In the structure of diamond, carbon atoms appears at

Question

In the structure of diamond, carbon atoms appears at (A) 0,0,0, and (1/2), (1/2), (1/2) (B) (1/4), (1/4), (1/4), and (1/2), (1/2), (1/2) (C) 0,0,0, and (1/4), (1/4), (1/4) (D) 0,0,0, and (3/4), (3/4), (3/4)

Tardigrade Admin · Accepted Answer

The space lattice of diamond is fcc. The primitive basis has two identical atoms at 0,0,0 and (1/4), (1/4), (1/4) associated with each point of the fcc lattice as shown in the figure

Q. In the structure of diamond, carbon atoms appears at

$0, 0, 0$ , and $\frac{1}{2}, \frac{1}{2}, \frac{1}{2}$

$\frac{1}{4}, \frac{1}{4}, \frac{1}{4}$ , and $\frac{1}{2}, \frac{1}{2}, \frac{1}{2}$

$0, 0, 0$ , and $\frac{1}{4}, \frac{1}{4}, \frac{1}{4}$

$0, 0, 0$ , and $\frac{3}{4}, \frac{3}{4}, \frac{3}{4}$

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

Q. In the structure of diamond, carbon atoms appears at

0,0,0, and 21​,21​,21​

41​,41​,41​, and 21​,21​,21​

0,0,0, and 41​,41​,41​

0,0,0, and 43​,43​,43​

The space lattice of diamond is fcc. The primitive basis has two identical atoms at 0,0,0 and 41​,41​,41​ associated with each point of the fcc lattice as shown in the figure

$0, 0, 0$ , and $\frac{1}{2}, \frac{1}{2}, \frac{1}{2}$

$\frac{1}{4}, \frac{1}{4}, \frac{1}{4}$ , and $\frac{1}{2}, \frac{1}{2}, \frac{1}{2}$

$0, 0, 0$ , and $\frac{1}{4}, \frac{1}{4}, \frac{1}{4}$

$0, 0, 0$ , and $\frac{3}{4}, \frac{3}{4}, \frac{3}{4}$

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