Question

Percentage of free space in the cubic close-packed structure and in body centered packed structure are, respectively:
A. {\text{30% and 26% }}
B. {\text{26% and 32% }}
C. {\text{32% and 48% }}
D. {\text{48% and 26% }}

Answer 1

The fraction of volume occupied by the spherical particles or atoms in an unit cell is known as packing fraction. It is basically the ratio volume of spherical particles and the volume of the cubic unit cell.

Complete step by step answer:
Packing fraction: In crystallography or we can say in the crystal structures, the fraction of volume occupied by the constituent particles is known as packing fraction. It is the fraction of volume occupied so it is a dimensionless quantity. Packing fraction has some other common names also such as atomic packing fraction and packing efficiency.
Calculation of packing fraction for cubic close packing :
As we all know that there is a unique relation between the radius of a spherical particle and the length of the side of a cube. The relation is as follows:
$a = 2\sqrt 2 r$ where, $a$ is the edge length and $r$ is the radius of spherical particles.
And the formula to calculate the packing efficiency is:
Packing efficiency(%) = $\dfrac{{{\text{volume of spheres in unit cell}}}}{{{\text{Total volume of unit cell}}}} \times 100$
And the volume of the sphere is calculated from the formula $\dfrac{4}{3}\pi {r^3}$.
So, Packing efficiency (%) = $\dfrac{{\dfrac{4}{3}\pi {r^3}}}{{{a^3}}} \times 100$
Now the relation which we have calculated between the radius of spheres and edge length will make our calculations easy. So substitute the value of edge length. The total number of atoms involved in cubic structure are four.
So, Packing efficiency (%) = $\dfrac{{4 \times \dfrac{4}{3}\pi {r^3}}}{{{{(2\sqrt 2 r)}^3}}} \times 100 = 74\%$
So the remaining empty space in structure is $(100 - 74 = 26\% )$ .
Similarly,
The relation between edge length and radius of sphere in BCC structure is as follows:
$a = \dfrac{{4r}}{{\sqrt 3 }}$ where, $a$ is the edge length and $r$ is the radius of spherical particles.
And the formula to calculate the packing efficiency is:
Packing efficiency(%) = $\dfrac{{{\text{volume of spheres in unit cell}}}}{{{\text{Total volume of unit cell}}}} \times 100$
And the volume of the sphere is calculated from the formula $\dfrac{4}{3}\pi {r^3}$.
So, Packing efficiency (%) = $\dfrac{{\dfrac{4}{3}\pi {r^3}}}{{{a^3}}} \times 100$
Now the relation which we have calculated between the radius of spheres and edge length will make our calculations easy. So substitute the value of edge length. The total number of atoms involved in BCC structure are two.
So, Packing efficiency (%) = $\dfrac{{2 \times \dfrac{4}{3}\pi {r^3}}}{{{{(\dfrac{{4r}}{{\sqrt 3 }})}^3}}} \times 100 = 68\%$
So the remaining empty space in structure is $(100 - 68 = 32\% )$ .
Hence the percentage of free space in the cubic close-packed structure and in body centered packed structure are, respectively {\text{26% and 32% }}.

So, the correct answer is Option B.

Note:
The remaining space or the unoccupied space from the spherical particles in the unit cell is known as free space or percentage of free space. It is basically calculated by subtracting the packing fraction (%) from $100\%$ .