Respuesta :
In a face centered cubic unit cell, the volume occupied by the particles of the substance is about 74% of the total unit cell. First, we find the volume of a single atom assuming it to be sphere:
Volume(atom) = 4/3 x π x r³
Volume(atom) = 4/3 x π x (197 x 10⁻¹²)³
Volume(atom) = 2.87 x 10⁻²⁹ m³
There are a total of 4 atoms in a FCC unit cell, so the total volume occupied by atoms is:
2.87 x 10⁻²⁹ x 4
= 1.15 x 10⁻²⁸ m³
Volume of cell = (1.15 x 10⁻²⁸ ) / 0.74
Volume of cell = 1.55 x 10⁻²⁸ m³
Volume(atom) = 4/3 x π x r³
Volume(atom) = 4/3 x π x (197 x 10⁻¹²)³
Volume(atom) = 2.87 x 10⁻²⁹ m³
There are a total of 4 atoms in a FCC unit cell, so the total volume occupied by atoms is:
2.87 x 10⁻²⁹ x 4
= 1.15 x 10⁻²⁸ m³
Volume of cell = (1.15 x 10⁻²⁸ ) / 0.74
Volume of cell = 1.55 x 10⁻²⁸ m³
The volume of unit cell in FCC has been [tex]\rm 1.53\;\times\;10^-^2^8\;m^3[/tex].
The face centered cubic lattice has been given as the arrangement of 4 atoms in the lattice of a cube of molecule. The fcc cell has been accomplished with 75% of the volume of the atoms.
Computation for the volume of cell
The given radius of the cell has been, [tex]r=197\;\rm pm[/tex]
The volume of the atom has been given by:
[tex]\rm Volume = \dfrac{4}{3}\;\pi\;\textit r^3\\ Volume= \dfrac{4}{3}\;\times\;3.14\;\times\;(197\;\times\;10^-^1^2)^3\;m^3\\ Volume =2.87\;\times\;10^-^2^9\;m^3[/tex]
The volume of a single atom in the FCC cell has been [tex]2.87\;\times\;10^-^2^9\;\rm m^3[/tex].
There are 4 atoms in a cell. The volume of 4 atoms has been:
[tex]\rm 1 \;atom=2.87\;\times\;10^-^2^9\;m^3\\ 4\;atoms=4\;\times\;2.87\;\times\;10^-^2^9\;m^3\\ 4\;atoms=1.15\;\times\;10^-^2^8\;m^3[/tex]
The volume of 4 atoms in a FCC cell has been [tex]1.15\;\times\;10^-^2^8\;\rm m^3[/tex].
The atoms have been contributing to 75% of the volume of cell. Thus, the volume of unit cell has been:
[tex]\rm Volume \;of\;unit\;cell=\dfrac{1.15\;\times\;10^-^2^8}{0.75} \;m^3\\ Volume \;of\;unit\;cell=1.53\;\times\;10^-^2^8\;m^3 [/tex]
The volume of unit cell in FCC has been [tex]\rm 1.53\;\times\;10^-^2^8\;m^3[/tex].
Learn more about the volume of FCC cell, here:
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