Chap. 3. The Structures of Crystalline Solids

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Introduction to Introduction to

Materials Science and Engineering

Chap 3 The Structures of Crystalline Solids

Chap. 3. The Structures of Crystalline Solids

¾ How do atoms assemble into solid structures?

¾ How does the density of a material depend on its

¾ How does the density of a material depend on its structure?

¾ When do material properties vary with the sample (i.e., orientation)?

1

( , )

http://bp.snu.ac.kr

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Content

1

Introduction Introduction

2

Unit Cell Unit Cell

3

Crystal Structure Crystal Structure

3

Crystal Structure Crystal Structure

4 Crystallographic Directions and Planes Crystallographic Directions and Planes y y g p g p

5 Interstitial Position & Coordination Number Interstitial Position & Coordination Number

6

Structure Determination Structure Determination

(X-Ray Diffraction)(X-Ray Diffraction)

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Introduction

¾ In chapter 2, you have learned the various types of atomic

bonding. In chapter 3, we will learn the atomic arrangements in g p , g the solid state.

¾ Solid materials are classified according to the regularity with which atoms and ions are arranged with respect to one another.

Solid Materials : Solid Materials

¾ So, how are they arranged ?

a) Periodically – having a long range order in 3-D:

Crystalline

b) Randomly – having a short range order with the characteristics of bonding type,

b t l i th l d

3

but losing the long range order:

Amorphous

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Introduction

¾ Materials and Packing 1 Crystalline materials 1. Crystalline materials

√ Atoms pack in periodic, 3D arrays.

√ Typical of - metals

- many ceramics

- some polymers

crystalline SiO crystalline SiO

²²

- some polymers

2. Noncrystalline materials...

√ At h i di ki

Si Oxygen

√ Atoms have no periodic packing.

√ Occurs for - complex structures id li

- rapid cooling

√ Amorphous = Noncrystalline

noncrystalline SiO

²²

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Introduction

¾ The arrangement of atoms in a material influence its physical and mechanical behavior:

physical and mechanical behavior:

- Remember that most of the properties of materials result from the bonding nature of atoms in solids, not the properties of atoms by itself.

Al provides good ductility. Polyethylene is easily deformed.

Fe provides good strength. Rubber can be elastically stretched.

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Content

1

Introduction Introduction

2

Unit Cell Unit Cell

3

Crystal Structure Crystal Structure

3

Crystal Structure Crystal Structure

4

Crystallographic Directions and Planes Crystallographic Directions and Planes y y g p g p

5

Interstitial Position & Coordination Number Interstitial Position & Coordination Number

6

Structure Determination (X-Ray) Structure Determination (X-Ray)

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Unit cell

¾ Lattice

√ 3D i t i h th t h i t h id ti l

√ 3D point array in space, such that each point has identical

surroundings. These points may or may not coincide with atom iti

positions.

(Fi 3 2)

example: iron (Fe) -- body centered cubic

(Fig. 3-2)

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Unit cell

¾ Unit cell

√ Smallest repetitive volume which contains the complete lattice pattern of a crystal.

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Unit cell

¾ Unit cell

9

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Unit cell

¾ Lattice parameters

length: a b c

length: a, b, c

angle: α β γ

angle: α, β, γ

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Unit cell

¾ 7 crystal systems (Table 3-2)

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Unit cell

¾ 7 crystal systems (continued) (Table 3-2)

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Unit cell

¾ 14 B i L i

¾ 14 Bravais Lattice -

Only 14 different types of unit cells are required to describe all lattices using symmetry

cubic hexagonal rhombohedral

(trigonal) tetragonal orthorhombic monoclinic triclinic

P I F

C

13

C

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Content

1

Introduction Introduction

2

Unit Cell Unit Cell

3

Crystal Structure Crystal Structure

3

Crystal Structure Crystal Structure

4

Crystallographic Directions, and Planes Crystallographic Directions, and Planes y y g p g p , ,

5

Interstitial Position & Coordination Number Interstitial Position & Coordination Number

6

Structure Determination (X-Ray) Structure Determination (X-Ray)

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Crystal system

¾ Hard Sphere model is used to describe st llin st t s

crystalline structures.

(Fig. 3-1)

f d bi

face-centered cubic (FCC)

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Crystal system

¾ Tend to be densely packed.

¾ Have several reasons for dense packing:

√ Typically several elements are present and

√ Typically, several elements are present, and each atomic radius is the same.

√ Metallic bonding is not directional.

√ Nearest neighbor distances tend to be small,

√ Nearest neighbor distances tend to be small, for lower bond energy.

¾ Have the simplest crystal structures.

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Crystal system

¾ Si l C bi St t (SC)

¾ Simple Cubic Structure (SC)

√ Rare due to low packing density

Only Po (polonium) has this structure.

√ Close-packed directions are cube edges

√ Close packed directions are cube edges

√ Coordination # = 6 (# of nearest neighbors)

√ 1 atom/unit cell: 8 corners X 1/8

17 (Courtesy P.M. Anderson)

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Crystal system

¾ Atomic Packing Factor (APF)

V l f t i it ll* (Eq. 3-2) APF = Volume of atoms in unit cell*

Volume of unit cell

*assume hard spheres

( q )

¾ APF for a simple cubic structure = 0.52

4 3

atoms / unit cell volume / atom APF =

3 4

3 π (0.5 a )3

a

1

R=0.5a

a ³

volume / unit cell

close-packed directions

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

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Crystal system

¾ B d C t d C bi St t (BCC)

¾ Body Centered Cubic Structure (BCC)

√ Atoms touch each other along cube diagonals

Cl s p k d di ti ns ub di n ls 111>

Close packed directions are cube diagonals <111>

√ Highest density plane {110}

√ ex: Cr W Fe ( ) Ta Mo

√ ex: Cr, W, Fe (α), Ta, Mo

√ Coordination # = 8 (# of nearest neighbors)

√ 2 atoms/unit cell: 1 center + 8 corners X 1/8

(Fig. 3-2)

19 (Courtesy P.M. Anderson)

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Crystal system

¾ APF of BCC structure = 0.68

3

a 3 a

a 2 a

R Close-packed directions: _{length = 4} R = 3 a a

4 ( 3 /4 )3 atoms / unit cell

l /

length 4 R 3 a

APF = 2 3 π ( 3 a /4 )3 3

m volume / atom

a ³ volume / unit cell

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Crystal system

¾ Face Centered Cubic Structure (FCC)

√ Atoms touch each other along face diagonals

Close packed directions are edge diagonals <110>

√ Highest density plane {111}

√ ex: Al, Cu, Au, Pb, Ni, Pt, Ag

√ Coordination # = 12 (# of nearest neighbors)

√ 4 atoms/unit cell: 6 face X 1/2 + 8 corners X 1/8

(Fig. 3-1)

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(Courtesy P.M. Anderson) http://bp.snu.ac.kr

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Crystal system

¾ APF of FCC structure = 0.74

Close-packed directions:

length = 4 R = 2 a a

g

a

4 ( 2 /4 )3 atoms / unit cell

APF =

4 3 π ( 2 a /4 )3 4

3 volume / atom a ³ volume / unit cell

- 2009-09-21

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Crystal system

¾ H l Cl d P ki S (HCP)

¾ Hexagonal Closed Packing Structure (HCP)

√ ex: Cd, Mg, Ti, Zn

√ Coordination # = 12 (# of nearest neighbors)

√ 2 atoms / unit cell (1 + 4 x 1/6 + 4 x 1/12)

√ a / c = 1.633

√ APF = 0.74

(Fig. 3-3)

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Unit cell

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Crystal system

¾ Hexagonal Closed Packing Structure (HCP)

(Fig. 3-3)

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Crystal system

25

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Crystal system

¾ Close Packing Crystal structure

k l

How can we stack metal atoms to minimize empty space?

close packed atomic arrangement in 2-D

Now stack these 2 D layers to make 3 D structures

Now stack these 2-D layers to make 3-D structures

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Crystal system

¾ Stacking sequence

(Fig. 3-13)

A A A A A

( g )

(Fig. 3-14) (Fig. 3-15)

A A A A A

B B B B

C C C C

A A A

A A A A A

B B B B

C C C C

A A A A A

B B B B

C C C C

A A A A A

B B B B

C C C C

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FCC

¾ A – B – C – A – B – C stacking sequence Æ FCC

(Fig 3-13)

A A A A A B

(Fig. 3 13) (Fig. 3-15)

A A A A A

B B B B

B B B

C B C B C B C B

C C C

C A C B

A A A A A

B B B B

C C C C

B B B B

C C C

C A B C

A A A A A

B B B B

C B C B C B C B

B B B B

C C C

C

B C A

A A A A A

B B B B

C C C C

B B B B

C C C

C A

B

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HCP

¾ A – B – A – B – A – B stacking sequence Æ HCP

(Fig. 3-14)

A A A A A A A A A A A

( g )

(Fig. 3-15)

A A A A A

B B B B

C B C B C B C B

A A A A A

A B A

A A A

A A A A A

B B B B

C C C C

B B B B

A A A A A

A B

A A A A A

B B B B

C B C B C B C B

B B B B

A A A A A

B A

A A A A A

B B B B

C C C C

B B B B

A A A A A A

B

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Crystal system

¾ FCC vs. HCP

FCC HCP

Both have the same APF.

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Crystal system

¾ Density

C ll U it

i At

f

Density = ρ =

M^Mass_Totalôf_VolumeÂtoms_ofⁱⁿ_UnitÛnit_Cell^Cell

V n

_C

A N

_A

ρ = (Eq. 3-5)

where

n = number of atoms/unit cell A = atomic weight (g/mol)

A atomic weight (g/mol)

V

_C

= Volume of unit cell = a

³

for cubic (cm

³

/unit cell) N = Avogadro’s number = 6 023 x 10

²³

(atoms/mol)

31

N

_A

= Avogadro s number = 6.023 x 10

²³

(atoms/mol)

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Crystal system

¾ D it

¾ Density

Example

¾ Copper (from Periodic Table)

√ Crystal structure = FCC: 4 atoms/unit cell

√ / /

√ Atomic weight = 63.55 g/mol (1 amu = 1 g/mol)

√ Lattice constant a = 0.361 nm (1 nm = 10^-9 m)

¾ ρ( Cu) = 8.97 g/cm

³

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Crystal system

¾ Density

√ Example - Cr (BCC)

√ Example Cr (BCC)

A = 52.00 g/mol R 0 125

R = 0.125 nm

n = 2

a = 0.2887 nm

R a a

52 00

ρ = 7.19 g/cm

³ 2

ρ

=

a ³

52.00 2

6.023 x 10²³

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Crystal system

¾ D iti f M t i l Cl

¾ Densities of Material Classes

Graphite/

Ceramics/

Metals/

All Composites/

Polymers Fib

ρ(metals) > ρ(ceramics) > ρ(polymers) _Ceramics/

Semicond

Alloys Polymers Fibers

20 30

*GFRE, CFRE, & AFRE are Glass, Carbon & Aramid Fiber-Reinforced T t l m

Gold, W Platinum

¾Why?

¾Metals have... Carbon, & Aramid Fiber-Reinforced Epoxy composites (values based on 60% volume fraction of aligned fibers

in an epoxy matrix).

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Steels Cu,Ni Tin Zinc Silver, Mo Tantalum

√close-packing (metallic bonding)

√large atomic mass

/cm )

3

3 45

Al i Titanium Tin, Zinc

Gl d

Si nitride Diamond Al oxide Zirconia

l

¾Ceramics have...

√less dense packing

ρ (g /

2 _Magnesium Aluminum

G raphite Silicon Glass -soda

Concrete PTFE PETPVC Silicone

AFRE * CFRE * GFRE*

Glass fibers Carbon fibers Aramid fibers

√often lighter elements

¾Polymers have...

√ k ( f h ) ₁

0.5

HDPE, PS PP, LDPE PCPET

W d

√poor packing (often amorphous)

√lighter elements (C, H, O)

¾Composites have _0.4^0.5 ^Wood

¾Composites have...

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Content

1

Introduction Introduction

2

Unit Cell Unit Cell

3

Crystal Structure Crystal Structure

3

Crystal Structure Crystal Structure

4

Crystallographic Directions and Planes Crystallographic Directions and Planes y y g p g p

5

Interstitial Position & Coordination Number Interstitial Position & Coordination Number

6

Structure Determination (X-Ray) Structure Determination (X-Ray)

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Crystallographic Directions and Planes

¾ Crystallographic Coordinates

√ Position: fractional multiples of unit cell edge lengths

√ ex) P: q, r, s

(Fi 3 5)

(Fig. 3-5)

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Crystallographic Directions and Planes

¾ Crystallographic Directions:

√ a line between two points or a vector

√ [uvw] square bracket, smallest integer

√ families of directions: <uvw> angle bracket

A (Fig 3-6)

A (Fig. 3 6)

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Crystallographic Directions and Planes

¾ Crystallographic Planes ( hkl ):

√ Miller Indices: Reciprocals of the three

z p

√ Miller Indices: Reciprocals of the three axial intercepts for a plane, cleared of

fractions & common multiples All parallel ur ur

c

ur

n y fractions & common multiples. All parallel

planes have the same Miller indices.

√ ( hkl )

a

^r

b

^r ^y

m

n

h 1 1

k 1

√ ( hkl )

l

¾ Algorithm

x ^m

h m k

n l

p

¾ E l

g

1. Read off intercepts of plane with axes in terms of a, b, & c.

k l f

¾ Example

1. let m = 2, n = 1, p = ∞ 2. reciprocals are 1/2, 1, 0 2. Take reciprocals of intercepts.

3. Reduce to smallest integer values.

4 Enclose in parentheses no commas i e

2. reciprocals are 1/2, 1, 0 3. then, h = 1, k = 2, l = 0 4. Miller index is (120) 4. Enclose in parentheses, no commas i.e.,

(hkl) .

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Crystallographic Directions and Planes

¾ Crystallographic Planes

(Fig. 3-9)

A B C D

E F

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Crystallographic Directions and Planes

¾ Crystallographic Planes

(Fig 3 9)

(Fig. 3-9)

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Crystallographic Directions and Planes

¾ Crystallographic Planes Family: ex. {110}

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Crystallographic Directions and Planes

¾ Crystallographic Planes Family: ex. {110}

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Crystallographic Directions and Planes

¾ Hexagonal Crystal System

√ Miller Bravais Scheme

√ Miller-Bravais Scheme

(skip)

[ uvtw ] ( hkil )

43

[ ]

( )

t = − + u v

( )

i = − + h k

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Crystallographic Directions and Planes

¾ Directions, Planes, and Family

√ d d

√ Line and direction

[111] square bracket

<111> angular bracket - family

√ Plane

(111) round bracket (Parentheses)

{111} braces - family y

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Crystallographic Directions and Planes

¾ Linear & Planar Density

number of atoms centered on direction vector LD= length of direction vector length of direction vector

number of atoms centered on a plane

PD= area of plane

110 110 2 2

2 atoms 1 2 atoms 1

ex) LD PD

4R 2 R 8 2R 4 2R

= = = = (Fig. 3-10)

4R 2 R 8 2R 4 2R

45

¾ Slip occurs on the most densely packed crystallographic planes, along directions having the greatest atomic packing in those planes

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Slip

(Fig. 7-9)

Plastically stretched y

zinc single crystal

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Content

1

Introduction Introduction

2

Unit Cell Unit Cell

3

Crystal Structure Crystal Structure

3

Crystal Structure Crystal Structure

4

Crystallographic Directions, and Planes Crystallographic Directions, and Planes y y g p g p , ,

5

Interstitial Position & Coordination Number Interstitial Position & Coordination Number

6

Structure Determination (X-Ray) Structure Determination (X-Ray)

47

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Interstitial Position & Coordination Number

¾ Interstitial sites in FCC

Octahedral sites Tetrahedral sites

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Content

1

Introduction Introduction

2

Unit Cell Unit Cell

3

Crystal Structure Crystal Structure

3

Crystal Structure Crystal Structure

4

Crystallographic Directions, and Planes Crystallographic Directions, and Planes y y g p g p , ,

5

Interstitial Position & Coordination Number Interstitial Position & Coordination Number

6

Structure Determination (X-Ray Diffraction) Structure Determination (X-Ray Diffraction)

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Crystals as Building Blocks

l l l

¾ Some engineering applications require single crystals:

- diamond single-crystal

powders for abrasives - turbine blades

(Fig. 8-33)

powders for abrasives

(Fig. 3-16)

¾ Crystal properties reveal

features of atomic structures.

√ Ex: Certain crystal planes in

quartz fracture more easily than others.

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Polycrystals

¾ Most engineering materials are polycrystals

1 mm

¾ Nb-Hf-W plate with an electron beam weld

√ Each "grain" is a single crystal.

√ If crystals are randomly oriented, overall properties are not directional.

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Crystalline nature of materials - Grains

¾ A grain is a region of distinct crystal orientation within a polycrystalline material

polycrystalline material.

¾ The grain boundary is the interface between individual grains

grains.

¾ Grain boundaries have significant influences on the ti f th t i l

properties of the material.

Silicon iron alloy

(53)

Various stages in the solidification of a polycrystalline material

p y y

(Fig 3 17) (Fig. 3-17)

Grain boundary

Grain

53

Grain

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Structure Determination (X-Ray)

¾ Diffraction of Light

Young’s experiment

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Structure Determination (X-Ray)

¾ Diffraction

√ The scattering centers (atoms), have spacing comparable in it d t th l th

magnitude to the wavelength.

√ Diffraction results from specific phase relationships established among the waves scattered by the scattering atoms (centers) among the waves scattered by the scattering atoms (centers).

55

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Structure Determination (X-Ray)

¾ Electromagnetic wave

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Structure Determination (X-Ray)

ff

^X-ray

¾ X-ray Diffraction

Detector

source

X-ray source Specimen

Detector

Goniometer

Al powder

hkl hkl

2d sinθ = λ ⁿ ^p

(Fig. 3-21)

57

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Structure Determination (X-Ray)

¾ Bragg’s Law

AB+BC = 2dsinθ = nλ

O

A C

B

(Fig 3-19)

¾ Interplanar spacing

hkl 2 2 2

d = a (in cubic)

(Fig. 3-19)

hkl

h +k +l

2 2 2

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Structure Determination (X-Ray)

¾ Determination of Crystal Structures

√ Pr blem: Determine the d sp cin f r the (111) pl ne

√ Problem: Determine the d-spacing for the (111) plane, the lattice parameter a

₀

and the atomic radius for Al.

) (

o

for

o

1542 0

111 19

38 2

λ

= θ

⇒

= θ

( )

. nm

sin

nm .

d sin sin

d

_o

0 237

19 2

1542 0

2

₁₁₁

2 =

= ⋅ θ

⋅

= λ

⇒ θ

⋅

= λ

l k

h d

l a k

h

d

_hkl

a

^o _o _hkl ² ² ²

2 2

2

⇒ = ⋅ + +

+

= +

a r

nm x

. nm

x .

a

_o

2 4

3 237

0 1

1 1

237

0

² ² ²

⋅

= +

+

=

59

a . nm

r r a

; nm .

a

_o _o ^o

0 144

4 2 2

410 4

0 ⋅ =

=

⋅ ⇒

=

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Structure Determination (X-Ray)

d l

¾ Indexing example:

√ polycrystalline α-iron (BCC) p y y

c z c z

a b y

c z

x b

a b y

x

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Scanning Tunneling Microscopy (STM)

61

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Scanning Tunneling Microscopy – Silicon (001) Surface

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The Si(001)-(2×1) Reconstruction

Unreconstructed Si(001)-(1×1) surface. The [110]

[010]

Si atoms of the topmost layer are highlighted in orange. These atoms are bonded only two other Si atoms, both of which are in the second

l ( h d d )

[110]

layer (shaded grey).

Reconstructed Si(001)-(2×1) surface. The Si [110]

[010]

( ) ( )

atoms of the topmost layer form a covalent bond with an adjacent surface atom are thus drawn together as pairs.

[110]

Z. Zhang et al. (Oak Ridge National Laboratory)

63 http://www.chem.qmul.ac.uk/surfaces/scc/scat1_6a.htm

Annual. Rev. Mater. Sci. 27, 525 (1997)

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Scanning Tunneling Microscopy

¾ Atoms can be arranged and imaged:

Carbon monoxide Iron atoms arranged on molecules arranged

on a platinum (111) surf c

a copper (111) surface. g

surface.

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Summary

¾ Atoms may assemble into crystalline or amorphous structures.

¾ We can predict the density of a material, provided we know the atomic weight, lattice constant, and crystal geometry

(e.g., FCC, BCC, or HCP).

¾ Material properties generally vary with the single-crystal orientation (i.e., they are anisotropic).

¾ In polycrystals with randomly-oriented grains, properties are generally non-directional (i.e., they are isotropic).

g y ( y p )

¾ Problems from Chap. 3 http://bp.snu.ac.kr

Prob 3-1 Prob 3-4 Prof 3-8 Prob 3-29 Prob. 3-1 Prob. 3-4 Prof. 3-8 Prob. 3-29 Prob. 3-30 Prob. 3-31 Prob. 3-32 Prob. 3-39 Prob. 3-40 Prob. 3-41 Prob. 3-42 Prob. 3-56

65

Prob. 3 40 Prob. 3 41 Prob. 3 42 Prob. 3 56 Prob. 3-60(a) Prob. 3-63 Prob. 3-66

http://bp.snu.ac.kr - 200-09-28