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Crystallography and Structure

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Presentation on theme: "Crystallography and Structure"— Presentation transcript:

1 Crystallography and Structure
Lec. (3) Crystallography and Structure

2 Overview: Crystal Structure – matter assumes a periodic shape
Non-Crystalline or Amorphous “structures” exhibit no long range periodic shapes Xtal Systems – not structures but potentials FCC, BCC and HCP – common Xtal Structures for metals Point, Direction and Planer ID’ing in Xtals X-Ray Diffraction and Xtal Structure

3 Energy and Packing • Non dense, random packing
typical neighbor bond length bond energy Energy r typical neighbor bond length bond energy • Dense, ordered packing Dense, ordered packed structures tend to have lower energies & thus are more stable.

4 CRYSTAL STRUCTURES Means: periodic arrangement of atoms/ions
over large atomic distances Leads to structure displaying long-range order that is Measurable and Quantifiable All metals, many ceramics, and some polymers exhibit this “High Bond Energy” and a More Closely Packed Structure

5 Are materials lacking long range order
Amorphous Materials Are materials lacking long range order These less densely packed lower bond energy “structures” can be found in Metals are observed in Ceramic GLASS and many “plastics”

6 Crystal Systems – Some Definitional information
Unit cell: smallest repetitive volume which contains the complete lattice pattern of a crystal. Fig. 3.4, Callister 7e. 7 crystal systems of varying symmetry are known These systems are built by changing the lattice parameters: a, b, and c are the edge lengths , , and  are interaxial angles

7 General Unit Cell Discussion
For any lattice, the unit cell &, thus, the entire lattice, is UNIQUELY determined by 6 constants (figure): a, b, c, α, β and γ which depend on lattice geometry. As we’ll see, we sometimes want to calculate the number of atoms in a unit cell. To do this, imagine stacking hard spheres centered at each lattice point & just touching each neighboring sphere. Then, for the cubic lattices, only 1/8 of each lattice point in a unit cell assigned to that cell. In the cubic lattice in the figure, each unit cell is associated with (8)  (1/8) = 1 lattice point.

8 Primitive Unit Cells & Primitive Lattice Vectors
In general, a Primitive Unit Cell is determined by the parallelepiped formed by the Primitive Vectors a1 ,a2, & a3 such that there is no cell of smaller volume that can be used as a building block for the crystal structure. As we’ve discussed, a Primitive Unit Cell can be repeated to fill space by periodic repetition of it through the translation vectors T = n1a1 + n2a2 + n3a3. The Primitive Unit Cell volume can be found by vector manipulation: V = a1(a2  a3) For the cubic unit cell in the figure, V = a3

9 Primitive Unit Cells A 2 Dimensional Example!
Note that, by definition, the Primitive Unit Cell must contain ONLY ONE lattice point. There can be different choices for the Primitive Lattice Vectors, but the Primitive Cell volume must be independent of that choice. A 2 Dimensional Example! P = Primitive Unit Cell NP = Non-Primitive Unit Cell

10 2-Dimensional Unit Cells
Artificial Example: “NaCl” Lattice points are points with identical environments.

11 2-Dimensional Unit Cells: “NaCl”
The choice of origin is arbitrary - lattice points need not be atoms - but the unit cell size must always be the same.

12 2-Dimensional Unit Cells: “NaCl”
These are also unit cells it doesn’t matter if the origin is at Na or Cl !

13 2-Dimensional Unit Cells: “NaCl”
These are also unit cells the origin does not have to be on an atom!

14 2-Dimensional Unit Cells: “NaCl”
These are NOT unit cells - empty space is not allowed!

15 2-Dimensional Unit Cells: “NaCl”
In 2 dimensions, these are unit cells – in 3 dimensions, they would not be.

16 Crystal Systems Crystal structures are divided into groups according to unit cell geometry (symmetry).

17 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 • Have the simplest crystal structures. We will examine three such structures (those of engineering importance) called: FCC, BCC and HCP – with a nod to Simple Cubic

18 Crystal Structure of Metals – of engineering interest

19 Simple Cubic Structure (SC)
• Rare due to low packing density (only Po – Polonium -- has this structure) • Close-packed directions are cube edges. • Coordination No. = 6 (# nearest neighbors) for each atom as seen (Courtesy P.M. Anderson)

20 Atomic Packing Factor (APF)
Volume of atoms in unit cell* APF = Volume of unit cell *assume hard spheres • APF for a simple cubic structure = 0.52 Adapted from Fig. 3.23, Callister 7e. close-packed directions a R=0.5a contains (8 x 1/8) = 1 atom/unit cell atom volume atoms unit cell 4 3 p (0.5a) 1 APF = 3 a unit cell volume Here: a = Rat×2 Where Rat is the ‘handbook’ atomic radius

21 Body Centered Cubic Structure (BCC)
• Atoms touch each other along cube diagonals within a unit cell. --Note: All atoms are identical; the center atom is shaded differently only for ease of viewing. ex: Cr, W, Fe (), Tantalum, Molybdenum Adapted from Fig. 3.2, Callister 7e. • Coordination # = 8 2 atoms/unit cell: (1 center) + (8 corners x 1/8) (Courtesy P.M. Anderson)

22 Atomic Packing Factor: BCC
3 a a 2 length = 4R = Close-packed directions: 3 a APF = 4 3 p ( a/4 ) 2 atoms unit cell atom volume a Adapted from Fig. 3.2(a), Callister 7e. • APF for a body-centered cubic structure = 0.68

23 Face Centered Cubic Structure (FCC)
• Atoms touch each other along face diagonals. --Note: All atoms are identical; the face-centered atoms are shaded differently only for ease of viewing. ex: Al, Cu, Au, Pb, Ni, Pt, Ag • Coordination # = 12 Adapted from Fig. 3.1, Callister 7e. 4 atoms/unit cell: (6 face x ½) + (8 corners x 1/8) (Courtesy P.M. Anderson)

24 Atomic Packing Factor: FCC
• APF for a face-centered cubic structure = 0.74 The maximum achievable APF! Close-packed directions: length = 4R = 2 a (a = 22*R) Unit cell contains: 6 x 1/2 + 8 x 1/8 = 4 atoms/unit cell Adapted from Fig. 3.1(a), Callister 7e. APF = 4 3 p ( 2 a/4 ) atoms unit cell atom volume a

25 Comparison of the 3 Cubic Lattice Systems
Unit Cell Contents Counting the number of atoms within the unit cell Atom Position Shared Between: Each atom counts: corner cells /8 face center cells /2 body center cell edge center cells /2 Lattice Type Atoms per Cell P (Primitive) [= 8  1/8] I (Body Centered) [= (8  1/8) + (1  1)] F (Face Centered) [= (8  1/8) + (6  1/2)] C (Side Centered) [= (8  1/8) + (2  1/2)] 25

In a Hexagonal Crystal System, three equal coplanar axes intersect at an angle of 60°, and another axis is perpendicular to the others and of a different length. The atoms are all the same.

27 Hexagonal Close-Packed Structure (HCP)
ex: Cd, Mg, Ti, Zn • ABAB... Stacking Sequence • 3D Projection • 2D Projection c a A sites B sites Bottom layer Middle layer Top layer Adapted from Fig. 3.3(a), Callister 7e. 6 atoms/unit cell • Coordination # = 12 • APF = 0.74 • c/a = (ideal)

28 Hexagonal Close Packed (HCP) Lattice
Bravais Lattice : Hexagonal Lattice He, Be, Mg, Hf, Re (Group II elements) ABABAB Type of Stacking  a = b Angle between a & b = 120° c = 1.633a,  basis: (0,0,0) (2/3a ,1/3a,1/2c) Crystal Structure 28

29 We find that both FCC & HCP are highest density packing schemes (APF =0.74) – this illustration shows their differences as the closest packed planes are “built-up”

30 Comments on Close Packing
B A B C Sequence AAAA… - simple cubic Sequence ABABAB.. hexagonal close pack Sequence ABAB… - body centered cubic Sequence ABCABCAB.. -face centered cubic close pack Crystal Structure 30

31 Theoretical Density, r Density =  = n A  = VC NA
Cell Unit of Volume Total in Atoms Mass Density =  = VC NA n A  = where n = number of atoms/unit cell A = atomic weight VC = Volume of unit cell = a3 for cubic NA = Avogadro’s number = x 1023 atoms/mol

32 Theoretical Density, r  = a R Ex: Cr (BCC) A = 52.00 g/mol
R = nm n = 2  a = 4R/3 = nm theoretical ractual = 7.18 g/cm3 = 7.19 g/cm3  = a 3 52.00 2 atoms unit cell mol g volume 6.023 x 1023

33 Locations in Lattices: Point Coordinates
z x y a b c 000 111 Point coordinates for unit cell center are a/2, b/2, c/ ½ ½ ½ Point coordinates for unit cell (body diagonal) corner are 111 Translation: integer multiple of lattice constants  identical position in another unit cell z 2c y b b

34 Crystallographic Directions
z Algorithm 1. Vector is repositioned (if necessary) to pass through the Unit Cell origin. 2. Read off line projections (to principal axes of U.C.) in terms of unit cell dimensions a, b, and c 3. Adjust to smallest integer values 4. Enclose in square brackets, no commas [uvw] y x ex: 1, 0, ½ Lecture 2 ended here => 2, 0, 1 => [ 201 ] -1, 1, 1 where ‘overbar’ represents a negative index [ 111 ] => families of directions <uvw>

35 x y z What is this Direction ????? Projections:
Projections in terms of a,b and c: Reduction: Enclosure [brackets] x y z 0c a/2 b 1 1/2 1 2 [120]

36 Linear Density – considers equivalance and is important in Slip
Linear Density of Atoms  LD =  Number of atoms Unit length of direction vector a [110] ex: linear density of Al in [110] direction  a = nm # atoms length 1 3.5 nm a 2 LD - = # atoms CENTERED on the direction of interest! Length is of the direction of interest within the Unit Cell

37 Defining 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 (in cubic lattices this is direct) 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)  families {hkl}

38 Crystallographic Planes
We want to examine the atomic packing of crystallographic planes – those with the same packing are equivalent and part of families Iron foil can be used as a catalyst. The atomic packing of the exposed planes is important. Draw (100) and (111) crystallographic planes for Fe. b) Calculate the planar density for each of these planes.

39 Planar Density of (100) Iron
Solution:  At T < 912C iron has the BCC structure. 2D repeat unit R 3 4 a = (100) Radius of iron R = nm = Planar Density = a 2 1 atoms 2D repeat unit nm2 12.1 m2 = 1.2 x 1019 R 3 4 area Atoms: wholly contained and centered in/on plane within U.C., area of plane in U.C.

40 Planar Density of (111) Iron
Solution (cont):  (111) plane 1/2 atom centered on plane/ unit cell 2 a atoms in plane atoms above plane atoms below plane 2D repeat unit 3 h = a 2 Area 2D Unit: ½ hb = ½*[(3/2)a][(2)a]=1/2(3)a2=8R2/(3) 3*1/6 = nm2 atoms 7.0 m2 0.70 x 1019 Planar Density = 2D repeat unit area

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