# 10.6 Lattice structures in crystalline solids  (Page 2/29)

 Page 2 / 29

In a simple cubic lattice, the unit cell that repeats in all directions is a cube defined by the centers of eight atoms, as shown in [link] . Atoms at adjacent corners of this unit cell contact each other, so the edge length of this cell is equal to two atomic radii, or one atomic diameter. A cubic unit cell contains only the parts of these atoms that are within it. Since an atom at a corner of a simple cubic unit cell is contained by a total of eight unit cells, only one-eighth of that atom is within a specific unit cell. And since each simple cubic unit cell has one atom at each of its eight “corners,” there is $8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{1}{8}\phantom{\rule{0.2em}{0ex}}=1$ atom within one simple cubic unit cell.

## Calculation of atomic radius and density for metals, part 1

The edge length of the unit cell of alpha polonium is 336 pm.

(a) Determine the radius of a polonium atom.

(b) Determine the density of alpha polonium.

## Solution

Alpha polonium crystallizes in a simple cubic unit cell:

(a) Two adjacent Po atoms contact each other, so the edge length of this cell is equal to two Po atomic radii: l = 2 r . Therefore, the radius of Po is $r\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\text{l}}{2}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\text{336 pm}}{2}\phantom{\rule{0.2em}{0ex}}=\text{168 pm}.$

(b) Density is given by $\text{density}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\text{mass}}{\text{volume}}.$ The density of polonium can be found by determining the density of its unit cell (the mass contained within a unit cell divided by the volume of the unit cell). Since a Po unit cell contains one-eighth of a Po atom at each of its eight corners, a unit cell contains one Po atom.

The mass of a Po unit cell can be found by:

$\text{1 Po unit cell}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{1 Po atom}}{\text{1 Po unit cell}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{1 mol Po}}{6.022\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{23}\phantom{\rule{0.2em}{0ex}}\text{Po atoms}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{208.998\phantom{\rule{0.2em}{0ex}}\text{g}}{\text{1 mol Po}}\phantom{\rule{0.2em}{0ex}}=3.47\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-22}\phantom{\rule{0.2em}{0ex}}\text{g}$

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

$V={l}^{3}={\left(336\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}\phantom{\rule{0.2em}{0ex}}\text{cm}\right)}^{3}=3.79\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-23}\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{3}$

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

Therefore, the density of $\text{Po}=\phantom{\rule{0.2em}{0ex}}\frac{3.471\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-22}\phantom{\rule{0.2em}{0ex}}\text{g}}{3.79\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-23}\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{3}}\phantom{\rule{0.2em}{0ex}}={\text{9.16 g/cm}}^{3}$

The edge length of the unit cell for nickel is 0.3524 nm. The density of Ni is 8.90 g/cm 3 . Does nickel crystallize in a simple cubic structure? Explain.

No. If Ni was simple cubic, its density would be given by:
$\text{1 Ni atom}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{1 mol Ni}}{6.022\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{23}\phantom{\rule{0.2em}{0ex}}\text{Ni atoms}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{58.693\phantom{\rule{0.2em}{0ex}}\text{g}}{\text{1 mol Ni}}\phantom{\rule{0.2em}{0ex}}=9.746\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-23}\phantom{\rule{0.2em}{0ex}}\text{g}$
$V={l}^{3}={\left(3.524\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}\phantom{\rule{0.2em}{0ex}}\text{cm}\right)}^{3}=4.376\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-23}\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{3}$
Then the density of Ni would be $=\phantom{\rule{0.2em}{0ex}}\frac{9.746\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-23}\phantom{\rule{0.2em}{0ex}}\text{g}}{4.376\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-23}\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{3}}\phantom{\rule{0.2em}{0ex}}={\text{2.23 g/cm}}^{3}$
Since the actual density of Ni is not close to this, Ni does not form a simple cubic structure.

Most metal crystals are one of the four major types of unit cells. For now, we will focus on the three cubic unit cells: simple cubic (which we have already seen), body-centered cubic unit cell    , and face-centered cubic unit cell    —all of which are illustrated in [link] . (Note that there are actually seven different lattice systems, some of which have more than one type of lattice, for a total of 14 different types of unit cells. We leave the more complicated geometries for later in this module.)

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