Question

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

Answer 1

Face-centered cubic unit cell is one kind of cubic lattice. This is also known as cubic close-packed Structure. In this unit cell, each face of the unit cell contains one atom at its center. And contains one atom at the eight corners of the unit cell. In the case of Body-centered packed structure an atom is at the center of the unit cell and contains one atom at the eight corners of the unit cell.

Complete step by step answer:
In the face-centered cubic unit cell. There are six atoms at the center of each face of the cube and eight atoms at the eight corners of the unit cell.
Now, the number of atoms in a unit cell can be calculated by calculation summation of the contribution of the atoms present in the unit cell.
The contribution of each corner atom is ${\dfrac{1}{8}^{}}$th of a total atom. So, the contribution of all eight atoms is, $\dfrac{1}{8} \times 8 = 1$ .
For the face center atoms, the contribution of each atom ${\dfrac{1}{2}^{}}$ th of a total atom. So, the total contribution of the total of six atoms is $\dfrac{1}{2} \times 6 = 3$ .
The total number of atoms is 4 in an F.C.C lattice.
Now the volume of one atom is $\dfrac{4}{3}\pi {r^3}$ ,
Therefore, for total 4 atoms, the total volume occupied is,

$$4 \times \dfrac{4}{3}\pi {r^3} \\\ = \dfrac{{16}}{3}\pi {r^3} \\\$$

Now, the percentage of the volume occupied by molecules is, $= \dfrac{{Volume\,occupied\,by\,all\,atoms}}{{total\,volume\,of\,unit\,cell}} \times 100$
The relation between edge length and radius of the atoms is, $r = \dfrac{{\sqrt 2 a}}{4}$
Therefore, the percentage of the volume occupied by molecules is,

$$= \dfrac{{Volume\,occupied\,by\,all\,atoms}}{{total\,volume\,of\,unit\,cell}} \times 100 \\\ = \dfrac{{\dfrac{{16}}{3}\pi {{\left( {\dfrac{{\sqrt 2 a}}{4}} \right)}^3}}}{{{a^3}}} \times 100 \\\ = \dfrac{{16}}{3}\pi {\left( {\dfrac{{\sqrt 2 }}{4}} \right)^3} \times 100 \\\ = 16.74 \times {\left( {\dfrac{1}{{2\sqrt 2 }}} \right)^3} \times 100 \\\ = \dfrac{{16.74}}{{22.6}} \times 100 \\\ = 74.1 \\\$$

So, the percentage of free space is, $100 - 74.1 = 25.9\%$
On the other hand,
In body-centered packed structure,
The contribution of each corner atom is ${\dfrac{1}{8}^{}}$th of a total atom. So, the contribution of all eight atoms is, $\dfrac{1}{8} \times 8 = 1$ .
For, the body-centered atom the contribution is 1.
The total number of atoms is 2 in a B.C.C lattice.
Now, the volume of one atom is $\dfrac{4}{3}\pi {r^3}$ ,
Therefore, for total 2 atoms, the total volume occupied is,

$$2 \times \dfrac{4}{3}\pi {r^3} \\\ = \dfrac{8}{3}\pi {r^3} \\\$$

The relation between edge length and the radius of the atoms is, $r = \dfrac{{\sqrt 3 a}}{4}$ .
Therefore, the percentage of the volume occupied by molecules is,

$$= \dfrac{{Volume\,occupied\,by\,all\,atoms}}{{total\,volume\,of\,unit\,cell}} \times 100 \\\ = \dfrac{{\dfrac{8}{3}\pi {{\left( {\dfrac{{\sqrt 3 a}}{4}} \right)}^3}}}{{{a^3}}} \times 100 \\\ = \dfrac{8}{3}\pi {\left( {\dfrac{{\sqrt 3 }}{4}} \right)^3} \times 100 \\\ = 8.37 \times {\left( {0.4330} \right)^3} \times 100 \\\ = 8.37 \times 0.0811 \times 100 \\\ = 67.9 \\\$$

So, the percentage of free space is, $100 - 67.9 = 32.1\%$

So, the correct option is B.

Note:
Number of atoms Per hexagonal close-packed Unit Cell is,
Each atom at the corner of the hexagonal face is shared by 6 unit cells i.e. each atom contributes ${\dfrac{1}{6}^{}}$ th to their mass. There are two such hexagonal faces with a total of 12 such atoms. Thus, the contribution of corner atoms = $\dfrac{1}{6} \times 12 = 2$
The atom at the center of the hexagonal faces is shared by two cells each. Therefore, the contribution of face-centered atoms = $\dfrac{1}{2} \times 2 = 1$
Three atoms are contained within the unit cell; therefore, their contribution is $3 \times 1 = 3$ .
So, the total number of atoms per unit cell is 2+1+3=6.