BCC Molybdenum — Atomic Radius, Density, and Plane Spacing: 04-BS-11, December 2014, Q1

Question 1: BCC Molybdenum — Atomic Radius, Density, and Plane Spacing (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.

Given. BCC molybdenum, lattice parameter $a_0=3.1468$ Å; $M_{Mo}=95.94$ g/mol (page 1 table); $N_A=6.02\times10^{23}$ mol$^{-1}$.

Find. Atomic radius $R$, density $\rho$, sketch showing the (112) plane and [011] direction, and the (102) interplanar spacing.

Fig. Q1 — BCC unit cell with the (112) plane (intercepts $a,a,a/2$, shaded) and the [011] direction (corner-to-corner diagonal on the $y$–$z$ face) marked.

Approach

In a BCC cell the corner and body-centre atoms touch along the cube's body diagonal, which fixes the atomic radius directly from $a_0$; density then follows from the standard mass-per-cell over volume-per-cell relation using the BCC atom count of 2. The (102) spacing follows from the general cubic interplanar-spacing formula, which is purely geometric and applies regardless of whether that reflection is diffraction-allowed.

Atomic radius from the body-diagonal contact. In BCC, atoms touch along the cube's body diagonal: $4R=a_0\sqrt3$, so $$R=\frac{a_0\sqrt3}{4}=\frac{3.1468\times1.7321}{4}=\boxed{1.363\ \text{Å}}.$$
Density. A BCC unit cell contains $n=2$ atoms (8 corners$\times\tfrac18$ + 1 body-centre); with $a_0=3.1468\times10^{-8}$ cm and $M_{Mo}=95.94$ g/mol, $$\rho=\frac{nM_{Mo}}{N_Aa_0^3}=\frac{2\times95.94}{6.02\times10^{23}\times(3.1468\times10^{-8})^3} =\boxed{10.23\ \text{g/cm}^3}.$$ (The accepted density of molybdenum is $10.28\ \text{g/cm}^3$ — the close agreement confirms the method.)
Unit-cell sketch. The (112) plane has Miller intercepts $1/1,1/1,1/2$, i.e. it cuts the axes at $a,a,a/2$ — drawn above as the shaded triangle connecting those three intercept points. The [011] direction runs from the origin to the corner at $(0,a,a)$, the face-diagonal of the $y$–$z$ face — drawn above as the arrow.
(102) interplanar spacing. For a cubic lattice, $$d_{hkl}=\frac{a_0}{\sqrt{h^2+k^2+l^2}}\ \Rightarrow\ d_{102}=\frac{3.1468}{\sqrt{1^2+0^2+2^2}}=\frac{3.1468}{\sqrt5}=\boxed{1.407\ \text{Å}}.$$ This formula is a purely geometric statement about the crystal lattice and holds regardless of whether (102) is an allowed X-ray reflection.

Quantity	Result
Atomic radius, $R$	1.363 Å
Density, $\rho$	10.23 g/cm³
(102) interplanar spacing, $d_{102}$	1.407 Å

Concept & Theory

Topic. BCC crystal geometry: atomic radius from the body-diagonal contact condition, density from the unit-cell mass/volume relation, and interplanar spacing from Miller indices.

Key relations. BCC contact: $4R=a\sqrt3$; density $\rho=nM/(N_Aa^3)$ ($n=2$ for BCC); cubic interplanar spacing $d_{hkl}=a/\sqrt{h^2+k^2+l^2}$.

Why this works. The body diagonal is the shortest line in a BCC cell along which the corner and body-centre atoms are nearest neighbours, so it is the geometric constraint that must be used (not the cube edge, which is where FCC atoms touch instead); the interplanar spacing formula falls straight out of the plane's Miller-index intercepts on the cube edges.

Common pitfall. Using the FCC contact condition ($4R=a\sqrt2$, edge-touching) or the FCC atom count ($n=4$) instead of BCC's body-diagonal contact and $n=2$ — and separately, assuming a plane with $h+k+l$ odd (like (102), or (100)/(111)) must be excluded from spacing calculations because it is a systematically-absent BCC X-ray reflection; that selection rule affects diffracted intensity, not the plane's geometric existence or spacing.

Source. Callister & Rethwisch, Materials Science and Engineering, 9th ed., Ch. 3 (crystal structure, Miller indices, X-ray diffraction).

Question 1 of 7: BCC Molybdenum — Atomic Radius, Density, and Plane Spacing

Question 1: BCC Molybdenum — Atomic Radius, Density, and Plane Spacing (20 marks)

Approach