The document discusses key concepts in X-ray diffraction (XRD) analysis including peak position, intensity, and width and how they relate to crystal structure properties. It explains how Miller indices are used to index diffraction peaks and determine the unit cell parameters and crystal structure. An example shows indexing the peaks for SrTiO3 to identify the crystal system as face-centered cubic based on consistent lattice parameters calculated from d-spacings. Crystallite size is calculated using the Scherrer formula applied to the nickel diffraction pattern peak widths.

1. Crystal Structure and
Crystallite Size
Determination from
XRD
-Nithyapremini 1

2. X-Ray Diffraction
2

3. XRD parameters
Peak position
Peak Intensity
Peak Width
Peak Intensity:
• Position of atoms within a lattice structure.
Peak Position:
• Distance between reflection planes (dhkl)
• Lattice parameter.
• Crystal structure.
Peak Width:
• Crystallite size.
• Perfection of lattice (crystalline/ amorphous).
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4. Crystal system & Bravais lattices
• Crystals possess a regular, repetitive internal structure.
• In general one can generate 14 basic crystal structures through
symmetries. These are called Bravais lattices.
• Any crystal structures can be reduced to one of these 14 Bravias lattices.
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5. Miller Indices (hkl)
• Symbolic vector representation for the orientation of
atomic plane in crystal lattice.
• It is defined as reciprocals fractional intercepts which
the plane makes with crystallographic axes.
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6. Interplanar spacing or d-spacing (dhkl)
• Interplanar spacing or interplanar distance is the perpendicular distance
between two successive planes in a family (h k l).
Example:
• Cubic crystals: highest symmetry (a=b=c)
fewest number of XRD peaks.
• Monoclinic crystals: lower symmetry
(a=b≠c) larger number of XRD peaks
As crystal symmetry
decreases, the number of
XRD peaks observed
increases
diffraction patterns of ZrO2
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7. Indexing the peaks
• It gives, Which planes are responsible for each reflection?
• Indexing is the process of determining the unit cell parameters from the peak positions.
• To index a powder diffraction pattern it is necessary to assign Miller indices, hkl, to each peak.
• A diffraction pattern cannot be analyzed until it has been indexed. It is always the first step in
analysis.
• Information in an XRD pattern is a direct result of two things:
 Size and shape of the unit cells, which determine the relative positions of diffraction peaks.
 Atomic positions within the unit cell, which determine the relative intensities.
• For lower symmetry structures (monoclinic, triclinic) it is usually necessary to use a computer
algorithm, called Autoindexing.
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8. Selection / Reflection rules
• Bragg’s law only defines the diffraction condition for primitive
unit cells, e.g. P cubic, P tetragonal, etc., where atoms are only at
unit cell corners.
• Crystal structures with non-primitive unit cells have atoms at
additional lattice (basis) sites.
• These extra scattering centres can cause out-of-phase scattering to
occur at certain Bragg angles.
• These out-of-phase arises because the centering leads to
destructive interference for some reflections and the missing
reflection are known as systematic absence.
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100 plane reflection from
FCC lattice

9. Indexing diffraction pattern from cubic materials
• Bragg’s Law tells us the location of a peak with indices hkl. θhkl is related to the interplanar
spacing, d, as follows:
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• We know that for a cubic phase the d-values can be calculated from the Miller indices (hkl):
Combining these two equations we get the following relationship
Sin2θ is proportional to h2+ k2 + l 2
i.e., planes with higher Miller indices will diffract at higher values of θ.

10. Indexing diffraction pattern from cubic materials
• In cubic systems, first XRD peak will be due to diffraction from
planes with lowest Miller indices
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• If the lattice is not primitive, hkl peaks will be missing. These are called
systematic absences.
 Primitive: All possible h, k and l values.
 Body-centered:
reflection is allowed when (h + k + l) is even.
reflection is not allowed when (h + k + l) is odd
 Face-centered:
reflection is allowed when h, k and l are either all even or all odd.
no reflection when h, k and l are mixed i.e., even and odd.
Conditions for allowed reflection for cubic lattices:

11. Steps to find Lattice parameter, Miller indices, Crystal
structure
• Determine dhkl for each of reflections. dhkl = /(2𝑠𝑖𝑛)
• Calculate lattice parameter for first 3 FCC and BCC reflections
using: a= dhkl1 ℎ1
2 + 𝑘1
2 + 𝑙1
2
b= dhkl2 ℎ2
2 + 𝑘2
2 + 𝑙2
2
c= dhkl3 ℎ3
2 + 𝑘3
2 + 𝑙3
2
• FCC: h,k,l need to be all even or odd(111),(200),(202)
• BCC: h+k+l= even(110),(200),(112)
• See whether lattice parameter is constant or changing to determine
crystal structure.
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For cubic system
Selection rule

12. Indexing example- SrTiO3
Determine: a) which planes are responsible for each reflection (indexing peak).
b) lattice parameters.
c) whether crystal structure is Simple cubic, FCC, BCC.
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Peak order Diffraction angle 2
1st 22.72
2nd 32.56
3rd 39.90
• Diffraction angles for first three peaks of XRD pattern for some cubic material.
• Monochromatic radiation having wavelength of 0.1542nm was used.

13. Indexing example- SrTiO3
Step1: Determine dhkl for each of reflections.
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2  2sin dhkl=

2sin
22.72 11.36 0.3939 0.391
32.56 16.28 0.5606 0.275
39.90 19.95 0.6823 0.226
dhkl=

2sin =0.1542
Step2: Use dhkl to calculate lattice parameter for first 3 reflections.
SC
(111) a = dhkl1 ℎ2 + 𝑘2 + 𝑙2
=0.391 12 + 12 + 12
a=0.391
(200) b= dhkl2 ℎ2 + 𝑘2 + 𝑙2
=0.275 22 + 02 + 02
b=0.55
(202) c= dhkl3 ℎ2 + 𝑘2 + 𝑙2
=0.226 22 + 02 + 22
c=0.639
(110) a= dhkl1 ℎ2 + 𝑘2 + 𝑙2
=0.391 12 + 12 + 02
a=0.55
(200) b= dhkl2 ℎ2 + 𝑘2 + 𝑙2
=0.275 22 + 02 + 02
b=0.55
(112) c= dhkl3 ℎ2 + 𝑘2 + 𝑙2
=0.226 12 + 12 + 02
c=0.613
BCC
FCC
(100) a = dhkl1 ℎ2 + 𝑘2 + 𝑙2
=0.391 12 + 02 + 02
a=0.391
(110) b= dhkl2 ℎ2 + 𝑘2 + 𝑙2
=0.275 12 + 12 + 02
b=0.391
(111) c= dhkl3 ℎ2 + 𝑘2 + 𝑙2
=0.226 12 + 12 + 12
c=0.391

14. Indexing example- SrTiO3
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Sin2θ is proportional to h2+ k2 + l 2
i.e., planes with higher Miller indices will
diffract at higher values of θ.

15. Indexing example- SrTiO3
.
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r=a/2
r=0.391/2=0.1955nm
SC

16. 16

17. Indexing example- Nickel
Step1: Determine dhkl for each of reflections.
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2  2sin dhkl=

2sin
44.5 22.25 0.7572 0.2034
51.84 25.92 0.8742 0.1762
76.37 38.18 1.2362 0.1246
dhkl=

2sin =0.1542
Step2: Use dhkl to calculate lattice parameter for first 3 reflections.
SC
(111) a = dhkl1 ℎ2 + 𝑘2 + 𝑙2
=0.2034 12 + 12 + 12
a=0.352
(200) b= dhkl2 ℎ2 + 𝑘2 + 𝑙2
=0.1762 22 + 02 + 02
b=0.352
(202) c= dhkl3 ℎ2 + 𝑘2 + 𝑙2
=0.1246 22 + 02 + 22
c=0.352
(110) a= dhkl1 ℎ2 + 𝑘2 + 𝑙2
=0.2034 12 + 12 + 02
a=0.287
(200) b= dhkl2 ℎ2 + 𝑘2 + 𝑙2
=0.1762 22 + 02 + 02
b=0.352
(112) c= dhkl3 ℎ2 + 𝑘2 + 𝑙2
=0.1246 12 + 12 + 02
c=0.1762
BCC
FCC
(100) a = dhkl1 ℎ2 + 𝑘2 + 𝑙2
=0.2034 12 + 02 + 02
a=0.2034
(110) b= dhkl2 ℎ2 + 𝑘2 + 𝑙2
=0.1762 12 + 12 + 02
b=0.249
(111) c= dhkl3 ℎ2 + 𝑘2 + 𝑙2
=0.1246 12 + 12 + 12
c=0.215

18. Indexing example- Nickel
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19. Indexing example- Nickel
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a=2r√2
a=0.3528
r=a/(2√2)=0.3524/2.8284
r=0.12459nm
FCC

20. Crystallite Size
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 CRYSTALLITE and GRAIN are both SINGLE CRYSTALS.
 CRYSTALLITE is a single crystal in POWDER form. The
crystallite size commonly determined by XRD.
 GRAIN is a single crystalline or polycrystalline material
within a BULK/THIN FILM form. Grain morphology is
commonly determined by SEM.
 Particle can have one or more crystallites with same size or
with a range of sizes.
• During annealing, smaller crystallites come closer and grow to become larger due to kinetics.
SEM images of CsPbBr3
(A) as-prepared (B) 90 •C
(C) 180 •C

21. Crystallite Size
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 Scherrer method (using FWHM).
 Stokes and Wilson method (using integral breadth).
 The variance method.
• Scherrer Formula to calculate Crystallite size:
• β= FWHM in radians, θ= Bragg angle, λ= X-Ray wavelength
• K= Scherrer constant which depends on shape:
• K actually varies from 0.62 to 2.08
• 0.94 for FWHM of spherical crystals with cubic symmetry
• 0.89 for integral breadth of spherical crystals with cubic.
𝐷 =
𝑘
𝛽 𝑐𝑜𝑠𝜃

22. Crystallite Size
22
• Scherrer Formula to calculate Crystallite size:
• Decrease in crystallite size causes an increase in width of the
diffraction.
𝐷 =
𝑘
𝛽 𝑐𝑜𝑠𝜃
SEM images of CsPbBr3 (A) as-prepared (B) 90 •C (C) 180 •C. (D) XRD patterns

23. Crystalline Size calculation example- Nickel
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24. Crystalline Size calculation example- Nickel
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25. Crystalline Size calculation example- Nickel
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X=44.46
Y=5435.819
=5435.819/2
=2717.909
X1=44.77
X2=44.325
X1-X2=44.77-44.325
=0.44

26. Crystalline Size calculation example- Nickel
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K-0.94
=0.1542
β= 0.44=0.44*π/180
= 0.007679Radians
D=
0.94∗0.1542
0.007679∗𝑐𝑜𝑠22.25
D=20.3 nm

27. Thank You!!!
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