Solid State
5.0 Calculation of Spaces occupied i.e., Packing Fraction
5.0 Calculation of Spaces occupied i.e., Packing Fraction
$1.$ Simple cubic
Edge length $= a$ 
Radius of sphere $= r$
As the spheres are touching each other, $$a = 2r$$
No. of spheres per unit cell $$ = \frac{1}{8} \times 8 = 1$$
Volume of the sphere $$ = \frac{4}{3} \times \pi {r^3}$$
Volume of the cube $$= a^3 = (2r)^3=8r^3$$
Therefore, fraction occupied (or packing fraction) is $$\begin{equation} \begin{aligned} = \frac{{\frac{4}{3}\pi {r^3}}}{{8{r^3}}} = 0.524 \\ \therefore \% \;occupied = 52.4\% \\\end{aligned} \end{equation} $$
$2.$ Face centred cubic
$PR=4r$
In right angled triangle $PQR$, $$\begin{equation} \begin{aligned} PR = \sqrt {P{Q^2} + Q{R^2}} = \sqrt {{a^2} + {a^2}} = a\sqrt 2 \\ \therefore a\sqrt 2 = 4r \\ \Rightarrow a = \frac{4}{{\sqrt 2 }}r \\\end{aligned} \end{equation} $$Therefore, Volume of the unit cell $ = {a^3} = {\left( {\frac{4}{{\sqrt 2 }}r} \right)^3} = \frac{{32}}{{\sqrt 2 }}{r^3}$
Number of spheres in the unit cell $$ = \left( {8 \times \frac{1}{8}} \right) + \left( {6 \times \frac{1}{2}} \right) = 4$$
Volume of $4$ spheres is $$ = 4 \times \frac{4}{3}\pi {r^3} = \frac{{16}}{3}\pi {r^3}$$
Therefore, fraction occupied (or packing fraction) is $$\begin{equation} \begin{aligned} = \frac{{\frac{{16}}{3}\pi {r^3}}}{{\frac{{32}}{{\sqrt 2 }}{r^3}}} = 0.74 \\ \therefore \% \;occupied = 74\% \\\end{aligned} \end{equation} $$
$3.$ Body centred cubic
The atom at the body centre touches the spheres at the corners,
$\therefore$ Body Diagonal $PS=4r$
Face diagonal is $$PR = \sqrt {P{Q^2} + Q{R^2}} = \sqrt 2 a$$ and body diagonal is $$PS = \sqrt {P{R^2} + R{S^2}} = \sqrt {2{a^2} + {a^2}} = \sqrt 3 a$$ Now, we can write $$\begin{equation} \begin{aligned} \sqrt 3 a = 4r \\ \Rightarrow a = \frac{4}{{\sqrt 3 }}r \\\end{aligned} \end{equation} $$
Volume of unit cell is $${a^3} = {\left( {\frac{4}{{\sqrt 3 }}r} \right)^3} = \frac{{64}}{{3\sqrt 3 }}{r^3}$$
No. of spheres per unit cell is $$\begin{equation} \begin{aligned} = \left( {8 \times \frac{1}{8}} \right) + \left( {1 \times 1} \right) \\ = 2 \\\end{aligned} \end{equation} $$
Volume of two spheres is $$ = 2 \times \frac{4}{3}\pi {r^3} = \frac{8}{3}\pi {r^3}$$Therefore, fraction occupied (or packing fraction) is $$\begin{equation} \begin{aligned} = \frac{{\frac{8}{3}\pi {r^3}}}{{\frac{{64}}{{3\sqrt 3 }}\pi {r^3}}} = 0.68 \\ \% \;occupied = 68\% \\\end{aligned} \end{equation} $$
$4.$ Hexagonal Close Packing
Each corner atom would be common to $6$ other unit cells, therefore their contribution to one unit cell would be $\frac{1}{6}$.
Total number of atom in $1$ hcp unit cell is $$\begin{equation} \begin{aligned} = \frac{{12}}{6}\left( {{\text{from 12 corners}}} \right) + \frac{2}{2}\left( {{\text{from 2 face centred}}} \right) + \frac{3}{1}\left( {{\text{from body centre}}} \right) \\ = 2 + 1 + 3 = 6 \\\end{aligned} \end{equation} $$
$ABCD$ is the base of hexagonal unit cell .
$AD=AB=a$. The sphere in the next layer has its centre $F$ vertically above $E$ it touches the three spheres whose centres are $A$, $B$ and $D$.
Hence, $$FE = \frac{h}{2} = \sqrt {{{\left( {2r} \right)}^2} - {{\left( {\frac{{2r}}{{\sqrt 3 }}} \right)}^2}} $$Therefore, the height of unit cell $(h)$ is $$h = 4r\sqrt {\frac{2}{3}} $$ The area of the base is equal to the area of six equilateral triangles is $ = 6 \times \frac{{\sqrt 3 }}{4}{\left( {2r} \right)^2}$.
The volume of the unit cell is $$ = 6 \times \frac{{\sqrt 3 }}{4}{\left( {2r} \right)^2} \times 4r\sqrt {\frac{2}{3}} $$ Therefore, $$PF = \frac{{6 \times \frac{4}{3}\pi {r^3}}}{{6 \times \frac{{\sqrt 3 }}{4}{{\left( {2r} \right)}^2} \times 4r\sqrt {\frac{2}{3}} }} = 0.74$$So, Volume fraction is $$VF = 1 - 0.74 = 0.26$$
