Fundamentals of Materials Science and Engineering Solutions Ch 2-3

CH A PTE R 2 ATOMIC STRU CTU RE A ND INTERATOMIC BOND ING 2.3 (a) In order to determine the number of grams in one amu of material, appropriate manipulation of the amu/atom, g/mol, and atom/mol relationships is all that is necessary, as 1 mol #g/amu = ×−24 atoms 6.023 1023 × 10 g/amu = 1.66  1 g/mol 1 amu/atom  2.14 (c) This portion of the problem asks that we determine for a K + -Cl− ion pair the interatomic spacing (ro ) and the bonding energy (E o ). From E quation (2.11) for E N A = 1.436 B = 5.86 × 10−6 n=9 Thus, using the solutions from Problem 2.13 A nB ro =  = and 1/(1−n) 1.436 (9)(5.86 × 10−6 ) 1.436 Eo = − 1/(1−9) = 0.279 nm 5.86 × 10−6 1/(1−9) +  (9)(5.86 × 10−6 ) 1.436  1.436 (9)(5.86 × 10−6 ) = −4.57 eV 9/(1−9) 2.19 The percent ionic character is a function of the electronegativities of the ions X A and X B according to E quation (2.10). The electronegativities of the elements are found in Figure 2.7. For TiO 2 , X Ti = 1.5 and X O = 3.5, and therefore, % IC =  1 − e (−0.25)(3.5−1.5) 1 2  × 100 = 63.2% CH A PTE R 3 STRU CTU RES OF METALS A ND CERA MICS 3.3 For this problem, we are asked to calculate the volume of a unit cell of aluminum. A luminum has an FCC crystal structure (Table 3.1). The FCC unit cell volume may be computed from E quation (3.4) as √ √ V C = 16R 3 2 = (16)(0.143 × 10−9 m) 3 2 = 6.62 × 10−29 m 3 3.7 This problem calls for a demonstration that the A PF for H CP is 0.74. A gain, the A PF is just the total sphere-unit cell volume ratio. For H CP, there are the equivalent of six spheres per unit cell, and thus VS = 6 4␲R 3  = 8␲R 3 3 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 ACD E shown below. D C a = 2R 30 A 60 B E a = 2R a = 2R The area of ACD E is just the length of CD times the height BC. But CD is just a or 2R, and BC = 2R cos(30◦ ) = √ 2R 3 2 Thus, the base area is just A R E A = (3)(CD )(BC) = (3)(2R ) √ 2R 3  2 √ = 6R 2 3 and since c = 1.633a = 2R (1.633) √ √ √ V C = (A R E A )(c) = 6R 2 c 3 = (6R 2 3)(2)(1.633)R = 12 3(1.633)R 3 2 Thus, A PF = VS = VC √ 8␲R 3 12 3(1.633)R 3 = 0.74 3.12. (a) This portion of the problem asks that we compute the volume of the unit cell for Z r. This volume may be computed using E quation (3.5) as VC = nA Zr ␳ NA Now, for H CP, n = 6 atoms/unit cell, and for Z r, A Zr = 91.2 g/mol. Thus, VC = (6 atoms/unit cell)(91.2 g/mol) (6.51 g/cm 3 )(6.023 × 1023 atoms/mol) = 1.396 × 10−22 cm 3 /unit cell = 1.396 × 10−28 m 3 /unit cell (b) We are now to compute the values of a and c, given that c/a = 1.593. From the solution to Problem 3.7, since a = 2R, then, for H CP VC = √ 3 3a 2 c 2 but, since c = 1.593a VC = √ 3 3(1.593)a 3 = 1.396 × 10−22 cm 3 /unit cell 2 Now, solving for a 1/3 (2)(1.396 × 10−22 cm 3 ) √ (3)( 3)(1.593) = 3.23 × 10−8 cm = 0.323 nm  a= A nd finally c = 1.593a = (1.593)(0.323 nm) = 0.515 nm 3.17 In this problem we are given that 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) G iven that the atomic packing factor and atomic radius are 0.547 and 0.177 nm, respectively we are to determine the number of atoms in each unit cell. From the definition of the A PF A PF = VS VC 4 ␲R 3 = n 3abc 3  we may solve for the number of atoms per unit cell, n, as n= = (A PF)abc 4 ␲R 3 3 (0.547)(4.79)(7.25)(9.78)(10−24 cm 3 ) 4 ␲(1.77 × 10−8 cm) 3 3 = 8.0 atoms/unit cell (b) In order to compute the density, we just employ E quation (3.5) as ␳ = = nA I abcN A (8 atoms/unit cell)(126.91 g/mol) [(4.79)(7.25)(9.78) × 10−24 cm 3 /unit cell](6.023 × 1023 atoms/mol) = 4.96 g/cm 3 3.22 This question asks that we generate a three-dimensional unit cell for AuCu 3 using the Molecule D efinition File on the CD -R O M. O ne set of directions that may be used to construct this unit cell and that are entered on the Notepad are as follows: [D isplayProps] R otatez=−30 R otatey=−15 [A tomProps] G old =LtR ed,0.14 Copper=LtYellow,0.13 [BondProps] SingleSolid =LtG ray [A toms] Au1=1,0,0,G old Au2=0,0,0,G old Au3=0,1,0,G old Au4=1,1,0,G old Au5=1,0,1,G old Au6=0,0,1,G old Au7=0,1,1,G old Au8=1,1,1,G old Cu1=0.5,0,0.5,Copper Cu2=0,0.5,0.5,Copper Cu3=0.5,1,0.5,Copper Cu4=1,0.5,0.5,Copper Cu5=0.5,0.5,1,Copper Cu6=0.5,0.5,0,Copper 4 [Bonds] B1=Au1,Au5,SingleSolid B2=Au5,Au6,SingleSolid B3=Au6,Au2,SingleSolid B4=Au2,Au1,SingleSolid B5=Au4,Au8,SingleSolid B6=Au8,Au7,SingleSolid B7=Au7,Au3,SingleSolid B8=Au3,Au4,SingleSolid B9=Au1,Au4,SingleSolid B10=Au8,Au5,SingleSolid B11=Au2,Au3,SingleSolid B12=Au6,Au7,SingleSolid When saving these instructions, the file name that is chosen should end with a period followed by mdf and the entire file name needs to be enclosed within quotation marks. For example, if one wants to name the file AuCu3, the name by which it should be saved is “AuCu3.mdf”. In addition, the file should be saved as a “Text D ocument.” 3.27 In this problem we are asked to show that the minimum cation-to-anion radius ratio for a coordination number of six is 0.414. Below is shown one of the faces of the rock salt crystal structure in which anions and cations just touch along the edges, and also the face diagonals. r r C A H G F From triangle FGH, G F = 2r A and FH = G H = r A + r C Since FGH is a right triangle (G H ) 2 + (FH ) 2 = (FG ) 2 or (r A + r C ) 2 + (r A + r C ) 2 = (2r A ) 2 which leads to 2r A rA + rC = √ 2 5 O r, solving for rC /rA rC rA  2 √ − 1 = 0.414 2 3.29 This problem calls for us to predict crystal structures for several ceramic materials on the basis of ionic charge and ionic radii. = (a) For CsI, from Table 3.4 0.170 nm r Cs+ = = 0.773 r I− 0.220 nm Now, from Table 3.3, the coordination number for each cation (Cs+ ) is eight, and, using Table 3.5, the predicted crystal structure is cesium chloride. (c) For KI, from Table 3.4 r K+ 0.138 nm = = 0.627 r I− 0.220 nm The coordination number is six (Table 3.3), and the predicted crystal structure is sodium chloride (Table 3.5). 3.36 This problem asks that we compute the theoretical density of diamond given that the C ––C distance and bond angle are 0.154 nm and 109.5◦ , respectively. The first thing we need do is to determine the unit cell edge length from the given C ––C distance. The drawing below shows the cubic unit cell with those carbon atoms that bond to one another in one-quarter of the unit cell. a y x From this figure, ␾ is one-half of the bond angle or ␾ = 109.5◦ /2 = 54.75◦ , which means that ␪ = 90◦ − 54.75◦ = 35.25◦ since the triangle shown is a right triangle. A lso, y = 0.154 nm, the carbon-carbon bond distance. Furthermore, x = a/4, and therefore, x= a 4 = y sin ␪ 6 Or a = 4y sin ␪ = (4)(0.154 nm)(sin 35.25◦ ) = 0.356 nm = 3.56 × 10−8 cm The unit cell volume, V C , is just a3 , that is V C = a 3 = (3.56 × 10−8 cm) 3 = 4.51 × 10−23 cm 3 We must now utilize a modified Equation (3.6) since there is only one atom type. There are eight equivalent atoms per unit cell (i.e., one equivalent corner, three equivalent faces, and four interior atoms), and therefore n′A C ␳ =V N C A = (8 atoms/unit cell)(12.01 g/g-atom) (4.51 × 10−23 cm 3 /unit cell)(6.023 × 1023 atoms/g-atom) = 3.54 g/cm 3 The measured density is 3.51 g/cm 3 . 3.39 (a) We are asked to compute the density of CsCl. Modifying the result of Problem 3.4, we get a= 2r Cs+ + 2r Cl− 2(0.170 nm) + 2(0.181 nm) = √ √ 3 3 = 0.405 nm = 4.05 × 10−8 cm From E quation (3.6) ␳= n ′ (A Cs + A Cl) n ′ (A Cs + A Cl) = a3NA VC NA For the CsCl crystal structure, n′ = 1 formula unit/unit cell, and thus ␳ = (1 formula unit/unit cell)(132.91 g/mol + 35.45 g/mol) (4.05 × 10−8 cm) 3 /unit cell(6.023 × 1023 formula units/mol) = 4.20 g/cm 3 (b) This value of the density is greater than the measured density. The reason for this discrepancy is that the ionic radii in Table 3.4, used for this computation, were for a coordination number of six, when, in fact, the coordination number of both Cs+ and Cl− is eight. U nder these circumstances, the actual ionic radii and unit cell volume (V C ) will be slightly greater than calculated values; consequently, the measured density is smaller than the calculated density. 3.45 We are asked in this problem to compute the atomic packing factor for the CsCl crystal structure. This requires that we take the ratio of the sphere volume within the unit cell and the total unit cell volume. From Figure 3.6 there is the equivalent of one Cs and one Cl ion per unit cell; the ionic radii of these two ions are 0.170 nm and 0.181 nm, respectively (Table 3.4). Thus, the sphere volume, V S , 7 is just VS = 4 (␲)[(0.170 nm) 3 + (0.181 nm) 3 ] = 0.0454 nm 3 3 For CsCl the unit cell edge length, a, in terms of the atomic radii is just 2r Cs+ + 2r Cl− 2(0.170 nm) + 2(0.181 nm) = √ √ 3 3 = 0.405 nm a= Since V C = a3 V C = (0.405 nm) 3 = 0.0664 nm 3 A nd, finally the atomic packing factor is just A PF = 0.0454 nm 3 VS = = 0.684 VC 0.0664 nm 3 3.50 (a) We are asked for the indices of the two directions sketched in the figure. 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 are b/2 and c, respectively. This is an [012] direction as indicated in the summary below. Projections Projections in terms of a, b, and c R eduction to integers E nclosure x y z 0a 0 0 b/2 1/2 1 [012] c 1 2 3.51 This problem asks for us to sketch several directions within a cubic unit cell. The [110], [121], and [012] directions are indicated below. _ [012] z __ [121] _ [110] y x 3.53 This problem asks that we determine indices for several directions that have been drawn within a cubic unit cell. D irection B is a [232] direction, the determination of which is summarized as follows. 8 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 Projections Projections in terms of a, b, and c R eduction to integers E nclosure 2a 3 2 3 2 y −b −1 −3 [232] z 2c 3 2 3 2 D irection D is a [136] direction, the determination of which 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 Projections Projections in terms of a, b, and c R eduction to integers E nclosure x y a 6 1 6 b 1 2 1 2 3 [136] z −c −1 −6 3.56 This problem asks that we determine the Miller indices for planes that have been drawn within a unit cell. 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; thus, this is a (112) plane, as summarized below. x y Intercepts −a −b Intercepts in terms of a, b, and c −1 −1 R eciprocals of intercepts E nclosure −1 −1 (112) z c 2 1 2 2 3.58 For plane B we will leave the origin at the unit cell as shown; this is a (221) plane, as summarized below. Intercepts Intercepts in terms of a, b, and c R eciprocals of intercepts E nclosure 9 x y a 2 1 2 2 b 2 1 2 2 (221) z c 1 1 3.59 The (1101) plane in a hexagonal unit cell is shown below. z a a 2 3 a _ (1101) 1 3.60 This problem asks that we specify the Miller indices for planes that have been drawn within hexagonal unit cells. (a) For this plane we will leave the origin of the coordinate system as shown; thus, this is a (1100) plane, as summarized below. Intercepts Intercepts in terms of a’s and c R eciprocals of intercepts E nclosure a1 a2 a3 z a 1 1 −a −1 −1 ∞a ∞ 0 (1100) ∞c ∞ 0 3.61 This problem asks for us to sketch several planes within a cubic unit cell. The (011) and (102) planes are indicated below. z __ (011) y x _ (102) 10