Crystal Structure 5a304m

CRYSTAL STRUCTURE and GEOMETRY Space lattice Unit cell Cubic system APF Density calculatio

Miller indices Positions Directions Planes Linear & planar density Structure mapping XRD, SEM, TEM

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INTRODUCTION • Significant property differences exist between the crystalline and non-crystalline materials having the same composition • Properties of some materials are directly related to their crystal structures. • Crystallography is the branch of science that deals with the geometric description of crystals and their internal arrangement. 2

Materials and Packing SOLID

CYSTALLINE -due to orderly structure of their atoms molecules or ions possess well defined shaped. Examples: metal, ceramic, polymer, alloy

AMORPHOUS -poor or no long range order and do not solidify with symmetry or cystalline solids

CRYSTALLINE -atoms/ion arranged in a pattern that repeat itself in 3 dimensions and they form a solid that has long range order (LRO) AMORPHOUS -atoms/ions are not arranged in a long range order, periodic and repeatable manner and possess only short range order (SRO)

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Materials and Packing Crystalline materials... • atoms pack in periodic, 3D arrays • typical of: -metals -many ceramics -some polymers

crystalline SiO2 Adapted from Fig. 3.22(a), Callister 7e.

Noncrystalline materials... • atoms have no periodic packing • occurs for: -complex structures -rapid cooling "Amorphous" = Noncrystalline

Si

Oxygen

noncrystalline SiO2 Adapted from Fig. 3.22(b), Callister 7e.

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Crystal Systems Unit cell: smallest repetitive volume which contains the complete lattice pattern of a crystal. 7 crystal systems (Table 3.1 Smith) *stress on CUBIC only 14 crystal lattices (Bravais lattices) a, b, and c are the lattice constants 6

An imaginary network of lines, with atoms at intersection of lines, representing the arrangement of atoms is called space lattice.

Unit cell is that block of atoms which repeats itself to form space lattice.

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• 7 different types of unit cells necessary to create all point lattices. • August Bravais (1811-1863) derived 14 possible unit cells to describe all possible lattice networks. *We learn 3 only. See Fig 3.2 Smith • In the cubic system, the 3 types of unit cells are  Simple Cubic  Body Centered (BCC)  Face Centered (FCC)

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The 14 Bravais

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7 Types of Unit Cells • Cubic Unit Cell a=b=c  α = β = γ = 900 

Simple

Body Centered

Figure 3.2

Face centered

• Tetragonal a

=b ≠ c  α = β = γ = 900

Simple

Body Centered 10

• Orthorhombic a≠ b≠ c  α = β = γ = 900 

Simple

Base Centered Body Centered

Face Centered

• Rhombohedral a

=b = c  α = β = γ ≠ 900

Simple 11

• Hexagonal a≠ b≠ c  α = β = γ = 900 

Simple

• Monoclinic a≠ b≠ c  α = β = γ = 900 

Simple

• Triclinic a≠ b≠ c  α = β = γ = 900

Figure 3.2



Simple 12

PRINCIPAL METALLIC CRYSTAL STRUCTURES • 90% of the metals have either: Body Centered Cubic (BCC), Face Centered Cubic (FCC) or Hexagonal Close Packed (H) crystal structure. **H is denser version of simple hexagonal crystal structure.

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Simple Cubic

• Number of atoms in a unit cell? 1/8th of an atom are at the cube corners = 8 corners × (1/8 atom) = 1 14

Atomic Packing Factor (APF) Volume of atoms in unit cell* APF = Volume of unit cell *assume hard spheres atoms unit cell

a R=0.5a close-packed directions contains 8 x 1/8 = 1 atom/unit cell Adapted from Fig. 3.23, Callister 7e.

APF =

volume atom 4 (0.5a) 3 1 3 a3

volume unit cell

**APF for a simple cubic structure = 0.52 15

ATOMIC PACKING FACTOR (APF) • It is the fraction of solid sphere volume in a unit cell, assuming the atomic hard sphere model. • It is the maximum packing possible for spheres all having the same diameter. Atomic Packing Factor =

Volume of atoms in unit cell, Va Volume of unit cell, Vc

**The higher the APF, Crystal structure is more packed. 16

SIMPLE CUBIC

where This means that SC 52% packed with 1 atom. (48% empty space). It is not a closed-packed structure. 17

Body Centered Cubic Structure (BCC) • Represented as one atom at each corner of cube and one at the center of cube.

(Courtesy P.M. Anderson)

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

Adapted from Fig. 3.2, Callister 7e.

2 atoms/unit cell: 1 center + 8 corners x 1/8 18

Body Centered Cubic (BCC) Crystal Structure

ex: Cr, W, Fe (), Tantalum, Molybdenum

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• Each unit cell has eight 1/8 atom at corners and 1 full atom at the center. • Therefore each unit cell has

(8x1/8 ) + 1 = 2 atoms •

Atoms each other at cube diagonal

Relationship between cube side length (a) and atomic radius (R) :

Therefore, lattice constant, a =

4R 3

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Formula check! Calculate √2 a Calculate √3 a X 2  a2  a2 X  2a

 4R  X 2  a 2  4 R  2  ( 2a ) 2  a 2  4 R  2  3a 2 2

a

4R 3

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Atomic Packing Factor: BCC 3a

a 2a

Adapted from Fig. 3.2(a), Callister 7e.

R

a

Close-packed directions: length = 4R = 3 a

atoms volume 4  ( 3a/4) 3 2 unit cell atom 3 APF = volume 3 a 22 unit cell

Example Question: BCC Calculate the APF for the BCC unit cell, assuming the atoms to be hard spheres

This means that BCC 68% packed with 2 atoms. (32% empty space). BCC is not a closedpacked structure. • APF for a body-centered cubic structure = 0.68 WHAT DOES THIS MEAN? 23

Example Problem Iron (Fe) at 20°C is a BCC with atoms of atomic radius 0.124 nm. Determine the lattice constant, a of the cube edge of the iron unit cell. 4R=√3 a  a = 4R/√3

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Solution Solution:

a =

4R 3

= 0.2864 nm

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Table 3.2 (Smith) Selected metals which have the BCC crystal structure at room temperature (20°C) and their lattice constants and atomic radii Metal

Lattice constant a, nm

*Atomic radii R, nm

Chromium

0.289

0.125

Iron

0.287

0.124

Molybdenum

0.315

0.136

Potassium

0.533

0.231

Sodium

0.429

0.186

Tantalum

0.330

0.143

Tungsten

0.316

0.137

Vanadium

0.304

0.132

*Calculated from lattice constant using equation

4R = 3 a

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Face Centered Cubic (FCC) Crystal Structure • FCC structure is represented as one atom each at the corner of cube and at the center of each cube face.

Adapted from Fig. 3.1, Callister 7e.

(Courtesy P.M. Anderson)

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

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Face Centered Cubic Structure (FCC) • Atoms touch each other along face diagonals. ex: Al, Cu, Au, Pb, Ni, Pt, Ag

4 atoms/unit cell  6 face x 1/2 atom + 8 corners x 1/8 atom 28

Face Centered Cubic (FCC) Crystal Structure • Each unit cell has eight 1/8th atom at corners and six ½ atoms at the center of six faces. • Therefore each unit cell has:

(8 x 1/8)+ (6 x ½) = 4 atoms/unit cell

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Relationship between cube edge length and atomic radius for FCC a2 + a2 = 2 a

a

4R 2 30

Atomic Packing Factor: FCC Close-packed directions: length = 4R = 2 a

Unit cell contains: 6 x 1/2 + 8 x 1/8 = 4 atoms/unit cell

atoms 2a

unit cell APF =

a Adapted from Fig. 3.1(a), Callister 7e.

4

4 3

( 2a/4) 3 a3

volume atom volume unit cell 31

Face Centered Cubic 4 3 Va  (4) R 3  4R 

Vc  a   3

3



 2 APF  0.74

APF for a FCC structure = 0.74 (maximum achievable APF)

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APF of FCC • APF = 0.74, which is greater than BCC (0.68) • This means that 74% of the FCC unit cell volume is filled by 4 atoms. • Atoms in FCC are packed as close together as possible. • It is a close-packed structure. • 26% is empty space. 33

Hexagonal Close-Packed Structure (H) • The H structure is represented as an atom at each of 12 corners of a hexagonal prism, 2 atoms at top and bottom face and 3 atoms in between top and bottom face.

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Hexagonal Close-Packed Structure (H) • ABAB... Stacking Sequence • 3D Projection

c

a

• APF = 0.74

• 2D Projection A sites

Top layer

B sites

Middle layer

A sites

Bottom layer Adapted from Fig. 3.3(a), Callister 7e.

6 atoms/unit cell ex: Cd, Mg, Ti, Zn 35

Hexagonal Close-Packed Structure (H) • Each unit cell has six 1/6th atoms at each of top and bottom layer, two half atoms at top and bottom layer and 3 full atoms at the middle layer. • Therefore each H unit cell has:

(2 x 6 x 1/6) + (2 x ½) + 3 = 6 atoms

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Class Exercise 1. Iron has an atomic radius of 0.124 nm and a BCC crystal structure. Show that APF for iron is 0.68. 2. Calculate the atomic radius of Nickel having the lattice constant of 0.352422 nm and FCC crystal structure. 37

Densities of Material Classes In general metals > ceramics > polymers 30 Why? Metals have...

Ceramics have... • less dense packing • often lighter elements

Polymers have...

 (g/cm3 )

• close-packing (metallic bonding) • often large atomic masses

• low packing density (often amorphous) • lighter elements (C,H,O)

Composites have... • intermediate values

Metals/ Alloys

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Platinum Gold, W Tantalum

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Silver, Mo Cu,Ni Steels Tin, Zinc

5 4 3 2 1

0.5 0.4 0.3

Titanium Aluminum Magnesium

Graphite/ Ceramics/ Semicond

Polymers

Composites/ fibers

Based on data in Table B1, Callister *GFRE, CFRE, & AFRE are Glass, Carbon, & Aramid Fiber-Reinforced Epoxy composites (values based on 60% volume fraction of aligned fibers in an epoxy matrix). Zirconia Al oxide Diamond Si nitride Glass -soda Concrete Silicon Graphite

PTFE Silicone PVC PET PC HDPE, PS PP, LDPE

Glass fibers GFRE* Carbon fibers CFRE* Aramid fibers AFRE*

Wood Data from Table B1, Callister 7e.

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Density Calculation 

mass unitcell 

volume unitcell 

 n  MW   Vc N A

Where: n = MW = Vc =

number of atoms per unit cell molecular/atomic weight (g/mol) Volume of a unit cell (m3)

NA

Avogadro’s No. (6.022 X 1023)

=

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Theoretical Density,  Density =  =  =

where

Mass of Atoms in Unit Cell Total Volume of Unit Cell

nA VC NA

n = number of atoms/unit cell A = atomic weight VC = Volume of unit cell = a3 for cubic NA = Avogadro’s number = 6.023 x 1023 atoms/mol 40

Theoretical Density,  • Ex: Cr (BCC) A = 52.00 g/mol R = 0.125 nm n=2 R atoms unit cell

= volume unit cell

a 2 52.00

a3 6.023 x 1023

a = 4R/ 3 = 0.2887 nm g mol

theoretical

= 7.18 g/cm3

actual

= 7.19 g/cm3

atoms mol

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Volume Density •

Volume density of metal =

v

=

Mass/Unit cell Volume/Unit cell

• Example:- Copper (FCC) has atomic mass of 63.54 g/mol and atomic radius of 0.1278 nm. Determine volume density of copper.

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Volume Density •

Volume density of metal =

v

=

Mass/Unit cell Volume/Unit cell

• Example:- Copper (FCC) has atomic mass of 63.54 g/mol and atomic radius of 0.1278 nm. a=

4R

=

4  0.1278nm

2

= 0.361 nm

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Volume of unit cell = V= a3 = (0.361nm)3 = 4.7 x 10-29 m3 FCC unit cell has 4 atoms.

v v



(4atoms / unitcell )(63.54 g / mol ) 4.7 10  29 (volume / unitcell )  6.023 10 23 atoms / mol



m g  8.98 3 V cm 43

Polymorphism and Allotropy • Some materials may exist in more than one crystal structure, this phenomenon is called polymorphism. • If the material is an elemental solid, it is called allotropy. • An example of allotropy is carbon, which can exist as diamond (at high pressure) and graphite (at ambient condition) 44

Polymorphism or Allotropy • Metals exist in more than one crystalline form. This is caller polymorphism or allotropy. • Temperature and pressure leads to change in crystalline forms. • Example:- Titanium, Iron, Cobalt - Iron exists in both BCC and FCC form Liquid depending on the temperature. Iron

9120C

-2730C

α Iron BCC

13940C 15390C

γ Iron FCC

δ Iron BCC 45

3-33

Tutorial 1. Molybdenum has a BCC crystal structure, an atomic radius of 0.1363 nm, and an atomic weight of 95.94 g/mol. Determine its density. 2. Calculate the radius of a palladium atom, given that Pd has an FCC crystal structure, a density of 12.0 g/cm3 , and an atomic weight of 106.4 g/mol. 3. Calculate the radius of a tantalum atom, given that Ta has a BCC crystal structure, a density of 16.6 g/cm3 , and an atomic weight of 180.9 g/mol 46

4.

Below are listed the atomic weight, density, and atomic radius for three hypothetical alloy. For each determine whether its crystal structure is FCC, BCC or simple cubic. Justify your determination. Alloy

Atomic weight (g/mol)

Density (g/cm3)

Atomic Radius (nm)

A

43.1

6.40

0.122

B

184.4

12.30

0.146

C

91.6

9.60

0.137 47