CHAPTER 3 THE STRUCTURE OF CRYSTALLINE SOLIDS PROBLEM SOLUTIONS

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1 HPTER THE STRUTURE OF RYSTLLINE SOLIDS PROBLEM SOLUTIONS Fundamental oncepts.1 What is the difference between atomic structure and crystal structure? tomic 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.

2 Unit ells Metallic rystal Structures.2 If the atomic radius of aluminum is 0.14 nm, calculate the volume of its unit cell in cubic meters. For this problem, we are asked to calculate the volume of a unit cell of aluminum. luminum has an F crystal structure (Table.1). The F unit cell volume may be computed from Equation.4 as V 16R 2 (16)( m) ( 2) m

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/. onsider the B unit cell shown below Using the triangle NOP ( NP) a + a 2a nd then for triangle NPQ, ( NQ) QP) NP ) ( + ( But NQ 4R, R being the atomic radius. lso, QP a. Therefore, (4 R) a + 2a or a 4R

4 .4 For the HP crystal structure, show that the ideal c/a ratio is 1.6. sketch of one-third of an HP unit cell is shown below. onsider 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. nd, since atoms at points J, K, and M, all touch one another, JM JK 2R a where R is the atomic radius. Furthermore, from triangle JHM, (JM ) 2 (JH) 2 (MH) 2 or a 2 (JH ) 2 + c 2 2

5 Now, we can determine the JH length by consideration of triangle JKL, which is an equilateral triangle, and a /2 cos 0 JH 2 JH a Substituting this value for JH in the above expression yields a 2 a 2 + c 2 a2 2 + c2 4 and, solving for c/a c a 8 1.6

6 .5 Show that the atomic packing factor for B is volume, or The atomic packing factor is defined as the ratio of sphere volume to the total unit cell PF V S V Since there are two spheres associated with each unit cell for B V S 4R 8R 2(sphere volume) 2 lso, the unit cell has cubic symmetry, that is V a. But a depends on R according to Equation., and Thus, V 4R 64 R VS 8 R / PF 0.68 V 64 R /

7 .6 Show that the atomic packing factor for HP is The PF is just the total sphere volume-unit cell volume ratio. For HP, there are the equivalent of six spheres per unit cell, and thus V S 4 R 6 8 R 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 DE shown below. The area of DE is just the length of D times the height B. But D is just a or 2R, and B 2 R cos (0 ) 2R 2 Thus, the base area is just RE ()(D)(B) ()(2 R) 2 R 6R 2 2 and since c 1.6a 2R(1.6) V (RE)(c) 6 R 2 c (.S1) (6 R 2 ) (2)(1.6)R 12 (1.6) R

8 Thus, VS 8 R PF 0.74 V 12 (1.6) R

9 Density omputations.7 Iron has a B crystal structure, an atomic radius of nm, and an atomic weight of g/mol. ompute and compare its theoretical density with the experimental value found in the front section of the book. This problem calls for a computation of the density of iron. ccording to Equation.5 n VN Fe For B, n 2 atoms/unit cell, and V 4 R Thus, n 4 R Fe N (2 atoms/unit cell)(55.85 g/mol) 7 2 ( ) ( ) (4) cm / /(unit cell) atoms/mol 7.90 g/cm The value given inside the front cover is 7.87 g/cm.

10 .8 alculate the radius of an iridium atom, given that Ir has an F crystal structure, a density of 22.4 g/cm, and an atomic weight of g/mol. We are asked to determine the radius of an iridium atom, given that Ir has an F crystal structure. For F, n 4 atoms/unit cell, and V 16R 2 (Equation.4). Now, nir V N n Ir ( 16R 2) N nd solving for R from the above expression yields n Ir R 16 N 2 1/ (4 atoms/unit cell) g/mol 2 ( )( )( ) (16) 22.4 g/cm atoms/mol 2 1/ cm 0.16 nm

11 .9 alculate the radius of a vanadium atom, given that V has a B crystal structure, a density of 5.96 g/cm, and an atomic weight of 50.9 g/mol. This problem asks for us to calculate the radius of a vanadium atom. For B, n 2 atoms/unit cell, and V 4 R 64 R Since, from Equation.5 nv V N nv 64 R N and solving for R the previous equation n V R 64 N 1/ and incorporating values of parameters given in the problem statement R ( )(2 atoms/unit cell)(50.9 g/mol) ( )( ) 2 (64) 5.96 g/cm atoms/mol 1/ cm 0.12 nm

12 .10 hypothetical metal has the simple cubic crystal structure shown in Figure.24. If its atomic weight is 70.6 g/mol and the atomic radius is nm, compute its density. For the simple cubic crystal structure, the value of n in Equation.5 is unity since there is only a single atom associated with each unit cell. Furthermore, for the unit cell edge length, a 2R (Figure.24). Therefore, employment of Equation.5 yields n V N n (2 R ) N and incorporating values of the other parameters provided in the problem statement leads to (1atom / unit cell)(70.6 g / mol) 8 2 (2) cm /(unit cell) atoms / mol ( ) ( ) 6.98g / cm

13 .11 Zirconium has an HP 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 1.59, compute the values of c and a. (a) The volume of the Zr unit cell may be computed using Equation.5 as V n N Zr Now, for HP, n 6 atoms/unit cell, and for Zr, Zr g/mol. Thus, V (6 atoms/unit cell)(91.22 g/mol) (6.51 g/cm )( atoms/mol) cm /unit cell m /unit cell (b) From Equation.S1 of the solution to Problem.6, for HP V R c 2 6 But, since a 2R, (i.e., R a/2) then V 2 2 a ac 6 c 2 2 but, since c 1.59a V (1.59) a cm /unit cell Now, solving for a 22 ( ) (2) cm a ()( )(1.59) cm 0.2 nm 1/ nd finally c 1.59a (1.59)(0.2 nm) nm

14 .12 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 Since Pb has an F crystal structure, n 4, and V 16R 2 (Equation.4). lso, R nm ( cm) and Pb g/mol. Employment of Equation.5 yields n V N Pb (4 atoms/unit cell)(207.2 g/mol) ( ) ( ) ( ) 8 2 (16) cm 2 / (unit cell) atoms/mol 11.5 g/cm The value given in the table inside the front cover is 11.5 g/cm. hromium has a B crystal structure for which n 2 and V a 4 R also r 52.00g/mol and R nm. Therefore, employment of Equation.5 leads to (Equation.); ( ) (2 atoms/unit cell)(52.00 g/mol) 8 (4) cm /(unit cell) atoms/mol ( ) 7.18 g/cm The value given in the table is 7.19 g/cm. opper also has an F crystal structure and therefore (4 atoms/unit cell)(6.55 g/mol) 8 2 ( )( ) ( ) (2) cm 2 /(unit cell) atoms/mol 8.90 g/cm The value given in the table is 8.90 g/cm.

15 obalt has an HP crystal structure, and from the solution to Problem.6 (Equation.S1), V R c 2 6 and, since c 1.62a and a 2R, c (1.62)(2R); hence V 6R 2 (1.62)(2R) (19.48)( )R (19.48)( )( cm) cm /unit cell lso, there are 6 atoms/unit cell for HP. Therefore the theoretical density is n V N o (6 atoms/unit cell)(58.9 g/mol) 2 2 ( cm /unit cell)( atoms/mol) 8.91 g/cm The value given in the table is 8.9 g/cm.

16 .1 Rhodium has an atomic radius of nm and a density of g/cm. Determine whether it has an F or B crystal structure. In order to determine whether Rh has an F or a B crystal structure, we need to compute its density for each of the crystal structures. For F, n 4, and a 2R 2 (Equation.1). lso, from Figure 2.6, its atomic weight is g/mol. Thus, for F (employing Equation.5) n a n Rh Rh N ( 2R 2) N (4 atoms/unit cell)( g/mol) ( ) 8 2 ( ) ( ) (2) cm 2 /(unit cell) atoms / mol g/cm which is the value provided in the problem statement. Therefore, Rh has the F crystal structure.

17 .14 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 F, B, or simple cubic and then justify your determination. simple cubic unit cell is shown in Figure.24. lloy tomic Weight Density tomic Radius (g/mol) (g/cm ) (nm) B For each of these three alloys we need, by trial and error, to calculate the density using Equation.5, and compare it to the value cited in the problem. For S, B, and F crystal structures, the respective values of n are 1, 2, and 4, whereas the expressions for a (since V a ) are 2R, 2 R 2, and 4R. For alloy, let us calculate assuming a simple cubic crystal structure. n V N n 2R N (1 atom/unit cell)(77.4 g/mol) (2)( ) /(unit cell) ( atoms/mol) 8.22 g/cm Therefore, its crystal structure is simple cubic. For alloy B, let us calculate assuming an F crystal structure. nb (2 2) R N (4 atoms/unit cell)(107.6 g/mol) 2 2( cm) /(unit cell) ( atoms/mol) 1.42 g/cm

18 Therefore, its crystal structure is F. For alloy, let us calculate assuming a simple cubic crystal structure. n 2R N (1 atom/unit cell)(127. g/mol) 8 2 ( ) ( ) (2) cm /(unit cell) atoms/mol Therefore, its crystal structure is simple cubic. 9.2 g/cm

19 .15 The unit cell for tin has tetragonal symmetry, with a and b lattice parameters of 0.58 and 0.18 nm, respectively. If its density, atomic weight, and atomic radius are 7.0 g/cm, g/mol, and nm, respectively, compute the atomic packing factor. In order to determine the PF for Sn, we need to compute both the unit cell volume (V ) which is just the a 2 c product, as well as the total sphere volume (V S ) 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.5 as n V N Sn ( )( ) (7.0 g/cm )(5.8) (.18) 10 cm atoms / mol g/mol Therefore 4.00 atoms/unit cell 4 (4) R V S PF V a c 2 ( ) ( ) 4 8 (4) ( )( cm) (5.810 cm) (.1810 cm) 0.544

20 .16 Iodine has an orthorhombic unit cell for which the a, b, and c lattice parameters are 0.481, 0.720, and nm, respectively. (a) If the atomic packing factor and atomic radius are and nm, respectively, determine the number of atoms in each unit cell. (b) The atomic weight of iodine is g/mol; compute its theoretical density. (a) For iodine, and from the definition of the PF VS PF V 4 n R abc we may solve for the number of atoms per unit cell, n, as (PF) abc n 4 R Incorporating values of the above parameters provided in the problem state leads to (0.547)( cm)( cm) cm cm 8 ( ) ( ) 8.0 atoms/unit cell (b) In order to compute the density, we just employ Equation.5 as ni abc N (8 atoms/unit cell)( g/mol) ( ) unit cell ( cm)( cm) cm / atoms/mol ( ) 4.96 g/cm

21 .17 Titanium has an HP unit cell for which the ratio of the lattice parameters c/a is If the radius of the Ti atom is nm, (a) determine the unit cell volume, and (b) calculate the density of Ti and compare it with the literature value. (a) We are asked to calculate the unit cell volume for Ti. For HP, from Equation.S1 (found in the solution to Problem.6) V 6 R 2 c But for Ti, c 1.58a, and a 2R, or c.16r, and V (6)(.16) R ( ) 8 2 (6)(.16) cm cm /unit cell (b) The theoretical density of Ti is determined, using Equation.5, as follows: n V N Ti For HP, n 6 atoms/unit cell, and for Ti, Ti g/mol (as noted inside the front cover). Thus, (6 atoms/unit cell)(47.87 g/mol) 2 2 ( cm /unit cell)( atoms/mol) 4.84 g/cm The value given in the literature is 4.51 g/cm.

22 .18 Zinc has an HP crystal structure, a c/a ratio of 1.856, and a density of 7.17 g/cm. ompute the atomic radius for Zn. In order to calculate the atomic radius for Zn, we must use Equation.5, as well as the expression which relates the atomic radius to the unit cell volume for HP; Equation.S1 (from Problem.6) is as follows: V 2 6R c In this case c 1.856a, but, for HP, a 2R, which means that V 6 R 2 (1.856)(2R) (1.856)(12 )R nd from Equation.5, the density is equal to n V N Zn n Zn ( ) (1.856) 12 R N nd, solving for R from the above equation leads to the following: R nzn (1.856) 12 ( ) N 1/ nd incorporating appropriate values for the parameters in this equation leads to (6atoms/unit cell)(65.41g/mol) R 2 (1.856) ( 12 )( 7.17 g/cm )( atoms/mol) 1/ cm 0.1 nm

23 .19 Rhenium has an HP crystal structure, an atomic radius of 0.14 nm, and a c/a ratio of ompute the volume of the unit cell for Re. In order to compute the volume of the unit cell for Re, it is necessary to use Equation.S1 (found in Problem.6), that is V 2 6R c The problem states that c 1.615a, and a 2R. Therefore V (1.615)(12 ) R ( ) (1.615) 12 ( cm ) cm nm 8 2 2

24 rystal Systems.20 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) alculate the density of the material, given that its atomic weight is 145 g/mol. (a) The unit cell shown in the problem statement belongs to the tetragonal crystal system since a b 0.0 nm, c 0.40 nm, and 90. (b) The crystal structure would be called body-centered tetragonal. (c) s with B, n 2 atoms/unit cell. lso, for this unit cell V ( cm) 2 ( cm) cm /unit cell Thus, using Equation.5, the density is equal to n V N (2 atoms/unit cell)(145 g/mol) 2 2 ( cm /unit cell)( atoms/mol) 1.8g / cm

25 .21 Sketch a unit cell for the body-centered orthorhombic crystal structure. unit cell for the body-centered orthorhombic crystal structure is presented below.

26 Point oordinates.22 List the point coordinates for all atoms that are associated with the F unit cell (Figure.1). From Figure.1b, the atom located of the origin of the unit cell has the coordinates 000. oordinates for other atoms in the bottom face are 100, 110, 010, and (The z coordinate for all 2 2 these points is zero.) For the top unit cell face, the coordinates are 001, 101, 111, 011, and 1 2 oordinates for those atoms that are positioned at the centers of both side faces, and centers of both front and back faces need to be specified. For the front and back-center face atoms, the coordinates are and 0 1 1, respectively. While for the left and right side center-face atoms, the 2 2 respective coordinates are and

27 .2 List the point coordinates of the titanium, barium, and oxygen ions for a unit cell of the perovskite crystal structure (Figure 12.6). 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. The oxygen ions are located at all face-centered positions; therefore, their coordinates are , , , , , and nd, finally, the titanium ion resides at the center of the cubic unit cell, with coordinates

28 .24 List the point coordinates of all atoms that are associated with the diamond cubic unit cell (Figure 12.15). 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. nother set of atoms reside on all of the face-centered positions, with the following coordinates: , , , , , and 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.4, the coordinates of the atom that lies toward the lower-leftfront of the unit cell has the coordinates 1 1, whereas the atom situated toward the lower-right-back of the unit cell has coordinates of 1 1. lso, the carbon atom that resides toward the upper-left-back of the unit cell has the 1 1 coordinates. nd, the coordinates of the final atom, located toward the upper-right-front of the unit cell, are

29 .25 Sketch a tetragonal unit cell, and within that cell indicate locations of the and point coordinates. tetragonal unit in which are shown the and point coordinates is presented below. 2 4