Question

Titanium metal has a body-centered cubic unit cell. The density of titanium is 4.50 g/cm³. Calculate the edge length of the unit cell and a value for the atomic radius of titanium. (Hint:In a body-centered arrangement of spheres, the spheres touch along the body diagonal.)

Short answer

The edge length of the unit cell is 328 pm, and the atomic radius of titanium is 142 pm.

Step-by-step solution

Body-centered unit cells

For a body-centered unit cell = 8 Corners × Corner

$\left(\frac{1}{8}\times 8\right)\mathrm{Ti}$+ 1 Ti at body center = 2 Ti atoms

All body-centered unit cells have two atoms per unit cell.

Finding the edge length of the unit cell

For a unit cell:

$$\begin{gathered} \mathrm{Density}=\frac{2\times 47.88\mathrm{gm}/\mathrm{mole}}{6.023\times {10}^{23}\mathrm{atoms}/\mathrm{mole}\times {\mathrm{L}}^3} \\ 4.50\mathrm{gm}/{\mathrm{cm}}^3=\frac{2\times 47.88\mathrm{gm}/\mathrm{mole}}{6.023\times {10}^{23}\mathrm{atoms}/\mathrm{mole}\times {\mathrm{L}}^3} \\ {\mathrm{L}}^3=\frac{2\times 47.88\mathrm{gm}/\mathrm{mole}}{6.023\times {10}^{23}\mathrm{atoms}/\mathrm{mole}\times 4.50\mathrm{gm}/{\mathrm{cm}}^3} \\ \mathrm{L}=3.28\times {10}^{-8}\mathrm{cm}\left(\frac{1\mathrm{pm}}{{10}^{-10}\mathrm{cm}}\right) \\ \mathrm{L}=328\mathrm{pm} \end{gathered}$$

Finding the radius of the atomic unit cell

For a body-centered cubic unit cell, the radius of the atom is related to the cube edge length by,

$$\begin{gathered} 4\mathrm{r}=\mathrm{L}\sqrt{3} \\ \mathrm{r}=\frac{\mathrm{L}\sqrt{3}}{4} \\ \mathrm{r}=\frac{328\mathrm{pm}\times \sqrt{3}}{4} \\ \mathrm{r}=142\mathrm{pm} \end{gathered}$$

The atomic radius of titanium is 142 pm.