Question

In a ccp type structure, if half of the face-centered atoms are removed, then percentage void in a unit cell is approximately:
A. $54\% $
B. $46.25\% $
C. $63\% $
D. $37\% $

Answer 1

Hint: We can calculate the percentage void by subtracting 100 from the packing efficiency. The packing efficiency is calculated by multiplying the number of atoms and volume occupied by one atom and is divided by the total volume of the cube. The final answer is again multiplied by 100 to get the overall packing efficiency.

Complete step by step answer:
For cubic close packing type structure, the atoms are found to bond the corners and face centre.
Therefore, the number of atoms present=$\dfrac{1}{8} \times 8 + \dfrac{1}{2} \times 6 = 4$
The number of atoms present in ccp is four.
When the half of the face-centered atoms is removed, the number of atoms present can be calculated.
The number of atoms present = $\dfrac{1}{8} \times 8 + \dfrac{1}{2} \times 3 = \dfrac{5}{2}$
The number of atoms present is $\dfrac{5}{2}$.
We know that the formula of volume of cube is ${a^3}$.
The edge of the cube is represented as a.
We can give the volume occupied by an atom as $\dfrac{4}{3}\pi {r^3}$.
Here the radius of the atoms is represented as r.
For FCC lattice,
$4r = \sqrt 2 a$
$ \Rightarrow $$a = 2\sqrt {2r} $
$ \Rightarrow $$r = \dfrac{a}{{2\sqrt 2 }}$
We can calculate the packing efficiency as the product of the number of atoms present to the volume occupied by an atom divided by the total volume of the cube and the final product is multiplied by 100.
Packing efficiency$ = \dfrac{{{\text{Number of atoms present}} \times {\text{Volume occupied by one atom}}}}{{{\text{Total volume of cube}}}} \times 100$
Packing efficiency = $\dfrac{{\dfrac{5}{2} \times \dfrac{4}{3}\pi {r^3}}}{{{a^3}}} \times 100$
Substituting the value of a we get,
$ \Rightarrow $Packing efficiency = $\dfrac{{\dfrac{5}{2} \times \dfrac{4}{3}\pi {r^3}}}{{{{\left( {2\sqrt 2 r} \right)}^3}}} \times 100$
$ \Rightarrow $Packing efficiency = $46.4\% $
The packing efficiency is $46.4\% $.
We can now calculate the percentage of the void.
The difference between 100 and the packing efficiency gives the percentage of the void.
Percentage of void = $100 - {\text{Packing efficiency}}$
Percentage of void = $100 - 46.4$
Percentage of void = $53.6\% $
Percentage of void = $ 54\% $
The percentage of the void is $54\% $.

So, the correct answer is Option A.

Note:
We must remember that the cubic, tetragonal, orthorhombic, hexagonal, monoclinic, triclinic, and rhombohedral are seven simple crystal structures. They vary in their arrangement of crystallographic axes and angles. We can relate hexagonal closed packed systems to face-centered cubic cells. An example of a compound that contains face-centered cubic lattice is sodium chloride. Examples of other compounds that have face-centered cubic