Rob Whittock

Assignment 3.

Question1.

(i). 200 reflection.

d200 = 5.638 / Ö 22+02+02 = 2.819

½ (2.819) = sin q / l = 0.177Å-1

Cosine part.

å n ¦ n cos {360 (hx+ky+lz)} = 4 (8.39 cos {360 (0) }) + 4(8.39 cos {360 (1) })

=33.56+33.56 = 67.12

Sin Part.

å n¦ nsin {2p (hx+ky+lz)}= 4 (8.39 {sin{2p (0)}) + 4 (8.39 {sin{2p (1)})

= 0 +3.67 = 3.67

F2002 = (67.12)2 +(3.67)2 = 4518 (to 0dp).

I200 = F2002 = 4518 units.

(ii). d100 = 5.638/Ö 12+02+02 = 5.638.

½ (5.638) = sin q /l = 0.0887Å-1

Cosine part.

å n¦ ncos {360(hx+ky+lz)} = 4 (9.05 cos {360(0)}) + 4(9.035 cos {360(1/2)})

=36.14 –35.14 = 0.

Sin Part.

å n¦ nsin {2p (hx+ky+lz)}= 4 (9.035 {sin{2p (0)}) + 4 (9.035 {sin{2p (1/2)})

= 0 +1.9806.

F1002 = 02+(1.9806)2 = 3.923.

I1002 = F2002 = 3.923 units.

Question 2.

J200 = 2.

2 sets of planes (200 and –200).

Question 3.

d200 = 5.638/Ö 22+02+02 = 5.638/2 = 2.819Å-1.

1.54 = 2 x 2.819 x sinq

sinq = 0.273

q = sin-1 (0.273) = 15.843

2q = 31.69o (to 2dp).

Cos (2q ) = 0.724

1+cos2(2q ) = 1.724

P200 = (1+cos2(2q ))/2 = 0.862.

Polarization produces a reduction in intensity of 86% in the case of the 200 reflection.

Question 4.

q = 15.84o

Sin q = 0.273

Sin2 q = 0.075

Cos q = 0.962

Sin2q cosq = 0.0722

L200 = 1/sin2q cosq = 13.85 (200 reflection).

Question 5.

I200 = j200xP200xL200xF2002

I200 = 2 x 0.862 x 13.85 x4518

I200 = 107, 878 c units (to 0dp).

I111/I200 = 52347 / 107878 = 0.485 (to 3dp).

Constant (c) = 0.485 (to 3dp).

Question 6.

Plausible factors (additional to j, p and l) which might be the cause of the disagreement between the measured and predicated I111 / I200 ratio:

(i). Sample preparation – if crystals within sample are not randomly orientated, then preferred orientation will give rise to a higher intensity value than the predicted value.

(ii). Texture – Presence of large crystals that are therefore not numerous enough, to provide a true powder average.

(iii). Spinning – if the spinning rate of the sample is too low, then not enough crystallites will satisfy the Bragg condition and diffraction intensities will be lower than predicted.

(iv). Transparency – If the sample is transparent, then diffraction peaks will be shifted to lower scattering angles, therefore intensities at predicted peaks will be different in the measured spectrum.

(v). Signal to noise ratio - the predicted value does not take into account the noise that appears in the measured spectrum.