MATERIALS SCIENCE &

BITSPilaniPilani Campus

MATERIALS SCIENCE & ENGINEERING

Dr. Subrata Bandhu Ghosh

Department of Mechanical Engg.

Imperfections in SolidsImperfections in Solids

In our pervious Lecture when

discussing Crystals we

ASSUMED PERFECT ORDER

In real materials we find: Crystalline Defects or lattice irregularity

Most real materials have one or more “errors in Most real materials have one or more “errors in perfection”

with dimensions on the order of an atomic diameter to many lattice sites

Defects can be classification:

1. according to geometry

(point, line or plane)

2. dimensions of the defect

Imperfections in Solids

• The properties of some materials are

profoundly influenced by the

presence of imperfections.

• It is important to have knowledge • It is important to have knowledge

about the types of imperfections that

exist and the roles they play in

affecting the behavior of materials.

Atom Purity and Crystal Perfection

• If we assume a perfect crystal structure

containing pure elements, then

anything that deviated from this

concept or intruded in this uniform

homogeneity would be an homogeneity would be an

imperfection.

1. There are no perfect crystals.

2. Many material properties are improved

by the presence of imperfections and

deliberately modified (alloying and

doping).

Polycrystalline Materials

Grain Boundaries

• regions between

crystals

• transition from lattice of

one region to that of the one region to that of the

other

• ‘slightly’ disordered

• low density in grain

boundaries

– high mobility

– high diffusivity

– high chemical reactivity Adapted from Fig. 4.7, Callister 7e.

• Vacancy atoms

• Interstitial atoms

• Substitutional atomsPoint defects

1-2 atoms

Types of Imperfections

• Dislocations Line defects1-dimensional

• Grain Boundaries Area defects2-dimensional

• Vacancies:-vacant atomic sites in a structure.

Point Defects

Vacancy

distortion

of planes

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

self-interstitial

distortion

of planes

Self Interstitials

• In metals, a self interstitial introduces

relatively large distortions (strain) in

the surrounding lattice since the atom

is substantially larger than the

interstitial site.

POINT DEFECTS

• The simplest of the point defect is a vacancy, or vacant lattice site.

• All crystalline solids contain vacancies.

• Principles of thermodynamics is used explain the necessity of the

existence of vacancies in crystalline solids.

• The presence of vacancies increases the entropy (randomness) of

the crystal.

• The equilibrium number of vacancies for a given quantity of

material depends on and increases with temperature as

follows:

Nv= N exp(-Qv/kT)

Equilibrium no. of vacancies

Total no. of atomic sites Energy required to form vacancy

T = absolute temperature in °Kelvin

k = gas or Boltzmann’s constant

Nv = exp − Qv

No. of defects

No. of potential

Activation energy –energy required for formation of vacancy

• Equilibrium concentration varies with temperature.

Equilibrium Concentration:

Point Defects

Boltzmann's constant

(1.38 x 10-23

J/atom-K)

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

N= exp

kT

No. of potential

defect sitesTemperature

Each lattice site is a potential vacancy site

• We can get Qv from

an experiment. Nv

N= exp

−Qv

kT

Measuring Activation Energy

• Measure this...

Nv

• Replot it...

Nv slope

note:

A El

El

NN

A

ρ=

Nv

N

T

exponential dependence!

defect concentration1/T

N

Nvln

-Qv /k

slope

Example Problem 4.1

Calculate the equilibrium number of vacancies per cubic meter for

copper at 1000°C. The energy for vacancy formation is 0.9 eV/atom; the

atomic weight and density (at 1000 ° C) for copper are 63.5 g/mol and

8.4 g/cm3, respectively.

Solution.

Use equation 4.1. Find the value of N, number of atomic sites per cubic

meter for copper, from its atomic weight Acu, its density, and meter for copper, from its atomic weight Acu, its density, and

Avogadro’s number NA.

toequal ie )1273( 1000at vacanciesofnumber theThus,

/8.0x10

/5.63

)/10)(/4.8)(/10023.6(

3 28

336323

KC

matoms

molg

mcmcmgmolatomsx

A

NN

Cu

A

o

=

==ρ

5

328

)1273)(/1062.8(

9.0(exp)/(8.0x10

exp

=

−=

− KKeVx

eVmatoms

kT

QNN v

v

Continuing:

325

5

/m vacancies2.2x10

)1273)(/1062.8(

=

− KKeVx

And Note: for MOST MATERIALS just

below Tm ���� Nv/N = 10-4

Here: 0.0022/8 = .000275 = 2.75*10-4

Two outcomes if impurity (B) added to host (A):• Solid solution of B in A (i.e., random dist. of point defects)

OR

Substitutional solid soln. Interstitial solid soln.

Point Defects in Alloys

• Solid solution of B in A plus particles of a new

phase (usually for a larger amount of B)

Substitutional solid soln.

(e.g., Cu in Ni)

Interstitial solid soln.

(e.g., C in Fe)

Second phase particle

--different composition

--often different structure.

Imperfections in Solids

Conditions for substitutional solid solution (S.S.)

• Hume – Rothery rules

– 1. ∆r (atomic radius) < 15%

– 2. Proximity in periodic table

• i.e., similar electronegativities• i.e., similar electronegativities

– 3. Same crystal structure for pure metals

– 4. Valency equality

• All else being equal, a metal will have a greater tendency

to dissolve a metal of higher valency than one of lower

valency (it provides more electrons to the “cloud”)

Imperfections in SolidsApplication of Hume–Rothery rules – Solid

Solutions

1. Would you predict

more Al or Ag

Element Atomic Crystal Electro- Valence

Radius Structure nega-

(nm) tivity

Cu 0.1278 FCC 1.9 +2

C 0.071

H 0.046more Al or Ag

to dissolve in Zn?

2. More Zn or Al

in Cu?

Table on p. 106, Callister 7e.

H 0.046

O 0.060

Ag 0.1445 FCC 1.9 +1

Al 0.1431 FCC 1.5 +3

Co 0.1253 HCP 1.8 +2

Cr 0.1249 BCC 1.6 +3

Fe 0.1241 BCC 1.8 +2

Ni 0.1246 FCC 1.8 +2

Pd 0.1376 FCC 2.2 +2

Zn 0.1332 HCP 1.6 +2

More Al because size is closer and val. Is

higher – but not too much – FCC in HCP

Surely Zn since size is closer thus

causing lower distortion (4% vs 12%)

Imperfections in Solids

• Specification of composition

– weight percent100x

21

11

mm

mC

+=

m1 = mass of component 1

100x 21

1'

1

mm

m

nn

nC

+=

nm1 = number of moles of component 1

– atom percent

Wt. % and At. % -- An example

Typically we work with a basis of 100g or 1000g

given: by weight -- 60% Cu, 40% Ni alloy

6009.44

63.55 /Cu

gn m

g m= =

'

'

4006.82

58.69 /

9.44.581 or 58.1%

9.44 6.82

6.82.419 or 41.9%

9.44 6.82

Ni

Cu

Ni

gn m

g m

C

C

= =

= ≈+

= ≈+

Converting Between: (Wt% and At%)

' 1 21

1 2 2 1

' 2 12

100

100

C AC

C A C A

C AC

C A C A

∗= ×

∗ + ∗

∗= ×

∗ + ∗

Converts

from wt% to

At% (Ai is 2

1 2 2 1

'

1 11 ' '

1 1 2 2

'

2 22 ' '

1 1 2 2

100

100

C A C A

C AC

C A C A

C AC

C A C A

∗ + ∗

∗= ×

∗ + ∗

∗= ×

∗ + ∗

At% (Ai is

atomic

weight)

Converts

from at% to

wt% (Ai is

atomic

weight)

Determining Mass of a Species per

Volume

" 311

1 2

10C

CC C

ρ ρ

= × +

• ρi is density of pure

element in g/cc

• Computed this way,

gives “concentration”

1 2

" 322

1 2

1 2

10C

CC C

ρ ρ

ρ ρ

+

= × +

gives “concentration”

of speciesi in kg/m3 of

the bulk mixture

(alloy)