X-RAY DIFFRACTION TECHNIQUE
Dr. Pramod Kumar SinghSchool of Basic Sciences & Research
School of Engineering and TechnologySharda University, Greater Noida, INDIA
E mail: [email protected]://pramodkumarsingh.weebly.com
X-ray Diffraction
There are number of physical methods for investigating the structures of solids.
Each technique has its own merit and weaknesses.
The most important X-ray diffraction *Fingerprint characterization of crystalline materials
*Determination of their crystal structures.
*Invisible
*Penetrate inside of materials
X-ray Diffraction Optical grating and diffraction of light
Diffraction of light by an optical grating.
An optical grating may consist of a piece of glass on which have been ruled a large number of accurately parallel and closely spaced lines.
The separation of lines should be a little larger than the wavelength of light, say 10,000 Å.
Consider what happens to a beam of light which hits the grating perpendicular to the plane.
A piece of glass without the lines would simply transmit the light
*In the grating lines act as a secondary point sources of light and radiate light in all the directions.
Interference then occurs between the waves originating from each lines source.
In certain directions, adjacent beams are in phase with each other and constructive interference occurs to give a resultant diffracted beam in the direction.
When the beams are out of phase, they cancel each other and the net intensity becomes zero
Production of X-raysX-rays are produced when high speed electrons are suddenly stopped by a solid object.Of the total energy supplied, only a fraction (about 1%) is converted to X-rays. Most of the energy is transferred into heat. To prevent the target from melting, it is cooled from behind by running water.
Thus the target material must be made of a high melting material which has good thermal conductivity. For the production of high intensity X-rays a target element should have a high atomic number.
Pure transition metals such as Mo, Cu, Cr etc. are typical target materials.
Imagine an atom consisting of a nucleus surrounded by a system of electrons contained in shells called K, L, M etc. as shown in Figure.
The electrons in K, L and M shells are the electrons with principal quantum number 1, 2 and 3 respectively.
An electron bombarding the target may have sufficient energy to completely displace a K electron from the target atom.
This produces an unstable ion, and 10-4 sec. later an electron from an outer shell drops into the vacant position.
This transition is accompanied by the emission of a characteristics X-ray
If the vacancy is filled by an M electron then the photon produced is called a Kb1 X-ray.
If the vacancy is filled by an L electron the a Ka1 or a Ka2 X-ray is produced, depending on the sub shell of the electron.
Diffraction of X-rays from crystalsBy 1912 the nature of X-rays – whether they were particles or waves was unresolved.
This was eventually achieved by von Laue using a crystal of copper sulphate as the diffraction grating.
X-ray diffraction by a crystal of beryl using the Laue method.
Diffraction of X-rays from crystalsCrystalline solids consist of regular arrays of atoms, ions or molecules with inter atomic spacing of the order of 100 pm.
For diffraction to take place, the wavelength of the incident light has to be of the same order of magnitude as the spacing of the grating.
Because of the periodic nature of the internal structure, it is possible for crystals to act as a three dimensional diffraction grating to light of a suitable wavelength: a Laue photograph is shown in the figure
X-ray diffraction by a crystal of beryl using the Laue method.
Figure illustrates the Bragg condition for the reflection of X-ray by a crystal.
The array of black points (section through a crystal)
Lines joining the dots mark a set of parallel planes with Miller indices hkl and interplanar spacing dhkl.
Path difference EF+FG
EF + FG = 2d Sin
The intensity of the diffracted beam will be maximum if the path difference is an integral multiple of wavelength
2d Sin = n
This is Braggs equation
Figure illustrates the Bragg condition for the reflection of X-ray by a crystal.
The ray A is scattered by the atom at B and the ray D is scattered by the atom at F.
For the reflected beams the
Path difference EF+FG
EF + FG = 2d Sin
2d Sin = n
This is Braggs equation
The intensity of the diffracted beam will be maximum if the path difference is an integral multiple of wavelength
Linear (1D) Diffraction Scans have Linear (1D) Diffraction Scans have better resolution and less noisebetter resolution and less noise
20 30 40 50 60 70 80 90 100 110 120 1302 (deg.)
Inte
nsity
(a.u
.)
[HighRes_run113_12hr scan.xml] run113 mono-Y2O3 OVD
59 60 61 62 63 64 65 66 67 68
Applications of X-ray-diffractionThe methods of X-ray diffraction has been used for more than 100 years to identify and characterize crystalline phases in solid materials
XRD technique is a powerful tool to distinguish between crystalline and amorphous materials
It is used for qualitative and quantitative analysis of crystalline materials
Applications of X-ray-diffractionX-ray diffraction from single crystals are used for crystal structure determinations
Powder X-ray diffraction is used for lattice parameter determination of crystal lattice
Powder X-ray diffraction is used for study of solid state reactions
Qualitative analysis For qualitative analysis powder X-ray diffraction technique is used. Modern instruments have computer programme for qualitative analysis
The pattern of each phase depends on its crystal structure and chemical composition
The pattern is treated simply as a set of peaks , each with a position and intensity
Qualitative analysisThe pattern for a mixture of phases is a simple combination of patterns of individual phases
Qualitative XRD analysis compares the pattern to a library of patterns of known materials
The most commonly used library is the Powder Diffraction File (PDF) or JCPDS-ICDD data
Joint Committee on Powder Diffraction Standards-International Center for Diffraction Data
Databases such as the Powder Diffraction File (PDF) Databases such as the Powder Diffraction File (PDF)
contain d, I lists for thousands of crystalline phases.contain d, I lists for thousands of crystalline phases. The PDF contains over
200,000 diffraction patterns.
Modern computer programs can help you determine what phases are present in your sample by quickly comparing your diffraction data to all of the patterns in the database.
The PDF card for an entry contains a lot of useful information, including literature references.
Quantitative analysis
Some suitable internal standard is used
The ratio of intensity of one phase of the material to the intensity of the internal standard is calculated and a calibration curve is plotted
From the calibration curve the quantity of the particular material is determined
Uses: different structuresUses: different structures
20 30 40 50 60 70
53816-ICSDLambda: 1.54178 Magnif: 1.0 FWHM: 0.200Space grp: F m -3 m Direct cell: 5.6400 5.6400 5.6400 90.00 90.00 90.00
20 30 40 50 60 70
53825-ICSDLambda: 1.54178 Magnif: 1.0 FWHM: 0.200Space grp: F m -3 m Direct cell: 6.2800 6.2800 6.2800 90.00 90.00 90.00
NaCl
KCl
Even if two structures are the same (and they are chemically similar) differences can be observed:Peak positions (unit cell changes) and relative intensities (atoms)
NaCl
KCl
Uses: different structuresUses: different structuresBUT, sometimes you can’t really see any changes on visual inspection…
This often happens in “open” structures where there is space for change of light atoms
Zeolite A
Zeolite A
Uses: polymorphsUses: polymorphs
Different polymorphs will have different powder patterns
e.g. Zn S
Peak BroadeningPeak BroadeningIn an X-ray diffraction pattern, peak width depends on the instrument
– radiation not pure monochromatic– Heisenberg uncertainty principle– focussing geometry
the sample…- a crystalline substance gives rise to sharp lines, whereas a truly amorphous material gives a broad “hump”.
What happens between the two?
Scherrer’s FormulaScherrer’s Formula
BcosBKt
t = thickness of crystallite K = constant dependent on
crystallite shape (0.89) = x-ray wavelength B = FWHM (full width at
half max) or integral breadth0 B= Bragg Angle
Peak BroadeningPeak Broadening
222SM BBB
B is the line broadening, by reference to a standard (Bs), and BM is that of materials, so that
When to Use Scherrer’s FormulaWhen to Use Scherrer’s Formula
Crystallite size <1000 Å Peak broadening by other factors
– Causes of broadening Size Strain Instrument
– If breadth consistent for each peak then assured broadening due to crystallite size
K depends on definition of t and B Within 20%-30% accuracy at best
Uses: particle size determinationUses: particle size determinationHere we see particle size increasing with temperature
0
50
100
150
200
15 20 25 30 35 40 45 50 55 6020 / o
30o
C
1050o
C
Particle size determination: ExampleParticle size determination: ExamplePeak at 28.2° 2 with FWHM of 0.36° 2 Standard material has FWHM of 0.16° 2 = CuK = 1.540 Å
0.36 ° = 0.36 x /180 = 0.0063 rad
0.16 ° = 0.16 x /180 = 0.0028 rad
B = 0.0056 rad
t = 255 Å = 0.0255 m
1.14cos0056.0540.19.0
t
Uses: more advancedUses: more advanced
Structure refinement – the Rietveld method
A refinement technique, not determination
Whole-pattern fitting - not just the Bragg reflections
Needs a MODEL - pattern calculated from model, compared point-by-point with observed pattern.
Originally developed (1967,1969) for use with neutron data- good reproducible peak shapes1977 - first report of application to X-ray data
Hugo Rietveld, b1932
http://home.wxs.nl/~rietv025/
Uses: Rietveld RefinementUses: Rietveld Refinementx y z
Ca/Ce 0.3333 0.6667 -0.0038(18)
Ce 0.2337(4) -0.0108 0.25
Si 0.403(3) 0.380(3) 0.25
O1 0.316(4) 0.467(4) 0.25
O2 0.597(5) 0.467(4) 0.25
O3 0.340(2) 0.252(3) 0.071(3)
O4 0 0 0.25
Here there was a similarity between the powder pattern of this phase and an existing one – also chemical composition similar.
2Th Degrees605040302010
Cou
nts
1,6001,400
1,2001,000
800600400200
0-200-400
HA 80.18 %b-TCP 19.82 %
Uses: more advancedUses: more advanced Quantitative phase analysis (how much of each)
Naïve approach - relative intensity of peak maxima? - Consider mixture of Ba,Si,O - Ba component would scatter more than Si component
(e.g. Ba2SiO4 c.f. SiO2)
Thus uses Rietveld method and takes into account relative scattering from each crystalline phase