21-Mat-A3 Structure and Characterization of Materials · Dec-12-Mtl-A3 2018
Question 1 of 8
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2018 — 12-Mtl-A3, Structure and Characterization of Materials. Three hours, open book, any non-communicating calculator permitted. Eight questions constitute a complete exam paper; all eight are solved here.
Reference texts. The answers below are keyed to the standard undergraduate materials-science references recommended for this syllabus code:
W. D. Callister Jr. and D. G. Rethwisch, Materials Science and Engineering: An Introduction, 10th ed. — crystal structure and defects (Ch. 4), mechanical properties and fracture (Ch. 6, 8), phase diagrams (Ch. 9), diffraction (Ch. 3), polymer structure and mechanical behaviour (Ch. 14–15).
B. D. Cullity and S. R. Stock, Elements of X-Ray Diffraction, 3rd ed. — structure-factor selection rules for cubic lattices.
Given. Host-atom (metal) radius $R$; interstitial (impurity) sites at the midpoint of every unit-cell edge, for both the FCC and BCC lattices.
Find. The maximum impurity radius $r$ (in terms of $R$) that just fits at an edge-center site, for FCC and for BCC.
Edge-center interstitial site: two host atoms of radius $R$ terminate the unit-cell edge; the impurity atom of radius $r$ sits at the midpoint, touching both. Close-packing (touching atoms) fixes the lattice parameter $a$ in terms of $R$, and geometry along the edge fixes $r$.
Approach. Relate $a$ to $R$ from the touching-atom condition on the relevant close-packed direction, identify the atom(s) nearest the edge midpoint, and set (host radius) + (impurity radius) equal to that nearest-neighbour distance.
Part (FCC) — lattice parameter from the face diagonal. In FCC, atoms touch along the face diagonal: $4R = \sqrt{2}\,a \;\Rightarrow\; a = 2\sqrt{2}\,R$. The edge-center site sits midway between the two corner atoms bounding that edge, at distance $a/2$ from each — and no other atom (face-center or corner of a neighbouring cell) is closer, since the nearest face-center atom lies at $a/\sqrt{2} \approx 0.707a > a/2$.
Part (FCC) — solve for r. Touching condition: $R + r = a/2$. Substituting $a = 2\sqrt2 R$: $$R + r = \sqrt{2}\,R \;\Rightarrow\; \boxed{r_{FCC} = (\sqrt{2}-1)R \approx 0.414R}$$ This is the classic FCC octahedral-void radius ratio.
Part (BCC) — lattice parameter from the body diagonal. In BCC, atoms touch along the body diagonal: $4R = \sqrt{3}\,a \;\Rightarrow\; a = 4R/\sqrt{3}$. For the edge-center site, checking all candidate neighbours — the two edge-terminating corner atoms at $a/2$, and the body-center atoms of the four cells sharing that edge at $a/\sqrt{2} \approx 0.707a$ — the corner atoms at $a/2$ are the closest.
Part (BCC) — solve for r. Touching condition: $R + r = a/2 = 2R/\sqrt{3}$. $$\boxed{r_{BCC} = \left(\frac{2}{\sqrt{3}}-1\right)R \approx 0.155R}$$ This is the BCC octahedral-void radius ratio (smaller than FCC's, and the site is non-cubic/distorted — the reason interstitial carbon strains the BCC ferrite lattice more anisotropically than the FCC austenite lattice).
