Question

Percentage of free space in a body centered cubic unit cell is:
A. $30$%
B. $32$%
C. $34$%
D. $28$%

Answer 1

Hint:In a body-centered cubic unit cell, the radius is one-fourth of diagonal length. The diagonal edge length of the body-centered cubic unit cell is$a\sqrt 3 $. 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.

Complete step by step solution:
The total space occupied by the particles in percentage is defined as the packing efficiency.
The formula to determine packing efficiency is as follows:
${\rm{packing}}\,{\rm{efficiency}}\,{\rm{ = }}\,\dfrac{{{\rm{volume}}\,{\rm{occupied}}\,{\rm{by}}\,{\rm{two}}\,{\rm{sphere}}\,{\rm{in}}\,{\rm{unit}}\,{\rm{cell}}}}{{\,{\rm{total}}\,{\rm{volume}}\,{\rm{of}}\,{\rm{unit}}\,{\rm{cell}}}}{\rm{ \times 100}}$
The volume of the sphere is, $\dfrac{{\rm{4}}}{{\rm{3}}}{\rm{\pi }}{{\rm{r}}^3}$
The formula of the volume of the cube is, ${a^3}$
So, the packing efficiency is,
${\rm{packing}}\,{\rm{efficiency}}\,{\rm{ = }}\,\dfrac{{2\, \times \dfrac{{\rm{4}}}{{\rm{3}}}{\rm{\pi }}{{\rm{r}}^3}}}{{\,{{\rm{a}}^3}}}{\rm{ \times 100}}$
The volume of the body-centred cubic unit cell ${{\rm{a}}^3}$ is as follows:
In the body-centred cubic unit cell, the relation between atomic radius edge length is as follows:
$r\, = \dfrac{{a\sqrt 3 }}{4}$
Where,
$r\,$ is the atomic radius.
$a$ is the edge length of the unit cell.
Rearrange for edge length, $a\, =\dfrac{{4\,r}}{{\sqrt 3 }}$
So, the volume body-centred cubic unit cell is, ${a^3}\, = {\left( {\dfrac{{4\,r}}{{\sqrt 3 }}} \right)^3}$
Substitute ${\left( {\dfrac{{4\,r}}{{\sqrt 3 }}} \right)^3}$for ${a^3}$ in packing efficiency formula.
$\Rightarrow {\rm{packing}}\,{\rm{efficiency}}\,{\rm{ = }}\,\dfrac{{2\, \times \dfrac{{\rm{4}}}{{\rm{3}}}{\rm{\pi }}{{\rm{r}}^3}}}{{\,{{\left( {\dfrac{{4\,r}}{{\sqrt 3 }}} \right)}^3}}}{\rm{ \times 100}}$
\[\Rightarrow {\rm{packing}}\,{\rm{efficiency}}\,{\rm{ = }}\,\dfrac{8}{{\rm{3}}}{\rm{\pi }}{{\rm{r}}^3}{\rm{ \times }}\dfrac{{3\sqrt 3 \,}}{{64{r^3}}}{\rm{ \times 100}}\]
\[\Rightarrow {\rm{packing}}\,{\rm{efficiency}}\,{\rm{ = }}\,68\]
So, the packing efficiency of a body-centred cubic unit cell is \[68\]%.
The total volume of the body centred cubic unit cell is \[100\]% out of which \[68\]% is occupied so, the free space is,
\[100\, - \,68 = \,32\]
So, the percentage of free space in a body-centered cubic unit cell is \[32\]%.

Therefore, option (B) \[32\]% is correct.

Note:
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 packing efficiency of the simple cubic unit cell is \[52.4\]% and the percentage of free space is \[47.6\]%. 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.