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Intensities Learning Outcomes By the end of this section you should: understand the factors that contribute to diffraction know and be able to use the.

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Presentation on theme: "Intensities Learning Outcomes By the end of this section you should: understand the factors that contribute to diffraction know and be able to use the."— Presentation transcript:

1 Intensities Learning Outcomes By the end of this section you should: understand the factors that contribute to diffraction know and be able to use the Structure Factor Equation be able to relate the structure factor equation to systematic absences be aware of the phase problem

2 The structure factor equation Is it as boring as it sounds? Yes and no! Its a fundamental equation in crystallography. Builds on concepts we have encountered already: Miller index fj Z Unit cells Positions of atoms (x,y,z) Symmetry (Wave equations)

3 What makes a diffraction pattern? Positions of peaks/spots –entirely due to size and shape of unit cell a,b,c,,, which gives d ( 2 ) Intensities of peaks –following section: why all different? Sample, instrumental factors

4 Intensities depend on… scattering power of atoms ( Z) position of atoms (x,y,z) structure factor (following sections) vibrations of atoms - temperature factor B Polarisation factor (function of sin / ) (see previous) Lorentz factor (geometry) absorption extinction preferred orientation (powders) multiplicities (i.e. 100=010=001 etc)

5 Scattering From before: the scattering from the plane will reflect which atoms are in the plane. hkl The scattering is the sum of all waves diffracted from the crystal.

6 Atomic scattering factor Again, from before: The atomic scattering factor, f j, depends on: the number of electrons in the atom (Z) the angle of scattering f varies as a function of angle, usually quoted as a function of (sin )/ f =0 = Z But we need the sum of all scattering Again we are considering interference effects.

7 Summing the waves The overall scattering intensity depends on Atom types (as above) - electron density Their position relative to one another. Or for simple (centrosymmetric) structures: See e.g. West, Basic Solid State Chemistry, for a derivation This is the sum of the (cos) waves, where: -f j is the atomic scattering factor for atom j -hkl are the Miller indices -x j, y j, z j are the atomic (fractional) coordinates

8 Centrosymmetric structure factor The expression 2 (hx+ky+lz) = phase difference aka Geometric structure factor Centrosymmetric means that there is a centre of symmetry, and for every atom at (x,y,z) there is an identical atom at (-x, -y, -z)

9 Intensity? We dont measure the structure factor We measure intensity Intensity of the wave is proportional to FF* (where F* is the complex conjugate of F) Thus we get: I f j 2 as the cos (or exp) terms cancel out. So something quite complex becomes simple, but….

10 Example: Polonium! Polonium is primitive cubic. Atoms at (0,0,0) All rest generated by symmetry/translation So F hkl = f j cos 2 (h0 + k0 + l0) = f j cos (0) = f j and I = k f j 2 (where k is a known constant) To finally get the diffraction pattern we would need to know the form of f j with (Z=84) and the unit cell parameters.

11 Polonium

12 Example: Iron ( -Fe) Iron is body centred cubic. Atoms at (0,0,0) (Fe1) and (½,½,½) (Fe2) All rest generated by symmetry/translation So F hkl = f Fe1 cos 2 (0) + f Fe2 cos 2 (½h + ½k + ½l) F hkl = f Fe + f Fe cos (h + k + l).Two cases: If h+k+l = 2n F hkl = f Fe [1 + 1] = 2f Fe I=4f Fe 2 If h+k+l = 2n+1 F hkl = f Fe [1 + (-1)] = 0 I=0 Thus, the odd reflections are systematically absent Generally true for all body centred structures

13 Iron (bcc)

14 Example: CsCl CsCl is primitive. Atoms at (0,0,0) (Cs) and (½,½,½) (Cl) All rest generated by symmetry/translation So F hkl = f Cs cos 2 (0) + f Cl cos 2 (½h + ½k + ½l) F hkl = f Cs + f Cl cos (h + k + l).Two cases: If h+k+l = 2n F hkl = f Cs + f Cl If h+k+l = 2n+1 F hkl = f Cs - f Cl So weak/strong reflections

15 CsCl cf CsCs – P vs I

16 Choice of origin Arbitrary, so we could have Cl at (0,0,0) and Cs at (½,½,½) What effect does this have on the structure factor equation? The intensities? (left as an exercise, Q1 in handout 12)

17 Example: Copper Copper is face centred cubic. Atoms at (0,0,0), (½,½,0), (½,0,½), (0,½,½) Three cases to consider h,k,l all odd h,k,l all even h,k,l mixed (2 odd, 1 even or 2 even, 1 odd) Thus, reflections present when … Generally true for all face centred structures

18 Example: NaCl NaCl is face centred cubic. Atoms at: Na1 (0,0,0), Na2 (½,½,0), Na3 (½,0,½), Na4 (0,½,½) Cl1 ((½,0,0), Cl2 (0,½,0), Cl3 (0,0,½), Cl4 (½,½,½) Show that F hkl = 4f Na + 4f Cl if h,k,l all even and F hkl = 4f Na - 4f Cl if h,k,l all odd Left as an example – but the result yields interesting consequences:

19 Comparison: NaCl vs KCl NaCl F hkl = 4f Na + 4f Cl if h,k,l all even F hkl = 4f Na - 4f Cl if h,k,l all odd KClAs mentioned before, K + and Cl - are isoelectronic So 4f K - 4f Cl ~ 0

20 Problem Most likely would index this incorrectly – as a primitive cube with a unit cell half the size. Can you see – from the structure - why?

21 The phase problem We can calculate the diffraction pattern (i.e. all F hkl ) from the structure using the structure factor equation Each F hkl depends on (hkl) (x,y,z) and f j f j depends primarily on Z, the number of electrons (or electron density) of atom j The structure factor is thus related to the electron density, so if we can measure the structure factor, we can tell where the atoms are. The structure factor is the Fourier transform of electron density (& vice versa)

22 Electron density We measure intensity I = F.F* so we know amplitude of F.….but phases lost. Several methods to help – complex but briefly

23 Helping us solve structures… Direct methods (Nobel Prize Hauptmann and Karle) Statistical trial and error method. F hkl s are interdependent so by guessing a few we can extrapolate H. Hauptmann & J. Karle b1917 b1918 Patterson Methods Uses an adapted electron density map where peaks correspond to vectors between atoms - peak height Z 1 Z 2 Heavy Atom Methods High Z atoms will dominate the electron density - easy to locate Use Patterson vectors to find other atoms.

24 Limitations of X-ray Structure determination gives average structure light atoms are difficult to detect (f Z) e.g. Li, H difficult to distinguish atoms of similar Z (e.g. Al, Si) need to grow single crystals ~ 0.5mm time for data collection and analysis (?) new instruments mean smaller crystals, shorter collection times! So in fact – data can pile up….


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