1. Chapter 3 -1 ISSUES TO ADDRESS... What is the difference in atomic arrangement between crystalline and noncrystalline solids? What features of a metal’s/ceramic’s atomic structure determine its density? Under what circumstances does a material property vary with the measurement direction? Chapter 3: Structures of Metals & Ceramics How do the crystal structures of ceramic materials differ from those for metals?

2. Chapter 3 -2 Non dense, random packing Dense, ordered packing Dense, ordered packed structures tend to have lower energies. Energy and Packing Energy r typical neighbor bond length typical neighbor bond energy Energy r typical neighbor bond length typical neighbor bond energy

3. Chapter 3 -3 atoms pack in periodic, 3D arrays Crystalline materials... -metals -many ceramics -some polymers atoms have no periodic packing Noncrystalline materials... -complex structures -rapid cooling crystalline SiO 2 noncrystalline SiO 2 "Amorphous" = Noncrystalline Adapted from Fig. 3.41(b), Callister & Rethwisch 4e. Adapted from Fig. 3.41(a), Callister & Rethwisch 4e. Materials and Packing SiOxygen typical of: occurs for:

4. Chapter 3 -4 Metallic Crystal Structures How can we stack metal atoms to minimize empty space? 2-dimensions vs. Now stack these 2-D layers to make 3-D structures

5. Chapter 3 -5 Tend to be densely packed. Reasons for dense packing: - Typically, only one element is present, so all atomic radii are the same. - Metallic bonding is not directional. - Nearest neighbor distances tend to be small in order to lower bond energy. - Electron cloud shields cores from each other Metals have the simplest crystal structures. We will examine three such structures... Metallic Crystal Structures

6. Chapter 3 -6 Rare due to low packing density (only Po has this structure) Close-packed directions are cube edges. Coordination # = 6 (# nearest neighbors) Simple Cubic Structure (SC) Click once on image to start animation (Courtesy P.M. Anderson)

7. Chapter 3 -7 APF for a simple cubic structure = 0.52 APF = a 3 4 3  (0.5a) 3 1 atoms unit cell atom volume unit cell volume Atomic Packing Factor (APF) APF = Volume of atoms in unit cell* Volume of unit cell *assume hard spheres Adapted from Fig. 3.43, Callister & Rethwisch 4e. close-packed directions a R=0.5a contains 8 x 1/8 = 1atom/unit cell

8. Chapter 3 -8 Coordination # = 8 Adapted from Fig. 3.2, Callister & Rethwisch 4e. Atoms touch each other along cube diagonals. --Note: All atoms are identical; the center atom is shaded differently only for ease of viewing. Body Centered Cubic Structure (BCC) ex: Cr, W, Fe (  ), Tantalum, Molybdenum 2 atoms/unit cell: 1 center + 8 corners x 1/8 Click once on image to start animation (Courtesy P.M. Anderson)

9. Chapter 3 - VMSE Screenshot – BCC Unit Cell 9

10. Chapter 3 -10 Atomic Packing Factor: BCC APF = 4 3  (3a/4) 3 2 atoms unit cell atom volume a 3 unit cell volume length = 4R = Close-packed directions: 3 a APF for a body-centered cubic structure = 0.68 a R Adapted from Fig. 3.2(a), Callister & Rethwisch 4e. a a 2 a 3

11. Chapter 3 -11 Coordination # = 12 Adapted from Fig. 3.1, Callister & Rethwisch 4e. Atoms touch each other along face diagonals. --Note: All atoms are identical; the face-centered atoms are shaded differently only for ease of viewing. Face Centered Cubic Structure (FCC) ex: Al, Cu, Au, Pb, Ni, Pt, Ag 4 atoms/unit cell: 6 face x 1/2 + 8 corners x 1/8 Click once on image to start animation (Courtesy P.M. Anderson)

12. Chapter 3 -12 APF for a face-centered cubic structure = 0.74 Atomic Packing Factor: FCC maximum achievable APF APF = 4 3  (2a/4) 3 4 atoms unit cell atom volume a 3 unit cell volume Close-packed directions: length = 4R = 2 a Unit cell contains: 6 x 1/2 + 8 x 1/8 =4 atoms/unit cell a 2 a Adapted from Fig. 3.1(a), Callister & Rethwisch 4e.

13. Chapter 3 -13 A sites B B B B B BB C sites C C C A B B ABCABC... Stacking Sequence 2D Projection FCC Unit Cell FCC Stacking Sequence B B B B B BB B sites C C C A C C C A

14. Chapter 3 -14

15. Chapter 3 -15 Coordination # = 12 ABAB... Stacking Sequence APF = 0.74 3D Projection 2D Projection Adapted from Fig. 3.3(a), Callister & Rethwisch 4e. Hexagonal Close-Packed Structure (HCP) 6 atoms/unit cell ex: Cd, Mg, Ti, Zn c/a = 1.633 c a A sites B sites A sites Bottom layer Middle layer Top layer

16. Chapter 3 - VMSE Screenshot – Stacking Sequence and Unit Cell for HCP 16

17. Chapter 3 -17 Theoretical Density,  where n = number of atoms/unit cell A = atomic weight V C = Volume of unit cell = a 3 for cubic N A = Avogadro’s number = 6.022 x 10 23 atoms/mol Density =  = VC NAVC NA n An A  = Cell Unit of VolumeTotal Cell Unit in Atomsof Mass

18. Chapter 3 -18 Ex: Cr (BCC) A = 52.00 g/mol R = 0.125 nm n = 2 atoms/unit cell  theoretical a = 4R/ 3 = 0.2887 nm  actual a R  = a 3 52.002 atoms unit cell mol g unit cell volume atoms mol 6.022 x 10 23 Theoretical Density,  = 7.18 g/cm 3 = 7.19 g/cm 3 Adapted from Fig. 3.2(a), Callister & Rethwisch 4e.

19. Chapter 3 -19 Bonding: -- Can be ionic and/or covalent in character. -- % ionic character increases with difference in electronegativity of atoms. Adapted from Fig. 2.7, Callister & Rethwisch 4e. (Fig. 2.7 is adapted from Linus Pauling, The Nature of the Chemical Bond, 3rd edition. Copyright 1939 and 1940, 3rd edition copyright © 1960 by Cornell University. Degree of ionic character may be large or small: Atomic Bonding in Ceramics SiC: small CaF 2 : large

20. Chapter 3 -20 Ceramic Crystal Structures Oxide structures –oxygen anions larger than metal cations –close packed oxygen in a lattice (usually FCC) –cations fit into interstitial sites among oxygen ions

21. Chapter 3 -21 Factors that Determine Crystal Structure 1. Relative sizes of ions – Formation of stable structures: --maximize the # of oppositely charged ion neighbors. Adapted from Fig. 3.4, Callister & Rethwisch 4e. - - -- + unstable - - - - + stable - - - - + 2. Maintenance of Charge Neutrality : --Net charge in ceramic should be zero. --Reflected in chemical formula: CaF 2 : Ca 2+ cation F - F - anions + A m X p m, p values to achieve charge neutrality

22. Chapter 3 -22 Coordination Number increases with Coordination Number and Ionic Radii Adapted from Table 3.3, Callister & Rethwisch 4e. 2 r cation r anion Coord. Number < 0.155 0.155 - 0.225 0.225 - 0.414 0.414 - 0.732 0.732 - 1.0 3 4 6 8 linear triangular tetrahedral octahedral cubic Adapted from Fig. 3.5, Callister & Rethwisch 4e. Adapted from Fig. 3.6, Callister & Rethwisch 4e. Adapted from Fig. 3.7, Callister & Rethwisch 4e. ZnS (zinc blende) NaCl (sodium chloride) CsCl (cesium chloride) r cation r anion To form a stable structure, how many anions can surround around a cation?

23. Chapter 3 -23 Computation of Minimum Cation-Anion Radius Ratio Determine minimum r cation /r anion for an octahedral site (C.N. = 6) a  2r anion

24. Chapter 3 -24 Bond Hybridization Bond Hybridization is possible when there is significant covalent bonding –hybrid electron orbitals form –For example for SiC X Si = 1.8 and X C = 2.5 ~ 89% covalent bonding Both Si and C prefer sp 3 hybridization Therefore, for SiC, Si atoms occupy tetrahedral sites

25. Chapter 3 -25 On the basis of ionic radii, what crystal structure would you predict for FeO? Answer: based on this ratio, -- coord # = 6 because 0.414 < 0.550 < 0.732 -- crystal structure is NaCl Data from Table 3.4, Callister & Rethwisch 4e. Example Problem: Predicting the Crystal Structure of FeO Ionic radius (nm) 0.053 0.077 0.069 0.100 0.140 0.181 0.133 Cation Anion Al 3+ Fe 2+ 3+ Ca 2+ O 2- Cl - F -

26. Chapter 3 -26 Rock Salt Structure Same concepts can be applied to ionic solids in general. Example: NaCl (rock salt) structure r Na = 0.102 nm r Na /r Cl = 0.564  cations (Na + ) prefer octahedral sites Adapted from Fig. 3.5, Callister & Rethwisch 4e. r Cl = 0.181 nm

27. Chapter 3 -27 MgO and FeO O 2- r O = 0.140 nm Mg 2+ r Mg = 0.072 nm r Mg /r O = 0.514  cations prefer octahedral sites So each Mg 2+ (or Fe 2+ ) has 6 neighbor oxygen atoms Adapted from Fig. 3.5, Callister & Rethwisch 4e. MgO and FeO also have the NaCl structure

28. Chapter 3 -28 AX Crystal Structures Adapted from Fig. 3.6, Callister & Rethwisch 4e. Cesium Chloride structure:  Since 0.732 < 0.939 < 1.0, cubic sites preferred So each Cs + has 8 neighbor Cl - AX–Type Crystal Structures include NaCl, CsCl, and zinc blende

29. Chapter 3 -29 AX 2 Crystal Structures Calcium Fluorite (CaF 2 ) Cations in cubic sites UO 2, ThO 2, ZrO 2, CeO 2 Antifluorite structure – positions of cations and anions reversed Adapted from Fig. 3.8, Callister & Rethwisch 4e. Fluorite structure

30. Chapter 3 -30 ABX 3 Crystal Structures Adapted from Fig. 3.9, Callister & Rethwisch 4e. Perovskite structure Ex: complex oxide BaTiO 3

31. Chapter 3 - VMSE Screenshot – Zinc Blende Unit Cell 31

32. Chapter 3 -32 Density Computations for Ceramics Number of formula units/unit cell Volume of unit cell Avogadro’s number = sum of atomic weights of all anions in formula unit = sum of atomic weights of all cations in formula unit

33. Chapter 3 -33 Densities of Material Classes  metals >  ceramics >  polymers Why? Data from Table B.1, Callister & Rethwisch, 4e.  (g/cm ) 3 Graphite/ Ceramics/ Semicond Metals/ Alloys Composites/ fibers Polymers 1 2 20 30 Based on data in Table B1, Callister *GFRE, CFRE, & AFRE are Glass, Carbon, & Aramid Fiber-Reinforced Epoxy composites (values based on 60% volume fraction of aligned fibers in an epoxy matrix). 10 3 4 5 0.3 0.4 0.5 Magnesium Aluminum Steels Titanium Cu,Ni Tin, Zinc Silver, Mo Tantalum Gold, W Platinum Graphite Silicon Glass-soda Concrete Si nitride Diamond Al oxide Zirconia HDPE, PS PP, LDPE PC PTFE PET PVC Silicone Wood AFRE* CFRE* GFRE* Glass fibers Carbonfibers Aramid fibers Metals have... close-packing (metallic bonding) often large atomic masses Ceramics have... less dense packing often lighter elements Polymers have... low packing density (often amorphous) lighter elements (C,H,O) Composites have... intermediate values In general

34. Chapter 3 -34 Silicate Ceramics Most common elements on earth are Si & O SiO 2 (silica) polymorphic forms are quartz, crystobalite, & tridymite The strong Si-O bonds lead to a high melting temperature (1710ºC) for this material Si 4+ O 2- Adapted from Figs. 3.10-11, Callister & Rethwisch 4e crystobalite

35. Chapter 3 -35 Bonding of adjacent SiO 4 4- accomplished by the sharing of common corners, edges, or faces Silicates Mg 2 SiO 4 Ca 2 MgSi 2 O 7 Adapted from Fig. 3.12, Callister & Rethwisch 4e. Presence of cations such as Ca 2+, Mg 2+, & Al 3+ 1. maintain charge neutrality, and 2. ionically bond SiO 4 4- to one another

36. Chapter 3 -36 Quartz is crystalline SiO 2 : Basic Unit: Glass is noncrystalline (amorphous) Fused silica is SiO 2 to which no impurities have been added Other common glasses contain impurity ions such as Na +, Ca 2+, Al 3+, and B 3+ (soda glass) Adapted from Fig. 3.42, Callister & Rethwisch 4e. Glass Structure Si0 4 tetrahedron 4- Si 4+ O 2- Si 4+ Na + O 2-

37. Chapter 3 -37 Layered Silicates Layered silicates (e.g., clays, mica, talc) –SiO 4 tetrahedra connected together to form 2-D plane A net negative charge is associated with each (Si 2 O 5 ) 2- unit Negative charge balanced by adjacent plane rich in positively charged cations Adapted from Fig. 3.13, Callister & Rethwisch 4e.

38. Chapter 3 -38 Kaolinite clay alternates (Si 2 O 5 ) 2- layer with Al 2 (OH) 4 2+ layer Layered Silicates (cont) Note: Adjacent sheets of this type are loosely bound to one another by van der Waal’s forces. Adapted from Fig. 3.14, Callister & Rethwisch 4e.

39. Chapter 3 -39 Polymorphic Forms of Carbon Diamond –tetrahedral bonding of carbon hardest material known very high thermal conductivity –large single crystals – gem stones –small crystals – used to grind/cut other materials –diamond thin films hard surface coatings – used for cutting tools, medical devices, etc. Adapted from Fig. 3.16, Callister & Rethwisch 4e.

40. Chapter 3 -40 Polymorphic Forms of Carbon (cont) Graphite –layered structure – parallel hexagonal arrays of carbon atoms –weak van der Waal’s forces between layers –planes slide easily over one another -- good lubricant Adapted from Fig. 3.17, Callister & Rethwisch 4e.

41. Chapter 3 -41 Polymorphic Forms of Carbon (cont) Fullerenes and Nanotubes Fullerenes – spherical cluster of 60 carbon atoms, C 60 –Like a soccer ball Carbon nanotubes – sheet of graphite rolled into a tube –Ends capped with fullerene hemispheres Adapted from Figs. 3.18 & 3.19, Callister & Rethwisch 4e.

42. Chapter 3 -42 Some engineering applications require single crystals: Properties of crystalline materials often related to crystal structure. (Courtesy P.M. Anderson) -- Ex: Quartz fractures more easily along some crystal planes than others. -- diamond single crystals for abrasives -- turbine blades Fig. 9.40(c), Callister & Rethwisch 4e. (Fig. 9.40(c) courtesy of Pratt and Whitney). (Courtesy Martin Deakins, GE Superabrasives, Worthington, OH. Used with permission.) Crystals as Building Blocks

43. Chapter 3 -43 Most engineering materials are polycrystals. Nb-Hf-W plate with an electron beam weld. Each "grain" is a single crystal. If grains are randomly oriented, overall component properties are not directional. Grain sizes typ. range from 1 nm to 2 cm (i.e., from a few to millions of atomic layers). Adapted from Fig. K, color inset pages of Callister 5e. (Fig. K is courtesy of Paul E. Danielson, Teledyne Wah Chang Albany) 1 mm Polycrystals Isotropic Anisotropic

44. Chapter 3 -44 Single Crystals -Properties vary with direction: anisotropic. -Example: the modulus of elasticity (E) in BCC iron: Data from Table 3.7, Callister & Rethwisch 4e. (Source of data is R.W. Hertzberg, Deformation and Fracture Mechanics of Engineering Materials, 3rd ed., John Wiley and Sons, 1989.) Polycrystals -Properties may/may not vary with direction. -If grains are randomly oriented: isotropic. (E poly iron = 210 GPa) -If grains are textured, anisotropic. 200  m Adapted from Fig. 5.19(b), Callister & Rethwisch 4e. (Fig. 5.19(b) is courtesy of L.C. Smith and C. Brady, the National Bureau of Standards, Washington, DC [now the National Institute of Standards and Technology, Gaithersburg, MD].) Single vs Polycrystals E (diagonal) = 273 GPa E (edge) = 125 GPa

45. Chapter 3 -45 Polymorphism Two or more distinct crystal structures for the same material (allotropy/polymorphism) titanium ,  -Ti carbon diamond, graphite BCC FCC BCC 1538ºC 1394ºC 912ºC  - Fe  - Fe  - Fe liquid iron system

46. Chapter 3 -46 Fig. 3.20, Callister & Rethwisch 4e. Crystal Systems 7 crystal systems 14 crystal lattices Unit cell: smallest repetitive volume which contains the complete lattice pattern of a crystal. a, b, and c are the lattice constants

47. Chapter 3 -47 Point Coordinates Point coordinates for unit cell center are a/2, b/2, c/2 ½ ½ ½ Point coordinates for unit cell corner are 111 Translation: integer multiple of lattice constants  identical position in another unit cell z x y a b c 000 111 y z 2c2c b b

48. Chapter 3 -48 Crystallographic Directions 1. Vector repositioned (if necessary) to pass through origin. 2. Read off projections in terms of unit cell dimensions a, b, and c 3. Adjust to smallest integer values 4. Enclose in square brackets, no commas [uvw] ex: 1, 0, ½=> 2, 0, 1=> [ 201 ] -1, 1, 1 families of directions z x Algorithm where overbar represents a negative index [ 111 ] => y

49. Chapter 3 - VMSE Screenshot – [101] Direction 49

50. Chapter 3 -50 ex: linear density of Al in [110] direction a = 0.405 nm Linear Density Linear Density of Atoms  LD = a [110] Adapted from Fig. 3.1(a), Callister & Rethwisch 4e. Unit length of direction vector Number of atoms # atoms length 1 3.5 nm a2 2 LD  

51. Chapter 3 -51 Drawing HCP Crystallographic Directions (i) 1. Remove brackets 2. Divide by largest integer so all values are ≤ 1 3. Multiply terms by appropriate unit cell dimension a (for a 1, a 2, and a 3 axes) or c (for z -axis) to produce projections 4. Construct vector by stepping off these projections Algorithm (Miller-Bravais coordinates) Adapted from Figure 3.25, Callister & Rethwisch 4e.

52. Chapter 3 -52 Drawing HCP Crystallographic Directions (ii) Draw the direction in a hexagonal unit cell. [1213] 4. Construct Vector 1. Remove brackets -1 -2 1 3 Algorithm a 1 a 2 a 3 z 2. Divide by 3 3. Projections proceed – a /3 units along a 1 axis to point p –2 a /3 units parallel to a 2 axis to point q a /3 units parallel to a 3 axis to point r c units parallel to z axis to point s p q r s start at point o Adapted from p. 62, Callister & Rethwisch 8e. [1213] direction represented by vector from point o to point s

53. Chapter 3 -53 1. Vector repositioned (if necessary) to pass through origin. 2. Read off projections in terms of three- axis ( a 1, a 2, and z ) unit cell dimensions a and c 3. Adjust to smallest integer values 4. Enclose in square brackets, no commas, for three-axis coordinates 5. Convert to four-axis Miller-Bravais lattice coordinates using equations below: 6. Adjust to smallest integer values and enclose in brackets [uvtw] Adapted from p. 74, Callister & Rethwisch 4e. Algorithm Determination of HCP Crystallographic Directions (ii)

54. Chapter 3 -54 4. Brackets [110] 1. Reposition not needed 2. Projections a a 0 c 1 1 0 3. Reduction 1 1 0 Example a 1 a 2 z 5. Convert to 4-axis parameters 1/3, 1/3, -2/3, 0 => 1, 1, -2, 0 => [ 1120 ] 6. Reduction & Brackets Adapted from p. 74, Callister & Rethwisch 4e. Determination of HCP Crystallographic Directions (ii) Determine indices for green vector

55. Chapter 3 -55 Crystallographic Planes Adapted from Fig. 3.26, Callister & Rethwisch 4e.

56. Chapter 3 -56 Crystallographic Planes Miller Indices: Reciprocals of the (three) axial intercepts for a plane, cleared of fractions & common multiples. All parallel planes have same Miller indices. Algorithm 1. Read off intercepts of plane with axes in terms of a, b, c 2. Take reciprocals of intercepts 3. Reduce to smallest integer values 4. Enclose in parentheses, no commas i.e., (hkl)

57. Chapter 3 -57 Crystallographic Planes z x y a b c 4. Miller Indices (110) examplea b c z x y a b c 4. Miller Indices (100) 1. Intercepts 1 1  2. Reciprocals 1/1 1/1 1/  1 1 0 3. Reduction 1 1 0 1. Intercepts 1/2   2. Reciprocals 1/½ 1/  1/  2 0 0 3. Reduction 2 0 0 examplea b c

58. Chapter 3 -58 Crystallographic Planes z x y a b c 4. Miller Indices (634) example 1. Intercepts 1/2 1 3/4 a b c 2. Reciprocals 1/½ 1/1 1/¾ 21 4/3 3. Reduction 63 4 (001)(010), Family of Planes {hkl} (100),(010),(001),Ex: {100} = (100),

59. Chapter 3 - VMSE Screenshot – Crystallographic Planes 59 Additional practice on indexing crystallographic planes

60. Chapter 3 -60 Crystallographic Planes (HCP) In hexagonal unit cells the same idea is used example a 1 a 2 a 3 c 4. Miller-Bravais Indices(1011) 1. Intercepts 1  1 2. Reciprocals 1 1/  1 0 1 1 3. Reduction1 0 1 a2a2 a3a3 a1a1 z Adapted from Fig. 3.24(b), Callister & Rethwisch 4e.

61. Chapter 3 -61 Crystallographic Planes We want to examine the atomic packing of crystallographic planes Iron foil can be used as a catalyst. The atomic packing of the exposed planes is important. a)Draw (100) and (111) crystallographic planes for Fe. b) Calculate the planar density for each of these planes.

62. Chapter 3 -62 Planar Density of (100) Iron Solution: At T < 912  C iron has the BCC structure. (100) Radius of iron R = 0.1241 nm R 3 34 a  Adapted from Fig. 3.2(c), Callister & Rethwisch 4e. 2D repeat unit = Planar Density = a 2 1 atoms 2D repeat unit = nm 2 atoms 12.1 m2m2 atoms = 1.2 x 10 19 1 2 R 3 34 area 2D repeat unit

63. Chapter 3 -63 Planar Density of (111) Iron Solution (cont): (111) plane 1 atom in plane/ unit surface cell 33 3 2 2 R 3 16 R 3 4 2 a3ah2area           atoms in plane atoms above plane atoms below plane ah 2 3  a 2 2D repeat unit 1 = = nm 2 atoms 7.0 m2m2 atoms 0.70 x 10 19 3 2 R 3 16 Planar Density = atoms 2D repeat unit area 2D repeat unit

64. Chapter 3 - VMSE Screenshot – Atomic Packing – (111) Plane for BCC 64

65. Chapter 3 -65 X-Ray Diffraction Diffraction gratings must have spacings comparable to the wavelength of diffracted radiation. Can’t resolve spacings  Spacing is the distance between parallel planes of atoms.

66. Chapter 3 -66 X-Rays to Determine Crystal Structure X-ray intensity (from detector)   c d d  n 2 sin  c Measurement of critical angle,  c, allows computation of planar spacing, d. Incoming X-rays diffract from crystal planes. Adapted from Fig. 3.38, Callister & Rethwisch 4e. reflections must be in phase for a detectable signal spacing between planes d incoming X-rays outgoing X-rays detector   extra distance travelled by wave “2” “1”“1” “2”“2” “1”“1” “2”“2”

67. Chapter 3 -67 X-Ray Diffraction Pattern Adapted from Fig. 3.40, Callister 4e. (110) (200) (211) z x y a b c Diffraction angle 2  Diffraction pattern for polycrystalline  -iron (BCC) Intensity (relative) z x y a b c z x y a b c

68. Chapter 3 -68 Atoms may assemble into crystalline or amorphous structures. We can predict the density of a material, provided we know the atomic weight, atomic radius, and crystal geometry (e.g., FCC, BCC, HCP). SUMMARY Common metallic crystal structures are FCC, BCC, and HCP. Coordination number and atomic packing factor are the same for both FCC and HCP crystal structures. Crystallographic points, directions and planes are specified in terms of indexing schemes. Crystallographic directions and planes are related to atomic linear densities and planar densities. Ceramic crystal structures are based on: -- maintaining charge neutrality -- cation-anion radii ratios. Interatomic bonding in ceramics is ionic and/or covalent.

69. Chapter 3 -69 Some materials can have more than one crystal structure. This is referred to as polymorphism (or allotropy). SUMMARY Materials can be single crystals or polycrystalline. Material properties generally vary with single crystal orientation (i.e., they are anisotropic), but are generally non-directional (i.e., they are isotropic) in polycrystals with randomly oriented grains. X-ray diffraction is used for crystal structure and interplanar spacing determinations.

70. Chapter 3 -70 Core Problems: Self-help Problems: ANNOUNCEMENTS Reading: