Question

location of atoms in body diagonal plane FCC

Answer 1

The locations of atoms on a body diagonal plane (such as a {110} plane) in an FCC structure form a rectangular arrangement. This rectangle has dimensions $a\sqrt{2} \times a$, where $a$ is the lattice constant. Atoms are situated at the four corners of this rectangle and at the midpoints of the two longer sides (of length $a\sqrt{2}$).

Answer 2

A common interpretation of a "body diagonal plane" in an FCC structure is a {110} crystallographic plane, such as the plane defined by the equation $x - y = 0$. This plane contains a body diagonal, for instance, the one connecting the corner at (0,0,0) to the opposite corner at (a,a,a). In an FCC unit cell with lattice constant $a$, atoms are located at the corners and face centers. We identify the positions of these atoms that lie on the plane $x=y$. The corner positions on this plane within the unit cell are (0,0,0) and (a,a,a). The face center positions on this plane are (a/2, a/2, 0) and (a/2, a/2, a). When considering the periodic arrangement of atoms in the crystal, the atoms on this plane form a specific pattern. The plane contains a rectangle defined by the points (0,0,0), (a,a,0), (a,a,a), and (0,0,a). This rectangle has dimensions $a\sqrt{2}$ (along the [110] direction) and $a$ (along the [001] direction). The atoms on the plane are located at the four corners of this rectangle and at the midpoints of the two longer sides of length $a\sqrt{2}$, which correspond to the face center positions (a/2, a/2, 0) and (a/2, a/2, a).