Question

Each edge of a cubic unit cell is 400 pm long. If the atomic mass of the element is 120 and its density is 6.25 g cm⁻³, the crystal lattice is:
(Use \(N_A = 6 \times 10^{23}\) mol⁻¹)

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

Hint:

BCC unit cell contains 2 atoms per unit cell.

Answer 1

The Correct Option is B

Step 1: Density formula:
\[ \rho = \frac{Z M}{a^3 N_A} \]
Step 2: Substitute values:
\[ 6.25 = \frac{Z \times 120}{(4 \times 10^{-8})^3 \times 6 \times 10^{23}} \quad \Rightarrow \quad Z = 2 \]
Step 3: Since \(Z = 2\), the crystal lattice is **BCC**.