Answer:

AIM:-TO STUDY THE CRYSTAL STRUCTURE OF A GIVEN SPECIMEN (B.C.C., F.C.C.,

H.C.P).

THEORY:- BCC:-

In body centered cubic structure each one atom is placed at the corner of

the CUBE and one atom is placed at the centre of the cube. Iron has BCC structure. At

room temperature the unit cell of iron has an atom at each corner and another at the body

centre of the cube. Each iron atom in BCC structure is surrounded by eight adjacent iron

atoms. The unit cell of a cubic cell contains eight atoms at corners which are shared by

the adjoining eight cubes.

Hence the share of each cube = ⅛ of each corner atoms

Total no of atoms = ⅛ × 8 = 1 atom

BCC crystal has one atom at center.

So, total no. of atoms in BCC = 2 atoms

F.C.C.:-

In this type of structure the unit cell contains one atom at center of each

corner plus at each face. Examples of such type of crystal structure are copper, silver,

gold etc. In FCC crystal the atom on each face is surrounded or shared by two cubes. Son

CONTRIBUTION of each towards crystal is ½, one atom at each corner. i.e. shared by eight

other cubes so that its contribution towards crystal is ⅛.

So total no of atoms = ⅛ × 8 + ½ ×6 = 4 atoms

H.C.P.:-

In case of hexagonal closed packing structure there are 12 atoms at corner. One

atom at the center of two hexagonal FACES and three atoms symmetrically arranged in the

body of unit cell.

Total no of atoms per unit cell = 1/6 × 6 + 1/6 × 6 + ½ ×2 + 3 = 6 atoms

Explanation: