X-ray Diffraction and Experimental Methods for Crystal Structure (20 marks): 21-Mat-A4 Deformation Behaviour and Properties of Materials, May 2014, Q5

Question V — X-ray Diffraction and Experimental Methods for Crystal Structure (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

V.1 — First four XRD peaks of LiMnPO4

Given. Primitive orthorhombic: $a=0.611$ nm, $b=1.04$ nm, $c=0.475$ nm; Co $K_\alpha$, $\lambda=0.179$ nm.

Find. $2\theta$ for the first four diffraction peaks (smallest angle = largest $d$-spacing), and a schematic pattern.

Approach. For orthorhombic, $1/d^2 = h^2/a^2+k^2/b^2+l^2/c^2$; since the lattice is primitive, every $(hkl)$ is allowed (no systematic absences), so rank candidate planes by increasing $1/d^2$ (decreasing $d$, increasing $2\theta$) and apply Bragg's law $\lambda = 2d\sin\theta$ to the four smallest.

Reciprocal-square terms. $1/a^2 = 2.679\ \text{nm}^{-2}$, $1/b^2=0.925\ \text{nm}^{-2}$, $1/c^2=4.432\ \text{nm}^{-2}$. Because $b$ is the longest axis, $(010)$ gives the smallest $1/d^2$ of all single-index planes and is the first peak.
Rank low-index planes by $1/d^2 = h^2/a^2+k^2/b^2+l^2/c^2$. Evaluating $(010),(100),(110),(020),(001),\ldots$ gives, in increasing order: $(010)=0.925$, $(100)=2.679$, $(110)=3.603$, $(020)=3.698\ \text{nm}^{-2}$ — the first four peaks.
Convert to $d$ and $2\theta$ via Bragg's law ($n=1$). $$2\theta = 2\arcsin\!\left(\frac{\lambda}{2d}\right),\qquad d=1/\sqrt{1/d^2}$$ $(010)$: $d=1.040$ nm $\Rightarrow \boxed{2\theta = 9.87^\circ}$
$(100)$: $d=0.611$ nm $\Rightarrow \boxed{2\theta = 16.85^\circ}$
$(110)$: $d=0.5268$ nm $\Rightarrow \boxed{2\theta = 19.56^\circ}$
$(020)$: $d=0.5200$ nm $\Rightarrow \boxed{2\theta = 19.82^\circ}$

Schematic powder pattern: the first peak (010) is well separated at low angle (large $d=b$); the (110) and (020) peaks fall only $0.26^\circ$ apart and would appear as a near-overlapping doublet at this resolution.

Question V — final results
Peak (hkl)	$d$-spacing	$2\theta$
(010)	1.040 nm	$9.87^\circ$
(100)	0.611 nm	$16.85^\circ$
(110)	0.527 nm	$19.56^\circ$
(020)	0.520 nm	$19.82^\circ$

Topic. Powder X-ray diffraction: interplanar spacing, Bragg's law, and peak indexing for orthorhombic lattices.

Key relations. $1/d^2 = h^2/a^2+k^2/b^2+l^2/c^2$ (orthorhombic); $n\lambda=2d\sin\theta$.

Why this works. A primitive lattice has a basis of one lattice point per cell, so there are no structure-factor cancellations from centering — every geometrically distinct $(hkl)$ family produces an allowed reflection, unlike body- or face-centred lattices where selection rules silently forbid many low-index peaks. Ranking by $1/d^2$ is equivalent to ranking by $2\theta$ because $\sin\theta$ increases monotonically with $1/d$.

Common pitfall. Forgetting to check permutations like $(001)$ and $(011)$ against $(020)$ before declaring the "first four" — a longer axis can push an apparently higher-index plane ahead of a lower-index one; solving for $\theta$ instead of doubling it to report $2\theta$.

Source. Callister & Rethwisch, Materials Science and Engineering, 10th ed., Ch. 3 (X-ray Diffraction: Determination of Crystal Structures).

Question 5 of 8: X-ray Diffraction and Experimental Methods for Crystal Structure (20 marks)

Question V — X-ray Diffraction and Experimental Methods for Crystal Structure (20 marks)

V.1 — First four XRD peaks of LiMnPO4