Question 1 of 7: BCC Molybdenum — Atomic Radius, Unit-Cell Sketch, Plane and Direction, Interplanar Spacing
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-11, Properties of Materials — May 2019 sitting (the cover page and every page footer read “04-BS-11, May2019”). 3 hours, closed-book examination (Casio/Sharp calculator only). Notes on the paper state that candidates are to
attempt five, and only five, questions, with only the first five appearing in the answer book marked and
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 and Miller indices; tensile testing and true strain;
hardness testing; solid solutions and grain size; phase diagrams and the lever rule; dislocations and
cold work; polymer molecular weight and viscoelastic behaviour; TTT diagrams and heat treatment;
fracture/fatigue).
Page-1 data used below: atomic masses (g/mol) H 1.01, C 12.01,
Mo 95.94; $N_A=0.602\times10^{24}$ mol$^{-1}$; cold work $CW=(A_0-A_f)/A_0$; grain size
$N=2^{n-1}$. Fig 1 (Al–Si diagram), Fig 2 (cold work vs. properties, iron and copper) and
Fig 3 (isothermal diagram, 0.8% C steel) are printed in the paper; the values used below were read off them.
Question 1: BCC Molybdenum — Atomic Radius, Unit-Cell Sketch, Plane and Direction, Interplanar Spacing (1 of 5)
Find. (a) Atomic radius $r$ and density $\rho$. (b) A unit-cell sketch showing the
$(112)$ plane and the $[011]$ direction. (c) The interplanar spacing $d_{102}$.
Approach
In BCC, atoms touch along the cube body diagonal, which fixes $r$ in terms of $a_0$. The $(112)$ plane
and $[011]$ direction are located on the sketch by the standard intercept/reciprocal and
tail-to-components constructions. The interplanar spacing for any $(hkl)$ in a cubic system follows the
closed-form $d_{hkl}=a_0/\sqrt{h^2+k^2+l^2}$.
Atomic radius from the BCC body-diagonal contact condition. The body diagonal has
length $a_0\sqrt3$ and is spanned by 4 atomic radii ($\tfrac14$ atom at each of two opposite corners,
one full atom at the body centre):
$$4r = a_0\sqrt3 \quad\Rightarrow\quad r = \frac{a_0\sqrt3}{4} = \frac{3.1468\times1.7321}{4}$$
$$\boxed{r \approx 1.363\ \text{Å}}$$
(This matches molybdenum’s accepted metallic radius of $\approx1.36$ Å, confirming the
BCC assignment.)
Density. A BCC cell contains $n=8\times\tfrac18+1=2$ atoms, and its volume is
$V_C=a_0^3=(3.1468\times10^{-8}\ \text{cm})^3=3.116\times10^{-23}$ cm$^3$:
$$\rho=\frac{nA_{Mo}}{V_CN_A}=\frac{2\times95.94}{(3.116\times10^{-23})(0.602\times10^{24})}=\frac{191.88}{18.76}$$
$$\boxed{\rho \approx 10.23\ \text{g/cm}^3}$$
(Molybdenum’s handbook density is $10.22$ g/cm$^3$.)
$(112)$ plane — intercepts and check. Intercepts in units of $a_0$: reciprocal
of $(1,1,2)$ gives intercepts $(1,\,1,\,\tfrac12)$ — the plane cuts the full cell edge on $x$ and
$y$ and half the edge on $z$. This is the shaded triangle in the sketch below.
$[011]$ direction. Components $0,1,1$ along $a_0,b_0,c_0$ from the origin: the line
from $(0,0,0)$ to $(0,a_0,a_0)$, i.e. the face diagonal of the $y$–$z$ face — shown dashed.
