Question

Packing efficiency of body centered unit cell is
A. $85$%
B. $68$%
C. $33.33$%
D. $45$%

Answer 1

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.
Formula used: ${\text{packing}}\,{\text{efficiency}}\,{\text{ = }}\,\dfrac{{{\text{volume}}\,{\text{occupied}}\,{\text{by}}\,{\text{two}}\,{\text{sphere}}\,{\text{in}}\,{\text{unit}}\,{\text{cell}}}}{{\,{\text{total}}\,{\text{volume}}\,{\text{of}}\,{\text{unit}}\,{\text{cell}}}}{{ \times 100}}$

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:
${\text{packing}}\,{\text{efficiency}}\,{\text{ = }}\,\dfrac{{{\text{volume}}\,{\text{occupied}}\,{\text{by}}\,{\text{two}}\,{\text{sphere}}\,{\text{in}}\,{\text{unit}}\,{\text{cell}}}}{{\,{\text{total}}\,{\text{volume}}\,{\text{of}}\,{\text{unit}}\,{\text{cell}}}}{{ \times 100}}$
The volume of the sphere is, $\dfrac{{\text{4}}}{{\text{3}}}{{\pi }}{{\text{r}}^3}$
The formula of the volume of the cube is, ${a^3}$
So, the packing efficiency is,
${\text{packing}}\,{\text{efficiency}}\,{\text{ = }}\,\dfrac{{2\, \times \dfrac{{\text{4}}}{{\text{3}}}{{\pi }}{{\text{r}}^3}}}{{\,{{\text{a}}^3}}}{{ \times 100}}$
The volume of the body-centred cubic unit cell ${{\text{a}}^3}$ 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.
${\text{packing}}\,{\text{efficiency}}\,{\text{ = }}\,\dfrac{{2\, \times \dfrac{{\text{4}}}{{\text{3}}}{{\pi }}{{\text{r}}^3}}}{{\,{{\left( {\dfrac{{4\,r}}{{\sqrt 3 }}} \right)}^3}}}{{ \times 100}}$
${\text{packing}}\,{\text{efficiency}}\,{\text{ = }}\,\dfrac{8}{{\text{3}}}{{\pi }}{{\text{r}}^3}{{ \times }}\dfrac{{3\sqrt 3 \,}}{{64{r^3}}}{{ \times 100}}$
${\text{packing}}\,{\text{efficiency}}\,{\text{ = }}\,68$
So, the packing efficiency of a body- centred cubic unit cell is$68$%.

Therefore, the correct answer is option (B) $68$%

Note: The total volume of the body centred cubic unit cell is $100$% out of which $68$% is occupied so, the free space is,$100\, - \,68 = \,32$. 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.