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The density of $KBr$ is $2.7$$gc{m^{ - 3}}$. The length of the unit cell is $654pm$. Atomic mass of amu $K = 39$, $Br = 80$. Then the solid is?
A.Face centered simple
B.Simple cubic system
C.Body centered cubic system
D.None of the above

Last updated date: 14th Jul 2024
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Hint :Solids are classified according to the orientation of their atoms. A unit cell is the smallest unit of the crystal lattice. Repeating unit cells forms a whole crystal solid.

Complete Step By Step Answer:
In a simple cubic system, only a single atom is located at the corner of the lattice. The number of unit cells present is $4$. Every atom is shared by $8$unit cells. The body center cube is similar to a simple cubic structure except it has a single atom in the center of the entire lattice. It has one atom at each corner. In the face centered cubic unit cell atoms are present on all corners and one atom is present on each plane as well. There is no center atom present at the center of the lattice.
Now density of the unit cell is defined by
$d = \dfrac{{z \times m}}{{{a^3} \times {N_o}}}$
$z = \dfrac{{d \times {a^3} \times {N_o}}}{m}$
Here $d = $density
$z = $Number of atoms per unit cell
$m = $ Molecular weight
$a = $ Edge length
${N_o} = $Avogadro’s number, which is $6.023 \times {10^{23}}$
The total molecular weight is $39 + 80 = 119$
So therefore we solve as
$z = $$\dfrac{{2.75 \times {{\left( {654 \times {{10}^{ - 10}}} \right)}^3} \times 6.023 \times {{10}^{23}}}}{{119}}$
$z = 4$
Therefore $4$ molecules of $KBr$ are present in the unit cell.
Hence it is a face-centered cubic unit cell.

Hence, option (C) body centered cubic center is correct.

Note :
In a simple cubic system, a unit cell has ${18^{th}}$of an atom. There are seven different types of a unit cell like cubic, tetragonal, orthorhombic, triclinic, monoclinic, hexagonal and rhombohedral. The difference between them is their angle and crystallographic axes.