Question

Packing fraction of simple cubic cell is:
A. $\dfrac{{{\pi }}}{{\text{2}}}$
B. $\dfrac{{{\pi }}}{{\text{6}}}$
C. $\dfrac{{{{3\pi }}}}{{\text{6}}}$
D. $\dfrac{{{\pi }}}{{{\text{2}}\sqrt {\text{2}} }}$

Answer 1

Hint: In a simple cubic cell, the radius is half of diagonal length. The diagonal edge length of the simple cubic cell is a. By this, we will determine the volume of the cube. Then divide the volume of two spheres by the volume of the cube to determine the packing efficiency.

Formula used: ${\text{packing}}\,{\text{fraction}}\,{\text{ = }}\,\dfrac{{{\text{volume}}\,{\text{occupied}}\,{\text{by}}\,{\text{sphere}}\,{\text{in}}\,{\text{unit}}\,{\text{cell}}}}{{\,{\text{total}}\,{\text{volume}}\,{\text{of}}\,{\text{unit}}\,{\text{cell}}}}$


Complete step-by-step solution
The total space occupied by the particles is defined as the packing fraction. In a simple cubic cell only one atom is present in the lattice, so volume occupied by the unit cell will be only by one atom.
The formula to determine packing efficiency is as follows:
${\text{packing}}\,{\text{fraction}}\,{\text{ = }}\,\dfrac{{{\text{volume}}\,{\text{occupied}}\,{\text{by}}\,{\text{sphere}}\,{\text{in}}\,{\text{unit}}\,{\text{cell}}}}{{\,{\text{total}}\,{\text{volume}}\,{\text{of}}\,{\text{unit}}\,{\text{cell}}}}$
The volume of the sphere is, $\dfrac{{\text{4}}}{{\text{3}}}{{\pi }}{{\text{r}}^3}$
The formula of the volume of the cube is, ${{\text{a}}^{\text{3}}}$
So, the packing fraction is,
${\text{packing}}\,{\text{fraction}}\,{\text{ = }}\,\dfrac{{\dfrac{{\text{4}}}{{\text{3}}}{{\pi }}{{\text{r}}^{\text{3}}}}}{{\,{{\text{a}}^{\text{3}}}}}$
The volume of the simple cubic cell ${{\text{a}}^3}$ is as follows:
In the simple cubic cell, the relation between atomic radius edge length is as follows:
${\text{r}}\,{\text{ = }}\dfrac{{\text{a}}}{{\text{2}}}$
Where,
${\text{r}}\,$is the atomic radius.
${\text{a}}$ is the edge length of the unit cell.
Rearrange for edge length, ${\text{a}}\,{\text{ = 2r}}$
So, the volume simple cubic cell is, ${{\text{a}}^{\text{3}}}\,{\text{ = }}{\left( {{\text{2r}}} \right)^{\text{3}}}$
Substitute ${\left( {{\text{2r}}} \right)^{\text{3}}}$for ${{\text{a}}^{\text{3}}}$ in packing fraction formula.
${\text{packing}}\,{\text{fraction}}\,{\text{ = }}\,\dfrac{{\dfrac{{\text{4}}}{{{3}}}{{\pi }}{{\text{r}}^{\text{3}}}}}{{\,{{\left( {{\text{2r}}} \right)}^{\text{3}}}}}$
\[{\text{Packing}}\,{\text{fraction}}\,{\text{ = }}\,\dfrac{4}{{\text{3}}}{{\pi }}{{\text{r}}^{\text{3}}}{{ \times }}\dfrac{{{\text{1}}\,}}{{{\text{8}}{{\text{r}}^{\text{3}}}}}\]
\[{\text{Packing}}\,{\text{fraction}}\,{\text{ = }}\,\dfrac{{{\pi }}}{6}\]
So, the packing fraction of a simple cubic cell is\[\dfrac{{{\pi }}}{6}\].

Therefore, option (B) \[\dfrac{{{\pi }}}{6}\] is correct.

Note:\[\dfrac{{{\pi }}}{6}\]is equal to $0.56$ so, the packing efficiency of a Simple cubic cell is $56$%.
The total volume of a cubic unit cell is\[100\]% out of which$56$% is occupied so, the free space is,\[100\, - \,52 = \,48\].The packing efficiency of the face-centered cubic unit cell which is found in hcp and ccp is\[78\]% and the percentage of free space is\[22\]%. The maximum packing efficiency is of the face-centered cubic unit cell. In the face-centered cubic lattice, the radius is one-fourth of the diagonal length. The diagonal edge length of the face-centered cubic unit cell is$a\sqrt 2 $. In a simple cubic unit cell, the edge length is double the radius of the unit cell.