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Why diffraction?


Title: Characterisation of Solids Author: University of Aberdeen Last modified by: Jan Skakle Created Date: 8/6/2007 3:01:31 PM Document presentation format – PowerPoint PPT presentation

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Title: Why diffraction?

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
  • 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/
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

Silicon single crystal
Graphite surface
Pictures from http//materials.usask.ca/photos/
Perfect Solids
  • Best-case scenario? Perfect crystalline solid.
  • Want to find the atom-level structure
  • Primary techniques DIFFRACTION

Revisiting Bragg
  • 1912 - Friedrich Knipping, under direction of
  • Extended by W. H. and W. L. Bragg (father and
  • Based on existing optical techniques

Max von Laue 1879 -1960 Nobel Prize 1914 for
his discovery of the diffraction of X-rays by
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
  • sin? XY/a
  • ? XY a sin ?

Possible Combination of waves
  • Destructive Waves combine and are exactly out
    of phase with each other cancelling. ??
  • 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
  • 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
Result for OPTICAL grating
  • What we see

General Diffraction
  • After the diffraction

tan ? D/L but if DltltL then we can write sin
? D/L But a sin ? ? So. a ?L/D
Summary of diffraction so far
  • Diffraction side a is related to ?
  • Observation side D is related to L
  • a sin ? ? so sin ? ?/a
  • This means that a must be gt ? or else sin ? is gt
  • If a gtgt ? then sin ? ? 0 and we see nothing
  • 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

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 ?0a) nx ? and for y z b (cos
?nb cos ?0b) ny ? c (cos ?nc cos ?0c)
nz ?
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
  • Braggs law represents a simpler construct for
    everyday use!

2d sin ? n? Make sure you (PX3012) can derive
this (Dr. Gibsons lectures)
But WHY do we need diffraction?
  • Why not just use a big microscope?
  • Cant focus X-rays (yet?!!)
  • Swift - Instruments - The X-Ray Telescope
  • Electron microscope not quite there yet, limited
    in application.

HREM image of gold Delft University of Technology
Tilt your head
Reflecting plane
  • 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

X-rays and solids
  • 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

Q. To what wavelength range does this frequency
range correspond?
?max 1.5 Å ?min 0.5 Å
Energy and Wavelength
Energy of photons usually measured in keV why?
Looking for wavelengths of the order of Å ?
therefore need keV
Production of X-rays
X-ray emission
  • Two processes lead to two forms of X-ray
  • Electrons stopped by target kinetic energy
    converted to X-rays
  • continuous 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 spectra
  • Wavelength characteristic of target

X-ray spectrum
  • Mixture of continuous

and line
Characteristic wavelengths
  • Thus, each element (target) has a characteristic
  • For copper, the ? are
  • CuK?1 1.540 Å
  • CuK?2 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)
  • 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.
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