Question 5 of 8: X-ray Diffraction and Microstructural Characterization (20 marks)

Question V — X-ray Diffraction and Microstructural Characterization (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 — TEM and SEM: principle and applications

Transmission electron microscopy (TEM) fires a high-energy (100–300 kV) electron beam through a very thin (<100 nm) sample; electrons are scattered/diffracted differently by different structural features (dislocations, grain boundaries, precipitates, crystal planes), and the transmitted beam is focused into a 2-D image or diffraction pattern. Because electron wavelengths are far shorter than visible light, TEM resolves individual atomic columns and dislocation cores — it is used for direct dislocation imaging, precipitate/second-phase characterisation, and (via selected-area diffraction) crystal-structure identification, at the cost of laborious thin-foil sample preparation.

Scanning electron microscopy (SEM) rasters a focused electron beam across a bulk sample's surface and collects secondary and/or backscattered electrons emitted at each point to build an image line by line. Because it images the surface rather than transmitting through the sample, specimen preparation is much simpler (bulk, minimally prepared samples), depth of field is large (giving the characteristic 3-D-looking topographic contrast), and typical resolution (a few nm to tens of nm) is coarser than TEM but far finer than optical microscopy. SEM is the workhorse for fractography (this exam's own Question VII territory), surface morphology, and (with an EDS detector) point chemical composition.

V.2 — Ni powder XRD peaks (Cu-K$\alpha$)

Given. Ni is FCC, $r = 1.25\times10^{-10}$ m; $\lambda = 1.542\times10^{-10}$ m (Cu-K$\alpha$); FCC reflection rule (appendix Bragg law $n\lambda=2d\sin\theta$, cubic $d$-spacing $d=a/\sqrt{h^2+k^2+l^2}$): only planes with unmixed indices (all even or all odd) reflect.

Find. The lattice parameter $a$, the first three allowed $(hkl)$, their $d$-spacings, and their $2\theta$ positions.

Approach. Get $a$ from the FCC touching condition, list candidate low-index planes and keep only unmixed ones (lowest $\sum h^2+k^2+l^2$ first), then apply Bragg's law with $n=1$.

1. Lattice parameter. FCC atoms touch along the face diagonal (as derived in III.2), $a=2\sqrt2r$: $$a = 2\sqrt2(1.25\times10^{-10}) = \boxed{3.536\times10^{-10}\ \text{m} = 0.3536\ \text{nm}}$$
2. Candidate planes and the reflection rule. Ranking by $\sqrt{h^2+k^2+l^2}$: (100) sum=1 — mixed parity (one odd, two even), forbidden; (110) sum=2 — mixed, forbidden; (111) sum=3 — all odd, allowed; (200) sum=4 — all even, allowed; (210) mixed, forbidden; (211) mixed, forbidden; (220) sum=8 — all even, allowed. First three allowed reflections: (111), (200), (220).
3. d-spacings. $d=a/\sqrt{h^2+k^2+l^2}$: $$d_{111}=\frac{0.3536}{\sqrt3}=0.2041\ \text{nm}, \quad d_{200}=\frac{0.3536}{2}=0.1768\ \text{nm}, \quad d_{220}=\frac{0.3536}{\sqrt8}=0.1250\ \text{nm}$$
4. Bragg angles. $\theta=\arcsin(\lambda/2d)$, then double: $$2\theta_{111}=2\arcsin\!\frac{0.1542}{2(0.2041)}=\boxed{44.4^{\circ}}, \quad 2\theta_{200}=2\arcsin\!\frac{0.1542}{2(0.1768)}=\boxed{51.7^{\circ}}, \quad 2\theta_{220}=2\arcsin\!\frac{0.1542}{2(0.1250)}=\boxed{76.2^{\circ}}$$

V.2(b) — Single crystal, (100) surface

A symmetric $\theta$–$2\theta$ scan of a single crystal only detects planes that are parallel to the sample surface (their normal along the scan/surface-normal direction) — unlike a randomly oriented powder, where every grain orientation is present and every allowed $(hkl)$ family contributes a peak regardless of specimen orientation. With the $(100)$ plane parallel to the surface, only the $(h00)$ family can satisfy the Bragg condition in this geometry. Of the three peaks found above, only $(200)$, at $2\theta\approx51.7^{\circ}$, is an $(h00)$-type reflection; $(111)$ and $(220)$ are not parallel to $(100)$ and would be absent from a single-crystal $(100)$-oriented scan. (Also note $(100)$ itself is forbidden by the FCC reflection rule even though it is the physically relevant surface plane — the observed peak from that orientation is its allowed higher-order relative, $(200)$.)