Lecture 1 - Imperfections in Solid

8/16/2019 Lecture 1 - Imperfections in Solid

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Imperfectionsin Solid

UNIVERSITI TUNKU ABDUL RAHMAN (UTAR)


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Outline

• Point Defects

• Impurities in Solids

• Dislocations

• Interfacial Defects

• Specification of Composition


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Imperfectionsin Solid

Point Defects

Dislocations(Linear Defects)

Interfacial Defects

Bulk or Volume Defects

Atomic Vibrations

Imperfections in Solid


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Definition of point defects

An atom missing or is in an irregular position

in the lattice structure.

• vacancies

• self interstitial atoms

• interstitial impurity atoms• substitutional atoms

Point Defects


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Vacancy – an empty space (vacant lattice site),where an atom should be, but is missing.• Vacancy is formed due to a missing atom• Vacancy is formed during crystallization/solidification or

mobility of atoms• Also caused due to plastic deformation, rapid cooling or

particle bombardment

Vacancy

distortionof planes

(a) (b)

Point Defects


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N V = num ber o f vacanc ies per m eter cub e

N = total number of atomic sites (usually per cubic meter)

Q n = energy required for the formation of a vacancy (J/mol or eV/atom)

T = absolute temperature in Kelvins

K = Boltzmann’s constant (1.38 x 10 -23 J/atom.K or 8.62 x10 -5 eV/atom.K)

Point Defects

N n = N exp ( – --- )Q n kT


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Point Defects

Self interstitial atom is an extra atom that has crowded

its way into an interstitial site (a small void space thatunder ordinary circumstances is not occupied) in thecrystal structure.

• Atom in a crystal, sometimes, occupies interstitial site

• This does not occur naturally• Can be induced by irradiation• This defects cause structural distortion

self-

interstitial

distortionof planes


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Impurities in Solids

Solid Solution – Solute atoms are added to thehost material (solvent or host atom), the crystalstructure is maintained and no new structures areformed. Example: Alloy


Substitutional Interstitial


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Substitutional impurity atom – atom of adifferent type than the bulk atoms, which hasreplaced one of the bulk atoms in the lattice.

Substitutional atom


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Interstitial impurity atom – atom of a differenttype than the bulk atoms; it is much smaller thanbulk atom. It fill the voids or interstices amongthe host atom.


Interstitial impurity atom


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• Atomic size factors – Appreciable quantities of asolute may be accommodated in this type of solidsolution only when the difference in atomic radiibetween the two atom types is less than about15%. Otherwise the solute atoms will createsubstantial lattice distortions and a new phase willform.

• Crystal structure – For appreciable solid solubilitythe crystal structures for metals of both atom typesmust be the same.

Condition for Substitutional Solid-solutionHume – Rothery ru le


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• Electronegativity – the more electropositive oneelement and the more electronegative the other, thegreater is the likelihood that they will form anintermetallic compound instead of a substitutionalsolid solution.

• Valences – Other factors being equal, a metal willhave more of a tendency to dissolve another metalof higher valency than one of a lower valency.

For example in aluminium-nickel alloy system,nickel (lower valence) dissolves 5 percentaluminium but aluminium (higher valence) dissolvesonly 0.04 percent nickel.


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Application of Hume –Rothery ru les

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

Cu 0.1278 FCC 1.9 +2

C 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 +2Cr 0.1249 BCC 1.6 +3Fe 0.1241 BCC 1.8 +2Ni 0.1246 FCC 1.8 +2Pb 0.1750 FCC 1.6 +2, +4Zn 0.1332 HCP 1.6 +2


Would you predictmore Al or Pbto dissolve in Cu?


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Dislocations :

Schematic of Zinc (HCP):

• before deformation• after tensile elongation

slip steps (slip bands)

• are line defects (linear defect around which some of the atoms aremisaligned; lattice distortions are centered around a line)• slip between crystal planes result when dislocations move• produce permanent (plastic) deformation• formed during: solidification, permanent deformation, atomic

mismatch in solid solution

Dislocations (Linear Defects)


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Dislocations (Linear Defects)

Different types of line defects are

• Edge dislocation

• Screw dislocation

• Mixed dislocation


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• Created due to shear stresses applied toregions of a perfect crystal separated by cuttingplane.

• Distortion of lattice in form of a spiral ramp.• Burgers vector is parallel to dislocation line.

Screw Dislocations


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Screw Dislocations

• A screw dislocation within a crystal. The atomic distortionassociated with a screw dislocation is also linear and along adislocation line, line AB.

• Atom positions above the slip plane are designated by opencircles, those below by solid circles.


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• Most crystal have components

of both edge and screwdislocation.

• Dislocation, since haveirregular atomic arrangement

will appear as dark lineswhen observed in electronmicroscope.

Mixed Dislocations

TEM analysis of Ti alloy


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Interfacial Defects

Interfacial Defects

External Surfaces

Grain Boundaries

Twin Boundaries

Planar Defects

Interfacial defects are boundaries that have two

dimensions and normally separate regions of materialsthat have different crystal structures and / or crystallographic orientations.


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External Surface Defects

• External Surfaces : The boundaries along which the

crystal structure terminates. Surface atoms are notbonded to maximum number of nearest neighbors. Theatoms are in higher energy state than atoms at interior positions.

• Free surface is also a defect : Bonded to atoms on onlyone side and hence has higher state of energy(highly reactive)

• Nanomaterials have small clusters of atoms and henceare highly reactive.

• At all possible, surface area tend to be minimum


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Grain Boundaries

Grain Boundary : The boundary separating two

grains or crystals having different crystallographicorientations (in polycrystalline materials).

3D view of grains


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Grain Boundaries

• Formed due to simultaneously growing

crystals meeting each other.• Width = 2-5 atomic diameters

• Within boundary region, there is someatomic mismatch

• Some atoms in grain boundaries havehigher energy (favor nucleation & growthof precipitates)

• Allow rapid diffusion of atoms due tolower atomic packing

• Restrict plastic flow and preventdislocation movement

Grain Boundaries in 1018 steel


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Twin Boundaries

• Twin Boundary : Grain boundary which there is a specific

mirror lattice symmetry. The atoms on one side of theboundary are located in mirror-image positions of the atomson the other side

• Formed during plastic deformation and recrystallization

• Strengthens the metal

Twin

Twin Plane


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Specification of Composition

C 1 = ---------------- x 100 m 1

m 1 + m 2

• Weight percent (wt%): weight of a particular element relative to the total alloy weight

m 1 & m2 weight (mass) of elements 1 and 2,

respectively


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• Atom percent (at%): number of moles of anelement in relation to the total moles of theelements in the alloy


C’ 1 = ---------------- x 100 nm1

nm1 + n m2

nm1 = ---------m 1 A1

nm1 = number of molesfor element 1

m 1 = mass in gram A1 = atomic weight

Atom percent of element 1


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Com pos i t ion Conv ers ion Conv ers ion of w t% to a t% for a tw o e lem ents a l loy

C’ 1 = --------------------- x 100 C 1 A2 C 1 A2 + C 2 A1

C’ 2

= --------------------- x 100 C 2 A1

C 1 A2 + C 2 A1

Weight percent = C 1 , C 2 Atoms percent = C’ 1, C’ 2 Atomic weight = A 1, A2


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Com pos i t ion Conv ers ion Conv ers ion of a t% to wt% fo r a two elem ents a lloy


C 1 = --------------------- x 100 C’ 1 A2 C’ 1 A1 + C’ 2 A2

C 2

= --------------------- x 100 C’ 2 A2

C’ 1 A1 + C’ 2 A2

Weight percent = C 1 , C 2 Atoms percent = C’ 1, C’ 2 Atomic weight = A 1, A2


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C’’ 1 = ----------------- x 1000 ------ +C 1 1

C 1

------C 2 2


To co nver t concen t rat ion f rom weigh t % to m ass o f one

com pon ent per un i t vo lum e of m ater ial (wt% to kg /m 3 )

C’’ 2 = ----------------- x 1000 ------ +C 1 1

C 2

------C 2 2

Density, = g/cm 3

E l 1


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Calculate the equilibrium number of vacancies per cubic

meter for copper at 1000 oC. The energy for vacancyformation is 0.9eV/atom; the atomic weight and density (at1000 oC) for copper are 63.5 g/mol and 8.4g/cm 3 respectively.

Solution:This problem may be solved by using equation

However it is necessary to determine the value N, number of atomic sites per cubic meter first

N n = N exp ( – --- )Q n kT

Example 1

E l 1


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Atomic weight A Cu , its density and Avogadro’s number N A.

N = -------- ACuN A

= ----------------------------------------63.5g/mol

(6.023 x10 23 atoms/mol)(8.4g/cm 3 )(10 6 cm 3 /m 3 )

= 8.0 x 10 28 atoms/m 3

Thus, the number of vacancies at 1000 oC (1273K) is equal to

N v

= N exp ( – ------)kT

Q v

= (8.0 x 10 28 atoms/m 3 exp[ – -----------------------------------] (0.9 eV)

= 2.2 x 10 25 vacancies/m 3(8.62 x10 -5 eV/K)(1273K)

Example 1

E l 2


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Determine the composition, in atom percent, of an

alloy that consist of 97 wt% aluminum and 3 wt%copper.

C Cu = 3 C Al = 97

C’ Cu = --------------------- x 100 C Cu A Al

C Cu A Al + C Al ACu

= ---------------------------------------------- x 100 (3)(26.98g/mol)

(3)(26.98g/mol) + (97)(63.55g/mol)= 1.30 at%

Example 2

E l 2


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Determine the composition, in atom percent, of an

alloy that consist of 97 wt% aluminum and 3 wt%copper.

C Cu = 3 C Al = 97

C’ Al = --------------------- x 100 C Al ACu

C Cu A Al + C Al ACu

= ---------------------------------------------- x 100 (97)(63.55g/mol)

(3)(26.98g/mol) + (97)(63.55g/mol)= 98.7 at%

Example 2

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Lecture 1 - Imperfections in Solid