Question

The cubic unit cell of a metal (molar mass = \(63.55\,g\,mol^{-1}\)) has an edge length of \(362\,pm\). Its density is \(8.92\,g\,cm^{-3}\). The type of unit cell is

• A. primitive
• B. face centred
• C. body centred
• D. end centred

Hint:

For cubic crystals: \(Z=1\) (simple), \(Z=2\) (BCC), \(Z=4\) (FCC). Calculate \(Z\) from density relation.

Answer 1

The Correct Option is B

Step 1: Use density formula for unit cell.
\[ \rho = \frac{Z \times M}{N_A \times a^3} \]
Where:
\(Z\) = number of atoms per unit cell
\(M = 63.55\,g/mol\)
\(a = 362\,pm = 362\times 10^{-10}\,cm = 3.62\times 10^{-8}\,cm\)
\(\rho = 8.92\,g/cm^3\)
Step 2: Substitute values.
\[ 8.92 = \frac{Z \times 63.55}{(6.022\times 10^{23})(3.62\times 10^{-8})^3} \]
Step 3: Compute \(a^3\).
\[ a^3 = (3.62\times 10^{-8})^3 \approx 4.74\times 10^{-23}\,cm^3 \]
Step 4: Solve for \(Z\).
\[ Z = \frac{8.92 \times 6.022\times 10^{23} \times 4.74\times 10^{-23}}{63.55} \]
\[ Z \approx \frac{8.92 \times 28.55}{63.55} \approx \frac{254.7}{63.55} \approx 4 \]
Step 5: Identify lattice type.
\[ Z=4 \Rightarrow \text{FCC (face centred cubic)} \]
Final Answer:
\[ \boxed{\text{Face centred cubic}} \]