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Lecture

ENG 2002 Lecture Notes - Diamond Cubic, Monochrome, Diffractometer

97 Pages
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Winter 2014

Department
Engineering
Course Code
ENG 2002
Professor
Hany E.Z.Farag

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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.
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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 3.4 as
V
C
= 16R32 = (16)(0.143 10-9 m)3(2)= 6.62 10-29 m3
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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/3.
Solution
Consider the BCC unit cell shown below
Using the triangle NOP
(NP)2= a2+ a2=2a2
And then for triangle NPQ,
(NQ)2=(
QP)2+(
NP)2
But
NQ = 4R, R being the atomic radius. Also,
QP = a. Therefore,
(4R)2 = a2 + 2a2
or
a = 4R
3

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Description
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 inthenucleusofanatom,aswellasthe 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 3.4 as 3 -9 3 -29 3 VC = 16R 2 = (16) (0.143  10 m ) ( 2 )= 6.62  10 m 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/ 3. Solution Consider the BCC unit cell shown below Using the triangle NOP 2 2 2 2 (NP) = a + a =2 a And then for triangle NPQ, (NQ) 2=( QP) 2 +( NP) 2 But NQ = 4R, R being the atomic radius. AlsQP = a. Therefore, (4R)2 = a +2 a 2 or 4R a = 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. 3.4 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 MH = c/2. And, since atoms at points J, K, and M, all touch one another, JM = JK =2 R = a where R is the atomic radius. Furthermore, from triangle JHM, 2 2 2 (JM ) =( JH)  (MH) 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. 2 2 2 c  a =( JH ) + 2 Now, we can determine the JH length by consideration of triangle JKL, which is an equilateral triangle, a/2 3 cos 30 = = JH 2 and JH = a 3 Substituting this value for JH in the above expression yields a  c  a 2 c2 a2 =   +  = +  3 2  3 4 and, solving for c/a c 8 a = 3 = 1.633 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 VS APF = V C Since there are two spheres associated with each unit cell for BCC 4 R3 8R 3 VS = 2(sphere volume) = 2  =  3  3 Also, the unit cell has cubic symmetry, that is V. But a depends on R according to Equation 3.3, and C 4R  64R 3 V =   = C  3 33 Thus, V 8R /3 APF = S = = 0.68 VC 64R /3 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. 3.6 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  3  R  3 VS =6  3 =8  R   Now, the unit cell volume is just the product of the base area times the cell height, c. This basearea isjust three times the area of the parallelepiped ACDE shown below. The area of ACDE is just the length of CD times the heightBC . But CD is just a or 2R, and BC =2 R cos (30) = 2R 3 2 Thus, the base area is just 2R 3  AREA = (3)(CD)(BC) = (3)(2R)  6 R 2 3  2  and since c = 1.633a = 2R(1.633) ) 1 S . V = (A3REA)(c) = 6R c 3 ( 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. = (6R 2 3) (2)(1.633)R = 12 3 (1.633)R3 Thus, V 8 R3 APF = S = = 0.74 VC 12 3 (1.633)R 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. Density Computations 7 . 3 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. Solution This problem calls for a computation of the density of iron. According to Equation 3.5 nAFe  = VCN A For BCC, n = 2 atoms/unit cell, and 3 4R  VC=  3  Thus, nAFe  = 3 4R   3 N A = (2 atoms/unit cell)(55.85 g/mol) (4)(0.124  10 -7 cm) /3 3 /(unit cel(6.022  10 23atoms/mol )   3 = 7.90 g/cm The value given inside the front cover is 7.87 g/cm . 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 . 3 Calculate the radius of an iridium atom, given that Ir has an FCC crystal structure, a density of 22.4 3 g/cm , 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 V = 16R 3 2 (Equation 3.4). Now, C nAIr  = V N C A = nAIr 16R 3 2 )N A And solving for R from the above expression yields 1/3  nA Ir  R =   6N A 2   /3 = (4 atoms/unit cell 92.2 g/mol (16)(22.4 g/cm3)(6.022  10 23 atoms/mol )( 2 )   -8 = 1.36  10 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 hasAny 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. 9 . 3 Calculate the radius of a vanadium atom, given that V has a BCC crystal structure, a density of 5.96 3 g/cm , 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 3 3 V = 4R  = 64R C  3  33 Since, from Equation 3.5 nA  = V VCN A nA = V 64 R 3 33  A   and solving for R the previous equation 1/3 33 nA V  R = 64 N   A  and incorporating values of parameters given in the problem statement  /3 R =  ( 33 )(2 atoms/unit cell)(50.9 g/mol)  64)(5.96 g/cm 3)(6.022  10 23 atoms/mol ) = 1.32  10 -8cm = 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.oduction 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 Some hypothetical metal has the simple cubic crystal structure shown in Figure 3.24. 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 3.5 is unity since there is only a single atom associated with each unit cell. Furthermore, for the unit cell edge length, a =2 R (Figure 3.24). Therefore, employment of Equation 3.5 yields  = nA = nA VCN A (2R) N A and incorporating values of the other parameters provided in the problem statement leads to  = (1 atom/unit cell)(70.4 g/mol)  -8   2)(1.26  10 cm )/(unit cel(6.022  10 23 atoms/mol )    7.31 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.production 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 3.11 Zirconium has an HCP crystal structure and a density of 6.51 g/cm . (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 3.5 as nA Zr V C  N A Now, for HCP, n = 6 atoms/unit cell, and for Zr, A = 91.22 g/mol. Thus, Zr (6 atoms/unit cell)(91.22 g/mol) VC  3 23 (6.51 g/cm )( 6.022  10 atoms/mol ) -22 3 -28 3 = 1.396  10 cm /unit cell = 1.396  10 m /unit cell (b) From Equation 3.S1 of the solution to Problem 3.6, for HCP V = 6R c 3 C But, since a = 2R, (i.e., R = a/2) then   33 a c V =6  c 3  C 2  2 but, since c = 1.593a 3 V = 33 (1.593)a = 1.396  10 -22cm /unit cell C 2 Now, solving for a 1/3 (2)1.396  10 -22cm 3) a =    (3)( 3)(1.593)  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.oduction 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 = 3.23  10 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. 3.12 Using atomic weight, crystal structure, and atomic radius data tabulated inside the front cover, compute the theoretical densities of lead, chromium, coppe r, and cobalt, and then compar e 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 = 16R 3 2 (Equation 3.4). Also, R = 0.175 nm (1.75 C  10 cm) and A = 207.2 g/mol. Employment of Equation 3.5 yields Pb nA  = Pb VCN A (4 atoms/unit cell)(207.2 g/mol)  -8 3 23  6)(1.75  10 cm ) ( 2 )/(unit cell)(6.022  10 atoms/mol ) 3 = 11.35 g/cm 3 The value given in the table inside the front cover is 11.35 g/cm . 3 3 4 R  Chromium has a BCC crystal structure for which n = 2 and V C = a =   (Equation 3.3); also ACr =  3  52.00g/mol and R = 0.125 nm. Therefore, employment of Equation 3.5 leads to   (2 atoms/unit cell)(52.00 g/mol)  -8   4)(1.25  10 cm )/(unit cel(6.022  10 23 atoms/mol )  3     3 = 7.18 g/cm The value given in the table is 7.19 g/cm . Copper also has an FCC crystal structure and therefore (4 atoms/unit cell)(63.55 g/mol)    -8 3  23  ( 1.28  10 cm )( 2) /(unit cell6.022  10 atoms/mol )   = 8.90 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. The value given in the table is 8.90 g/cm Cobalt has an HCP crystal structure, and from the solution to Problem 3.6 (Equation 3.S1), 2 VC =6 R c 3 and, since c = 1.623a and a = 2R, c = (1.623)(2R); hence V  6R 2(1.623)(R )3  (19.48)( 3)R 3 C  (19.48)( 3)(1.25 108 cm) 3 23 3  6.59  10 cm /unit cell Also, there are 6 atoms/unit cell for HCP. Therefore the theoretical density is  = nACo VCN A (6 atoms/unit cell)(58.93 g/mol)  (6.59  10-23cm /unit cel)(6.022  10 23atoms/mol ) = 8.91 g/cm3 The value given in the table is 8.9 g/cm 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.ction 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 3.13 Rhodium has an atomic radius of 0.1345 nm and a density of 12.41 g/cm. 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 aR= 2 (Equation 3.1). Also, from Figure 2.6, its atomic weight is 102.91 g/mol. Thus, for FCC (employing Equation 3.5) nA nA   Rh  Rh a N (2R 2 )3N A A (4 atoms/unit cell)(102.91 g/mol) =  -8 3  23  )1.345  10 cm )( 2 ) /(unit ce6.022  10 atoms/mol ) = 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. 4 1 . 3 Below are listed the atomic weight, density, and atomic radius for three hypothetical alloys. For each determine whether its crystal stru cture is FCC, BCC, or simple cubic and then justify your determination. A simple cubic unit cell is shown in Figure 3.24. Alloy Atomeiigcht Density Atomic Radius ( g/mol) (g/cm ) (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 V = a ) are 2R, , and 4R . C 3 For alloy A, let us calculate  assuming a simple cubic crystal structure. nA  = A VCN A nA = A R 3N A = (1 atom/unit cell)(77.4 g/mol)  8 3  23  )(1.25  10 ) /(unit cel(6.022  10 atoms/mol) = 8.22 g/cm3 Therefore, its crystal structure is simple cubic. For alloy B, let us calculate  assuming an FCC crystal structure. nA  = B (2R 2) N 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. = (4 atoms/unit cell)(107.6 g/mol)  -8 3  23  2 (1.33  10 cm ) /(unit cel(6.022  10 atoms/mol )   = 13.42 g/cm 3 Therefore, its crystal structure is FCC. For alloy C, let us calculate  assuming a simple cubic crystal structure. nA C = 3 R N A = (1 atom/unit cell)(127.3 g/mol)  -8 3  23  )(1.42  10 cm ) /(unit cel(6.022  10 atoms/mol ) 3 = 9.23 g/cm 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. 3.15 The unit cell for tin has tetragonal symmetry, with a and b lattice parameters of 0.583 and 0.318 nm, 3 respectively. If its density, atomic weight, and atomic radius are 7.30 g/cm , 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 (V C) which is just the 2 a c 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 3.5 as VCN A n = ASn (7.30 g/cm 3)(5.83) (3.18)( 10-24 cm 3)(6.022  10 23 atoms/mol ) = 118.69 g/mol = 4.00 atoms/unit cell Therefore 4 3  V (4)3  R  APF = S = V C (a) (c) 4  (4) ()(1.51  10 -8 cm) 3 3  = (5.83  10 -8 cm) (3.18  10 -8 cm) = 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.roduction 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 Iodine has an orthorhombic unit cell for which th e 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 indium, and from the definition of the APF   n  R3 VS 3  APF = V = abc C we may solve for the number of atoms per unit cell, n, as n = (APF)abc 4 3 3 R Incorporating values of the above parameters provided in the problem state leads to (0.547)(4.79  10-8 cm)(7.25  10 -8cm) (9.78  10 -8cm ) = 4 (1.77  10 -8cm )3 3 = 8.0 atoms/unit cell (b) In order to compute the density, we just employ Equation 3.5 as nAI  = abc N A = (8 atoms/unit cell)(126.91 g/mol) (4.79  10 -8cm)(7.25  10 -8 cm) (9.78  10-8 cm ) /unit cel(6.022  10 23 atoms/mol)    3 = 4.96 g/cm 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 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) VC=6 R c 32 But for Ti, c = 1.58a, and a = 2R, or c = 3.16R, and 3 VC = (6)(3.16) R 3 -8 3 23 3 = (6)(3.16)( 3)  45  10 cm = 9.91  10 cm /unit cell (b) The theoretical density of Ti is determined, using Equation 3.5, as follows: nA Ti  = V N C A For HCP, n = 6 atoms/unit cell, and for Ti, A= 47.87 g/mol (as noted inside the front cover). Thus, Ti  = (6 atoms/unit cell)(47.87 g/mol) ( 9.91  10 -23cm /unit cel)(6.022  10 23 atoms/mol ) = 4.81 g/cm 3 The value given in the literature is 4.51 g/cm . 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.oduction 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 3.18 Zinc has an HCP crystal structure, a c/a ratio of 1.856, and a density of 7.13 g/cm . Compute the atomic radius for Zn. Solution In order to calculate the atomic radius for Zn, we must use Equation 3.5, 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: V =6 R c 32 C In this case c = 1.856a, but, for HCP, a = 2R, which means that VC=6 R (1.856)(2R)3  (1.856)(12 3 )R 3 And from Equation 3.5, the density is equal to nA nA  = Zn = Zn 3 V C A (1.856)(12 3 )R N A And, solving for R from the above equation leads to the following:  /3 R =  nAZn  1.856)(12 3 ) N  A And incorporating appropriate values for the parameters in this equation leads to 1/3  (6 atoms/unit cell)(65.41 g/mol)  R =   1.856)(12 3 )(7.13 g/cm 3)(6.022  10 23 atoms/mol ) = 1.33  10-8cm = 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 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 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 V =6 R c 32 C The problem states that c = 1.615a, and a = 2R. Therefore VC = (1.615)(12 3 ) R3 = (1.615)(12 3 )(1.37  10 -8cm )3 = 8.63  10 -23cm 3 = 8.63  10 -2nm 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 haAny 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 3.20 Below is 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 8 2 8 VC= (3.0  10 cm ) (4.0  10 cm ) 23 3 = 3.60  10 cm /unit cell Thus, using Equation 3.5, the density is equal to nA  = VCN A (2 atoms/unit cell)(141 g/mol) = (3.60  10 -23cm /unit cel)(6.022  10 23 atoms/mol ) = 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 haAny 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. 3.21 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 adopteAny 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 3.22 List the point coordinates for all atoms that are associated with the FCC unit cell (Figure 3.1). Solution From Figure 3.1b, the atom located of the origin of the unit cell has the coordinates 000. Coordinates for 1 1 other atoms in the bottom face are 100, 110, 010, and 0 . (The z coordinate for all these points is zero.) 2 2 For the top unit cell face, the coordinates are 001, 101, 111, 011, and 1 1. 2 2 Coordinates for those atoms that are positioned at the centers of both side faces, and centers of both front 1 1 1 1 and back faces need to be specified. For the front and back-center f ace atoms, the coordinates are 1 and 0 , 2 2 2 2 respectively. While for the left and right side center-face atoms, the respective coordinates are0 1 and 1 1 . 2 2 2 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 bAny 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 List the point coordinates of the titanium, barium, and oxygen ions for a unit cell of the perovskite crystal structure (Figure 12.6). Solution In Figure 12.6, the barium ions are situated at all corner positions. The point coordinates for these ions are as follows: 000, 100, 110, 010, 001, 101, 111, and 011. 1 1 1 1 The oxygen ions are located at all face-centered positions; therefore, their coordinates are 0 , 1, 2 2 2 2 1 1 1 ,0 1 1 , 1 0 1, and 111 . 2 2 2 2 2 2 2 2 1 1 1 And, finally, the titanium ion resides at the center of the cubic unit cell, with coordinates . 2 2 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 beenAny 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. 3.24 List the point coordinates of all atoms that are associated with the diamond cubic unit cell (Figure 12.15). Solution First of all, one set of carbon atoms occupy all corner positions of the cubic unit cell; the coordinates of these atoms are as follows: 000, 100, 110, 010, 001, 101, 111, and 011. 1 1 Another set of atoms reside on all of the face-centered positions, with the following coordinates: 2 2 0, 1 1 1 1 1 1 1 1 1 1 1 ,1 ,0 , 0 , and 1 . 2 2 2 2 2 2 2 2 2 2 The third set of carbon atoms are positioned within the interior of the unit cell. Using an x-y-z coordinate system oriented as in Figure 3.4, the coordinates of the atom that lies toward the lower-left-front of the unit cell has 3 1 1 the coordinates 4 4 4, whereas the atom situated toward the lower-right-back of the unit cell has coordinates of 1 3 1. Also, the carbon atom that resides toward the upper-left-back of the unit cell has the 1 1 3 coordinates. 4 4 4 4 4 4 3 3 3 And, the coordinates of the final atom, located toward the upper-right-front of the unit cell, are . 4 4 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 beenAny 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. 3.25Sketch a tetragonal unit cell , and within that cell indicate locations of t1e1 and 1 1 3 point 2 2 4 2 4 coordinates. Solution 1 1 1 1 3 A tetragonal unit in which are shown the 1 and point coordinates is presented below. 2 2 4 2 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 hasAny 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 Using the Molecule Definition Utility found in both “Metallic Crystal Structures and Crystallography” and “Ceramic Crystal Structures” m odules of VMSE, located on the book’s web site [www.wiley.com/college/Callister (Student Companion Site)], generate a three-dimensional unit cell for the intermetallic compound AuCu 3given 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), and “Cu” as the name for the copper atoms. Next it is necessary to choose a color for each atom type from the selections that appear in the pull-down menu—for example, “Yellow” for Au and “Red” for Cu. In the “Atom Size” window, it is necessary to enter an atom/ion size. In the instructions for this step, it is suggested that the atom/ion diameter in nanometers be used. From the table found inside the front cover of the textbook, the atomic radii for gold and copper are 0.144 nm and 0.128 nm, respectively, and, therefore, their ionic diameters are twice these values (i.e., 0.288 nm and 0.256 nm); therefore, we enter the values “0.288” and “0.256” for the two atom types. Now click on the “Register” button, followed by clicking on the “Go to Step 2” button. In the “Step 2” window we specify positions for all of the atoms within the unit cell; their point coordinates are specified in the problem statement. Let’s begin with gold. Click on the yellow sphere that is located to the right of the “Molecule Definition Utility” box. 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 coordinates 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. 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 x-axis (i.e., at the 100 point coordinate). 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. 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 click on the red sphere that is located next to the “Molecule Definition Utility” box. The point coordinates for some of the Cu atoms are fractional ones; in these instances, the a 1 1 unit cell length (i.e., 0.374) is multiplied by the fraction. For example, one Cu atom is located 1 2 2 coordinate. 1 1 Therefore, the x, y, and z atoms positions are (1)(0.374) = 0.374, (0.374) = 0.187, and (0.374) = 0.187, 2 2 respectively. For the gold atoms, the x, y, and z atom position entries for all 8 sets of point coordinates are 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 beenAny 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. 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 Step3, wemayspecifywhichatomsaretoberepresentedas being bonded to one another, and which type of bond(s) to use (single solid, single dashed, double, and triple 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, probably the best thing to do is to represent unit cell edges as bonds. This image 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 screen shot) 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 3.27 Draw an orthorhombic unit cell, and within that cell a [121 ] direction. Solution This problem calls for us to draw a [121 ] direction within an orthorhombic unit cell (a ≠ b ≠ c,  =  =  = 90). Such a unit cell with its origin positioned at point O is shown below. We first 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 [121 ] 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. 3.28 Sketch a monoclinic unit cell, and within that cell a [01 1] direction. Solution This problem asks that a [0 11] 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. For this di rection, there is no projection along the x-axis since the first index is zero; thus, the direction lies in the y-z plane. We next move from the origin along the minus y-axis b units (from point O to point R). Since the final index is a one, move from point R parallel to the z-axis, c units (to point P). Thus, the [0 11] direction corresponds to the vector passing from the origin (point O) 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. 3.29 What are the indices for the directions indicated by the two vectors in the sketch below? Solution For direction 1, the projection on the x-axis is zero (since it lies in the y-z plane), while projections on the y- and z-axes, b/2 and c, respectively. This is[012] direction as indicated in the summary below. x y z Projections 0a b/2 c Projections in terms of a, b, and c 0 1/2 1 Reduction to integers 0 1 2 e r u s o l c n E [012] Direction 2 is [112]as summarized below. x y z Projections a/2 b/2 -c Projections in terms of, b, and c 1/2 1/2 -1 Reduction to integers 1 1 -2 Enclosure [112] 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. 3.30 Within a cubic unit cell, sketch the following directions: (a) [1 10] , (e) [11 1] , (b) [12 1] , (f) [1 22] , (c) [01 2] , (g) [123] , (d) [13 3] , (h) [103] . Solution The directions asked for are indicated in the cubic unit cells 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. 3.31 Determine the indices for the directions shown in the following cubic unit cell: Solution Direction A is a [011]direction, which determination is summarized as follows. We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system x y z Projections 0a – b – c Projections in terms of a, b, and c 0 –1 –1 Reduction to integers not necessary Enclosure [01 1] Direction B is a [210] direction, which determination is summarized as follows. We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system x y z Projections –a b 0 c 2 1 Projections in terms of a, b, and c –1 0 2 Reduction to integers –2 1 0 Enclosure [210] Direction C is a [112] direction, which determination is summarized as follows. We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system 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 y z a b Projections c 2 2 Projections in terms of a, b, and c 1 1 1 2 2 Reduction to integers 1 1 2 Enclosure [112] Direction D is a[112] direction, which determination is summarized as follows. We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system x y z Projections a b – c 2 2 1 1 Projections in terms of a, b, and c –1 2 2 Reduction to integers 1 1 –2 Enclosure [112] 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 Determine the indices for the directionsshown in the following cubic unit cell: Solution Direction A is a [30] direction, which determination is summarized as follows. We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system x y z Projections – 2a b 0 c 3 2 2 1 Projections in terms of a, b, and c – 0 3 2 Reduction to integers –4 3 0 Enclosure [430] Direction B is a [232] direction, which determination is summarized as follows. We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system x y z 2a 2c Projections – b 3 3 Projections in terms of a, b, and c 2 –1 2 3 3 Reduction to integers 2 –3 2 Enclosure [232] Direction C is a [133] direction, which determination is summarized as follows. We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system 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 y z a Projections – b – c 3 Projections in terms of a, b, and c 1 –1 –1 3 Reduction to integers 1 –3 –3 Enclosure [133] Direction D is a [136] direction, which determination is summarized as follows. We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system x y z a b Projections – c 6 2 Projections in terms of a, b, and c 1 1 –1 6 2 Reduction to integers 1 3 –6 Enclosure [136] 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 For tetragonal crystals, cite the indices of directi ons 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: [00 1]. (b) For the [110] direction, equivalent directions are as follows: [1 10], [110], and [110] (b) Also, for the [010] direction, equivalent directions are the following: [010], [100] , and [100]. 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 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 3.6 1 1 2 u = (2u' v' ) = [(2)(1) 0] = 3 3 3 v = 1(2vΥ uΥ) = 1 [(2)(0) 1] =  1 3 3 3 2 1  1 t=  (u + v) =     3 3  3 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] = [21 10]. For [111], u' = 1, v' = 1, and w' = 1; therefore, 1 1 u = 3[(2)(1) 1] = 3 1 1 v = [(2)(1) 1] = 3 3 t =   1 2 3 3 3 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
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