10.6 Lattice structures in crystalline solids  (Page 4/29)

Calculation of atomic radius and density for metals, part 2

(a) What is the atomic radius of Ca in this structure?

(b) Calculate the density of Ca.

Solution

(a) In an FCC structure, Ca atoms contact each other across the diagonal of the face, so the length of the diagonal is equal to four Ca atomic radii (d = 4 r ). Two adjacent edges and the diagonal of the face form a right triangle, with the length of each side equal to 558.8 pm and the length of the hypotenuse equal to four Ca atomic radii:

Solving this gives $r=\sqrt{\frac{{\left(558.8\text{pm}\right)}^2+{\left(558.5\text{pm}\right)}^2}{16}}=\text{197.6 pmg for a Ca radius}.$

(b) Density is given by $\text{density}=\frac{\text{mass}}{\text{volume}}.$ The density of calcium can be found by determining the density of its unit cell: for example, the mass contained within a unit cell divided by the volume of the unit cell. A face-centered Ca unit cell has one-eighth of an atom at each of the eight corners $(8\times \frac{1}{8}=1$ atom) and one-half of an atom on each of the six faces $6\times \frac{1}{2}=3$ atoms), for a total of four atoms in the unit cell.

The mass of the unit cell can be found by:

The volume of a Ca unit cell can be found by:

(Note that the edge length was converted from pm to cm to get the usual volume units for density.)

Then, the density of $\text{Po}=\frac{2.662\times {10}^{\mathrm{-22}}\text{g}}{1.745\times {10}^{\mathrm{-22}}{\text{cm}}^3}={\text{1.53 g/cm}}^3$

Check your learning

(a) What is the atomic radius of Ag in this structure?

(b) Calculate the density of Ag.