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Spectroscopy
XRD
(X-Ray Diffraction on powders)
Dr. Chris
UP, Feb. 2016
Energy regions of electromagnetic
waves
Different principles
(1) Energy absorption
E1 and E2 are different states of a molecule:
 Vibrational states -> IR spectroscopy
 Nuclear spin states -> NMR spectroscopy
 electronic states -> UV/VIS spectroscopy
(2) Energy emission
Raman (infrared)
Fluorescence (uv)
(3) XRD “spectroscopy”
Different principle: reflection of X-Rays on a sample
Generation of X-Rays
The target metal
defines the energy of
the x-rays
Excitation of INNER electrons, falling back emits X-Rays
Typical metals are Mo (λ = 0.07 nm) and Cu (λ = 0.154 nm)
XRD Principle
Different planes in a crystal give different signals =
positive interference of waves
Why 2 ϴ ?
The Bragg diffraction condition contains only one
factor of θ: 2dsinθ=nλ
It should be noted that θ refers to the incidence angle
of the x-ray beam, and the beam is actually deflected
by an angle of 2θ, as illustrated in the image below:
Bragg’s Law
Spectroscopy XRD
PLANES IN CRYSTALS
MILLER INDICES
Crystal lattices – 3 cubic structures
NaCl type
Movie clip:
youtube.com/watch?v=pMTA_wiY784
Indices h k l
http://slideplayer.org/slide/792387/#
Spectroscopy XRD
Identify crystal layers
Miller indices h k l
http://www.iue.tuwien.ac.at/phd/dhar/node17.html
Practice Miller layers
½  2
Negative indices
Spectroscopy XRD
Examples
Miller Indices Example
http://www.wolframalpha.com/widget/widgetPopup.jsp?p
=v&id=cc011df99e5873930ccb659743a221b&title=Calculat
e%20Miller%20Indices%20for%20Planes&theme=purple&i
0=1&i1=2&i2=0&podSelect=&showAssumptions=1&show
Warnings=1
Cubic structure – interplanar distances
Calculate plane distances
Examples
Examples
http://www.slideshare.net/meonly21Icandependonh
imallmylifeAA/xrd-lecture-1
What is the unit cell dimension a of NaCl ?
Use plane 111 with 2x theta = 27 deg
and λ = 1.54 nm (Cu-Kα)
Which plane will give a signal at
2x theta = 46 deg when the cubic constant
a = 0.5 nm and λ = 1.54 nm ?
X-rays with wavelength 1.54 Å are “reflected” from the
(1 1 0) planes of a cubic crystal with unit cell a = 6 Å.
Calculate the Bragg angle, ϴ, for orders of reflection, n =
1-5.
Use Braggs Law to calculate possible values for ϴ :
Solution:
Indexing Example
constant
Find out which hkl combinations using in this
formula will give a constant value.
Spectroscopy XRD
ESTIMATED PARTICLE SIZES
http://mahendrakoppolu.blogspot.com/2013/07
/online-crystallite-size-calculator.html
AG NANOPARTICLES
From jcpds database
http://comptech.compres.us/tools/jcpds/
In Angstrom =
10 x nm
VERSION: 4
COMMENT: Silver (04-0783, shock wave)
K0: 120.800
K0P: 4.84000
SYMMETRY: CUBIC
A: 4.08620
ALPHAT: 0.000000
DIHKL: 2.3590 100. 1.00 1.00 1.00
DIHKL: 2.0440 52. 2.00 0.00 0.00
DIHKL: 1.4450 32. 2.00 2.00 0.00
DIHKL: 1.2310 36. 3.00 1.00 1.00
DIHKL: 1.1760 12. 2.00 2.00 2.00
DIHKL: 1.0215 6. 4.00 0.00 0.00
DIHKL: 0.9375 23. 3.00 3.00 1.00
DIHKL: 0.9137 22. 4.00 2.00 0.00
DIHKL: 0.8341 23. 4.00 2.00 2.00
Ag Nanoparticles XRD
http://www.azonano.com/article.aspx?ArticleID=2318#5
Particle size estimation
Debye-Scherrer Formula:
λ = 0.154 nm , W = width at half maximum = 0.011 rad,
Theta = 45 deg
Plane distance
http://pubs.rsc.org/en/content/articlehtml/2013/ce/c3ce40497h
a = b = c = 0.4081 nm
 Distance between 111 planes
 between 100 planes
GRAPHITE AND GRAPHENE OXIDE
(GO)
JCPDS DatabaseVERSION: 4
COMMENT: Graphite
K0: 100.000
K0P: 4.00000
SYMMETRY: HEXAGONAL
A: 2.456
C: 6.696
VOLUME: 34.9786
ALPHAT: 00.00E0
DIHKL: 3.3480 100. 0 0 2
DIHKL: 2.1270 3. 1 0 0
DIHKL: 2.0271 17. 1 0 1
DIHKL: 1.7953 3. 1 0 2
DIHKL: 1.6740 7. 0 0 4
DIHKL: 1.5398 5. 1 0 3
DIHKL: 1.2280 2. 1 1 0
DIHKL: 1.1529 3. 1 1 2
DIHKL: 1.1333 2. 1 0 5
DIHKL: 1.1160 2. 0 0 6
It needs 4 indices to
describe the planes in
hexagonal structure
Indices for HCP structures
http://www.materials.ac.uk/elearning/matter
/crystallography/indexingdirectionsandplanes
/indexing-of-hexagonal-systems.html
https://www.youtube.com/watch?v=vK913oWl_XI
Graphite and Graphene Oxide
http://www.rsc.org/suppdata/cc/c
0/c0cc01259a/c0cc01259a.pdf
Graphite structure
Spectroscopy XRD
http://bgcryst.com/symp14/papers/291_Shalaby_
BCC_47-1_2015.pdf
Reduced GO: 2θ = 26.29 degree
With λ = 0.154 nm the distance between the planes:
The close d-spacing of RGO to pristine graphite and
disappearance of peak at 2θ = 12.43 degree indicate
that the oxygen containing group of graphite oxide
have been efficiently removed
From: Nanomaterials 2015, 5, 826-834
Graphene Oxide Synthesis from Agro Waste
The peak at 2θ = 11.6° indicates an interlayer distance
of 0.79 nm  fully oxidized graphite sheets
FT-IR spectrum of GO
In the IR spectrum typical peaks of functional groups
can be identified:
Around 3500 cm-1: O-H stretching
1700 cm-1: typical for C=O stretching
1600 cm-1: C-C vibrations of graphite
1210 cm-1: C-OH stretching
From: Chem. Commun., 2011,47, 12370-12372
One-pot reduction of graphene oxide at subzero
temperatures
From: J. Chil. Chem. Soc. vol.58 no.4 Concepción dic. 2013
http://dx.doi.org/10.4067/S0717-97072013000400067
GREEN SYNTHESIS AND CHARACTERIZATION OF
GRAPHITE OXIDE BY ORTHOGONAL EXPERIMENT
Different
oxidation
parameters for
graphite
Spectroscopy XRD
From: SENSORS AND ACTUATORS B CHEMICAL
199:190–200 · AUGUST 2014
ZINC OXIDE NANOPARTICLES
Hexagonal Closest Packing
Zinc oxide XRD – Wurzite Structure
http://www.hindawi.com/journals/isrn/2012/372505/
Estimate particle sizes
Nanoscience and Nanotechnology 2015, 5(1): 1-6
Synthesis of Zinc Oxide Nanoparticles via Sol – Gel
Route and Their Characterization
Spectroscopy XRD
Hope this was helpful !
Please follow Ajarn Chris on Facebook

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