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MME 2001 MATERIALS SCIENCE 1 27.10.2015

outline Planar density

Interplanar spacing

Structure-property relations

X-ray diffraction

Imperfections in solids

point defects

vacancies

impurities

solid solutions

line defects

dislocations

at 14:50

planar densities The parameter corresponding to the linear density

for crystallographic planes is planar density, and

planes having the same planar density values are

also equivalent.

(001) / {001}

planar density Planar Density of Atoms

area of the plane

number of atoms inside the plane

(110) Although six atoms have

centers that lie on this plane,

only one-quarter of each of

atoms A, C, D, and F, and one-

half of atoms B and E, for a

total equivalence of just 2

atoms, are on that plane.

planar density

[110] (110) area of (110) plane is

equal to the product of

its length and width.

length = 4R

width (vertical

dimension)= 2R2

Planar density of (110) planes of FCC crystal

area of this plane =

(4R)( 2R2) = 8R2 2

area of (110) plane = 8R2 2

PD110 = 2 atoms / 8R2 2

= 1 / 4R22

planar density Planar Density of Atoms PD =

[110] Area of the plane

Number of atoms

Corner atoms (A, C, D and F) only quarter

of each is in the plane: 4 x ¼ = 1

Face atoms (B and E): only half of each is

in the plane: 2x1/2: 1

Total number of atoms inside the plane = 2

planar density

[110] (110)

Calculate the planar atomic density for the (110)

plane in FCC copper (a = 0.3615 nm) in

atoms/cm2.

A, C, D, F: corner atoms;

shared by 4 unit cells

B and E: edge atoms; shared

by 2 unit cells

planar density

[110] (110)

Number of atoms on the (110) plane in FCC:

4 atoms at the corners, each shared by 4 unit cells,

2 atoms at the edges, each shared by 2 unit cells

4x1/4 + 2x1/2 = 2 atoms

area of (110) plane = a x a2

a = 0.3615 nm

PD110 = 2 atoms /a22

= 2/a2

= 1.082 x 1015 atoms/cm2

Planar Density of (100) Iron Solution: At T < 912C iron has the BCC structure

Radius of iron R = 0.1241 nm

(100) plane

= PD= a 2

1 atoms

atoms 12.1

atoms = 1.2 x 1019

4 area

Planar Density of (111) Iron (111) plane 1 atom in plane/unit surface cell

3 3 3 2

4 2 a 3 ah 2 area =

atoms in plane

atoms above plane

atoms below plane

atoms 7.0

atoms 0.70 x 1019

3 2 R 3

16 Planar Density =

area Radius of iron R = 0.1241 nm

planar density (111) Plane in a FCC crystal:

(3 corner atoms x 1/6) + (3 side atoms x ½) = 2 atoms

Planar density of (111) plane in the FCC:

PD(111) = 2

2.a x 2.a

3.a2 =

PD(111) > PD (110)

(111) planes are more

densely packed

PD110 = 2/a2

Linear and planar densities

Linear and planar densities are important

considerations relative to the process of

slip—that is, the mechanism by which metals

plastically deform.

Slip occurs on the most densely packed

crystallographic planes and, in those planes,

along directions having the greatest atomic

packing.

Spacing between crystal planes

Relation between planes and directions In the cubic system planes and directions having same

indices are perpendicular to each other,i.e. İf [uvw] direction

is perpendicular to (hkl) plane, then h=u, k=v and l=w

Ex: {100} planes and <100> directions are perpendicular to

each other.

If [uvw] direction is parallel to (hkl), that is if [uvw] lies in

the plane (hkl) then hu + kv + lw = 0.

For example, _

[110] lies in the plane (111)

Since 1x(-1) + 1x1 + 1x0 = 0

Structure property correlation

Al: ductile Fe: not ductile Mg: not ductile

FCC BCC HCP

Plastic deformation via SLIP

Sliding of crystal planes over one another!

SLIP occurs on most densely packed planes in the

most closely packed planes!

slip plane + slip direction: SLIP SYSTEM

Structure property correlation In FCC,

{111} planes are closely packed and there are

4 unique {111} planes.

Each of these planes contains

3 closely packed <111> directions.

In HCP,

The basal plane (0001) is the close packed.

It contains 3 close packed <1120> directions.

4x3=12

systems

Higher number of slip systems allows greater plastic

deformation before fracture imparting ductility to FCC

metals!

Structure property correlation

4x3=12 slip systems

1x3=3 slip systems

Structure property correlation Close packed planes are planes with greatest

interplanar spacing. This allows slip to takeplace

easily on these planes.

BCC structure has 48 possible slip systems.

However, there is no close packed plane. Hence

plastic deformation before fracture is not

substantial.

Slip might occur in {110}, {112} and {123} planes in

the <111> directions.

Learning check

Sketch within a cubic unit cell the plane (312):

Which one

shows the

plane (221)

correctly

drawn?

learning check

Which one

shows the

(212) plane

correctly

drawn?

learning check

Which one

shows the

(111) plane

correctly

Drawn?

learning check

Sketch within a cubic unit cell the following

planes:

Determine the Miller indices for the planes in the

following unit cell:

A: (403) B: (112)

learning check Determine the Miller indices for the planes in the

following unit cell:

A: (322)

B: (202)

B: (221) A: (324)

X-ray diffraction waves 1 and 2 have the same wavelength and

remain in phase after a scattering event.

The amplitudes of the scattered waves add together

in the resultant wave.

X-ray diffraction Waves 3 and 4 have the same wavelength and

become out of phase after a scattering event.

The amplitudes of the two scattered waves

cancel one another.

X-ray diffraction n = SQ + QT

n = dhkl sin + dhkl sin

n = 2dhkl sin

X-Rays to Determine Crystal Structure

• Incoming X-rays diffract from crystal planes.

reflections must be in phase for a detectable signal

spacing between planes

extra distance travelled by wave “2”

X-ray diffraction The magnitude of the distance between two

adjacent and parallel planes of atoms (i.e., the

interplanar spacing d) is a function of the Miller

indices (h, k, and l) as well as the lattice

parameter(s).

For cubic symmetry,

in which a is the lattice parameter (unit cell edge

length).

X-ray diffraction Schematic diagram of an

x-ray diffractometer;

T x-ray source,

S specimen,

C detector, and

O the axis around which

the specimen and

detector rotate.

interplanar spacing computations

diffraction angle computations For BCC iron, compute the diffraction angle for the (220) set

of planes. The lattice parameter for Fe is 0.2866 nm. Also,

assume that monochromatic radiation having wavelength of

0.1790 nm is used and the order of reflection is 1.

X-Ray Diffraction

Diffraction gratings must have spacings comparable to the wavelength of diffracted radiation.

Can’t resolve spacings

Spacing is the distance between parallel planes of atoms.

X-Ray Diffraction Pattern

Diffraction pattern for polycrystalline a-iron (BCC)

Diffraction angle 2

imperfections in solids

homogenization/annealing

rolling/extrusion Atomic

composition

bonding

X’tal structure

Addition and manipulaton

of defects

Microstructure:

material

properties Therm

mechanic

● perfect order is assumed to exist throughout

crystalline materials on an atomic scale.

● However, such an idealized solid does not

exist; all contain large numbers of various

defects.

● Crystalline defect refers to a lattice

irregularity having one or more of its

dimensions on the order of an atomic

diameter.

● Many of the properties of materials are

profoundly sensitive to deviations from

crystalline perfection!

● the impact is not always adverse!

● often specific characteristics are

deliberately fashioned by the introduction

of controlled quantity of defects.

• vacancy atoms

• interstitial atoms

• substitutional atoms

Point defects

0-dimensional

Types of Imperfections

• dislocations Line defects

1-dimensional

• grain boundaries Area defects

2-dimensional

• cavities/porosity volume defects

3-dimensional

vacant atomic site/missing atom

Point Defects / vacancies

Vacancy

distortion of planes

All crystalline solids contain vacancies.

vacancies increase the entropy (the randomness) of the

crystal: thermodynamically favorable!

the number of vacancies increases exponentially with T!

For most metals, Nv/N at Tm is on the order of 104; one

lattice site out of 10,000 will be empty.

Point Defects / self-interstitial

self- interstitial

distortion of planes

● own atom crowded into an interstitial site!

● introduces relatively large distortions in the

surrounding lattice as the atom is much larger than

the interstitial position.

● is not highly probable!

● exists in very small concentrations

● significantly lower than for vacancies.

Most common defects in crystalline solids are point

defects.

At high temperatures,

atoms frequently and

randomly change their

positions leaving behind

empty lattice sites.

In general, diffusion

(mass transport by atomic

motion) can only occur

because of vacancies.

Point Defects

distance

vacancy atom

energy

Point Defects in ionic crystals Charge neutrality must be

maintained!

Schottky imperfection

Formation of equivalent

(not necessarily equal)

numbers of cationic and

anionic vacancies

Frenkel imperfection

Formation of an ion

vacancy and an ion

interstitial

Schottky defect

Frenkel defect

Boltzmann's constant

(1.38 x 10 -23 J/atom-K)

(8.62 x 10 -5 eV/atom-K)

N = exp

No. of defects

No. of potential

defect sites

Activation energy

Temperature

Each lattice site is a potential vacancy site

• Equilibrium concentration varies with temperature!

Equilibrium Concentration: Point Defects

• We can get Qv from

an experiment. N v

N = exp

Measuring Activation Energy

• Measure this...

exponential dependence!

defect concentration

• Replot it...

N v ln

- Q v /k

N = Q v

Equilibrium # of vacancies in 1 m3 of Cu at 1000C?

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

Q v = 0.9 eV/atom N A = 6.02 x 1023 atoms/mol

Estimating Vacancy Concentration

For 1 m3 , N = N A

A Cu x x 106 cm3 = 8.0 x 1028 sites/

8.62 x 10-5 eV/atom-K

0.9 eV/atom

N = exp

= 2.7 x 10-4

N v = (2.7 x 10-4)(8.0 x 1028) sites = 2.2 x 1025 vacancies

1 m3 m3

Equilibrium # of vacancies in 1 m3 of Fe at 850C?

A Fe = 55.85 g/mol = 7.65 g / cm 3

8.62 x 10-5 eV/atom-K

1.08 eV/atom

N v N = exp Q v

N v = 1.18 x 1024 vacancies

A Fe = exp

N = (6.022x1023 atoms/mol) (7.65g/cm3)

55.85 g/mol

1.08 eV/atom

(8.62x10-5 eV/atom-K)(1123K) exp -

fraction of atom sites that are vacant for lead at its

melting temperature of 327°C?

8.62 x 10-5 eV/atom-K

0.55 eV/atom

N = exp

= 2.41 x 10-5

see you next week!

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