Why diffraction? Learning Outcomes By the end of this section you should: • understand what we are looking at with diffraction and why we need diffraction in crystallography • be able to compare optical and X-ray diffraction • be able to outline the factors which are important in diffraction • understand the processes of X-ray emission and the basic outline of an X-ray tube
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Why diffraction? Learning Outcomes By the end of this section you should: understand what we are looking at with diffraction and why we need diffraction.
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Why diffraction?
Learning Outcomes
By the end of this section you should:• understand what we are looking at with diffraction
and why we need diffraction in crystallography• be able to compare optical and X-ray diffraction • be able to outline the factors which are important in
diffraction• understand the processes of X-ray emission and the
basic outline of an X-ray tube
Characterisation of Solids
What is it?• Powder• Single crystal• Glass/amorphous• Polymer• Inorganic/Organic• Composite material
Insulin crystals, Nasa.gov
Characterisation of Solids
What scale are we interested in?• Bulk/Macro – overall structure• Micro (microstructure) – grains, defects• Nano – crystal structure
SiC screw disclocation, from http://focus.aps.org/story/v20/st3
Open porous structure in lava flow
Characterisation of Solids
What part are we interested in?• Surface vs bulk - • Defects vs “perfection” ---semiconductors
Properties?• Mechanical• Magnetic/electronic/ionic • Chemical (e.g. catalytic, pharmaceutical….)
Obviously many techniques are required to fully characterise a material
Pictures from http://materials.usask.ca/photos/
Silicon single crystal
Graphite surface
“Perfect Solids”
• Best-case scenario? “Perfect” crystalline solid.• Want to find the atom-level structure• Primary techniques: DIFFRACTION
Single crystal Powder
X-ray
neutron
electron
diffraction
Revisiting Bragg
1912 - Friedrich & Knipping, under direction of LaueExtended by W. H. and W. L. Bragg (father and son)Based on existing optical techniques
Max von Laue
1879 -1960
Nobel Prize 1914 “for his discovery of the diffraction
of X-rays by crystals”
W. H. Bragg
1862 -1942
W. L. Bragg
1890 -1971 Nobel Prize 1915 “for their services in the analysis of crystal structure by means of
X-rays"
Optical grating – a 1d analogue
Path difference XY between diffracted beams 1 and 2:
sin = XY/a
XY = a sin
XY
1
2
a
Coherent incident light Diffracted light
Possible Combination of waves
• Destructive: Waves combine and are exactly “out of phase” with each other – cancelling. = /2
• Constructive: Waves combine and are exactly “in phase” with each other – adding together to give maximum possible. =
• Partial: Somewhere between the two.
Result for OPTICAL grating
Path difference XY between diffracted beams 1 and 2:
sin = XY/a
XY = a sin
For constructive interference, we want XY to be a whole number of wavelengths
So for this set-up, a sin = for first order diffraction
XY
1
2
a
Coherent incident light Diffracted light
Result for OPTICAL grating
What we see:
General Diffraction
After the diffraction
D
L
tan = D/L
but if D<<L then we can write
sin ~ D/L
But a sin =
So….
a ~ L/D
Summary of diffraction so far…
1. Diffraction side: a is related to Observation side: D is related to L
2. a sin = so sin = /a
This means that a must be > or else sin is > 1
If a >> then sin 0 and we see nothing
3. D is related to 1/a, so the closer the slits, the further apart the diffraction lines. You can see this nicely in this applet: Diffraction Applet
Optical X-ray
• With optical diffraction we can observe effects from a couple of slits
• With X-rays, the interaction with matter is very weak – most pass straight through
• Therefore we need many (100-1000s) of waves
Laue Equations – 3d
• By analogy with the above:
For constructive interference:
(AB – CD) = a (cos na – cos 0
a) = nx and for y & z
b (cos nb – cos 0
b) = ny
c (cos nc – cos 0
c) = nz
0a
na
BA
CD
Laue equations – in reality
• These work well and describe the interactions• Basic idea is still the constructive interference which
occurs at an integer no. of wavelengths• However, not routinely used
• Bragg’s law represents a simpler construct for everyday use!
XY
Z
d
Incident radiation “Reflected” radiation
Transmitted radiation
1
22d sin = n
Make sure you (PX3012)can derive this (Dr. Gibson’s lectures)
• Electron microscope… not quite there yet, limited in application.
HREM image of gold
Delft University of Technology (2007)
Tilt your head…
• If we draw the Bragg construction in the same way as the optical grating, we can clearly see that the diffracted angle is 2. The plane of “reflection” bisects this angle.
• Thus we measure 2 in the experiment – next section…
“Reflecting plane”
X-rays and solids
X-rays - electromagnetic waves
So X-ray photon has E = hX-ray wavelengths vary from .01 - 10Å; those used in
crystallography have frequencies 2 - 6 x 1018 Hz
Q. To what wavelength range does this frequency range correspond?
c =
c
max = 1.5 Å
min = 0.5 Å
Energy and Wavelength
Energy of photons usually measured in keV – why?
E
hc
)eV(E
12400 (Å)
Looking for wavelengths of the order of Å
therefore need keV
Production of X-rays
Anode Cathode
Beryllium windowFilament
e.g. tungsten
Pyrex glass envelope (under vacuum)
X-ray emission
Two processes lead to two forms of X-ray emission:
Electrons stopped by target; kinetic energy converted to X-rayscontinuous spectrum of “white” radiation, with cut-off at short (according to h=½mv2)Wavelength not characteristic of target
Incident electrons displace inner shell electrons, intershell electron transitions from outer shell to inner shell vacancy.
“line” spectraWavelength characteristic of target
X-ray spectrum
Mixture of continuous and line
Characteristic wavelengths
• Thus, each element (target) has a characteristic wavelength.
For copper, the are:
• CuK1 = 1.540 Å
• CuK2 = 1.544 Å
• CuK = 1.39 Å
Typical emission spectrum
Energy transitions
Many intershell transitions can occur - the common transitions encountered are:
2p (L) - 1s (K), known as the K line
3p (M) - 1s (K), known as the K line
(in fact K is a close doublet, associated with the two spin states of 2p electrons)
Example
Copper K X-rays have a wavelength of 1.54 Å and are produced when an electron falls from the L shell to a vacant site in the K shell of a copper atom. Calculate the difference in the energy levels between the K and L shells of copper.
E = h
= c/ = (3 x108) / (1.54 x 10-10)
= 1.95 x 1018 Hz
E = h = 6.626 x 10-34 x 1.95 x 1018
= 1.29 x 10-15 J
~ 8 keV
..and vice versa - each transition has its own wavelength.