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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.

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Presentation on theme: "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."— Presentation transcript:

1 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

2 Characterisation of Solids What is it? Powder Single crystal Glass/amorphous Polymer Inorganic/Organic Composite material Insulin crystals, Nasa.govNasa.gov

3 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

4 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

5 Perfect Solids Best-case scenario? Perfect crystalline solid. Want to find the atom-level structure Primary techniques: DIFFRACTION Single crystalPowder X-ray neutron electron diffraction

6 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"

7 Optical grating – a 1d analogue Path difference XY between diffracted beams 1 and 2: sin = XY/a XY = a sin

8 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.

9 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

10 Result for OPTICAL grating What we see:

11 General Diffraction After the diffraction D L tan = D/L but if D< { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/3/778867/slides/slide_11.jpg", "name": "General Diffraction After the diffraction D L tan = D/L but if D<

12 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 AppletDiffraction Applet

13 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

14 Laue Equations – 3d By analogy with the above: For constructive interference: (AB – CD) =a (cos n a – cos 0 a ) = n x and for y & z b (cos n b – cos 0 b ) = n y c (cos n c – cos 0 c ) = n z 0 a n a B A C D

15 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)

16 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)

17 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

18 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 10 18 Hz Q. To what wavelength range does this frequency range correspond? c = max = 1.5 Å min = 0.5 Å

19 Energy and Wavelength Energy of photons usually measured in keV – why? (Å) Looking for wavelengths of the order of Å therefore need keV

20 Production of X-rays

21 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 =½mv 2 ) 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

22 X-ray spectrum Mixture of continuous and line

23 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

24 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)

25 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 x10 8 ) / (1.54 x 10 -10) = 1.95 x 10 18 Hz E = h = 6.626 x 10 -34 x 1.95 x 10 18 = 1.29 x 10 -15 J ~ 8 keV..and vice versa - each transition has its own wavelength.


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