Packing in crystals

The fact that atoms have a finite size, as determined by the electron distribution, means that atoms take up a finite ``volume'' in the crystal. When a substance crystallizes, the lattice vectors must adjust to accomodate the atoms in a minimum energy structure, which will be the preferred crystal structure for the particular substance. The particular crystal lattice that results allows us to determine an effective ``atomic radius'' for atoms in the crystal, which will be a measure of how much space they take up.

Consider, first, the simple cubic structure, which has atoms at the corners of a repeating lattice of cubes. The distance between nearest neighbors in such a crystal is simple the cubic edge length $L$. Thus, the effective atomic radius in a simple cubic structure is $L/2$.

In a bcc lattice, with atoms at ${\bf R}_1 = (0,0,0)$ and ${\bf R}_2 = (L/2,L/2,L/2)$, the nearest neighbor distance is $\vert{\bf R}_1-{\bf R}_2\vert$, which is $\sqrt{(L/2)^2 + (L/2)^2 + (L/2)^2} = L\sqrt{3}/2$. Thus, the effective atomic radius is $L\sqrt{3}/4$.

In an fcc lattice, with atoms at ${\bf R}_1 = (0,0,0)$, ${\bf R}_2 = (L/2,L/2,0)$, ${\bf R}_3 = (L/2,0,L/2)$ and ${\bf R}_4 = (0,L/2,L/2)$, the nearest neighbor distance can be determined from $\vert{\bf R}_1-{\bf R}_2\vert = \sqrt{(L/2)^2 + (L/2)^2} = L\sqrt{2}/2$. Hence, the effective atomic radius is $L\sqrt{2}/4$.

In order to have a measure of the amount of space taken up by the atoms in the unit cell, we introduce a quantity known as the packing fraction. This is defined to be

$${\rm Packing\ Fraction} = {{\rm Volume\ of\ the\ atoms\ in\ the\ cell} \over {\rm Volume\ of\ the\ unit\ cell}}$$

Let us determine the packing fractions for the simple cubic, bcc and fcc structures:

Simple Cubic: The atomic radius is $L/2$, and there is just one atom in the unit cell. Hence the packing fraction is

$${\rm Packing\ Fraction} = {{4 \over 3}\pi\left({L \over 2}\right)^3 \over L^3} = {\pi \over 6} \approx 0.524$$

bcc: There are two atoms in the unit cell, the atomic radius of each is $L\sqrt{3}/4$. Hence, the packing fraction is

$${\rm Packing\ Fraction} = {2{4 \over 3}\pi \left({L\sqrt{3}... ...r 4}\right)^3 \over L^3} = {\pi\sqrt{3} \over 8} \approx 0.68$$

fcc: There are four atoms in the unit cell, and the atomic radius of each is $L\sqrt{2}/4$. Hence, the packing fraction is

$${\rm Packing\ Fraction} = {4{4 \over 3}\pi\left({L\sqrt{2} \... ... 4}\right)^3 \over L^3} = {\pi \sqrt{2} \over 6} \approx 0.74$$

Hence, the fcc lattice involves the largest packing fraction, following by bcc, followed by simple cubic.