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Why diffraction?

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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

What is it?

- Powder
- Single crystal
- Glass/amorphous
- Polymer
- Inorganic/Organic
- Composite material

Insulin crystals, Nasa.gov

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

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/

Single crystal

Powder

diffraction

X-ray

electron

neutron

- Best-case scenario? “Perfect” crystalline solid.
- Want to find the atom-level structure
- Primary techniques: DIFFRACTION

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

- 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.

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

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

- Diffraction side: a is related to
Observation side: D is related to L

- 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

- 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

- 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

na

0a

D

C

B

A

- 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

- 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!

2d sin = n

Make sure you (PX3012)

can derive this (Dr. Gibson’s lectures)

Why not just use a big microscope?

- “Can’t” 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)

- 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”

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:

- 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

Mixture of continuous

and line

- 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

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.