Question 2 of 8: Silver — FCC Density, Atomic Radius, and Planar Packing
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-11, Properties of Materials — December 2013. 3 hours,
closed-book examination (approved Casio or Sharp calculator only). Any five questions constitute
a complete paper; only the first five questions as they appear in the answer book are marked. All
eight questions are solved below for completeness.
Silver's molar mass ($M_{Ag}=107.87$ g/mol) is not listed in this paper's atomic-mass table
(page 1 lists only H, Be, C, N, O, F, Al, Si, Cl, Fe, Ni, Cu) — the standard periodic-table
value is used below, as the question cannot be solved without it.
Find. (a) Density $\rho$ and atomic radius $r$. (b) Planar density and packing
fraction on (100), (110), (111); identify the close-packed plane.
Approach
For an FCC unit cell, the atomic radius follows from the face-diagonal contact condition
($4r=a\sqrt2$), and density follows from the mass and volume of one unit cell (4 atoms/cell).
Planar density and packing fraction on each low-index plane follow from counting the atoms whose
centres lie in that plane within one repeat area, and comparing the area actually covered by
atoms to the total plane area.
(a) Atomic radius from the FCC face-diagonal contact. Atoms touch along the
face diagonal, so $4r=a\sqrt2$:
$$r=\frac{a\sqrt2}{4}=\frac{0.4073\sqrt2}{4}=\boxed{0.1440\ \text{nm}}\ (144.0\ \text{pm}).$$
Density. An FCC cell contains $n=4$ atoms; with $a=4.073\times10^{-8}$ cm,
$$\rho=\frac{n\,M_{Ag}}{N_A\,a^3}=\frac{4\times107.87}{6.02\times10^{23}\times(4.073\times10^{-8})^3}=\boxed{10.61\ \text{g/cm}^3}.$$
(b) (100) plane. The cube face contains $4\times\tfrac14$ corner atoms $+\,1$
face-centre atom $=2$ atoms over area $a^2$:
$$\text{PD}_{100}=\frac{2}{a^2}=12.06\ \text{atoms/nm}^2,\qquad
\text{PF}_{100}=\text{PD}_{100}\times\pi r^2=\boxed{0.785}\ (=\pi/4).$$
(110) plane. The diagonal rectangular face ($a\times a\sqrt2$) contains 2
atoms:
$$\text{PD}_{110}=\frac{2}{a^2\sqrt2}=8.52\ \text{atoms/nm}^2,\qquad
\text{PF}_{110}=\boxed{0.555}\ (=\pi\sqrt2/8).$$
(111) plane. The close-packed triangular plane has nearest-neighbour spacing
$d=a/\sqrt2$ and 2-D hexagonal packing (area/atom $=\tfrac{\sqrt3}{2}d^2$):
$$\text{PD}_{111}=\frac{4}{a^2\sqrt3}=13.92\ \text{atoms/nm}^2,\qquad
\text{PF}_{111}=\boxed{0.907}\ (=\pi/(2\sqrt3)).$$
Since $0.907$ is the maximum possible 2-D circle-packing fraction, (111) is the close-packed
plane (the other two are not: (100) $=0.785$, (110) $=0.555$).
