Dear Reader,

It is the return of the cubic unit cells ! Now before we get going on how to calculate the radius of a metal atom from the density of the solid we should look at how to make a sodium chloride unit cell. We start by drawing a cube which we are going to decorate with atoms.

Firstly we add the chlorines at the corners of the cube, we have a total of eight chlorines but it is important to note that all eight are shared between eight cells. So in fact we only have one chlorine in the cell.

Next we add the sodium atoms which are half way along the edges of the cube, as each sodium is shared between four cells we have 12 / 4 sodium atoms inside the cell. This works out as three sodium atoms. Here is the cell with the extra atoms added.

Next we add the chlorine atoms which are at the centre of the faces of the cube. As these atoms are shared between two cells we now have 6 / 2 = 3 chlorine atoms to add to the cell. Here is a picture of the cell with the new chlorine atoms added.

Finally we add the sodium which is right in the centre of the cell, as it is fully within the cell it is not shared with any other cells so it is 1 / 1 = 1 atom in the cell.

Now we should have four chlorines and four sodiums in the unit cell, this gives us a 1:1 ratio which is in agreement with the formula of sodium chloride (NaCl).

Now we should consider the density and radius of atoms in metals.

But back to unit cells, the most simple cubic phase for a metal is the primitive cubic lattice.

The distance along an edge of this cubic cell will be twice the radius of the atom, hence the volume will be given by

V = (2r)^{3} = 8r^{3}

The cell contains only one atom so the mass will be given by the equation

Mass = M / N_{a}

Combine the two equations to give us

ρ = M / 8 r^{3} N_{a}

Rearrange for r

ρ r^{3} = M / 8 N_{a}

r^{3} = M / ρ 8 N_{a}

r = (M / ρ 8 N_{a})^{ ⅓}

The body centred cubic plutonium has a distance long the edge of a cell which is 3.6375 Å and a density of 16.84

The distance between fractional coordinates 0 0 0 and 1 1 1 will be 4 times the radius (r) of the atoms.

Hence the distance along an edge of the cell will be 4r/3^{½}

Which means the volume of the cell will be given by

V = (4r/3^{½})^{3} = 64 r^{3} / (3^{1.5})

The cell contains two atoms so the mass of the atoms in the cell are given by

Mass = 2M / N_{a}

Combine the two equations to give

ρ = 2M 3^{1.5} / N_{a} 64 r^{3}

ρ = M 3^{1.5} / N_{a} 32 r^{3}

Now rearrange for r

ρ r^{3} = M 3^{1.5} / N_{a} 32

r^{3} = M 3^{1.5} / N_{a} ρ 32

r = (M 3^{1.5} / N_{a} ρ 32)^{ ⅓}

The delta plutonium is face centred cubic and has a distance along the cell edge of 4.6347 and a density of 16.28

The distance between fractional coordinates 0 0 0 and 1 1 0 will be four times the radius of the plutonium atoms. So the distance along the side of a cell will be

4r/(2^{½}) = r8^{½}

So the volume of the cell will be given by

[4r/(2^{½})]^{3} = V = 8^{1.5}r^{3}

As the cell contains four atoms the density of the solid will be given by

ρ = 4M / N_{a}8^{1.5}r^{3}

we can now rearrange it to give us the radius of the atoms

ρ r^{3 }= 4M / N_{a}8^{1.5}

r^{3 }= 4M / ρ N_{a}8^{1.5}

r = (4M / ρ N_{a}8^{1.5})^{⅓}

What I suggest that you might want to do is to calculate what the density would be of a primitive cubic plutonium (if it did exist).

Filed under: crystallography, crystals, cubic unit cells | Tagged: cubic unit cells |

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