What is the percentage of free space in FCC and BCC unit cell respect…

The correct answer is option 3. 26% and 32%

Face centered cubic unit cell (FCC) is present in CCP Lattice structure

Let the unit cell edge length be ‘a’ and face diagonal AC = b.

In ∆ ABC

AC² = b² = BC² + AB² = a² + a² = 2a² or b = \(\sqrt{2}\)a

If r is the radius of the sphere, we find b = 4r = \(\sqrt{2}\)a or a = \(\frac{4r}{\sqrt{2}}\) 

a = 2\(\sqrt{2}\)r

we can also write, r = \(\frac{a}{2\sqrt{2}}\)

We know that each unit cell in ccp structure i.e., FCC unit cell has effectively 4 spheres.

Total volume of four spheres is equal to 4 x \(\frac{4}{3}\)πr³ and volume of the cube is a³ or (2\(\sqrt{2}\)r)³.

Therefore,

Packing efficiency = \(\frac{\text{Volume occupied by four spheres in the unit cell}}{\text{Total volume of the unit cell}}\) x 100%

Packing efficiency = \(\frac{4 × \frac{4}{3}πr^3}{(2\sqrt{2}r)^3}\) x 100% = 74%

% of free space in FCC unit cell i.e., Void efficiency = 100 - 74 = 26%

Body centered cubic unit cell (BCC)

Atom at the centre will be in touch with the other two atoms diagonally arranged.

In ∆ EFD,

b² = a² + a² = 2a²

b = \(\sqrt{2}\)a

Now in ∆ AFD

c² = a² + b² = a² + 2a² = 3a²

c = \(\sqrt{3}\)a

The length of the body diagonal c is equal to 4r, where r is the radius of the sphere (atom), as all the three spheres along the diagonal touch each other.

Therefore, \(\sqrt{3}\)a = 4r

a = \(\frac{4r}{\sqrt{3}}\)

Also we can write, r = \(\frac{\sqrt{3}}{4}\)a

In this type of structure, total number of atoms is 2 and their volume is 2 x \(\frac{4}{3}\)πr³

Volume of the cube, a³ will be equal to (\(\frac{4}{\sqrt{3}}\)r)³ or a³ = (\(\frac{4}{\sqrt{3}}\)r)³

Packing efficiency = \(\frac{\text{Volume occupied by four spheres in the unit cell}}{\text{Total volume of the unit cell}}\) x 100%

P.E. = \(\frac{2 × \frac{4}{3}πr^3}{(\frac{4}{\sqrt{3}})^3}\) x 100 = 68%

% of free space in BCC unit cell i.e., Void efficiency = 100 - 68 = 32%