The correct order of the packing efficiency in different types of the unit cells:
(A) \[FCC{\text{ }} < {\text{ }}BCC\]
(B) \[FCC{\text{ }} > {\text{ }}BCC{\text{ }} > \]Simple cubic
(C) \[FCC{\text{ }} < {\text{ }}BCC{\text{ }} > \]Simple cubic
(D) \[BCC{\text{ }} < {\text{ }}FCC{\text{ }} > \]Simple cubic

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Hint: To calculate the packing efficiency, we need to note three factors: volume of the unit cell, number of atoms in a structure and volume occupied by those atoms or spheres.

Formula used: \[packing{\text{ }}efficiency{\text{ }} = {\text{ }}\dfrac{{volume{\text{ }}occupied{\text{ }}by{\text{ }}number{\text{ }}of{\text{ }}atoms}}{{total{\text{ }}volume{\text{ }}of{\text{ }}unit{\text{ }}cell}}{\text{ }} \times {\text{ }}100\]

Complete step-by-step solution:
Packing efficiency of a lattice can be understood as the percentage of space occupied by the constituent particles packed in the crystal lattice. It depends on arrangements of atoms and the type of packing done. For the three types of unit cells, it can be calculated depending on their structure.
For Face-centred cubic lattice,
Total number of atoms are 4
The total volume occupied by these 4 atoms \[ = 4{\text{ }} \times {\text{ }}\dfrac{4}{3}\pi {r^3}\]
Total volume of the unit cell or cube is ${a^3}$which is equal to \[{\left( {2\sqrt 2 r} \right)^3}\]
Using the below formula and putting the above values in it, we get
\[packing{\text{ }}efficiency{\text{ }} = {\text{ }}\dfrac{{volume{\text{ }}occupied{\text{ }}by{\text{ }}4{\text{ }}atoms}}{{total{\text{ }}volume{\text{ }}of{\text{ }}unit{\text{ }}cell}}{\text{ }} \times {\text{ }}100\]
Packing efficiency $ = \dfrac{{4 \times \dfrac{4}{3}\pi {r^3}}}{{{{(2\sqrt 2 r)}^3}}} \times 100$
$\therefore $Packing efficiency of fcc lattice is $74\% $
Now let us look at Body-centred cubic lattice,
Total number of atoms are 2
The total volume occupied by these 2 atoms \[ = 2{\text{ }} \times {\text{ }}\dfrac{4}{3}\pi {r^3}\]
Total volume of the unit cell or cube is \[{(4/\sqrt {3r{\text{ }}} )^3}\]
Using the below formula and putting the above values in it, we get
\[Packing{\text{ }}efficiency{\text{ }} = {\text{ }}\dfrac{{volume{\text{ }}occupied{\text{ }}by{\text{ }}2{\text{ }}atoms}}{{total{\text{ }}volume{\text{ }}of{\text{ }}unit{\text{ }}cell}}{\text{ }} \times {\text{ }}100\]
Packing efficiency $ = \dfrac{{4 \times \dfrac{4}{3}\pi {r^3}}}{{{{(2\sqrt 2 r)}^3}}} \times 100$
$\therefore $Packing efficiency of fcc lattice is $68\% $
In simple cubic crystal lattice,
Total number of atoms are 1
The total volume occupied by the atom \[ = \dfrac{4}{3}\pi {r^3}\]
Total volume of the unit cell or cube is \[{{\text{a}}^{\text{3}}}\]which is equal to \[8{r^3}\]
Using the below formula and putting the above values in it, we get
\[Packing{\text{ }}efficiency{\text{ }} = {\text{ }}\dfrac{{volume{\text{ }}occupied{\text{ }}by{\text{ }}the{\text{ }}atom}}{{total{\text{ }}volume{\text{ }}of{\text{ }}unit{\text{ }}cell}}{\text{ }} \times {\text{ }}100\]
Packing efficiency $ = \dfrac{{\dfrac{4}{3}\pi {r^3}}}{{8{r^3}}} \times 100$
$\therefore $Packing efficiency of fcc lattice is $52.4\% $

Hence, the correct option is (B).

Note: Irrespective of the packing, some voids are always present in the cell. This is known as void space. Percentage of void space is 100 – packing efficiency. It means that low packing efficiency depicts large void space. And therefore the order gets reversed for void space.