Kennedy Miller

Advanced Certificate in Powder Diffraction on the Web - Assignment 2

1. The fundamental Structure factor equation is given by F(S) = ⁿSf_n e ^{2p i (hx+ky+lz)}, which translates in to the equation,

F(S) = Sⁿf_n cos{2p (hx+ky+lz)}+ iSⁿf_n 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

d_hkl = 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, f_Na+ and f_Cl- were calculated from the International Crystallographic Tables, via a section from the course notes. This produced the following table.

Comment: Correct.
5 Marks

ⁿSf_n cos{2p (hx+ky+lz)}+ iSⁿf_n 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)}

ⁿSf_n cos{2p (hx+ky+lz)}+ iSⁿf_n sin{2p (hx+ky+lz)} or F200 = -17.548 + 0

\F2002 = -17.5482

\F₂₀₀² = 307.949 units or ~308 units

Comment: Your general method of calculation is correct but you seem to have put h,k,l=1,1,1 (as in original Diffraction Theory Release) everywhere in the calculation rather than h,k,l=2,0,0 as for this assignment. Hence you obtain a value of -17.54 rather than 87.08.
3 Mark

(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

d_hkl = 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.

Comment: It is correct that the intensity for the 1,0,0 plane is zero, but the reason is because of cancellation of scattering from the 4 Na's and 4 Cl's, as explained in the released model answer.
1 Mark

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

Comment: The logic is correct as far as putting h=2 or h=-2; however being cubic, the other two axes are identical so 020, 0-20, 002, 00-2 are also permissable giving a final multiplicity of 6.
1 Mark

3. Polarisation

Using the case of unpolarised X-rays, the equation for Polarisation is given as

P = (1 + cos² 2q) / 2

d_hkl = 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 = 0.273^o

q = sin^-1 (0.273) = 15.85^o

2q = 31.70^o

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

Comment: This is correct.
4 Marks

4. Lorentz Factor

Using the relationship L = 1 / sin2q cosq and using the value of q from Q3 (15.85^o)

Sin (q) = 0.273

Sin² (q) = 0.075

Cos (q) = 0.962

Sin² (q) Cos (q) = 0.072

1 / Sin² (q) Cos (q) = 13.94^o

\ L = 13.94^o

Comment: This is correct. Note that the value is purely numerical and has no units.
3 Marks

5. Agreement

I_hkl = c*j_hkl*P_hkl*L_hkl*A*F_hkl²

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

Comment: The method is correct but of course two of the values (F² and j) are wrong so the numerical result obtained will not be correct.
3 Marks

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

Comment: These factors will affect the positions and widths of the peak but have little effect on the intensities of the peaks.

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

1 Mark

iii.) Detector type: Scintillation has poor energy resolution, gas-filled can lose linearity used instead of a position sensitive detector

Comment: The resolution won't have much effect on the intensity, but the non-linearity is a good candidate because it affects the recorded intensity of each peak differently.
1 Mark

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

Comment: Agreed it is a possibility though the relative effect will be small since the peaks are fairly close to each other.
1 Mark

vi.) Pressure: Changes in atmospheric pressure can lead to positional changes in the structure

Comment: But the relative intensities will be unaffected.

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.

Comment: The direct effect is small, but in the extreme (e.g. sample way off correct position) there will be an effect.
1 Mark

viii.) Sample preparation: poor sample preparation i.e damage by grinding the sample can lead to lump powders and distorted data

1 Mark

ix.) Sample holder: Capiliary or flat plate can make a difference between results. Alignment of the sample holder is also a factor.

Comment: Any relative intensity effect will be insignificant.

x.) Moisture: moisture within the sample

Comment: This can in general give rise to relative changes in intensity, though the effect would be small in this particular case (see model answer).
1 Mark

6 Marks overall for this question

Comment: A good solid assignment but two early mistakes propagated errors into the later parts.