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