Module-03-01-Crystalography.pdf

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RSET, 2011-12 Engineering PhysicsCrystallography

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Crystallography


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0. Introduction

Crystallography Branch of solid state physics dealing with the structure ofcrystalline solids

Tools of crystal analysis: X-ray, electron beam Properties of solids depend on their crystal structure

E.g. Elasticity, electrical conductivity, magnetism

Classification of solids Crystalline solids Solids with atoms/molecules/groups arranged in a regular fashion such

that their positions are exactly periodic

E.g. NaCl Polycrystalline solids

Composed of many microscopic crystals, called crystallites or grains Solids with the periodicity of spiecies interupted at grain boundaries

Grain: Region within which particular periodicity exists

E.g.: Metals, Ceramics Amorphous solids

Solids with atoms distributed irregularly or randomly E.g. Glass
http://localhost/var/www/apps/conversion/tmp/scratch_1//upload.wikimedia.org/wikipedia/commons/2/2c/Crystalline_polycrystalline_amorphous.svg


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1. Crystal Structure: Space lattice

Crystal structure = Space lattice + basis Space Lattice It is the geometry of a set of points in space

Space lattice represents an array of points in space such that theenvironment about each point is the same

Mathematical representation Using basic vectors and their translation or parallel displacement Position vector of any point P in a space lattice can be represented as

sum of integral multiples of basic vectors

2D:

3D:

, and are non-collinear vectors or basic vectors of the

lattice Lattice has translational symmetry (non-variability under

displacement) specified by the lattice vectors

Crystal looks the same when viewed from any of lattice points

bnanT

21

cnbnanT

321

a

b

c

a

b

p

T

2D Lattice


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2. Crystal Structure: Basis

Basis The atom, ions, or group which constitute the geometrical

pattern of a crystal lattice

It may be monoatomic (e.g. Al, Ba), diatomic (e.g. NaCl, KCl)or triatomic (e.g. CaF2), etc.

Basis is identical in composition, arrangement and orientation

Crystal structure = Space lattice + Basis

Basis with 3 atoms


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3. Crystal Structure: Unit Cell

Primitive cell Volume of the space that when translated through all the vectors in aspace lattice, fill the space without overlapping or leaving void

It is the minimum volume cell which contains precisely one lattice point If a, b and c are primitive vectors, then the volume of the primitive cell is

given by Vcell = a x b.c (Box product); a, b and c are lattice vectors

Unit cell The smallest geometric unit, repetition of which in 3-D gives the actual

crystal structure

Fundamental pattern of minimum atoms, ions or groups of atoms whichrepresent all the characteristics of the crystal

Unit cell may contain more than one lattice point Unit cells may be identical to a primitive cell, but all primitive cells are not

unit cells

Volume of a unit cell is an integral multiple of the volume of the primitivecell


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4. Unit Cell Parameters

Lattice/Unit Cell parameters Full set of lattice parameters

Three lattice constants (a, b, c)

Three angles between them (, , )

E.g. Unit cell definition using parallelepiped withlengths a, b, c and angles between the sides given by

, ,

Cubic crystal structures All three lattice constants are equal and we only refer to a

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http://en.wikipedia.org/wiki/File:UnitCell.pnghttp://en.wikipedia.org/wiki/File:UnitCell.png


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5. Crystal Systems (Bravais Lattices)

Bravais space lattices (Auguste Bravais, France, 1848) There are 14 unique lattices (ways of arranging points) in 3-Dcrystal systems (x Frankheim, 1845: 15 unique lattices)

7 crystal systems and 6 lattice centerings Crystal systems

Triclinic, Monoclinic, Ortorhombic, Tetragonal, Rhombohedral(trigonal), Hexagonal, Cubic

Defined by different relations between unit cell parameters (a, b, c,and , , )

Lattice centerings Primitive centering (P): Lattice points on cell corners only

Body centered (I): One additional lattice point at the center of thecell

Face-centered (F): One additional lattice point at the center of eachof the faces of the cell

End-centered (A, B or C centering): One additional lattice point atthe center of one pair of the cell faces

Number of unique lattices Total number of combinations: 7 x 6 = 42

Several are but equivalent to each other. E.g. All A and B centeredlattice can be described by either C or P centering

There are only 14 unique lattices: Bravais lattices


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5. Crystal Systems (Bravais Lattices)

Crystal System Possible Variations Axial Distances (edgelengths) Axial Angles Examples

CubicPrimitive, Body centred,

Face centreda = b = c = = = 90 NaCl, Zinc Blende, Cu

Tetragonal Primitive, Body centred a = b c = = = 90White tin, SnO2, TiO2,

CaSO4

OrthorhombicPrimitive, Body centred,

Face centred,End centreda b c = = = 90

Rhombic Sulphur, KNO3,

BaSO4

Hexagonal Primitive a = b c = = 90, = 120 Graphite, ZnO, CdS

Rhombohedral (trigonal) Primitive a = b = c = = 90

Calcite (CaCO3, Cinnabar

(HgS)

Monoclinic Pr imitive, End centred a b c = = 90, 90Monoclinic Sulphur,

Na2SO4.10H2O

Triclinic Primitive a b c 90K2Cr2O7, CuSO4.5H2O,

H3BO3

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http://en.wikipedia.org/wiki/NaClhttp://en.wikipedia.org/wiki/Zinc_Blendehttp://en.wikipedia.org/wiki/Copperhttp://en.wikipedia.org/wiki/White_tinhttp://en.wikipedia.org/wiki/SnO2http://en.wikipedia.org/wiki/SnO2http://en.wikipedia.org/wiki/TiO2http://en.wikipedia.org/wiki/CaSO4http://en.wikipedia.org/wiki/File:Orthorhombic-body-centered.svghttp://en.wikipedia.org/wiki/Allotropes_of_sulfurhttp://en.wikipedia.org/wiki/KNO3http://en.wikipedia.org/wiki/Barium_sulfatehttp://en.wikipedia.org/wiki/Graphitehttp://en.wikipedia.org/wiki/ZnOhttp://en.wikipedia.org/wiki/CdShttp://en.wikipedia.org/wiki/Calcitehttp://en.wikipedia.org/wiki/Cinnabarhttp://en.wikipedia.org/wiki/File:Monoclinic.svghttp://en.wikipedia.org/wiki/File:Monoclinic-base-centered.svghttp://en.wikipedia.org/wiki/Allotropes_of_sulfurhttp://en.wikipedia.org/wiki/K2Cr2O7http://en.wikipedia.org/wiki/K2Cr2O7http://en.wikipedia.org/wiki/K2Cr2O7http://en.wikipedia.org/wiki/K2Cr2O7http://en.wikipedia.org/wiki/K2Cr2O7http://en.wikipedia.org/wiki/K2Cr2O7http://en.wikipedia.org/wiki/H3Bo3http://en.wikipedia.org/wiki/H3Bo3http://en.wikipedia.org/wiki/H3Bo3http://en.wikipedia.org/wiki/H3Bo3http://en.wikipedia.org/wiki/File:Triclinic.svghttp://en.wikipedia.org/wiki/File:Monoclinic-base-centered.svghttp://en.wikipedia.org/wiki/File:Monoclinic.svghttp://en.wikipedia.org/wiki/File:Rhombohedral.svghttp://en.wikipedia.org/wiki/File:Hexagonal_lattice.svghttp://en.wikipedia.org/wiki/File:Orthorhombic-face-centered.svghttp://en.wikipedia.org/wiki/File:Orthorhombic-body-centered.svghttp://en.wikipedia.org/wiki/File:Orthorhombic-base-centered.svghttp://en.wikipedia.org/wiki/File:Orthorhombic.svghttp://en.wikipedia.org/wiki/File:Tetragonal-body-centered.svghttp://en.wikipedia.org/wiki/File:Tetragonal.svghttp://en.wikipedia.org/wiki/H3Bo3http://en.wikipedia.org/wiki/H3Bo3http://en.wikipedia.org/wiki/H3Bo3http://en.wikipedia.org/wiki/H3Bo3http://en.wikipedia.org/wiki/K2Cr2O7http://en.wikipedia.org/wiki/K2Cr2O7http://en.wikipedia.org/wiki/K2Cr2O7http://en.wikipedia.org/wiki/K2Cr2O7http://en.wikipedia.org/wiki/K2Cr2O7http://en.wikipedia.org/wiki/K2Cr2O7http://en.wikipedia.org/wiki/Allotropes_of_sulfurhttp://en.wikipedia.org/wiki/Cinnabarhttp://en.wikipedia.org/wiki/Calcitehttp://en.wikipedia.org/wiki/CdShttp://en.wikipedia.org/wiki/ZnOhttp://en.wikipedia.org/wiki/Graphitehttp://en.wikipedia.org/wiki/Barium_sulfatehttp://en.wikipedia.org/wiki/KNO3http://en.wikipedia.org/wiki/Allotropes_of_sulfurhttp://en.wikipedia.org/wiki/Allotropes_of_sulfurhttp://en.wikipedia.org/wiki/CaSO4http://en.wikipedia.org/wiki/TiO2http://en.wikipedia.org/wiki/SnO2http://en.wikipedia.org/wiki/SnO2http://en.wikipedia.org/wiki/White_tinhttp://en.wikipedia.org/wiki/Copperhttp://en.wikipedia.org/wiki/Zinc_Blendehttp://en.wikipedia.org/wiki/NaCl


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7. Number of atoms per unit cell

Number of atoms per unit cell Simple cubic (sc) E.g. Alpha-polonium Lattice point at the corner is shared by 8 unit cells

Share of one cell = 1/8 Lattice points at the 8 corners = 8

Lattice point/unit cell = 1/8 x 8 = 1 Body-centered cubic (bcc) E.g. Li, Na, Ba, Cr, Mo Contribution from 8 corner points = 1/8 x 8 = 1 Lattice point at the center of the cell = 1 Lattice point/unit cell = 1+1 = 2

Face-centered cubic (fcc) E.g. Ca, Ni, Ag, NaCl, Pb, Pt, Au Contribution from corner points = 1/8 x 8 =1 Contribution from the face-centered lattice points = 6 x = 3 Lattice point/unit cell = 1+3 = 4


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8. Co-ordination number

Co-ordination number Number of nearest neighbours in a given structure Characteristic feature of a given Bravais lattice Co-ordination number Packing density

Simple cubic 4 nearest neighbours in the same plane

2 nearest neighbours exactly above and below in a vertical plane Co-ordination number = 4+2 = 6

Body-centered cubic Every centered atom has 8 nearest neighbours Co-ordination number = 8

Face-centered cubic Nearest neighbours in the same plane = 4 No of perpendicular planes = 2

Number of atoms in one perpendicular plane = 4 Total neighbours in two perpendicular planes = 2 x 4 = 8

Co-ordination number = 4 + 8 = 12

p


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9. Atomic Radius

Atomic radius (R) Assumption: Atoms as spheres in contact in a crystal Definition: R is half the distance between the nearest neighbours in a crystal R = R(a)

a = side of simple cubic unit cell Simple cubic

R = a/2 or a = 2R Body-centered cubic (bcc)

(DF)2 = (DA)2 + (AF)2

(4R)2 = (a)2 + (a2)2

(4R)2 = 3a2 ; R = 3 a/4 or a = 4R/(3) Face-centered cubic (fcc)

(DB)2 = (DC)2 + (CB)2

(4R)2 = (a)2 + (a)2 = 2a2

R = (2 a)/4 or a = 4R/ 2

R R

a

Simple cubic

4R

a2

A B

D C

E F

GH

bcc

R

2R

R

DC

B

a

a

fcc


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10. Packing Factor (Density/Fraction)

Packing density (fraction) Ratio of the volume of the atoms per unit cell to thetotal volume of the cell

Simple cubic

No. of atoms/unit cell = 1 Volume of one atom = 4/3R3

Side of the unit cell, a = 2R Volume of the cell = a3

Packing density = (4/3R3)/a3= (4/3R3)/(2R)3 = /6 = 0.52


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10. Packing Factor (Density/Fraction)

Body centered cubic (bcc) No. of atoms/unit cell = 2 Volume of two atoms = 2 x 4/3R3

Side of the unit cell, a = 4R/3 Volume of the cell = a3

Packing density =(2 x 4/3R3)/a3 = 2 x 4/3R3/(4R/3)3 = 3 x/8 =0.68

Face centered cubic (fcc) No. of atoms/unit cell = 4 Volume of four atoms = 4 x 4/3R3

Side of the unit cell, a = 4R/2 Volume of the cell = a3

Packing density =(4 x 4/3R3)

/a3

= 4 x 4/3R3

/(4R/2)3

= 22 x /6= 0.74 Utilization of space is maximum for fcc and simple cubic

structure is least dense.


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11. Density and crystal lattice constants

Relation between density and crystal latticeconstants

Cubic crystal of lattice constant, a

Number of atoms per unit cell, n

Volume occupied by n atoms Density,

Atomic weight of material, MA Avogadro Number, NA

Volume occupied by NA atoms = [MA / ] m3 Volume occupied by 1 atom = [MA/ . NA]

Volume occupied by n atoms = [MA.n / . NA] = a3

= [MA. n / NA.a3]

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12. Lattice Planes and Miller Indices Miller indices (William Hallowes Miller, UK)

A treatise on Crystallography, 1839 A designation scheme to describe the orientation of crystallographic

planes and crystal faces relative to crystallographic axes They are symbolic vector representation for the orientation of an atomic

plane in a crystal lattice

Method of finding Miller indices:

Step 1: Ascertain the fractional intercepts that the plane/face makes witheach crystallographic axis. Find how far along the unit cell lengths does the plane intersect the axis

Step 2: Take the reciprocal of the fractional intercept of each unit length foreach axis.

Step 3: Clear the fractions by multiplying with least common multiple(LCM). The cleared fractions result in 3 integer values, designated h, k,and l.

Step 4: These integers are paranthetically enclosed (hkl) to indicate thespecific crystallographic plane within the lattice. Since the unit cell repeats in space, the (hkl) notation actually represents a

family of planes, all with the same orientation.

Miller Indices are defined as the reciprocals of the fractional interceptswhich a crystal plane makes with the crystallographic axes reduced tosmallest integer numbers


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12. Lattice Planes and Miller Indices

Problem 1: Find Miller indices of the plane whoseintercepts are 1a, 2b, and 3c.

Step 1: Find fractional intercepts: 1a/a, 2b/b, 3c/c or 1,2,3

Step 2: Take reciprocals: 1/1, , 1/3

Step 3: Clear fractions or Multiply with LCM: 6/1, 6/2, 6/3

Step 4: Miller indices are (632)

Problem 2: The crystal cuts the intercepts 2, 3, and 4along the three axes. Find the Miller indices of the plane

Step 1: Find fractional intercepts: 2, 3, 4

Step 2: Take reciprocals: 1/2, 1/3, 1/4

Step 3: Clear fractions or Multiply with LCM: 12/2, 12/3, 12/4

Step 4: Miller indices are (643)


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12. Lattice Planes and Miller Indices

Problem 3: Find Miller indices of three principal planes x = a; y = 0; z = 0 Step 1: Fractional intercepts: 1, , (When a face is parallel to the axis, its

intercepts on the axis is taken as infinity)

Step 2: Take reciprocals: 1/1, 1/, 1/ Step 3: Clear fractions or Multiply with LCM: 1/1, 0, 0

Step 4: Miller indices are (100) x = a; y = b; z = 0 Step 1: Fractional intercepts: 1, 1, Step 2: Take reciprocals: 1/1, 1/1, 1/ Step 3: Clear fractions or Multiply with LCM: 1/1, 1/1, 0 Step 4: Miller indices are (110)

x = a; y = b; z = c Step 1: Fractional intercepts: 1, 1, 1 Step 2: Take reciprocals: 1/1, 1/1, 1/1 Step 3: Clear fractions or Multiply with LCM: 1/1, 1/!, 1/1 Step 4: Miller indices are (111)

y

x

z

(100) y

x

z

(110)

y

x

z

(111)


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13. Separation between lattice planes

[For orthogonal crystal systems, i.e. cubic, orthorhombic and tetragonal crystalsystems for which = = = 90 )

Suppose the plane ABC, belonging to a family of planes whose Miller indices are (hkl) Sides of lattice a, b and c respectively Intercepts OA = a/h (since, h = a/OA); OB = b/k, OC = c/l Let OP be perpendicular drawn from the origin to the plane ABC. Distance OP represents interplanar spacing d of the family of planes Let , and be the angles between coordinate axes x, y, z and OP respectively

For an orthogonal cartesian coordinate system, there is an identity: cos2 + cos2 +cos2 = 1 Direction cosines of OP are: cos = d/OA, cos = d/OB, cos = d/OC Substitute for cos , cos and cos (d/OA)2 + (d/OB)2 + (d/OC)2 or (dh/a)2 + (dk/b)2 + (dl/c)2 = 1 or d = 1/[h2/a2 + k2/b2 +

l2/c2]1/2

For cubic system a = b = c: d = a/[h2 + k2 + l2]1/2

y

A

B

C

P

A

B

C

O yP


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13. Separation between lattice planes

Find interplanar distance for simple cubic lattice 1) d100 = a /(h2 +k2 +l2) = a/1 = a 2) d110 = a /(h2 +k2 +l2) = a/(12 +12) = a/2 2) d111 = a /(h2 +k2 +l2) = a/(12 +12 +12) = a/3 Ratio of the reciprocals of interplanar distances:

1/d100: 1/d110: 1/d111 = 1: 2: 3

Find interplanar distance of fcc 1) (d100)fcc = (d100)sc = a/2 Find interplanar distance of bcc

1) 1) (d100)bcc = (d100)sc = a/2

y

x

z

d100

bcc

y

x

z

d100

fcc


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14. Braggs Law Braggs law of x-ray diffraction

[William Henry Bragg (Spectrometer) and son W.Lowerence Bragg (Braggs law) UK, Nobelprize: 1915: Analysis of crystal structure using x-rays]

Crystal as diffraction grating for x-rays Laue (1912)

X-ray is of very short Grating with 40 million lines per cm required for producing

diffraction pattern

Crystals can act as a space grating (not a plane grating) Interatomic separation is of the same order of magnitude as

wavelength of x-rays

Laue spots : Diffraction pattern from NaCl crystal W.L. Braggs Explanation

Assumption: x-rays are not refracted (mirror-like or specularreflection)

Diffraction pattern are due to reflections of x-rays from varioussets of parallel crystal planes which contain large number of atoms

Each atom acts as a source of secondary wavelets

As different from ordinary mirror, x-rays penetrate through crystalto encounter thousands of planes parallel to one another

Radiations reflecting from two successive parallel planes producesdiffraction pattern


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14. Braggs Law

Braggs law Consider a ray PA gets reflected from atom A in the direction AR from plane 1 Another ray QB gets reflected at another atom B in the direction BS Draw perpendiculars AC and AD on the incident ray QB and reflected ray BS

respectively

P.d between two reflected rays = CB + BD Condition for constructive interference (reinforcement): p.d = n CB + BD = n CB = BD = dsin Or 2dhklsin = n, n = 1,2,3,... for first order, second order or third order ..maxima

respectively

The intensity goes on decreasing as the order of the spectrum increases Only first few orders are of importance

Plane 1

Plane 2

Plane 3

Q

P R

S

C D

B

A

d

d


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Braggs X-ray spectrometer [W.H. Bragg][A single crystal method or rotating crystal method]

X-ray tube To produce x-ray

Matched filters S1 and S2

To collimate incident x-ray into a pencil of beam onto the cleavage plane of the crystal C X-ray absorbers, e.g. Lead

Rotatable table It is provided with a vernier, V1 and graduated circlar platform, S To mount crystal to be analyzed Rotating crystal method: Sample crystal is rotated to get diffraction maxima

Spectrometer couples the rotation of the crystal and the detector Angle of rotation of detector is twice the angle of rotation of crystal

Slit S3 To collimate the diffracted beam of x-ray

S1 S2

S

v1

v2

S3

G

= Glancing angle

w

1 2 3

Sin1: Sin2: Sin3 =1/d1:1/d2:1/d3

15. Braggs X-ray spectrometer & Crystal Analysis


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Braggs X-ray spectrometer ctd... Ionisation chamber (detector) Filled with ethyl bromide which is highly sensitive to x-ray ionisation Anode A

Long wire insulated from the chamber kept away from axis:To avoid direct hit withx-ray and hence photoelectric emission

Cathod: The chamber Window, W: Aluminium foil

Ionization current measurement Ionisation current: Measured by Electrometer G The current is a measure of the intensity of x-ray

X-ray produces high energy electrons by photoelectric effect and Compton effect(reduction of energy of x-rays due to scattering at matter).

High energy electrons ionize the gas Peaks in Ionisation current refers to incidence angles corresponding to

different interference maxima

2dsin = n: from knowledge of of x-ray and , lattice parameter, d can bedetermined

RSET 2011 12


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Sample analysis (KCl crystal with (100) plane Peak intensities at 1 = 5.22 ; 2 = 7.30 ; 3 = 9.05

Braggs angles for first order of diffraction

2d1sin 1 = ; 2d2sin 2 = ; 2d3sin 3 =

1/d1 : 1/d2: 1/d3 = sin 5.22 : sin 7.30 : sin 9.05=0.091:0.0127:0.157 = 1:1.4: 1.73 = 1:2:3

The crystal is simple cubic

Comments Single crystal method: Useful when perfectly ordered

geometrical array of atoms is investigated (Single crystal freeof distortions)

Determination of crystal orientation

Determination of crystal quality


RSET 2011 12 C ll h


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9. Powder Crystal Diffraction

Powder Crystal method (Debye and Scherrer) Suitable for analysis of materials that are available in micro-crystalline

or powder nature Though atoms in individual crystals are ordered, the orientations of crystals

themselves are disordered

No sigle crystal requirement Suitable for metals and alloys

Crystal structure (lattice parameters), presence of impurities, distortions,preferred crystal orientation, grain size, stresses present in the material etc.are studied

Debye-Scherrer camera

X-ray

sourceFilter Collimator

Specimen

Photographic film

Transmitted beam2

RSET 2011 12 E i i Ph iC t ll h


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Working X-ray Filtering To obtain monochromatic radiation

X-ray is incident on powder sample mounted at the centre of the camera Photographic film is placed cylindrically about the sample

Powdered crystal statistically contains crystallites which are so oriented that

reflection from a particular plane (hkl) is possible Debye-Scherrer pattern

Large number of curved surfaces corresponding to the Bragg planes truncated bythe edges of the film due to its limited width

Diffracted beams lie on a circular cone of half angle 2 Beams are bent through all angles upto nearly 180 are recorded

Rays diffracted through small angles make arcs around the central spot on

the film When incident beam is diffracted through 90 the cone becomes a flat sheet

and the trace is a straight line

Then onwards curvature will be reversed

Spacing between pairs of arcs Bragg angle corresponding to diffractionfrom a particular plane


r

45

0 1 212

entrance exit

l2l1

RSET 2011 12 E i i Ph iC t ll h


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Theory Angle 2 is formed between the diffracted rays and the directionof incident beam

There exists one cone corresponding to each solution of the Braggequation

Let l1, l2, l3, etc. be the central distances between symmetrical lines onthe streched photograph and r, the radius of the cylindrical film

l1/( 2r) = 4/360 = l1/2r x 90 / ....(1) sin = n/2dhkl .....(2)

Interplanar distance dhkl can be calculated from (1) and (2)

2

Reflecting

plane

l2

2

r

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Module-03-01-Crystalography.pdf