Kennedy Miller
Advanced Certificate in Powder Diffraction on the Web - Assignment 2
1. The fundamental Structure factor equation is given by F(S) = nSfn e 2p i (hx+ky+lz), which translates in to the equation,
F(S) = Snfn cos{2p (hx+ky+lz)}+ iSnfn sin{2p (hx+ky+lz)}
(2,0,0) reflection
Using miller indices to find d in the (2,0,0) plane, where a = 5.638, h=2, k=0, l=0
dhkl = a / square root (h2 + k2 + l2)
\ d200 = 5.638/ square root (22 + 02 + 02)
\ d200 = 2.819Å
From Bragg's law,
l = 2d sinq
so, sinq/l = 1/2d
sinq/l = 1/(2x2.819)
sinq/l = 0.177Å -1
The atomic scattering factors, fNa+ and fCl- were calculated from the International Crystallographic Tables, via a section from the course notes. This produced the following table.
| sinq/l | 0.15 | 0.177 | 0.20 |
| fNa+ | 9.035 | 8.682 | 8.39 |
| fCl- | 14.12 | 13.069 | 12.20 |
nSfn cos{2p (hx+ky+lz)}+ iSnfn sin{2p (hx+ky+lz)} = 8.682 cos{2p (1x0 + 1x0 + lx0)} + 8.682 cos{2p (1x0 + 1x1/2 + lx1/2)} + 8.682 cos{2p (1x1/2 + 1x0 + lx1/2)} + 8.682 cos{2p (1x1/2 + 1x1/2 + lx0)} + 13.069 cos{2p (1x1/2 + 1x0 + lx0)} + 13.069 cos{2p (1x0 + 1x1/2 + lx0)} + 13.069 cos{2p (1x0 + 1x0 + lx1/2)} + 13.069 cos{2p (1x1/2 + 1x1/2 + lx1/2)} + 8.682 sin{2p (1x0 + 1x0 + lx0)} + 8.682 sin{2p (1x0 + 1x1/2 + lx1/2)} + 8.682 sin{2p (1x1/2 + 1x0 + lx1/2)} + 8.682 sin{2p (1x1/2 + 1x1/2 + lx0)} + 13.069 sin{2p (1x1/2 + 1x0 + lx0)} + 13.069 sin{2p (1x0 + 1x1/2 + lx0)} + 13.069 sin{2p (1x0 + 1x0 + lx1/2)} + 13.069 sin{2p (1x1/2 + 1x1/2 + lx1/2)}
nSfn cos{2p (hx+ky+lz)}+ iSnfn sin{2p (hx+ky+lz)} or F200 = -17.548 + 0
\F2002 = -17.5482
\F2002 = 307.949 units or ~308 units
(1,0,0) reflection
Using miller indices to find d in the (1,0,0) plane, where a = 5.638, h=1, k=0, l=0
dhkl = a / square root (h2 + k2 + l2)
\ d100 = 5.638/ square root (12 + 02 + 02)
\ d100 = 5.638 Å
No displacement occurs along this plane since it will be in alignment with the incident radiation.
2. Multiplicity
There are 8 plausible planes with the same d spacing to the (2,0,0) plane (including (2,0,0)) itself
These planes are
(2,0,0)(2,-0,0) (2,0,-0) (2,-0,-0)
(-2,0,0) (-2,-0,0) (-2,0,-0) (-2,-0,-0)
However, since +0 and 0 are identical, all the planes on the top row are equivalent to each other and all the planes on the 2nd row are equivalent to each other
This equality of planes leads to a multiplicity j200 = 2
3. Polarisation
Using the case of unpolarised X-rays, the equation for Polarisation is given as
P = (1 + cos2 2q) / 2
dhkl = a / (h2 + k2 + l2) ½
d200 = 5.638 / (22 + 02 + 02 ) ½
d200 = 2.819Å
Copper K-a radiation with the wavelength 1.54Å, adds to Braggs equation to give
l = 2d sinq
sinq = 0.273o
q = sin-1 (0.273) = 15.85o
2q = 31.70o
cos (2q) = 0.851
cos2 (2q) = 0.724
1 + cos2 (2q) = 1.724
P200 = 1 + cos2 (2q) / 2 = 0.862
Therefore, polarisation reduces the intensity to 86% in the case of the 200 plane
4. Lorentz Factor
Using the relationship L = 1 / sin2q cosq and using the value of q from Q3 (15.85o)
Sin (q) = 0.273
Sin2 (q) = 0.075
Cos (q) = 0.962
Sin2 (q) Cos (q) = 0.072
1 / Sin2 (q) Cos (q) = 13.94o
\ L = 13.94o
5. Agreement
Ihkl = c*jhkl*Phkl*Lhkl*A*Fhkl2
I111 / I200 = c*j111*P111*L111*A*F1112 / c*j200*P200*L200*A*F2002
Absorption, c, is a constant and can be cancelling out in the equation
I111 / I200 = c*8*0.894*18.39*398 / c*2*0.862*13.94*308
I111 / I200 = 7.07
6. Theory vs Experimental
The factors are in order of severity i) being the most severe, but this severity will certainly depend on the magnitude of the source.
i.) Equipment resolution/beam focusing: The resolution and focus of the incident of the equipment can incur experimental error
ii.) Crystal defects/impurities The crystal itself might not be a purely grown as possible and any defects/impurities within the structure will lead to an effect
iii.) Detector type: Scintillation has poor energy resolution, gas-filled can lose linearity used instead of a position sensitive detector
iv.) Magnetism: Magnetism from another instrument or source near the experimental area could alter structural properties and/or the alignments of the crystallites
v.) Temperature: Fluctuation can change the position of atoms within the structure to alter diffraction readings
vi.) Pressure: Changes in atmospheric pressure can lead to positional changes in the structure
vii.) Instrument Calibration: If any section of the equipment has been poorly / not or fell out of calibration will lead to errors. This include alignment of the beams, sample holders, detector etc.
viii.) Sample preparation: poor sample preparation i.e damage by grinding the sample can lead to lump powders and distorted data
ix.) Sample holder: Capiliary or flat plate can make a difference between results. Alignment of the sample holder is also a factor.
x.) Moisture: moisture within the sample