Solution Manual for Materials Science and Engineering: An Introduction, 10th Edition

CHAPTER 3 THE STRUCTURE OF CRYSTALLINE SOLIDS PROBLEM SOLUTIONS Fundamental Concepts 3.1 What is the difference between atomic structure and crystal structure? Solution Atomic structure relates to the number of protons and neutrons in the nucleus of an atom, as well as the number and probability distributions of the constituent electrons. On the other hand, crystal structure pertains to the arrangement of atoms in the crystalline solid material. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Unit Cells Metallic Crystal Structures 3.2 If the atomic radius of aluminum is 0.143 nm, calculate the volume of its unit cell in cubic meters. Solution For this problem, we are asked to calculate the volume of a unit cell of aluminum. Aluminum has an FCC crystal structure (Table 3.1). The FCC unit cell volume may be computed from Equation 3.6 as Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.3 Show for the body-centered cubic crystal structure that the unit cell edge length a and the atomic radius R are related through a =4R/ . Solution Consider the BCC unit cell shown below Using the triangle NOP And then for triangle NPQ, But NQ = 4R, R being the atomic radius. Also, QP = a. Therefore, or Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.4 For the HCP crystal structure, show that the ideal c/a ratio is 1.633. Solution A sketch of one-third of an HCP unit cell is shown below. Consider the tetrahedron labeled as JKLM, which is reconstructed as The atom at point M is midway between the top and bottom faces of the unit cell–that is = c/2. And, since atoms at points J, K, and M, all touch one another, where R is the atomic radius. Furthermore, from triangle JHM, Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. or Now, we can determine the length by consideration of triangle JKL, which is an equilateral triangle, and Substituting this value for in the above expression yields and, solving for c/a Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.5 Show that the atomic packing factor for BCC is 0.68. Solution The atomic packing factor is defined as the ratio of sphere volume to the total unit cell volume, or Since there are two spheres associated with each unit cell for BCC Also, the unit cell has cubic symmetry, that is VC = a3. But a depends on R according to Equation 3.4, and Thus, Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.6 Show that the atomic packing factor for HCP is 0.74. Solution The APF is just the total sphere volume-unit cell volume ratio. For HCP, there are the equivalent of six spheres per unit cell, and thus Now, the unit cell volume is just the product of the base area times the cell height, c. This base area is just three times the area of the parallelepiped ACDE shown below. The area of ACDE is just the length of times the height . But is just a or 2R, and Thus, the base area is just and since c = 1.633a = 2R(1.633) (3.S1) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Thus, Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Density Computations 3.7 Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomic weight of 55.85 g/mol. Compute and compare its theoretical density with the experimental value found inside the front cover of the book. Solution This problem calls for a computation of the density of iron. According to Equation 3.8 For BCC, n = 2 atoms/unit cell, and Thus, = 7.90 g/cm3 The value given inside the front cover is 7.87 g/cm3. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.8 Niobium (Nb) has a BCC crystal structure, an atomic radius of 0.143 nm and an atomic weight of 92.91 g/mol. Calculate the theoretical density for Nb. Solution According to Equation 3.8 For BCC, n = 2 atoms/unit cell. Furthermore, because VC = a3, and (Equation 3.4), then Thus, realizing that ANb = 92.91 g/mol, using the above version of Equation 3.8, we compute the theoretical density of Nb as follows: = 8.48 g/cm3 The experimental density for Nb is 8.57 g/cm 3. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.9 Calculate the radius of an iridium atom, given that Ir has an FCC crystal structure, a density of 22.4 g/cm3, and an atomic weight of 192.2 g/mol. Solution We are asked to determine the radius of an iridium atom, given that Ir has an FCC crystal structure. For FCC, n = 4 atoms/unit cell, and VC = (Equation 3.6). Now, And solving for R from the above expression yields = 1.36  10-8 cm = 0.136 nm Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.10 Calculate the radius of a vanadium atom, given that V has a BCC crystal structure, a density of 5.96 g/cm3, and an atomic weight of 50.9 g/mol. Solution This problem asks for us to calculate the radius of a vanadium atom. For BCC, n = 2 atoms/unit cell, and Since, from Equation 3.8 and solving for R the previous equation and incorporating values of parameters given in the problem statement = 1.32  10-8 cm = 0.132 nm Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.11 Some hypothetical metal has the simple cubic crystal structure shown in Figure 3.3. If its atomic weight is 70.4 g/mol and the atomic radius is 0.126 nm, compute its density. Solution For the simple cubic crystal structure, the value of n in Equation 3.8 is unity since there is only a single atom associated with each unit cell. Furthermore, for the unit cell edge length, a = 2R (Figure 3.3). Therefore, employment of Equation 3.8 yields and incorporating values of the other parameters provided in the problem statement leads to 7.31 g/cm3 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.12 Zirconium has an HCP crystal structure and a density of 6.51 g/cm3. (a) What is the volume of its unit cell in cubic meters? (b) If the c/a ratio is 1.593, compute the values of c and a. Solution (a) The volume of the Zr unit cell may be computed using Equation 3.8 as Now, for HCP, n = 6 atoms/unit cell, and for Zr, AZr = 91.22 g/mol. Thus, = 1.396  10-22 cm3/unit cell = 1.396  10-28 m3/unit cell (b) From Equation 3.S1 of the solution to Problem 3.6, for HCP But, since a = 2R, (i.e., R = a/2) then but, since c = 1.593a Now, solving for a Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. = 3.23  10-8 cm = 0.323 nm And finally c = 1.593a = (1.593)(0.323 nm) = 0.515 nm Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.13 Using atomic weight, crystal structure, and atomic radius data tabulated inside the front cover, compute the theoretical densities of lead, chromium, copper, and cobalt, and then compare these values with the measured densities listed in this same table. The c/a ratio for cobalt is 1.623. Solution Since Pb has an FCC crystal structure, n = 4, and V = C (Equation 3.6). Also, R = 0.175 nm (1.75  10 cm) and APb = 207.2 g/mol. Employment of Equation 3.8 yields -8 = 11.35 g/cm3 The value given in the table inside the front cover is 11.35 g/cm 3. Chromium has a BCC crystal structure for which n = 2 and VC = a3 = (Equation 3.4); also, ACr = 52.00 g/mol and R = 0.125 nm. Therefore, employment of Equation 3.8 leads to = 7.18 g/cm3 The value given in the table is 7.19 g/cm3. Copper also has an FCC crystal structure and therefore Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. = 8.90 g/cm3 The value given in the table is 8.90 g/cm3. Cobalt has an HCP crystal structure, and from the solution to Problem 3.6 (Equation 3.S1), and, since c = 1.623a and a = 2R, c = (1.623)(2R); hence Also, there are 6 atoms/unit cell for HCP. Therefore, the theoretical density is = 8.91 g/cm3 The value given in the table is 8.9 g/cm3. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.14 Rhodium has an atomic radius of 0.1345 nm and a density of 12.41 g/cm3. Determine whether it has an FCC or BCC crystal structure. Solution In order to determine whether Rh has an FCC or a BCC crystal structure, we need to compute its density for each of the crystal structures. For FCC, n = 4, and a = (Equation 3.1). Also, from Figure 2.8, its atomic weight is 102.91 g/mol. Thus, for FCC (employing Equation 3.8) = 12.41 g/cm3 which is the value provided in the problem statement. Therefore, Rh has the FCC crystal structure. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.15 The atomic weight, density, and atomic radius for three hypothetical alloys are listed in the following table. For each determine whether its crystal structure is FCC, BCC, or simple cubic and then justify your determination. Alloy Atomic Weight (g/mol) Density (g/cm3) Atomic Radius (nm) A 77.4 8.22 0.125 B 107.6 13.42 0.133 C 127.3 9.23 0.142 Solution For each of these three alloys we need, by trial and error, to calculate the density using Equation 3.5, and compare it to the value cited in the problem. For SC, BCC, and FCC crystal structures, the respective values of n are 1, 2, and 4, whereas the expressions for a (since VC = a3) are 2R, , and . For alloy A, let us calculate  assuming a simple cubic crystal structure. = 8.22 g/cm3 Therefore, its crystal structure is simple cubic. For alloy B, let us calculate  assuming an FCC crystal structure. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. = 13.42 g/cm3 Therefore, its crystal structure is FCC. For alloy C, let us calculate  assuming a simple cubic crystal structure. = 9.23 g/cm3 Therefore, its crystal structure is simple cubic. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.16 The unit cell for tin (Sn) has tetragonal symmetry, with a and b lattice parameters of 0.583 and 0.318 nm, respectively. If its density, atomic weight, and atomic radius are 7.30 g/cm 3, 118.69 g/mol, and 0.151 nm, respectively, compute the atomic packing factor. Solution In order to determine the APF for Sn, we need to compute both the unit cell volume (VC) which is just the a2c product, as well as the total sphere volume (VS) which is just the product of the volume of a single sphere and the number of spheres in the unit cell (n). The value of n may be calculated from Equation 3.8 as = 4.00 atoms/unit cell Therefore = 0.534 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.17 Iodine has an orthorhombic unit cell for which the a, b, and c lattice parameters are 0.479, 0.725, and 0.978 nm, respectively. (a) If the atomic packing factor and atomic radius are 0.547 and 0.177 nm, respectively, determine the number of atoms in each unit cell. (b) The atomic weight of iodine is 126.91 g/mol; compute its theoretical density. Solution (a) For iodine, and from the definition of the APF we may solve for the number of atoms per unit cell, n, as Incorporating values of the above parameters provided in the problem state leads to = 8.0 atoms/unit cell (b) In order to compute the density, we just employ Equation 3.8 as = 4.96 g/cm3 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3. 18 Titanium has an HCP unit cell for which the ratio of the lattice parameters c/a is 1.58. If the radius of the Ti atom is 0.1445 nm, (a) determine the unit cell volume, and (b) calculate the density of Ti and compare it with the literature value. Solution (a) We are asked to calculate the unit cell volume for Ti. For HCP, from Equation 3.S1 (found in the solution to Problem 3.6) But for Ti, c = 1.58a, and a = 2R, or c = 3.16R, and (b) The theoretical density of Ti is determined, using Equation 3.8, as follows: For HCP, n = 6 atoms/unit cell, and for Ti, ATi = 47.88 g/mol (as noted inside the front cover). Thus, = 4.81 g/cm3 The value given in the literature is 4.51 g/cm3. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.19 Zinc has an HCP crystal structure, a c/a ratio of 1.856, and a density of 7.13 g/cm3. Compute the atomic radius for Zn. Solution In order to calculate the atomic radius for Zn, we must use Equation 3.8, as well as the expression which relates the atomic radius to the unit cell volume for HCP; Equation 3.S1 (from Problem 3.6) is as follows: In this case c = 1.856a, but, for HCP, a = 2R, which means that And from Equation 3.8, the density is equal to And, solving for R from the above equation leads to the following: And incorporating appropriate values for the parameters in this equation leads to = 1.33  10-8 cm = 0.133 nm Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.20 Rhenium has an HCP crystal structure, an atomic radius of 0.137 nm, and a c/a ratio of 1.615. Compute the volume of the unit cell for Re. Solution In order to compute the volume of the unit cell for Re, it is necessary to use Equation 3.S1 (found in Problem 3.6), that is The problem states that c = 1.615a, and a = 2R. Therefore Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Polymorphism and Allotropy 3.21 Iron (Fe) undergoes an allotropic transformation at 912C: upon heating from a BCC ( phase) to an FCC ( phase). Accompanying this transformation is a change in the atomic radius of Fe—from rBCC = 0.12584 nm to rFCC = 0.12894 nm—and, in addition a change in density (and volume). Compute the percent volume change associated with this reaction. Does the volume increase or decrease? Solution To solve this problem let us first compute the density of each phase using Equation 3.8, and then determine the volumes per unit mass (the reciprocals of densities). From these values it is possible to calculate the percent volume change. The density of each phase may be computed using Equation 3.8—i.e., The atomic weight of Fe will be the same for both BCC and FCC structures (55.85 g/mol); however, values of n and VC will be different. For BCC iron, n = 2 atoms/unit cell, whereas the volume of the cubic unit cell is the cell edge length a cubed–VC = a3. However, a and the atomic radius (rBCC) are related according to Equation 3.4—that is which means that The value of rBCC is given in the problem statement as 0.12584 nm = 1.2584  10-8 cm. It is now possible to calculate the density of BCC iron as follows: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. For FCC iron, n = 4 atoms/unit cell, (Equation 3.6), and, as noted in the problem statement, rFCC = 0.12894 nm = 1.2894  10-8 cm. We now calculate the density of FCC iron as follows: Because we are interested in volumes of these two phases, for each we take the reciprocal of it density, which is equivalent to volume per unit mass (or specific volume), v. The percent volume change experienced by iron during this phase transformation, upon heating is equal to The volume decreases because vFCC < vBCC –i.e., since . Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Crystal Systems 3.22 The accompanying figure shows a unit cell for a hypothetical metal. (a) To which crystal system does this unit cell belong? (b) What would this crystal structure be called? (c) Calculate the density of the material, given that its atomic weight is 141 g/mol. Solution (a) The unit cell shown in the problem statement belongs to the tetragonal crystal system since a = b = 0.30 nm, c = 0.40 nm, and  =  =  = 90. (b) The crystal structure would be called body-centered tetragonal. (c) As with BCC, n = 2 atoms/unit cell. Also, for this unit cell Thus, using Equation 3.8, the density is equal to Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. = 13.0 g/cm3 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.23 Sketch a unit cell for the body-centered orthorhombic crystal structure. Solution A unit cell for the body-centered orthorhombic crystal structure is presented below. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Point Coordinates 3.24 List the point indices for all atoms that are associated with the BCC unit cell (Figure 3.2). Solution This problem asks that we list the point indices for all of the atoms associated with the BCC unit cell. The reduced-sphere BCC unit cell, Figure 3.2b, is shown below on which is superimposed an x-y-z coordinate axis system. Also, each atom in the unit cell is labeled with a number. Of course, because the unit cell is cubic, the unit cell edge length along each of the x, y, and z axes is a. Indices for each of these points is determined in a manner similar to that demonstrated in Example Problem 3.6. For the atom labeled 1 in the BCC unit cell, its lattice position coordinates referenced to the x, y, and z axes are 0a, 0b, and 0c, respectively. Therefore, its lattice position coordinates from Equations 3.9a through 3.9c are as follows: And, solving these three expressions for q, r, and s, leads to the following: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Therefore, the q r s indices for point 1 are 0 0 0. For point 3, the lattice position coordinates referenced to the x, y, and z axes are a, b, and 0c, respectively. Therefore, its lattice position coordinates from Equations 3.9a through 3.9c are as follows: And, solving these three expressions for q, r, and s, leads to the following: Therefore, the q r s indices for point 3 are 1 1 0. This same procedure is carried out for the remaining the points in the unit cell; indices for all nine points are listed in the following table. Point number q r s 1 2 3 4 5 6 7 8 0 1 1 0 0 1 1 0 0 0 1 1 0 0 1 1 0 0 0 0 1 1 1 1 9 1 2 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.25 List the point indices of the titanium, barium, and oxygen ions for a unit cell of the perovskite crystal structure (Figure 12.6). Solution This problem asks that we list the point indices for all of the ions associated with a unit cell for the perovskite crystal structure. The reduced-sphere perovskite unit cell (Figure 12.6) is shown below; the unit cell is cubic with Ba2+ ions positions on each of the eight corners, with an O2- ion on each of the six faces, and a single Ti4+ ion at the unit cell center. Also, an x-y-z coordinate axis system is superimposed on the unit cell, and each ion is labeled with a number—Ba2+ ions, 1 through 8; O2- ions, 9 through 14; and the Ti4+ ion, 15. Of course, because the unit cell is cubic, the unit cell edge length along each of the x, y, and z axes is a. Indices for each of these points is determined in a manner similar to that demonstrated in Example Problem 3.6. For example, the Ba2+ ion labeled 8 in this unit cell, its lattice position coordinates referenced to the x, y, and z axes are 0a, b, and c, respectively. Therefore, its lattice position coordinates from Equations 3.9a through 3.9c are as follows: And, solving these three expressions for q, r, and s, leads to the following: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Therefore, the q r s indices for point 8 are 0 1 1. Next let us consider the O2- ion labeled 11 in the above figure. Its lattice position coordinates referenced to the x, y, and z axes are a, b/2, and c/2, respectively. Therefore, its lattice position coordinates from Equations 3.9a through 3.9c are as follows: And, solving these three expressions for q, r, and s, leads to the following: Therefore, the q r s indices for point 11 are 1 . And finally, consider the Ti4+ ion labeled 15 in the above figure. Its lattice position coordinates referenced to the x, y, and z axes are a/2, b/2, and c/2, respectively. Therefore, its lattice position coordinates from Equations 3.9a through 3.9c are as follows: And, solving these three expressions for q, r, and s, leads to the following: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Therefore, the q r s indices for point 15 are . This same procedure is carried out for the remaining the points in the unit cell; indices for all fifteen points are listed in the following table. Point number q r s 1 2 3 4 5 6 7 8 0 1 1 0 0 1 1 0 0 0 1 1 0 0 1 1 0 0 0 0 1 1 1 1 9 10 11 12 13 14 15 1 0 2 1 2 0 1 1 2 1 0 1 2 1 1 2 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.26 List the point indices of all atoms that are associated with the diamond cubic unit cell (Figure 12.16). Solution This problem asks that we list the point indices for all of the atoms associated with the diamond cubic unit cell. The reduced-sphere diamond cubic unit cell (Figure 12.16) is shown below; the unit cell is cubic with C atoms positioned on each of the eight corners (labeled 1 through 8), on each of the six faces (labeled 9 through 14), and four atoms within the interior of the unit cell (labeled 15 through 18). Also, an x-y-z coordinate axis system is superimposed on the unit cell. Of course, because the unit cell is cubic, the unit cell edge length along each of the x, y, and z axes is a. Indices for each of these points is determined in a manner similar to that demonstrated in Example Problem 3.6. For example, the C atom labeled 8 in this unit cell, its lattice position coordinates referenced to the x, y, and z axes are 0a, b, and c, respectively. Therefore, its lattice position coordinates from Equations 3.9a through 3.9c are as follows: And, solving these three expressions for q, r, and s, leads to the following: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Therefore, the q r s indices for point 8 are 0 1 1. Next let us consider the C atom labeled 12 in the above figure. Its lattice position coordinates referenced to the x, y, and z axes are a/2, b, and c/2, respectively. Therefore, its lattice position coordinates from Equations 3.9a through 3.9c are as follows: And, solving these three expressions for q, r, and s, leads to the following: Therefore, the q r s indices for point 12 are 1 . And finally, consider the C labeled 16 in the above figure. Its lattice position coordinates referenced to the x, y, and z axes are a/4, 3b/4, and c/4, respectively. Therefore, its lattice position coordinates from Equations 3.9a through 3.9c are as follows: And, solving these three expressions for q, r, and s, leads to the following: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Therefore, the q r s indices for point 15 are . This same procedure is carried out for the remaining the points in the unit cell; indices for all eighteen points are listed in the following table. Point number q r s 1 2 3 4 5 6 7 8 0 1 1 0 0 1 1 0 0 0 1 1 0 0 1 1 0 0 0 0 1 1 1 1 9 10 11 12 13 14 1 0 2 1 2 0 1 1 2 1 0 1 2 1 15 16 17 18 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.27 Sketch a tetragonal unit cell, and within that cell indicate locations of the 1 2 1 1 2 and 1 1 3 4 2 4 point indices. Solution A tetragonal unit in which is shown the and point indices is presented below. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.28 Sketch an orthorhombic unit cell, and within that cell indicate locations of the and 1 1 1 4 3 4 point indices. Solution An orthorhombic unit cell in which is shown the location of the An orthorhombic unit cell in which is shown location of the point indices is presented below. point indices is presented below. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.29 Using the Molecule Definition Utility found in both “Metallic Crystal Structures and Crystallography” and “Ceramic Crystal Structures” modules of VMSE, located on the book’s website [www.wiley.com/college/Callister (Student Companion Site)], generate a three-dimensional unit cell for the intermetallic compound AuCu3 given the following: (1) the unit cell is cubic with an edge length of 0.374 nm, (2) gold atoms are situated at all cube corners, and (3) copper atoms are positioned at the centers of all unit cell faces. Solution First of all, open the “Molecular Definition Utility”; it may be found in either of “Metallic Crystal Structures and Crystallography” or “Ceramic Crystal Structures” modules. In the “Step 1” window, it is necessary to define the atom types, colors for the spheres (atoms), and specify atom sizes. Let us enter “Au” as the name for the gold atoms (since “Au” the symbol for gold). It is next necessary to choose a color for gold; we will select “Yellow”. We now need to specify the atom size for gold. It is suggested that atom/ion diameters in nanometers be used. From inside the front cover of the book, the atomic radius for gold is 0.144 nm–therefore, its atomic diameter is double this value, or 0.288 nm. After this value has been entered, it is necessary to click on the “Register Atom Type” button. In the window below the “Start Over” button, the symbol we have chosen for the gold atom (Au), its atomic diameter (in nm), as well as the selected color (Yellow) are displayed. This same procedure is followed for the copper atoms. Let us choose “Cu” to denote the atom type, “Red” for the atom color, and “0.256” for the atom diameter (since its atomic radius is 0.128 nm). This information also needs to be registered. In the “Step 2” window we specify positions for all of the atoms within the unit cell; their point indices are specified in the problem statement. Let’s begin with gold. First click on “Au[Yellow]” in the “Select atom” window. Again, Au atoms are situated at all eight corners of the cubic unit cell. One Au will be positioned at the origin of the coordinate system—i.e., its point indices are 000, and, therefore, we enter a “0” (zero) in each of the “X”, “Y”, and “Z” atom position boxes. Next we click on the “Register Atom Position” button, which displays a yellow sphere at the 000 location in the X-Y-Z coordinate system. Now we enter the coordinates of another gold atom; let us arbitrarily select the one that resides at the corner of the unit cell that is one unit-cell length along the Xaxis (i.e., at the 100 point indices). Inasmuch as it is located a distance of a units along the X-axis the value of “0.374” is entered in the “X” atom position box (since this is the value of a given in the problem statement); zeros are entered in each of the “Y” and “Z” position boxes; this atom position is registered. We repeat this procedure for the remaining six Au atoms. After this step has been completed, it is necessary to specify positions for the copper atoms, which are located at all six face-centered sites. To begin, we select “Cu[Red]” in the “Select atom” box. The point coordinates for some of the Cu atoms are fractional ones; in these instances, the a unit cell length (i.e., 0.374) is Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. multiplied by the fraction. For example, one Cu atom is located positions are (1)(0.374) = 0.374, (0.374) = 0.187, and indices. Therefore, the X, Y, and Z atom (0.374) = 0.187, respectively. For the gold atoms, the X, Y, and Z atom position entries for all 8 sets of point indices are as follows: 0, 0, and 0 0.374, 0, and 0 0, 0.374, and 0 0, 0, and 0.374 0, 0.374, 0.374 0.374, 0, 0.374 0.374, 0.374, 0 0.374, 0.374, 0.374 Now, for the copper atoms, the X, Y, and Z atom position entries for all 6 sets of point coordinates are as follows: 0.187, 0.187, 0 0.187, 0, 0.187 0, 0.187, 0.187 0.374, 0.187, 0.187 0.187, 0.374, 0.187 0.187, 0.187, 0.374 In Step 3, we may specify which atoms are to be represented as being bonded to one another, and which type of bond(s) to use (single, double, triple dashed, and dotted are possibilities), or we may elect to not represent any bonds at all (in which case we are finished). If it is decided to show bonds, it is first necessary to select one of the five bond types, and then one of the bond colors. Next, the two atoms between which the bond is to appear are selected one at a time by double clicking on them (this causes the two atoms to become highlighted. Finally, the bond appears upon clicking on the “Register Bond” button. The image you have created may be rotated by using mouse click-and-drag Your image should appear as the following screen shot. Here the gold atoms appear lighter than the copper atoms. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. [Note: Unfortunately, with this version of the Molecular Definition Utility, it is not possible to save either the data or the image that you have generated. You may use screen capture (or screenshot) software to record and store your image.] Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Crystallographic Directions 3.30 Draw an orthorhombic unit cell, and within that cell a direction. Solution This problem calls for us to draw a direction within an orthorhombic unit cell (a ≠ b ≠ c,  =  =  = 90). This direction may be plotted using the procedure outlined in Example Problem 3.8, with which rearranged forms of Equations 3.10a-3.10c are used. Let us position the tail of the direction vector at the origin of our coordinate axes; this means that tail vector indices for this direction are u=1 v=2 w = −1 Because the tail of the direction vector is positioned at the origin, its coordinates are as follows: x1 = 0a y1 = 0b z1 = 0c Head coordinates are determined using the rearranged forms of Equations 3.10a-3.10c, as follows: Therefore, coordinates for the vector head are a, 2b, and −c. To locate the vector head, we start at the origin, point O, and move along the +x-axis a units (from point O to point A), then parallel to the +y-axis 2b units (from point A to point B). Finally, we proceed parallel to the z-axis -c units (from point B to point C). The direction is the vector from the origin (point O) to point C as shown. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.31 Sketch a monoclinic unit cell, and within that cell a direction. Solution This problem asks that a direction be drawn within a monoclinic unit cell (a ≠ b ≠ c, and  =  = 90º ≠ ). One such unit cell with its origin at point O is sketched below. This direction may be plotted using the procedure outlined in Example Problem 3.8, with which rearranged forms of Equations 3.10a-3.10c are used. Let us position the tail of the direction vector at the origin of our coordinate axes; this means that vector head indices for this direction are u=0 v = −1 w=1 Because the tail of the direction vector is positioned at the origin, its coordinates are as follows: x1 = 0a y1 = 0b z1 = 0c Head coordinates are determined using the rearranged forms of Equations 3.10a-3.10c, as follows: Therefore, coordinates for the vector head are 0a, −b, and c. To locate the vector head, we start at the origin, point O. Since the x index is zero, it is not necessary to move from point O along the x-axis. We now move from point O b units along the minus y-axis (from point O to point R). Since the final coordinate is c, we move from point R parallel to the z-axis, c units (to point P). Thus, the direction corresponds to the vector passing from the origin to point P, as indicated in the figure. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.32 What are the indices for the directions indicated by the two vectors in the sketch below? Solution We are asked for the indices of the two directions sketched in the figure. Unit cell edge lengths are a = 0.4 nm, b = 0.5 nm, and c = 0.3 nm. In solving this problem we will use the symbols a, b, and c rather than the magnitudes of these parameters The tail of the Direction 1 vector passes through the origin, therefore, its tail coordinates are x1 =0a y1 =0b z1 =0c And the vector head coordinates are as follows: x2 = 0a y2 = b/2 z2 = c To determine the directional indices we employ Equations 3.10a, 3.10b, and 3.10c. Because there is a 2 in the denominator for y2 we will assume n = 2. Hence Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Therefore, Direction 1 is a [012] direction. We use the same procedure to determine the indices for Direction 2. Again, because the vector tail passes through the origin of the coordinate system, the values of x1, y1, and z1 are the same as for Direction 1. Furthermore, vector head coordinates are as follows: x2 = a/2 y2 = b/2 z2 = −c We again choose a value of 2 for n because of the 2 in the denominator of the x2 and y2coordinates. Therefore, Therefore, Direction 2 is a direction. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.33 Within a cubic unit cell, sketch the following directions: (a) (e) (b) (f) (c) (g) (d) (h) Solution (a) For the direction, it is the case that u = −1 v=1 w=0 lf we select the origin of the coordinate system as the position of the vector tail, then x1 = 0a y1 = 0b z1 = 0c It is now possible to determine values of x2, y2, and z2 using rearranged forms of Equations 3.10a through 3.10c as follows: Thus, the vector head is located at −a, b, and 0c, and the direction vector having these head coordinates is plotted below. [Note: even though the unit cell is cubic, which means that the unit cell edge lengths are the same (i.e., a), in order to clarify construction of the direction vector, we have chosen to use b and c to designate edge lengths along y and z axes, respectively.] Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. (b) For the u = −1 direction, it is the case that v = −2 w=1 lf we select the origin of the coordinate system as the position of the vector tail, then x1 = 0a y1 = 0b z1 = 0c It is now possible to determine values of x2, y2, and z2 using rearranged forms of Equations 3.10a through 3.10c as follows: Thus, the vector head is located at −a, −2b, and c, and the direction vector having these head coordinates is plotted below. [Note: even though the unit cell is cubic, which means that the unit cell edge lengths are the same (i.e., a), in order to clarify construction of the direction vector, we have chosen to use b and c to designate edge lengths along y and z axes, respectively.] (c) For the direction, it is the case that u=0 v = −1 w=2 lf we select the origin of the coordinate system as the position of the vector tail, then x1 = 0a y1 = 0b z1 = 0c It is now possible to determine values of x2, y2, and z2 using rearranged forms of Equations 3.10a through 3.10c as follows: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Thus, the vector head is located at 0a, −b, and 2c, and the direction vector having these head coordinates is plotted below. [Note: even though the unit cell is cubic, which means that the unit cell edge lengths are the same (i.e., a), in order to clarify construction of the direction vector, we have chosen to use b and c to designate edge lengths along y and z axes, respectively.] (d) For the direction, it is the case that u=1 v = −3 w=3 lf we select the origin of the coordinate system as the position of the vector tail, then x1 = 0a y1 = 0b z1 = 0c It is now possible to determine values of x2, y2, and z2 using rearranged forms of Equations 3.10a through 3.10c as follows: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Thus, the vector head is located at a, −3b, and 3c. However, in order to reduce the vector length, we have multiplied these coordinates by 1/3, which gives the new set of head coordinates as a/3, −b, and c; the direction vector having these head coordinates is plotted below. [Note: even though the unit cell is cubic, which means that the unit cell edge lengths are the same (i.e., a), in order to clarify construction of the direction vector, we have chosen to use b and c to designate edge lengths along y and z axes, respectively.] (e) For the direction, it is the case that u = −1 v = −1 w=1 lf we select the origin of the coordinate system as the position of the vector tail, then x1 = 0a y1 = 0b z1 = 0c It is now possible to determine values of x2, y2, and z2 using rearranged forms of Equations 3.10a through 3.10c as follows: Thus, the vector head is located at −a, −b, and c, and the direction vector having these head coordinates is plotted below. [Note: even though the unit cell is cubic, which means that the unit cell edge lengths are the same (i.e., a), in order to clarify construction of the direction vector, we have chosen to use b and c to designate edge lengths along y and z axes, respectively.] Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. (f) For the direction, it is the case that u = −1 v=2 w=2 lf we select the origin of the coordinate system as the position of the vector tail, then x1 = 0a y1 = 0b z1 = 0c It is now possible to determine values of x2, y2, and z2 using rearranged forms of Equations 3.10a through 3.10c as follows: Thus, the vector head is located at −a, 2b, and 2c. However, in order to reduce the vector length, we have multiplied these coordinates by 1/2, which gives the new set of head coordinates as −a/2, b, and c; the direction vector having these head coordinates is plotted below. [Note: even though the unit cell is cubic, which means that the unit cell edge lengths are the same (i.e., a), in order to clarify construction of the direction vector, we have chosen to use b and c to designate edge lengths along y and z axes, respectively.] Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. (g) For the direction, it is the case that u=1 v = −2 w = −3 lf we select the origin of the coordinate system as the position of the vector tail, then x1 = 0a y1 = 0b z1 = 0c It is now possible to determine values of x2, y2, and z2 using rearranged forms of Equations 3.10a through 3.10c as follows: Thus, the vector head is located at a, −2b, and −3c, and the direction vector having these head coordinates is plotted below. [Note: even though the unit cell is cubic, which means that the unit cell edge lengths are the same (i.e., a), in order to clarify construction of the direction vector, we have chosen to use b and c to designate edge lengths along y and z axes, respectively.] Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. (h) For the direction, it is the case that u = −1 v=0 w=3 lf we select the origin of the coordinate system as the position of the vector tail, then x1 = 0a y1 = 0b z1 = 0c It is now possible to determine values of x2, y2, and z2 using rearranged forms of Equations 3.10a through 3.10c as follows: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Thus, the vector head is located at −a, 0b, and 3c. However, in order to reduce the vector length, we have multiplied these coordinates by 1/3, which gives the new set of head coordinates as −a/3, 0b, and c; the direction vector having these head coordinates is plotted below. [Note: even though the unit cell is cubic, which means that the unit cell edge lengths are the same (i.e., a), in order to clarify construction of the direction vector, we have chosen to use b and c to designate edge lengths along y and z axes, respectively.] Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.34 Determine the indices for the directions shown in the following cubic unit cell: For Direction A We determine the indices of this direction vector using Equations 3.10a-3.10c—that is, by subtracting vector tail coordinates from head coordinates. Tail coordinates are as follows: x1 = 0a y1 = b z1 = c Whereas head coordinates are as follows: x2 = 0a y2 = 0b z2 = 0c From Equations 3.10a, 3.10b, and 3.10c assuming a value of 1 for the parameter n Therefore, Direction A is . For Direction B Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. We determine the indices of this direction vector using Equations 3.10a-3.10c—that is, by subtracting vector tail coordinates from head coordinates. Tail coordinates are as follows: x1 = a y1 = b/2 z1 = c Whereas head coordinates are as follows: x2 = 0a y2 = b z2 = c From Equations 3.10a, 3.10b, and 3.10c assuming a value of 2 for the parameter n Therefore, Direction B is . For Direction C Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. We determine the indices of this direction vector using Equations 3.10a-3.10c—that is, by subtracting vector tail coordinates from head coordinates. Tail coordinates are as follows: x1 = a/2 y1 = b/2 z1 = 0c Whereas head coordinates are as follows: x2 = a y2 = b z2 = c From Equations 3.10a, 3.10b, and 3.10c assuming a value of 2 for the parameter n Therefore, Direction C is . For Direction D Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. We determine the indices of this direction vector using Equations 3.10a-3.10c—that is, by subtracting vector tail coordinates from head coordinates. Tail coordinates are as follows: x1 = a/2 y1 = 0b z1 = c Whereas head coordinates are as follows: x2 = a y2 = b/2 z2 = 0c From Equations 3.10a, 3.10b, and 3.10c assuming a value of 2 for the parameter n Therefore, Direction D is . Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.35 Determine the indices for the directions shown in the following cubic unit cell: For Direction A We determine the indices of this direction vector using Equations 3.10a-3.10c—that is, by subtracting vector tail coordinates from head coordinates. Tail coordinates are as follows: x1 = 2a/3 y1 = 0b z1 = c Whereas head coordinates are as follows: x2 = 0a y2 = b/2 z2 = c From Equations 3.10a, 3.10b, and 3.10c assuming a value of 3 for the parameter n Because the v index is 3/2, it is necessary to multiply each index by the factor 2 to eliminate the fraction. Therefore, Direction A is . For Direction B Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. We determine the indices of this direction vector using Equations 3.10a-3.10c—that is, by subtracting vector tail coordinates from head coordinates. Tail coordinates are as follows: x1 = a/3 y1 = b z1 = 0c Whereas head coordinates are as follows: x2 = a y2 = 0b z2 = 2c/3 From Equations 3.10a, 3.10b, and 3.10c assuming a value of 3 for the parameter n Therefore, Direction B is . For Direction C Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. We determine the indices of this direction vector using Equations 3.10a-3.10c—that is, by subtracting vector tail coordinates from head coordinates. Tail coordinates are as follows: x1 = a/3 y1 = b z1 = c Whereas head coordinates are as follows: x2 = 2a/3 y2 = 0b z2 = 0c From Equations 3.10a, 3.10b, and 3.10c assuming a value of 3 for the parameter n Therefore, Direction C is . For Direction D Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. We determine the indices of this direction vector using Equations 3.10a-3.10c—that is, by subtracting vector tail coordinates from head coordinates. Tail coordinates are as follows: x1 = a/3 y1 = 0b z1 = c Whereas head coordinates are as follows: x2 = a/2 y2 = b/2 z2 = 0c From Equations 3.10a, 3.10b, and 3.10c assuming a value of 3 for the parameter n In order to eliminate the factor 2 in the denominators of u and v it is necessary to multiply each index by 2. Therefore, Direction D is . Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.36 (a) What are the direction indices for a vector that passes from point to point in cubic unit cell? (b) Repeat part (a) for a monoclinic unit cell. Solution (a) Point indices for the vector tail, r=1 , means that s=1 or that, using Equations 3.9a-3.9c, lattice position coordinates to the three axis are determine as follows: Similarly for the vector head: q=0 r=1 And we determine lattice position coordinates for using Equations 3.9a-3.9c, in a similar manner: And, finally, determination of the u, v, and w direction indices is possible using Equations 3.10a-3.10c; because there is a 3 in the denominator and a 2 in another denominator of two of the lattice position coordinates, let us assume a value of 6 for the parameter n in these equations. Therefore, Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Therefore, this vector points in a direction. (b) For a monoclinic unit cell, the direction indices will also be . Lattice position coordinates for both vector head and tail will be the same as for cubic. Likewise, incorporating these lattice position coordinates into Equations 3.10a-3.10c also yield u = −4 v=0 w = −3 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.37 (a) What are the direction indices for a vector that passes from point to point in a tetragonal unit cell? (b) Repeat part (a) for a rhombohedral unit cell. Solution (a) Point indices for the vector tail, , means that r=0 or that, using Equations 3.9a-3.9c, lattice position coordinates to the three axes are determine as follows: Similarly for the vector head: r=1 And we determine lattice position coordinates for using Equations 3.9a-3.9c, in a similar manner: And, finally, determination of the u, v, and w direction indices is possible using Equations 3.10a-3.10c; because there is a 3 in the denominator and a 2 in another denominator of two of the lattice position coordinates, let us assume a value of 6 for the parameter n in these equations. Therefore, Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Therefore, this vector points in a direction. (b) For a rhombohedral unit cell, the direction indices will also be . Lattice position coordinates for both vector head and tail will be the same as for cubic. Likewise, incorporating these lattice position coordinates into Equations 3.10a-3.10c also yield u = −3 v=6 w=1 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.38 For tetragonal crystals, cite the indices of directions that are equivalent to each of the following directions: (a) [001] (b) [110] (c) [010] Solution For tetragonal crystals, a = b ≠ c and  =  =  = 90; therefore, projections along the x and y axes are equivalent, which are not equivalent to projections along the z axis. (a) Therefore, for the [001] direction, there is only one equivalent direction: (b) For the [110] direction, equivalent directions are as follows: , (b) Also, for the [010] direction, equivalent directions are the following: . , and . , , and . Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.39 Convert the [100] and [111] directions into the four-index Miller–Bravais scheme for hexagonal unit cells. Solution For [100] U=1 V=0 W=0 From Equations 3.11a-3.11d w=W=0 It is necessary to multiply these numbers by 3 in order to reduce them to the lowest set of integers. Thus, the direction is represented as [uvtw] = . For [111], U = 1, V = 1, and W = 1; therefore, t= w=1 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. If we again multiply these numbers by 3, then u = 1, v = 1, t = -2, and w = 3. Thus, the direction is represented as Thus, the direction is represented as [uvtw] = . Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.40 Determine indices for the directions shown in the following hexagonal unit cells: Solutions (a) One way solve this problem is to begin by determining the U, V, and W indices for the vector referenced to the three-axis scheme; this is possible using Equations 3.12a through 3.12c. Because the tail of the vector passes through the origin, a1¢ = a2¢ = 0a , and z ¢ = 0c . Furthermore, coordinates for the vector head are as follows: Now, solving for U, V, and W using Equations 3.12a, 3.12b, and 3.12c assuming a value of 2 for the parameter n Now, it becomes necessary to convert these indices into an index set referenced to the four-axis scheme. This requires the use of Equations 3.11a through 3.11d, as follows: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. w=W=1 No reduction is necessary inasmuch as all of these indices are integers; therefore, this direction in the four-index scheme is (b) We solve this problem by first determining the U, V, and W indices for the vector referenced to the threeaxis scheme; this is possible using Equations 3.12a through 3.12c. Because the tail of the vector passes through the origin, a1¢ = a2¢ = 0a , and z ¢ = 0c . Furthermore, coordinates for the vector head are as follows: Now, solving for U, V, and W using Equations 3.12a, 3.12b, and 3.12c assuming a value of 2 for the parameter n Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Now, it becomes necessary to convert these indices into an index set referenced to the four-axis scheme. This requires the use of Equations 3.11a through 3.11d, as follows: w=W=0 No reduction is necessary inasmuch as all of these indices are integers; therefore, this direction in the four-index scheme is (c) We solve this problem by first determining the U, V, and W indices for the vector referenced to the threeaxis scheme; this is possible using Equations 3.12a through 3.12c. Because the tail of the vector passes through the origin, , and . Furthermore, coordinates for the vector head are as follows: Now, solving for U, V, and W using Equations 3.12a, 3.12b, and 3.12c assuming a value of 2 for the parameter n Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Now, it becomes necessary to convert these indices into an index set referenced to the four-axis scheme. This requires the use of Equations 3.11a through 3.11d, as follows: w=W=1 Now, in order to get the lowest set of integers, it is necessary to multiply all indices by the factor 3, with the result that this direction is a direction. (d) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. We solve this problem by first determining the U, V, and W indices for the vector referenced to the threeaxis scheme; this is possible using Equations 3.12a through 3.12c. Because the tail of the vector passes through the origin, , and . Furthermore, coordinates for the vector head are as follows: Now, solving for U, V, and W using Equations 3.12a, 3.12b, and 3.12c assuming a value of 1 for the parameter n Now, it becomes necessary to convert these indices into an index set referenced to the four-axis scheme. This requires the use of Equations 3.11a through 3.11d, as follows: w=W=0 Now, in order to get the lowest set of integers, it is necessary to multiply all indices by the factor 3, with the result that this is a direction. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.41 Using Equations 3.11a, 3.11b, 3.11c, and 3.11d, derive expressions for each of the three U, V, and W indices in terms of the four u, v, t, and w indices. Solution It is first necessary to do an expansion of Equation 3.11a as And solving this expression for V yields (S3.41a) Now, substitution of this expression into Equation 3.11b gives Or (S3.41b) And, solving for v from Equation 3.11c leads to which, when substituted into the above expression for U (Equation S3.41b), yields (S3.41c) In solving for an expression for V, we begin with the above Equation S3.41a—i.e., Now, substitution of the above expression for U (Equation S3.41c) into Equation S3.41a leads to (S3.41d) And solving for u from Equation 3.11c gives Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. which, when substituted in Equation S3.41d results in the following expression for V And, finally, from Equation 3.11d W=w Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Crystallographic Planes 3.42 (a) Draw an orthorhombic unit cell, and within that cell a (210) plane. (b) Draw a monoclinic unit cell, and within that cell a (002) plane. Solution (a) We are asked to draw a (210) plane within an orthorhombic unit cell. For the orthorhombic crystal system, relationships among the lattice parameters are as follows: a≠b≠c  =  =  = 90 Thus, the three coordinate axes are parallel to one another. In order to construct the (210) plane it is necessary to determine intersections with the coordinate axes. The A, B, and C intercepts are computed using rearranged forms of Equations 3.13a, 3.13b, and 3.13c with h = 2, k = 1, and l = 0. Thus, intercept values are as follows (assuming n = 1): Thus, intercepts with the x, y, and z axes are respectively, a/2, b, and ∞c; this plane parallels the z axis inasmuch as its intercept is infinity. The plane that satisfies these requirements has been drawn within the orthorhombic unit cell below. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. (b) In this part of the problem we are asked to draw a (002) plane within a monoclinic unit cell. For the monoclinic crystal system, relationships among the lattice parameters are as follows: a≠b≠c  =  = 90   In order to construct the (002) plane it is necessary to determine intersections with the coordinate axes. The A, B, and C intercepts are computed using rearranged forms of Equations 3.13a, 3.13b, and 3.13c with h = 0, k = 0, and l = 2. Thus, intercept values are as follows (assuming n = 1): Thus, intercepts with the x, y, and z axes are respectively, ∞a, ∞b, and c/2; this plane parallels both the x and y axes inasmuch their intercepts are infinity. The plane that satisfies these requirements has been drawn within the monoclinic unit cell below. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.43 What are the indices for the two planes drawn in the sketch below? Solution In order to solve for the h, k, and l indices for these two crystallographic planes it is necessary to use Equations 3.13a, 3.13b, and 3.13c. Plane 1, is parallel to both the the x and z axes and intercepts the y axis b/2—that is A = a C = c B = b/2 Thus, from Equations 3.13a through 3.13c (assuming n = 1) we have the following: Therefore, Plane 1 is a (020) plane. Plane 2 the intercepts the x, y, and z axes at a/2 (0.2 nm), −b/2 (−0.2 nm), and c (0.2 nm)—that is A = a/2 B = −b/2 C=c And when we incorporate these values into Equations 3.13a, 3.13b, and 3.13c, computations of the h, k, and l indices are as follows (assuming that n = 1): Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Therefore, Plane 2 is a plane. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.44 Sketch within a cubic unit cell the following planes: (a) (e) (b) (f) (c) (g) (d) (h) Solutions In order to plot each of these planes it is necessary to determine the axial intercepts using rearranged forms of Equations 3.13a, 3.13b, and 3.13c. The indices for plane (a) are as follows: h=0 k = −1 l = −1 Thus, we solve for the A, B, and C intercepts using rearranged Equations 3.13 as follows (assuming that n = 1): Thus, this plane intersects the y and z axes at −b and −c, respectively, and parallels the x axis. This plane has been drawn in the following sketch. (b) The indices for the h=1 plane are as follows: k=1 l = −2 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Thus, we solve for the A, B, and C intercepts using rearranged Equations 3.13 as follows (assuming that n = 1): Thus, this plane intersects the x, y, and z axes at a, b, and −c/2, respectively. This plane has been drawn in the following sketch. (c) The indices for the h=1 plane are as follows: k=0 l = −2 Thus, we solve for the A, B, and C intercepts using rearranged Equations 3.13 as follows (assuming that n = 1): Thus, this plane intersects the x and z axes at a and −c/2, respectively, and parallels the y axis. This plane has been drawn in the following sketch. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. (d) The indices for the h=1 plane are as follows: k = −3 l=1 Thus, we solve for the A, B, and C intercepts using rearranged Equations 3.13 as follows (assuming that n = 1): Thus, this plane intersects the x, y, and z axes at a, −b/3, and c, respectively. This plane has been drawn in the following sketch. (e) The indices for the h = −1 plane are as follows: k=1 l = −1 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Thus, we solve for the A, B, and C intercepts using rearranged Equations 3.13 as follows (assuming that n = 1): Thus, this plane intersects the x, y, and z axes at −a, b, and −c, respectively. This plane has been drawn in the following sketch. (f) The indices for the h=1 plane are as follows: k = −2 l = −2 Thus, we solve for the A, B, and C intercepts using rearranged Equations 3.13 as follows (assuming that n = 1): Thus, this plane intersects the x, y, and z axes at a, −b/2, and −c/2, respectively. This plane has been drawn in the following sketch. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. (g) The indices for the h = −1 plane are as follows: k=2 l = −3 Thus, we solve for the A, B, and C intercepts using rearranged Equations 3.13 as follows (assuming that n = 1): Thus, this plane intersects the x, y, and z axes at −a, b/2, and −c/3, respectively. This plane has been drawn in the following sketch. (h) The indices for the h=0 plane are as follows: k = −1 l = −3 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Thus, we solve for the A, B, and C intercepts using rearranged Equations 3.13 as follows (assuming that n = 1): Thus, this plane intersects the y and z axes at −b and −c/3, respectively, and parallels the x axis. This plane has been drawn in the following sketch. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.45 Determine the Miller indices for the planes shown in the following unit cell: Solution For plane A, its intersections with with the x, y, and z axes are a/2, b, and 2c/3 (because this plane parallels the y axis)—these three values are equal to A, B, and C, respectively. If we assume that the value of n is 1, the values of h, k, and l are determined using Equations 3.13a, 3.13b, and 3.13c as follows: In order to make these indices integers, we need to multiply each index by the factor 2. Therefore, plane A is a plane. For plane B we will move the origin of the unit cell one unit cell distance to the right along the y axis, and one unit cell distance parallel to the x axis. Using this new origin, intersections with with the x, y, and z axes are, respectively, as follows: −a, −b, and c/2; these are the values of A, B, and C, respectively. If we assume that the value of n is 1, the values of h, k, and l are determined using Equations 3.13a, 3.13b, and 3.13c as follows: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Therefore, plane B is a plane. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.46 Determine the Miller indices for the planes shown in the following unit cell: Solution For plane A we will move the origin of the unit cell one unit cell distance upward along the z axis. Using this new origin, intersections with with the x, y, and z axes are, respectively, as follows: a/3, b/2, and −c/2; these are the values of A, B, and C, respectively. If we assume that the value of n is 1, the values of h, k, and l are determined using Equations 3.13a, 3.13b, and 3.13c as follows: Therefore, plane A is a plane. For plane B we will move the original of the coordinate system on unit cell distance along the x axis. Using this new origin, intersections with with the x, y, and z axes are, respectively, as follows: −a/2, b (since the plane parallels the y axis), and c/2; these are the values of A, B, and C, respectively. If we assume that the value of n is 1, the values of h, k, and l are determined using Equations 3.13a, 3.13b, and 3.13c as follows: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. We can reduce these indices to the lowest set of integers by dividing each by the factor 2. Therefore, plane B is a plane. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.47 Determine the Miller indices for the planes shown in the following unit cell: Solution For plane A, since the plane passes through the origin of the coordinate system as shown, we will move the origin of the coordinate system one unit cell distance to the right along the y axis. Using this new origin, intersections with with the x, y, and z axes are, respectively, as follows: 2a/3, −b, and c/2; these are the values of A, B, and C, respectively. If we assume that the value of n is 1, the values of h, k, and l are determined using Equations 3.13a, 3.13b, and 3.13c as follows: We can reduce these indices to the lowest set of integers by dividing each by the factor 2. Therefore, plane A is a plane. For plane B, its intersections with with the x, y, and z axes are a/2, b/2, and c, which are the values of A, B, and C, respectively. If we assume that the value of n is 1, the values of h, k, and l are determined using Equations 3.13a, 3.13b, and 3.13c as follows: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Therefore, plane B is a (221) plane. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.48 Cite the indices of the direction that results from the intersection of each of the following pair of planes within a cubic crystal: (a) (100) and (010) planes, (b) (111) and planes, and (c) and (001) planes. Solution (a) In the figure below is shown (100) and (010) planes, and, as indicated, their intersection results in a [001], or equivalently, a direction. (b) In the figure below is shown (111) and , or equivalently, a planes, and, as indicated, their intersection results in a direction. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. (c) In the figure below is shown [010], or equivalently, a and (001) planes, and, as indicated, their intersection results in a direction. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.49 Sketch the atomic packing of (a) the (100) plane for the BCC crystal structure, and (b) the (201) plane for the FCC crystal structure (similar to Figures 3.11b and 3.12b). Solution (a) A BCC unit cell, its (100) plane, and the atomic packing of this plane are indicated below. Corresponding atom positions in the two drawings are indicated by letters W, X, Y, and Z. (b) An FCC unit cell, its (201) plane, and the atomic packing of this plane are indicated below. Corresponding atom positions in the two drawing are indicated by the letters A, B, and C. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.50 Consider the reduced-sphere unit cell shown in Problem 3.22, having an origin of the coordinate system positioned at the atom labeled with an O. For the following sets of planes, determine which are equivalent: (a) , (010), and, (b) , , , and (c) , , , and Solution (a) The unit cell in Problem 3.22 is body-centered tetragonal. Of the three planes given in the problem statement the and (010) are equivalent—that is, have the same atomic packing. The atomic packing for these two planes as well as the are shown in the figure below. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. (b) Of the four planes cited in the problem statement, and are equivalent to one another— have the same atomic packing. The atomic arrangement of these planes is shown in the left drawing below. Furthermore, the and are equivalent to each other (but not to the other pair of planes); their atomic arrangement is represented in the other drawing. Note: the 0.424 nm dimension in the left-most drawing comes from the relationship . Likewise, the 0.500 nm dimension found in the right-most drawing comes from (c) All of the . , , , and planes are equivalent, that is, have the same atomic packing as illustrated in the following figure: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.51 The accompanying figure shows the atomic packing schemes for several different crystallographic directions for some hypothetical metal. For each direction the circles represent only those atoms contained within a unit cell, which circles are reduced from their actual size. (a) To what crystal system does the unit cell belong? (b) What would this crystal structure be called? Solution Below is constructed a unit cell using the six crystallographic directions that were provided in the problem. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. (a) This unit cell belongs to the tetragonal system since a = b = 0.40 nm, c = 0.50 nm, and  =  =  = 90. (b) This crystal structure would be called face-centered tetragonal since the unit cell has tetragonal symmetry, and an atom is located at each of the corners, as well as at the centers of all six unit cell faces. In the figure above, atoms are only shown at the centers of three faces; however, atoms would also be situated at opposite faces. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.52 The accompanying figure shows three different crystallographic planes for a unit cell of some hypothetical metal. The circles represent atoms: (a) To what crystal system does the unit cell belong? (b) What would this crystal structure be called? (c) If the density of this metal is 8.95 g/cm3, determine its atomic weight. Solution The unit cells constructed below show the three crystallographic planes that were provided in the problem statement. (a) This unit cell belongs to the orthorhombic crystal system since a = 0.30 nm, b = 0.40 nm, c = 0.35 nm, and  =  =  = 90. (b) This crystal structure would be called body-centered orthorhombic since the unit cell has orthorhombic symmetry, and an atom is located at each of the corners, as well as at the cell center. (c) In order to compute its atomic weight, we employ Equation 3.8, with n = 2; thus Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. = 113.2 g/mol Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.53 Convert the (010) and (101) planes into the four-index Miller–Bravais scheme for hexagonal unit cells. Solution For (010), h = 0, k = 1, and l = 0, and, from Equation 3.14, the value of i is equal to Therefore, the (010) plane becomes . Now for the (101) plane, h = 1, k = 0, and l = 1, and computation of i using Equation 3.14 leads to such that (101) becomes . Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.54 Determine the indices for the planes shown in the hexagonal unit cells below: Solution (a) For this plane, intersections with a1, a2, and z axes are a, a, and c/2 (the plane parallels both a1 and a2 axes). Therefore, these three values are equal to A, B, and C, respectively. If we assume that the value of n is 1, the values of h, k, and l are determined using Equations 3.13a, 3.13b, and 3.13c as follows: Now, from Equation 3.14, the value of i is Hence, this is a plane. (b) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. This plane passes through the origin of the coordinate axis system; therefore, we translate this plane one unit distance along the x axis, per the sketch shown below: For this plane, intersections with a1, a2, and z axes are a, a, and c (the plane parallels both a2 and z axes). Therefore, these three values are equal to A, B, and C, respectively. If we assume that the value of n is 1, the values of h, k, and l are determined using Equations 3.13a, 3.13b, and 3.13c as follows: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Now, from Equation 3.14, the value of i is Hence, this is a plane. (c) For this plane, intersections with a1, a2, and z axes are −a, a, and c. Therefore, these three values are equal to A, B, and C, respectively. If we assume that the value of n is 1, the values of h, k, and l are determined using Equations 3.13a, 3.13b, and 3.13c as follows: Now, from Equation 3.14, the value of i is Hence, this is a plane. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. (d) For this plane, intersections with a1, a2, and z axes are −a/2, a, and c/2. Therefore, these three values are equal to A, B, and C, respectively. If we assume that the value of n is 1, the values of h, k, and l are determined using Equations 3.13a, 3.13b, and 3.13c as follows: Now, from Equation 3.14, the value of i is Hence, this is a plane. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.55 Sketch the and planes in a hexagonal unit cell. Solution For the values of h, k, i, and l are, respectively, 1, –1, 0, and 1. Now, for h, k, and l, we solve for values of intersections with the a1, a2, and z axes (i.e., A, B, and C) using rearranged forms of Equations 3.13a, 3.13b, and 3.13c (assuming a value of 1 for the parameter n) as follows: C= nc (1)c = =c l 1 Hence, this plane intersects the a1 axis at a, the a2 axis at –a, and the z axis at c. The plane having these intersections is shown in the figure below. For the values of h, k, i, and l are, respectively, 1, 1, –2, and 0. Now, for h, k, and l, we solve for values of intersections with the a1, a2, and z axes (i.e., A, B, and C) using rearranged forms of Equations 3.13a, 3.13b, and 3.13c (assuming a value of 1 for the parameter n) as follows: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Hence, this plane intersects the a1 axis at a, the a2 axis at a, and parallels the z axis. The plane having these intersections is shown in the figure below. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Linear and Planar Densities 3.56 (a) Derive linear density expressions for FCC [100] and [111] directions in terms of the atomic radius R. (b) Compute and compare linear density values for these same two directions for silver. Solution (a) In the figure below is shown a [100] direction within an FCC unit cell. For this [100] direction there is one atom at each of the two unit cell corners, and, thus, there is the equivalent of 1 atom that is centered on the direction vector. The length of this direction vector is just the unit cell edge length, (Equation 3.1). Therefore, the expression for the linear density of this plane is An FCC unit cell within which is drawn a [111] direction is shown below. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. For this [111] direction, the vector shown passes through only the centers of the single atom at each of its ends, and, thus, there is the equivalence of 1 atom that is centered on the direction vector. The length of this direction vector is denoted by z in this figure, which is equal to where x is the length of the bottom face diagonal, which is equal to 4R. Furthermore, y is the unit cell edge length, which is equal to (Equation 3.1). Thus, using the above equation, the length z may be calculated as follows: Therefore, the expression for the linear density of this direction is (b) From the table inside the front cover, the atomic radius for silver is 0.144 nm. Therefore, the linear density for the [100] direction is While for the [111] direction Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.57 (a) Derive linear density expressions for BCC [110] and [111] directions in terms of the atomic radius R. (b) Compute and compare linear density values for these same two directions for tungsten. Solution (a) In the figure below is shown a [110] direction within a BCC unit cell. For this [110] direction there is one atom at each of the two unit cell corners, and, thus, there is the equivalence of 1 atom that is centered on the direction vector. The length of this direction vector is denoted by x in this figure, which is equal to where y is the unit cell edge length, which, from Equation 3.4 is equal to . Furthermore, z is the length of the unit cell diagonal, which is equal to 4R. Thus, using the above equation, the length x may be calculated as follows: Therefore, the expression for the linear density of this direction is Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. A BCC unit cell within which is drawn a [111] direction is shown below. For although the [111] direction vector shown passes through the centers of three atoms, there is an equivalence of only two atoms associated with this unit cell—one-half of each of the two atoms at the end of the vector, in addition to the center atom belongs entirely to the unit cell. Furthermore, the length of the vector shown is equal to 4R, since all of the atoms whose centers the vector passes through touch one another. Therefore, the linear density is equal to (b) From the table inside the front cover, the atomic radius for tungsten is 0.137 nm. Therefore, the linear density for the [110] direction is While for the [111] direction Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.58 (a) Derive planar density expressions for FCC (100) and (111) planes in terms of the atomic radius R. (b) Compute and compare planar density values for these same two planes for nickel. Solution (a) In the figure below is shown a (100) plane for an FCC unit cell. For this (100) plane there is one atom at each of the four cube corners, each of which is shared with four adjacent unit cells, while the center atom lies entirely within the unit cell. Thus, there is the equivalence of 2 atoms associated with this FCC (100) plane. The planar section represented in the above figure is a square, wherein the side lengths are equal to the unit cell edge length, (Equation 3.1); and, thus, the area of this square is just 2 = 8R . Hence, the planar density for this (100) plane is just That portion of an FCC (111) plane contained within a unit cell is shown below. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. There are six atoms whose centers lie on this plane, which are labeled A through F. One-sixth of each of atoms A, D, and F are associated with this plane (yielding an equivalence of one-half atom), with one-half of each of atoms B, C, and E (or an equivalence of one and one-half atoms) for a total equivalence of two atoms. Now, the area of the triangle shown in the above figure is equal to one-half of the product of the base length and the height, h. If we consider half of the triangle, then which leads to h = . Thus, the area is equal to And, thus, the planar density is (b) From the table inside the front cover, the atomic radius for nickel is 0.125 nm. Therefore, the planar density for the (100) plane is Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. While for the (111) plane Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.59 (a) Derive planar density expressions for BCC (100) and (110) planes in terms of the atomic radius R. (b) Compute and compare planar density values for these same two planes for vanadium. Solution (a) A BCC unit cell within which is drawn a (100) plane is shown below. For this (100) plane there is one atom at each of the four cube corners, each of which is shared with four adjacent unit cells. Thus, there is the equivalence of 1 atom associated with this BCC (100) plane. The planar section represented in the above figure is a square, wherein the side lengths are equal to the unit cell edge length, (Equation 3.4); and, thus, the area of this square is just = . Hence, the planar density for this (100) plane is just A BCC unit cell within which is drawn a (110) plane is shown below. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. For this (110) plane there is one atom at each of the four cube corners through which it passes, each of which is shared with four adjacent unit cells, while the center atom lies entirely within the unit cell. Thus, there is the equivalence of 2 atoms associated with this BCC (110) plane. The planar section represented in the above figure is a rectangle, as noted in the figure below. From this figure, the area of the rectangle is the product of x and y. The length x is just the unit cell edge length, which for BCC (Equation 3.4) is . Now, the diagonal length z is equal to 4R. For the triangle bounded by the lengths x, y, and z Or Thus, in terms of R, the area of this (110) plane is just Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. And, finally, the planar density for this (110) plane is just (b) From the table inside the front cover, the atomic radius for vanadium is 0.132 nm. Therefore, the planar density for the (100) plane is While for the (110) plane Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.60 (a) Derive the planar density expression for the HCP (0001) plane in terms of the atomic radius R. (b) Compute the planar density value for this same plane for magnesium. Solution (a) A (0001) plane for an HCP unit cell is show below. Each of the 6 perimeter atoms in this plane is shared with three other unit cells, whereas the center atom is shared with no other unit cells; this gives rise to three equivalent atoms belonging to this plane. In terms of the atomic radius R, the area of each of the 6 equilateral triangles that have been drawn is , or the total area of the plane shown is . And the planar density for this (0001) plane is equal to (b) From the table inside the front cover, the atomic radius for magnesium is 0.160 nm. Therefore, the planar density for the (0001) plane is Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Polycrystalline Materials 3.61 Explain why the properties of polycrystalline materials are most often isotropic. Solution Although each individual grain in a polycrystalline material may be anisotropic, if the grains have random orientations, then the solid aggregate of the many anisotropic grains will behave isotropically. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. X-ray Diffraction: Determination of Crystal Structures 3.62 The interplanar spacing dhkl for planes in a unit cell having orthorhombic geometry is given by where a, b, and c are the lattice parameters. (a) To what equation does this expression reduce for crystals having cubic symmetry? (b) For crystals having tetragonal symmetry? Solution (a) For the crystals having cubic symmetry, a = b = c. Making this substitution into the above equation leads to (b) For crystals having tetragonal symmetry, a = b  c. Replacing b with a in the equation found in the problem statement leads to Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.63 Using the data for molybdenum in Table 3.1, compute the interplanar spacing for the (111) set of planes. Solution From the Table 3.1, molybdenum has a BCC crystal structure and an atomic radius of 0.1363 nm. Using Equation (3.4), the lattice parameter a may be computed as Now, the interplanar spacing d111 maybe determined using Equation 3.21 as Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.64 Using the data for aluminum in Table 3.1, compute the interplanar spacings for the (110) and (221) sets of planes. Solution From the table, aluminum has an FCC crystal structure and an atomic radius of 0.1431 nm. Using Equation 3.1 the lattice parameter, a, may be computed as Now, the d110 interplanar spacing may be determined using Equation 3.21 as And, similarly for d221 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.65 Determine the expected diffraction angle for the first-order reflection from the (113) set of planes for FCC platinum when monochromatic radiation of wavelength 0.1542 nm is used. Solution We first calculate the lattice parameter using Equation 3.1 and the value of R (0.1387 nm) cited in Table 3.1, as follows: Next, the interplanar spacing for the (113) set of planes may be determined using Equation 3.21 according to And finally, employment of Equation 3.20 yields the diffraction angle as Which leads to And, finally Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.66 The 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 (a) From the data given in the problem, and realizing that 69.22 = 2, the interplanar spacing for the (220) set of planes for iridium may be computed using Equation 3.20 as (b) In order to compute the atomic radius we must first determine the lattice parameter, a, using Equation 3.21, and then R from Equation 3.1 since Ir has an FCC crystal structure. Therefore, And, from Equation 3.1 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.67 The metal rubidium has a BCC crystal structure. If the angle of diffraction for the (321) set of planes occurs at 27.00 (first-order reflection) when monochromatic x-radiation having a wavelength of 0.0711 nm is used, compute (a) the interplanar spacing for this set of planes, and (b) the atomic radius for the rubidium atom. Solution (a) From the data given in the problem, and realizing that 27.00 = 2, the interplanar spacing for the (321) set of planes for Rb may be computed using Equation 3.20 as follows: (b) In order to compute the atomic radius we must first determine the lattice parameter, a, using Equation 3.21, and then R from Equation 3.4 since Rb has a BCC crystal structure. Therefore, And, from Equation 3.4 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.68 For which set of crystallographic planes will a first-order diffraction peak occur at a diffraction angle of 46.21 for BCC iron when monochromatic radiation having a wavelength of 0.0711 nm is used? Solution The first step to solve this problem is to compute the interplanar spacing using Equation 3.20. Thus, Now, employment of both Equations 3.21 and 3.4 (since Fe’s crystal structure is BCC), and the value of R for iron from Table 3.1 (0.1241 nm) leads to This means that By trial and error, the only three integers having a sum that is even, and the sum of the squares of which equals 10.0 are 3, 1, and 0. Therefore, the set of planes responsible for this diffraction peak are the (310) ones. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.69 Figure 3.25 shows the first four peaks of the x-ray diffraction pattern for copper, which has an FCC crystal structure; monochromatic x-radiation having a wavelength of 0.1542 nm was used. (a) Index (i.e., give h, k, and l indices) for each of these peaks. (b) Determine the interplanar spacing for each of the peaks. (c) For each peak, determine the atomic radius for Cu and compare these with the value presented in Table 3.1. Solution (a) Since Cu has an FCC crystal structure, only those peaks for which h, k, and l are all either odd or even will appear. Therefore, the first peak results by diffraction from (111) planes. (b) For each peak, in order to calculate the interplanar spacing we must employ Equation 3.20. For the first peak which occurs at 43.8 (c) Employment of Equations 3.21 and 3.1 is necessary for the computation of R for Cu as = 0.1266 nm Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Similar computations are made for the other peaks, which results are tabulated below: Peak Index 2 dhkl(nm) R (nm) 200 50.8 0.1797 0.1271 220 74.4 0.1275 0.1275 311 90.4 0.1087 0.1274 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.70 Below are listed diffraction angles for the first four peaks (first-order) of the x-ray diffraction pattern for chromium, which has a BCC crystal structure; monochromatic x-radiation having a wavelength of 0.0711 nm was used. Plane Indices Diffraction Angle (2) (110) 20.1 (200) 28.5 (211) 35.1 (220) 40.7 (a) Determine the interplanar spacing for each of the peaks. (b) For each peak, determine the atomic radius for Cr, and compare these with the value presented in Table 3.1. Solution (a) For each peak, in order to calculate the interplanar spacing we must employ Equation 3.20. For the first peak [which occurs by diffraction from the (110) set of planes] and occurs at 20.1–that is, 2 = 20.1; this means that  = 20.1/2 = 10.05. Using Equation 3.20, the interplanar spacing is computed as follows: (b) Employment of Equations 3.21 and 3.4 is necessary for the computation of R for Cr as = ( 3)(0.204 nm) (1)2 + (1)2 + (0)2 4 = 0.125 nm Similar computations are made for the other peaks which results are tabulated below: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Peak Index 2 dhkl(nm) R (nm) 200 28.5 0.144 0.125 211 35.1 0.118 0.125 220 40.7 0.102 0.125 The atomic radius for chromium cited in Table 3.1 is 0.1249 nm. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.71 Below are listed diffraction angles for the first three peaks (first-order) of the x-ray diffraction pattern for some metal. Monochromatic x-radiation having a wavelength of 0.1254 nm was used. (a) Determine whether this metal's crystal structure is FCC, BCC, or neither FCC or BCC and explain the reason for your choice. (b) If the crystal structure is either BCC or FCC, identify which of the metals in Table 3.1 gives this diffraction pattern. Justify your decision. Peak Number Diffraction Angle (2) 1 31.2 2 44.6 3 55.4 Solution (a) The steps in solving this part of the problem are as follows: 1. For each of these peaks compute the value of dhkl using Equation 3.20 in the form (P.1) taking n = 1 since this is a first-order reflection, and  = 0.1254 nm (as given in the problem statement). 2. Using the value of dhkl for each peak, determine the value of a from Equation 3.21 for both BCC and FCC crystal structures—that is (P.2) For BCC the planar indices for the first three peaks are (110), (200), and (211), which yield the respective h2 + k2 + l2 values of 2, 4, and 6. On the other hand, for FCC planar indices for the first three peaks are (111), (200), and (220), which yield the respective h2 + k2 + l2 values of 3, 4, and 8. 3. If the three values of a are the same (or nearly the same) for either BCC or FCC then the crystal structure is which of BCC or FCC has the same a value. 4. If none of the set of a values for both FCC and BCC are the same (or nearly the same) then the crystal structure is neither BCC or FCC. Step 1 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Using Equation P.1, the three values of dhkl are computed as follows: Step 2 Using Equation P.2 let us first compute values of a for BCC. For the (110) set of planes For the (200) set of planes: And for the (211) set of planes: Inasmuch as these three values of a are nearly the same, the crystal structure is BCC. (b) Since we know the value of a for this BCC crystal structure, it is possible to calculate the value of the atomic radius R using a rearranged form of Equation 3.4; and once we know the value of R, we just find that metal in Table 3.1 that has this atomic radius. Thus, we calculate the value of R (using Equation 3.4) as follows (using a value of 0.3302 for a): From Table 3.1 the only BCC metal that has this atomic radius is tantalum. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Noncrystalline Solids 3.72 Would you expect a material in which the atomic bonding is predominantly ionic in nature to be more or less likely to form a noncrystalline solid upon solidification than a covalent material? Why? (See Section 2.6.) Solution A material in which atomic bonding is predominantly ionic in nature is less likely to form a noncrystalline solid upon solidification than a covalent material because covalent bonds are directional whereas ionic bonds are nondirectional; it is more difficult for the atoms in a covalent material to assume positions giving rise to an ordered structure. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. FUNDAMENTALS OF ENGINEERING QUESTIONS AND PROBLEMS 3.1FE A hypothetical metal has the BCC crystal structure, a density of 7.24 g/cm3, and an atomic weight of 48.9 g/mol. The atomic radius of this metal is: (A) 0.122 nm (C) 0.0997 nm (B) 1.22 nm (D) 0.154 nm Solution The volume of a BCC unit cell is calculated using Equation 3.4 as follows Now, using Equation 3.8, we may determine the density as follows: And, from this expression, solving for R, leads to Since there are two atoms per unit cell (n = 2) and incorporating values for the density () and atomic weight (A) provided in the problem statement, we calculate the value of R as follows: = 1.22  10−8 cm = 0.122 nm which is answer A. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.2FE In the following unit cell, which vector represents the [121] direction? Solution In order to solve this problem, let us take the position of the [121] direction vector as the origin of the coordinate system, and then, using Equations 3.10a, 3.10b, and 3.10c, determine the head coordinates of the vector. The vector in the illustration that coincides with this vector or is parallel to it corresponds to the [121] direction. Using this scheme, vector tail coordinates are as follows: x1 = 0a y1 = 0b z1 = 0c For this direction, values of the u, v, and w indices are as follows: u=1 v=2 w =1 Now, assuming a value of 1 for the parameter n, values of the head coordinates (x2, y2 , and z2) are determined (using Equations 3.10) as follows: If we now divide these three head coordinates by a factor of 2, it is possible, in a stepwise manner, locate the location of the vector head as follows: move a/2 units along the x axis, then b units parallel to the y axis, and from here c/2 units parallel to the z axis. The vector from the origin to this point corresponds to B. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.3FE What are the Miller indices for the plane shown in the following cubic unit cell? (C) (10 12 ) (A) (201) ( (B) 1 ∞ 12 ) (D) (102) Solution The Miller indices for this direction may be determined using Equations 3.13a, 3.13b, and 3.13c. However, it is first necessary to note intersections of this plane with the x, y, and z coordinate axes. These respective intercepts are a, a (since the plane is parallel to the y axis), and a/2; that is A=a B = a C = a/2 Thus, using Equations 3.13, values of the h, k, and l indices (assuming that n = 1) are as follows: Therefore this is a (102) plane, which means that D is the correct answer. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.