Solveeit Logo

Question

Question: The correct order of the packing efficiency in different types of the unit cells: (A) \[FCC{\text{...

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

Explanation

Solution

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 efficiency = volume occupied by number of atomstotal volume of unit cell × 100packing{\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 × 43πr3 = 4{\text{ }} \times {\text{ }}\dfrac{4}{3}\pi {r^3}
Total volume of the unit cell or cube is a3{a^3}which is equal to (22r)3{\left( {2\sqrt 2 r} \right)^3}
Using the below formula and putting the above values in it, we get
packing efficiency = volume occupied by 4 atomstotal volume of unit cell × 100packing{\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 =4×43πr3(22r)3×100 = \dfrac{{4 \times \dfrac{4}{3}\pi {r^3}}}{{{{(2\sqrt 2 r)}^3}}} \times 100
\therefore Packing efficiency of fcc lattice is 74%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 × 43πr3 = 2{\text{ }} \times {\text{ }}\dfrac{4}{3}\pi {r^3}
Total volume of the unit cell or cube is (4/3r )3{(4/\sqrt {3r{\text{ }}} )^3}
Using the below formula and putting the above values in it, we get
Packing efficiency = volume occupied by 2 atomstotal volume of unit cell × 100Packing{\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 =4×43πr3(22r)3×100 = \dfrac{{4 \times \dfrac{4}{3}\pi {r^3}}}{{{{(2\sqrt 2 r)}^3}}} \times 100
\therefore Packing efficiency of fcc lattice is 68%68\%
In simple cubic crystal lattice,
Total number of atoms are 1
The total volume occupied by the atom =43πr3 = \dfrac{4}{3}\pi {r^3}
Total volume of the unit cell or cube is a3{{\text{a}}^{\text{3}}}which is equal to 8r38{r^3}
Using the below formula and putting the above values in it, we get
Packing efficiency = volume occupied by the atomtotal volume of unit cell × 100Packing{\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 =43πr38r3×100 = \dfrac{{\dfrac{4}{3}\pi {r^3}}}{{8{r^3}}} \times 100
\therefore Packing efficiency of fcc lattice is 52.4%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.