Question 7 of 8: Question VII — X-ray Diffraction and Experimental Methods (20 marks)

Question VII — X-ray Diffraction and Experimental Methods (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.

VII.1 — Distinguishing FCC from BCC by XRD

Bragg's law, $n\lambda = 2d\sin\theta$, relates each diffraction peak's angle to the spacing $d_{hkl}$ of the reflecting plane family; combined with the cubic interplanar-spacing relation $d_{hkl}=a/\sqrt{h^2+k^2+l^2}$, every measured $2\theta$ converts to a value of $\sqrt{h^2+k^2+l^2}$ once the lattice parameter is known (or, conversely, the ratios of $\sin^2\theta$ between peaks give the ratios of $h^2+k^2+l^2$ directly, without needing $a$ first). The key discriminator is the structure-factor selection rule, which forbids different sets of reflections in FCC and BCC because of destructive interference from the extra (non-primitive) lattice points:

• BCC (one extra atom at the body centre) reflects only when $h+k+l$ is even — allowed peaks: $\{110\},\{200\},\{211\},\{220\},\dots$
• FCC (extra atoms at all face centres) reflects only when $h,k,l$ are all even or all odd (unmixed parity) — allowed peaks: $\{111\},\{200\},\{220\},\{311\},\dots$

In practice: index the observed peaks by their $\sin^2\theta$ ratios (which are proportional to $h^2+k^2+l^2$), reduce those ratios to the smallest integer set, and compare the pattern of allowed $(hkl)$ against the two selection rules above — e.g. a first peak with $h^2+k^2+l^2=2$ (i.e. (110)) identifies BCC, whereas a first peak with $h^2+k^2+l^2=3$ (i.e. (111)) identifies FCC. This is exactly the method applied to niobium below.

VII.2 — Niobium XRD indexing

Given. BCC niobium; $(211)$ reflection at $2\theta=75.99^\circ$ (first order, $n=1$); $\lambda=0.1659\ \text{nm}$.

Find. $d_{211}$, the atomic radius of Nb, and the lowest-angle allowed peak (its $(hkl)$ and $2\theta$).

Approach. Get $d_{211}$ from Bragg's law, then the lattice parameter from the cubic $d$-spacing relation, then the atomic radius from the BCC body-diagonal contact condition; finally identify the smallest allowed $h^2+k^2+l^2$ for BCC to locate the lowest-angle peak.

1. Interplanar spacing. $\theta = 75.99^\circ/2 = 37.995^\circ$: $$d_{211} = \frac{n\lambda}{2\sin\theta} = \frac{0.1659}{2\sin(37.995^\circ)} = \boxed{0.1347\ \text{nm}}$$
2. Lattice parameter and atomic radius. $a = d_{211}\sqrt{2^2+1^2+1^2} = d_{211}\sqrt6$: $$a = (0.1347)(2.449) = \boxed{0.3301\ \text{nm}}$$ BCC contact is along the body diagonal, $a\sqrt3 = 4R$: $$R_{Nb} = \frac{a\sqrt3}{4} = \frac{(0.3301)(1.732)}{4} = \boxed{0.1429\ \text{nm}}$$ (Close to the accepted metallic radius of niobium, ≈0.143 nm.)
3. Lowest-angle allowed peak. For BCC ($h+k+l$ even), the smallest $h^2+k^2+l^2$ among allowed indices is $(110)$, with $h^2+k^2+l^2=2$ — this gives the largest $d$-spacing and hence the smallest diffraction angle: $$d_{110} = \frac{a}{\sqrt2} = \frac{0.3301}{1.414} = 0.2334\ \text{nm}$$ $$\sin\theta_{110} = \frac{\lambda}{2d_{110}} = \frac{0.1659}{2(0.2334)} = 0.3554 \;\Rightarrow\; \theta_{110}=20.82^\circ$$ $$2\theta_{110} = \boxed{41.6^\circ,\ (hkl)=(110)}$$