# Number of (atoms) constituent particles in cubic unit cell Chemistry Notes

## Number of (atoms) constituent particles in cubic unit cell Chemistry Notes

→ We know that a crystal lattice is formed by maximum number of unit cells and a constituent particle exists on each lattice points. In different types of unit cell, these constituent particles are present at corners, centre of faces or centre of body. It is also clear that each unit cell of crystalline solid is attached with other unit cell. So, the constituent particles of a unit cell are also shared by neighbouring unit cells. As a result, some part of constituent particles is related to a unit cell which are further represented by figures.

→ Simple (primitive) Cubic Unit Cell: A constituent particle is present at each corner of simple or primitive cubic unit cell. According to figure, any one constituent particle is distributed in eight simple cubic unit cells among which four are situated towards lower end and four are situated towards upper end. Hence infact a constituent particle is related to only 1/8th part of specific unit cell. → Simple cubic unit cell can be represented in different types as shown in figure 1.23. Figure (a) represents only nuclei of constituent particles which are present at corners. Figure (b) represents actual shape of constituent particles while figure (c) represents that part of constituent particles which exist in that unit cell actually So, the constituent particles per unit cell is $$\frac{1}{8}$$ × 8 = 1 → Boay-centred Uudic unit cel In body-centred Cubic (bcc) unit cell, one constituent particle is present at each corner and one is also present in center of body. This unit cell can be represented in different types in figure 1.24. Figure (a) represents centre of constituent particles only, figure (b) represents actual shape and figure (c) represents total part of constituent particles exist in body-centred cubic unit cell.

• 8 corners × $$\frac{1}{8}$$ particle per corner = 1 particle
• 1 particle is body-centre = 1 particle
• Total nunber of particle per unit cell = 2 particles → Face-centred Cubic unit cell: This structure is also known as cubic close packing. In face-centred cubic unit cell, the constituent particles are present at all corners and one constituent particle is also present at centre of each face (six faces). This unit cell is represented in different types in figure 1.25 Figure (a) represents only centre of constituent particles. Figure (b) represents actual shape and figure (c) represents total part of constituent particles exist in each unit cell.
8 corners × $$\frac{1}{8}$$ particle per unit cell = 1 atom or particle
6 face-centred atoms × $$\frac{1}{2}$$ = 3 atoms or particles
∴ Total number of particles per unit cell (1 + 3)
= 4 atoms or particles Rules for calculation of Number of Atoms :

• Only $$\frac{1}{8}$$th part of atoms (or molecules or ions) present at corners of unit cell is related to specific unit cell because it is shared by eight other unit cells.
• Only $$\frac{1}{2}$$nd part of atoms (or molecular or ions) present at centers of faces of unit cell is
• related to specific unit cell because it is shared by two other unit cells on faces.
• Total part of atom (or molecule or ion) present at centre of body of unit cell is related to specific unit cell because it is not shared with other unit cells.
• Only $$\frac{1}{4}$$th part of atoms (or molecular or ions) present at edges of unit cell is related to specific unit cell because it is participated by four other unit cells. Calculation of Number of Constituent Particles :

→ Number of Atoms in Simple Cubic Unit cell :

Number of constituent particles (atoms)
= 8 (corners) × $$\frac{1}{8}$$ (atoms per corner) + 1
= $$\left(8 \times \frac{1}{8}\right)$$ + (1 × 1) = 1 + 1 = 2 atoms
= 8 x = 1 atom (2) Number of Atoms in Body-centred Cubic Unit Cell Number of constituent

→ Number of Atoms in Face-centred Cubic unit cell :
Number of constituent particles (atoms)  → Number of Atoms in End-Centred Cubic Unit Cell :

Number of constituent particles (atoms)
= 8(corners) × $$\frac{1}{8}$$(atoms per corner) + 2 (faces)
× $$\frac{1}{8}$$(atoms) per face
= 8 × $$\frac{1}{8}$$ + 2 × $$\frac{1}{2}$$+ 1 + 1 = 2 atoms