Question

In the structure of diamond, carbon atoms appears at

• A. $0,0,0$, and $\frac{1}{2}, \frac{1}{2}, \frac{1}{2}$
• B. $\frac{1}{4}, \frac{1}{4}, \frac{1}{4}$, and $\frac{1}{2}, \frac{1}{2}, \frac{1}{2}$
• C. $0,0,0$, and $\frac{1}{4}, \frac{1}{4}, \frac{1}{4}$
• D. $0,0,0$, and $\frac{3}{4}, \frac{3}{4}, \frac{3}{4}$

Answer 1

Answer: (A)

The space lattice of diamond is fcc. The primitive basis has two identical atoms at $0,0,0$ and $\frac{1}{4}, \frac{1}{4}, \frac{1}{4}$ associated with each point of the fcc lattice as shown in the figure