Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-11, Properties of Materials — December 2016. 3 hours,
closed-book examination (approved Casio or Sharp calculator only). Candidates attempt any
five of the seven questions for a complete paper, all questions of equal value. All seven
questions are solved below for completeness.
Reference texts: Callister & Rethwisch, Materials Science and
Engineering: An Introduction, 9th ed. (crystal structure, mechanical behaviour, phase
diagrams, polymer viscoelasticity, composites, electrochemistry, fatigue/fracture, heat
treatment and hardenability, concrete).
Given. BCC chromium, atomic radius $r=0.1249$ nm; atomic mass
$M_{Cr}=52.0$ g/mol (page-1 table); Avogadro's number $N_A=6.02\times10^{23}$ mol$^{-1}$
(page-1 gives this as $0.602\times10^{24}$ — the same value in a shifted mantissa/exponent
split, a house style of this subject’s constants table, not an error).
Find. Theoretical density $\rho$; a sketch of the BCC unit cell showing the
(112) plane and the [011] direction; the interplanar spacing $d_{102}$.
Approach
For BCC, the atoms touch along the cube’s body diagonal, which fixes the lattice
parameter $a$ in terms of $r$; density then follows from the standard $n,M,N_A,a^3$ relation
($n=2$ atoms/cell for BCC). Interplanar spacing for a cubic system depends only on $a$ and the
Miller indices.
Lattice parameter from the touching-atoms condition. In BCC the body
diagonal $a\sqrt3$ spans 4 atomic radii:
$$a=\frac{4r}{\sqrt3}=\frac{4(0.1249)}{1.7321}=0.2884\ \text{nm}$$
Theoretical density. BCC has $n=2$ atoms per unit cell:
$$\rho=\frac{nM}{N_Aa^3}=\frac{2(52.0)}{(6.02\times10^{23})(0.2884\times10^{-7}\,\text{cm})^3}
=\frac{104.0}{6.02\times10^{23}\times2.401\times10^{-23}\,\text{cm}^3}$$
$$\boxed{\rho\approx7.20\ \text{g/cm}^3}$$
(the real, handbook density of chromium is 7.19 g/cm$^3$ — excellent agreement, a
strong check that $r$ and the BCC assumption were used correctly).
Unit-cell sketch. The figure below shows the BCC unit cell (8 corner atoms
$+$ 1 body-centre atom), with the (112) plane shaded and the [011] direction drawn as an arrow
from the origin corner to the $(0,1,1)$ corner. The (112) plane has intercepts $1,1,\tfrac12$ on
the $x,y,z$ edges, so within the cell it is the triangle joining $(1,0,0)$, $(0,1,0)$ and
$(0,0,\tfrac12)$.
Fig. Q1 — BCC unit cell (corner + body-centre atoms) with the (112)
plane shaded and the [011] direction arrowed.
Interplanar spacing $d_{102}$. For a cubic system,
$$d_{hkl}=\frac{a}{\sqrt{h^2+k^2+l^2}}$$
$$d_{102}=\frac{0.2884}{\sqrt{1^2+0^2+2^2}}=\frac{0.2884}{\sqrt5}=\frac{0.2884}{2.2361}$$
$$\boxed{d_{102}\approx0.1290\ \text{nm}}$$
(this is the purely geometric spacing between successive (102) planes; it is unaffected by the
BCC structure-factor selection rule, which governs which planes give a diffracted
reflection, not the plane spacing itself — a distinction worth stating explicitly since (102)
has $h+k+l=3$, odd, so it is in fact a systematically-absent BCC diffraction reflection even
though the geometric spacing above is perfectly well defined).
