ENGR 151: Materials of Engineering MIDTERM 1 REVIEW MATERIAL

ENGR 151: Materials of Engineering MIDTERM 1 REVIEW MATERIAL

MIDTERM 1 General properties of materials Bonding (primary, secondary and sub-types) Properties of different kinds of bonds Types of materials (metals, non-metals, ceramics, polymers, semiconductors) and their characteristics Different cell structures and properties (APF, coordination number, relationship between a and R, etc.) Directions and planes Geometric problems, sketching and interpreting

MIDTERM 1 Different coordinate systems (e.g. Miller- Bravais) Conversion from one system to the other Density calculations (linear, planar) Bragg s law problems Read Chapter 1 from textbook

CHAPTER 1 - INTRODUCTION Metals used in structures & machinery Plastics used in packaging, medical devices, consumer goods, clothing Ceramics used in electronics (insulative) Composites are novel materials for all applications listed above 4

CHAPTER 2 OBJECTIVES Understand the elements used to make engineering materials Review basic chemistry and physics principles Overview of the materials classes: Metals: good conductors, strong, lustrous Polymers: organic, low densities, flexible Ceramics: clay, cement, glass; insulators Composites: fiberglass; strength and flexibility 5

ATOMIC STRUCTURE Elements Atomic number (Z, number of protons in nucleus) Protons, neutrons, electrons (masses) m p = m n = 1.67 x 10-27 kg, m e = 9.11 x 10-31 kg Atomic Mass (A) = sum of proton and neutron masses in nucleus Isotope = same element, differing atomic masses E.g. Hydrogen (P = 1, N = 0), Deuterium (P = 1, N = 1), Tritium (P = 1, N = 2). Atomic Weight = Average atomic mass of all naturallyoccurring isotopes. amu (atomic mass unit) = 1/12 of atomic mass of carbon 12 One mole = 6.023 x 10 23 (Avogadro s number) atoms 6

ELECTRON CONFIGURATIONS Rules of electron configuration (Table 2.1, pg. 23) Electrons are quantized (have specific energies discrete energy levels) Quantum numbers (4) Principal: Position (n, distance of an electron from nucleus) Azimuthal: Subshell (l) Determines orbital angular momentum s, p, d, or f (shape of electron subshell) Magnetic: Number of energy states per subshell (m l ) s-1, p-3, d-5, f-7 Spin: Spin moment (m s ) +1/2, -1/2 7

ELECTRON CONFIGURATIONS CONTD. 8

ELECTRON CONFIGURATIONS CONTD. Pauli Exclusion Principle: No more than two electrons per electron state Number of electron states per shell determined by magnetic quantum number Examples: 3p shell has 3 states (-1, 0, +1), therefore can accommodate up to 6 electrons (2 electrons per state). 3d shell has 5 states (-2, -1, 0, +1, +2), therefore can accommodate up to 10 electrons (2 electrons per state). 9

QUICK REVIEW (TABLE 2.2, PG. 22) How many valence electrons do they have? Hydrogen, 1s 1 Aluminum, 1s 2 2s 2 2p 6 3s 2 3p 1 Chlorine, 1s 2 2s 2 2p 6 3s 2 3p 5 Answer: 1, 3, 7 10

ELECTRONEGATIVITY Measures the tendency of an element to give up or accept valence electrons Electropositive elements (e.g. alkali metals) Capable of giving up few valence electrons to become positively charged (e -, negative charge) Electronegative elements (e.g. halogens) Readily accept electrons to form negatively charged ions. Also share electrons (covalent bonding) Electronegativity increases left to right, bottom to top Atoms accept electrons if shells are closer to nucleus Example: Na gives up one electron, Cl accepts the electron to form NaCl

ATOMIC BONDING To understand the physical properties behind materials, we must have an understanding of interatomic forces that bind atoms together. At large distances, the interactions between two atoms are negligible BUT as they come closer to each other they start to exert a force on each other.

ATOMIC BONDING There are two types of forces that are both functions of the distance between two atoms: 1) Attractive Force (F A ) Depends on bonding between atoms 2) Repulsive Force (F R ) Originates due to repulsion between atoms individual (negatively-charged) electron clouds

ATOMIC BONDING Magnitude of an attractive force varies with distance. The Net Force (F N ) is the sum of the attractive and repulsive forces:

ATOMIC BONDING CONTD. When FA = FR the net force is zero: (State of equilibrium) In a state of equilibrium, the two atoms will remain separated by the distance, ro. Attractive force is the same as repulsive force at ro. For many atoms, ro is approximately.3 nm or 3 angstroms (Å)

ATOMIC BONDING CONTD. Another way to represent this relationship in attractive and repulsive forces is to look at potential energy relationships. Force-energy relationships: Both force and energy are functions of distance r Measure of amount of work done to move an atom from infinity (zero force) to a distance r. Alternatively:

ATOMIC BONDING CONTD. Energy relationships: EN = net energy EA = attractive energy ER = repulsive energy

ATOMIC BONDING CONTD. Energy relationships:

Why does zero force correspond to minimum energy?

ATOMIC BONDING CONTD. The net potential energy curve has a trough around its minimum. The potential energy minimum is ro away from the origin. Force is the derivative of energy.

ATOMIC BONDING CONTD. The Bonding Energy, Eo, refers to the vertical distance between the minimum potential energy and the x-axis. This is the energy that would be required to separate the atoms to an infinite separation. Force and Energy plots become more complex in actual materials. Why?

ATOMIC BONDING ENERGY Magnitude of bonding energy and shape of energy-versus-interatomic separation curve vary from material to material AND depend on the type of bonding that is taking place between atoms.

ATOMIC BONDING ENERGY (EO) Materials with large Bonding Energies also have high MELTING POINTS At room temperature, materials with high bonding energy are in the SOLID STATE; those with small bonding energy are in the GASEOUS STATE; LIQUID STATE for intermediate bonding energy FLEXIBILITY: stiff materials have a steep slope at the r=ro position; the slope is less steep for more flexible materials. Flexibility or hardness of a material is gauged by the MODULUS of ELASTICITY. How much a material expands upon heating or contracts upon cooling is related to the potential energy curve. This is called LINEAR COEFFICIENT of THERMAL EXPANSION.

TYPES OF ATOMIC BONDING Three types of PRIMARY bonds: IONIC, COVALENT, METALLIC. Nature of the bond depends on the electron structures of the bonding atoms AND type of bond depends on the tendency of atoms to assume stable electron structures (completely filling outermost electron shell)

IONIC BONDING Always found in compounds that are formed by reactions between metallic and nonmetallic elements (elements at the horizontal extremities of the periodic table). Metallic elements easily give up valence electrons to nonmetallic atoms. In this process, all atoms acquire stable or inert gas configurations and electrical charges (becoming ions)

IONIC BONDING The attractive bonding forces within ionic bonds are COULOMBIC (positive and negative ions attracting one another). E.g. Na + Cl Na + + Cl - NaCl Notice the crystalline structure in the diagrams below.

IONIC BONDING CONTD. Attractive energy is a function of interatomic distance (inversely proportional):

IONIC BONDING CONTD. Constant A is expressed as:

IONIC BONDING CONTD. Repulsive energy:

IONIC BONDING CONTD. Equation for attractive force:

COVALENT BONDING Stable electron configurations are assumed by the sharing of electrons between adjacent atoms. Atoms contribute at least one electron to the bond, and the shared electrons are considered to belong to both atoms Covalent bonds are directional (between specific atoms)

COVALENT BONDING CONTD. Examples include: Hydrogen (H 2 ), Chloride (Cl 2 ), Fluoride (F 2 ), Water (H 2 0), diamond, silicon, germanium, elemental solids located on right hand side of the periodic table, Gallium Arsenide (GaAs), indium antimonide (InSb), and silicon carbide (SiC). Covalent bond in Methane (CH 4 )

COVALENT BONDING The number of covalent bonds for an atom is determined by the number of valence electrons. For N' valence electrons, an atom can covalently bond with at most 8 - N' other atoms.

METALLIC BONDING Metallic materials have one, two, or at most three valence electrons which are not bound to any atom within the solid and can drift throughout the metal Creates a sea of electrons (electrons that belong to the entire metal) Remaining nonvalence electrons and their atomic nuclei are called ION CORES (have net positive charge equal in magnitude to total valence charge per atom)

METALLIC BONDING CONTD. Free electrons shield ion cores from repulsive forces and act as glue to hold ion cores together Metallic bond is nondirectional Free electrons allow metal to be good conductors of heat and electricity

METALLIC BONDING ENERGIES Can be strong or weak (Mercury, 68 kj/mol, - 39 degrees Celsius; Tungsten, 850 kj/mol, 3410 degrees Celsius)

SECONDARY BONDING Very weak bonds (10 kj/mol) Exist between all atoms and molecules but obscured if other primary bonding is occurring Found in: inert gases because of their stable electron structures between molecules that are covalently bonded arise from dipoles (separation of positive and negative portions of an atom or molecule). Also known as van der Waals bonding or physical bonding. Physical, as opposed to chemical bonding weaker bonds.

SECONDARY BONDING CONTD. + - + - Atomic or Molecular Dipoles

FLUCTUATING INDUCED DIPOLE BONDS Atom structure is not symmetric and positively charged nucleus creates dipole with electron cloud on opposite side of atom. Instantaneous and short-lived distortions of electric field. Atomic Nucleus Electron Cloud Atomic Nucleus Electron Cloud + - Electrically Symmetric Atom Induced Atomic Dipole

POLAR MOLECULE-INDUCED DIPOLE BONDS Polar molecules: asymmetrical arrangements of positively and negatively charged regions within molecules. Can also induce dipoles in adjacent nonpolar molecules to form a bond. H Cl + -

PERMANENT DIPOLE BONDS Hydrogen Bonding: a type of secondary bonding occurring in molecules that have hydrogen covalently bonded to fluorine, oxygen, and nitrogen. Stronger than induced dipole bond Adjacent polar molecules attracting each other H F H F

SUMMARY OF BONDING TYPES Bonding Primary Secondary Ionic Covalent Metallic Fluctuating Induced Dipole Bonds Polar Molecule-Induced Dipole Bonds Permanent Dipole Bonds

CRYSTAL STRUCTURE Material properties depend on crystal structure of the material Atoms are thought of as being solid spheres having well-defined diameters (atomic hard sphere model, atoms touching each other) Lattice: three-dimensional array of points coinciding with atom positions (sphere centers) 43

CRYSTAL STRUCTURE CONTD. 44

UNIT CELLS Subdivide the crystal structure into small repeating entities called UNIT CELLS These cells are mainly in cubes, prisms, sixsided figures regular, repeating geometric structures Geometric symmetry The unit cell is the basic structural unit or building block of crystal structure 45

METALLIC CRYSTAL STRUCTURES No restrictions as to the number and position of nearest-neighbor atoms (dense atomic packing) Each sphere in atomic hard sphere model equates to an ion core 46

METALLIC CRYSTAL STRUCTURES CONTD. Four simple crystal structures are found in metals: Simple Cubic (SC) Face-Centered Cubic (FCC) Body-Centered Cubic (BCC) Hexagonal Close-Packed (HCP) 47

ANALYSIS TECNIQUES Analysis of crystal structures gives insight into properties of material such as strength, density, how the material may behave under physicals stress, etc. Analysis steps: Identify spatial geometries associated with unit cell Relate dimensions of unit cell to atomic radius Characterize/calculate required properties of crystal structure 48

METALLIC CRYSTAL STRUCTURES 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 four such structures... 49

SIMPLE CUBIC STRUCTURE (SC) Rare due to low packing density (only Po has this structure) Close-packed directions are cube edges. Coordination # = 6 (# nearest neighbors) Fig. 3.3, Callister & Rethwisch 9e. 50

ATOMIC PACKING FACTOR (APF) APF = Volume of atoms in unit cell* Volume of unit cell APF for a simple cubic structure = 0.52 a *assume hard spheres close-packed directions contains 8 x 1/8 = 1 atom/unit cell Adapted from Fig. 3.3 (a), Callister & Rethwisch 9e. R = 0.5a atoms unit cell APF = 1 volume 4 3 π (0.5a) 3 atom a 3 volume unit cell 51

FACE-CENTERED CUBIC (FCC) The UNIT CELL is CUBIC Atoms located at each of the corners and the centers of all the cubic faces Aluminum, Copper, Gold, Lead, Nickel, Platinum, Silver 52

FACE-CENTERED CUBIC The cube edge length a and the atomic radius are related through: a = 2R 2 53

ATOMIC PACKING FACTOR: FCC APF for a face-centered cubic structure = 0.74 maximum achievable APF 2 a a Adapted from Fig. 3.1(a), Callister & Rethwisch 9e. atoms unit cell APF = Close-packed directions: length = 4R = 2 a Unit cell contains: 6 x 1/2 + 8 x 1/8 = 4 atoms/unit cell 4 4 3 π ( 2 a/4 ) 3 a 3 volume atom volume unit cell 54

FCC Each corner atom is shared by eight unit cells (Eight 1/8 portions per unit cell) Each face atom is shared by two unit cells (six ½ portions per unit cell) Grand total of four whole atoms per unit cell 55

COORDINATION NUMBER The number of nearest neighbor or touching atoms The coordination number for FCC is?? 56

COORDINATION NUMBER The number of nearest neighbor or touching atoms The coordination number for FCC is 12 57

ATOMIC PACKING FACTOR (APF) The fraction of solid sphere volume in a unit cell: APF = Volume of atoms in a unit cell Total unit cell volume Using the APF equation and the volume of an FCC unit cell, find the APF of a FCC crystal structure 58

BODY-CENTERED CUBIC (BCC) 59

BCC Atoms located in all eight corners and a single atom at the cube center Derive an expression for the unit cell edge length (a) using the atomic radius (R) 60

ATOMIC PACKING FACTOR: BCC APF for a body-centered cubic structure = 0.68 3 a a 2 a Adapted from Fig. 3.2(a), Callister & Rethwisch 9e. atoms R unit cell APF = a 2 4 3 π ( 3 a/4 ) 3 a 3 Close-packed directions: length = 4R = 3 a volume unit cell volume atom 61

BCC 8 x 1/8 = 1 (each corner) 1 atom in center 2 atoms per unit cell What is the coordination number and APF? 62

BCC 8 x 1/8 = 1 (each corner) 1 atom in center 2 atoms per unit cell What is the coordination number and APF? C.N. = 8 APF =.68 63

HEXAGONAL CLOSE-PACKED (HCP) Unit cell is hexagonal Six atoms form a regular hexagon and surround a single atom Another plane is situated in between top and bottom plane (provides three additional atoms) 64

HCP Six atoms altogether within unit cell 65

HCP Six atoms altogether within unit cell 1/6 x 12 (top and bottom plane portions ½ x 2 (center face atoms) 3 midplane interior atoms C.N. = 12 APF =.74 c/a = 1.633 66

OVERVIEW Crystal Structure Relationship between a and R (cubic structures) Number of atoms per unit cell Coordination Number APF FCC 4 12 0.74 BCC 2 8 0.68 HCP ------ 6 12 0.74 67

THEORETICAL DENSITY, R Density = = = Mass of Atoms in Unit Cell Total Volume of Unit Cell n A V C N A 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 68

THEORETICAL DENSITY, R Ex: Cr (BCC) A (atomic weight) = 52.00 g/mol n = 2 atoms/unit cell R = 0.125 nm atoms unit cell = volume unit cell R a 2 52.00 a 3 6.022 x 10 23 a = 4R/ 3 = 0.2887 nm g mol theoretical actual atoms mol = 7.18 g/cm 3 = 7.19 g/cm 3 69

Point Coordinates Coordinates of selected points in the unit cell. The number refers to the distance from the origin in terms of lattice parameters.

Point Coordinates Contd. Each unit cell is a reference or basis. The length of an edge is normalized as a unit of measurement. E.g. if the length of the edge of the unit cell along the X-axis is a, then ALL measurements in the X-direction are referenced to a (e.g. a/2).

c z Point Coordinates 111 Point coordinates for unit cell center are a/2, b/2, c/2 ½ ½ ½ x a z 000 b 2c y Point coordinates for unit cell corner are 111 b y Translation: integer multiple of lattice constants identical position in another unit cell b 72

EXAMPLE PROBLEMS 73

EXAMPLE PROBLEMS CONTD. 74

EXAMPLE PROBLEMS CONTD. 75

Miller Indices, Directions Determine the Miller indices of directions A, B, and C. (c) 2003 Brooks/Cole Publishing / Thomson Learning

GENERAL APPROACH FOR MILLER INDICES 77

SOLUTION Direction A 1. Two points are 1, 0, 0, and 0, 0, 0 2. 1, 0, 0, -0, 0, 0 = 1, 0, 0 3. No fractions to clear or integers to reduce 4. [100] Direction B 1. Two points are 1, 1, 1 and 0, 0, 0 2. 1, 1, 1, -0, 0, 0 = 1, 1, 1 3. No fractions to clear or integers to reduce 4. [111] Direction C 1. Two points are 0, 0, 1 and 1/2, 1, 0 2. 0, 0, 1-1/2, 1, 0 = -1/2, -1, 1 3. 2(-1/2, -1, 1) = -1, -2, 2 4. [1 22] (c) 2003 Brooks/Cole Publishing / Thomson Learning

CRYSTALLOGRAPHIC DIRECTIONS x z ex: 1, 0, ½ => 2, 0, 1 => [ 201 ] y Algorithm 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] -1, 1, 1 => [ 111 ] where overbar represents a negative index Problem 3.6, pg. 58 79

EXAMPLE PROBLEMS 80

EXAMPLE PROBLEMS, CONTD. 81

EXAMPLE PROBLEMS, CONTD. Exam problems make sure you translate and scale vector to fit WITHIN unit cube if asked! 82

HCP CRYSTALLOGRAPHIC DIRECTIONS Hexagonal Crystals 4 parameter Miller-Bravais lattice coordinates are related to the direction indices (i.e., u'v'w') as follows. a 3 z a 2 a 1 Fig. 3.8(a), Callister & Rethwisch 8e. - [ u ' v ' w '] u v t w = = = = 1 ( 2 u '- v ') 3 1 ( 2 v '- u ') 3 -( u + v ) w ' [ uvtw ] 83

HCP CRYSTALLOGRAPHIC DIRECTIONS z a 2 a 3 - a 1 Fig. 3.8(a), Callister & Rethwisch 8e. 84

HCP CRYSTALLOGRAPHIC DIRECTIONS a 3 z a 2 Adapted from Fig. 3.8(a), Callister & Rethwisch 8e. a 1 ex: ½, ½, -1, 0 => - Algorithm 1. Vector repositioned (if necessary) to pass through origin. 2. Read off projections in terms of unit cell dimensions a 1, a 2, a 3, or c 3. Adjust to smallest integer values 4. Enclose in square brackets, no commas [uvtw] [ 1120 ] a 3 a 2 2 a 1 2 dashed red lines indicate projections onto a 1 and a 2 axes a 1 a 2 -a 3 85

REDUCED-SCALE COORDINATE AXIS 86

PROBLEM 3.8 87

PROBLEM 3.8, CONTD. 88

PROBLEM 3.8, CONTD. 89

PROBLEM 3.8, CONTD. 90

PROBLEM 3.8, CONTD. 91

PROBLEM 3.8, CONTD. 92

PROBLEM 3.9 93

PROBLEM 3.9, CONTD. 94

DRAWING HCP CRYSTALLOGRAPHIC DIRECTIONS (I) Algorithm (Miller-Bravais coordinates) 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 placing tail at origin and stepping off these projections to locate the head Adapted from Figure 3.10, Callister & Rethwisch 9e. 95

DRAWING HCP CRYSTALLOGRAPHIC DIRECTIONS (II) Draw the Adapted from p. 72, Callister & Rethwisch 9e. s [1213] [1 2 13] direction in a hexagonal unit cell. Algorithm a 1 a 2 a 3 z 1. Remove brackets -1-2 1 3 1 2. Divide by 3 1 3. Projections 3 2 3 1 3 r q p 4. Construct Vector start at point o proceed a/3 units along a 1 axis to point p 2a/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 [1213] direction represented by vector from point o to point s 96

DETERMINATION OF HCP CRYSTALLOGRAPHIC DIRECTIONS (II) Algorithm 1. Determine coordinates of vector tail, pt. 1: x 1, y 1, & z 1 ; and vector head, pt. 2: x 2, y 2, & z 2. in terms of three axis (a 1, a 2, and z) 2. Tail point coordinates subtracted from head point coordinates and normalized by 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: Adapted from p. 72, Callister & Rethwisch 9e. 6. Adjust to smallest integer values and enclose in brackets [uvtw] 97

DETERMINATION OF HCP CRYSTALLOGRAPHIC DIRECTIONS (II) Determine indices for green vector Adapted from p. 72, Callister & Rethwisch 9e. Example a 1 a 2 z 1. Tail location 0 0 0 Head location a a 0c 2. Normalized 1 1 0 3. Reduction 1 1 0 4. Brackets [110] 6. Reduction & Brackets 1/3, 1/3, -2/3, 0 => 1, 1, -2, 0 => [ 1120 ] 98

FAMILIES OF DIRECTIONS <UVW> For some crystal structures, several nonparallel directions with different indices are crystallographically equivalent; this means that atom spacing along each direction is the same. 99

CRYSTALLOGRAPHIC PLANES Adapted from Fig. 3.10, Callister & Rethwisch 8e. 100

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) 101

CRYSTALLOGRAPHIC PLANES z example a b c 1. Intercepts 1 1 2. Reciprocals 1/1 1/1 1/ 1 1 0 3. Reduction 1 1 0 4. Miller Indices (110) example a b c 1. Intercepts 1/2 2. Reciprocals 1/½ 1/ 1/ 2 0 0 3. Reduction 1 0 0 4. Miller Indices (100) a x a c c z b b y y x 102

CRYSTALLOGRAPHIC PLANES z example a b c 1. Intercepts 1/2 1 3/4 2. Reciprocals 1/½ 1/1 1/¾ 2 1 4/3 3. Reduction 6 3 4 4. Miller Indices (634) x a c b y Family of Planes {hkl} Ex: {100} = (100), (010), (001), (100), (010), (001) 103

FAMILY OF PLANES Planes that are crystallographically equivalent have the same atomic packing. Also, in cubic systems only, planes having the same indices, regardless of order and sign, are equivalent. Ex: {111} = (111), (111), (111), (111), (111), (111), (111), (111) Ex: {100} = (100), (010), (001), (100), (010), (001) 104

FCC UNIT CELL WITH (110) PLANE 105

BCC UNIT CELL WITH (110) PLANE 106

CRYSTALLOGRAPHIC PLANES (HCP) In hexagonal unit cells the same idea is used z example a 1 a 2 a 3 c 1. Intercepts 1-1 1 2. Reciprocals 1 1/ -1 1 1 0-1 1 3. Reduction 1 0-1 1 a 2 a 3 4. Miller-Bravais Indices (1011) (hkil) a 1 Adapted from Fig. 3.8(b), Callister & Rethwisch 8e. 107

DISTINCTION BETWEEN CRYSTALLOGRAPHIC DIRECTIONS AND MILLER INDICES Crystallographic Direction Crystallographic Planes Associated with point coordinates in 3D space Associated with planar structures in 3D space Denoted by enclosing in square brackets as [abc] Denoted by enclosing in regular parentheses (hkl) 108

EXAMPLE PROBLEMS 109

EXAMPLE PROBLEMS, CONTD. Example of family of planes 110

EXAMPLE PROBLEMS, CONTD. Don t forget to slide plane back within unit cell! 111

EXAMPLE PROBLEMS, CONTD. Strategy Find answer in 3 coordinates and then convert to 4- coordinate representation 112

EXAMPLE PROBLEMS, CONTD. 113

LINEAR DENSITY Linear Density of Atoms LD = [110] Number of atoms Unit length of direction vector ex: linear density of Al in [110] direction a = 0.405 nm a 2 atoms per line, sharing computed along vector length Adapted from Fig. 3.1(a), Callister & Rethwisch 8e. # atoms LD = 2 = length 2 a 3.5 atoms/nm 114

PLANAR DENSITY OF (100) IRON Solution: At T < 912ºC iron has the BCC structure. 2D repeat unit (100) a = 4 3 3 R Adapted from Fig. 3.2(c), Callister & Rethwisch 8e. atoms 2D repeat unit 1 Planar Density = a 2 area 2D repeat unit = 4 3 1 3 Radius of iron R = 0.1241 nm R 2 = atoms 12.1 nm 2 = 1.2 x 10 19 atoms m 2 115

PLANAR DENSITY OF (111) IRON Solution (cont): (111) plane 1 atom in plane/ unit surface cell 2 a atoms in plane atoms above plane atoms below plane atoms 2D repeat unit Planar Density = area 2D repeat unit 4 3 2 area = 2 ah = 3 a = 3 R = 3 16 3 1 3 2 R atoms nm 2 = 7.0 = h 2 = 0.70 x 10 19 3 a 2 16 3 3 R atoms m 2 2 116

CONSTRUCTIVE INTERFERENCE Occurs at angle θ to the planes, if path length difference is equal to a whole number, n, of wavelengths (n equals order of reflection): Bragg s Law

BRAGG S LAW If Bragg s Law is not satisfied then we get nonconstructive interference, which will yield a very low-intensity diffracted beam. For cubic unit cells:

BRAGG S LAW Specifies when diffraction will occur for unit cells having atoms positioned only at the cell corners Atoms at different locations act as extra scattering centers and results in the absence of some diffracted beams.

EXAMPLE PROBLEM