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CYL729: Materials Characterization. Diffraction Microscopy Thermal Analysis A. Ramanan Department of Chemistry [email protected]. Reference Books. George M. Crankovic (Editor). Electro-magnetic Spectrum. History of X-rays. - PowerPoint PPT Presentation
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CYL729: Materials CharacterizationCYL729: Materials Characterization
DiffractionDiffraction
MicroscopyMicroscopy
Thermal AnalysisThermal Analysis
A. RamananA. RamananDepartment of ChemistryDepartment of Chemistry
[email protected]@chemistry.iitd.ac.in
Reference Books
George M. Crankovic (Editor)
Electro-magnetic Spectrum
History of X-raysHistory of X-rays
1885-1895 Wm. Crookes sought unsuccessfully the cause of 1885-1895 Wm. Crookes sought unsuccessfully the cause of repeated fogging of photographic plates stored near his cathode repeated fogging of photographic plates stored near his cathode ray tubes.ray tubes.
X-rays discovered in 1895 by Roentgen, using ~40 keV electrons X-rays discovered in 1895 by Roentgen, using ~40 keV electrons (1st Nobel Prize in Physics 1901)(1st Nobel Prize in Physics 1901)
1909 Barkla and Sadler discovered characteristic X-rays, in 1909 Barkla and Sadler discovered characteristic X-rays, in studying fluorescence spectra (though Barkla incorrectly studying fluorescence spectra (though Barkla incorrectly understood origin) (Barkla got 1917 Nobel Prize) understood origin) (Barkla got 1917 Nobel Prize)
1909 Kaye excited pure element spectra by electron 1909 Kaye excited pure element spectra by electron bombardmentbombardment
History of X-rays - cont’dHistory of X-rays - cont’d
1912 von Laue, Friedrich and Knipping observe X-ray diffraction 1912 von Laue, Friedrich and Knipping observe X-ray diffraction (Nobel Prize to von Laue in 1914) (Nobel Prize to von Laue in 1914)
1912-13 Beatty demonstrated that electrons directly produced 1912-13 Beatty demonstrated that electrons directly produced two radiations: (a) independent radiation, Bremsstrahlung, and (b) two radiations: (a) independent radiation, Bremsstrahlung, and (b) characteristic radiation only when the electrons had high enough characteristic radiation only when the electrons had high enough energy to ionize inner electron shells.energy to ionize inner electron shells.
1913 WH + WL Bragg build X-ray spectrometer, using NaCl to 1913 WH + WL Bragg build X-ray spectrometer, using NaCl to resolve Pt X-rays. Braggs’ Law. (Nobel Prize 1915)resolve Pt X-rays. Braggs’ Law. (Nobel Prize 1915)
n = 2d sin
History of X-rays - cont’dHistory of X-rays - cont’d
1913 Moseley constructed an x-1913 Moseley constructed an x-ray spectrometer covering Zn to ray spectrometer covering Zn to Ca (later to Al), using an x-ray Ca (later to Al), using an x-ray tube with changeable targets, a tube with changeable targets, a potassium ferrocyanide crystal, potassium ferrocyanide crystal, slits and photographic platesslits and photographic plates
1914, figure at right is the first 1914, figure at right is the first electron probe analysis of a man-electron probe analysis of a man-made alloymade alloy
T. Mulvey Fig 1.5 (in Scott & Love, 1983). Note impurity lines in Co and Ni spectra
History of X-rays - cont’dHistory of X-rays - cont’dMoseley found that wavelength of characteristic X-Moseley found that wavelength of characteristic X-rays varied systematically (inversely) with atomic rays varied systematically (inversely) with atomic number number
Z
• Using wavelengths, Moseley developed the concept of atomic number and how elements were arranged in the periodic table.
• The next year, he was killed in Turkey in WWI. “In view of what he might still have accomplished (he was only 27 when he died), his death might well have been the most costly single death of the war to mankind generally,” says Isaac Asimov (Biographical Encyclopedia of Science &Technology).
1923 Manne Siegbahn published 1923 Manne Siegbahn published The Spectroscopy of X-raysThe Spectroscopy of X-rays in in which he shows that the Bragg equation must be revised to take which he shows that the Bragg equation must be revised to take refraction into account, and he lays out the “Siegbahn notation” refraction into account, and he lays out the “Siegbahn notation” for X-raysfor X-rays1931 Johann developed bent crystal spectrometer (higher 1931 Johann developed bent crystal spectrometer (higher efficiency)efficiency)
Historical Summary of X-raysHistorical Summary of X-rays1859 Kirchhoff and Bunsen showed patterns of lines given off by 1859 Kirchhoff and Bunsen showed patterns of lines given off by incandescent solid or liquid are characteristic of that substanceincandescent solid or liquid are characteristic of that substance 1904 Barkla showed each element could emit ≥1 characteristic 1904 Barkla showed each element could emit ≥1 characteristic groups (K,L,M) of X-rays when a specimen was bombarded with groups (K,L,M) of X-rays when a specimen was bombarded with beam of x-raysbeam of x-rays 1909 Kaye showed same happened with bombardment of cathode 1909 Kaye showed same happened with bombardment of cathode rays (electrons)rays (electrons)1913 Moseley found systematic variation of wavelength of 1913 Moseley found systematic variation of wavelength of characteristic X-rays of different elementscharacteristic X-rays of different elements1922 Mineral analysis using X-ray spectra (Hadding)1922 Mineral analysis using X-ray spectra (Hadding)
1923 Hf discovered by von Hevesy (gap in Moseley plot at1923 Hf discovered by von Hevesy (gap in Moseley plot at Z=72). Z=72). Proposed XRF (secondary X-ray fluorescence)Proposed XRF (secondary X-ray fluorescence)
Summary of X-ray PropertiesSummary of X-ray PropertiesX-rays are considered both particles and waves, X-rays are considered both particles and waves, i.e., consisting of small packets of electromagnetic i.e., consisting of small packets of electromagnetic waves, or photons.waves, or photons.
X-rays produced by accelerating HV electrons in X-rays produced by accelerating HV electrons in a vacuum and colliding them with a target.a vacuum and colliding them with a target.
The resulting spectrum contains (1) continuous The resulting spectrum contains (1) continuous background (Bremsstrahlung;“white X-rays”), (2) background (Bremsstrahlung;“white X-rays”), (2) occurrence of sharp lines (characteristic X-rays), occurrence of sharp lines (characteristic X-rays), and (3) a cutoff of continuum at a short and (3) a cutoff of continuum at a short wavelength.wavelength.
X-rays have no mass, no charge (vs. electrons)X-rays have no mass, no charge (vs. electrons)
X-ray CrystallographyX-ray Crystallography
DIFFRACTIONDIFFRACTION
What is a Unit Cell?What is a Unit Cell?
Unit cell can be chosen in different Unit cell can be chosen in different ways!ways!
Unit Cells?Unit Cells?
White and black birds by the artist, M. C. Escher.
A unit cell chosen such that it contains minimum volume but exhibit maximum symmetry
{R = n1 a1 + n2 a2 + n3 a3}
Translationalvector
Crystal Structure
Ideal Crystal: Contain periodical array of atoms/ionsRepresented by a simple lattice of pointsA group of atoms attached to each lattice points
Basis
LATTICE = An infinite array of points in space, in which each point has identical surroundings to all others.
CRYSTAL STRUCTURE = The periodic arrangement of atoms in the crystal.
It can be described by associating with each lattice point a group of atoms called the MOTIF (BASIS)
Lattice parameters: a, b, c;
7 Crystal Systems
Bravais Lattice: an infinite array of discrete points with an arrangement and orientation that appears exactly the same from whichever of the points the array is viewed.
Crystal SystemCrystal System Bravais Bravais LatticeLattice
Essential Essential Symmetry Symmetry
ConditionsConditions
CubicCubic P, F, IP, F, I 4 C4 C33 aa==bb==cc
====90900 0
TetragonalTetragonal P, IP, I 1 C1 C44 along [ along [cc-axis]-axis] aa==bb==cc
====909000
HexagonalHexagonal PP 1 C1 C66 along [ along [cc-axis]-axis] aa==bb==cc
====909000
RhombohedralRhombohedral RR 1 C1 C33 along body along body
diagonaldiagonalaa==bb==cc
= = = = 90 9000
OrthorhombicOrthorhombic P, F, I, CP, F, I, C 3 C3 C22 mutually mutually
perpedicular along perpedicular along the three axesthe three axes
a a b b cc
====909000
MonoclinicMonoclinic P, CP, C 1 C1 C22 along [ along [bb-axis]-axis] a a bb cc
===90=9000 & & 90 9000
TriclinicTriclinic PP CC22 or inversion centre or inversion centre a a b b cc
909000
14 Bravais lattices14 Bravais lattices
Unit cell symmetries - cubicUnit cell symmetries - cubic
3 C3 C4 4 - passes through pairs of opposite - passes through pairs of opposite face centers, parallel to cell axes face centers, parallel to cell axes
4 C4 C3 3 - passes through cubic - passes through cubic diagonalsdiagonals
A cube need not have C4 !!
Copper metal is Copper metal is face-centered face-centered cubiccubic
Identical atoms at corners and at Identical atoms at corners and at face centers face centers
Lattice type FLattice type F
also Ag, Au, Al, Ni...also Ag, Au, Al, Ni...
-Iron is body-centered cubic
Identical atoms at corners and body center (nothing at face centers)
Lattice type I
Also Nb, Ta, Ba, Mo...
periodic tableperiodic table
Hexagonal closedpacked (hcp)
face-centered cubic (fcc)
body-centered cubic (bcc)
Caesium Chloride (CsCl) isCaesium Chloride (CsCl) is primitive primitive cubiccubic
Different atoms at corners and body Different atoms at corners and body center. NOT body centered, therefore.center. NOT body centered, therefore.
Lattice type PLattice type P
Also CuZn, CsBr, LiAgAlso CuZn, CsBr, LiAg
Sodium Chloride (NaCl) - Na is much smaller than Cs
Face Centered Cubic
Rocksalt structure
Lattice type F
Also NaF, KBr, MgO….
Diamond Structure: two sets of FCC Lattices
Z = 8 C atoms per unit cell
one C4
Why not F tetragonal?
Tetragonal: P, I
Yellow and green colors represents same atoms but different depths.
ExampleExample
CaCCaC22 - has a rocksalt-like structure but with - has a rocksalt-like structure but with
non-spherical carbidesnon-spherical carbides
2-C
C
Carbide ions are aligned parallel to c
c > a,b tetragonal symmetry
Orthorhombic: P, I, F, C
C F
Side centering
Side centered unit cell
Notation:
A-centered if atom in bc plane
B-centered if atom in ac plane
C-centered if atom in ab plane
Trigonal: P : 3-fold rotation
Hexagonal
Monoclinic Triclinic
Unit cell contentsUnit cell contents Counting the number of atoms Counting the number of atoms withinwithin the unit cell the unit cell
Many atoms are shared between unit cells
Atoms Shared Between: Each atom counts:corner 8 cells 1/8face center 2 cells 1/2body center 1 cell 1edge center 4 cells 1/4
lattice type cell contentsP 1 [=8 x 1/8]I 2 [=(8 x 1/8) + (1 x 1)]F 4 [=(8 x 1/8) + (6 x 1/2)]C 2 [=(8 x 1/8) + (2 x 1/2)]
e.g. NaCl Na at corners: (8 1/8) = 1 Na at face centres (6 1/2) = 3Cl at edge centres (12 1/4) = 3 Cl at body centre = 1
Unit cell contents are 4(Na+Cl-)
(0,0,0)(0, ½, ½)(½, ½, 0)(½, 0, ½)
Fractional Coordinates
Cs (0,0,0)Cl (½, ½, ½)
Density Calculation
AC NV
nA
n: number of atoms/unit cell
A: atomic mass
VC: volume of the unit cell
NA: Avogadro’s number (6.023x1023 atoms/mole)
Calculate the density of copper.
RCu =0.128nm, Crystal structure: FCC, ACu= 63.5 g/mole
n = 4 atoms/cell, 333 216)22( RRaVC
3
2338/89.8
]10023.6)1028.1(216[
)5.63)(4(cmg
8.94 g/cm3 in the literature
Miller IndicesMiller Indices
describe which plane of atom isdescribe which plane of atom isinteracting with the x-raysinteracting with the x-rays
How to Identify Miller indices (hkl)?How to Identify Miller indices (hkl)?
direction: [hkl]family of directions: <hkl>planes: (hkl)family of planes: {hkl}
[001]
[010]
[001]
ab
c
to identify planes:Step 1 : Identify the intercepts on the x- , y- and z- axes.Step 2 : Specify the intercepts in fractional coordinatesStep 3 : Take the reciprocals of the fractional intercepts
Miller indices (hkl)Miller indices (hkl)
e.g.: cubic system:
to identify planes:Step 1 : Identify the intercepts on the x- , y- and z- axes (a/2, ∞, ∞)Step 2 : Specify the intercepts in fractional co-ordinates (a/2a, ∞, ∞) = (1/2,0,0) Step 3 : Take the reciprocals of the fractional intercepts (2, 0, 0)
(210)
(110) (111) (100)
Miller IndicesMiller Indices
Miller IndicesMiller Indices
Crystallographic Directions And Planes
Lattice DirectionsIndividual directions: [uvw][uvw]Symmetry-related directions: <uvw><uvw>
Miller Indices:1. Find the intercepts on the axes in terms of the lattice
constant a, b, c2. Take the reciprocals of these numbers, reduce to the
three integers having the same ratio(hkl)(hkl)
Set of symmetry-related planes: {hkl}{hkl}
(100) (111)
(200) (110)
In cubic system,
[hkl] direction perpendicular to (hkl) plane
Wilhelm Conrad Röntgen
Wilhelm Conrad Röntgen discovered 1895 the X-rays. 1901 he was honoured by the Noble prize for physics. In 1995 the German Post edited a stamp, dedicated to W.C. Röntgen.
The Principles of an X-ray Tube
Anode
focus
Fast electronsCathode
X-Ray
The Principle of Generation of X-ray
X-ray
Fast incident electron
nucleus
Atom of the anodematerial
electrons
Ejected electron(slowed
down and changed
direction)
The Principle of Generation the Characteristic Radiation
K
L
K
K
L
M
EmissionPhotoelectron
Electron
The Generating of X-rays
Bohr`s model
The Generating of X-rays
M
K
L
K K K K
energy levels (schematic) of the electrons
Intensity ratios KKK
The Generating of X-rays
Anode
Mo
Cu
Co
Fe
(kV)
20.0
9.0
7.7
7.1
K1 : 0,70926
K2 : 0,71354
K1 : 0,63225
Filter
K1 : 1,5405
K2 : 1,54434
K1 : 1,39217
K1 : 1,78890
K2 : 1,79279
K1 : 1,62073
K1 : 1,93597
K2 : 1,93991
K1 : 1,75654
Zr0,08mm
Mn0,011mm
Fe0,012mm
Ni0,015mm
Wavelength (Angström)
The Generating of X-rays
Emission Spectrum of aMolybdenum X-Ray Tube
Bremsstrahlung = continuous spectra
characteristic radiation = line spectra
Interaction between X-ray and Matter
d
wavelength Pr
intensity Io
incoherent scattering
Co (Compton-Scattering)
coherent scattering
Pr(Bragg´s-scattering)
absorbtionBeer´s law I = I0*e-µd
fluorescense
> Pr
photoelectrons
C. Gordon Darwin
C. Gordon Darwin, grandson of C. Robert Darwin developed 1912 dynamic theory of scattering of X-rays at crystal lattice
P. P. Ewald
P. P. Ewald 1916 published a simple and more elegant theory of X-ray diffraction by introducing the reciprocal lattice concept. Compare Bragg’s law (left), modified Bragg’s law (middle) and Ewald’s law (right).
sin2
n
d
2
1sin d
12
sin
Bragg’s Description
• The incident beam will be scattered at all scattering centres, which lay on lattice planes.
• The beam scattered at different lattice planes must be scattered coherent, to give an maximum in intensity.
• The angle between incident beam and the lattice planes is called .
• The angle between incident and scattered beam is 2 .
• The angle 2 of maximum intensity is called the Bragg angle.W.H. Bragg (father) and William
Lawrence.Bragg (son) developed a simple relation for scattering angles, now call Bragg’s law.
sin2 n
d
Bragg’s Law
• A powder sample results in cones with high intensity of scattered beam.
• Above conditions result in the Bragg equation
• or
sin2 dns
sin2
n
d
/hchE
35KeV ~ 0.1-1.4ACu K 1.54 A
Mo:
X-Ray Diffraction
Structure Determination
High Voltage
X-Ray DiffractionX-ray Tube
Lead Screen
X-ray Beam
Crystal
Photographic Plate
Projection Screen
Visible Light Laser 35mm slide
Optical Transforms
L
X
LightInterference fringes
Constructive
Destructive
Diffraction
Diffraction Conditions
Diffraction Conditions
Fraunhofer diffraction Bragg diffraction
For constructive interference, d sin = n
For constructive interference, 2(d sin ) = n
}d
d }}
d
d sin
}
}d
d sin d sin
2
222
2
1
a
lkh
dhkl
For cubic system
Lattice spacing
Bragg’s Law
For cubic system: But not all planes have the diffraction !!!
sin2
sinsin
hkl
hklhkl
d
dd
QTSQn
222 lkh
adhkl
(200)(211)
Powder diffraction
X-Ray
Powder X-ray Diffraction
Tube
Powder
Film
The Elementary Cell
a
b
c
a = b = c = = = 9
0
o
Relationship between d-value and the Lattice Constants
=2dsin Bragg´s law The wavelength is known
Theta is the half value of the peak position
d will be calculated
1/d2= (h2 + k2)/a2 + l2/c2 Equation for the determination of the d-value of a tetragonal elementary cell
h,k and l are the Miller indices of the peaks
a and c are lattice parameter of the elementary cell
if a and c are known it is possible to calculate the peak position
if the peak position is known it is possible to calculate the lattice parameter
D8 ADVANCE Bragg-Brentano Diffractometer
• A scintillation counter may be used as detector instead of film to yield exact intensity data.
• Using automated goniometers step by step scattered intensity may be measured and stored digitally.
• The digitised intensity may be very detailed discussed by programs.
• More powerful methods may be used to determine lots of information about the specimen.
The Bragg-Brentano Geometry
Tube
measurement circle
focusing-circle
2
Detector
Sample
The Bragg-Brentano Geometry
Divergence slit
Detector-
slitTube
Antiscatter-slit
Sample
Mono-chromat
or
Powder Diffraction Pattern
What is a Powder Diffraction Pattern?
A powder diffractogram is the result of a convolution of a) the diffraction capability of the sample (Fhkl) and b) a complex system function.
The observed intensity yoi at the data point i is the result of
yoi = of intensity of "neighbouring" Bragg peaks + background
The calculated intensity yci at the data point i is the result of
yci = structure model + sample model + diffractometer model + background model
Which Information does a Powder Pattern offer?
peak position - dimension of the elementary cell
peak intensity - content of the elementary cell
peak broadening - strain/crystallite size/nanostr.
Powder Pattern and Structure
• The d-spacings of lattice planes depend on the size of the elementary cell and determine the position of the peaks.
• The intensity of each peak is caused by the crystallographic structure, the position of the atoms within the elementary cell and their thermal vibration.
• The line width and shape of the peaks may be derived from conditions of measuring and properties - like particle size - of the sample material.
(110)
(200)(211)
Powder diffraction
X-Ray
)(75.0);(5.0sin
sin2
2
FCCBCCB
A
Example: layered silicatesmica
2*theta d
7.2 12.1
14.4 6.1
22 4.0
nd sin2
growth oriented along c-axis
(hkl)
(001)
(002)
(003)
C~12.2 A
(00l)
What we will see in XRD of simple cubic, BCC, FCC?
What we will see in XRD of simple cubic, BCC, FCC?
)(75.0);(5.0sin
sin2
2
FCCBCCB
A
222 lkh
adhkl
Observable diffraction peaks
222 lkh Ratio
Simple cubic
SC: 1,2,3,4,5,6,8,9,10,11,12..
BCC: 2,4,6,8,10, 12….
FCC: 3,4,8,11,12,16,24….
222 lkh
adhkl
nd sin2
Ex: An element, BCC or FCC, shows diffraction peaks at 2: 40, 58, 73, 86.8,100.4 and 114.7. Determine:(a) Crystal structure?(b) Lattice constant?(c) What is the element?
2theta theta (hkl)
40.0 20 0.117 1 (110)
58.0 29 0.235 2 (200)
73.0 36.5 0.3538 3 (211)
86.8 43.4 0.4721 4 (220)
100.4 50.2 0.5903 5 (310)
114.7 57.35 0.7090 6 (222)
2sin 222 lkh
a =3.18Å, BCC, W