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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

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

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"

Optical grating a 1d analogue

- Path difference XY between diffracted beams 1 and

2 - sin? XY/a
- ? XY a sin ?

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

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

1 - 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

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 ?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

(2007)

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

section

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

Hz

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?

(Å)

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

emission

- 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

wavelength. - 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)

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.