Question

Find the following:
Packing fraction of FCC.

Answer 1

Hint: The packing efficiency can be calculated by the percent of space occupied by spheres present in a unit cell. Here we will draw a general face centered unit cell with the spheres in it and proceed further by evaluating the volume of spheres in the unit cell and the total volume of the unit cell.
We will use the following figure of the face centered unit cell to solve the problem.
 

 

Let the side of a unit cell = a
And diagonal AC = b
Now, in right triangle ABC,
Let us use the Pythagoras theorem.
AD is perpendicular, DC is base and AC is diagonal
$
  \because A{C^2} = A{D^2} + D{C^2} \\
   \Rightarrow {b^2} = {a^2} + {a^2} \\
   \Rightarrow {b^2} = 2{a^2} \\
   \Rightarrow b = \sqrt 2 a \\
 $
Let r is the radius of sphere, so b = 4r,
Thus,
\[
  b = 4r = a\sqrt 2 \\
   \Rightarrow a = \dfrac{{4r}}{{\sqrt 2 }} \\
   \Rightarrow a = \dfrac{{4r}}{{\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }} \\
   \Rightarrow a = \dfrac{{4\sqrt 2 r}}{2} \\
   \Rightarrow a = 2\sqrt 2 r.........(1) \\
 \]
Now, volume of cube $ = {\left( {{\text{side}}} \right)^3} = {a^3}$
Substituting the value of a from equation (i) we get,
Volume of cube
$
   = {a^3} = {\left( {2\sqrt 2 r} \right)^3} \\
   = 8 \times 2\sqrt 2 \times {r^3} \\
   = 16\sqrt 2 {r^3} \\
 $
Volume of cube $ = 16\sqrt 2 {r^3}$ - (2)
Now, volume of sphere
$ = \dfrac{4}{3}\pi {r^3}..........(3)$
Since one unit cell of face entered cell has 4 spheres
$ = \dfrac{1}{8} \times 8 + \dfrac{1}{2} \times 6 = 1 + 3 = 4$
As the contribution of the corner sphere in a cell is one eighth and that of the face sphere is half.
Therefore, volume of 4 atoms, i.e. 4 spheres:
$
   = 4 \times \dfrac{4}{3}\pi {r^3} \\
   = \dfrac{{16}}{3}\pi {r^3} \\
 $
We know that
Packing efficiency = (Volume of spheres in unit cell/ total volume of unit cell) × 100%
Since there are 4 atoms in the unit cell of face centered cell
Therefore, packing efficiency of face centered cell
Packing efficiency = (Volume of 4 spheres in unit cell/ total volume of unit cell) × 100%
Now, packing efficiency (in %)
$
   = \dfrac{{{\text{volume of 4 spheres in unit cell}}}}{{{\text{total volume of unit cell}}}} \times 100 \\
   = \dfrac{{\dfrac{{16}}{3}\pi {r^3}}}{{16\sqrt 2 {r^3}}} \times 100 \\
 $
Let us solve the equation by cancelling the common term to find the percentage.
\[
   = \dfrac{{16\pi }}{{3 \times 16\sqrt 2 }} \times 100 \\
   = \dfrac{{\pi \times 100}}{{3 \times \sqrt 2 }} \\
   = \dfrac{{3.14 \times 100}}{{3 \times 1.414}} \\
   = 74.048\% \\
   = 74.05\% \\
 \]
We know that the packing fraction and packing efficiency are the same.
Hence, packing fraction for face centered unit cell is 74.05%

Note- The packing efficiency is defined as the percentage of space filled by constituent particles contained within the lattice. This can be calculated using geometry in three structures namely: HCP and CCP. Simple packing efficiency for cubic lattice is 52.4 per cent. And the packing efficiency of body-centered cubic lattice (bcc) is 68 per cent.