Question

Face centred cubic crystal lattice of copper has density of $8.996\;{\rm{g}} \cdot {\rm{c}}{{\rm{m}}^{ - 3}}$. Calculate the volume of the unit cell.
Given molar mass of copper is $63.5\;{\rm{g}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}$ and Avogadro number ${N_A}$ is $6.022 \times {10^{23}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}$.

Answer 1

Hint: An unit cell that possesses atoms at the centre of all faces of the cube and at all the corners of the crystal lattice is termed as face centred unit cell.

Complete answer:
To calculate the volume of unit cell, we first derive the formula, that is, $V = \dfrac{{Z \times {\rm{Atomic}}\;{\rm{mass}}}}{{{N_A} \times d}}$, then we substitute all the given values to calculate the volume of unit cell.
We know that density is the ratio of mass and volume. Similarly, density of a unit cell can be calculated as follows:

$\begin{array}{c}{\rm{Density}}\;{\rm{of}}\;{\rm{unit}}\;{\rm{cell}}\left( d \right) = \dfrac{{{\rm{Mass}}\;{\rm{of}}\;{\rm{unit}}\;{\rm{cell}}\left( M \right)}}{{{\rm{Volume}}\;{\rm{of}}\;{\rm{unit}}\;{\rm{cell}}\left( V \right)}}\\d = \dfrac{M}{V}\end{array}$ …… (1)

Now, we take Z as the number of atoms in the unit cell and m is the mass of each atom in the unit cell. Then, mass of atoms in the unit cell would be product of number of atoms and mass of each atom, that is,
$M = Z \times m$ …… (2)
And we know that mass of each atom is equal to $m = \dfrac{{{\rm{Atomic}}\;{\rm{mass}}}}{{{\rm{Avogadro's}}\;{\rm{number}}\left( {{N_A}} \right)}}$.

Now, we substitute the value of m in equation (2).

$M = \dfrac{{Z \times {\rm{Atomic}}\;{\rm{mass}}}}{{{N_A}}}$

Now, we substitute the value of M in equation (1). So, equation (1) becomes,

$d = \dfrac{{Z \times {\rm{Atomic}}\;{\rm{mass}}}}{{{N_A} \times V}}$ …… (3)

Now, we have to find the number of atoms of the unit cell.
In face centred cubic unit cell,

(i) $8\;{\rm{corner}}\;{\rm{atoms}} \times \dfrac{1}{8}\;{\rm{atom}}\;{\rm{per}}\;{\rm{unit}}\;{\rm{cell}} = {\rm{8}} \times \dfrac{1}{8} = 1\;{\rm{atom}}$
(ii) $6\;{\rm{face}}\;{\rm{centred}}\;{\rm{atom}} \times \dfrac{1}{2}\;{\rm{atom}}\;{\rm{per}}\;{\rm{unit}}\;{\rm{cell = 6}} \times \dfrac{1}{2} = 3\;{\rm{atoms}}$

So, In FCC unit cells, the number of atoms is 4.

Now, we rearrange equation (3) to calculate the value of V.

$V = \dfrac{{Z \times {\rm{Atomic}}\;{\rm{mass}}}}{{{N_A} \times d}}$ ……. (4)

Now, we substitute all the values in equation (3). Number of atoms (Z) is equal to 4, density is equal to $8.996\;{\rm{g}} \cdot {\rm{c}}{{\rm{m}}^{ - 3}}$, ${N_A}$ is $6.022 \times {10^{23}}$ and atomic mass of copper is $63.5\;{\rm{g}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}$.

$\begin{array}{c}V = \dfrac{{Z \times {\rm{Atomic}}\;{\rm{mass}}}}{{{N_A} \times d}}\\ = \dfrac{{4 \times 63.5\;{\rm{g}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}}}{{6.022 \times {{10}^{23}}\;{\rm{mo}}{{\rm{l}}^{ - 1}} \times 8.996\;{\rm{g}} \cdot {\rm{c}}{{\rm{m}}^{ - 3}}}}\\ = \dfrac{{254}}{{54.17 \times {{10}^{23}}}}\;{\rm{c}}{{\rm{m}}^{\rm{3}}}\\ = 4.7 \times {10^{ - 23}}\;{\rm{c}}{{\rm{m}}^{\rm{3}}}\end{array}$

Hence, volume of unit cell is $4.7 \times {10^{ - 23}}\;{\rm{c}}{{\rm{m}}^{\rm{3}}}$.

Note: Students might confuse the number of atoms of different unit cells. In a cubic unit cell the number of atoms is only one, in the body centered cubic unit the number of atoms is 2 and in the face centered cubic unit, the number of atoms is four.