Majmaah University
Department of Physics
College of Science, Al-Zulfi
KINGDOM OF SAUDI ARABIA
1. X-Ray Diffraction
2. Electron Diffraction
3. Neutron Diffraction
Friday, July 1, 2016
Prof. Dr. Abdul Majid
1
Majmaah University
Department of Physics
College of Science, Al-Zulfi
KINGDOM OF SAUDI ARABIA
Crystal Diffraction
X-ray Diffraction (XRD)
•
•
•
•
Friday, July 1, 2016
What is X-ray Diffraction
Basics of Crystallography
Production of X-rays
Applications of XRD
Prof. Dr. Abdul Majid
2
Majmaah University
Department of Physics
College of Science, Al-Zulfi
KINGDOM OF SAUDI ARABIA
Crystal Diffraction
Friday, July 1, 2016
Prof. Dr. Abdul Majid
3
Majmaah University
Department of Physics
College of Science, Al-Zulfi
KINGDOM OF SAUDI ARABIA
Crystal Diffraction
Friday, July 1, 2016
Prof. Dr. Abdul Majid
4
Majmaah University
Department of Physics
College of Science, Al-Zulfi
KINGDOM OF SAUDI ARABIA
Crystal Diffraction
Bragg’s Law
𝒏𝝀 = 𝟐𝒅𝑺𝒊𝒏 𝜽
English physicists Sir W.H. Bragg and his son Sir W.L. Bragg developed
a relationship in 1913 to explain why the cleavage faces of crystals
appear to reflect X-ray beams at certain angles of incidence (theta, q).
The variable d is the distance between atomic layers in a crystal, and the
variable lambda l is the wavelength of the incident X-ray beam; n is an
integer. This observation is an example of X-ray wave interference,
commonly known as X-ray diffraction (XRD), and was direct evidence for
the periodic atomic structure of crystals postulated for several centuries
Friday, July 1, 2016
Prof. Dr. Abdul Majid
5
Majmaah University
Department of Physics
College of Science, Al-Zulfi
KINGDOM OF SAUDI ARABIA
Crystal Diffraction
Friday, July 1, 2016
Prof. Dr. Abdul Majid
6
Majmaah University
Department of Physics
College of Science, Al-Zulfi
KINGDOM OF SAUDI ARABIA
Crystal Diffraction
Friday, July 1, 2016
Prof. Dr. Abdul Majid
7
Majmaah University
Department of Physics
College of Science, Al-Zulfi
KINGDOM OF SAUDI ARABIA
Crystal Diffraction
Friday, July 1, 2016
Prof. Dr. Abdul Majid
8
Majmaah University
Department of Physics
College of Science, Al-Zulfi
KINGDOM OF SAUDI ARABIA
Crystal Diffraction
Powder diffraction
Debye-Scherrer Powder Camera
Friday, July 1, 2016
Prof. Dr. Abdul Majid
9
Majmaah University
Department of Physics
College of Science, Al-Zulfi
KINGDOM OF SAUDI ARABIA
Crystal Diffraction
Powder diffraction
Debye-Scherrer Powder Camera
Friday, July 1, 2016
Prof. Dr. Abdul Majid
10
Majmaah University
Department of Physics
College of Science, Al-Zulfi
KINGDOM OF SAUDI ARABIA
Crystal Diffraction
Friday, July 1, 2016
Prof. Dr. Abdul Majid
11
Majmaah University
Department of Physics
College of Science, Al-Zulfi
KINGDOM OF SAUDI ARABIA
Crystal Diffraction
Friday, July 1, 2016
Prof. Dr. Abdul Majid
12
Majmaah University
Department of Physics
College of Science, Al-Zulfi
KINGDOM OF SAUDI ARABIA
Bragg’s Crystal Diffraction
I
C’
II
D
A
C
d
B
Path Difference (P.d.) = AB + BC - AC’ …………………….
(1)
Majmaah University
Department of Physics
College of Science, Al-Zulfi
KINGDOM OF SAUDI ARABIA
Bragg’s Crystal Diffraction
I
C’
II
D
A
C
d
B
D
ABD
A
𝐴𝐵
𝐵𝐷
B
1
𝑆𝑖𝑛𝜃
𝐵𝐷
= 𝑆𝑖𝑛𝜃
=
𝐴𝐵
But BD = d Distance between two layers
AB = d/Sin …………………….
(2)
Majmaah University
Department of Physics
College of Science, Al-Zulfi
KINGDOM OF SAUDI ARABIA
Bragg’s Crystal Diffraction
I
C’
II
D
A
C
d
B
D
C
B
Similarly in
BCD
BC/BD= 1/Sin
BC = BD/Sin
But BD = d Distance between two layers
BC = d/Sin …………………….
(3)
Majmaah University
Department of Physics
College of Science, Al-Zulfi
KINGDOM OF SAUDI ARABIA
Bragg’s Crystal Diffraction
I
C’
II
D
A
C
d
C’
B
A
C
Similarly in
ACC’
AC’/AC = Cos
AC’ = AC Cos
AC’ = (AD+DC) Cos
Here AD/BD = 1/tan
AD = BD/tan = d/tan
Similarly in DC = d/tan
Therefore AC’ = (AD+DC) Cos will be
AC’ = (d/tan + d/tan) Cos 
AC’ = (2d/tan) Cos  ……………… (4)
Majmaah University
Department of Physics
College of Science, Al-Zulfi
KINGDOM OF SAUDI ARABIA
Bragg’s Crystal Diffraction
Therefore equation 1 become by putting equations 2, 3, and 4
Path Difference (P.d.) = AB + BC - AC’ …………………….
= d/Sin + d/Sin - (2d/tan) Cos 
= 2d/Sin - (2d/tan) Cos 
= 2d/Sin - (2d/Sin) Cos  Cos
= 2d/Sin - (2d/Sin) Cos2
= 2d/Sin (1 - Cos2)
(1)
= 2d/Sin (Sin2)
P.d = 2d Sin
For contractive interference the 2d Sin should be equal to integral
number n of wave length Lambda so
P.d = n = 2d Sin
n = 2d Sin
This is our required Bragg ‘s Law
Therefore equation 1 become by putting equations 2, 3, and 4
Path Difference (P.d.) = AB + BC - AC’ …………………….
= d/Sin + d/Sin - (2d/tan) Cos 
= 2d/Sin - (2d/tan) Cos 
= 2d/Sin - (2d/Sin) Cos  Cos
= 2d/Sin - (2d/Sin) Cos2
= 2d/Sin (1 - Cos2)
(1)
= 2d/Sin (Sin2)
P.d = 2d Sin
For contractive interference the 2d Sin should be equal to integral
number n of wave length Lambda so
P.d = n = 2d Sin
n = 2d Sin
This is our required Bragg ‘s Law where d give the lattice constant as under
Majmaah University
Department of Physics
College of Science, Al-Zulfi
KINGDOM OF SAUDI ARABIA
Crystal Diffraction
2
Friday, July 1, 2016

Sine 
Prof. Dr. Abdul Majid
d
a=b=c
a
19
Majmaah University
Department of Physics
College of Science, Al-Zulfi
KINGDOM OF SAUDI ARABIA
Crystal Diffraction
2
Friday, July 1, 2016

Sine 
d
Prof. Dr. Abdul Majid
(hkl)
a=b=c
a
20