CHAPTER 3 THE STRUCTURE OF CRYSTALLINE SOLIDS PROBLEM SOLUTIONS

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1 CHAPTER THE STRUCTURE OF CRYSTALLINE SOLIDS PROBLEM SOLUTIONS Fundamental Concepts.6 Show that the atomic packing factor for HCP is 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 4π R V S 6 8π 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 ACDE shown below. The area of ACDE is just the length of CD times the height BC. But CD is just a or R, and BC R cos (0 ) R

2 Thus, the base area is just AREA ()(CD)(BC) ()( R) R 6R and since c.6a R(.6) V C (AREA)(c) 6 R c (.S) (6 R ) ()(.6)R (.6) R Thus, APF V S V C 8π R (.6) R 0.74

3 Density Computations.8 Calculate the radius of an iridium atom, given that Ir has an FCC crystal structure, a density of.4 g/cm, and an atomic weight of 9. g/mol. 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 C 6R (Equation.4). Now, ρ na Ir V C N A na Ir (6R )N A And solving for R from the above expression yields R na / Ir 6ρN A (4 atoms/unit cell) ( 9. g/mol) / (6)(.4 g/cm )(6.0 0 atoms/mol)( ) cm 0.6 nm

4 . 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.6. Since Pb has an FCC crystal structure, n 4, and V C 6R (Equation.4). Also, R 0.75 nm ( cm) and A Pb 07. g/mol. Employment of Equation.5 yields ρ na Pb V C N A (4 atoms/unit cell)(07. g/mol) {[ (6)( cm) ( ) ]/(unit cell) }(6.0 0 atoms/mol).5 g/cm The value given in the table inside the front cover is.5 g/cm. Chromium has a BCC crystal structure for which n and V C a 4 R (Equation.); also A Cr 5.00g/mol and R 0.5 nm. Therefore, employment of Equation.5 leads to ρ ( atoms/unit cell)(5.00 g/mol) (4)( cm) /(unit cell) (6.0 0 atoms/mol) 7.8 g/cm The value given in the table is 7.9 g/cm. Copper also has an FCC crystal structure and therefore ρ (4 atoms/unit cell)(6.55 g/mol) (6.0 0 atoms/mol) [()( cm)( ) ] /(unit cell) 8.90 g/cm

5 The value given in the table is 8.90 g/cm. Cobalt has an HCP crystal structure, and from the solution to Problem.6 (Equation.S), V C 6R c and, since c.6a and a R, c (.6)(R); hence V C 6R (.6)(R) (9.48)( )R (9.48)( )( cm) cm /unit cell Also, there are 6 atoms/unit cell for HCP. Therefore the theoretical density is ρ na Co V C N A (6 atoms/unit cell)(58.9 g/mol) ( cm /unit cell)(6.0 0 atoms/mol) 8.9 g/cm The value given in the table is 8.9 g/cm.

6 .6 Iodine has an orthorhombic unit cell for which the a, b, and c lattice parameters are 0.479, 0.75, and nm, respectively. (a) If the atomic packing factor and atomic radius are and 0.77 nm, respectively, determine the number of atoms in each unit cell. (b) The atomic weight of iodine is 6.9 g/mol; compute its theoretical density. (a) For indium, and from the definition of the APF APF V S V C n 4 π R abc we may solve for the number of atoms per unit cell, n, as n (APF) abc 4 π R Incorporating values of the above parameters provided in the problem state leads to (0.547)( cm)( cm)( cm) 4 π ( cm) 8.0 atoms/unit cell (b) In order to compute the density, we just employ Equation.5 as ρ na I abc N A (8 atoms/unit cell)(6.9 g/mol) {[ ( cm)( cm)( cm) ]/unit cell}(6.0 0 atoms/mol) 4.96 g/cm

7 Crystal Systems.0 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 4 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 α β γ 90. (b) The crystal structure would be called body-centered tetragonal. (c) As with BCC, n atoms/unit cell. Also, for this unit cell V C ( cm) ( cm).60 0 cm /unit cell Thus, using Equation.5, the density is equal to ρ na V C N A ( atoms/unit cell) (4 g/mol) ( cm /unit cell)(6.0 0 atoms/mol).0 g/cm

8 . Determine the indices for the directions shown in the following cubic unit cell: Direction A is a [0 ]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 Reduction to integers not necessary Enclosure [0 ] Direction B is a [0] 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 0c Projections in terms of a, b, and c 0 Reduction to integers 0 Enclosure [0] Direction C is a [] 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

9 x y z a b Projections c Projections in terms of a, b, and c Reduction to integers Enclosure [] Direction D is a [] 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 Projections in terms of a, b, and c Reduction to integers Enclosure []

10 .4 Determine the Miller indices for the planes shown in the following unit cell: 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; thus, this is a (4) plane, as summarized below. x y z Intercepts a c b Intercepts in terms of a, b, and c Reciprocals of intercepts Reduction 4 Enclosure (4) For plane B we will leave the origin at the unit cell as shown; this is a () plane, as summarized below. x y z a b Intercepts c Intercepts in terms of a, b, and c Reciprocals of intercepts Reduction not necessary Enclosure ()

11 .46 Consider the reduced-sphere unit cell shown in Problem.0, 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) (00), (00), and, ( 00) (b) ( 0), (0), (0 ), and ( 0) (c) ( ), ( ), ( ), and ( ) (a) The unit cell in Problem.0 is body-centered tetragonal. Of the three planes given in the problem statement the (00) and (00) are equivalent that is, have the same atomic packing. The atomic packing for these two planes as well as the (00) are shown in the figure below.

12 (b) Of the four planes cited in the problem statement, ( 0) and ( 0) 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 (0) and (0) are equivalent to each other (but not to the other pair of planes); their atomic arrangement is represented in the other drawing. Note: the 0.44 nm dimension in the left-most drawing comes from the relationship (0.0 nm) + (0.0 nm) [ ] /. Likewise, the nm dimension found in the right-most drawing comes from [(0.0 nm) + (0.40 nm) ] /. (c) All of the ( ), ( ), ( ), and ( ) planes are equivalent, that is, have the same atomic packing as illustrated in the following figure:

13 .47 Here are shown 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? Below is constructed a unit cell using the six crystallographic directions that were provided in the problem.

14 (a) This unit cell belongs to the tetragonal system since a b 0.40 nm, c 0.50 nm, and α β γ 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.

15 R..55 (a) Derive planar density expressions for BCC (00) and (0) planes in terms of the atomic radius (b) Compute and compare planar density values for these same two planes for vanadium. (a) A BCC unit cell within which is drawn a (00) plane is shown below. For this (00) 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 atom associated with this BCC (00) plane. The planar section represented in the above figure is a square, wherein the side lengths are equal to the unit cell edge length, 4 R (Equation.); and, thus, the area of this square is just plane is just 4R 6 R. Hence, the planar density for this (00) PD 00 number of atoms centered on (00) plane area of (00) plane atom 6 R 6 R A BCC unit cell within which is drawn a (0) plane is shown below.

16 For this (0) 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 atoms associated with this BCC (0) 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.) is 4 R. Now, the diagonal length z is equal to 4R. For the triangle bounded by the lengths x, y, and z Or y z x y (4 R) 4R 4 R Thus, in terms of R, the area of this (0) plane is just

17 Area(0) xy 4 R 4 R 6 R And, finally, the planar density for this (0) plane is just PD 0 number of atoms centered on (0) plane area of (0) plane atoms 6 R 8 R (b) From the table inside the front cover, the atomic radius for vanadium is 0. nm. Therefore, the planar density for the (00) plane is PD 00 (V) 6 R 6 (0. nm) 0.76 nm m While for the (0) plane PD 0 (V) 8 R 8 (0. nm) 5. nm m

18 .66 Figure.5 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 0.54 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.. (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 () planes. (b) For each peak, in order to calculate the interplanar spacing we must employ Equation.. For the first peak which occurs at 4.8 d nλ sin θ ()(0.54 nm) () sin nm (c) Employment of Equations.4 and. is necessary for the computation of R for Cu as R a (d hkl) (h) + (k) + (l) (0.067 nm) () + () + () 0.66 nm

19 Similar computations are made for the other peaks which results are tabulated below: Peak Index θ d hkl (nm) R (nm)

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