Question

In CCP of n spheres of radius r, n spheres of radius 0.414 r and 2n spheres of radius 0.225 r are introduced in the octahedral and tetrahedral void respectively. Therefore, packing fraction of the lattice is:

• A. 0.7
• B. 0.81
• C. 0.68
• D. 0.9

Answer 1

The correct option is B 0.81

The cubic closed packing, the number of atoms per unit cell is 4
$\therefore n=4$
To this packing arrangement, 4 spheres of radius 0.414r and 8 spheres of radius 0.225r are introduced in voids.

We know for cubic close packing,
$4r=\sqrt{2}a$
$a=2\sqrt{2}r$
$\text{Volune of unit cell is}=(2\sqrt{2}r{)}^3$

$\text{Packing fraction}=\frac{\text{Total volume occupied by spheres}}{\text{Volume of unit cell}}$
$\therefore$
$P.F.=\frac{4\times \frac{4}{3}\pi r^3+4\times \frac{4}{3}\pi \times (0.414r{)}^3+8\times \frac{4}{3}\pi \times (0.225r{)}^3}{(2\sqrt{2}r{)}^3}$
$P.F.=\frac{4\times \frac{4}{3}\pi r^3[1+(0.414{)}^3+2\times \left(0.225{)}^3\right]}{16\sqrt{2}{r}^3}$
$P.F.=\frac{\pi }{3\sqrt{2}}[1+(0.414{)}^3+2\times (0.225{)}^3$
$P.F.=\frac{\pi }{3\sqrt{2}}(1+0.071+0.024)P.F.=0.811\simeq 0.81$

Thus, option (b) is correct.