# Packing factor

A packing factor of 100% means that the particles occupy the entire unit cell volume. Physically it is impossible for this to happen, because it would imply, for example, that the atoms deform their radii and dissolve as if they were an “electronic liquid”. The geometry of the atoms, spherical for convenience, always gives rise to empty spaces during packing.

In the definition of the packing factor, it is assumed that the atoms consist of rigid spheres, like the balls of gum or candy from a dispensing machine (image above). Between the spheres there will always be hollow spaces through which smaller spheres (impurities or additives) can be cast.

If we increase the packing factor, the spheres will become squeezed, making the crystal more compact and dense; or on the other hand, more deformable, as happens with malleable and ductile metals.

The packing factor applies to any type of glass. However, its calculation can be a bit tedious, so it will only be considered here for atomic crystals with simple structures.

__Packing factor formula__

The packing factor is usually expressed as percentages. For example, if its value is 40%, it means that the particles only occupy 40% of the total space of the unit cell; or what is the same as affirming that 60% of the glass is “empty”.

The above said clarifies what is the formula to calculate this factor:

- FEA = (Volume of atoms) / (Unit cell volume)

Where FEA stands *for Atomic Packing Factor* , which are the simplest crystals.

The volume of the unit cell depends on its parameters (such as the length of its sides), with which we proceed by simple geometry to calculate its volume. The atoms, on the other hand, are the ones that define said cell, so it is possible to express its dimensions from the atomic radii, as will be seen in the next sections.

With regard to the volume of the atoms, the total number of them that is present in the unit cell (1, 2, 3, etc.) must be considered, as well as their spherical geometry. So the formula is modified a bit:

FEA = (No. Atoms) (Atom Volume) / (Unit Cell Volume)

To calculate FEA, it is necessary to determine No., V _{atom} and V _{unit cell} .

__Simple cubic__

__Simple cubic__

The simplest unit cell of all is the simple cubic one. In it we have some portions of atoms in each of the corners. If we see, we will notice that the length *a* of this cell is equal to 2r, since it is the atoms that define the cell. So the volume of the unit cell will be equal to:

V _{unit cell} = *a *^{3} (volume of a cube)

= (2r) ^{3}

= 8r ^{3}

Meanwhile, the volume of the atom will be equal to:

V _{atom} = (4/3) πr ^{3} (volume of a sphere)

Each of the corners is shared by 8 other neighboring unit cells. Therefore, we have a fraction of 1/8 in each corner, and since there are 8 of them, we do not equal 1 atom per unit cell (1/8 x 8 = 1).

The packing factor is:

FEA = (1) (4/3) πr ^{3} / 8r ^{3}

= π / 6 ≈ 52%

That is, in a simple cubic cell the atoms occupy 52% of the entire volume of the crystal.

__Body-centered cubic__

__Body-centered cubic__

**Determination of cell volume**

Now let’s look at the body-centered cubic cell. Side *a* can no longer be equal to 2r, because we have an empty space between the two corner atoms. Therefore, we must consider a diagonal *d* equal to 4r (green color) that crosses the cell through the center and touches the opposite corners, and another diagonal *d* of the face (black color).

Sides *a* , *d,* and 4r draw a right triangle to which we can apply trigonometry to calculate the value of *a* :

(4r) ^{2} = d ^{2} + a ^{2}

But on the other hand, at the base of the unit cell we have another triangle ( *a* , *a* and *d* ) to which we can calculate the hypotenuse:

d ^{2 }= a ^{2} + a ^{2}

= 2a ^{2}

Substituting then we will have:

(4r) ^{2} = (2a ^{2} ) + a ^{2}

(4r) ^{2} = 3a ^{2}

a = (4 / √3) r

The _{unit cell} V is equal to:

V _{unit cell} = a ^{3}

= ((4 / √3) r) ^{3}

**Determination of the packing factor**

Note that regarding the number of atoms we have 1 atom within this cell, following the same deduction made for the simple cubic cell, and another additional atom that is located in the center of the cell. Thus, there are a total of 2 atoms for each cubic cell centered in the body.

The packing factor then becomes:

FEA = (2) (4/3) πr ^{3} / ((4 / √3) r) ^{3}

= (√3 / 8) π ≈ 68%

In other words, in a cubic cell centered in the body, 68% of the volume of the crystal is occupied by atoms. Consequently, this crystalline arrangement is more compact (or dense) than the simple cubic one.

__Face-centered cubic__

__Face-centered cubic__

**Determination of cell volume**

Let’s now look at the face-centered cubic unit cell, very common in inorganic salts and some metals, such as gold and silver. To determine its packing factor we must start, following the previous examples, by finding what the volume of its unit cell is. It is therefore necessary to calculate again the side *a* and thus the volume of the cube *a *^{3} .

This time the procedure is easier and more direct, since we have a frontal diagonal *d* that, together with the sides *a* , form a right triangle to which we can apply trigonometry:

d ^{2} = a ^{2} + a ^{2}

= 2a ^{2}

Solving *for a* we will have:

*a* = d / √2

But, visually we see that *d* is equal to 4r, so we make a substitution:

*a* = 4r / √2

= 2r 2 ^{1-1 / 2}

= (2√2) r

Being V _{unit cell} equal to:

*a *^{3} = ((2√2) r) ^{3}

= (16√2) r ^{3}

Regarding the number of atoms per cell, we again have eight portions of an atom in each corner, and also one half of an atom for each of the six faces, which is shared by another neighboring cell. Therefore, the number of atoms is equal to:

No. atoms = 1/8 (8) + 1/2 (6) = 1 + 3 = 4

**Determination of the packing factor**

Given that there are 4 atoms in each face-centered cubic cell, and also its volume, which is equal to (16√2) r ^{3} , we can then calculate the packing factor:

FEA = (No. Atoms) (Atom Volume) / (Unit Cell Volume)

= (4) (4/3) πr ^{3} / (16√2) r ^{3}

= π / (3√2) ≈ 74%

Note that this cell is even more compact than the previous ones: 74% of the total volume of the cell is occupied by atoms. In a perfect and pure crystal, this would be equivalent to saying that 26% of its volume is available to accommodate other host atoms.

__Compact hex__

__Compact hex__

Finally, among the simplest and most compact unit cells we have the compact hexagonal. Unlike the previous ones, the calculation of its volume is a bit more cumbersome. As can be seen, it is not cubic, so it has two parameters *a* and *c* , the latter being the height of the cell.

**Determination of cell height**

The hexagonal cell can be decomposed into three rhombohedral cells, and from one of them we proceed to calculate *a* and *c* . Side *a* , although not so obvious in the image above, is equal to 2r. To calculate *c* , instead, we use the triangle and the red dot product of the internal triangular prism of the same cell.

We need to calculate the distance *d* in order to determine what *c is* worth . On the floor the red triangle is equilateral, with an angle of 60º. But if we consider another internal right triangle with sides *a* / 2 and *d* , and an angle of 30º (half), then by trigonometry we can determine *d* :

Cos (30º) = ( *a* / 2) / *d*

*d* = *a* / √3

And now we consider the right triangle composed of the sides *c* / 2 (green), *a* (black) and *d* (dotted):

*a *^{2} = ( *a* / √3) ^{2} + (c / 2) ^{2}

Solving for *c* we would have:

*c* = √ (8/3) *a*

And replacing *a* by 2r:

*c* = √ (8/3) (2r)

* *= √ (4 · 2/3) (2r)

= 4√ (2/3) r

**Determination of cell volume**

To determine the volume of the hexagonal cell, multiply the area of the hexagon by its height. Knowing that equilateral triangles have sides *a* , their height *h* is calculated . Thus, finding the area of a triangle, which in this case is √3 / 4 *to *^{2} , we multiply this value by 6 to obtain the area of the hexagon: 3 (√3 / 2) *to *^{2}

The volume is therefore:

V _{unit cell} = Hexagon Area x Height

= 3 (√3 / 2) *a *^{2} x 4√ (2/3) r

And once again replacing *a* by 2r:

V _{unit cell }= 3 (√3 / 2) (2r) ^{2} x 4√ (2/3) r

= 24√2 r ^{3}

**Determination of the packing factor**

In the hexagonal cell there are 12 atoms in the corners, which have 1/6 of their volumes inside. Likewise, there are 3 internal atoms whose volumes are complete, and another 2 atoms on the upper and lower faces whose half of their volumes are inside the cell.

Therefore, the number of atoms is equal to:

No. atom = 1/6 (12) + 1 (3) + 1/2 (2) = 6 atoms

And the packing factor finally becomes:

FEA = (No. Atoms) (Atom Volume) / (Unit Cell Volume)

= (6) (4/3) πr ^{3} / 24√2 r ^{3}

= π / (3√2) ≈ 74%

Note that the compartment factor for the hexagonal cell is the same as for the face-centered cubic cell. That is, both are equally compact.