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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
0
50
100
20 30 40 50 602 /o
Re
lati
ve
inte
ns
ity
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”.
hkl
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:
I fj2
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 1
1
0 0
2 1 1
2
0 2
2
0 1
3
2 2
2
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 0
1
0 1
1
1 1
1
0 0
2
0 1
2 1 1
2
0 2
2
0 0
31 2
2
0 1
3
1 1
3
2 2
2
0 2
3
1 2
3
0 0
4
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 0
1
0 1
1
1 1
1 0 0
2
0 1
2
1 1
2
0 2
2
0 0
31 2
2
0 1
3
1 1
3
2 2
2
0 2
3 1 2
3
0 0
4
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
1 1
0 0
2
0 2
2
1 1
3
2 2
2
0 0
4
1 3
3
0 2
4
2 2
4
1 1
53
3 3
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
1 1
1
0 0
2
0 2
2
1 1
3
2 2
2
0 0
4
1 3
3
0 2
4
2 2
4
1 1
53
3 3
0 4
4
1 3
5
0 0
62 4
4
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?
1 1
1
0 0
2
0 2
2
1 1
3
2 2
2
0 0
4
1 3
3
0 2
4
2 2
4
1 1
53
3 3
0 4
4
1 3
5
0 0
62 4
4
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….