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
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Learning Outcomes
By the end of this section you should:
What is it?
Insulin crystals, Nasa.gov
What scale are we interested in?
SiC screw disclocation, from http://focus.aps.org/story/v20/st3
Open porous structure in lava flow
What part are we interested in?
Properties?
Obviously many techniques are required to fully characterise a material
Silicon single crystal
Graphite surface
Pictures from http://materials.usask.ca/photos/
Single crystal
Powder
diffraction
X-ray
electron
neutron
1912 - Friedrich & Knipping, under direction of Laue
Extended 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"
Path difference XY between diffracted beams 1 and 2:
sin = XY/a
XY = a sin
Path difference XY between diffracted beams 1 and 2:
sin = XY/a
For constructive interference, we want XY to be a whole number of wavelengths
So for this set-up, a sin = for first order diffraction
What we see:
D
L
After the diffraction
tan = D/L
but if D<<L then we can write
sin ~ D/L
But a sin =
So….
a ~ L/D
Observation side: D is related to L
na
0a
D
C
B
A
For constructive interference:
(AB – CD) =a (cos na – cos 0a) = nx and for y & z
b (cos nb – cos 0b) = ny
c (cos nc – cos 0c) = nz
2d sin = n
Make sure you (PX3012)
can derive this (Dr. Gibson’s lectures)
Why not just use a big microscope?
Swift: - Instruments - The X-Ray Telescope
HREM image of gold
Delft University of Technology (2007)
“Reflecting plane”
c =
X-rays - electromagnetic waves
So X-ray photon has E = h
X-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?
max = 1.5 Å
min = 0.5 Å
Energy of photons usually measured in keV – why?
(Å)
Looking for wavelengths of the order of Å
therefore need keV
Two processes lead to two forms of X-ray emission:
Mixture of continuous
and line
For copper, the are:
Typical emission spectrum
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)
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 10-34x 1.95 x 1018
= 1.29 x 10-15 J
~ 8 keV
..and vice versa - each transition has its own wavelength.