# Question c9ca2

Dec 27, 2016

$\textsf{\rho = 1.06 \textcolor{w h i t e}{x} {\text{g/cm}}^{3}}$

#### Explanation:

From the data on the BCC structure I will work out the mass of 1 sodium atom.

Then I will work out the volume of the FCC unit cell, from which I will work out the density.

BCC refers to "body centred cubic". The unit cell is the smallest unit that, when repeatedly stacked together, will generate the entire structure.

A ball and stick model looks like this: The lines joining the atoms are not bonds in the usual sense. They represent points of contact. Not all the atoms lie inside the unit cell.

There is 1 atom in the centre and at each vertex there is 1/8 of an atom.

This means that the unit cell contains 1 + (8 x 1/8) = 2 atoms.

The side of the unit cell is of length a which we can now calculate.

This will enable us to find the volume of the unit cell which will be $\textsf{{a}^{3}}$.

The direction from a corner of a cube to the farthest corner is called the body diagonal (bd).

The face diagonal (fd) is a line drawn from one vertex to the opposite corner of the same face.

Imagine looking at the unit cell in the plane of the screen from the direction shown: You can see that a right - angled triangle is formed.

Using Pythagoras:

$\textsf{f {d}^{2} = {a}^{2} + {a}^{2} = 2 {a}^{2}} \text{ } \textcolor{red}{\left(1\right)}$

From the diagram we can say also that:

$\textsf{b {d}^{2} = f {d}^{2} + {a}^{2}} \text{ } \textcolor{red}{\left(2\right)}$

From $\textsf{\textcolor{red}{\left(1\right)}}$ we can substitute $\textsf{f {d}^{2}}$ so $\textsf{\textcolor{red}{\left(2\right)}}$ becomes:

$\textsf{b {d}^{2} = {a}^{2} + {a}^{2} + {a}^{2}}$

$\therefore$$\textsf{b {d}^{2} = 3 {a}^{2}}$

The atoms that lie along the body diagonal (bd) touch each other. From the geometry we can say:

$\textsf{b d = 4 R}$

Where R is the radius of the atom.

$\therefore$$\textsf{{\left(4 R\right)}^{2} = 3 {a}^{2}}$

$\therefore$$\textsf{4 R = \sqrt{3 {a}^{2}}}$

$\textsf{4 R = \sqrt{3} . a}$

$\therefore$sf(a=(4R)/(sqrt(3))#

Putting in the numbers:

$\textsf{a = \frac{4 \times 180.6 \times {10}^{- 12}}{1.732} = 417.09 \times {10}^{- 12} \textcolor{w h i t e}{x} m}$

$\textsf{a = 417.09 \textcolor{w h i t e}{x} \text{pm}}$

This is the length of the unit cell.

So the volume V of the unit cell is given by:

$\textsf{V = {a}^{3} = {\left(417.09 \times {10}^{- 12}\right)}^{3} = 7.2558 \times {10}^{- 29} \textcolor{w h i t e}{x} {m}^{3}}$

The mass m of the unit cell is given by:

$\textsf{m = \rho \times V}$

$\therefore$$\textsf{m = 0.971 \times {10}^{3} \times 7.255 \times {10}^{- 29} = 7.044 \times {10}^{- 26} \textcolor{w h i t e}{x} k g}$

We have already worked out that there are 2 atoms in the unit cell so the mass $\textsf{{m}_{N a}}$ of 1 atom of sodium is given by:

$\textsf{{m}_{N a} = \frac{7.044}{2} \times {10}^{- 26} = 3.522 \times {10}^{- 26} \textcolor{w h i t e}{x} k g}$

The unit cell for face centred cubic (FCC) looks like this: You can see that there is an atom at the centre of each face of the unit cell.

Each vertex contains 1/8 of an atom. Each face contains 1/2 of an atom.

So the total no. of atoms in the unit cell will be (8 x 1/8) + (6 x 1/2) = 1 + 3 = 4 atoms.

This can also be seen in the space filling model: In the diagram the face diagonal is labelled b.

We can use Pythagoras to find a

Since the atoms are touching we can say that:

$\textsf{b = r + 2 r + r = 4 r}$

Where r is the atomic radius.

From Pythagoras:

$\textsf{{a}^{2} + {a}^{2} = {\left(4 r\right)}^{2}}$

$\textsf{2 {a}^{2} = 16 {r}^{2}}$

$\textsf{{a}^{2} = 8 {r}^{2}}$

$\textsf{a = \sqrt{8} . r}$

$\therefore$$\textsf{a = \sqrt{8} \times 180.6 \times {10}^{- 12} = 510.81 \times {10}^{- 12} \textcolor{w h i t e}{x} m}$

$\therefore$$\textsf{{V}_{c e l l} = {\left(519.81 \times {10}^{- 12}\right)}^{3} = 1.3329 \times {10}^{- 28} \textcolor{w h i t e}{x} {m}^{3}}$

$\textsf{\rho = {m}_{c e l l} / {V}_{c e l l}}$

We have shown that there are 4 atoms in the unit cell so we can write:

$\textsf{\rho = \frac{4 \times 3.522 \times {10}^{- 26}}{1.3329 \times {10}^{- 28}} = 1.0569 \times {10}^{3} \textcolor{w h i t e}{x} {\text{kg/m}}^{3}}$

$\textsf{\rho = 1.06 \textcolor{w h i t e}{x} {\text{g/cm}}^{3}}$

You can see that there is a slight increase in density at this very high pressure as FCC is a more closely packed arrangement.