Practicing Success
What is the percentage of free space in FCC and BCC unit cell respectively? |
32% and 24% 48% and 26% 26% and 32% 74% and 26% |
26% and 32% |
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% |