Question

How many atoms is there in a cube based unit cell having one atom on each corner and two atoms on each body diagonal of the cube?
A) $8$
B) $6$
C) $4$
D) $9$

Answer 1

Hint: We know that a unit is the smallest representation of a whole crystal. During a unit, an atom's coordination number is the number of atoms it's touching. The coordination number of face-centered cubic (FCC) is twelve and contains four atoms per unit. The coordination of the body-centered cubic (bcc) is eight and contains two atoms per unit.

Complete step by step answer:
There are four body diagonals during a cube. As there are two atoms on each diagonal, there is a complete of \[4 \times 2 = 8\] atoms on the diagonals. These eight atoms are within the body of the cubic unit and aren't shared by other units.
The contribution of eight corner atoms, \[8 \times \left( {1/8} \right) = 1atom\]
Thus, the total number of atoms within the unit \[ = 8 + 1 = 9atoms\].

Therefore, the option D is correct.

Additional Information:
Now we can classify the cubic lattice into three types as,
The atom in the simple cubic is shared equally between eight neighboring cubes, and the unit cell therefore contains in total one atom.
Body-centered cubic (bcc) has one lattice point in the center of the unit cell in addition to the eight corner points and the unit cell contains in total two atoms.
Face-centered cubic (FCC) are eight atoms at corners of the unit cell and one atom centered in each of the faces. The atom in the face centered cubic lattice is shared with the adjacent cell.

Note:
Don’t confuse between the amount of atoms within the primitive cell and therefore the body center unit cells. The entire number of atoms within the body centered unit is 2. One among the atoms is present at the middle of the middle of the unit whereas the entire number of atoms within the primitive cell is one.