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17-Phys-B7 Structure of Materials · December 2018

Question 5 of 7: Microstructural Characterization

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. 17-Phys-B7 Structure of Materials, National Examination December 2018 — a closed-book examination (Casio or Sharp approved calculators only; all necessary equations, constants and diagrams supplied in the paper's own appendix). Candidates attempt any five of the seven questions, each worth 20 marks; every question is nonetheless answered in full below so the paper remains a complete study resource.

Reference texts. W. D. Callister Jr. & D. G. Rethwisch, Materials Science and Engineering: An Introduction, 10th ed. (atomic bonding, crystal structure and packing, point defects, diffusion, dislocations and slip, mechanical properties, phase diagrams and the lever rule, precipitation hardening, X-ray diffraction).

Check — two points on the printed paper. (1) Question I.2 prints the ion as "Cl- (Z = 16)"; Z = 16 is sulfur, not chlorine (Cl is Z = 17) — a printed typo. The electronic structure below uses the correct Z = 17 for chlorine. (2) Question III.1(d) prints the hexagonal plane as $(2\bar{2}10)$, i.e. Miller–Bravais indices $h=2,\,k=-2,\,i=1$; the third index of a valid Miller–Bravais symbol is never independent — it is fixed by $i=-(h+k)$, here $i=-(2-2)=0$, not 1. Reducing the self-consistent index $(2\bar200)$ by its common factor of 2 gives $(1\bar100)$, a standard prism-type plane. The drawing below uses the symmetry-equivalent, non-degenerate face $(10\bar10)$ of the same $\{1\bar100\}$ family (chosen because it renders visibly in the cell projection used here) and states this substitution explicitly.

Question V: 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.

Given. X-ray diffraction peaks at $2\theta=40^\circ,58^\circ,73^\circ, 86.8^\circ,100.4^\circ,114.7^\circ$; wavelength $\lambda=0.154$ nm; element is either BCC or FCC.

Given data — XRD peaks
Peak123456
2θ (deg)40587386.8100.4114.7

Find. (a) the cubic structure (BCC or FCC); (b) the lattice constant $a$.

Approach. Compute $\sin^2\theta$ for each peak and form the RATIO of each to the first; a BCC pattern's allowed reflections have $h^2+k^2+l^2=2,4,6,8,10,12,\ldots$ (ratios $1:2:3:4:5:6$), while an FCC pattern's have $h^2+k^2+l^2=3,4,8,11,12,16,\ldots$ (ratios $1:1.33:2.67:3.67:4:5.33$) — whichever integer-ratio sequence the data matches identifies the structure; then get $a$ from Bragg's law, $d=\lambda/(2\sin\theta)$ and $a=d\sqrt{h^2+k^2+l^2}$, for each peak's assigned $(hkl)$.

  1. $\sin^2\theta$ and its ratio to the first peak. $$\sin^2\theta = 0.1170,\ 0.2350,\ 0.3538,\ 0.4721,\ 0.5903,\ 0.7089 \;\;\Longrightarrow\;\; \text{ratio} = 1.00,\ 2.01,\ 3.02,\ 4.04,\ 5.05,\ 6.06.$$ This is an almost exact $1:2:3:4:5:6$ progression, matching the BCC $h^2+k^2+l^2$ sequence $2,4,6,8,10,12$ (the FCC sequence $1:1.33:2.67:3.67:4:5.33$ does not fit at all) $\Longrightarrow\boxed{\text{BCC}}$.
  2. Assign $(hkl)$ and get $d$-spacings. BCC allows $h^2+k^2+l^2=2,4,6,8,10,12$, i.e. $(110),(200),(211),(220),(310),(222)$; from Bragg's law $d=\lambda/(2\sin\theta)$: $d_1=0.2251$, $d_2=0.1588$, $d_3=0.1294$, $d_4=0.1121$, $d_5=0.1002$, $d_6=0.0914$ nm.
  3. Lattice constant from each peak, then average. $$a_i = d_i\sqrt{h^2+k^2+l^2}\ \Longrightarrow\ 0.3184,\,0.3177,\,0.3171,\,0.3170,\,0.3169,\, 0.3168\ \text{nm},$$ $$\bar a = \boxed{0.317\ \text{nm}}$$ (the six values agree to within 0.5%, confirming the BCC indexing; a lattice constant this size is consistent with a BCC transition metal such as molybdenum, $a\approx0.315$ nm, or tungsten, $a\approx0.317$ nm — a useful independent plausibility check).
Final results — Question V.1
QuantityValue
Crystal structureBCC
Lattice constant $a$0.317 nm

Part 2 — SEM vs. TEM.

SEM vs. TEM comparison
AspectScanning electron microscopy (SEM)Transmission electron microscopy (TEM)
(a) Physical principleA finely focused electron beam is RASTER-SCANNED across the sample surface; secondary and backscattered electrons ejected from a thin near-surface layer are collected point-by-point to build a (mostly topographic) image. A broad, high-energy electron beam is TRANSMITTED through an electron-transparent sample (typically <100 nm thick); the transmitted/diffracted beam is focused by electromagnetic lenses to form a projected 2-D image (and diffraction pattern) of the internal structure.
(b) Typical electron energy$\approx1$–30 keV (often 5–20 keV for materials work)$\approx100$–300 keV (some up to 1 MeV)
(c) Resolution / magnification$\approx1$–10 nm resolution; $\approx10\times$–$100{,}000\times$ magnification $\approx0.1$–0.2 nm resolution (atomic-scale); up to $\approx1{,}000{,}000\times$ magnification

(d) TEM is the technique that reveals sub-surface dislocation activity in a metallic thin sample: the transmitted electrons diffract off the strain field surrounding each dislocation line running through the specimen's thickness, producing dark dislocation-line contrast against the bright diffracting matrix — this is the standard method for imaging dislocation density, tangles, and slip-band structure. SEM only images the (already-deformed) outer surface topography and cannot see internal/sub-surface dislocations.