Question

The density of a face-centered cubic (FCC) crystal is:

• A. \( \frac{4M}{\sqrt{2}a^3} \)
• B. \( \frac{4M}{a^3} \)
• C. \( \frac{6M}{a^3} \)
• D. \( \frac{2M}{a^3} \)

Hint:

Understanding the relationship between atomic mass and unit cell dimensions is crucial for calculating the density of crystalline materials, which has implications in material science and engineering.

Answer 1

The Correct Option is B

To determine the density of a face-centered cubic (FCC) crystal structure, we follow these steps: 
Step 1: Identify the number of atoms per unit cell in FCC. In an FCC structure, each unit cell contains 4 atoms. This is because each corner atom is shared among 8 adjacent unit cells and each face-centered atom is shared between 2 unit cells: \[ \text{Total atoms per unit cell} = \frac{8 \times \frac{1}{8} + 6 \times \frac{1}{2}}{1} = 1 + 3 = 4. \] 
Step 2: Express the mass of the unit cell. The mass of the unit cell is the product of the number of atoms per unit cell and the atomic mass \( M \): \[ \text{Mass of unit cell} = 4M. \] 
Step 3: Calculate the volume of the unit cell. The volume \( V \) of the cubic unit cell is given by the cube of the edge length \( a \): \[ V = a^3. \]
Step 4: Formulate the density equation. Density \( \rho \) is defined as mass per unit volume. Substituting the mass and volume of the unit cell: \[ \rho = \frac{\text{Mass of unit cell}}{\text{Volume of unit cell}} = \frac{4M}{a^3}. \] 

Conclusion: The density \( \rho \) of an FCC crystal structure is given by: \[ \rho = \frac{4M}{a^3}, \] where \( M \) is the atomic mass and \( a \) is the edge length of the unit cell.