X-RAY DIFFRACTION

X- Ray Sources

Diffraction: Bragg’s Law

Crystal Structure Determination

Elements of X-Ray DiffractionB.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/Diffraction.htm

MATERIALS SCIENCE

&

ENGINEERING

Anandh Subramaniam & Kantesh Balani

Materials Science and Engineering (MSE)

Indian Institute of Technology, Kanpur- 208016

Email: anandh@iitk.ac.in, URL: home.iitk.ac.in/~anandh

AN INTRODUCTORY E-BOOK

Part of

http://home.iitk.ac.in/~anandh/E-book.htm

A Learner’s Guide

How to produce monochromatic X-rays?

How does a crystal scatter these X-rays to give a diffraction pattern?

Bragg’s equation

What determines the position of the XRD peaks? Answer) the lattice.

What determines the intensity of the XRD peaks? Answer) the motif.

How to analyze a powder pattern to get information about the lattice type?(Cubic crystal types).

What other uses can XRD be put to apart from crystal structure determination?

Grain size determination Strain in the material…

What will you learn in this ‘sub-chapter’?

For electromagnetic radiation to be diffracted* the spacing in the grating (~a series

of obstacles or a series of scatterers) should be of the same order as the

wavelength.

In crystals the typical interatomic spacing ~ 2-3 Å** so the suitable radiation

for the diffraction study of crystals is X-rays.

Hence, X-rays are used for the investigation of crystal structures.

Neutrons and Electrons are also used for diffraction studies from materials.

Neutron diffraction is especially useful for studying the magnetic ordering in

materials.

Some Basics

** Lattice parameter of Cu (aCu) = 3.61 Å

dhkl is equal to aCu or less than that (e.g. d111 = aCu/3 = 2.08 Å)

** If the wavelength is of the order of the lattice spacing, then diffraction effects will be prominent.Click here to know more about this

Beam of electrons TargetX-rays

An accelerating (or decelerating) charge radiates electromagnetic radiation

X-rays can be generated by decelerating electrons.

Hence, X-rays are generated by bombarding a target (say Cu) with an electron beam.

The resultant spectrum of X-rays generated (i.e. X-rays versus Intensity plot) is shown in

the next slide. The pattern shows intense peaks on a ‘broad’ background.

The intense peaks can be ‘thought of’ as monochromatic radiation and be used for X-ray

diffraction studies.

Generation of X-rays

Mo Target impacted by electrons accelerated by a 35 kV potential shows the emission

spectrum as in the figure below (schematic)

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

Inte

nsi

ty

Wavelength ()0.2 0.6 1.0 1.4

White

radiation

Characteristic radiation →

due to energy transitions

in the atom

K

K

Intense peak, nearly

monochromatic

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

Absorption (Heat)

Incident X-rays

SPECIMEN

Transmitted beam

Fluorescent X-raysElectrons

Compton recoil PhotoelectronsScattered X-rays

Coherent

From bound charges

X-rays can also be refracted (refractive index slightly less than 1) and reflected (at very small angles)

• 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.

Incoherent (Compton modified)

From loosely bound charges

Click here to know more

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).

Let us consider a special case of diffraction → a case where we get ‘sharp[1]

diffraction peaks’.

Diffraction (with sharp peaks) (with XRD being a specific case) requires three important conditions to

be satisfied:

Coherent, monochromatic, parallel waves (with wavelength ).

Crystalline array of scatterers* with spacing of the order of (~) .

Fraunhofer diffraction geometry

[1] The intensity- plot looks like a ‘’ function.

* A quasicrystalline array will also lead to diffraction with sharp peaks (which we shall not consider in this text).

** Amorphous material will give diffuse peak.

Diffraction Click here to “Understand Diffraction”

Coherent, monochromatic, parallel wave

Fraunhofer geometry

Diffraction pattern

with sharp peaksCrystalline*,**

Aspects related to the wave

Aspects related to the material

Aspects related to the diffraction set-up

(diffraction geometry)

The waves could be:

electromagnetic waves (light, X-rays…),

matter waves** (electrons, neutrons…) or

mechanical waves (sound, waves on water surface…).

Not all objects act like scatterers for all kinds of radiation.

If wavelength is not of the order of the spacing of the scatterers, then the number

of peaks obtained may be highly restricted (i.e. we may even not even get a

single diffraction peak!).

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)

interference.To give an analogy the results of Young’s double slit experiment is interpreted as interference, while the result of

multiple slits (large number) is categorized under diffraction.

Fraunhofer diffraction geometry implies that parallel waves are impinging on the scatteres

(the object), and the screen (to capture the diffraction pattern) is placed far away from the

object.

** With a de Broglie wavelength

Some comments and notes

Click here to know more about Fraunhofer and Fresnel diffraction geometries

Sets Electron cloud into oscillationSets nucleus into oscillation

Small effect neglected

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.

XRD the first step

Schematics

Incoming X-rays

Secondary

emission

Oscillating charge re-radiates In phase with

the incoming x-rays

We can get a better physical picture of diffraction by using Laue’s formalism (leading to the Laue’s

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 Bragg’s equation (Bragg’s 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.

Incident and scattered

waves are in phase if

Scattering from across planes is in phase

In plane scattering is in phase

Some points to recon with

Extra path traveled by incoming waves AY

A B

X Y

Atomic Planes

Extra path traveled by scattered waves XB

These can be in phase if

incident = scattered

A B

X Y

But this is still reinforced scattering

and NOT reflection

Let us consider in-plane scattering

There is more to this

Click here to know more and get

introduced to Laue equations describing

diffraction

BRAGG’s EQUATION

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 Bragg’s equation. (More about this sooner).

The path difference between Ray-1 and Ray-3 is = 2(2d.Sin) = 2n = 2n. This implies that if Ray-1

and Ray-2 constructively interfere Ray-1 and Ray-3 will also constructively interfere. (And so forth).

Let us consider scattering across planes

Click here to visualize

constructive and

destructive interference

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!

Reflection versus Diffraction

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 ALSO be reflected at very small angles of incidence

Though diffraction (according to Bragg’s 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).

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).

Bragg’s equation is a negative statement

If Bragg’s eq. is NOT satisfied NO ‘reflection’ can occur

If Bragg’s eq. is satisfied ‘reflection’ MAY occur

(How?- we shall see this a little later).

The interplanar spacing appears in the Bragg’s 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.

Understanding the Bragg’s equation

2 SinCu K hkl hkln d

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

2 2 2

Cubic crystal

hkl

ad

h k l

8220

ad

2110

ad

2

1

110

220 d

d

Relation between dnh nk nl and dhkl

e.g.

2 2 2( ) ( ) ( )nhnk nl

ad

nh nk nl

2 2 2

hklnhnk nl

dad

nn h k l

Order of the reflection (n)

sin2 hkldn

In XRD nth order reflection from (h k l) is considered as 1st order reflection from (nh nk nl)

sin2n

dhkl

sin2 n n n lkhd

1nhnk nl

hkl

d

d n

300

100

1

3

d

d

200

100

1

2

d

d

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).

How is it that we are able to get information about lattice parameters of the order

of Angstroms (atoms which are so closely spaced) using XRD?Funda Check

Diffraction is a process in which

‘linear information’ (the d-spacing of the planes)

is converted to ‘angular information’ (the angle of diffraction, Bragg).

If the detector is placed ‘far away’ from the sample (i.e. ‘R’ in the figure below is large) the

distances along the arc of a circle (the detection circle) get amplified and hence we can make

‘easy’ measurements.

Forward and Back Diffraction

Here a guide for quick visualization of forward and backward scattering (diffraction) is presented

Funda Check What is (theta) in the Bragg’s 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).

We had mentioned that Bragg’s equation is a negative statement: i.e. just because Bragg’s

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).

The missing ‘reflections’

100 1002d Sin 100

100

1.540.244

2 2(3.15)Sin

d

100 14.149

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

The wave scattered from the middle plane is out of phase

with the ones scattered from top and bottom planes. I.e. if

the green rays are in phase (path difference of ) then the

red ray will be exactly out of phase with the green rays

(path difference of /2).

However, the second order reflection from (100) planes (which is equivalent to the first order

reflection from the (200) planes is observed

100

100

2 1.540.488

2 3.15Sin

d

2 1

100 200~ 29.267nd ndorder order

This is because if the green rays have a path difference of 2 then the red ray will have path

difference of → which will still lead to constructive interference!

Continuing with the case of BCC Mo…

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

Important

points

Position of the ‘reflections’/‘peaks’ tells us about the lattice type.

The Intensities tells us about the motif.

Intensity of the Scattered waves

Electron

Atom

Unit cell (uc)

Scattering by a crystal can be understood in three steps

A

B

C

Polarization factor

Atomic scattering factor (f)

Structure factor (F)

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

Structure factor calculations

&

Intensity in powder patterns

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

Click here to know more about

Bragg’s 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.

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 constructionClick here to know more about

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

DCEither, h, k and l are all odd or

all are even & (h + k + l) divisible by 4

Selection / Extinction Rules

As we have noted before even if Bragg’s 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. Click here to see the derivations

Structure factor calculations

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

15

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

Cannot be expressed as (h2+k2+l2)

The ratio of (h2 + k2 + l2) derived from extinction rules (previous page)

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 …

Note that we have to consider the ratio of only two lines to distinguish FCC and DC. I.e. if the

ratios are 3:4 then the lattice is FCC.

But, to distinguish between SC and BCC we have to go to 7 lines!

Crystal structure determination

Monochromatic X-rays

Panchromatic X-rays

Monochromatic X-rays

Many s (orientations)

Powder specimenPOWDER

METHOD

Single LAUE

TECHNIQUE

Varied by rotation

ROTATING

CRYSTAL

METHOD

λ fixed

θ variable

λ fixed

θ rotated

λ variable

θ fixed

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.

Only the powder method (which is commonly used in materials science) will be considered in this text.

THE POWDER METHOD

2222 sin)( lkh

2

2

2222 sin

4)(

alkh

)(sin4

222

2

22 lkha

222 lkh

adCubic

2d Sin

222

222 sin4

lkh

a

Cubic crystal

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.

(1) (2)

(2) in (1)

* In reality this is true only to an extent

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)

POWDER METHOD

Different cones

for different

reflections

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].

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()

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

In the power diffraction method a 2 versus intensity (I) plot is obtained from the

diffractometer (and associated instrumentation).

The ‘intensity’ is the area under the peak in such a plot (NOT the height of the peak).

The information of importance obtained from such a pattern is the ‘relative intensities’

and the absolute value of the intensities is of little importance (for now).

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).

Determination of Crystal Structure from 2 versus Intensity Data in Powder Method

Powder diffraction pattern from Al

Radiation: Cu K, = 1.54 Å

Increasing

Increasing d

n 2→ Intensity Sin Sin2 ratio

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).

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)

Powder diffraction pattern from Al

111

20

0

22

0

311

22

2

40

0

K1 & K2 peaks resolved in high angle peaks

(in 222 and 400 peaks this can be seen)

Radiation: Cu K, = 1.54 Å

Note:

Peaks or not idealized peaks broadened

Increasing splitting of peaks with g

Peaks are all not of same intensity

In low angle peaks K1 & K2 peaks merged

What is the maximum value of possible (experimentally)?

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 (microstrain), grain size etc. broadening.

Peaks could be shifted from their ideal positions

Could be due to uniform strain→ macrostrain.

Relative intensities of the peaks could be altered

Could be due to texture in the sample.

Funda Check 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.

2max is 180.

Funda Check What will determine how many peaks I will get?

1) smaller the wavelength of the X-rays, more will be the number of peaks possible.

From Bragg’s equation: [=2dSin], (Sin)max will correspond to dmin. (Sin)max=1.

Hence, dmin=/2. Hence, if is small then planes with smaller d spacing (i.e. those which

occur at higher 2 values) will also show up in a XRD patter (powder pattern). Given that

experimentally cannot be greater than 90.

2) Lattice type in SC we will get more peaks as compared to (say) FCC/DC. Other things being

equal.

3) Lower the symmetry of the crystal, more the number of peaks (e.g., in tetragonal crystal

the 100 peak will lie at a different 2 as compared to the 001 peak).

2dSin

min

max

2dSin

min2

d

# 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

Determination of Crystal Structure (lattice type) from 2 versus Intensity Data

From the ratios in column 6 we conclude that FCC

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)

Solved example

2d Sin 111 1111.54 2 2 0.33

3

ad Sin

o

4.04Aa Al

Using

We can get the lattice parameter which correspond to that for Al

1

Note: Error in d spacing decreases with → so we should use high angle lines for lattice parameter calculation

Click here to know more

Note that Sin cannot be > 1

XRD_lattice_parameter_calculation.ppt

2222 sin)( lkh

Note

2→ Sin Sin2 Ratios

of Sin2

Dividing Sin2 by

0.134/3 = 0.044667

Whole

number

ratios

1 21.5 0.366 0.134 1 3

2 25 0.422 0.178 1.33 3.99 4

3 37 0.60 0.362 2.70 8.10 8

4 45 0.707 0.500 3.73 11.19 11

5 47 0.731 0.535 4 11.98 12

6 58 0.848 0.719 5.37 16.10 16

7 68 0.927 0.859 6.41 19.23 19

FCC

Another example

Given the positions of the Bragg peaks we find the lattice type

Solved example

2

More Solved

Examples on XRDClick here

Comparison of diffraction patterns of SC, BCC & B2 structures

Click here

Aluminium = 1.54 Å = 3 Å = 0.1 Å

hkl d Sin() 2 Sin() 2 Sin() 2

111 2.34 0.33 19.26 38.52 0.64 39.87 79.74 0.02 1.22 2.45

200 2.03 0.38 22.38 44.76 0.74 47.64 95.28 0.02 1.41 2.82

220 1.43 0.54 32.57 65.14 1.05 - - 0.03 2.00 4.01

311 1.22 0.63 39.13 78.26 1.23 - - 0.04 2.35 4.70

222 1.17 0.66 41.24 82.47 1.28 - - 0.04 2.45 4.90

400 1.01 0.76 49.56 99.11 1.49 - - 0.05 2.84 5.68

331 0.93 0.83 56.02 112.03 1.61 - - 0.05 3.08 6.16

420 0.91 0.85 58.30 116.60 1.65 - - 0.05 3.15 6.30

422 0.83 0.93 68.74 137.47 1.81 - - 0.06 3.45 6.91

333 0.78 0.99 81.89 163.78 1.92 - - 0.06 3.68 7.35

Funda CheckWhat happens when we increase or decrease ?

We had pointed out that ~ a is preferred for diffraction. Let us see what happens if we ‘drastically’ increase or decrease .

(This is only a thought experiment!!)

If we ~double → we get too

few peaks

If we make small→

all the peaks get

crowded to small

angles

With CuK = 1.54 Å

And the detector may not be able to resolve

these peaks if they come too close!

Bravais lattice determination

Lattice parameter determination

Determination of solvus line in phase diagrams

Long range order

Applications of XRD

Crystallite size and Strain

Determine if the material is amorphous or crystalline

We have already seen these applications

Click here to know more

Next slide

Diffraction angle (2) →

Inte

nsi

ty→

90 1800

Crystal

90 1800

Diffraction angle (2) →

Inte

nsi

ty→ Liquid / Amorphous solid

90 1800

Diffraction angle (2) →

Inte

nsi

ty→

Monoatomic gas

Schematic of difference between

the diffraction patterns of various phases

Sharp peaks

Diffuse Peak

No peak

Diffuse peak from

Cu-Zr-Ni-Al-Si

Metallic glass

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

Actual diffraction pattern from an amorphous solid

Note

Sharp peaks are missing

Broad diffuse peak survives → the peak corresponds to the average spacing between atoms which the diffraction experiment ‘picks

out’

Funda Check What is the minimum spacing between planes possible in a crystal?

How many diffraction peaks can we get from a powder pattern?

2 2 2

Cubic crystal

hklad

h k l

Let us consider a cubic crystal (without loss in generality)

As h,k, l increases, ‘d’ decreases we could have planes with infinitesimal spacing

101

ad a

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With increasing indices the

interplanar spacing decreases

The number of peaks we obtain in a powder

diffraction pattern depends on the wavelength

of x-ray we are using. Planes with ‘d’ < /2 are

not captured in the diffraction pattern.

These peaks with small ‘d’ occur at high angles

in diffraction pattern.