00 Crystallography Basics

Brief introduction to Crystallography

Brief introduction to Crystallography

•• • • • • • • • • Crystallography Crystallography Structure Structure Symmetry Symmetry d-spacing d-spacing

Defines results one achieves with XRD, TEM, SEM, etc.

Crystalline vs. Amorphous Materials

Crystalline vs. Amorphous Materials

Crystalline

Crystalline mamaterial –terial – atatoms aroms are situae situated in ated in a periodic array over large distances.

periodic array over large distances. Amorphous

Crystalline vs. Amorphous Materials

Crystalline vs. Amorphous Materials

Crystalline

Crystalline mamaterial –terial – atatoms aroms are situae situated in ated in a periodic array over large distances.

periodic array over large distances. Amorphous

Amorphous (or(or non-crystallinenon-crystalline) ma) mateteriarial –l – whwhereree long range order is absent. Therefore no real

long range order is absent. Therefore no real symmetry.

•

• Close packClose packed dired directions are ections are cube edgescube edges..

SIMPLE CUBIC STRUCTURE (SCC)

SIMPLE CUBIC STRUCTURE (SCC)

•

• Coordination # = 8

Adapted from Fig. 3.2, Callister 6e.

(Courtesy P.M. Anderson)

• Close packed directions are cube diagonals.

--Note: All atoms are identical; the center atom is shaded differently only for ease of viewing.

• Close packed directions are face diagonals.

--Note: All atoms are identical; the face-centered atoms are shaded differently only for ease of viewing.

FACE CENTERED CUBIC STRUCTURE (FCC)

Adapted from Fig. 3.1(a),

• Coordination Number = 12 • Atomic packing factor

takenfrombdhuey

• ABCABC... Stacking Sequence along cube diagonal • 2D Projection: A sites B sites C sites B _B B B B B B C C C A A

A

B

C

FCC STACKING SEQUENCE

• Coordination # = 12 • ABAB... Stacking Sequence

• Only difference from FCC is in the stacking (and the

orientation we generally consider when describing/drawing the stacking):

• 3D Projection • 2D Projection

A sites

B sites

A sites _{Bottom layer} Middle layer Top layer

Adapted from Fig. 3.3, Callister 6e.

Lattices and Unit Cells

• What if we want to consider more complicated crystals? • Take advantage of symmetry and well known structures

 –  A “ 

identical surroundings (nearest neighbors, etc).  –  A

• strictly the unit cell is the smallest repeating cell (though this is

sometimes totally impractical and thus more convenient, but larger, unit cells are employed).

 –  A lattice does not define the crystal, only its symmetry.

 –  Atoms (or ions, molecules, etc.) will be placed at any number of lattice points . ONLY THEN do we have a crystal.

lattice” is a repeating array of special points in space that have

unit cell defines a repeating unit within the lattice 

2-d Lattices

• A lattice repeats in all directions, and defines the symmetry.

• The unit cell is the simplest repeating structure.

Square unit cell

“Basis”

• BUT, a lattice only identifies repeating points and their symmetry.

 – What’s the lattice for the wall or floor in the classroom?

• Must be combined with a “basis” to know where atoms are in the crystal

 – What kind of bricks are hung at each lattice point?  – Are there smaller stones patterned regularly in

between the bricks?

• Only by knowing a lattice AND a basis can a crystal structure (or anything else symmetric) truly be built.

2-d Lattice

• Anions at lattice positions (0,0) • Cations at (½,0) • Cations at (0,½) square Lattice: Basis: tetragonal

and Basi

• Lattice:

 – A minimized pattern that will be followed • Basis:

 – a) A vector that defines the position of hooks on

which something (atoms, molecules, bricks) will be hung with relation to the underlying pattern

(lattice) 

 – and b) What exactly will be hung on each hook (atom, molecule)

• Rules:

 – At least one hook (basis) per lattice point 

• (but can be more than 1)

 – Basis vectors must put atoms within the unit cell (not in an adjacent cell)

 – Each basis vector applies to every single lattice point 

Seven Crystal Systems Cubic a=b=c ===90° Tetragonal a=b≠_{c ===90°} Orthorhombic a≠ _b≠_{c ===90°} Hexagonal a=b≠_{c ==90° =120°} Monoclinic a≠ _b≠_{c ==90°}≠_ Triclinic a≠ _b≠_{c }≠_≠_ Rhombohedral a=b=c ==≠_90°

In 3 dimensions, there are only 7 basic crystal systems. Easily defined by: 3 edge lengths (a, b and c) and 3

interaxial angles (  and ).

This class

Cubic Bravais Lattices

/

Tetragonal a=b≠_c

°

/

Orthorhombic Bravais Lattices / / / / / / b / / / /

Other Bravais Lattices

a=b°c

Defining Basis Positions Using

Fractional Co-ordinates

To locate atom positions in cubic unit cells we use rectangular x, y, and z axes.

The FCC Structure

Note: All atoms are not always shown because technically there are only 4/unit cell (for FCC)

Lattice=Face Centered Cubic Basis=red atoms at (0,0,0)

-of course it looks like there are more atoms, but these simply result from the repeating lattice and symmetry; ie they are not distinct

Each position might be provided, but probably the fewest number of positions will be given—the rest fill in as long as you know the Bravais lattice (ie the basic symmetry).

Simple unit cells

Based on Simple Cubic Lattice with a basis of: 1) anions at the SC points 2) cations at the center

[offset by (½, ½, ½)]

Based on FCC Lattice with a basis of: 1) anions at the FCC points

More common unit cells

Based on FCC Lattice with a basis of: 1) anions at the FCC points

2) cations at (¼,¼,¼)

Based on SC Lattice with a basis of: 1) Big cation at the SC points

2) Small cation at (½,½,½)

• Consider CaF2 : based on the CsCl structure.

• BUT: there are half as many Ca2+ _{as F}- _{ions—how can}

we accommodate this?

CsCl structure w/only half of cation sites occupied.

The initial ‘unit cell’ is no longer valid since every lattice site doesn’t have a basis attached to it (some are void). Instead, we use a ‘supercell.’ 2 possibilities here:

takenfrombdhuey

A

X

STRUCTURES

• Consider CaF₂ : Based on the CsCl structure, BUT: there are half as many Ca2+ _{as F}- _ions

 – This is accommodate with ‘empty cubes’.

 – i.e. CsCl based structure w/only half of cation sites occupied.

• Thus, the proposed CsCl ‘unit cell’ is no longer valid since every Simple Cubic lattice site doesn’t have a basis attached to it (half are void!).

 – Instead, use a ‘supercell.’

 – Other

possibilities:

Crystallographic Directions and Planes

• Terminology

 – Positions (,,,)

 – Directions [] note no commas  – Planes () note no commas  – Families of directions <> note no

commas

 – Families of planes {} note no commas

and NOT “one bar zero zero”  100)=“bar one zero zero”

Directions in Cubic Unit Cells

Position co-ordinates of vector OM = 1, 0.5, 0 (but these aren’t integers)

Position co-ordinates must be multiplied by 2 to obtain integers. Direction indices of OM = 2 (1, 0.5, 0) = [2, 1, 0]

A negative index direction (e.g. see vector ON) is written with a bar over the index

• For cubic crystals, the crystallographic direction indices are the vector components of the direction resolved along each of the co-ordinate axes (x,y,z)

[Directions]

1. Draw cell + origin 2. Draw vector  3. “ reduce” to smallest integer values 4. [xyz] Group exercise

more [Directions]

For negative directions : a. Add more unit cells.

OR

Directions in Cubic Unit Cells

The origin may be shifted. Thus:

<Directions> =family of directions

 A <family of directions> includes all possible directions with the same basic coordinates.

(Planes)

z 1. Draw the origin, cell,

and normal vector. 2. Draw the plane at a distance from the origin of 1/sqrt(a2_+b2_+c2_).

{Planes} =family of planes

in each family (“multiplicity”)? multiplici ty 3 6 ? ?

For a simple cubic lattice and (0,0,0) basis:

• Along (xyz) plane • (100) • (110) x y z ± ±

Should be able to draw atoms in, and above/below, a given plane in a crystal.

For CsCl (simple cubic lattice, basis of Cl-(0,0,0) and Cs-(½½,½)) • Along (xyz) plane • (100) • (110) x y z ± ± ± ± ± Group exercise

Crystallographic Planes – 

Miller Indices

Crystallographic Planes – Miller

Indices

When planes are more complicated, follow these rules:

1. Choose a plane that does not pass through the origin at (0,0,0)

2. Determine the intercepts of the plane in terms of the x, y, and z axes of the cube. 1/3,2/3,1 3. Form the reciprocals of these

intercepts 3, 3/2, 1 4. Bring to smallest integers by

multiplying or dividing through by a common factor. 6, 3, 2 5. Enclose integers in parentheses