Question 3 of 7: Question III: Crystal Structure I
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 98-Phys-B7 Structure of Materials, National Examination
December 2016 — a closed-book examination (Casio or Sharp approved calculators only; all
necessary equations, constants, the error-function table and the Cu–Ag phase diagram are
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. This sitting numbers its questions with Roman numerals (Question
I–VII) while sub-items inside each question use Arabic numerals (1., 2., 3.).
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, X-ray diffraction); D. J. Griffiths, Introduction to Quantum Mechanics,
3rd ed. (Bohr model, de Broglie wavelength, Heisenberg uncertainty).
Part 1 — Given. BCC iron, lattice parameter $a=0.2866$ nm, molar
mass $M=55.847$ g/mol, $N_A=6.023\times10^{23}$ /mol; BCC has $n=2$ atoms per unit
cell.
Find. Theoretical density $\rho$.
Unit cell volume. $a=2.866\times10^{-8}\ \text{cm}$,
$V_c=a^3=2.355\times10^{-23}\ \text{cm}^3$.
Density. $$\rho=\frac{nM}{N_A\,V_c}=\frac{2\times55.847}
{6.023\times10^{23}\times2.355\times10^{-23}}=\boxed{7.88\ \text{g/cm}^3}.$$ This matches the
accepted density of iron ($7.87$–$7.88$ $\text{g/cm}^{3}$), confirming the BCC assumption.
Part 2 — Given. Ideal HCP axial ratio $c/a=1.633$; 6 atoms per hexagonal
unit cell, each of radius $R=a/2$; cell volume $V_c=\tfrac{3\sqrt3}{2}a^2c$.
Find. The atomic packing factor (APF).
Sphere volume in the cell. $6\times\dfrac{4}{3}\pi R^3
=8\pi(a/2)^3=\pi a^3$ (using $R=a/2$).
Packing factor. $$\text{APF}=\frac{\pi a^3}{\frac{3\sqrt3}{2}a^2c}
=\frac{2\pi}{3\sqrt3}\cdot\frac{a}{c}=\frac{2\pi}{3\sqrt3\,(c/a)}
=\frac{2\pi}{3\sqrt3\times1.633}=\boxed{0.740}.$$ This is the well-known ideal close-packing
factor shared by HCP and FCC.
Part 3. Miller-index planes are drawn from their axis intercepts $1/h,1/k,1/l$;
Miller–Bravais directions in the hexagonal system are drawn as the vector sum
$u\,\vec a_1+v\,\vec a_2+t\,\vec a_3+w\,\vec c$ with $\vec a_1,\vec a_2,\vec a_3$ at
$120^\circ$ to each other in the basal plane ($t=-(u+v)$).
(a-i) $(1\,\bar1\,0)$: intercepts $x=1$, $y=-1$, parallel to $z$—the
plane is drawn one cell over in $-y$ and shaded as the diagonal rectangle it forms with the
$z$-axis.
(a-ii) $(2\,2\,1)$: intercepts $x=1/2$, $y=1/2$, $z=1$—a triangle cutting
the near-top corner of the cell.
(b-i) $[1\,\bar1\,0\,0]$ (top-down basal-plane view): $\vec a_1-\vec a_2$
points to an EDGE midpoint of the hexagon, the family that is perpendicular to the close-packed
directions.
(b-ii) $[1\,1\,\bar2\,0]$: $\vec a_1+\vec a_2-2\vec a_3$ points to a hexagon
CORNER—a close-packed $\langle11\bar20\rangle$ direction.
