Question

An element has a cubic structure with a cell edge of 288 pm. The density of the element is $7.2~g~cm^{-3}$. 208 g of the element has $24.16\times10^{23}$ numbers of atoms. The unit cell of this cubic structure is ________.

• A. primitive
• B. body-centered
• C. face-centered
• D. hexagonal

Hint:

Z values: Primitive (1), BCC (2), FCC (4).

Answer 1

The Correct Option is B

Step 1: Concept
Use the density formula: $\rho = \frac{Z \times M}{N_A \times a^3}$ or find the number of unit cells.
Step 2: Analysis
Volume of unit cell ($a^3$) = $(288 \times 10^{-10}~cm)^3 = 2.39 \times 10^{-23}~cm^3$. Volume of 208 g of element = $\frac{Mass}{Density} = \frac{208~g}{7.2~g~cm^{-3}} = 28.88~cm^3$.
Step 3: Calculation
Number of unit cells = $\frac{Total~Volume}{Volume~of~one~unit~cell} = \frac{28.88}{2.39 \times 10^{-23}} \approx 12.08 \times 10^{23}$. Number of atoms per unit cell ($Z$) = $\frac{Total~Atoms}{Total~Unit~Cells} = \frac{24.16 \times 10^{23}}{12.08 \times 10^{23}} = 2$.
Step 4: Conclusion
A cubic structure with $Z = 2$ is body-centered cubic (BCC).
Final Answer:(B)