Intensities Learning Outcomes By the end of this section you should: understand the factors that...

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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 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.

Builds on concepts we have encountered already:• Miller index• fj Z• Unit cells• Positions of atoms (x,y,z)• Symmetry• (Wave equations)

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

20 30 40 50 602 /o

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)

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.

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

But we need the sum of all scatteringAgain we are considering interference effects.

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:

)(2expF hkl j

jjjj lzkyhxif

See e.g. West, Basic Solid State Chemistry, for a “derivation”

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

)lzkyhx(2cosfj

jjjj hklF

Centrosymmetric structure factor

The expression 2(hx+ky+lz) = phase difference aka Geometric structure factor

)lzkyhx(2cosfj

jjjj hklF

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)

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)

Thus we get:

as the cos (or exp) terms cancel out.

So something quite complex becomes simple, but….

Example: Polonium!

• Polonium is primitive cubic. • Atoms at (0,0,0)• All rest generated by symmetry/translation

)lzkyhx(2cosfj

jjjj hklF

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 (Z=84) and the unit cell

parameters.

Polonium

0 10 20 30 40 50 60 70 80 90 100 110 120

Polonium (Desando, R.J.;Lange, R.C. Journal of Inorganic and Nuclear Chemistry)Lambda: 1.54180 Magnif: 1.0 FWHM: 0.300Space grp: P m -3 m Direct cell: 3.3590 3.3590 3.3590 90.00 90.00 90.00

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 h+k+l = 2n Fhkl = fFe[1 + 1] = 2fFe I=4fFe2

If h+k+l = 2n+1 Fhkl = fFe[1 + (-1)] = 0 I=0

Thus, the odd reflections are systematically absent

Generally true for all body centred structures

Iron (bcc)

0 10 20 30 40 50 60 70 80 90 100 110 120 130 140

alpha-Iron (Hull, A.W. (1917) Phys Rev 10, 661)Lambda: 1.54180 Magnif: 1.0 FWHM: 0.300Space grp: I m -3 m Direct cell: 2.8660 2.8660 2.8660 90.00 90.00 90.00

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 h+k+l = 2n Fhkl = fCs + fCl

If h+k+l = 2n+1 Fhkl = fCs - fCl

So weak/strong reflections

CsCl cf “CsCs” – P vs I

0 10 20 30 40 50 60 70 80 90 100

CsCl (Davey, W.P.;Wick, F.G. (1921) Physical Review 17, 403-404)Lambda: 1.54180 Magnif: 1.0 FWHM: 0.300Space grp: P m -3 m Direct cell: 4.1200 4.1200 4.1200 90.00 90.00 90.00

0 10 20 30 40 50 60 70 80 90 100

"CsCs" - mythical bcc materialLambda: 1.54180 Magnif: 1.0 FWHM: 0.300Space grp: P m -3 m Direct cell: 4.1200 4.1200 4.1200 90.00 90.00 90.00

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)

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

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:

Comparison: NaCl vs KCl1

20 30 40 50 60 70 80 90 100

NaCl (Hull, A.W. 1919)Lambda: 1.54178 Magnif: 1.0 FWHM: 0.200Space grp: F m -3 m Direct cell: 5.6400 5.6400 5.6400 90.00 90.00 90.00

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

Fhkl = 4fNa - 4fCl if h,k,l all odd

20 30 40 50 60 70 80 90 100

KCl (Hull, A.W. 1919)Lambda: 1.54178 Magnif: 1.0 FWHM: 0.200Space grp: F m -3 m Direct cell: 6.2800 6.2800 6.2800 90.00 90.00 90.00

KCl As mentioned before, K+ and Cl- are isoelectronic

So 4fK - 4fCl ~ 0

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?

20 30 40 50 60 70 80 90 100

KCl (Hull, A.W. 1919)Lambda: 1.54178 Magnif: 1.0 FWHM: 0.200Space grp: F m -3 m Direct cell: 6.2800 6.2800 6.2800 90.00 90.00 90.00

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)

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

H. Hauptmann & J. Karle b1917 b1918

• Patterson MethodsUses an adapted electron density map where peaks correspond to vectors between atoms - peak height Z1Z2

• Heavy Atom MethodsHigh Z atoms will dominate the electron density - “easy” to locate Use Patterson vectors to find other atoms.

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….

Intensities Learning Outcomes By the end of this section you should: understand the factors that...

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