Question

Predict the type of cubic lattice of a solid element having edge length of 400 pm and density of 6.25 g/ml. 
(Atomic mass of element = 60)

Hint:

For cubic lattices: A simple cubic structure has ( Z = 1 ), a body-centered cubic (BCC) lattice has ( Z = 2 ), and a face-centered cubic (FCC) lattice has ( Z = 4 ).

Answer 1

The type of cubic lattice can be identified using the following formula: \[ Z = \frac{\rho N_A a^3}{M} \] where: - \( Z \) = Number of atoms per unit cell, 
- \( \rho \) = Density \( (6.25 \, \text{g/cm}^3) \), 
- \( N_A \) = Avogadro’s number \( (6.022 \times 10^{23} \, \text{mol}^{-1}) \), 
- \( a \) = Edge length \( (400 \, \text{pm} = 4.0 \times 10^{-8} \, \text{cm}) \), 
- \( M \) = Atomic mass \( (60 \, \text{g/mol}) \). 
Substituting the values into the formula: 
\[ Z = \frac{(6.25) \times (6.022 \times 10^{23}) \times (4.0 \times 10^{-8})^3}{60} \] After solving, we obtain \( Z = 4 \), which corresponds to a Face-Centered Cubic (FCC) lattice.