Question

Density of a lithium atom is $0.53{\text{ g/c}}{{\text{m}}^3}$. The edge length of ${\text{Li}}$ is ${\text{3}}{\text{.5 {$A^0$}}}$. What is the number of lithium atoms in a unit cell? (Atomic mass of lithium is ${\text{6}}{\text{.94}}$)

Answer 1

Hint: The distance from the centre of a solid to its vertices is known as the edge length. The ratio of the mass of the unit cell and the volume of the unit cell is known as the density of the unit cell.

Formula Used: $\rho = \dfrac{{Z \times M}}{{{a^3} \times {N_0}}}$

Complete step by step answer:
The edge length of the unit cell is ${\text{3}}{\text{.5 {$A^0$}}}$.
But $1{\text{ {$A^0$}}} = 1 \times {10^{ - 8}}{\text{ cm}}$
Thus, the edge length of the unit cell is ${\text{3}}{\text{.5}} \times {\text{1}}{{\text{0}}^{ - 8}}{\text{ cm}}$.
Calculate the number of lithium atoms in a unit cell using the equation as follows:
$\rho = \dfrac{{Z \times M}}{{{a^3} \times {N_0}}}$
Where $\rho $ is the density of the unit cell,
Z is the number of atoms in the unit cell,
M is the atomic mass,
a is the edge length of the unit cell,
${N_0}$ is Avogadro's number.
Thus,
$Z = \dfrac{{\rho \times {a^3} \times {N_0}}}{M}$
Substitute $0.53{\text{ g/c}}{{\text{m}}^3}$ for the density of the unit cell, ${\text{3}}{\text{.5}} \times {\text{1}}{{\text{0}}^{ - 8}}{\text{ cm}}$ for the edge length of the unit cell, $6.023 \times {10^{23}}{\text{ mo}}{{\text{l}}^{ - 1}}$ for the Avogadro’s number, ${\text{6}}{\text{.94 g mo}}{{\text{l}}^{ - 1}}$ for the atomic mass. Thus,
$Z = \dfrac{{0.53{\text{ g/c}}{{\text{m}}^3} \times {{\left( {{\text{3}}{\text{.5}} \times {\text{1}}{{\text{0}}^{ - 8}}{\text{ cm}}} \right)}^3} \times 6.023 \times {{10}^{23}}{\text{ mo}}{{\text{l}}^{ - 1}}}}{{{\text{6}}{\text{.94 g mo}}{{\text{l}}^{ - 1}}}}$
Z = 1.97
$Z \approx 2$

Thus, the number of lithium atoms in a unit cell is 2.

Note: The number of atoms in a unit cell is directly proportional to the density and volume of the unit cell and inversely proportional to the atomic mass. The mass of the unit cell is equal to the product of the mass of each atom in the unit cell and the number of atoms in the unit cell.
A body centered cubic unit cell has a total of 2 atoms in its unit cell. One atom from the eight corners of the cube and one atom placed in the centre of the cube. Thus, the unit cell of lithium is a body centered cubic (BCC) unit cell.