Document 6458618

X-Ray Diffraction
HOW IT WORKS WHAT IT CAN AND WHAT IT CANNOT TELL US
Hanno zur Loye
X-rays are electromagnetic radiation of wavelength about
1 Å (10-10 m), which is about the same size as an atom.
The discovery of X-rays in 1895 enabled scientists to probe crystalline structure at the atomic level. X-ray diffraction has been in use in two main areas, for the fingerprint characterization of crystalline materials and the determination of their structure. Each crystalline solid has its unique characteristic X-ray powder pattern
which may be used as a "fingerprint" for its identification. Once the material has been identified, X-ray crystallography may be used to determine its structure, i.e. how the atoms pack together in the crystalline state and what the interatomic distance and angle are etc. X-ray diffraction is one of the most important characterization tools used in solid state chemistry and materials science.
We can determine the size and the shape of the unit cell for any compound most easily using X-ray diffraction. X-ray Diffraction
­ Structural Analysis
¬ X-ray diffraction provides most definitive
structural information
¬ Interatomic distances and bond angles
­ X-rays
¬ To provide information about structures we
need to probe atomic distances - this requires a
probe wavelength of 1 x 10-10 m ~Angstroms
Production of X-rays
­ X-rays are produced by bombarding a metal
target (Cu, Mo usually) with a beam of
electrons emitted from a hot filament (often
tungsten). The incident beam will ionize
electrons from the K-shell (1s) of the target
atom and X-rays are emitted as the resultant
vacancies are filled by electrons dropping
down from the L (2P) or M (3p) levels.
This gives rise to Ka and Kb lines.
X-rays e -
Cu, Mo
target
Cu Kα = 1.5418 Å
Mo Kα = 0.7107 Å
20 - 50 kV
3p
M
2p
L
1s
K
electromagnetic
radiation
X-ray Generation
­ Broad background is
called Bremsstrahlung.
Electrons are slowed
down and loose energy
in the form of X-rays
X-ray Production
­ As the atomic number Z of the target element
increases, the energy of the characteristic emission
increases and the wavelength decreases.
­ Moseley’s Law (c/l)1/2 ∝ Z
¬ Cu
¬ Mo
Ka = 1.54178 Å
Ka = 0.71069 Å
­ We can select a monochromatic beam of one
wavelength by:
¬ Crystal monochromator
Bragg equation
¬ Filter - use element (Z-1) or (Z-2), i.e. Ni for Copper
and Zr for Molybdenum. X-ray Generation
Brookhaven National Synchrotron Light Source
Single Crystal Diffraction
A single crystal at random orientations and its corresponding diffraction
pattern. Just as the crystal is rotated by a random angle, the diffraction pattern
calculated for this crystal is rotated by the same angle
A 'powder' composed from 4 single crystals in random orientation (left)
and the corresponding diffraction pattern (middle). The individual
diffraction patterns plotted in the same color as the corresponding crystal
start to add up to rings of reflections. With just four reflection its difficult
though to recognize the rings. The right image shows a diffraction pattern
of 40 single crystal grains (black). The colored spots are the peaks from the
4 grain 'powder' shown in the middle image. As we have more grains, the diffraction pattern looks more and more continuous and we get the expected powder
pattern shown on the left.
Diffraction from several randomly
oriented single crystals (powder)
Diffraction from one
single crystal
Sample of hundreds of randomly
oriented single crystals (powder)
and film used to collect the data.
Our XRD instruments
Crystal Systems
Axis
1 Cubic
A=B=C
2 Tetragonal
A=B≠C
3 Orthorhombic A ≠ B ≠ C
4a Hexagonal
A=B≠C
4b Rhombohedral A = B = C
5 Monoclinic
A≠B≠C
6 Triclinic
A≠B≠C
Axis Angles
α = β = γ = 90
α = β = γ = 90
α = β = γ = 90
α = β = 90, γ = 120
α = β = γ ≠ 90 < 120
α = γ = 90, β > 90
α ≠ β ≠ γ ≠ 90
The higher the symmetry, the easier it is to index the pattern
and the fewer lines there are in the pattern.
Miller Indices
a
The origin is 0, 0, 0
c
b
c/3
O
If we drew a third
plane, it would pass
through the origin.
a/2
(1 0 0) (2 0 0) (3 0 0)
The plane cuts the x-axis at a/2, the b-axis at b, and the c-axis at c/3. The the
reciprocals of the fractions, (2 1 3), which
are the miller indices.
A plane that is parallel to an axis will intersect that axis at infinity. One over infinity = 0. I.e. the
(100) plane. Parallel to b and c and intersects a at 1.
The Bragg Equation
Reflection of X-rays from two planes of atoms in a solid. x = dsinθ
The path difference between two waves: 2 x wavelength = 2dsin(theta)
Bragg Equation: nλ = 2dsinθ d-spacing in different crystal systems
­ Cubic
­ Tetragonal
­ Orthorhombic
­ Hexagonal
­ Monoclinic
­ Triclinic - 1 h 2 + k 2 + l2
=
2
d
a2
1 h 2 + k 2 l2
=
+ 2
2
2
d
a
c
1 h 2 k 2 l2
= 2+ 2+ 2
2
d
a
b
c
1 4 ⎛ h 2 + hk + k 2 ⎞ l 2
= ⎜
⎟+ 2
2
2
d
3⎝
a
⎠ c
1
1 ⎛ h 2 k 2 sin 2 β l 2 2hlcos β ⎞
= 2 ⎜ 2+
+ 2−
⎟
2
2
d
sin β ⎝ a
b
c
ac ⎠
Diffraction pattern
•  Intensity (I) is the
total area under a
peak I
2θ (deg)
Indexing Patterns
­ Indexing is the process of determining the
unit cell dimensions from the peak positions
¬ Manual indexing (time consuming...but still
useful)
¬ Pattern matching/auto indexing (JADE or other
computer based indexing software)
Cubic pattern
Relationship between diffraction peaks,
miller indices and lattice spacings
Simple cubic material a = 5.0 Å
hkl
100
110
111
d(Å)
5.00
3.54
2.89
h2
2Θ
17.72
25.15
30.94
k2
+l2
1 h 2 + k 2 + l2
=
2
d
a2
Bragg Equation: nλ = 2dsinθ
Use 1 =
+
, and nλ = 2dsinΘ
d2 a2 b2 c2
How many lattice planes are possible?
How many d-spacings?
The number is large but finite.
nλ = 2dsinθ so if theta = 180, then d = λ/2. For Cu radiation
that means that we can only see d-spacings down to 0.77 Å
for Mo radiation, down to about 0.35 Å
Tetragonal pattern
1 h 2 + k 2 l2
=
+ 2
2
2
d
a
c
sin θ =
2
λ2
4
[
h 2 +k 2
a2
+
l2
c2
]
Orthorhombic a ≠ b ≠ c, all
angles are 90°
Multiplicity is
further
decreased as
the symmetry
decreases.
1 h 2 k 2 l2
= 2+ 2+ 2
2
d
a
b
c
sin θ =
2
λ2
4
[
h2
a2
+
k2
b2
+
l2
c2
]
Systematic Absences
Cesium Metal
c
2πn
c
c'
c
c'
Body
Centered
Cubic
Structure
a = 2πn
a = 2πn
c'
c'
c
c
c
In order to see diffraction from the (100) plane, the phase
difference must be a multiple of 2p. However, the c and c’
planes are out of phase. Therefore - cancellation of the diffracted peak. The (200) plane, however, does not have this problem.
What Information Do We Get or Can We
Get From Powder X-ray Diffraction
­ Lattice parameters
­ Phase identity
­ Phase purity
­ Crystallinity
­ Crystal structure
­ Percent phase composition
What Information Do We NOT Get
From Powder X-ray Diffraction
­ Elemental analysis - ¬ How much lithium is in this sample?
¬ Is there iron in this sample
¬ What elements are in this sample
­ Tell me what this sample is ????
¬ Unless you know something about this sample,
powder XRD won’t have answers !!!
Powder Preparation
­ It needs to be a powder
­ It needs to be a pure powder
­ Its nice to have about 1/2 g of sample, but
one can work with less
­ The powder needs to be packed tightly in
the sample holder. Lose powders will give
poor intensities.
Data Collection
­ The scattering intensity drops as 1/2(1+cos22θ)
­ This means that you don’t get much intensity
past 70 ° 2θ. A good range is 10-70 ° 2θ.
­ How long should you collect (time per step)?
­ Depends on what you want to do!
¬ Routine analysis may only take 30-60 min.
¬ Data for Rietveld analysis may take 12-18 hours to
collect
Rietveld Refinement
• employs a least squares matching algorithm
• refinement based on sample parameters and instrumental
parameters
Data Analysis
­ If you are trying to confirm that you have
made a known material, do a search/match
using the JCPDS data base. They have
about 100,000 patterns on file.
­ For a new material, you need to index the
pattern. Unit cell, lattice parameters and
symmetry. Data Bases
Preferred Orientation
The top image shows 200 random crystallites. The
bottom picture shows 200 oriented crystallites. Despite the identical number of reflections,several powder lines are completely missing and the intensity of other lines is very misleading. Preferred orientation can substantially alter the appearance of the powder pattern. It is a
serious problem in experimental powder diffraction. Structure Refinement: The Rietveld Method