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)

But WHY do we need diffraction?

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)

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.