Question

An element has a bcc structure having unit cells $12.08 \times {10^{23}}$. Find the number of atoms in these cells.

Answer 1

Hint: the bcc structure is one of the type of unit cell in which the atoms are present at the corners of the cube as well as at the center of the structure. So each 8 corners of the cube share the 1/8th part of one atom and the whole one atom is present at the center which means the bcc structure has two atoms in it. The other unit cells are the primitive cubic unit cell and face centered cubic unit cell.

Complete answer:
The unit cell is defined as the repeating unit of the crystal lattice. These unit cells are identical in shape and form a sort of 3D arrangement of atoms, molecules or the ions which are present in the crystal lattice. So as we know that the bcc structure is made up of 2 atoms and we have to find the number of atoms for $12.08 \times {10^{23}}$ unit cells. So,
1 unit cell of bcc structure has = 2 atoms present in it so,
$12.08 \times {10^{23}}$ unit cells of bcc structure has = $2 \times 12.08 \times {10^{23}} = 24.16 \times {10^{23}}$ .
So the number of atoms present in $12.08 \times {10^{23}}$ unit cells is $24.16 \times {10^{23}}$ atoms.

Note: In the primitive cubic unit cell the atoms are present at the corners only so it has only one atom in its structure. Whereas in a face centered cubic unit cell we have the atoms at the corners as well as at the center of each face of the cube. So it has 4 atoms in total in its structure, the 1 atom is shared at the corners and the other three atoms share the ½ th portion with the 6 faces of the cube.