Practicing Success
Efficiency of packing in body centered cubic structures is found to be: |
33% 74% 52.4% 68% |
68% |
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% |