X-RAY DIFFRACTION

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X-RAY DIFFRACTION

• X- Ray Sources
• Diffraction Braggs Law
• Crystal Structure Determination

Elements of X-Ray Diffraction B.D. Cullity
S.R. Stock Prentice Hall, Upper Saddle River
(2001)

• Recommended websites
• http//www.matter.org.uk/diffraction/
• http//www.ngsir.netfirms.com/englishhtm/Diffract
  ion.htm

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What will you learn in this sub-chapter?

• How to produce monochromatic X-rays?
• How does a crystal scatter these X-rays to give a
  diffraction pattern? ? Braggs equation
• What determines the position of the XRD peaks? ?
  Answer) the lattice
• What determines the intensity of the XRD peaks? ?
  Answer) the motif
• What other uses can XRD be put to apart from
  crystal structure determination?? Grain size
  determination ? Strain in the material

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Some Basics

• For electromagnetic radiation to be diffracted
  the spacing in the grating (a series of
  obstacles) should be of the same order as the
  wavelength.
• In crystals the typical interatomic spacing 2-3
  Å so the suitable radiation is X-rays.
• Hence, X-rays can be used for the study of
  crystal structures.

aCu 3.61 Å ? dhkl is equal to aCu or less
than that (e.g. d111 aCu/?3 2.08 Å)
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Generation of X-rays

• X-rays can be generated by decelerating electrons

Target
X-rays
Beam of electrons
A accelerating (or decelerating) charge radiates
electromagnetic radiation
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Mo Target impacted by electrons accelerated by a
35 kV potential shows the emission spectrum as in
the figure below (schematic)
X-ray sources with different ? for doing XRD
studies
Target Metal ? Of K? radiation (Å)
Mo 0.71
Cu 1.54
Co 1.79
Fe 1.94
Cr 2.29
The high intensity nearly monochromatic K? x-rays
can be used as a radiation source for X-ray
diffraction (XRD) studies ? a monochromator can
be used to further decrease the spread of
wavelengths in the X-ray
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• When X-rays hit a specimen, the interaction can
  result in various signals/emissions/effects
• The coherently scattered X-rays are the ones
  important from a XRD perspective

Incident X-rays
Absorption (Heat)
SPECIMEN
Fluorescent X-rays
Electrons
Scattered X-rays
Compton recoil
Photoelectrons
Coherent From bound charges
Incoherent (Compton modified) From loosely bound
charges
Click here to know more
Transmitted beam
X-rays can also be refracted (refractive index
slightly less than 1) and reflected (at very
small angles)
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Diffraction

• Now we shall consider the important topic as to
  how X-rays interact with a crystalline array (of
  atoms, ions etc.) to give rise to the phenomenon
  known as X-ray diffraction (XRD).
• Diffraction (with sharp peaks) (with XRD being a
  specific case) requires two important conditions
  to be met? Coherent waves (with wavelength ?)
  on a? Crystalline array with spacing of the
  order of () ?
• The waves could be ? electromagnetic waves
  (light, X-rays), ? matter waves (electrons,
  neutrons) or ? mechanical waves (sound, waves
  on water surface)
• In short diffraction is coherent reinforced
  scattering (or reinforced scattering of coherent
  waves)
• In a sense diffraction is nothing but a special
  case of constructive ( destructive)
  interferenceTo give an analogy ? the results of
  Youngs double slit experiment is interpreted as
  interference, while the result of multiple slits
  is categorized under diffraction

A quasicrystalline array will also lead to
diffraction (which we shall not consider in this
text) With a de Broglie wavelength
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XRD ? the first step

• A beam of X-rays directed at a crystal interacts
  with the electrons of the atoms in the crystal
• The electrons oscillate under the influence of
  the incoming X-Rays and become secondary sources
  of EM radiation
• The secondary radiation is in all directions
• The waves emitted by the electrons have the same
  frequency as the incoming X-rays ? coherent
• The emission can undergo constructive or
  destructive interference

Schematics
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Some points to recon with

• We can get a better physical picture of
  diffraction by using Laues formalism (leading to
  the Laues equations)
• However, a parallel approach to diffraction is
  via the method of Bragg, wherein diffraction can
  be visualized as reflections from a set of
  planes
• As the approach of Bragg is easier to grasp we
  shall use that in this elementary text
• We shall do some intriguing mental experiments to
  utilize the Braggs equation (Braggs model) with
  caution

• Let us consider a coherent wave of X-rays
  impinging on a crystal with atomic planes at an
  angle ? to the rays
• Incident and scattered waves are in phase if the
  i) in-plane scattering is in phase and ii)
  scattering from across the planes is in phase

In plane scattering is in phase
Incident and scattered waves are in phase if
Scattering from across planes is in phase
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Let us consider in-plane scattering
There is more to this Click here to know more
and get introduced to Laue equations describing
diffraction
Extra path traveled by incoming waves ? AY
These can be in phase if ? ?incident ?scattered
Extra path traveled by scattered waves ? XB
But this is still reinforced scatteringand NOT
reflection
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BRAGGs EQUATION
Let us consider scattering across planes
Click here to visualize constructive and
destructive interference

• A portion of the crystal is shown for clarity-
  actually, for destructive interference to occur
  many planes are required (and the interaction
  volume of x-rays is large as compared to that
  shown in the schematic).
• The scattering planes have a spacing d.
• Ray-2 travels an extra path as compared to Ray-1
  ( ABC). The path difference between Ray-1 and
  Ray-2 ABC (d Sin? d Sin?) (2d.Sin?).
• For constructive interference, this path
  difference should be an integral multiple of ?
  n? 2d Sin? ? the Braggs equation. (More about
  this sooner).
• The path difference between Ray-1 and Ray-3 is
  2?(2d.Sin?) 2?n? 2n?. This implies that if
  Ray-1 and Ray-2 constructively interfere Ray-1
  and Ray-3 will also constructively interfere.
  (And so forth).

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• The previous page explained how constructive
  interference occurs. How about the rays just of
  Bragg angle? Obviously the path difference would
  be just off ? as in the figure below. How come
  these rays go missing?

Click here to understand how destructive
interference of just of-Bragg rays occur
Interference of Ray-1 with Ray-2
Note that they almost constructively interfere!
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Reflection versus Diffraction

• Though diffraction (according to Braggs picture)
  has been visualized as a reflection from a set of
  planes with interplanar spacing d ? diffraction
  should not be confused with reflection (specular
  reflection)

Reflection Diffraction
Occurs from surface Occurs throughout the bulk
Takes place at any angle Takes place only at Bragg angles
100 of the intensity may be reflected Small fraction of intensity is diffracted
Note X-rays can be reflected at very small
angles of incidence
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Understanding the Braggs equation

• n? 2d Sin?The equation is written better with
  some descriptive subscripts
• n is an integer and is the order of the
  reflection (i.e. how many wavelengths of the
  X-ray go on to make the path difference between
  planes)
• Braggs equation is a negative statement? If
  Braggs eq. is NOT satisfied ? NO reflection
  can occur? If Braggs eq. is satisfied ?
  reflection MAY occur (How?- we shall see this
  a little later)
• The interplanar spacing appears in the Braggs
  equation, but not the interatomic spacing a
  along the plane (which had forced ?incident
  ?scattered) but we are not free to move the
  atoms along the plane randomly ? click here to
  know more
• ? For large interplanar spacing the angle of
  reflection tends towards zero ? as d increases,
  Sin? decreases (and so does ?)? The smallest
  interplanar spacing from which Bragg diffraction
  can be obtained is ?/2 ? maximum value of ? is
  90?, Sin? is 1 ? from Bragg equation d ?/2

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Order of the reflection (n)

• For Cu K? radiation (? 1.54 Å) and d110 2.22
  Å

n Sin? n?/2d ?
1 0.34 20.7º First order reflection from (110) ? 110
2 0.69 43.92º Second order reflection from (110) planes ? 110 Also considered as first order reflection from (220) planes ? 220
Relation between dnh nk nl and dhkl
e.g.
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In XRD nth order reflection from (h k l) is
considered as 1st order reflection from (nh nk nl)
Hence, (100) planes are a subset of (200) planes
Important point to note In a simple cubic
crystal, 100, 200, 300 are all allowed
reflections. But, there are no atoms in the
planes lying within the unit cell! Though, first
order reflection from 200 planes is equivalent
(mathematically) to the second order reflection
from 100 planes for visualization purposes of
scattering, this is better thought of as the
later process (i.e. second order reflection from
(100) planes)
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Forward and Back Diffraction
Here a guide for quick visualization of forward
and backward scattering (diffraction) is presented
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Funda Check

• What is ? (theta) in the Braggs equation?

• ? is the angle between the incident x-rays and
  the set of parallel atomic planes (which have a
  spacing dhkl). Which is 10? in the above figure.
• It is NOT the angle between the x-rays and the
  sample surface (note specimens could be
  spherical or could have a rough surface)

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The missing reflections

• We had mentioned that Braggs equation is a
  negative statement i.e. just because Braggs
  equation is satisfied a reflection may not be
  observed
• Let us consider the case of Cu K? radiation (?
  1.54 Å) being diffracted from (100) planes of Mo
  (BCC, a 3.15 Å d100)

But this reflection is absent in BCC Mo
The missing reflection is due to the presence of
additional atoms in the unit cell (which are
positions at lattice points) ? which we shall
consider next
However, the second order reflection from (100)
planes (which is equivalent to the first order
reflection from the (200) planes is observed
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Important points

• Presence of additional atoms/ions/molecules in
  the UC ? at lattice points? or as a part of the
  motif can alter the intensities of some
  of the reflections
• Some of the reflections may even go missing

• Position of the reflections/peaks tells us
  about the lattice type.
• The Intensities tells us about the motif.

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Intensity of the Scattered waves

• Braggs equation tells us about the position of
  the intensity peaks (in terms of ?) ? but tells
  us nothing about the intensities. The intensities
  of the peaks depend on many factors as considered
  here.

Scattering by a crystal can be understood in
three steps
To understand the scattering from a crystal
leading to the intensity of reflections (and
why some reflections go missing), three levels of
scattering have to be considered 1) scattering
from electrons 2) scattering from an atom 3)
scattering from a unit cell Click here to know
the details
A
Electron
Polarization factor
B

• Structure Factor (F) The resultant wave
  scattered by all atoms of the unit cell
• The Structure Factor is independent of the shape
  and size of the unit cell but is dependent on
  the position of the atoms/ions etc. within the
  cell

Atom
Atomic scattering factor (f)
C
Structure factor calculations Intensity in
powder patterns
Unit cell (uc)
Structure factor (F)
Click here to know more about
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• The concept of a Reciprocal lattice and the Ewald
  Sphere construction
• Reciprocal lattice and Ewald sphere constructions
  are important tools towards understanding
  diffraction (especially diffraction in a
  Transmission Electron Microscope (TEM))
• A lattice in which planes in the real lattice
  become points in the reciprocal lattice is a very
  useful one in understanding diffraction
• ? click here to go to a detailed description of
  these topics

Reciprocal Lattice Ewald Sphere construction
Click here to know more about
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Selection / Extinction Rules

• As we have noted before even if Braggs equation
  is satisfied, reflections may go missing ?
  this is due to the presence of additional atoms
  in the unit cell.
• The reflections present and the missing
  reflections due to additional atoms in the unit
  cell are listed in the table below.

Bravais Lattice Reflections which may be present Reflections necessarily absent
Simple all None
Body centred (h k l) even (h k l) odd
Face centred h, k and l unmixed h, k and l mixed
End centred (C centred) h and k unmixed h and k mixed
Bravais Lattice Allowed Reflections
SC All
BCC (h k l) even
FCC h, k and l unmixed
DC h, k and l are all odd or all are even (h k l) divisible by 4
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h2 k2 l2 SC FCC BCC DC
1 100
2 110 110
3 111 111 111
4 200 200 200
5 210
6 211 211
7
8 220 220 220 220
9 300, 221
10 310 310
11 311 311 311
12 222 222 222
13 320
14 321 321
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16 400 400 400 400
17 410, 322
18 411, 330 411, 330
19 331 331 331
Allowed reflections in SC, FCC, BCC DC
crystals
lattice decorated with monoatomic/monoionic
motif
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The ratio of (h2 k2 l2) derived from
extinction rules As we shall see soon the ratios
of (h2 k2 l2) is proportional to Sin2? ?
which can be used in the determination of the
lattice type
SC 1 2 3 4 5 6 8
BCC 1 2 3 4 5 6 7
FCC 3 4 8 11 12
DC 3 8 11 16
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Crystal structure determination

• As diffraction occurs only at specific Bragg
  angles, the chance that a reflection is observed
  when a crystal is irradiated with monochromatic
  X-rays at a particular angle is small (added to
  this the diffracted intensity is a small fraction
  of the beam used for irradiation).
• The probability to get a diffracted beam (with
  sufficient intensity) is increased by either
  varying the wavelength (?) or having many
  orientations (rotating the crystal or having
  multiple crystallites in many orientations).
• The three methods used to achieve high
  probability of diffraction are shown below.

Many ?s (orientations) Powder specimen
POWDER METHOD
Monochromatic X-rays
Single ?
LAUETECHNIQUE
Panchromatic X-rays
ROTATINGCRYSTALMETHOD
? Varied by rotation
Monochromatic X-rays
Only the powder method (which is commonly used in
materials science) will be considered in this
text.
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THE POWDER METHOD

• In the powder method the specimen has
  crystallites (or grains) in many orientations
  (usually random).
• Monochromatic X-rays are irradiated on the
  specimen and the intensity of the diffracted
  beams is measured as a function of the diffracted
  angle.
• In this elementary text we shall consider cubic
  crystals.

Cubic crystal
(2)
(1)
(2) in (1)
?
?
In reality this is true only to an extent
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POWDER METHOD

• In the powder sample there are crystallites in
  different random orientations (a
  polycrystalline sample too has grains in
  different orientations)
• The coherent x-ray beam is diffracted by these
  crystallites at various angles to the incident
  direction
• All the diffracted beams (called reflections)
  from a single plane, but from different
  crystallites lie on a cone.
• Depending on the angle there are forward and back
  reflection cones.
• A diffractometer can record the angle of these
  reflections along with the intensities of the
  reflection
• The X-ray source and diffractometer move in arcs
  of a circle- maintaining the Bragg reflection
  geometry as in the figure (right)

Different cones for different reflections
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How to visualize the occurrence of peaks at
various angles
It is somewhat difficult to actually visualize
a random assembly of crystallites giving peaks at
various angels in a XRD scan. The figures below
are expected to give a visual feel for the
same. Hypothetical crystal with a 4Å is
assumed with ?1.54Å. Only planes of the type xx0
(like (100,110)are considered.
The sample is not rotating only the source and
detector move in arcs of a circle
Random assemblage of crystallites in a material
As the scan takes place at increasing angles,
planes with suitable d, which diffract are
picked out from favourably oriented crystallites
?h2 hkl d Sin(q) q
1 100 4.00 0.19 11.10
2 110 2.83 0.27 15.80
3 111 2.31 0.33 19.48
4 200 2.00 0.39 22.64
5 210 1.79 0.43 25.50
6 211 1.63 0.47 28.13
8 220 1.41 0.54 32.99
9 300 1.33 0.58 35.27
10 310 1.26 0.61 37.50
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Determination of Crystal Structure from 2? versus
Intensity Data in Powder Method

• In the power diffraction method a 2? versus
  intensity (I) plot is obtained from the
  diffractometer (and associated instrumentation)
• The intensity is relative intensity and is
  the area under the peak in such a plot (NOT the
  height of the peak)? I is really diffracted
  energy (as Intensity is Energy/area/time)
• A table is prepared as in the next slide to
  tabulate the data and make calculations to find
  the crystal structure (restricting ourselves to
  cubic crystals for the present)

Powder diffraction pattern from Al
Radiation Cu K?, ? 1.54 Å
Increasing d
Increasing ?
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Determination of Crystal Structure from 2? versus
Intensity Data
The following table is made from the 2? versus
Intensity data (obtained from a XRD experiment on
a powder sample (empty starting table of columns
is shown below- completed table shown later).
n 2?? ? Intensity Sin? Sin2 ? ratio





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Powder diffraction pattern from Al
Radiation Cu K?, ? 1.54 Å

• Note
• This is a schematic pattern
• In real patterns peaks or not idealized ? peaks ?
  broadened
• Increasing splitting of peaks with ?g ?(?1 ?2
  peaks get resolved in the high angle peaks)
• Peaks are all not of same intensity
• No brackets are used around the indexed
  numbers(the peaks correspond to planes in the
  real space)

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Radiation Cu K?, ? 1.54 Å
Powder diffraction pattern from Al

• Note
• Peaks or not idealized ? peaks ? broadened
• Increasing splitting of peaks with ?g ?
• Peaks are all not of same intensity

111
200
220
311
222
400
K?1 K?2 peaks resolved in high angle peaks(in
222 and 400 peaks this can be seen)
In low angle peaks K?1 K?2 peaks merged
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Funda Check
How are real diffraction patterns different from
the ideal computed ones?

• We have seen real and ideal diffraction patterns.
  In ideal patterns the peaks are ? functions.
• Real diffraction patterns are different from
  ideal ones in the following ways? Peaks are
  broadened (could be due to instrumental,
  residual non-uniform strain, grain size etc.
  broadening)? Peaks could be shifted from their
  ideal positions (could be due to uniform
  strain)? Relative intensities of the peaks could
  be altered (could be due to texture in the
  sample)

Funda Check
What is the maximum value of ? possible?
Ans 90?

• At ? 90? the reflected ray is opposite in
  direction to the incident ray.
• Beyond this angle, it is as if the source and
  detector positions are switched.
• ? 2?max is 180?.

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Solved example
Determination of Crystal Structure (lattice type)
from 2? versus Intensity Data
1
Let us assume that we have the 2? versus
intensity plot from a diffractometer ? To know
the lattice type we need only the position of the
peaks (as tabulated below)
2? ? Sin? Sin2 ? ratio Index d
1 38.52 19.26 0.33 0.11 3 111 2.34
2 44.76 22.38 0.38 0.14 4 200 2.03
3 65.14 32.57 0.54 0.29 8 220 1.43
4 78.26 39.13 0.63 0.40 11 311 1.22
5 82.47 41.235 0.66 0.43 12 222 1.17
6 99.11 49.555 0.76 0.58 16 400 1.01
7 112.03 56.015 0.83 0.69 19 331 0.93
8 116.60 58.3 0.85 0.72 20 420 0.91
9 137.47 68.735 0.93 0.87 24 422 0.83
10 163.78 81.89 0.99 0.98 27 333 0.78
Note that Sin? cannot be gt 1
Note
From the ratios in column 6 we conclude that
FCC
Using
We can get the lattice parameter ? which
correspond to that for Al
Note Error in d spacing decreases with ? ? so we
should use high angle lines for lattice parameter
calculation
Click here to know more
XRD_lattice_parameter_calculation.ppt
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Solved example
2
Another example Given the positions of the Bragg
peaks we find the lattice type
2?? ? Sin? Sin2 ? Ratiosof Sin2? Whole number ratios
1 21.5 0.366 0.134 1 3
2 25 0.422 0.178 1.33 4
3 37 0.60 0.362 2.66 8
4 45 0.707 0.500 3.66 11
5 47 0.731 0.535 4 12
6 58 0.848 0.719 5.33 16
7 68 0.927 0.859 6.33 19
FCC
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Comparison of diffraction patterns of SC, BCC
B2 structures
Click here
More Solved Examples on XRD
Click here
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What happens when we increase or decrease ??
Funda Check
We had pointed out that ? a is preferred for
diffraction. Let us see what happens if we
drastically increase or decrease ?.
If we make ? small? all the peaks get crowded to
small angles
If we double ? ? we get too few peaks
With CuK? ? 1.54 Å
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Applications of XRD
Bravais lattice determination
We have already seen these applications
Lattice parameter determination
Determination of solvus line in phase diagrams
Long range order
Crystallite size and Strain
Click here to know more
Next slide
Determine if the material is amorphous or
crystalline
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Crystal
Schematic of difference between the diffraction
patterns of various phases
Intensity ?
Sharp peaks
90
0
180
Diffraction angle (2?) ?
Monoatomic gas
No peak
Intensity ?
Diffraction angle (2?) ?
Diffraction angle (2?) ?
Liquid / Amorphous solid
180
Intensity ?
90
0
Diffuse Peak
0
90
180
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Actual diffraction pattern from an amorphous solid
Diffuse peak from Cu-Zr-Ni-Al-Si Metallic glass

• Note
• Sharp peaks are missing
• Broad diffuse peak survives ? the peak
  corresponds to the average spacing between atoms
  which the diffraction experiment picks out

(XRD patterns) courtesy Dr. Kallol Mondal, MSE,
IITK
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Funda Check

• What is the minimum spacing between planes
  possible in a crystal?
• How many diffraction peaks can we get from a
  powder pattern?

Let us consider a cubic crystal (without loss in
generality)
As h,k, l increases, d decreases ? we could
have planes with infinitesimal spacing
The number of peaks we obtain in a powder
diffraction pattern depends on the wavelength of
x-ray we are using. Planes with d lt ?/2 are not
captured in the diffraction pattern. These peaks
with small d occur at high angles in
diffraction pattern.
With increasing indices the interplanar spacing
decreases
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