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X-ray Diffraction
X-ray diffraction is a method
of determining the structure
of a crystal from its
diffraction pattern. X-ray
diffraction techniques are
based on the elastic
scattering of x-rays from
structures that have long
range order.
Max von Laue was awarded the 1914 Nobel Prize in Physics
for his discovery of the diffraction of X-rays by crystals.
He theorized that if X-rays were waves, the wavelengths
must be extremely small (on the order of 10-10
meters)
If true, the regular structure of crystalline materials should
be “viewable” using X-rays
Solvay Conference 1913
STANDING: HASSENOHRL VERSCHAFFELT JEANS BRAGG LAUE RUBENS Mme CURIE GOLDSCHMIDT
HERZEN EINSTEIN LINDEMANN de BROGLIE POPE GRUNEISEN KNUDSEN HOSTELET LANGEVIN
SEATED: NERNST RUTHERFORD WIEN J.J. THOMSON WARBURG LORENTZ BRILLOUIN BARLOW
KAMERLINGH ONNES WOOD GOUY WEISS
STANDING: HASSENOHRL VERSCHAFFELT JEANS BRAGG LAUE RUBENS Mme CURIE GOLDSCHMIDT
HERZEN EINSTEIN LINDEMANN de BROGLIE POPE GRUNEISEN KNUDSEN HOSTELET LANGEVIN
SEATED: NERNST RUTHERFORD WIEN J.J. THOMSON WARBURG LORENTZ BRILLOUIN BARLOW
KAMERLINGH ONNES WOOD GOUY WEISS
Some notable names here: Come across these?
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Generation of X-rays
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Anode
Cathode
X-rays
Once the switch is closed a high potential difference is in the
circuit. This causes the cathode to heat up and high energy
electrons are attracted to the anode.
Discoverer, Wilhelm Conrad Roentgen
(1845-1923) in 1896. Ironically (given the
nature of his discovery), Roentgen was not
fond of being photographed. There are
relatively few images of him after the
discovery, most in the same rigid and
solemn pose.
Site of the discovery, the Physical Institute
of the University of Wurzburg, taken in
1896. The Roentgens lived in apartments
on the upper storey, with laboratories and
classrooms in the basement and first floor
X-rays
Early experimental tubes like
those used by Roentgen and
others to investigate the nature
of light.
The famous radiograph made by
Roentgen on 22 December 1895, and
sent to physicist Franz Exner in
Vienna. This is traditionally known as
"the first X-ray picture" and "the
radiograph of Mrs. Roentgen's hand."
X-rays
Generation of X-rays (Classical View)
The free electron collides with an atom in the Anode, knocking an
electron out of a lower orbital. A higher orbital electron fills the
empty position, releasing its excess energy as a photon
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Anode Atom Energy Levels
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Generation of X-rays (Quantum View)
As in the classical view the free electron collides with an atom in
the Anode. An electron is knocked out of a lower energy level. A
electron in a higher level then fills the empty position, releasing
its excess energy as a photon.
h
Substituting yields the
relationship between
wavelength and energy.
c hE 1
hcE 1
398.121 E
As energy of primary electron beam is usually measured in eV
(or KeV), we use these units for the estimate of required voltage.
X-rays may be described as waves and particles, having both
wavelength () and energy (E1).
E1 is the energy of the X-ray photon
(in KeV)
h is Planck’s constant
(4.135 x 10-15
eVs)
c is the speed of light
(3 x 1018
Å/s)
is the wavelength (in Å)
OR
Generation of X-rays
Why X-Rays?
Typical inter-atomic distances in a
solid are of the order of 10-10
m (1
Angstrom). Therefore, to probe the
structures of solids on the atomic
scale once must have a wavelength
close to this size.
eVhc
PhotonofEnergyhc
3
10103.12
10
Photons at the energy we need are X-rays
Calculating
The image created showed:
1. The lattice of the crystal
produced a series of regular
spots from concentration of
the x-ray intensity as it
passed through the crystal.
2. Demonstrated the wave
character of the x-rays
3. Proved that x-rays could be
diffracted by crystalline
materials
Von Laue’s results were
published in 1912
Von Laue’s experiment
In the early 20th
Century Von Laue conducted experiments
using x-rays. One experiment used an X-ray source
directed into a lead box containing an oriented crystal with
a photographic plate behind the box
Bragg’s Law
In 1915, William Henry Bragg and
William Lawrence Bragg were awarded
the Nobel Prize.
They discovered that
diffraction of X-rays by
solids could be treated
as reflection from
evenly spaced planes if
monochromatic x-rays
were used.
nλ = 2dSinθ
Bragg’s Law nλ = 2dSinθ
Bragg’s model assumes that the crystal is made of
parallel planes of ions, spaced an equal distance, d,
apart.
RECALL: There is more than one way to separate a lattice into
planes! Each such set of planes will produce its own reflection
Conditions for forming sharp peaks of the scattered radiation;
1. X-rays are specularly reflected by ions in any one plane
2. X-rays from successive planes interfere constructively
- The path difference between the two rays
reflected from the two successive planes is 2dSinθ
Bragg’s Law nλ = 2dSinθ
where n is an integer, the order of the corresponding reflection
is the wavelength of the X-radiation
d is the interplanar spacing in the crystalline material
is the diffraction angle
dSinn 2=
Constructive interference occurs when the path difference
between the reflected X-rays is integer multiples of the
wavelength. This leads to the Bragg Condition;
Bragg’s Law nλ = 2dSinθ
The Bragg Angle, θ, is defined as half of the total angle by which
the incident beam is deflected.
But there is an important thing to remember when measuring
the angles…
Bragg’s Law nλ = 2dSinθ
The sample may be of irregular shape. The direction of the atomic
planes has nothing to do with the way it is cut or the shape.
Bragg’s Law nλ = 2dSinθ
You can see here that
the planes of the
crystal are not aligned
with the flat of the
sample.
Von Laue formulation of X-ray Diffraction
Von Laue’s approach had the advantage that it does not require
sectioning the crystal into lattice planes and no ad hoc
assumption of specular reflection is imposed as in Bragg
reflection. Instead, one regards the crystal as composed of
identical microscopic objects (ions, atoms, etc) placed at sites R
of a Bravais lattice. Each such object can radiate the incident
radiation in all directions.
Incident Beam
A scattered ray will be
observed in a direction n’ with
wavelength λ, and wave vector
k’ = 2πn’/λ provided that the
path difference between the
rays scattered by each of the
two ions is an integral number
of wavelengths.
Sharp peaks will be observed only in directions and at
wavelengths for which the x-rays scattered from all lattice points
interfere constructively. Consider first just two scatterers,
separated by a displacement vector d.
d
Let an X-ray be incident from very far
away, along a direction n, with
wavelength λ, wave vector k = 2πn/λ
n
k
k
n’
k’
k’
θ θ’
Path difference between the rays
scattered by the two ions is:
d cos θ’ = - d ∙ n’
)'('coscos nnd dd
Von Laue formulation of X-ray Diffraction
Von Laue formulation of X-ray Diffraction
)'('coscos nnd dd
The condition for interference is that the path difference is an
integral number of wavelengths.
Path difference between the
rays scattered by the two ions
Zmm ;)'( nnd
Multiplying both sides of the above equation by 2π/λ results in a
condition on the incident and scattered wave vectors
Zmm ;2)'( kkd
d
n k
k
n’
k’
k’
θ θ’
d cos θ’ = - d ∙ n’
Taking this condition, now consider not just two scatterers but
an array of scatterers at all sites of a Bravais lattice. Let us put
the zero coordinate at one of the points of the array. Since the
lattice sites are displaced from one another by the Bravais lattice
vectors R, all scattered rays interfere constructively is when the
condition, d(k – k’) = 2m, holds simultaneously for all values of
R that are Bravais lattice vectors:
Von Laue formulation of X-ray Diffraction
2m; ) ' ( k k R m Z and all Bravais
lattice vectors R
An equivalent statement is to say that;
By comparing this equation with the definition of a reciprocal
vector, we see that constructive interference occurs if
K = k - k’ is a vector of the reciprocal lattice
Von Laue formulation of X-ray Diffraction
2m; ) ' ( k k R m Z and all Bravais
lattice vectors R
.
; 1 ) ' (
R
R k k
vectors lattice
Bravais all for e i
Bragg planes
If we notice that k and k’ have the same magnitude, we can see
that Von Laue condition can be formulated as : The tip of the
incident wave vector k must lie in the plane that is the
perpendicular bisector of a line joining the origin of k-space to
a reciprocal lattice point K. Such k-space planes are called
Bragg planes.
Exercise
You wish to study a material with a cubic crystal structure with
three orthogonal sides of length a = 0.3nm. What is the
longest possible wavelength of the X-ray source you could use
to reveal the (211) Bragg peak of the crystal?
If you were interested in the (422) peak, what would change?
Problems/Questions?
Are you comfortable with all the equations and constants used in
this lecture?
Did you follow the Von Laue Formulation?
Did you follow the Bragg Formulation?
I would urge you to know the answers to these questions before
next time.
Good resources
Solid State Physics ~ Ashcroft, Ch. 6
Introduction to Solid State Physics ~ Kittel, Ch. 2
Solid State Physics ~ Hook & Hall, Ch. 1
