
Packing fraction in face centered cubic unit cell is:
(A) 0.7406
(B) 0.6802
(C) 0.5236
(D) None of the above
Answer
446.4k+ views
Hint: For a better solution it is very important to understand what is packing fraction So Packing fraction is defined as a way of expressing the variation of isotopic mass from the whole mass number (atomic mass). This fraction can have positive or can have negative sign. This difference is due to the transformation of mass into energy in the formation of nucleus.
Complete step by step solution:
Let us know what is face centered cubic unit cell and then solve for its packing fraction:
Packing faction or Packing efficiency is the percentage of total space filled by the particles.
The face centered unit cell (FCC) contains atoms at all the corners of the crystal lattice and at the center of all the faces of the cube. The atom present at the face centered is shared between 2 adjacent unit cells and only 1/2 of each atom belongs to an individual cell. The packing efficiency of FCC lattice is 74%.
Let r be the radius of the sphere and $ a $ be the edge length of the cube and the number of atoms or spheres is n that is equal to 4.
As there are 4 sphere in FCC unit cell
Volume of four spheres is given by
$ V = 4 \times \dfrac{4}{3}\pi {r^3} $
In FCC, the corner spheres are in touch with the face centered sphere so the relation between the edge length and radius of the sphere is given by
$ r = \dfrac{a}{{4\sqrt 2 }} $
Substituting the above values in the below equation and solving:
$ packing{\text{ }}efficiency = \dfrac{{volume{\text{ }}occupied{\text{ }}by{\text{ }}all{\text{ }}the{\text{ }}spheres{\text{ in }}unit{\text{ }}cell}}{{Total{\text{ }}volume{\text{ }}of{\text{ }}the{\text{ }}unit{\text{ }}cell}} \times 100 $
$ Packing{\text{ }}fraction = \dfrac{{n \times \dfrac{4}{3}\pi {r^3}}}{{{a^3}}} $
$ packing{\text{ }}fraction = \dfrac{{4 \times \dfrac{4}{3} \times {{\left( {\dfrac{a}{{4\sqrt 2 }}} \right)}^3}}}{{{a^3}}} = \dfrac{\pi }{{3\sqrt 2 }} = 0.74 $
Hence the correct option is (A).
Note:
The relationship between the edge length of the cube unit and the radius of the sphere can be properly derived by a proper diagram and mathematical theorems. The packing efficiency of different cubic unit cells vary according to their structure and shape.
Complete step by step solution:
Let us know what is face centered cubic unit cell and then solve for its packing fraction:
Packing faction or Packing efficiency is the percentage of total space filled by the particles.
The face centered unit cell (FCC) contains atoms at all the corners of the crystal lattice and at the center of all the faces of the cube. The atom present at the face centered is shared between 2 adjacent unit cells and only 1/2 of each atom belongs to an individual cell. The packing efficiency of FCC lattice is 74%.
Let r be the radius of the sphere and $ a $ be the edge length of the cube and the number of atoms or spheres is n that is equal to 4.
As there are 4 sphere in FCC unit cell
Volume of four spheres is given by
$ V = 4 \times \dfrac{4}{3}\pi {r^3} $
In FCC, the corner spheres are in touch with the face centered sphere so the relation between the edge length and radius of the sphere is given by
$ r = \dfrac{a}{{4\sqrt 2 }} $
Substituting the above values in the below equation and solving:
$ packing{\text{ }}efficiency = \dfrac{{volume{\text{ }}occupied{\text{ }}by{\text{ }}all{\text{ }}the{\text{ }}spheres{\text{ in }}unit{\text{ }}cell}}{{Total{\text{ }}volume{\text{ }}of{\text{ }}the{\text{ }}unit{\text{ }}cell}} \times 100 $
$ Packing{\text{ }}fraction = \dfrac{{n \times \dfrac{4}{3}\pi {r^3}}}{{{a^3}}} $
$ packing{\text{ }}fraction = \dfrac{{4 \times \dfrac{4}{3} \times {{\left( {\dfrac{a}{{4\sqrt 2 }}} \right)}^3}}}{{{a^3}}} = \dfrac{\pi }{{3\sqrt 2 }} = 0.74 $
Hence the correct option is (A).
Note:
The relationship between the edge length of the cube unit and the radius of the sphere can be properly derived by a proper diagram and mathematical theorems. The packing efficiency of different cubic unit cells vary according to their structure and shape.
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