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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

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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 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
Characterisation of Solids

What is it?

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

Insulin crystals, Nasa.gov

characterisation of solids1
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 solids2
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

Single crystal

Powder

diffraction

X-ray

electron

neutron

“Perfect Solids”
  • Best-case scenario? “Perfect” crystalline solid.
  • Want to find the atom-level structure
  • Primary techniques: DIFFRACTION
revisiting bragg
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
Optical grating – a 1d analogue

Path difference XY between diffracted beams 1 and 2:

sin = XY/a

 XY = a sin 

possible combination of waves
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
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

general diffraction

D

L

General Diffraction

After the diffraction

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
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 >  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
optical x ray
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

na

0a

D

C

B

A

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
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!

2d sin  = n

Make sure you (PX3012)

can derive this (Dr. Gibson’s lectures)

but why do we need diffraction
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
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

c = 

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 and Wavelength

Energy of photons usually measured in keV – why?

(Å)

Looking for wavelengths of the order of Å

 therefore need keV

x ray emission
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
X-ray spectrum

Mixture of continuous

and line

characteristic wavelengths
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
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
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 10-34x 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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