Question 7 of 8: Question VII: X-ray Diffraction and Experimental Methods for Crystal Structure

Question VII: 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.

Part 1. Bragg's law, $n\lambda=2d_{hkl}\sin\theta$, converts each diffracted peak's angle into an interplanar spacing $d_{hkl}$, and for a cubic crystal $d_{hkl}=a/\sqrt{h^2+ k^2+l^2}$ links every peak back to a single lattice parameter $a$. The key diagnostic is which $(hkl)$ combinations are allowed to diffract at all — the structure factor for a body-centred lattice vanishes unless $h+k+l$ is even (so the first BCC peaks are (110), (200), (211), (220)…), while for a face-centred lattice it vanishes unless $h$, $k$, $l$ are all odd or all even (so the first FCC peaks are (111), (200), (220), (311)…). In practice: index the observed peaks by their $\sin^2\theta$ ratios (which scale as $h^2+k^2+l^2$ for a cubic cell), then compare the resulting integer sequence against the two allowed lists — a sequence $1,2,3,4,5,6,8,\ldots$ (in units of the smallest ratio) identifies BCC, while $3,4,8,11,12,16,\ldots$ identifies FCC. Matching the observed pattern to one of these two sequences determines the structure without needing to know $a$ or $\lambda$ in advance.

Part 2 — Given. Niobium (BCC); $(211)$ reflection at diffraction angle $2\theta=75.99^{\circ}$ (first order, $n=1$); $\lambda=0.1659$ nm.

Find. (a) $d_{211}$, (b) the niobium atomic radius, (c) the lowest-angle peak and its $(hkl)$.

Approach. Get $d_{211}$ from Bragg's law, recover $a$ from the cubic $d$-spacing formula, then convert to atomic radius via the BCC body-diagonal contact condition; for (c), the lowest allowed $(h+k+l)$-even BCC reflection is always $(110)$.

1. (a) Interplanar spacing. With $\theta=75.99^{\circ}/2=37.995^{\circ}$, $$d_{211}=\frac{n\lambda}{2\sin\theta}=\frac{0.1659}{2\sin(37.995^{\circ})} =\frac{0.1659}{2(0.6155)}=\boxed{0.1347\ \text{nm}}.$$
2. (b) Lattice parameter and atomic radius. For cubic $(211)$, $d=a/\sqrt{2^2+1^2+1^2}=a/\sqrt6$, so $$a=d_{211}\sqrt6=0.1347\times2.449=\boxed{0.3301\ \text{nm}}.$$ BCC atoms touch along the body diagonal, $a\sqrt3=4R$, so $$R=\frac{a\sqrt3}{4}=\frac{0.3301\times1.732}{4}=\boxed{0.1429\ \text{nm}\ (142.9\ \text{pm})}.$$ (The accepted metallic radius of niobium is $\approx$143–146 pm — close agreement confirms 75.99° was correctly read as $2\theta$, not $\theta$.)
3. (c) Lowest-angle peak. For a BCC lattice the allowed reflections need $h+k+l$ even; the smallest-index combination satisfying this is $(110)$, which (having the largest $d$-spacing of any allowed plane) always produces the lowest-angle peak. $$d_{110}=\frac{a}{\sqrt2}=\frac{0.3301}{1.414}=0.2334\ \text{nm},\qquad \sin\theta_{110}=\frac{\lambda}{2d_{110}}=\frac{0.1659}{0.4668}=0.3554,$$ $$\theta_{110}=20.82^{\circ}\ \Rightarrow\ 2\theta_{110}=\boxed{41.64^{\circ},\ (110)}.$$