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**Simple cubic (SC) unit cell**

A simple cubic unit cell consists of eight corner atoms. In a simple cubic unit cell, a corner atom touches with another corner atom. The simple cubic unit cell is shown

**Number of atoms in a simple cubic unit cell**

A simple cubic unit cell consists of 8 corner atoms. Each and every corner atom is shared by eight adjacent unit cells. Therefore, one corner atom contributes 1/8th of its parts to one unit cell. Since, there are eight corner atoms in a unit cell, the total number of atoms is (1/8)* 8=1. Therefore, the number of atoms in a simple cubic unit cell is one.

**Atomic radius**

In a simple cubic lattice, a corner atom touches with another corner atom.

Therefore, 2r=a. So, the atomic radius of an atom in a simple cubic unit cell is a/2.

**Coordination number**

Consider a corner atom in a simple cubic unit cell. It has four nearest neighbours in its own plane. In a lower plane, it has one more nearest neighbour and in an upper plane, it has one more nearest neighbour. Therefore, the total number of nearest neighbour is six.

**Packing density**

The packing density of a simple cubic unit cell is calculated as follows:

theunit cell Number of atoms p

Substituting, r=a/2, we get,

The packing density of a simple cubic unit cell is 0.52. It means, 52% of the volume of the unit cell is occupied by atoms and the remaining 48% volume is vacant.**Primitive unit cell:**

In a Primitive unit cell, atoms are present at the corner only. An atom is shared by 8 unit cell. Hence the only 1/8th of the atom belonging to the particular unit cell. There are total 8 corners in each unit cell, therefore,1/8 part of 8 atoms would give the value of the number of atoms per unit cell.

In brief 8 corners, 1 atom at each corner, 1/8 of each atom in unit cell

(b) space-filling structure

(c) actual portions of atoms belonging to one unit cell

**Centred unit cell (body- centred):**

**Body- centred unit cell:**

The number of atoms present at the different position is as follows:

(a) 8 corners, 1 atom at each corner (1/8 of each atom in the unit cell)

(b) 1 atom at the centre of the body (1 atom completely present in the unit cell)

**The total number of atom present in each unit cell**

= Contribution of atoms at corner + contribution of atom at the centre of the body

**Packing Efficiency**

**Face-centred unit cell:**

The number of atoms present at the different position is as follows:

(a) 8 corners, 1 atom at each corner (1/8 of each atom in unit cell)

(b) 6 faces, 1 atom at each face of the unit cell (1/6; of each atom in the unit cell)

**The total number of atom present in each unit cell **= Contribution of atoms at corner + contribution of atom at the face of the body

**Packing Efficiency**

**Q.1. An element crystallizes into a structure which may be described by a cubic type of unit cell having one atom in each corner of the cube and two atoms on one of its face diagonals. If the volume of this unit cell is 24 x 10 ^{-24 }cm^{3} and density of the element is 7.20 gm/cm^{3}, calculate no. of atoms present in 200 gm of the element.**

Mass of one unit cell = volume Ã— its density

= 24 Ã— 10

= 172.8 Ã— 10

âˆ´ 172.8 10

âˆ´ 200 gm is the mass = 2 Ã— 200 / 172.8 Ã— 10

**Q.2. Calculate the void fraction for the structure formed by A and B atoms such that A form hexagonal closed packed structure and B occupies 2/3 of octahedral voids. Assuming that B atoms exactly fitting into octahedral voids in the HCP formed by A.****Ans: **Total volume of A atom = 6 Ã— 4 / 3 Ï€rA^{3}

Total volume of B atoms = 4 Ã— 4/3 Ï€rA^{3}

=4 Ã— 4/3 Ï€(0.414rA)^{3}

Since rB/rA as B is in octahedral void of A

Volume of HCP = 24âˆš2rA^{3}

Packing fraction = 6 Ã— 4/3 Ï€rA^{3} + 4 Ã— 4/3 Ï€ (0.414rA)^{3} / 24âˆš2rA^{3} = 0.7756

Void fraction = 1-0.7756 = 0.2244

**Q.3: ****An element with molar mass 2.7Ã—10**^{-}^{2}**kg mol**^{-1}** forms a cubic unit ****cell with edge length 405 pm****.**** ****If its density is**** ****2.7Ã—10**^{3}kg^{-3}**,**** what is the nature of the**** ****cubic unit cell?**** ****Ans: **d = 2.7Ã—10^{3}kg^{-3}

M = 2.7Ã—10 ^{-}^{2 }kg mol^{-1}

a= 405 pm = 405 X 10^{-12}

N_{A}= 6.023 X 10^{23}

Using the formula

Z=4

Unit cell is fcc unit cell.

Volume of 54 g of the element = 0.054/(2.7 x 10^{3}) = 2 X 10^{-6}

Number of unit cell in this volume = volume of 554 g of element/volume of each unit cell = 2 x 10^{-6}/(405 x 10^{-12})^{3} = 3.012 x 10^{22}

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