Question

The number of atoms per unit cell in a simple cubic, bcc and fcc system are respectively:
A.1, 2 and 4
B.8, 6 and 10
C.1, 4 and 2
D.2, 4 and 1

Answer 1

To answer this question you should recall the concept of close packing in a solid crystal. Voids refer to the gaps between the constituent particles. These voids in solid crystals mean the vacant space between the constituent particles

Complete step by step answer:
In case of the primitive cubic unit cell, the atoms are present only at the corners. Every atom at the corner is shared among 8 adjacent unit cells. There are 4 unit cells in the same layer and 4 in the upper layer. Therefore, a particular unit cell has the only $1/{8^{th}}\;$of an atom. Each small sphere in the following figure represents the center of a particle that occupies that particular position and not its size. This structure is known as an open structure. In each cubic unit cell, there are 8 atoms at the corners. Therefore, the total number of atoms in one-unit cell is $8{\text{ }} \times {\text{ }}1/8{\text{ }} = \;{\mathbf{1}}\,{\text{atom}}$.
In case of a BCC unit cell has atoms at each corner of the cube and an atom at the centre of the structure. The diagram shown below is an open structure. According to this structure, the atom at the body centre wholly belongs to the unit cell in which it is present. Thus, in a BCC cell, we have 8 corners and $1/{8^{th}}\;$ per corner atom = $8{\text{ }} \times {\text{ }}1/8{\text{ }} = \;{\mathbf{1}}\,{\text{atom}}$and 1 body centre atom $= {\text{ }}1{\text{ }} \times {\text{ }}1{\text{ }} = {\text{ }}1\,{\text{atom}}$. Therefore, the total number of atoms present per unit cell = 2 atoms.
On the other hand, an FCC unit cell contains atoms at all the corners of the crystal lattice and the center of all the faces of the cube. The atom present at the face-centered is shared between 2 adjacent unit cells and only 1/2 of each atom belongs to an individual cell. Thus, in a face-centred cubic unit cell, we have 8 corners and $1/8$ per corner atom = $8{\text{ }} \times {\text{ }}1/8{\text{ }} = \;{\mathbf{1}}\,{\text{atom}}$ and 6 face-centred atoms and $1/2$atom per unit cell = 3 atoms. Therefore, the total number of atoms in a unit cell = 4 atoms.

Hence, the correct option is A.

Note:
Tetrahedral voids: In case of a cubic close-packed structure, the second layer of spheres is present over the triangular voids of the first layer. This results in each sphere touching the three spheres of the first layer. When we join the centre of these four spheres, we get a tetrahedron and the space left over by joining the centre of these spheres forms a tetrahedral void.
Octahedral voids: Adjacent to tetrahedral voids you can find octahedral voids. When the triangular voids of the first layer coincide with the triangular voids of the layer above or below it, we get a void that is formed by enclosing six spheres. This vacant space which is formed by the combination of the initial formed triangular voids of the first layer and that of the second layer is known as Octahedral Voids.