Question 1 of 7: Silver Crystal Structure, Density, and Planar Packing

Question 1: Silver Crystal Structure, Density, and Planar Packing (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. Silver (Ag) is face-centered cubic (FCC) with lattice constant $a=0.4073$ nm. Atomic mass $M_{Ag}=107.87$ g/mol (page-1 table); Avogadro's number $N_A=0.602\times10^{24}$ mol$^{-1}=6.02\times10^{23}$ mol$^{-1}$ (page-1 constants table, same value with the exponent split differently).

Find. (a) Theoretical density $\rho$ and atomic radius $R$ of Ag. (b) Planar density and 2-D packing fraction on $(100)$, $(110)$, $(111)$; identify the close-packed plane.

Approach

The FCC unit cell holds $Z=4$ atoms, which fixes both the density (mass/cell ÷ cell volume) and the atomic radius (from the face-diagonal touching condition). For the planar questions, each $(hkl)$ plane is intersected with the cell to find the 2-D repeat area and the number of atom centres it contains; dividing atom count by area gives the planar density, and multiplying by the atomic cross-section $\pi R^2$ gives the packing fraction.

1. (a) Atomic radius from the FCC touching condition. In FCC, atoms touch along the face diagonal, so $4R=a\sqrt2$: $$R=\frac{a\sqrt2}{4}=\frac{(0.4073)(1.41421)}{4}$$ $$\boxed{R\approx0.1440\ \text{nm}}$$ This matches the accepted metallic radius of silver almost exactly, a good check on the method.
2. (a) Theoretical density. Converting the lattice constant to cm, $a=4.073\times10^{-8}$ cm, so $a^3=6.757\times10^{-23}$ cm$^3$: $$\rho=\frac{ZM_{Ag}}{N_Aa^3}=\frac{4(107.87)}{(6.02\times10^{23})(6.757\times10^{-23}\,\text{cm}^3)}$$ $$\boxed{\rho\approx10.61\ \text{g/cm}^3}$$ (handbook silver is $10.49$ g/cm$^3$ — the small excess reflects the idealized hard-sphere/rigid-lattice model used here).
3. (b) $(100)$ plane. The square face of the cell (side $a$) contains 4 corner atoms ($\tfrac14$ each, in-plane) $+$ 1 face-centred atom lying exactly on that face $=2$ atoms per $a^2$ of area: $$\text{PD}_{100}=\frac{2}{a^2}=\frac{2}{(0.4073)^2}\approx\boxed{12.06\ \text{atoms/nm}^2}$$ The packing fraction is atom count $\times\,\pi R^2$ over the same area; because $4R=a\sqrt2$, this reduces to a pure number independent of $a$: $$\text{PF}_{100}=\frac{2\pi R^2}{a^2}=\frac{\pi}{4}\approx\boxed{0.7854}$$
4. (b) $(110)$ plane. The rectangular section through the cell (sides $a$ and $a\sqrt2$, area $a^2\sqrt2$) contains 4 corner atoms ($\tfrac14$ each $=1$), 2 long-edge face-centred atoms ($\tfrac12$ each $=1$), for 2 atoms total: $$\text{PD}_{110}=\frac{2}{a^2\sqrt2}\approx\boxed{8.52\ \text{atoms/nm}^2}$$ $$\text{PF}_{110}=\frac{2\pi R^2}{a^2\sqrt2}=\frac{\pi}{4\sqrt2}\approx\boxed{0.5554}$$
5. (b) $(111)$ plane. The equilateral-triangle section through the cell (area $\tfrac{\sqrt3}{4}a^2$ per 1 corner-shared + 3 half-edge atoms, reducing to 2 atoms per $\tfrac{a^2\sqrt3}{2}$) gives: $$\text{PD}_{111}=\frac{4}{a^2\sqrt3}\approx\boxed{13.92\ \text{atoms/nm}^2}$$ $$\text{PF}_{111}=\frac{4\pi R^2}{a^2\sqrt3}=\frac{\pi}{2\sqrt3}\approx\boxed{0.9069}$$ Since $0.9069>0.7854>0.5554$, the $(111)$ plane is the most densely packed — the classic FCC result: $\{111\}$ planes are the close-packed planes (stacked ABCABC…, the cubic close-packed sequence), while $(110)$ is the most open of the three low-index planes.