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 calculate the density of a face-centered cubic (FCC) crystal structure, follow these steps:
Step 1: Determine the number of atoms within an FCC unit cell. An FCC structure has 4 atoms per unit cell. This is derived from corner atoms shared by 8 cells and face-centered atoms shared by 2 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. This is calculated by multiplying the number of atoms per unit cell by the atomic mass \( M \): \[ \text{Mass of unit cell} = 4M. \] 
Step 3: Calculate the volume of the unit cell. The volume \( V \) of a cubic unit cell is the cube of its edge length \( a \): \[ V = a^3. \]
Step 4: Formulate the density equation. Density \( \rho \) is defined as mass divided by volume. Substituting the unit cell's mass and volume yields: \[ \rho = \frac{\text{Mass of unit cell}}{\text{Volume of unit cell}} = \frac{4M}{a^3}. \] 

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