CH2 - The structure of crystalline solids - İstanbul Kültür...

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CHAPTER 2

The Structure of Crystalline Solids

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OUTLINE� 1.0 Energy & Packing

� 2.0 Materials & Packing

� 3.0 Crystal Structures

� 4.0 Theoretical Density, r

� 5.0.Polymorphism and Allotropy

� 6.0 Close - Packed Crystal Structures

� 7.0 Structure of Compounds: NaCl

� 8.0 X-Ray Diffraction From Crystals

� 9.0 Crystallographic Points, Directions and Planes: Miller

Indices

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1.0 Energy & Packing

i. Non-Dense Random Packing

ii. Dense Regular Packing

Energy

r

typical neighbor bond length

typical neighbor bond energy

Energy

r

typical neighbor bond length

typical neighbor bond energy

�Dense, regular-packed structures tend to have lower energy.

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2.0 Materials & Packing

Si Oxygen •the atoms pack in periodic, 3D arrays in a

long range

•dense, regular packing

•typical of:

1. Crystalline materials...

-metals

-many ceramics

-some polymers

• atoms have no periodic packing in long

range

• occurs for:

2. Noncrystalline materials... "Amorphous"

-complex structures

-rapid cooling

crystalline SiO2

noncrystalline SiO2

Adapted from Fig. 3.18(b),Callister 6e.

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Examples

Basic Unit

Si0 4 tetrahedron4-

Si4+

O2-

• Crystalline materials: Quartz(SiO2)

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Examples• Noncrystalline materials(Amorphous): Glass

• Amorphous structureoccurs by adding impurities

(Na+,Mg2+,Ca2+, Al3+)

• Impurities:

interfere with formation of

crystalline structure.

Si4+

Na+

O2-

(soda glass)

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Type of Crystals:1. Single Crystals:

When the regular arrangement of atoms is perfect and extends throughout the specimen without interruption, the result is single crystal. They are difficult to grow.

GRAPHITE

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• Some engineering applications require single crystals:

2)turbine blades

Fig. 8.30(c), Callister 6e.

3)silicon single crystals are grown for

quantum electronics

1)diamond single crystals for abrasives

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2. Polycrystalline Minerals and Materials

� A collection of many small crystals or grains

� Random orientations

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Polycrystalline Minerals and Materials

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� Most engineering materials are polycrystalline.

� Nb-Hf-W plate with an electron beam weld.

� Each "grain" is a single crystal.

� If crystals are randomly oriented, overall component properties are not directional.

� Crystal sizes range from 1 nm to 2 cm (i.e., from a few atoms to millions of atomic layers).

Callister 6e.1 mm

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Single vs Polycrystals

• Single Crystals

1. Properties vary with direction: anisotropic

Example: the modulus

of elasticity (E) in BCC iron:

• Polycrystals

1. Properties may/may not vary with

direction.

1. If grains are randomly

oriented: isotropic(Epoly iron = 210 GPa)

2. If grains are textured:anisotropic

E (diagonal) = 273 GPa

E (edge) = 125 GPa

200 mm

Data from Table 3.3,

Callister 6e.

Adapted from Fig. 4.12(b), Callister 6e.

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3.0 Crystal Structures

� Diffraction pattern of electron beams produced for a single crystal of Ga As (Gallium Arsenide) using Transmission Electron Microscope

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Definitions about Crystal Structures

� Lattice: 3D array of regularly spaced points

� (a) Hard sphere representation:atoms denoted by hard, touching spheres

� (b) Reduced sphere representation

� (c)Unit cell: basic building block unit (such as a flooring tile) that repeats in space to create the crystal structure

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Definition of Crystal Systems

� Lattice Parameters:� The unit cell geometry is defined in terms of six parameters.

� the three edge lengths a, b, c, and

� the three interaxial angels α, β, γ

� Seven possible combinations of a, b, c& α, β, γ, resulting in seven crystal systems

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Crystal Systems

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•Atomic Packing Factor

APF = Volume of atoms in unit cell*

Volume of unit cell

*assume hard spheres

•Coordination Number

The number of closest atoms to a selected atom in a unit cell

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Cubic Crystal Structures

1. Simple Cubic (SC)

2. Face Centered Cubic ( FCC )

3. Body Centered Cubic ( BCC )

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Simple Cubic Structure (SC)

� Rare due to poor packing (only Po has this structure)

� Close-packed directions are cube edges.

Coordination # = 6(# nearest neighbors)

(Courtesy P.M. Anderson)

Click on image to animate

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Volume of Atoms In Unit Cell(SC)

Fig. 3.19,Callister 6e.

a3Volume of unit cell=

atoms

unit cell= 8 x 1/8 = 1 atom/unit cell

close-packed directions

a

R=0.5a

R=0.5aStep1:

4

3π (0.5a) 3Volume of atom =

Step2:

Step3:

Step4:

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Atomic Packing Factor: SC

APF = Volume of atoms in unit cell*

Volume of unit cell

*assume hard spheres

APF for a simple cubic structure = 0.52

APF =

a3

4

3π (0.5a)31

atoms

unit cellatom

volume

unit cell

volume

a

R=0.5a

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Face Centered Cubic Structure (FCC)

• Coordination # = 12

• Close packed directions are face diagonals.

Note: All atoms are identical; the face-

centered atoms are shaded

differently only for ease of viewing.


This structure is exhibited by metals Ag, Al,

Au, Cu, Ir, Ni, Pb, Pd, Pt, and Rh, as well

as the noble gases Ar, Kr, Ne, and Xe.

This arrangement (but without actual contact

by atoms of the same element) is also

assumed by the anions (or, less often, the

cations) in numerous ionic compounds.

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Atomic Packing Factor: FCC

APF =

a3

4

3π ( 2a/4)34

atoms

unit cell atom

volume

unit cell

volume

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

• APF for a body-centered cubic structure = 0.74

Close-packed directions: length = 4R = 2 a

Adapted from

Fig. 3.1(a),

Callister 6e.

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Body Centered Cubic Structure (BCC)

Coordination # = 8

The metals Ba, Cr, Cs, α-Fe, K, Li, Mo, Na, Rb,

Ta, V, and W adopt this structure. If the atom

in the center of the cube is different than

those at the corners, the cesium chloride

structure results.

• Close packed directions are cube diagonals.


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

APF =

a3

4

3π ( 3a/4)32

atoms

unit cell atom

volume

unit cell

volume

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

Atomic Packing Factor: BCC

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4.0 Theoretical Density, ρρρρDensity = mass/volume

mass = (number of atoms per unit cell) x (mass of each atom)

mass of each atom = Molar Mass /Avogadro’s number

ρ = nA

VcNA

# atoms/unit cell Atomic weight (g/mol)

Volume/unit cell

(cm3/unit cell)

Avogadro's number

(6.023 x 1023 atoms/mol)

Molar Mass

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Theoretical Density, ρρρρ

Example: Copper

• crystal structure = FCC: 4 atoms/unit cell

• molar mass = 63.55 g/mol (1 amu = 1 g/mol)

• atomic radius R = 0.128 nm (1 nm = 10 cm)-7

Vc = a3 ; For FCC, a = 4R/ 2 ; Vc = 4.75 x 10

-23cm3

Compare to actual: ρCu = 8.94 g/cm3

Result: theoretical ρCu = 8.89 g/cm3

ρ = nA

VcNA

# atoms/unit cell Atomic weight (g/mol)

Volume/unit cell

(cm3/unit cell)

Avogadro's number

(6.023 x 1023 atoms/mol)

Molar Mass

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Symbol Al Ar Ba Be B Br Cd Ca C Cs Cl Cr Co Cu F Ga Ge Au He H

Atomic radius (nm) 0.143 ------ 0.217 0.114 ------ ------ 0.149 0.197 0.071 0.265 ------ 0.125 0.125 0.128 ------ 0.122 0.122 0.144 ------ ------

Crystal Structure FCC ------ BCC HCP Rhomb ------ HCP FCC Hex BCC ------ BCC HCP FCC ------ Ortho. Dia. cubic FCC ------ ------

Callister 6e.

Characteristics of Selected Elements at 200C

Element Aluminum Argon Barium Beryllium Boron Bromine Cadmium Calcium Carbon Cesium Chlorine Chromium Cobalt Copper Flourine Gallium Germanium Gold Helium Hydrogen

Molarmass

(g/mol) 26.98 39.95 137.33 9.012 10.81 79.90 112.41 40.08 12.011 132.91 35.45 52.00 58.9363.5519.00 69.72 72.59 196.97 4.003 1.008

19.32 ------

Density

(g/cm3)

2.71 ------3.5 1.85 2.34 ------8.65 1.55 2.25 1.87 ------7.19 8.9 8.94------5.90 5.32

------

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ρ (g/cm3)

Graphite/ Ceramics/ Semicond

Metals/ Alloys

Composites/ fibers

Polymers

1

2

20

30Based 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). 10

3

4

5

0.3

0.4 0.5

Magnesium

Aluminum

Steels

Titanium

Cu,Ni

Tin, Zinc

Silver, Mo

Tantalum Gold, W Platinum

Graphite

Silicon

Glass-soda Concrete

Si nitride Diamond Al oxide

Zirconia

HDPE, PS PP, LDPE

PC

PTFE

PET PVC Silicone

Wood

AFRE*

CFRE*

GFRE*

Glass fibers

Carbon fibers

Aramid fibers

Why?Metals have...• close-packing

(metallic bonding)

• large atomic mass

Ceramics have...• less dense packing

(covalent bonding)

• often lighter elements

Polymers have...• poor packing

(often amorphous)

• lighter elements (C,H,O)

Composites have...• intermediate values Data from Table B1, Callister 6e.

Densities of Material Classes

ρmetals> ρceramics> ρpolymers

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5.0.Polymorphism and Allotropy

� HEATING AND COOLING OF AN IRON WIRE

Demonstrates "polymorphism"

The same atoms can

have more than one

crystal structure.Temperature, C

BCC Stable

FCC Stable

914

1391

1536

shorter

longer!shorter!

longer

Tc 768 magnet falls off

BCC Stable

Liquid

heat up

cool down

Polymorphism

Same compound occurring in more than one crystal structure.

Crystal structure depends on Temperature and Pressure.

Allotropy

Polymorphism in elemental solids (e.g., sulphur,carbon, oxygen)

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Shape memory alloys

� materials that "remember" their original shape and return to it when heated,

� even if apparent residual deformation was introduced below a certain temperature.

� The original shape can be set easily by heat treatment.

� These alloys are used in orthopedic implants and other medical applications, as well as in air conditioners and other home appliances, automobiles, etc.

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6.0 Close - Packed Crystal StructuresClose packed planes of atoms:Planes

having a maximum atom density.

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Close - Packed Crystal Structures

Close packed planes of atoms:Planes

having a maximum atom density.

Compare with open pack

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7.0 Structure of Compounds: NaCl

• Compounds: Often have similar close-packed structures.

• Structure of NaCl



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8.0 X-ray Diffraction From Crystals

� Diffraction based techniques.� X-ray :X-rays interact with electrons in matter.

X-rays are scattered by the electron clouds of atoms.

� Neutron

� Electron

� Microscopy.

� Thermal Analysis.

� Spectroscopy.

There are many experimental techniques that can be

used to study the properties and structure of solids.

These include:

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X-ray Diffraction From Crystals

W. H. Bragg (father) and William Lawrence Bragg (son)

developed a simple relation for scattering angles,

nowadays known as Bragg’s law.

X Rays

λ :0.01-10 nm

ν:1018 Hz

(visible light 1015 Hz)

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� nλ = SQ + QT

� nλ = dhkl sinθ + dhkl sinθ=2 dhkl sinθ

� nλ = 2 dhkl sinθ

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A Modern Diffractometer

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Powder Diffraction

� Powder diffraction experiment requires only as small quantity of a mineral:

� 10-500mg

� Sample preparation is very simple and fast

� Reliable accurate results are obtained in a relatively short time:10 minutes to 2 hours.

� The ideal powder size is 5-10 microns.

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Applications of XRD

� Identification of unknowns.

� Phase purity.

� Determination of lattice parameters.

� Determination of crystal size.

� Variable temperature studies.

� Structure determination and refinement.

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9.0 Crystallographic Points,

Directions And Planes: Miller Indices

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Crystallographic Directions

� It is a vector between two points.

1. The length of the vector projection on the

three axis is determined as a, b, c.

2. These three numbers are multiplied or by

divided by a common number to reduce to

the smallest integers.

3. Three indices are written as [u, v, w]

4. Negative indices are represented by a bar

over the index.

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Crystallographic Directions And Planes: Miller Indices

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Crystallographic Directions And Planes: Miller Indices

1. If the plane passes through the selected

origin, either another parallel plane or a

new origin must be established.

2. Find intersects of the plane with the three

axis

3. Take the reciprocals. A plane that

parallels and axis has an infinite intercept

4. If necessary find the smallest integers

5. Write the indices as (h k l)

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Miller Indices of Planes:Examples

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Miller Indices of Planes:Examples

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FCC Unit Cell With (110) Plane

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Problem 2.10

� (a) The tetragonal crystal system since a = b= 0.35

nm, c = 0.45 nm, and α = β = γ = 90°.

� (b) The crystal structure can be called body-centered

tetragonal.

� (c) BCC, n = 2 atoms/unit cell. A=141g/mol

VC= (3.5 x 10−8 cm)2(4.5 x 10−8

cm)

=5.51 x 10−23cm3/unit cell

ρ = nA / VCNA

= (2 atoms/unit cell)(141 g/mol) /(5.51 x 10−23cm3/unit

cell )(6.023 x 1023 atoms/mol)

ρ = 8.49 g/cm3

Below is a unit cell for a

hypothetical metal.

a) To which

crystal system

does this unit

cell belong?

b) What would this

crystal structure

be called?

c) Calculate the

density of the

material, given

that its molar

mass is 141

g/mol.

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Problem 2.11 What are the indices for the direction indicated by the two

vectors in the sketch?

� For direction 1,

� the projection on the x-axis is zero (since it lies in

the y-z plane),

� while projections on the y- and z-axes are b/2

and c,

For direction 2

For direction 1

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Problem about direction(Callister 3.52,p.88) Determine the indices for the directions shown.

Direction A : We position the origin of

the coordinate system at the tail of

the direction vector; then in terms of

this new coordinate system

[ -110] direction

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Problem 2.13 What are the indices for the two planes drawn in the

sketch below?

� Plane 1 :(020)

� Plane 2 :

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Problem2.12 Sketch within a cubic unit cell the following planes:

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Problem 2.15 Determine the Miller indices for the planes shown in the

following unit cell.

Plane A

Plane B

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CH2 - The structure of crystalline solids - İstanbul Kültür...

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