Question

Calculate the number of atoms present in 2 grams of a crystal which has Face Centred Cubic or FCC crystal lattice having edge length $\text{100 pm}$ and density 10 $\text{g c}{{\text{m}}^{\text{-3}}}$ .

Answer 1

In crystallography there are different types of crystal systems among which the most common one is the “cubic crystal system” which has three categories that include the simple cubic, body-centred cubic, face-centred cubic, and end-centred cubic. In the face-centred cubic crystal system there are a total four atoms in the unit cells.

Complete step by step solution:
Here it is given that the length of the edge is $\text{100 pm}$ = $1\times {{10}^{-8}}$ cm
The density of the Face Centred Cubic or FCC crystal lattice = 10 $\text{g c}{{\text{m}}^{\text{-3}}}$
The number of atoms in the unit cell is 4.
As per the formula, the density of the unit cell,
$\text{ }\rho\text{ =}\dfrac{\text{z }\times \text{M}}{{{\text{a}}^{\text{3}}}\times {{\text{N}}_{\text{A}}}}$
After rearranging we get, $\text{M =}\dfrac{\text{ }\rho\text{ }\times {{\text{a}}^{\text{3}}}\times {{\text{N}}_{\text{A}}}}{\text{z}}$
After putting the values of the values of the given parameters,
$\text{M =}\dfrac{10\times {{\left( 1\times {{10}^{-8}} \right)}^{\text{3}}}\times \left( 6.023\times {{10}^{23}} \right)}{4}$
$\text{M = 1}\text{.5 g/mol}$
Therefore, $\text{1}\text{.5 g}$ of the substance contains the Avogadro’s number of atoms = $\left( 6.023\times {{10}^{23}} \right)$ atoms of the substance.
So 2 grams of the substance contains $\dfrac{6.023\times {{10}^{23}}}{1.5}\times 2=8.03\times {{10}^{23}}$ atoms of the substance.
Hence the number of atoms present in 2 grams of a crystal which has Face Centred Cubic or FCC crystal lattice having edge length $\text{100 pm}$ and density 10 $\text{g c}{{\text{m}}^{\text{-3}}}$ is $8.03\times {{10}^{23}}$ .

Note:
In the face-centred cubic lattice, there are eight atoms at the eight corners of the cube and six atoms on the faces of the cube. Those atoms which at the corners are shared by eight other unit cells and hence each unit cell gets $\dfrac{1}{8}$ atom and those atoms which are located on the face are shared by another unit cell located beside it. Hence each unit cell gets three atoms from the faces. So in total each unit cell in an FCC lattice has four atoms.