Systematic absences

Systematic absences are ‘missing’ diffraction intensities on a powder X-ray diffraction spectrum of intensity versus 2θ. Consider the possible values of h, k and l that satisfy Bragg’s equation of $sin\theta = \frac{\lambda \sqrt{h^2+k^2+l^2}}{2a}$ for a primitive cubic unit cell:

hkl	100	110	111	200	210	211	220	300	221
h²+ k²+ l²	1	2	3	4	5	6	8	9	9

7 is missing in the h²+ k²+ l² sequence, as h, k and l are integers. An extension of the sequence reveals that 15, 23, etc. are missing as well. Hence, the spectrum for a primitive cubic unit cell has a specific set of missing intensities. Likewise, different missing intensity sets are expected for other unit cells. For example, a sample with grains that are composed of face-centred cubic unit cells (figure I below) has scattered rays from {200} planes that are out of phase with rays from {100} (figure II), resulting in destructive interference with no {100} intensity peak on the spectrum.

In fact, for all face-centred cubic unit cells (FCC), intensity peaks are observed when the integers h, k and l are either all odd or all even, e.g. {111}, {200}, {220}, …. A mathematical treatment of this requires the knowledge of the structure factor, which is described in a later section. The structure factor for a body-centred cubic unit cell (BCC) also shows that systematic absences do not appear when h + k + l = 2n, i.e. when the sum of the Miller indices is even, e.g. {110}, {200}, {211}, ….

Question

Identify the cubic unit cells of polonium and aluminium and determine their dimensions using the following powder X-ray diffraction spectral data:

2θ_i	2θ₁	2θ₂	2θ₃	2θ₄	2θ₅	2θ₆	2θ₇	2θ₈
Po	12.1^o	17.1^o	21.0^o	24.3^o	27.2^o	29.9^o	34.7^o	36.9^o
Al	38.4^o	44.7^o	65.0^o	78.1^o	82.3^o	98.9^o	111.8^o	116.4^o

λ_Al= 71.0 pm and λ_Po= 154.1 pm

Answer

Let’s tabulate the systematic absences of the three cubic unit cells for reference, knowing that i) 7 is missing for a primitive cubic cell, ii) the integers h, k and l for an FCC cell are either all odd or all even, and iii) that h + k + l = 2n for a BCC cell.

hkl		100	110	111	200	210	211	220	300	221
h²+ k² + l²	Primitive	1	2	3	4	5	6	8	9	9
	BCC	–	2	–	4	–	6	8	–	–
	FCC	–	–	3	4	–	–	8	–	–

Squaring both sides of eq13, we have

$sin^2\theta =\frac{\lambda ^2}{4a^2}(h^2+k^2+l^2)\; \; \; \; \; \; \; (14)$

If the unit cell is primitive, the successive ratios of $\frac{sin^2\theta _i}{sin^2\theta _1}$ will produce the sequence 1,2,3,4,5,6,8,9… since from eq14, $\frac{sin^2\theta _i}{sin^2\theta _1}=\frac{({h}^{2}+{k}^{2}+{l}^{2})_i}{(h^{2}+{k}^{2}+{{l}^{2})_1}}$ and ${(h^{2}+{k}^{2}+{{l}^{2})_1}}=1$ . If the unit cell is BCC, we need to multiply the successive ratios of $\frac{sin^2\theta _i}{sin^2\theta _1}$ by the factor 2 to obtain the fingerprint sequence of 2,4,6,8… since ${(h^{2}+{k}^{2}+{{l}^{2})_1}}=2$ . By the same logic, the multiplication factor is 3 if the unit cell is FCC. Tabulating the ratios,

	$\frac{sin^2\theta _1}{sin^2\theta _1}$	$\frac{sin^2\theta _2}{sin^2\theta _1}$	$\frac{sin^2\theta _3}{sin^2\theta _1}$	$\frac{sin^2\theta _4}{sin^2\theta _1}$	$\frac{sin^2\theta _5}{sin^2\theta _1}$	$\frac{sin^2\theta _6}{sin^2\theta _1}$	$\frac{sin^2\theta _7}{sin^2\theta _1}$	$\frac{sin^2\theta _8}{sin^2\theta _1}$
Po	1.00	1.99	2.99	3.99	4.98	5.99	8.01	9.02
Al	1.00	1.34	2.67	3.67	4.00	5.34	6.34	6.68

it is clear that the unit cell for polonium is primitive cubic while that for aluminium is FCC. To verify the fingerprint sequence of an FCC unit cell, we multiply the aluminium ratios by the factor of 3:

	$\frac{sin^2\theta _1}{sin^2\theta _1}$	$\frac{sin^2\theta _2}{sin^2\theta _1}$	$\frac{sin^2\theta _3}{sin^2\theta _1}$	$\frac{sin^2\theta _4}{sin^2\theta _1}$	$\frac{sin^2\theta _5}{sin^2\theta _1}$	$\frac{sin^2\theta _6}{sin^2\theta _1}$	$\frac{sin^2\theta _7}{sin^2\theta _1}$	$\frac{sin^2\theta _8}{sin^2\theta _1}$
*Al x3*	3.00	4.01	8.01	11.01	12.01	16.01	19.02	20.04

Finally, the dimensions of the unit cell of each sample is obtained by substituting any corresponding set of θ and hkl values in eq14:

	a
Po	337 pm
Al	406 pm

As mentioned above, to mathematically show that the integers h, k and l for an FCC cell are either all odd or all even, and that h + k + l = 2n for a BCC cell, we have to explain what structure factor is. To do that, we need to first understand the Laue equations and the scattering factor.

Systematic absences

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