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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*a*^{3}

The volume of the unit cell is therefore

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:

Click

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

Faces: 6 × ½ = 3

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*

However, we want to have the r in terms of

Knowing the close-packed directions will help us find

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 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

\( \frac{\sqrt{2}}{24}\pi a^3 \) 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%**

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selection page

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*a*^{2}sin(120°) so the
volume of the unit cell is

*a*^{2}*c* sin(120°)

The area of the rhombus is

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

\(\sqrt {2}a^3 \)

\(c = \sqrt \frac{{8}}{3}a \)

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:

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

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

where

We want to have

Following one of the close-packed directions we find
that

2*r* = *a*

or

*r* = *a* / 2

2

or

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 \[ \frac{a^3 \pi}{3} \]

\[ V = \frac{a^3 \pi}{6} \] The volume occupied by the

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%

Click Back to return to structure
selection page

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*a*^{3}

The volume of the unit cell is therefore

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:

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

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.

where

We want to have

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 2*r*, so \[ \frac{\sqrt{3}}{2} a = 2r \] \[ r
= \frac{\sqrt{3}}{4} a \]

\( \frac{\sqrt{3}}{2} a\)

This is equal to 2

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

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%**

\[ \frac{\sqrt{3}}{8}\pi a^3 ÷ a^3 = 0.68 \]

Converting this to a percentage gives the packing efficiency of

Click Back to return to structure
selection page