Question

The unit cell of a metallic element of atomic mass $108{\text{g mol}}^{-1}$and density $10.5{\text{g/cm}}^3$ is a cube with edge length of $410\text{pm}$. The structure of the crystal lattice is:

Answer 1

Density of the unit cell,
$\rho =\frac{Z\times M}{N_A(a^3\times {10}^{-30})}{\text{g cm}}^{-3}$
where,
$Z=\text{No. of atoms in a unit cell}M=\text{Molar mass}{N}_A=\text{Avogadro number}a=\text{Edge length in pm}$

Here,
$M=108{\text{g mol}}^{-1},N_A=6.023\times {10}^{23}$
$a=410\text{pm}$
$\rho =10.5{\text{g/cm}}^3$
Substituting these values, we get
$Z=\frac{\rho \times N_A(a^3\times {10}^{-30})}{M}$
$Z=\frac{10.5\times 6.023\times {10}^{23}({410}^3\times {10}^{-30})}{108}$
$Z\approx 4=$ number of atoms per unit cell
So, The structure of the crystal lattice is fcc.