Question

The BCC iron, which has a lattice parameter of 0.2866 nm, has a density of

• A. \( 5.882 \, \text{g/cm}^3 \)
• B. \( 6.882 \, \text{g/cm}^3 \)
• C. \( 7.882 \, \text{g/cm}^3 \)
• D. \( 8.882 \, \text{g/cm}^3 \)

Hint:

For BCC structures, always use 2 atoms per unit cell in density calculations. BCC is less dense than FCC because it has a lower packing efficiency (68%), leaving more empty space in the unit cell.

Answer 1

The Correct Option is C

The density for a Body-Centered Cubic (BCC) structure is calculated as:
\[\rho = \frac{\text{Number of atoms per unit cell} \times \text{Atomic weight}}{\text{Volume of the unit cell} \times N_A}\]
A BCC unit cell contains 2 atoms. Using the provided values:
\[\rho = \frac{2 \times 55.85}{(0.2866 \, \text{nm})^3 \times 10^{-21} \, \text{cm}^3 \times 6.022 \times 10^{23}} \approx 7.882 \, \text{g/cm}^3\]