Q. 24

# The percentage of empty space in a body centred cubic arrangement is ________.

A. 74

B. 68

C. 32

D. 26

Answer :

• In a body-centred cubic arrangement of lattices, the constituent atoms are located on the 8 corners of the cube with one atom at the centre of the cube.

Let the edge length of the cube be ‘**a**’ (ED or BC in the fig) and radius of each constituent spherical particle be ‘**r**’ and the arrangement is such as that atom at the centre is in touch with other two atoms diagonally.

• Hence the body diagonal **AF** will be equal to = r +2r + r = 4r (1)

In the right angle ∆ EFD,

FD^{2} or b^{2=} FE^{2} + ED^{2} ( by Pythagoras’s theorem ) = a^{2} + a^{2} = 2a^{2}

Or, FD or b = √2a

• In the right-angled triangle, ∆ AFD

AF^{2} or c^{2}= AD^{2} + FD^{2} = a^{2} + b^{2} = a^{2} + 2a^{2} (as b^{2}= 2a^{2} ) =3a^{2}

Or, AF or c =√3a (2)

Hence, from equations (1) and (2) we can get ,

AF = 4r =√3a a

Or, r = √3a/4

And a = 4r/√3

• Then, the volume of 1 cubic unit cell is a^{3} = (4r/√3) = 64r^{3}/ 3√3

• Since in body centred arrangement , there are 2 atoms per unit cell

Volume occupied by the atoms = 2 X 4/3 π r^{3} = 8/3 π r^{3}

Since, **Packing efficiency = Volume occupied by two spheres in the unit cell X 100 %**

**Total volume of the unit cell**

**=** __8/3 π r ^{3}__

**X 100 % = 68.04 % = 68%**

_{appx.}**64r ^{3}/ 3√3**

Therefore , Packing efficiency in a body centred cubic lattice is 68 %

So, empty space or void space will be = (100 -68 )% = 32 %

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