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X-RAY DIFFERECTION
For electromagnetic radiation to be diffracted the spacing
in the grating 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 are used to the study of crystal structures
Beam of electrons TargetX-rays
Generation of x-ray
X-ray
Fast incident electron
nucleus
Atom of the anodematerial
electrons
Ejectedelectron
(slowed down and changed
direction)
THE PRINCIPLE OF GENERATION THE CHARACTERISTIC RADIATION
K
L
K
K
L
M
EmissionPhotoelectron
Electron
Inte
nsity
Wavelength ()
Mo Target impacted by electrons accelerated by a 35 kV potential
0.2 0.6 1.0 1.4
Characteristic radiation →
due to energy transitions
in the atom
K
K
Target Metal Of K radiation (Å)
Mo 0.71
Cu 1.54
Co 1.79
Fe 1.94
Cr 2.29
Anode Filters
Mo Zr
Cu Ni
Co Fe
Fe Mn
That atom will act as filter which is 1
position earlier in the periodic table
Constructive interference
occurs only when path
difference between ray 1
and ray 2
AB=BC
Deriving Bragg’s Law:
X-ray 1
X-ray 2
dSinAB
dSinn 2
dSinn 2
Powder Diffraction Pattern
CRYSTAL SYSTEMS
Crystal systems Axes system
cubic a = b = c , = = = 90°
Tetragonal a = b c , = = = 90°
Hexagonal a = b c , = = 90°, = 120°
Rhomboedric a = b = c , = = 90°
Orthorhombic a b c , = = = 90°
Monoclinic a b c , = = 90° , 90°
Triclinic a b c , °
Bravais lattice determination
Lattice parameter determination
Volume of the unit cell
Applications of XRD
Crystallite size and Strain
Density of the unit cell
Powder diffraction pattern from Al
42
0
111
20
0
22
0
311
22
2
40
0 33
1
42
2
Radiation: Cu K, = 1.54056 Å
X-Ray Diffraction: A Practical Approach, C. Suryanarayana & M. Grant Norton, Plenum Press, New York (1998)
n 2 SinSin2
(C)
Ratio
(L)Index a (nm)
1 38.52 19.26 0.33 0.11 3 111 0.40448
2 44.76 22.38 0.38 0.14 4 200 0.40457
3 65.14 32.57 0.54 0.29 8 220 0.40471
4 78.26 39.13 0.63 0.40 11 311 0.40480
5 82.47 41.235 0.66 0.43 12 222 0.40480
6 99.11 49.555 0.76 0.58 16 400 0.40485
7 112.03 56.015 0.83 0.69 19 331 0.40491
8 116.60 58.3 0.85 0.72 20 420 0.40491
9 137.47 68.735 0.93 0.87 24 422 0.40494
Determination of Crystal Structure from 2 versus Intensity
n C C1*L C2*L C3*L
1 0.11 0.11*1 0.055*2 0.036*3
2 0.14 0.11*1.27 0.055*2.54 0.036*4
222 lkh
adCubic
Edge length for Cubic crystal
dSin2
222
222 sin4
lkh
a
)(sin4
222
2
22 lkha
h k l
Lattice type
h+k+l = even Body centered (I)
h+k = even c centered (C)
h+l = even b centered (B)
k+l = even a centered (A)
If all are even or all are odd phase centered (F)
no systematic presence or absence promitive unit cell (P)
Glide plane (reflection + translation)
h k 0
h 0 l
0 k l
Types of glide plane
Axial glide plane (h =2n) (a glide plane with a/2 translation
Diagonal glide plane (h+k = 2n)
Diamond glide plane (h+k = 4n)
220
230
240
glide plane : Axial type is present
Screw axis (rotation + translation)
h00
0k0
00l
e.g. 200
h= 2
Diad with ½ translation
Volume of unit cell
Volume of cubic unit cell
Density of cubic crystal
M = atomic mass
Z = no. of atoms in a unit cell
V = volume of unit cell
Analyze XRD data
X-ray diffraction of copper is done using
x-ray with a wavelength of 0.154 nm, one
peak of XRD is at 2θ = 43.2 ⁰
What are Miller indices for this peak ?
Use Bragg’s law
d = 0.209 nm
dSin2
Spacing between the plane
As Cu is FCC
r = 0.128 nm
= 1.732
= 3.0
222lkh
adhkl
++=
For FCC
principle diffraction plane are those whose
indices are all even or all odd ( e.g. 111,
200, 220 )
only possibility is
h = k = l = 1
So, diffraction peak is (111) plane