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Intensities

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Title: Intensities


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
? structure factor (following sections)
  • scattering power of atoms (? Z)
  • position of atoms (x,y,z)
  • vibrations of atoms - temperature factor B
  • Polarisation factor (function of sin? /?) (see
    previous)
  • Lorentz factor (geometry)
  • absorption
  • extinction
  • preferred orientation (powders)
  • multiplicities (i.e. 100010001 etc)

5
Scattering
  • From before the scattering from the plane will
    reflect which atoms are in the plane.

The scattering is the sum of all waves diffracted
from the crystal.
6
Atomic scattering factor
  • Again, from before
  • The atomic scattering factor, fj, 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

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
This is the sum of the (cos) waves, where - fj
is the atomic scattering factor for atom j - hkl
are the Miller indices - xj, yj, zj are the
atomic (fractional) coordinates
See e.g. West, Basic Solid State Chemistry, for a
derivation
8
Centrosymmetric structure factor
  • The expression 2?(hxkylz) ? 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 ? fj2 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 Fhkl fj cos 2? (h0 k0 l0) fj cos (0)
fj and I k fj2 (where k is a known
constant) To finally get the diffraction pattern
we would need to know the form of fj with ?
(Z84) 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 Fhkl fFe1 cos 2? (0) fFe2 cos 2? (½h ½k
½l) Fhkl fFe fFe cos ? (h k l). Two
cases
If hkl 2n Fhkl fFe1 1
2fFe I4fFe2 If hkl 2n1 Fhkl
fFe1 (-1) 0 I0
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 Fhkl fCs cos 2? (0) fCl cos 2? (½h ½k
½l) Fhkl fCs fCl cos ? (h k l). Two
cases
If hkl 2n Fhkl fCs fCl If hkl
2n1 Fhkl fCs - fCl
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 Fhkl 4fNa 4fCl if h,k,l all
even and Fhkl 4fNa - 4fCl if h,k,l all
odd
Left as an example but the result yields
interesting consequences
19
Comparison NaCl vs KCl
  • Fhkl 4fNa 4fCl if h,k,l all even
  • Fhkl 4fNa - 4fCl if h,k,l all odd

NaCl
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 Fhkl) from the structure using the structure
    factor equation
  • Each Fhkl depends on (hkl) (x,y,z) and fj
  • fj 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 1985 - Hauptmann and Karle)
  • Statistical trial and error method. Fhkls 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
    ? Z1Z2
  • 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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