2Crystal Structure Objectives Relationships between structures-bonding-properties of engineering materials.Arrangements in crystalline solidsGive examples of each: Lattice, Crystal Structure, Unit Cell and Coordination NumbersDescribe hard-sphere packing and identify cell symmetry.Define directions and planes (Miller indices) for crystalsOutlineStructure of the Atom and Atomic BondingElectronic Structure of the AtomLattice, Unit Cells, Basis, and Crystal StructuresPoints, Directions, and Planes in the Unit CellCrystal Structures of Ionic MaterialsCovalent Structures
3Crystal Structure Crystal Structure Lattice- A collection of points that divide space into smaller equally sized segments.Unit cell - A subdivision of the lattice that still retains the overall characteristics of the entire lattice.Atomic radius - The apparent radius of an atom, typically calculated from the dimensions of the unit cell, using close-packed directions (depends upon coordination number).Packing factor - The fraction of space in a unit cell occupied by atoms.Types of Crystal StructureBody centered cubic (BCC)Face centered cubic (FCC)Hexagonal close packed (HCP)
4Crystal StructureA number of metals are shown below with their room temperature crystal structure indicated. There are substances without crystalline structure at room temperature; for example, glass and silicone. All metals and alloys are crystalline solids, and most metals assume one of three different lattice, or crystalline, structures as they form: body-centered cubic (BCC), face-centered cubic (FCC), or hexagonal close-packed (HCP).Aluminum (FCC)Chromium (BCC)Copper (FCC)Iron (alpha) (FCC)Iron (gamma) (BCC)Iron (delta) (BCC)Lead (FCC)Nickel (FCC)Silver (FCC)Titanium (HCP)Tungsten (BCC)Zinc (HCP)
5Number of Lattice Points in Cubic Crystal Systems Crystal StructureNumber of Lattice Points in Cubic Crystal SystemsIn the SC unit cell: point / unit cell = (8 corners)1/8 = 1In BCC unit cells: point / unit cell = (8 corners)1/8 + (1 center)(1) = 2In FCC unit cells: point / unit cell = (8 corners)1/8 + (6 faces)(1/2) = 4Relationship between Atomic Radius and Lattice ParametersIn SC, BCC, and FCC structures when one atom is located at each lattice point.
6Crystal Structure Packing Factor In a FCC cell, there are four lattice points per cell; if there is one atom per lattice point, there are also four atoms per cell. The volume of one atom is 4πr3/3 and the volume of the unit cell is a0 3
7Crystal Structure Density Density of BCC iron, which has a lattice parameter of nm.Atoms/cell = 2, a0 = nm = 10-8 cmAtomic mass = g/molVolume of unit cell = = (2.866 10-8 cm)3 = cm3/cellAvogadro’s number NA = 6.02 1023 atoms/mol
8Crystal Structure Points, Directions, and Planes in the Unit Cell GeometryUnit Cell: The basic structural unit of a crystal structure. A unit cell is the smallest component of the crystal that reproduces the whole crystal when stacked together with purely translational repetition.RRRRaAtomic configuration in Face-Centered-CubicAtomic configuration in Simple Cubic
10Crystal Structure Unit Cells Types A unit cell is the smallest component of the crystal that reproduces the whole crystal when stacked together with purely translational repetition.Primitive (P) unit cells contain only a single lattice point.Internal (I) unit cell contains an atom in the body center.Face (F) unit cell contains atoms in the all faces of the planes composing the cell.Centered (C) unit cell contains atoms centered on the sides of the unit cell.Face-CenteredPrimitiveBody-CenteredEnd-CenteredCrystal Classes (cubic, tetragonal, orthorhombic, hexagonal, monclinic, triclinic, trigonal) with 4 unit cell types (P, I, F, C) symmetry allows for only 14 types of 3-D lattice.
11Crystal Structure Counting Number of Atoms Per Unit Cell Counting Atoms in 3D CellsAtoms in different positions are shared by differing numbers of unit cells.• Vertex atom shared by 8 cells => 1/8 atom per cell.• Edge atom shared by 4 cells => 1/4 atom per cell.• Face atom shared by 2 cells => 1/2 atom per cell.• Body unique to 1 cell => 1 atom per cell.Simple Cubic8 atoms but shared by 8 unit cells. So,8 atoms/8 cells = 1 atom/unit cellHow many atoms/cell forBody-Centered Cubic?And, Face-Centered Cubic?
12Crystal Structure Atomic Packing Fraction for FCC Face-Centered-Cubic ArrangementAPF = vol. of atomic spheres in unit celltotal unit cell vol.No. of atoms per unit cell =Volume of one atom=Volume of cubic cell =“R” related to “a” by4/cell4R3/3a3= 0.74
13Crystal Structure APF for BCC Again, For BCC BCC: a = b = c = a and angles a = b =g= 90°.2 atoms in the cubic cell: (0, 0, 0) and (1/2, 1/2, 1/2).
14Crystal Structure A B C FCC Stacking Highlighting the stacking the faces
15Crystal Structure FCC Stacking ABCABC.... repeat along <111> direction gives Cubic Close-Packing (CCP)Face-Centered-Cubic (FCC) is the most efficient packing of hard-spheres of any lattice.Unit cell showing the full symmetry of the FCC arrangement : a = b =c, angles all 90°4 atoms in the unit cell: (0, 0, 0) (0, 1/2, 1/2) (1/2, 0, 1/2) (1/2, 1/2, 0)
16Crystal Structure A B A HCP Stacking Highlighting the stacking the cellLayer ALayer BLayer A
17Crystal Structure HCP Stacking ABABAB.... repeat along <111> direction gives Hexagonal Close-Packing (HCP)Unit cell showing the full symmetry of the HCP arrangement is hexagonalHexagonal: a = b, c = 1.633a and angles a = b = 90°, g = 120°2 atoms in the smallest cell: (0, 0, 0) and (2/3, 1/3, 1/2).
18Crystal Structure Crystallographic Points, Directions, and Planes. To define a point within a unit cell….Express the coordinates uvw as fractions of unit cell vectors a, b, and c(so that the axes x, y, and z do not have to be orthogonal).pt. coord.pt.x (a) y (b) z (c)0 0 0origin1 0 01 1 11/2 0 1/2
19Crystal Structure Crystallographic Points, Directions, and Planes. Crystallographic directions and coordinates.Direction B1. Two points are 1, 1, 1 and 0, 0, 02. 1, 1, 1, -0, 0, 0 = 1, 1, 13. No fractions to clear or integers to reduce4. Direction A1. Two points are 1, 0, 0, and 0, 0, 02. 1, 0, 0, -0, 0, 0 = 1, 0, 04. Direction C1. Two points are 0, 0, 1 and 1/2, 1, 02. 0, 0, 1 -1/2, 1, 0 = -1/2, -1, 13. 2(-1/2, -1, 1) = -1, -2, 2
20Crystal Structure Crystallographic Points, Directions, and Planes. Procedure:Any line (or vector direction) is specified by 2 points.The first point is, typically, at the origin (000).Determine length of vector projection in each of 3 axes in units (or fractions) of a, b, and c.X (a), Y(b), Z(c)Multiply or divide by a common factor to reduce the lengths to the smallest integer values, u v w.Enclose in square brackets: [u v w]:  direction.abcDIRECTIONS will help define PLANES (Miller Indices or plane normal).5. Designate negative numbers by a barPronounced “bar 1”, “bar 1”, “zero” direction.6. “Family” of  directions is designated as <110>.
21Crystal Structure Crystallographic Points, Directions, and Planes. Figure 2.9 Crystallographic planes and interceptsPlane B1. The plane never intercepts the z axis, so x = 1, y = 2, and z =2.1/x = 1, 1/y =1/2, 1/z = 03. Clear fractions:1/x = 2, 1/y = 1, 1/z = 04. (210)Plane A1. x = 1, y = 1, z = 12.1/x = 1, 1/y = 1,1 /z = 13. No fractions to clear4. (111)Plane C1. We must move the origin, since the plane passes through 0, 0, 0. Let’s move the origin one lattice parameter in the y-direction. Then, x = ∞ , y = -1, and z = ∞2.1/x = 0, 1/y = 1, 1/z = 03. No fractions to clear.4 (o1-o)
22Crystal Structure t r s Miller Indices for HCP Planes 4-index notation is more important for planes in HCP, in order to distinguish similar planes rotated by 120o.tAs soon as you see , you will know that it is HCP, and not  cubic!Find Miller Indices for HCP:Find the intercepts, r and s, of the plane with any two of the basal plane axes (a1, a2, or a3), as well as the intercept, t, with the c axes.Get reciprocals 1/r, 1/s, and 1/t.Convert reciprocals to smallest integers in same ratios.Get h, k, i , l via relation i = - (h+k), where h is associated with a1, k with a3, i with a2, and l with c.Enclose 4-indices in parenthesis: (h k i l) .rs
23Crystal Structure Miller Indices for HCP Planes What is the Miller Index of the pink plane?The plane’s intercept a1, a3 and c at r=1, s=1 and t= , respectively.The reciprocals are 1/r = 1, 1/s = 1, and 1/t = 0.They are already smallest integers.We can write (h k i l) = (1 ? 1 0).Using i = - (h+k) relation, k=–2.Miller Index is
24Crystal Structure Linear Density in FCC LD =Number of atoms centered on a direction vectorLength of the direction vectorExample: Calculate the linear density of an FCC crystal along [1 1 0].ASKa. How many spheres along blue line?b. What is length of blue line?ANSWER2 atoms along [1 1 0] in the cube.Length = 4RXZ = 1i + 1j + 0k = 
25Planar Packing Density in FCC Crystal StructurePlanar Packing Density in FCCPD =Area of atoms centered on a given planeArea of the planeExample: Calculate the PPD on (1 1 0) plane of an FCC crystal.Find area filled by atoms in plane: 2R2Find Area of Plane: 8√2 R2Hence,Always independent of R!
27Crystal Structures of Ionic Materials Factors need to be considered in order to understand crystal structures of ionically bonded solids are:Ionic RadiiElectrical NeutralityConnection between Anion PolyhedraVisualization of Crystal Structures Using ComputersThe cesium chloride structure, a SC unit cell with two ions (Cs+ and CI-) per lattice point. (b) The sodium chloride structure, a FCC unit cell with two ions (Na+ + CI-) per lattice point
28Crystal Structures of Ionic Materials The zinc blende unit cell, (b) plan view.Fluorite unit cell, (b) plan view.
29Crystal Structures of Ionic Materials The perovskite unit cell showing the A and B site cations and oxygen ions occupying the face-center positions of the unit cell.
30Crystal Structure Covalent Structures Tetrahedron and (b) the diamond cubic (DC) unit cell. This open structure is produced because of the requirements of covalent bondingThe silicon-oxygen tetrahedron and the resultant β-cristobalite form of silica
31Crystal Structure Covalent Structures Tetrahedron and (b) the diamond cubic (DC) unit cell. This open structure is produced because of the requirements of covalent bonding