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21-Mat-A4 Deformation Behaviour and Properties of Materials · December 2019

Question 3 of 7: Crystal Structure

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

Notes on this paper

Paper format. National Exams, December 2019 — 10-Met-A4, Structure of Materials. Three hours, closed book, one approved calculator (Casio or Sharp). Seven questions of 20 marks each; the rubric asks for any five, with only the first five in the answer book marked. All seven are solved here, because this set is a study resource rather than an exam script. All necessary equations, constants and an error-function table are provided in the exam's own appendix (reproduced where used below).

Note on the exam title. The printed exam header reads 10-Met-A4, Structure of Materials. Only Question 6 (dislocations, slip, resolved shear stress) touches mechanical/deformation properties directly; the paper as a whole is a broad introductory materials-science survey — atomic structure, bonding, crystal structure/directions/planes, point defects, XRD, dislocations and phase diagrams — and is answered as such below.

Reference texts. The answers below are keyed to the works normally recommended for this syllabus code:

Question 3: 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.

3.1 — Crystal Directions and Planes

[121](232)intercepts a/2, a/3, a/2simple cubic
Simple cubic cell: direction [121] (through the point $(a,2a,a)$) and plane (232), intercepts $a/2,\,a/3,\,a/2$ on the $x,y,z$ axes.

[121]: a line from the origin through the lattice point at coordinates $(1,2,1)$ in units of $a$ — one edge-length along $x$, two along $y$, one along $z$.

(232): a plane whose axis intercepts are the reciprocals of the indices, $1/2,\,1/3,\,1/2$ (in units of $a$) — it cuts the $x$ and $z$ edges at the half-way point and the $y$ edge at one-third, forming the shaded triangle above.

A bare $(1210)$ fails the Miller–Bravais closure identity $i=-(h+k)$ (would need $i=-3$, not $1$). Testing $(1\bar2 10)$ instead ($h=1,k=-2,i=1,l=0$) gives $i=-(1-2)=1$ — consistent — so the plane is read as $(1\bar2 10)$, a member of the $\{11\bar20\}$ second-order-prism family in the hexagonal system.

carhombic-prism unit cell (2 atoms); the full hexagonalprism (6 atoms) tiles 3 of these about the c-axis
Hexagonal unit cell context for $(1\bar2 10)$: a second-order prism plane of the $\{11\bar20\}$ family, containing the $c$-axis and cutting the basal $a_1,a_2$ axes at equal, opposite-signed intercepts.

3.2 — Perovskite Structure of CaTiO₃

Ca²⁺ (corners)Ti⁴⁺ (body centre)O²⁻ (face centres)
Cubic perovskite ABO₃ unit cell for CaTiO₃: Ca²⁺ (A-site) at the 8 cube corners, Ti⁴⁺ (B-site) at the body centre, O²⁻ at the 6 face centres.

The perovskite structure places the larger, lower-valence A cation (Ca²⁺) at the corners of the cube (12-fold coordination with the O sublattice), the smaller, higher-valence B cation (Ti⁴⁺) at the body centre, and O²⁻ anions at the six face centres. The six face-centre oxygens surrounding the body-centre Ti⁴⁺ form a regular TiO₆ octahedron (6-fold coordination) — the defining structural motif of the perovskite family. One formula unit (1 Ca, 1 Ti, 3 O) belongs to the cell: 8 corner Ca × 1/8 = 1 Ca; 1 body-centre Ti = 1 Ti; 6 face-centre O × 1/2 = 3 O.

3.3 — HCP Unit Cell Volume and Packing Factor

carhombic-prism unit cell (2 atoms); the full hexagonalprism (6 atoms) tiles 3 of these about the c-axis
Full HCP unit cell (hexagonal prism, 6 atoms): base a regular hexagon of side $a$, height $c$.

Given. HCP lattice parameters $a$ (basal edge) and $c$ (prism height); part (d) uses Re: $R=0.137$ nm, $c/a=1.615$.

Find. (a) $V_C(a,c)$; (b) $V_C(R,c)$; (c) APF at $c=1.633a$; (d) $V_C$ for Re.

Approach. Treat the conventional HCP unit cell as the full hexagonal prism (6 atoms): base area = area of a regular hexagon of side $a$, times height $c$. Substitute $a=2R$ (basal-plane spheres touching) to get $V_C(R,c)$, then use $\text{APF}=V_{\text{spheres}}/V_C$ with $n=6$ atoms per cell.

  1. (a) Volume in terms of $a,c$. A regular hexagon of side $a$ has area $6\times\left(\tfrac{\sqrt3}{4}a^2\right)=\tfrac{3\sqrt3}{2}a^2$, so $$V_C=\boxed{\dfrac{3\sqrt3}{2}\,a^2 c}$$
  2. (b) Volume in terms of $R,c$. Atoms touch along the basal edge, $a=2R$: $$V_C=\frac{3\sqrt3}{2}(2R)^2c=\boxed{6\sqrt3\,R^2 c}$$
  3. (c) Atomic packing factor at $c=1.633a=3.266R$. Six atoms occupy the full hexagonal prism: $$\text{APF}=\frac{6\times\tfrac43\pi R^3}{6\sqrt3\,R^2(3.266R)}=\frac{8\pi}{3\sqrt3(3.266)}=\boxed{0.740}$$ — the ideal HCP packing factor, identical to FCC, confirming $c/a=1.633$ is the hard-sphere close-packed ratio.
  4. (d) Unit-cell volume for rhenium. $a=2R=0.274$ nm, $c=1.615a=0.4425$ nm: $$V_C=6\sqrt3\,R^2c=6\sqrt3(0.137)^2(0.4425)=\boxed{0.0863\ \text{nm}^3}$$
Question 3 — summary
ItemResult
3.3(a) $V_C(a,c)$$\tfrac{3\sqrt3}{2}a^2c$
3.3(b) $V_C(R,c)$$6\sqrt3\,R^2c$
3.3(c) APF ($c=1.633a$)$0.740$
3.3(d) $V_C$ (Re)$0.0863$ nm³