ppt slides - 2

7/27/2019 PPT Slides - 2

1/61

1

Crystalline and amorphous materials

Solids can be divided into two categories.

CrystallineAmorphous

Crystalline has long range order

Amorphous materials have short range order

Effect of Crystallinity on Physical properties - ex. Polyethylene

7/27/2019 PPT Slides - 2

2/61

2

CrystalType

Particles InterparticleForces

Physical Behaviour Examples

Atomic

Molecular

Metallic

Ionic

Network

Atoms

Molecules

Atoms

Positive andnegative ions

Atoms

Dispersion

DispersionDipole-dipoleH-bonds

Metallic bond

Ion-ionattraction

Covalent

Soft Very low mp Poor thermal and electrical conductors

Fairly softLow to moderate mpPoor thermal and electrical conductors Soft to hardLow to very high mpMellable and ductileExcellent thermal and electrical

conductorsHard and brittle

High mpGood thermal and electrical conductors

in molten condition Very hard Very high mp Poor thermal and electrical conductors

Group 8A

Ne to Rn

O2, P4, H 2O,Sucrose

Na, Cu, Fe

NaCl, CaF 2,MgO

SiO 2(Quartz)

C (Diamond)

TYPES OF CRYSTALLINE SOLIDS

7/27/2019 PPT Slides - 2

3/61

3

CRYSTAL STRUCTURE

Crystal structure is the periodic arrangement of atoms in thecrystal. Association of each lattice point with a group of atoms (Basis or Motif) .Lattice : Infinite array of points in space, in which each point hasidentical surroundings to all others.

Space Lattice Arrangements of atoms= Lattice of points onto which the atoms are hung.

Elemental solids (Argon): Basis = single atom.Polyatomic Elements: Basis = two or four atoms .Complex organic compounds: Basis = thousands of atoms .

+

Space Lattice + Basis = Crystal Structure

=

7/27/2019 PPT Slides - 2

4/61

4

THREE DIMENTIONAL UNIT CELLS / UNIT CELL SHAPES

1

2

3

4

5

6

7

7/27/2019 PPT Slides - 2

5/61

5

Primitive ( P ) Body Centered ( I )

Face Centered ( F ) C-Centered ( C )

LATTICE TYPES

7/27/2019 PPT Slides - 2

6/61

6

BRAVAIS LATTICES

7 UNIT CELL TYPES + 4

LATTICE TYPES = 14

BRAVAIS LATTICES

7/27/2019 PPT Slides - 2

7/61

7/27/2019 PPT Slides - 2

8/61

8

Close-packing- HEXAGONAL coordination of each sphere

SINGLE LAYER PACKING

SQUARE PACKING CLOSE PACKING

7/27/2019 PPT Slides - 2

9/61

9

NON-CLOSE-PACKED STRUCTURES

68% of space is occupied

Coordination number = 8

a) Body centered cubic ( BCC ) b) Primitive cubic ( P)

52% of space is occupiedCoordination number = 6

7/27/2019 PPT Slides - 2

10/61

10

6

ABCABC 12Cubic close packed

ABABAB 12Hexagonal

close packed

ABABAB 8Body-centered

Cubic

AAAAA PrimitiveCubic

Stacking pattern

Coordinationnumber

Structure

Non-close

packing

Closepacking

7/27/2019 PPT Slides - 2

11/61

11

TETRAHEDRAL HOLES

OCTAHEDRAL HOLES

TYPE OF HOLES IN CLOSE PACKING

7/27/2019 PPT Slides - 2

12/61

12

Ionic structures may be derived from the

occupation of holes by oppositely charged

ions (interstitial sites) in the close-packed

arrangements of ions.

IONIC CRYSTAL STRUCTURES

7/27/2019 PPT Slides - 2

13/61

7/27/2019 PPT Slides - 2

14/61

14

IONIC CRYSTAL TYPES

Ionic crystaltype

Co-ordinationnumber

A X

Structure type

AX

AX 2

AX 3

6 68 8

6 3

8 4

6 2

NaClCsCl

Rutile(TiO 2)

Fluorite (CaF2)

ReO 3

7/27/2019 PPT Slides - 2

15/61

15

Formula Type and fraction of sitesoccupied

CCP HCP

AX All octahedral

Half tetrahedral (T+ or T-)

Rock salt (NaCl)

Zinc Blend (ZnS)

Nickel Arsenide (NiAs)

Wurtzite (ZnS)

AX2

All Tetrahedral

Half octahedral (orderedframework)

Half octahedral (Alternatelayers full/ empty)

Fluorite (CaF2),

Anti-Fluorite (Na 2O)

Anatase (TiO 2)

Cadmium Chloride(CdCl 2)

Not known

Rutile (TiO 2)

Cadmium iodide (CdI 2)

A3X All octahedral & AllTetrahedral

Li3Bi Not known

AX 3 One third octahedral YCl 3 BiI 3

SUMMARY OF IONIC CRYSTAL STRUCTURE TYPES

7/27/2019 PPT Slides - 2

16/61

16

Crystallographic notation of atomic planes and directions

7/27/2019 PPT Slides - 2

17/61

7/27/2019 PPT Slides - 2

18/61

18

POINT Coordinates

To define a point within a unit cell. Express the coordinates uvw as fractions of unit cell vectors a , b , and c (so that the axes x, y, and z do not have to be orthogonal).

a

b

c

origin

pt. coord.

x ( a ) y ( b ) z ( c )

0 0 0

1 0 0

1 1 1

1/2 0 1/2

pt.

7/27/2019 PPT Slides - 2

19/61

7/27/2019 PPT Slides - 2

20/61

20

Self-Assessment Example 1: What is crystallographic direction?

ab

c Along x: 1 a

Along y: 1 b

Along z: 1 c

[1 1 1]DIRECTION =

Magnitude alongX

Y

Z

7/27/2019 PPT Slides - 2

21/61

7/27/2019 PPT Slides - 2

22/61

22

Families and Symmetry: Cubic Symmetry

x

y

z

(100)

Rotate 90 o about z-axis

x

y

z

(010)

x

y

z

(001)Rotate 90 o about y-axis

Similarly for other equivalent directions

Symmetry operation cangenerate all the directionswithin in a family.

7/27/2019 PPT Slides - 2

23/61

7/27/2019 PPT Slides - 2

24/61

24

How Do We Designate Lattice Planes?

Example 1

Planes intersects axes at: a axis at r= 2 b axis at s= 4/3 c axis at t= 1/2

How do we symbolically designate planes in a lattice?

Possibility #1: Enclose the values of r, s, and t in parentheses (r s t) Advantages:

r, s, and t uniquely specify the plane in the lattice, relative to the origin. Parentheses designate planes, as opposed to directions given by [...]

Disadvantage: What happens if the plane is parallel to --- i.e. does not intersect--- one of the axes? Then we would say that the plane intersects that axis at ! This designation is unwieldy and inconvenient.

7/27/2019 PPT Slides - 2

25/61

25

How Do We Designate Lattice Planes?

Planes intersects axes at: a axis at r= 2

b axis at s= 4/3 c axis at t= 1/2

How do we symbolically designate planes in a lattice?

Possibility #2: THE ACCEPTED ONE

1. Take the reciprocal of r, s, and t . Here: 1/r = 1/2 , 1/s = 3/4 , and 1/r = 2

2. Find the least common multipl e that converts all reciprocals to integers . With LCM = 4, h = 4/r = 2 , k= 4/s = 3 , and l= 4/r = 8

3. Enclose the new triple (h,k,l) in parentheses : (238) 4. This notation is called the Miller Index .

* Note: If a plane does not intercept an axes (I.e., it is at ), then you get 0.* Note: All parallel planes at similar staggered distances have the same Miller index.

7/27/2019 PPT Slides - 2

26/61

7/27/2019 PPT Slides - 2

27/61

27

Families of Lattice Planes

The Miller indices (hkl) usually refer to the plane thatis nearest to the origin without passing through it.

You must always shift the origin or move the planeparallel, otherwise a Miller index integer is 1/0, i.e.,!

Sometimes (hkl) will be used to refer to any other planein the family, or to the family taken together.

Importantly, the Miller indices (hkl) is the same vector asthe plane normal!

Given any plane in a lattice, there is a infinite set of parallel lattice planes(or family of planes) that are equally spaced from each other.

One of the planes in any family always passes through the origin.

7/27/2019 PPT Slides - 2

28/61

28

z

x

y

Look down this direction(perpendicular to the plane)

Crystallographic Planes in FCC: (100)

d 100

aDistance between (100) planes

Distance to (200) plane d 200a

2

7/27/2019 PPT Slides - 2

29/61

7/27/2019 PPT Slides - 2

30/61

30

Crystallographic Planes in FCC: (111)

z

x

y

Look down this direction(perpendicular to the plane)

d 111

a 33

Distance between (111) planes

7/27/2019 PPT Slides - 2

31/61

31

Note: similar to crystallographic directions, planes that are parallel toeach other, are equivalent

i i i C C l

7/27/2019 PPT Slides - 2

32/61

32

Directions in HCP Crystals

1. To emphasize that they are equal, a and b is changed to a 1 and a 2.2. The unit cell is outlined in blue .

3. A fourth axis is introduced (a 3) to show symmetry. Symmetry about c axis makes a 3 equivalent to a 1 and a 2. Vector addition gives a 3 = ( a 1 + a 2).

4. This 4-coordinate system is used: [a 1 a 2 ( a 1 + a 2) c]

Di i i HCP C l d

7/27/2019 PPT Slides - 2

33/61

33

Directions in HCP Crystals: 4-index notation

Example What is 4-index notation for vector D?

Projecting the vector onto the basal plane, it liesbetween a 1 and a 2 (vector B is projection).

Vector B = (a 1 + a 2), so the direction is [110] incoordinates of [a 1 a 2 c], where c-intercept is 0.

In 4-index notation, because a 3 = ( a 1 + a 2), thevector B is since it is 3x farther out.

In 4-index notation c = [0001], which must beadded to get D (reduced to integers ) D = [1123]

Self-Assessment Test: What is vector C?

Easiest to remember: Find the coordinate axes that straddle the vector of interest, and follow along those axes (but divide the a 1, a 2 , a 3 part of vector by 3 because you are now three times farther out!) .

13

[112 0]

Check w/ Eq. 3.7

or just use Eq. 3.7

a 2 2a 3 B without 1/3

7/27/2019 PPT Slides - 2

34/61

Mill I di f HCP Pl

7/27/2019 PPT Slides - 2

35/61

35

Miller Indices for HCP Planes

As soon as you see [1100], you will knowthat it is HCP, and not [110] cubic!

4-index notation is more important for planes in HCP, in order to distinguish similar planes rotated by 120 o.

1. Find the intercepts , r and s , of the plane with any two of the basal plane axes ( a 1, a 2, or a 3), as well as theintercept, t , with the c axes.

2. Get reciprocals 1/r, 1/s, and 1/t.3. Convert reciprocals to smallest integers in same ratios .

4. Get h, k, i , l via relation i = - (h +k), where h isassociated with a 1, k with a 2, i with a 3, and l with c .5. Enclose 4-indices in parenthesis: (h k i l ) .

Find Miller Indices for HCP:

r s

t

Mill I di f HCP Pl

7/27/2019 PPT Slides - 2

36/61

36

Miller Indices for HCP Planes

What is the Miller Index of the pink plane?

1. The planes intercept a 1, a 3 and c at r=1, s=1 and t= , respectively.

1. The reciprocals are 1/r = 1, 1/s = 1, and 1/t = 0.

2. They are already smallest integers.

3. We can write (h k i l ) = ( 1 ? 1 0 ).

4. Using i = - (h+k) relatio n, k= 2 .

5. Miller Index is (1210)

7/27/2019 PPT Slides - 2

37/61

37

Yes, Yes.we can get it without a 3!

1. The planes intercept a 1, a 2 and c at r=1, s= 1/2 and t= , respectively.

2. The reciprocals are 1/r = 1, 1/s = 2, and 1/t = 0.

3. They are already smallest integers.

4. We can write (h k i l ) =

5. Using i = - (h+k ) relation , i=1 .

6. Miller Index is (12 10)

(12 ?0)

But note that the 4- index notation is unique.Consider all 4 intercepts: plane intercept a 1, a 2, a 3 and c at 1, 1/2, 1, and , respectively. Reciprocals are 1, 2, 1, and 0. So, there is only 1 possible Miller Index is (12 10)

7/27/2019 PPT Slides - 2

38/61

38

(1 1 1) plane of FCCz

x

y

z

a 1

a 2

a 3

(0 0 0 1) plane of HCPSAME THING!*

7/27/2019 PPT Slides - 2

39/61

7/27/2019 PPT Slides - 2

40/61

40

Defects in Solids

There is no such thing as a perfect crystal.

What are these imperfections? Why are they important?

Many of the important properties of materials

are due to the presence of imperfections.

7/27/2019 PPT Slides - 2

41/61

41

Vacancy atoms Interstitial atoms Substitutional atoms

Point defects

Types of Imperfections

Dislocations Line defects

Grain Boundaries Area defects

7/27/2019 PPT Slides - 2

42/61

42

Vacancies :-vacant atomic sites in a structure.

Self-Interstitials :-"extra" atoms positioned between atomic sites.

Point Defects

Vacancy distortion

of planes

self- interstitial

distortionof planes

7/27/2019 PPT Slides - 2

43/61

43

Boltzmann's constant

(1.38 x 10 -23 J/atom-K)(8.62 x 10 -5 eV/atom-K)

N v N

exp Q v k T

No. of defects

No. of potentialdefect sites.

Activation energy

AbsoluteTemperature

Each lattice siteis a potentialvacancy site

Equilibrium concentration varies with temperature!

Equilibrium Concentration:Point Defects

1 eV = 1.6 x 10 -19 J

unitless; probability

7/27/2019 PPT Slides - 2

44/61

7/27/2019 PPT Slides - 2

45/61

45

Find the equil. # of vacancies in 1 m 3 of Cu at 1000 C. Given:

A Cu = 63.5 g/mol r = 8.4 g / cm 3

Q v = 0.9 eV/atom N A = 6.02 x 10 23 atoms/mol

Estimating Vacancy Concentration

For 1 m 3 , N = N A A Cu

r x x 1 m 3 = 8.0 x 10 28 sites8.62 x 10 -5 eV/atom-K

0.9 eV/atom

1273K

N v N

exp Q v k T

= 2.7 x 10 -4

Answer:

N v

= (2.7 x 10 -4)(8.0 x 10 28 ) sites = 2.2 x 10 25 vacancies

Cu T m=1083 C, so at T~T m~ 1/10 4 sites are vacant(general rule)

7/27/2019 PPT Slides - 2

46/61

7/27/2019 PPT Slides - 2

47/61

47

Imperfections in Solids

Conditions for substitutional solid solution (S.S.) W. Hume Rothery rule

1. r (atomic radius) < 15% 2. Proximity in periodic table

i.e., similar electronegativities 3. Same crystal structure for pure metals 4. Valency

All else being equal, a metal will have a greater tendency to dissolve a metal of higher valency thanone of lower valency.

7/27/2019 PPT Slides - 2

48/61

48


Application of Hume Rothery rules Solid Solutions

1. Would you predictmore Al or Agto dissolve in Zn?

2. More Zn or Al

in Cu?

Element Atomic Crystal Electro- ValenceRadius Structure nega-(nm) tivity

Cu 0.1278 FCC 1.9 +2C 0.071H 0.046O 0.060

Ag 0.1445 FCC 1.9 +1 Al 0.1431 FCC 1.5 +3Co 0.1253 HCP 1.8 +2

Cr 0.1249 BCC 1.6 +3Fe 0.1241 BCC 1.8 +2Ni 0.1246 FCC 1.8 +2Pd 0.1376 FCC 2.2 +2Zn 0.1332 HCP 1.6 +2

7/27/2019 PPT Slides - 2

49/61

49


Specification of composition

weight percent 100x21

11 mm

mC

m 1 = mass of component 1

100x21

1'1

mm

m

nn

nC

nm1 = number of moles of component 1

atom percent

7/27/2019 PPT Slides - 2

50/61

50

Line Defects in Solids

Edge Dislocation

One dimensional defect in which atoms are mispositioned.

7/27/2019 PPT Slides - 2

51/61

51


Linear Defects ( Dislocations ) Are one-dimensional defects around which atoms are misaligned

Edge dislocation: extra half-plane of atoms inserted in a crystal structure

b to dislocation line Screw dislocation:

spiral planar ramp resulting from shear deformation b to dislocation line

Burgers vector, b : measure of lattice distortion

7/27/2019 PPT Slides - 2

52/61

52

Dislocation motion requires the successive bumping of a half plane of atoms (from left to right here).

Bonds across the slipping planes are broken and remade in succession.

materials can be deformed rather easily

Atomic view of edgedislocation motion fromleft to right as a crystalis sheared.

Motion of Edge Dislocation

7/27/2019 PPT Slides - 2

53/61

53


Screw Dislocation

Burgers vector,b

Dislocationline

b

(a)

(b)

Screw Dislocation

Edge Screw and Mixed

7/27/2019 PPT Slides - 2

54/61

54

Edge, Screw, and MixedDislocations

Edge

Screw

Mixed

7/27/2019 PPT Slides - 2

55/61

55


Dislocations are visible in electron micrographs

7/27/2019 PPT Slides - 2

56/61

56

Planar Defects in Solids One case is a twin boundary (plane)

Essentially a reflection of atom positions across the twin plane .

Stacking faults For FCC metals an error in ABCABC packing sequence Ex: ABCABABC

Adapted from Fig. 4.9, Callister 7e.

7/27/2019 PPT Slides - 2

57/61

57

Polycrystalline Materials

Grain Boundaries regions between crystals transition from lattice of

one region to that of theother

slightly disordered low density in grain

boundaries high mobility high diffusivity high chemical reactivity

Adapted from Fig. 4.7, Callister 7e.

7/27/2019 PPT Slides - 2

58/61

58

Microscopic Examination Crystallites (grains) and grain boundaries. Vary

considerably in size. Can be quite large ex: Large single crystal of quartz or diamond or Si ex: Aluminum light post or garbage can - see the

individual grains Crystallites (grains) can be quite small (mm or

less) necessary to observe with a microscope.

O ti l Mi

7/27/2019 PPT Slides - 2

59/61

59

Useful up to 2000X magnification.

Polishing removes surface features (e.g., scratches) Etching changes reflectance, depending on crystal orientation.

Micrograph of brass (a Cu-Zn alloy)

0.75mm

Optical Microscopy

Adapted from Fig. 4.13(b) and (c), Callister 7e. (Fig. 4.13(c) is courtesyof J.E. Burke, General Electric Co.

crystallographic planes

O ti l Mi

7/27/2019 PPT Slides - 2

60/61

60

Grain boundaries...

are imperfections, are more susceptible

to etching, may be revealed as

dark lines,

change in crystalorientation acrossboundary. Adapted from Fig. 4.14(a)

and (b), Callister 7e. (Fig. 4.14(b) is courtesyof L.C. Smith and C. Brady,

the National Bureau of Standards, Washington, DC[now the National Institute of Standards and Technology,Gaithersburg, MD].)

Optical Microscopy

ASTM grainsize number

N = 2 n -1

number of grains/in 2 at 100x

magnification

Fe-Cr alloy(b)

grain boundary surface groove

polished surface

(a)

7/27/2019 PPT Slides - 2

61/61

ppt slides - 2

Documents