Intensities Learning Outcomes

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

The structure factor equation

• Is it as boring as it sounds?

• Yes and no! It’s a fundamental equation in crystallography.

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

Intensities depend on…

• scattering power of atoms ( Z)

• position of atoms (x,y,z)

Scattering

• From before: “the scattering from the plane will reflect which atoms are in the plane”.

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

Summing the waves

• The overall scattering intensity depends on

• Atom types (as above) - “electron density”

• Their position relative to one another.

Centrosymmetric structure factor

• The expression 2(hx+ky+lz) =  phase difference

• aka Geometric structure factor

Intensity?

• We don’t measure the structure factor

• We measure intensity

• Intensity of the wave is proportional to FF* (where F* is the complex conjugate of F)

Example: Polonium!

• Polonium is primitive cubic.

• Atoms at (0,0,0)

• All rest generated by symmetry/translation

Polonium

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

Iron (bcc)

Example: CsCl

• CsCl is primitive.

• Atoms at (0,0,0) (Cs) and (½,½,½) (Cl)

• All rest generated by symmetry/translation

CsCl cf “CsCs” – P vs I

Choice of origin

• Arbitrary, so we could have Cl at (0,0,0) and Cs at (½,½,½)

Example: Copper

• Copper is face centred cubic.

• Atoms at (0,0,0), (½,½,0), (½,0,½), (0,½,½)

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)

Comparison: NaCl vs KCl

• Fhkl = 4fNa + 4fCl if h,k,l all even

• Fhkl = 4fNa - 4fCl if h,k,l all odd

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?

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.

Electron density

• We measure intensity I = F.F*

• so we know amplitude of F.….but phases lost.

• Several methods to help – complex but briefly 

Helping us solve structures…

• Direct methods

• (Nobel Prize 1985 - Hauptmann and Karle)

• Statistical trial and error method. Fhkl’s are interdependent so by “guessing” a few we can extrapolate

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….