Chap 3 (1).pdf

Chapter 3 Spring 2024/2025 Dr. Zafar Said SREE University of Sharjah 1 The Structure of Crystalline Solids 2 ISSUES TO EXPLORE... What is the difference in atomic arrangement between crystalline and noncrystalline solids? What ar...

Chapter 3 Spring 2024/2025 Dr. Zafar Said SREE University of Sharjah 1 The Structure of Crystalline Solids 2 ISSUES TO EXPLORE... What is the difference in atomic arrangement between crystalline and noncrystalline solids? What are the crystal structures of metals? What are the characteristics of crystal structures? What characteristics of a material’s atomic structure determine its density? 3 From Individual Atoms and Atomic Bonding to Atomic Arrangements We’ve seen: Electronic structure of atoms → Periodic table → Atomic bonding We will see: 1 Atom arrangements in solid state 2 Volume of unit cell, atomic density in a unit cell 3 Miller indices 4 Bragg’s Law to convert between diffraction angle and interplanar spac- ing 5 Experimental identification of crystalline structures from X-ray diffrac- tion patterns 6 Defects in solids 4 Definitions Definition A crystalline material is one in which the atoms are situated in a three-dimensional repeating or periodic array (order) over large atomic distances, in which each atom is bonded to its nearest-neighbor atoms. A material that does not crystallize (disorder) is called noncrystalline or amorphous. The crystalline structure of a material is a description of the spatial arrangement of its atoms, ions, or molecules. Sometimes the term lattice is used in the context of crystal structures; in this sense lattice means a three-dimensional array of points coinciding with atom positions (or sphere centers). Small repeat entities in a structure are called unit cells. 5 6 Materials and Atomic Arrangements Crystalline materials... atoms arranged in periodic, 3D arrays typical of: -metals -many ceramics -some polymers crystalline SiO2 Adapted from Fig. 3.24(a), Callister & Rethwisch 10e. Si Oxygen Noncrystalline materials... atoms have no periodic arrangement occurs for: -complex structures -rapid cooling "Amorphous" = Noncrystalline noncrystalline SiO2 Adapted from Fig. 3.24(b), Callister & Rethwisch 10e. 7  UNIT CELLS Small repeat entities in a structure are called unit cells. The unit cell is chosen to represent the symmetry of the crystal structure. Thus, the unit cell is the basic structural unit or building block of the crystal structure and defines the crystal structure by virtue of its geometry and the atom positions within. unit cell 8  METALLIC CRYSTAL STRUCTURES Three relatively simple crystal structures are found for most of the common metals: 1. Face-centered cubic 2. Body-centered cubic 3. Hexagonal close-packed. 9  Definitions Coordination Number Coordination Number = number of nearest-neighbor or touching atoms Atomic Packing Factor (APF) Volume of atoms in unit cell* APF = Volume of unit cell *assume hard spheres 10  Simple Cubic (SC) Crystal Structure Centers of atoms located at the eight corners of a cube Rare due to low packing density (only Po has this structure) Close-packed directions are cube edges. ex: Po (polonium) Coordination # = ? (# nearest or touching neighbors) Adapted from Fig. 3.3, Callister & Rethwisch 10e. # atoms/unit cell: ? 11 12  Simple Cubic (SC) Crystal Structure Centers of atoms located at the eight corners of a cube Rare due to low packing density (only Po has this structure) Close-packed directions are cube edges. ex: Po (polonium) Coordination # = 6 (# nearest or touching neighbors) Adapted from Fig. 3.3, Callister & Rethwisch 10e. 1 atoms/unit cell: 8 corners x 1/8 13 Now we need to count how many atoms are in each unit cell. It may look like there are 8 atoms because there are 8 corners, but actually the cell only intersects 1/8th of each atom. 8.1/8=1, so there is one atom per unit cell. 14  Atomic Packing Factor (APF) for Simple Cubic Crystal Structure a = cube edge length R = atomic radius volume atoms atom 4 3 2R a unit cell 1 π (0.5a) 3 R = 0.5a APF = = 0.52 a3 volume unit cell close-packed directions About 52% of the volume of the cube is occupied by the atoms or Unit cell contains 1 atom = 8 x 1/8 = 1 atom/unit cell empty space is 48%. Volume of atoms in unit cell* APF = Volume of unit cell 15 Body-Centered Cubic Structure (BCC) Atoms located at 8 cube corners with a single atom at cube center. --Note: All atoms in the animation are identical; the center atom is shaded differently for ease of viewing. ex: Cr, W, Fe (g), Ta, Mo Coordination # = ? Adapted from Fig. 3.2, Callister & Rethwisch 10e. # atoms/unit cell: ? 16 17 18 Body-Centered Cubic Structure (BCC) Atoms located at 8 cube corners with a single atom at cube center. --Note: All atoms in the animation are identical; the center atom is shaded differently for ease of viewing. ex: Cr, W, Fe (g), Ta, Mo Coordination # = 8 Adapted from Fig. 3.2, Callister & Rethwisch 10e. 2 atoms/unit cell: 1 center + (8 corners x 1/8) 19 VMSE Screenshot – BCC Unit Cell 20  Atomic Packing Factor: BCC APF for the body-centered cubic structure = 0.68 R 4R = 3a 2R a R a 2a For close-packed directions R R= 3 a/4 a a volume atoms 4 atom unit cell 2 π ( 3 a/4 ) 3 3 APF = volume a3 unit cell 21 22 Face-Centered Cubic Structure (FCC) Atoms located at 8 cube corners and at the centers of the 6 faces. --Note: All atoms in the animation are identical; the face-centered atoms are shaded differently for ease of viewing. ex: Al, Cu, Au, Pb, Ni, Pt, Ag Coordination # = ? Adapted from Fig. 3.1, Callister & Rethwisch 10e. # atoms/unit cell: ? 23 24 25 Face-Centered Cubic Structure (FCC) Atoms located at 8 cube corners and at the centers of the 6 faces. --Note: All atoms in the animation are identical; the face-centered atoms are shaded differently for ease of viewing. ex: Al, Cu, Au, Pb, Ni, Pt, Ag Coordination # = 12 Adapted from Fig. 3.1, Callister & Rethwisch 10e. 4 atoms/unit cell: (6 face x 1/2) + (8 corners x 1/8) 26  Atomic Packing Factor: FCC APF for the face-centered cubic structure = 0.74 maximum achievable APF For close-packed directions: æ 2a ö 4R = 2a çi.e., R = ÷ 2a ç 4 ÷ è ø Unit cell contains: (6 x ½) + (8 x 1/8) a = 4 atoms/unit cell volume atoms 4 π ( 2 a/4 ) 3 atom unit cell 4 3 APF = = 0.74 a3 volume unit cell 27 28 Adapted from Callister & Rethwisch 10e. 29 Adapted from Callister & Rethwisch 10e. 30  FCC Plane Stacking Sequence ABCABC... Stacking Sequence–Close-Packed Planes of Atoms 2D Projection B B C A A sites B B B C C B sites B B C sites Stacking Sequence referenced to an FCC A Close-Packed Unit Cell, showing (111) plane B Plane C 31  FCC Plane Stacking Sequence ABCABC... Stacking Sequence–Close-Packed Planes of Atoms The triangle shows a (111) plane 32 Hexagonal Close-Packed Structure (HCP) Adapted from Fig. 3.4, Callister & Rethwisch 10e. 33 34 Close-packed Structures: ABAB vs. ABCABC H G F E J D B C A Figure 2.3: The extended unit cell (repeated entity) of the HCP structure, adapted from. The 3 interior atoms in the mid-plane of the cell are completely enclosed, Each of the 2 center face atoms (one from each of the top and bottom faces) are half-shared with once HCP unit cell, either above or below, Each of the 12 remaining face corner atoms are shared with 6 HCP unit cells. ⇒ Total = 3 ×1 + 2 ×1/ 2 + 12 ×1/ 6 = 6 atoms in unit cell 35 36 Hexagonal Close-Packed Structure (HCP) ABAB... Stacking Sequence–Close-Packed Planes of Atoms 3D Projection 2D Projection A sites Top layer c B sites Middle layer A sites Bottom layer a Coordination # = ? # atoms/unit cell = ? APF = 0.74 ex: Cd, Mg, Ti, Zn Ideal c/a = 1.633 37 Hexagonal Close-Packed Structure (HCP) ABAB... Stacking Sequence–Close-Packed Planes of Atoms 3D Projection 2D Projection A sites Top layer c B sites Middle layer A sites Bottom layer a Coordination # = 12 # atoms/unit cell = 6 APF = 0.74 ex: Cd, Mg, Ti, Zn Ideal c/a = 1.633 If a and c represent, respectively, the short and long unit cell dimensions 38 https://www.youtube.com/watch?v=_OhQv5gsoSQ Hexagonal Close-Packed Structure (HCP) ABAB... Stacking Sequence–Close-Packed Planes of Atoms 39 Determination of BCC Unit Cell Volume Molybdenum (Mo) has a BCC structure with an atomic radius of 1.36 Å. Calculate the lattice parameter for BCC Mo. body diagonal in BCC: H b G E E G a 2 (4r) 2 2 √( a2 a2 ) = 3a 2 a = a + + C A a A C a a 2 B Therefore Figure 2.6: BCC unit cell. a = √4r= 3.14 Å Molybdenum atoms touch along the 3 40 Theoretical Density for Metals Density = ρ = Mass of Atoms in Unit Cell = Total Volume of Unit Cell ρ= nA nA ρ= VC NA VC NA where n = number of atoms/unit cell A = atomic weight VC = Volume of unit cell = a3 for cubic NA = Avogadro’s number = 6.022 x 1023 atoms/mol 41 Theoretical Density for Chromium Cr has BCC crystal structure A = 52.00 g/mol R = 0.125 nm n = ? atoms/unit cell a = 4R/ 3 = 0.2887 nm R a VC = a3 = ? cm3 n A ρ= = =? VC NA 42 Theoretical Density for Chromium Cr has BCC crystal structure A = 52.00 g/mol R = 0.125 nm n = 2 atoms/unit cell a = 4R/ 3 = 0.2887 nm R a VC = a3 = 2.406 x 10-23 cm3 atoms g unit cell mol n A 2 52.00 ρ= = = 7.19 g/cm3 VC NA 2.406 x 10-23 6.022 x 1023 atoms volume mol unit cell ρactual = 7.15 g/cm3 43  SEE EXAMPLE PROBLEM 3.4 Theoretical Density Copper has an atomic radius of 0.128 nm, an FCC crystal structure, and an atomic weight of 63.5 g/mol. Compute its theoretical density and compare the answer with its measured density (8.94 g/cm3). From the crystal structure of a metal- volume of the unit cell, and Avo- lic solid we can compute its theoreti- gadro’s number, respectively. cal density: ρ= nA ρ= 4× 63.5 16 √2(0.128·10 −7 ) 3 ×6.022·10 23 VC N A = 8.88 g/cm3 where n, A , VC , and N represent the number of atoms associated with Very close to the measured density of each unit cell, the atomic weight, and the value. 44  Densities Comparison for Four Material Types In general Graphite/ ρ ρ ρ Metals/ Composites/ Ceramics/ Polymers metals > ceramics > polymers Alloys Semicond fibers 30 Why? Based on data in Table B1, Callister 20 Platinum *GFRE, CFRE, & AFRE are Glass, Gold, W Metals have... Tantalum Carbon, & Aramid Fiber-Reinforced Epoxy composites (values based on close-packing 60% volume fraction of aligned fibers 10 Silver, Mo in an epoxy matrix). (metallic bonding) Cu,Ni Steels often large atomic masses Tin, Zinc Zirconia (g/cm3 ) 5 Ceramics have... 4 Titanium Al oxide often lighter elements 3 Diamond Si nitride Aluminum Glass -soda Glass fibers Polymers have... 2 Concrete Silicon PTFE GFRE* Carbon fibers low packing density Magnesium ρ Graphite CFRE* Silicone Aramid fibers PVC AFRE* (often amorphous) 1 PET PS lighter elements (C,H,O) PE Composites have... 0.5 moderate to low densities 0.4 Wood 0.3 45 46 Crystal Systems 47 Table 1: Crystal structures for some common metals (mainly FCC, BCC, and HCP), adapted from. Atomic Atomic Crystal Radiusb Crystal Radius Metal Structurea (nm) Metal Structure (nm) Aluminum FCC 0.1431 Molybdenum BCC 0.1363 Cadmium HCP 0.1490 Nickel FCC 0.1246 Chromium BCC 0.1249 Platinum FCC 0.1387 Cobalt HCP 0.1253 Silver FCC 0.1445 Copper FCC 0.1278 Tantalum BCC 0.1430 Gold FCC 0.1442 Titanium (α) HCP 0.1445 Iron (α) BCC 0.1241 Tungsten BCC 0.1371 Lead FCC 0.1750 Zinc HCP 0.1332 a FCC = face-centered cubic; HCP = hexagonal close-packed; BCC = body-centered cubic. 48  1 2 H He Struct. hcp fcc ρ, g/cc n/a n/a a, A˚ 4.70 4.24 3 4 5 6 7 8 9 10 Li Be B C N O F Ne Struct. bcc hcp rhom diam hcp mon mon fcc ρ, g/cc 0.533 1.85 2.47 3.51 n/a n/a n/a n/a a, A˚ 3.51 2.29 5.06 3.56 3.86 5.40 5.50 4.42 11 12 13 14 15 16 17 18 Na Mg Al Si P S Cl Ar Struct. bcc hcp fcc diam tricl orth orth fcc ρ, g/cc 0.966 1.74 2.70 2.33 1.82 2.09 n/a n/a a, A˚ 4.29 3.21 4.05 5.43 11.5 10.4 6.22 5.26 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Struct. bcc fcc hcp hcp bcc bcc cubic bcc hcp fcc fcc hcp orth diam rhom mon orth fcc ρ, g/cc 0.862 1.53 2.99 4.51 6.09 7.19 7.47 7.87 8.8 8.91 8.93 7.13 5.91 5.32 5.78 4.81 n/a n/a a, A˚ 5.33 5.56 3.31 2.95 3.03 2.88 8.91 2.86 2.51 3.52 3.61 2.66 4.52 5.65 3.76 9.05 6.72 5.71 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe Struct. bcc fcc hcp hcp bcc bcc hcp hcp fcc fcc fcc hcp tetrag tetrag trig trig orth fcc ρ, g/cc 1.53 2.58 4.48 6.51 8.58 10.22 11.5 12.36 12.42 12.0 10.5 8.65 7.29 7.29 6.69 6.25 4.95 n/a a, A˚ 5.59 6.08 3.65 3.61 3.30 3.14 2.74 2.71 3.80 3.88 4.08 2.98 3.25 5.83 4.31 4.46 7.18 6.20 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Struct. bcc bcc hcp hcp bcc bcc hcp hcp fcc fcc fcc rhom hcp fcc rhom cub n/a n/a ρ, g/cc 1.91 3.59 6.17 13.3 16.7 19.3 21.0 22.58 22.55 21.4 19.28 n/a 11.87 11.34 9.80 9.2 n/a n/a a, A˚ 6.14 5.01 3.77 3.20 3.30 3.17 2.76 2.73 3.84 3.92 4.07 3.01 3.46 4.95 4.74 3.36 n/a n/a 49 Single vs. Polycrystalline Materials I Most crystalline solids are composed of many small crystals: polycrystalline. (a) (b) (c) (d) Figure 4.1: Stages of solidification of a polycrystalline material; the square grids depict unit cells. (a) Small crystallite nuclei. (b) Growth of the crystallites. (c) Upon completion of solidification, grains having irregular shapes have formed. (d) The grain structure as it would appear under the microscope; dark lines are the grain boundaries.. 50 Single vs. Polycrystalline Materials II When the periodic arrangement of atoms is perfect4 throughout the whole material specimen, the result is a single crystal (see mono-Si in figure 4.2). Figure 4.2: Comparison solar cell poly-Si (see grain boundaries separating grains with different crys- tallographic orientation) vs mono-Si, from Creative Commons.. 51 Single Crystals When the periodic arrangement of atoms (crystal structure) extends without interruption throughout the entire specimen. -- diamond single -- single crystal for crystals for abrasives turbine blade (Courtesy Martin Deakins, GE Superabrasives, Worthington, OH. Used with permission.) -- Quartz single crystal (Courtesy P.M. Anderson) Fig. 8.35(c), Callister & Rethwisch 10e. (courtesy of Pratt and Whitney) 52  The World's Largest Crystals: Mexico's Cave of the Giant Selenite Crystals https://interestingengineering.com/science/the-worlds-largest-crystals-mexicos-cave-of-the-giant-selenite-crystals 53 Polycrystalline Materials Most engineering materials are composed of many small, single crystals (i.e., are polycrystalline). large grain Courtesy of Paul E. Danielson, Teledyne Wah 1 mm Chang Albany small grain Nb-Hf-W plate with an electron beam weld. Each "grain" is a single crystal. Grain sizes typically range from 1 nm to 2 cm (i.e., from a few to millions of atomic layers). 54 Single vs. Polycrystalline Materials III The physical properties (electric conductivity, index of reflection, elastic modulus, etc.) of single crystals of some substances depend on the crystal- lographic direction (atomic and ionic spacing) in which measurements are taken: anisotropy (opposite to isotropy). Anisotropy is important in low-symmetry crystals (triclinic are highly anisotropic). Table 3: Modulus of elasticity values (GPa) for several metals at various crystallographic orientations. Metal G G G Al 63.7 72.6 76.1 Cu 66.7 130.3 191.1 W 384.6 384.6 384.6 55 Single vs. Polycrystalline Materials IV Remark Some materials may exist in more than one crystal structure, this is called polymorphism. If the material is an elemental solid, it is called allotropy. Figure 4.3: Carbon allotropes. 4 All unit are oriented the same way 56 Anisotropy Anisotropy — Property value depends on crystallographic direction of measurement. - Observed in single crystals. E (diagonal) = 273 GPa - Example: modulus of elasticity (E) in BCC iron E(edge) ≠ E(diagonal) E (edge) = 125 GPa Unit cell of BCC iron 57 Isotropy Polycrystals 200 μm - Properties may/may not vary with direction. - If grains randomly oriented: properties are isotropic. (Epoly iron = 210 GPa) - If grains textured (e.g., deformed grains have preferential crystallographic orientation): properties are anisotropic. Fig. 4.15(b), Callister & Rethwisch 10e. [Fig. 4.15(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).] 58 Polymorphism/Allotropy Some metals, as well as nonmetals, may have more than one crystal structure, a phenomenon known as polymorphism. When found in elemental solids, the condition is often termed allotropy. The prevailing crystal structure depends on both the temperature and the external pressure. One familiar example is found in carbon: graphite is the stable polymorph at ambient conditions, whereas diamond is formed at extremely high pressures. Also, pure iron has a BCC crystal structure at room temperature, which changes to FCC iron at 912°C. Most often a modification of the density and other physical properties accompanies a polymorphic transformation. 59 Polymorphism/Allotropy Two or more distinct crystal structures for the same material (allotropy/polymorphism) Iron system Titanium: α or β forms T liquid Carbon: 1538°C Temperature diamond, graphite δ -Fe BCC 1394°C γ -Fe FCC 912°C α -Fe BCC 60  Polymorphism/Allotropy Remark Table 2: Some metal allotropes. Allotropy: change of crystal R.T. Crystal Structure at structures depending on temperature Metal Structure Other Temperatures and pressure (see table 2): Ca FCC BCC (>447◦C) Co HCP FCC (>427◦C) Carbon: graphite at ambient Hf HCP BCC (>1742◦C) conditions, diamond at high pres- Fe BCC FCC (>912◦C) BCC (>1394◦C) sures and temperature, Li BCC BCC (< −193 ◦C) Na BCC BCC (< −233 ◦C) Pure iron: BCC at room temper- Sn BCT Cubic (234◦C) ature, FCC at 912 ℃ and back to Ti HCP BCC (>883◦C) BCC (>1481◦C) BCC at 1394 ℃, and Y Zr HCP HCP BCC (>872◦C) titanium: HCP to BCC at 882 ℃. 61 62 Point Coordinates A point coordinate is a lattice position in a unit cell Determined as fractional multiples of a, b, and c unit cell edge lengths Example: Unit cell upper corner 1. Lattice position is z 111 a, b, c a, b, c c 011 2. Divide by unit cell edge lengths (a, b, and c) and remove commas 000 y a b x 3. Point coordinates for unit cell corner are 111 63 Point Coordinates 64 Common Crystallographic Directions Adapted from Fig. 3.7, Callister & Rethwisch 10e. 65 Some Working Examples I Example Specify point coordinates for all atom positions for a BCC unit cell (see figure. 3.1) z a Point number Point coordinates a 9 6 8 1 000 7 5 2 100 a 3 110 y 1 4 4 010 2 3 x 5 ?? Figure 3.1 66 Some Working Examples I 67 Point Coordinates: FCC 68 Some Working Examples II Example Locate the point 1 11 4 2 in figure. 3.2 z z 0.46 nm 1 11 4 2 P 0.40 nm 0.20 nm y 0.12 nm M y N 0.46 nm O x x Figure 3.2 Figure 3.3 69 Linear Density of Atoms (LD) Planar Density of Atoms (PD) 70 Linear Density of Atoms (LD) LD = number of atoms centered on direction vector length of direction vector ex: linear density of Al in direction (FCC structure) There are 2 half atoms and 1 full atom = 2 atoms centered on vector # atoms LD = = = ? nm-1 a length a = 0.405 nm 71 Linear Density of Atoms (LD) LD = number of atoms centered on direction vector length of direction vector ex: linear density of Al in direction (FCC structure) There are 2 half atoms and 1 full atom = 2 atoms centered on vector # atoms 2 2 LD = = = 3.5 nm-1 a 2a 2 (0.405 nm) length a = 0.405 nm 72  Planar Density of Atoms (PD) PD = number of atoms centered on a plane area of plane 2D repeat unit ex: planar density of (100) plane of BCC Fe There are 4 quarter atoms 4 a= R = 1 atom centered on plane 3 4 4 a= R = (0.1241 nm) = 0.287 nm 3 3 # atoms Radius of iron, atoms R = 0.1241 nm PD = = = ? nm2 area 73  Planar Density of Atoms (PD) PD = number of atoms centered on a plane area of plane 2D repeat unit ex: planar density of (100) plane of BCC Fe There are 4 quarter atoms 4 a= R = 1 atom centered on plane 3 4 4 a= R = (0.1241 nm) = 0.287 nm 3 3 # atoms Radius of iron, 1 1 atom R = 0.1241 nm PD = = = 12.1 atoms a2 (0.287 nm)2 nm2 area 74 Miller indices (hkl) 75 Miller indices (hkl) The orientations of planes for crystal structures are represented again using the unit cell and the (xyz) coordinate system. Crystallographic planes are specified by the three Miller indices (hkl) (expect for hexagonal systems) and are determined as follows: 1 Find intercepts3 along three axes of the crystal system. 2 Take the reciprocals of the intercepts in terms of a, b, and c. 3 Multiply reciprocals by the smallest integer necessary to convert them into a set of integers. 4 Enclose resulting integers in parentheses, (hkl). 76 Miller indices (hkl) The equation of an arbitrary plane with intercepts A, B , and C , relative to the lattice parameters is given by: 1x 1y 1𝑥 + + = 1 (4) A a B b C c We designate a plane by Miller indices, h, k, and l, which are simply the reciprocals of the intercepts, A, B , and C: 1 1 1 = (hkl) (5) AB C 3 Planes should have nonzero intercepts. If the plane passes through the selected origin, either another parallel plane must be constructed within the unit cell by an appropriate translation, or a new origin must be established at the corner of another unit cell. 77 Miller indices (hkl) 78 Miller indices (hkl) 79 Miller indices (hkl) 80 81 82 83 84 The plane ABC’ is then said to be (1 1 2). 85 Atomic Arrangements I Atomic arrangement/packing for a crystallographic planes depend on the crystal structure. A family of plane contains all planes with the same atomic packing (see figures 3.4 and 3.5) such as {111} family in cubic crystals that would contain: (111), (¯1¯1¯1),(¯111), (11¯1¯), (111¯), (¯11¯1), (1¯11¯), and (1¯11) C A B A B C D E F E F D Figure 3.4: Atomic packing of an FCC (110) plane, adapted from. 86 Atomic Arrangements II B’ A’ A’ B’ C’ C’ D’ E’ E’ D’ Figure 3.5: Atomic packing of an BCC (110) plane, adapted from. 87 Some Working Examples I Example Determine the indices for the direction shown in figure 3.5 z x y z Projection on x axis (a/2) Projection on Projections a/2 b 0 y axis (b) c y In terms of a, 1/2 1 0 a b and c b Reduction 1 2 0 x Figure 3.5 ⇒ Direction: 88 Some Working Examples II Example Draw a [1¯10] direction within a cubic unit cell. The overbar designate a negative direction. z 1 Draw unit cell (cubic, for exam- ple) + coordinate system a 2 Draw projections a, −a and 0 a O on x, y and 𝑥 respectively –y +y Direction a 3 Draw vector passing by 0 and P –a a a point P x Figure 3.6 89 X-Ray Diffraction I A lot of what we know, and what is being developed about atomic and molecular arrangements is a result of X-ray diffraction investigations. Figure 5.1: Diffraction phenomena. 90 X-Ray Diffraction II O Scattering Wave 1 Wave 1' event λ λ λ A A 2A Amplitud λ λ + e A A Wave 2 Wave 2' O' Position (a) P Scattering Wave 3 Wave 3' event λ λ A A Amplitud λ + e A A λ Wave 4 P' Wave 4' Position (b) Figure 5.2: Constructive vs. destructive interference of waves after a scattering.. 91 X-Ray Diffraction III To diffract light, the diffraction grating spacing must be comparable to the light wavelength. X-rays are diffracted by planes of atoms. Interplanar spacing is the distance between parallel planes of atoms. 92 X-ray Diffraction IV X-rays are a form of electromagnetic radiation that have high energies and short wavelengths (on the order of the atomic spacings for solids). The interplanar spacing, d, between adjacent planes (same hkl Miller in- dices) allows us to use XRD for structural determination. 1 1' Incident Diffracted beam beam 2 2' A θ P θ A' Sinq = SQ/PQ S θ T dhkl = SQ/dhkl B Q B' SQ=dhkl Sinq Figure 5.3: Diffraction of monochromatic and coherent X-ray beams by planes of atoms ( A A ' ) and (BB'). 93  X-Rays to Determine Crystal Structure Crystallographic planes diffract incoming X-rays reflections must be in phase for a detectable signal, extra λ nq = 2d sinq distance travelled θ θ by wave “2” spacing = 2 (d sinq) d between planes Measurement of X-ray nλ diffraction angle, qc, intensity d= (measured 2 sin θc allows computation of interplanar spacing, d. By detector) Diffraction occurs when q = qc θ θc 94 X-ray Diffraction V Constructive interference of scattered rays 1 ' and 2 ' at angle θ if : n λ = S Q + QT = dh k l sin θ + dh k l sin θ = 2dh k l sin θ (6) where S Q + QT is the difference of distance covered by beam 2 vs beam 1 (SQ) and difference of distance covered by beam 2 ' vs 1 ' , and n an integer. This is know as Bragg’s law relating the X-ray wavelength and interatomic spacing, to the angle of the diffracted beam. If Bragg’s law is not satisfied, then the interference will be non-constructive so as to yield a very low-intensity diffracted beam. 95 X-ray Diffraction VI Remark Bragg’s law says that: if we bombard a crystal lattice with X-rays of a known wavelength, λ, and at a known diffraction angle θ, we will be able to detect diffracted X-rays of various intensities that represent a specific interplanar spacing, d, in the lattice. The result of scanning through different angles at a fixed wavelength and counting the number (intensity) of diffracted X-rays is a diffraction pattern. 96  X-Ray Diffraction Pattern z z z c c c (110) y plane y y a b a b a b Intensity (relative) x x x (211) plane (200) plane Diffraction angle 2θ Diffraction pattern for polycrystalline α-iron (BCC) 97 X-ray Diffraction VII Table 4: D-spacings and Miller indices for some crystal symmetries. Symmetry D-spacing relation 1 h2 + k2 + l 2 Cubic d2 = a2 1 h 2+ k 2 l2 Tetragonal d2 = a2 + c2 1 4 h 2+ h k + k 2 l2 Hexagonal d2 = 3 a2 + c2 1 h2 k2 l2 Orthorhombic d2 = a2 + b2 + c2 1 1 h2 k 2 sin 2β l2 2h l cos β Monoclinic d2 = sin 2 β a2 + b2 + c2 − ac 98 X-ray Diffraction VIII Excerpt from published paper: "The N i O (200) corresponding XRD patterns obtained from the as-synthesized powder are N i O (111) shown in figure 5.4. Sharp, well- I n t e n s i t y / a.u. defined reflection peaks, with strong N i O (220) intensities indicating the high crys- tallinity of the analyzed powder, are N i O (311) N i O (222) perfectly indexed with the peaks of the PDF-4+ (ICDD) 04-011-8441 file. These peaks correspond to the crystal 30 40 50 60 2θ / d e g. 70 80 hkl planes (111), (200), (220), (311) and (222) of standard cubic NiO (bun- Figure 5.4: X-ray diffraction patterns of senite, NaCl type structure)." NiO nanomaterials.. 99 X-ray Diffraction IX Example For BCC iron, compute 1 the interplanar spacing and 2 the diffraction angle for the (220) set of planes. The lattice parameter for Fe is 0.2866 nm. Also, assume that monochromatic radiation having a wavelength of 0.1790 nm is used, and the order of reflection is 1. 100 X-ray Diffraction X From table 4, the d-spacing is giving The angle θ would be obtained from: by: 1 h2 + k 2 + l 2 = nλ 1 ×0.1790 d2 a2 sin θ = = = 0.884 2d 2 ×0.1013 With a = 0.2866 nm, h = 2, k = 2 and l = 0 0.2866 Diffraction angle 2θ is therefore 2 × d= √ = 0.1013 nm sin −1 (0.884) = 124.26◦ 2 2 2 + 2 + 0 2 101 X-ray Diffraction XI Example Figure 5.5 shows and X-ray diffraction pattern for FCC Fe taken using a diffractometer and monochromatic x-radiation having a wavelength of 0.1542 nm. Each diffraction peak on the pattern has been indexed with its Miller plane. Compute the interplanar spacing for each set of planes indexed; and determine the lattice parameter of Fe for each of the peaks. Find the radius of the atom of iron. 102 X-ray Diffraction XII (110) Intensity (relative) (211) (200) 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0 100.0 Diffraction angle 2θ Figure 5.5 103 X-ray Diffraction XIII Applying Bragg’s law: nλ d(h k l ) = 2 sin θ with n = 1 (monochromatic), λ =0.1542 nm, and θ equals to 45/2 = 22.5◦ for (110) plane, 32.5◦ for (200) plane, and 41.5◦ for (211) plane. We should find: d (110) = 0.201 nm d(200) = 0.143 nm d (211) = 0.116 nm In all cases, a should be constant and equal to = 0.286 nm. Solving for R using a = 2√ 2 R gives R = 0.101 nm. 104 X-ray Diffractometer XIV O S θ 0 T 2θ C Figure 5.6: Schematic diagram of an X-ray diffractometer; T: X-ray source, S: specimen, C: detector, and O: the axis around which the specimen and detector rotate.. https://www.youtube.com/watch?v=HrWT2M63DbU 105 Summary Atoms may assemble into crystalline (ordered) or amorphous (disordered) structures. 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. We can calculate the theoretical density of a metal, given its crystal structure, atomic weight, and unit cell lattice parameters. Atomic and planar densities are related to crystallographic directions and planes, respectively. 106 Summary (continued) Materials can exist as single crystals or polycrystalline. For most single crystals, properties vary with crystallographic orientation (i.e., are anisotropic). For polycrystalline materials having randomly oriented grains, properties are independent of crystallographic orientation (i.e., they are isotropic). Some materials can have more than one crystal structure. This is referred to as polymorphism (or allotropy). X-ray diffraction is used for crystal structure and interplanar spacing determinations. 107 Summary Make sure you understand language and concepts related to: Allotropy, Amorphous, Anisotropy, Atomic packing factor (APF), Body- centered cubic (BCC), Crystal structure, Crystalline, Face-centered cubic (FCC), Grain, Grain boundary, Hexagonal close-packed (HCP), Isotropic, Lattice parameter, Non-crystalline, Polycrystalline, Polymorphism, Single crystal, Unit cell. 108 References 1 Brian S. Mitchell. An introduction to materials engineering and science: for chemical and materials engineers. John Wiley & Sons, 2004. 2 William D. Callister and David G. Rethwisch. Materials Science and Engineering: an introduction. John Wiley & Sons, 8th edition, 2010. 109