Question 3 of 7: Critical Radius Ratio for 4-Fold Coordination; FCC Lead Lattice Constant and Atomic Densities

Question 3: Critical Radius Ratio for 4-Fold Coordination; FCC Lead Lattice Constant and Atomic Densities (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. (b) Pb is FCC ($n=4$ atoms/cell); $M_{Pb}=207.2$ g/mol (page-1 table); $N_A=0.602\times10^{24}$ mol$^{-1}$ (page-1 constants); $\rho_{Pb}=11.36$ g/cm$^3$.

Find. (a) Derivation that the critical (minimum) $r_c/r_a$ for coordination number 4 is 0.225. (b) $a_0$ (nm), planar density on (110) (atoms/mm$^2$), linear density along [111] (atoms/mm).

Approach

(a) The critical ratio is the geometry at which the cation just touches all four surrounding anions and the anions just touch each other — below this ratio the anions would overlap before the cation could reach them, which is geometrically forbidden. The standard construction places four anions on alternating corners of a cube (they form a regular tetrahedron this way) with the cation at the cube centre. (b) is a direct density → lattice constant back-calculation, followed by standard FCC planar/linear density geometry.

1. (a) Set up the cube construction. Let the cube have edge length $a$. Anions sit on four alternating corners (a regular tetrahedron, edge = a cube face diagonal); the cation sits at the cube’s centre. At the critical ratio, adjacent anions touch along a face diagonal, and the cation touches every anion along half the body diagonal: $$2r_a=a\sqrt{2}\ \Rightarrow\ r_a=\frac{a}{\sqrt{2}},\qquad r_a+r_c=\frac{a\sqrt{3}}{2}$$
2. Solve for the ratio. $$r_c=\frac{a\sqrt3}{2}-\frac{a}{\sqrt2}=a\left(\frac{\sqrt3}{2}-\frac{1}{\sqrt2}\right) \ \Rightarrow\ \frac{r_c}{r_a}=\sqrt2\left(\frac{\sqrt3}{2}-\frac{1}{\sqrt2}\right) =\frac{\sqrt6}{2}-1$$ $$\boxed{\frac{r_c}{r_a}=\frac{\sqrt6}{2}-1\approx0.225}$$
3. (b) Lattice constant from density. For FCC, $n=4$ atoms per conventional cell: $$\rho=\frac{nM}{V_{cell}N_A}\ \Rightarrow\ V_{cell}=\frac{nM}{\rho N_A}=\frac{4(207.2)}{(11.36)(0.602\times10^{24})}=1.212\times10^{-22}\ \text{cm}^3$$ $$a_0=V_{cell}^{1/3}=\boxed{0.495\ \text{nm}}$$ (matches the well-documented real lattice constant of lead, $a_0\approx0.495$ nm — a strong sanity check).
4. Planar density on (110). The (110) plane cuts a rectangle of sides $a_0$ and $a_0\sqrt2$ through the FCC cell, containing 4 corner atoms (each $\tfrac14$ inside the rectangle) plus 1 atom lying fully inside it (a neighbouring cell’s face-centring atom that sits exactly in this plane) $=4(\tfrac14)+1=2$ atoms per $a_0^2\sqrt2$ of area. With $a_0=4.953\times10^{-7}$ mm: $$PD_{(110)}=\frac{2}{a_0^2\sqrt2}=\frac{2}{(4.953\times10^{-7})^2(1.414)} =\boxed{5.76\times10^{12}\ \text{atoms/mm}^2}$$
5. Linear density along [111]. In FCC, only the two corner atoms lie exactly on the body diagonal (unlike BCC, no atom sits at the diagonal’s midpoint), each shared $\tfrac12$ into this repeat length $a_0\sqrt3$: $$LD_{[111]}=\frac{1}{a_0\sqrt3}=\frac{1}{(4.953\times10^{-7})(1.732)} =\boxed{1.17\times10^{6}\ \text{atoms/mm}}$$