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b 2 – c Intercepts in terms of a, b, and c ∞ 1 2 – 1 Reciprocals of intercepts 0 2 – 1 Reduction not necessary Enclosure (021 ) 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 3.43 (a) In the figure below is shown (110) and (111) planes, and, as indicated, their intersection results in a [1 10] , or equivalently, a [11 0] direction. (b) In the figure below is shown (110) and (11 0) planes, and, as indicated, their intersection results in a [001], or equivalently, a [001 ] direction. (c) In the figure below is shown (111 ) and (001) planes, and, as indicated, their intersection results in a [1 10] , or equivalently, a [11 0] 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. 3-51 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 3.44 (a) The atomic packing of the (100) plane for the FCC crystal structure is called for. An FCC unit cell, its (100) plane, and the atomic packing of this plane are indicated below. (b) For this part of the problem we are to show the atomic packing of the (111) plane for the BCC crystal structure. A BCC unit cell, its (111) plane, and the atomic packing of this plane are indicated 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. 3-53 3.45 (a) The unit cell in Problem 3.20 is body-centered tetragonal. Of the three planes given in the problem statement the (100) and (01 0) are equivalent—that is, have the same atomic packing. The atomic packing for these two planes as well as the (001) are shown in the figure below. (b) Of the four planes cited in the problem statement, only (101), (011), and (1 01) are equivalent—have the same atomic packing. The atomic arrangement of these planes as well as the (110) are presented in the figure below. Note: the 0.495 nm dimension for the (110) plane comes from the relationship (0.35 nm)2 + (0.35 nm)2[ 1] /2 . Likewise, the 0.570 nm dimension for the (101), (011), and (1 01) planes comes from (0.35 nm)2 + (0.45 nm)2[ ]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. 3-54 (c) All of the (111), (11 1) , (111 ) , and (1 11 ) 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. 3-55 3.46 Unit cells are constructed below from the three crystallographic planes provided in the problem statement. (a) This unit cell belongs to the tetragonal system since a = b = 0.40 nm, c = 0.55 nm, and α = β = γ = 90°. (b) This crystal structure would be called body-centered tetragonal since the unit cell has tetragonal symmetry, and an atom is located at each of the corners, as well as the cell center. 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. 3-56 3.47 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 = 0.25 nm, b = 0.30 nm, c = 0.20 nm, and α = β = γ = 90°. (b) This crystal structure would be called face-centered orthorhombic since the unit cell has orthorhombic symmetry, and an atom is located at each of the corners, as well as at each of the face centers. (c) In order to compute its atomic weight, we employ Equation 3.5, with n = 4; thus A = ρVC N A n = (18.91 g/cm3) (2.0)(2.5)(3.0) (× 10-24 cm3/unit cell)(6.023 × 10 23 atoms/mol) 4 atoms/unit cell = 42.7 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. 3-57 3.48 This problem asks that we convert (111) and (01 2) planes into the four-index Miller-Bravais scheme, (hkil), for hexagonal cells. For (111), h = 1, k = 1, and l = 1, and, from Equation 3.7, the value of i is equal to i = − (h + k) = − (1 + 1) = − 2 Therefore, the (111) plane becomes (112 1) . Now for the (01 2) plane, h = 0, k = -1, and l = 2, and computation of i using Equation 3.7 leads to i = − (h + k) = −[0 + (−1)] = 1 such that (01 2) becomes (01 12) . 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 3.49 This problem asks for the determination of Bravais-Miller indices for several planes in hexagonal unit cells. (a) For this plane, intersections with the a1, a2, and z axes are ∞a, –a, and ∞c (the plane parallels both a1 and z axes). In terms of a and c these intersections are ∞, –1, and ∞, the respective reciprocals of which are 0, –1, and 0. This means that h = 0 k = –1 l = 0 Now, from Equation 3.7, the value