨯=0.186nm so its crystal structure is BCC.3. (P80 3-5) the density of potassium, which has the BCC structure and oneatom per lattice point, is 0.855g/cm 3.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 twoatoms in per unit cell,atomic mass is 39.09g/mol,density ρ=0.855g/cm 3Avogadro ’s number N A =6.022310⨯atoms/mol())'()()(/snumber Avogadro cell unit of volume mass atomic cell atoms of numbers =ρ0.855g/cm 3=)/()1002.6()/(09.392233mol atoms a mol g atoms ⨯⨯⨯ So a=3231002.6855.009.392⨯⨯⨯=0.53cm710-⨯=5.3Å(b)then r=a 43=0.229710-⨯cm=2.29Å4. (P81 3-20) determine the indices forthe directions in the cubic unit cell shown in Figure 3-32.The procedure for finding the Miller indices for directions is as follows:ing a right-handed coordinatesystem, 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 oflattice 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.[]011Direction B 1.Two points are 1,0,1 and 21,1,0 2.1,0,1-21,1,0=21,-1,1 3.2(21,-1,1)=1,-2,24.[]221Direction C 1.Two points are 1,0,0 and 0,43,1 2.1,0,0-0,43,1=1, -43,-1 3.4(1, -43,-1)=4, -3, -4 4.[]434Direction D 1.Two points are 0,1,21 and 1,0,0 2.0,1,21 -1,0,0=-1,1, 21 3.2(-1,1, 21)=-2,2,1 4.[]2125.(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 throughthe 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 A 1. x=-1, y=21, z=43 2. 341,21,11==-=z y x3. Clear fractions: -3, 6, 44. (643)Plane B 1. x=1, y=-43, z=∞ 2. 01,341,11=-==z y x3. Clear fractions: 3, -4, 04. ()043Plane C 1. x=2, y=23, z=1 2. 11,321,211===z y x3. Clear fractions: 3, 4, 64. (346)6.(P82 3-23) Sketch the following planes and directions within a cubic unit cell:(a) [101] (b) [010] (c) [122] (d)[301] (e) [201] (f) [213](g) (011) (h) (102) (i) (002) (j) (130) (k) (212) (l) (321)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] as31111001101011''''''cos 222222222222=+++++⨯+⨯+⨯=+++++++=w v u w v u ww vv uu θ73.54=θ8. Use a calculation to verify that theatomic 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. Thevolume of one atom is 4πr 3/3 and thevolume of the unit cell is a 3:Packing factor = 4×4πr 3/3 a 3Since for FCC unit cell, a=4r/2, packingfactor =o.749. 写出溶解在γ-Fe 中碳原子所处的位置，若此类位置全部被碳原子占据，那么试问在这种情况下，γ-Fe 能溶解多少重量百分数的碳？而实际上在γ-Fe 中最大溶解度是多少？两者在数值上有差异的原因是什么？解答：γ-Fe 是fcc ，八面体间隙尺寸大，故C 的存在位置为八面体间隙。