Aluminium Has an FCC structure. Given that the atomic radius of an aluminum atom is 0.143nm, calculate the theoretical density of Al.

Solution:

Use

\rho=\frac{n\cdot A}{V_{c}\cdot N_{a}}

Where n is the number of atoms in the structure, A is the atomic mass, V_{c} represents the volume of the cube, and N_{a} is Avogadro’s number.

To do this question, it is important to remember the structure of a FCC crystal.

Using geometry, we can see that a=2R\sqrt{2}. To find the volume, we multiple height by length by width, which is equal to a^{3}

Now we simply plug the values in to get the theoretical density.

\rho=\frac{n\cdot A}{V_{c}\cdot N_{a}} = \frac{(4)\cdot(26.982)}{a^{3}\cdot(6.023\cdot10^{23})} (there are 4 atoms in the FCC crystal structure, thus n = 4)

We know that a=2R\sqrt{2} so V_{c}=[2R\sqrt{2}]^{3}

therefore \rho=\frac{n\cdot A}{V_{c}\cdot N_{a}} = \frac{(4)\cdot(26.982)}{(2\sqrt{2}\cdot0.143\cdot10^{-9})^{3}\cdot(6.023\cdot10^{23})} (note that the radius must be in meters so we convert nm to m by multiplying it by 10^{-9})