Solid State: Edge Cases and Subtle Traps (5)

Question

A face-centred cubic unit cell of an element has edge length $a = 400\text{ pm}$. Calculate the radius of the atom and the density given atomic mass $M = 60\text{ g/mol}$.

Solution — Step by Step

Why This Works

In FCC, atoms sit at the 8 corners (each shared by 8 cells, contributing $1/8$) and 6 face centres (each shared by 2 cells, contributing $1/2$). Total: $8 \times 1/8 + 6 \times 1/2 = 4$ atoms per unit cell.

Atoms touch along the face diagonal, not the edge — this is the key geometry trap. Edge contains only 1 corner atom and the touch is at the face centre, so the relation $4r = a\sqrt{2}$ comes from the face diagonal.

For BCC: $4r = a\sqrt{3}$ (body diagonal), $Z = 2$. For simple cubic: $2r = a$, $Z = 1$.

Alternative Method

Use packing efficiency. FCC has 74% packing. Volume of 4 atoms = $4 \times (4/3)\pi r^3$. This equals $0.74 \times a^3$. Solve for $r$ — same answer.

For density, use the alternative formula $\rho = (Z \times M)/(N_A \times a^3)$ but in SI units: $a = 4 \times 10^{-10}\text{ m}$, $a^3 = 6.4 \times 10^{-29}\text{ m}^3$. Use molar mass in kg/mol, divide by $N_A$ (per mol). Density comes out in kg/m³ — divide by 1000 to get g/cm³.

Common Mistake

Using $a = 2r$ (simple cubic relation) for FCC. The atoms touch along the face diagonal, not the edge — using the wrong relation gives a radius too large by a factor of $\sqrt{2}$.