X-RAY DIFFRACTION and Crystal Defects

7/30/2019 X-RAY DIFFRACTION and Crystal Defects

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Introduction:

X-ray diffraction techniques are very useful for crystalstructure analysis and identification of different types of

crystals.

Experimental study of crystalline materials becamepossible only after the discovery of x-rays.

Diffraction occurs when waves traveling through anaperture whose dimensions are order of integralmultiples of wavelength.

Typical inter atomic spacing in crystals is 2-3A. Thesewavelengths lies in x-ray spectra.

Hence x-rays diffraction is used to study the crystalstructures.


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Braggs law:

It states that, the path difference between the

two reflected rays from parallel planes of acrystal should be always equal to integralmultiples of wavelength of x-rays so as toproduce maxima or constructive interference.

The crystal acts like a series of parallelreflecting planes.

Consider a ray PA reflected at an atom A, inthe direction AR from plane 1 and anotherray QB reflected from plane 2 at another atomB in the direction BS.


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Therefore the path difference between these tworays = CB+BD

When path difference is equal to the integral multipleof wave length, the two rays will reinforce with eachother subsequently an intense spot is produced.

The path length traveled by x-rays along QCBDS isgreater than that of the path length traveled alongPAR.

Thus the path difference = QCBDS PAR.

= (QC + CB + BD + DS) (PA + AR)

since QC = PA and DS = AR (from fig.)


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CB+BD=n

CB=BD=d sin (from Fig)

2d sin= n

Where n = 1,2,3,.. represents the first

order, second order nth

order spectra. For 1st order sin1 = /2d.

For 2nd order sin2 = 2 / 2d.

For 3rd order sin3 = 3 / 2d. where 1, 2 and 3 are the glancing

angles for n=1,2 and 3 respectively.


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Let d1 0 0, d1 1 0 and d1 1 1 represents the inter planar distances of (1 0

0) , (1 1 0) and (1 1 1 ) planes. Let the corresponding glancing angles

are 1, 2 and 3 respectively for the same wavelength using

Braggs law we get

2d1 0 0 sin1 = n

2d1 1 0 sin2 = n and

2d1 1 1 sin3 = n

From above relations it is very clear that sin 1/d

Therefore sin1 : sin2 : sin3 = 1/d1 0 0 : 1/d1 1 0 : 1/d1 1 1

When above ratio satisfies then it is

conformed to be simple cubic system.If it satisfies then it is BCC.

And for the ratio it is FCC.

Thus, the Braggs law can help in determining the crystal structures2

3:2:1

3:2

1:1

3:2:1


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Crystal Defects


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1.Vacancies or Schottky

2.Interstitialcies or Frenke

3. Compositional defects.

a. Substitutional

b. interstitial4. Electronic defects

Defects

Point defects

(0-dimensional)

Line defects

(1-dimensional)

Surface defects

(2-dimensional)

Volume defects

(3-dimensional)

1.Edge dislocation

2.Screw dislocation

1.Grain boundaries

2.Tilt boundaries

3.Twin boundaries

4.Stacking faults

1.Cracks

2.Voids or air bubbles


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Point defectsThese are the places where an atom is missing or

irregularly placed in the lattice structure. Point defects

include lattice vacancies, self-interstitial atoms, substitutionimpurity atoms, and interstitial impurity atoms.Linear defectsMissing of groups of atoms ina plane is calledLine defects

are commonly known as dislocations.

Planar defects

These are interfaces between homogeneous regions of thematerial. Planar defects include tilt, twin and grain

boundaries.

The stacking faults and external surfaces also

comes under these defects.

Volume defects

Cracks, voids or air bubbles represents these defects.


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Different types of point defects are shown in figure below.

(a)Self interstitial atom. (c) Substitutional atom

(b) Interstitial impurity atom. (d) a vacancy.


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vacancy Interstitialimpurity

Substitutional

impurity

Point Defects


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Vacancies: Missing of atoms at their regular sites in crystals is

called Vacancies. These are common, especially at high

temperatures when atoms are frequently and randomly change theirpositions leaving behind empty lattice sites. Thus the vacancies are

not permanent sites. They can be created and destroyed.

Self interstitialcy: An extra atom that occupies the vacancy

position where an atom is normally missed at the lattice point. Self

interstitial atoms occur only in the low packing concentrations of

metals.

Interstitial Impurity: It is the position where an extra atom thatcan be introduced which is not of the regular atomic site. If the

size of the impurity atom is substantially small then the structure

can not be dislodged. The converse is true.


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These are much smaller than the atoms in the bulk matrix. Thus

the interstitial impurity atoms fit into the open space between

the bulk atoms of the lattice structure.

Example: The carbon atoms that are added to iron to makesteel. Carbon atoms, with a radius of 0.071 nm, fit nicely in the

open spaces between the larger (0.124 nm) iron atoms.

Substitutional impurity : It is an atom of a different typewhich replaces one of the parent atoms. Substitutional impurity

atoms are usually with approximately 15% of the parent atoms.

Example:- Zinc atoms in brass.

In brass, zinc atoms with a radius of 0.133 nm have replacedsome of the copper atoms, which are of radius 0.128 nm.


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Defects in ionic

solids


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N b f i i


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Number of vacancies at any temperature in

elemental solids:

(1)

From eqn. (1)

If nN

In general, the number of vacancies n is always very small

compared to the total number of metal ions/ atoms. Hence theequation (3) can be used to calculate the number of vacancies

when temperature is much below the melting point.

(2)

(3)


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F k l d f t i i i t l t E ilib i


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Frenkel defects in ionic crystals at Equilibrium

Let Ef be the average energy required to displace a cation from the

normal lattice position to an interstial site and Ni be the number ofinterstitial sites. If N be the total number of cations (in an ionic

crystal A+, B-), the number of Frenkel defects is

Hence at low temperature, the number of Frenkel defects is small.


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Calculation of number of vacancies at a giventemperature.

All most in all crystals vacancies are present and the main causefor these defects is thermal agitation.

Let us consider Ev is the energy required to move an atom fromlattice site inside the crystal to lattice site on the surface.

Therefore the amount of energy required to produce n number ofisolated vacancies can be written as

vnEU


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At thermal equilibrium, free energy is constant

and minimum with respect to n, hence

}TK

ENexp{n

Nnif

}TK

Eexp{

n

nN

}n

nNTlog{KE

logn}1n)log(NT{1KE

0nlogn})n)n)log(N(NT{NlogNK(nEdn

d

odn

dF

B

v

B

v

Bv

Bv

Bv

Hence equilibrium concentration of vacancies

decreases with increase of temperature.


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The total number of ways to move n numbers of

ion pairs out of N number of ionic molecules in a

crystal on to the surface will be

2

2

]!)!(

!log[

log

]!)!(

![

nnN

NKS

PKS

nnN

NP

B

B


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The free energy

2

Bp ]n!n)!(N

N!Tlog[KnEF

TSUF

Using stirling approximation xxxx log!log

nlogn]n)n)log(N(NT[NlogN2KnEF

n]nlognn)n)log(N(N2[NlogN]n!n)!(N

N!log[

n]nlognn)(Nn)n)log(N(NN2[NlogN]n!n)!(N

N!log[

Bv

2

2


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At thermal equilibrium, free energy is constant and

minimum with respect to n, hence

}T2K

ENexp{n

Nnif

}T2K

En)exp{(Nn

]n

nNlog[

T2K

E

]n

nNTlog[2KE

0]dn

dF[

B

p

B

p

B

p

BP

T


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Let us consider Ei is the energy required to

move an atom from lattice site inside the crystal

to a lattice site on the surface.

The amount of energy required to produce nnumber of isolated vacancies

inEU


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Line defects

Line defects are one dimensional

imperfections in the geometrical sense.

These are also called dislocations. The

dislocations are of two types .

1. Edge dislocation

2. Screw dislocation


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Edge dislocation

In a perfect crystal, atoms are arranged in both vertical and

horizontal planes parallel to the side faces.

If one of these vertical planes does not extended to fulllength but ends in between, within the crystal as shown in

figure, it is called edge dislocation.

Edge dislocations are symbolically represented by or depending on whether the incomplete plane starts from thetop or from the bottom of the crystal.

These two configurations are referred to as positive andnegative edge dislocations.


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Perfect crystal

Edge dislocated crystal

Extra half plane

Slip plane


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Positive and negative dislocations


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Burgers vector

The magnitude and the direction of the

displacement are defined by a vector called

the Burgers vector.

Consider two crystals one perfect and

another with edge dislocation.


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From fig 1


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From fig. 1.

Starting from the point P, we go up by 6 steps, then movetowards right by 5 steps, and move down by 6 steps andfinally move towards left by 5 steps to reach the starting

point P, the burgers circuit gets closed.

From fig 2.

We end up at Q instead of the starting point P.

Now we have to move an extra step QP to return to P inorder to close the burgers circuit.

The magnitude and the direction of the step defines theBurgers vector (BV)

BV = QP = b

The Burgers vector is perpendicular to the edge

dislocation line.

SCREW DISLOCATION


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SCREW DISLOCATION

Displacement of the atoms in one part of a crystalrelative to the rest of the crystal, forming a spiral ramp

around the dislocation is called Screw dislocation.

In this, the atoms are displaced in two separate planesperpendicular to each other.

In a figure the plane ABCD is the slipped area.

The upper portion of the crystal has been sheared by an

atomic distance to the right relative to the lowerportion.

No slip has taken place to the right of AD and AD is a

dislocation line.


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Here, the dislocation is parallel to its Burgers

vector or shear vector.

The dislocation is terminated at AD


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An extra half plane


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An extra half plane

or a missing half plane

An extra half plane


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p

or a missing half plane

EdgeDislocation


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The atom positions around an edge dislocation; extra half plane of atoms shown


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The atom positions around an edge dislocation; extra half-plane of atoms shown

in perspective.

Burgers Vector


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Burgers Vector

The magnitude and the direction of the displacement of atoms in

planes are defined by a vector called the Burgers vector.

Burger vector is perpendicular to the edge dislocation.

Burger vector = FS = b


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Perfect crystal An incomplete plane in aC l l i d


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P

Crystal results in an edgedislocation

PQ

b

Starting from the point P,we go up by 6 steps, thenmove towards right by 5

steps, and move down by6 steps and finally movetowards left by 5 steps toreach the starting point P,the burgers circuit gets

closed.

From above fig it is very clear that weend up at Q instead of the starting pointP.Therefore we have to move an extra

step QP to reach the original position Pin order to close the burgers circuit.

The magnitude and the direction of thestep defines the Burgers vector (BV)

BV = QP = b


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1 2 3 4 5 6 7 8 9


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1 2 3 4 5 6 7 8 9

1 2 3 4 5 6 7 8 9


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1 2 3 4 5 6 7 8 9

1 2 3 4 5 6 7 8 9

1 2 3 4 5 6 7 8 9


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1 2 3 4 5 6 7 8 9

1 2 3 4 5 6 7 8 9

slipno slip

boundary = edge dislocation

Slip planeb

Burgers vector

1 2 3 4 5 6 7 8 9


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1 2 3 4 5 6 7 8 9

3 5 6 8 9

slipno slip

boundary = edge dislocation

Slip planeb

Burgers vector

t


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Slip plane

slip no slip

dislocatio

n

b

t

Dislocation: slip/no

slip boundary

b: Burgers vectormagnitude and

direction of the slip

t: unit vector tangent

to the dislocation line

Di l i Li


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Dislocation Line:A dislocation line is the boundary between slip and no slip regions of a

crystal

Burgers vector:The magnitude and the direction of the slip is represented by a vector

b called the Burgers vector,

Line vectorA unit vectort tangent to the dislocation line is called a tangent vectoror the line vector.


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In general, there can be any angle between the Burgers vector b(magnitude and the direction of slip) and the line vector t (unit vectortangent to the dislocation line)

b t Edge dislocation

b t Screw dislocation

b t , b t Mixed dislocation


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X-RAY DIFFRACTION and Crystal Defects