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Packing efficiency
Pick a lattice type to see how its packing efficiency is calculated
Step 1: Calculate volume of unit cell
The unit cell is cubic with sides of length a (also known as the lattice parameter).

The volume of the unit cell is therefore a3
Step 2: Calculate the number of atoms per unit cell
Be aware that in this case the atoms are only located at the lattice points. If there were atoms located at other positions we would need to take them into account
Step 2: Calculate the number of atoms per unit cell

How many atoms are there in this unit cell?

Click Give up to see how to find the answer

Ans:

Step 2: Calculate the number of atoms per unit cell
Remember: Lattice points without a height given are considered to located at both the top and the bottom of the unit cell. Therefore they are counted twice
Corners: 8 × 1/8 = 1
Faces: 6 × ½ = 3
Total = 4
Step 3: Calculate the volume of the atoms
Each atom has a volume of $$\frac{4}{3} \pi r^{3}$$

However, we want to have the r in terms of a, the lattice parameter

Knowing the close-packed directions will help us find r in terms of a
The close packed directions for ccp are the <110> directions. These are the diagonals along the faces of the unit cell

The length across the diagonal is equal to 4r
(The spheres in the diagram represent the centres of the atoms rather than the atoms themself)

From the diagram we see that $r = \frac{\sqrt{2}}{4}a$
volume = $$\frac{4}{3}\pi r^3$$ therefore volume = $$\frac{4}{3}\pi \left ( \frac{\sqrt{2}}{4}a \right )^3 = \frac{\sqrt{2}}{24}\pi a^3$$

$$\frac{\sqrt{2}}{24}\pi a^3$$ is the volume of a single atom

There are 4 atoms per unit cell so

$$4 \times \frac{\sqrt{2}}{24}\pi a^3 = \frac{\sqrt{2}}{6}\pi a^3$$ is the volume occupied by the atoms
Step 4: Divide volume of atoms by volume of unit cell $\frac{\sqrt{2}}{6}\pi a^3 ÷ a^3= 0.74$ Converting this to a percentage gives the packing efficiency of ccp as 74%
Step 1: Calculate volume of unit cell
To find the volume of the unit cell, we consider the cell to be a rhombus of side a and angle 120°, which extends to a height c. (a and c are known as the lattice parameters).

The area of the rhombus is a2sin(120°) so the volume of the unit cell is
a2c sin(120°)
Step 1: Calculate volume of unit cell
In a hexagonally close-packed structure the lattice parameter c is related to a by
$$c = \sqrt \frac{{8}}{3}a$$
It shall be left as an exercise for the reader to show that this is true.
Thus the volume of the unit cell is
$$\sqrt {2}a^3$$
Step 2: Calculate the number of atoms per unit cell
Be aware that in this case the atoms are only located at the lattice points. If there were atoms located at other positions we would need to take them into account
Step 2: Calculate the number of atoms per unit cell

How many atoms are there in this unit cell?
Click Give up to see how to find the answer

Ans:

Step 2: Calculate the number of atoms per unit cell

Remember: lattice points without a height given are considered to be located at both the top and the botton of the unit cell. Therefore, they are counted twice.
Corners: 8 & 1/8 = 1

Inside: 1

Total: 2
Step 3: Calculate the volume of the atoms
Each atom has a volume of $$\frac{4}{3}\pi r^3$$

where r is the radius of the atoms

We want to have r in terms of a so that we can compare the volume of the unit cell to the volume occupied by the atoms
Following one of the close-packed directions we find that

2r = a

or

r = a / 2
So the volume of a single atom is $$V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \left ( \frac{a}{2} \right )^3$$

$V = \frac{a^3 \pi}{6}$ The volume occupied by the two atoms in the unit cell is $\frac{a^3 \pi}{3}$
Step 4: Divide volume of atoms by volume of unit cell $\frac{a^3 \pi}{3} ÷ \sqrt{2}a^3 = 0.74$ Converting this to a percentage gives the packing efficiency of hcp to be 74%
Step 1: Calculate volume of unit cell
The unit cell is cubic with sides of length a (also known as the lattice parameter).

The volume of the unit cell is therefore a3
Step 2: Calculate the number of atoms per unit cell
Be aware that in this case the atoms are only located at the lattice points. If there were atoms located at other positions we would need to take them into account
Step 2: Calculate the number of atoms per unit cell
How many atoms are there in this unit cell?
Click Give up to see how to find the answer

Ans:

Step 2: Calculate the number of atoms per unit cell

Remember: lattice points without a height given are considered to be located at both the top and the botton of the unit cell. Therefore, they are counted twice.
Corners: 8 & 1/8 = 1

Inside: 1

Total: 2
Step 3: Calculate the volume of the atoms
Each atom has a volume of $$\frac{4}{3} \pi r^3$$

where r is the radius of the atoms

We want to have r in terms of a so that we can compare the volume of the unit cell to the volume occupied by the atoms.
The directions in which the atoms are touching are the <111> directions.
$\frac{\sqrt{3}}{2} a$
From the diagram we can see that the distance between the centres of the two atoms is
$$\frac{\sqrt{3}}{2} a$$
This is equal to 2r, so $\frac{\sqrt{3}}{2} a = 2r$ $r = \frac{\sqrt{3}}{4} a$

The volume of a single atom is $$V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \left ( \frac{\sqrt{3}}{4}a \right )^3$$ $V = \frac{\sqrt{3}}{16}\pi a^3$
The volume of the two atoms in the unit cell is $\frac{\sqrt{3}}{8}\pi a^3$
Step 4: Divide volume of atoms by volume of unit dell

$\frac{\sqrt{3}}{8}\pi a^3 ÷ a^3 = 0.68$
Converting this to a percentage gives the packing efficiency of bcc to be 68%