Introduction to Solid State Physics...Introduction to Solid State Physics Prof. Igor Shvets...
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Transcript of Introduction to Solid State Physics...Introduction to Solid State Physics Prof. Igor Shvets...
Equivalence of the Bragg and von Laue formulations
In the next few slides we will prove that both the von Laue
and Bragg formulations of X-ray diffraction are indeed
equivalent.
Von Laue states that d(k – k’) = 2pm as the condition for
constructive interference for reflection by points.
Bragg formulation states nλ = 2dSinθ as the condition for
constructive interference for specular reflection by planes.
To show von Laue is equivalent to the Bragg formulation we
will demonstrate how one could construct the family of Bragg
planes.
Suppose the incident and scattered wave vectors, k and k’, satisfy
the Laue condition such that K = k’ – k is a reciprocal lattice vector.
k
k’
(-k) K = k’ - k
θ θ
Consider the lattice plane perpendicular to K. Both the incident
and scattered wave vectors make the same angle θ with this
plane.
Due to elastic scattering the magnitudes of k and k’ are equal.
So the direction of K = k’ – k bisects the angle between k and k’.
φ φ
So the scattering can be seen as
Bragg reflection from the family of
lattice planes perpendicular to K
Since the reciprocal lattice is also a
Bravais lattice, K is an integral
multiple of Ko, the shortest
reciprocal lattice vector in this
direction. K = nKo
From an earlier theorem we know
that
dnhenceanddo pp 22 KK
Equivalence of the Bragg and von Laue formulations
k
k’
(-k)
K
θ θ
dnp2K
From the previous slide we know that the magnitude of K is given by;
Now consider the situation
geometrically.
By drawing a new line,
parallel to the line indicating
the plane, we can see that
the angles here are also θ. θ
θ kSinθ
kSinθ
And so the magnitude of K
can also be given by;
kSin2K
Equating the two we get;
p kSindn
Which, using the definition of the wave vector, k, gives the Bragg
condition dSinn 2
Equivalence of the Bragg and von Laue formulations
Getting an accurate picture of the structure of a crystalline
material requires X-radiation that is as close to monochromatic
as possible. Another requirement to observe clear reflected
peaks is that the beam must be collinear.
In an X-ray tube, the interactions are between the electrons and
the target. The wavelength of the X-rays depends on the
structure of the electron orbitals and hence the material of the
anode.
Generating X-rays for Diffraction
Elem. At.
#
K
(Å)
Char
Min
(keV)
Opt
kV Advantages (Disadvantages)
Cr 24 2.291 5.99 40
High resolution for large d-
spacings, particularly organics
(High attenuation in air)
Fe 26 1.937 7.11 40
Most useful for Fe-rich materials
where Fe fluorescence is a
problem (Strongly fluoresces Cr in
specimens)
Cu 29 1.542 8.98 45
Best overall for most inorganic
materials (Fluoresces Fe and Co K
and these elements in specimens can
be problematic)
Mo 42 0.710 20.00 80
Short wavelength good for small unit
cells, particularly metal alloys (Poor
resolution of large d-spacings;
optimal kV exceeds capabilities of
most HV power supplies.)
Common Anode Materials
Following the Bragg law, each component wavelength of a
polychromatic beam of radiation directed at a single crystal of
known orientation and d-spacing will be diffracted at a discrete
angle.
Monochromators make use of this fact to selectively remove
radiation outside of a tunable energy range, and pass only the
radiation of interest.
Monochromatic X-rays
Generating X-rays for Diffraction
An incident wave vector k will lead
to a diffraction peak if and only if
the tip of the wave vector lies on a
k-space Bragg plane (Laue
condition).
Since the set of all Bragg planes is a discrete family of planes, it
cannot begin to fill up three-dimensional k-space, and in general
the tip of k will not lie on a Bragg plane. Thus for a fixed X-ray
wavelength and fixed incident direction relative to the crystal axes
there will be in general no diffraction peaks at all.
In order to search experimentally for Bragg peaks, one must vary
either the magnitude of k (by changing the wavelength) or the
direction of k (usually done by changing the orientation of the
crystal).
Experimental geometries suggested by the Laue condition
k’
The Ewald Construction – (Reciprocal space)
A sphere with radius k is drawn
about the tip of the vector k in
reciprocal space.
Consider the incident wave vector k in reciprocal space.
k
O
Diffraction peaks will be observed
only if the sphere cuts through a
point of the reciprocal lattice.
So diffraction peaks corresponding
to reciprocal vectors k’ - k = K will
be observed only if K gives a
reciprocal lattice point on the
surface of the sphere.
K
In general this will not result in a
peak, but one can observe peaks
either by sample alignment or by
powder method (see below).
Experimental geometries suggested by the Laue condition
If a non-monochromatic X-ray
beam is used, this will result in a
range of wavelengths from λ0 to λ
1
k0
Draw two spheres based on the
two extreme wave vectors
k0=2πn/λ
0 and k
1=2πn/λ
1.
k1 Bragg peaks will be observed
corresponding to any reciprocal
lattice vectors lying within the
region between the two spheres.
What diffraction peaks will be
observed?
However, this is not a practical
proposition for material analysis.
Experimental geometries suggested by the Laue condition
Experimental use of the Ewald Sphere
The Rotation Method
This method uses monochromatic X-rays but allows the angle of
incidence to vary.
θ - 2θ Scan; Both the sample
and detector are rotated.
2θ Scan; The sample is fixed
and the detector is rotated.
2θ 2θ θ
θ
2θ θ
Bede-D1 High Resolution x-ray diffractometer
Equipment
Inside view of Bede-D1 diffractometer
Source
optics
Sample stage
Detector
stage
Equipment
Beam conditioner
Specimen
Incident beam
rotation
Tilt
Set analyzer
Detector
axis - 2
CCC Detector
Schematic of X-ray diffractometer
The D1 System aligned for a triple axis scan, with the CCCs in
high resolution mode. The detector arm shown is a Triple Axis
Stage fitted with a single channel Si (111) Analyzer Crystal.
Beam Divergence: 20”
Schematic of Bede-D1 diffractometer
Schematic triple axis maps arising from the structure shown. Both
structures could give rise to similar double axis rocking curves.
DAD = double axis diffraction, TAD = triple axis diffraction
X-ray Reciprocal Space Mapping (Contour map)
43//// OFeMgO kk
Film maintains its fully
strained state even after
160 min annealing in air.
Films did not relax!
The volume of the
unit cell changes.
Reciprocal space map for magnetite films annealed for 160 min
X-ray Reciprocal Space Mapping (Contour map)
MgO
Fe3O
4
MgAl2O
4
Fe3O
4
Splitting of diffraction spots
confirms the relaxation in
Fe3O
4/MgAl
2O
4
TEM studies - Selected Area Diffraction patterns
RHEED is a technique used to characterize the surface of
crystalline materials. RHEED systems gather information only
from the surface layer of the sample.
Reflection High-Energy Electron Diffraction (RHEED) –
Recap
A RHEED system requires an electron source (gun), a
photoluminescent detector screen and a sample with a clean
surface all shown here.
The process is as follows:
The electron gun generates a beam of electrons which strike the
sample at a very small angle relative to the sample surface.
Incident electrons diffract from atoms at the surface of the sample.
A small fraction of the diffracted electrons interfere constructively
at specific angles and form regular patterns on the detector.
The electrons interfere according to the position of atoms on the
sample surface.
The diffraction pattern at the detector is therefore a function of the
sample surface.
Reflection high-energy electron diffraction (RHEED)
RHEED pattern after the
growth of 10 monolayer
of Magnetite on the MgO
substrate.
RHEED pattern of a MgO
single crystalline substrate
along <100> azimuth after
cleaning.
Reflection high-energy electron diffraction (RHEED)
Reflection high-energy electron diffraction (RHEED)
RHEED pattern of a MgO
single crystalline substrate
along <110> azimuth after
cleaning.
RHEED pattern after the
growth of 10 monolayer
of Magnetite on the MgO
substrate.
Exercise
A compound material composed of two fractions, both SC
lattice, with lattice constants of 0.399 and 0.401nm
respectively is characterised by an X-ray diffractometer with a
source operating at = 0.2020nm. Find the difference
between the angles of the (221) Bragg peaks produced by
each fraction.
Problems/Questions?
Did you follow the process showing the equivalence of von Laue
and Bragg diffraction?
What is Ewald construction?
Why is it important to vary angle and/or wavelength of X-rays when
investigating diffraction peaks?
I would urge you to know the answers to these questions before
next time.
Good resources
Solid State Physics ~ Ashcroft, Ch. 6
Solid State Physics ~ Hook & Hall, Ch. 1
Introduction to Solid State Physics ~ Kittel, Ch. 2
The Physics and Chemistry of Solids ~ Elliott, Ch. 2
