Question

Copper crystal has a face centred cubic structure. Atomic radius of copper metal? Atomic mass of copper is \[63.5\].
A) \[6.7gc{m^{ - 3}}\]
B) \[7.8gc{m^{ - 3}}\]
C) \[8.9gc{m^{ - 3}}\]
D) \[9.10gc{m^{ - 3}}\]

Answer 1

Hint: In matters are divided into three types. There are solids, liquids and gas. Solids are divided into types. They are crystalline and amorphous. The crystalline is further divided into ionic solids, covalent solids, molecular solids and metallic solids. In crystalline solid, the basic repeating structural unit is called a unit cell.
Formula used:
Number of atoms in a face centered cubic unit cell is dependent on the number of atoms in the corner of the unit cell and face center atoms in the cube.
Number of atoms in a face centered cubic unit cell is equal to the sum of the number of atoms in the corner of the unit cell divided by eight, , because each corner is divided by eight unit cell and the number of atoms in the face of the unit cell divided by two, because each face is divided by two unit cell and totally six faces in the unit cell.
\[\begin{gathered}
  {\text{number of atoms in face cubic unit cell = }}\dfrac{{{\text{Nc}}}}{{\text{8}}}{\text{ + }}\dfrac{{{\text{Nf}}}}{2} \\
    \\
\end{gathered} \]
Here, Nc is represented as the number of atoms in the corner of the unit cell.
Nf is represent as the number of atoms in the face of the unit cell
Density of the unit cell,
\[{\text{Density of the unit cell = }}\dfrac{{{\text{nM}}}}{{{{\text{a}}^{\text{3}}}{{\text{N}}_{\text{A}}}}}\]
Volume of unit cell is a3
Here,
M is molar mass.
NA is Avogadro number.
a is the edge length of its unit cell.
n is the number of atoms in a unit cell.

Complete answer:
Calculate the number of atoms in the face centered cubic cell,
\[{\text{number of atoms in face cubic unit cell = }}\dfrac{{{\text{Nc}}}}{{\text{8}}}{\text{ + }}\dfrac{{{\text{Nf}}}}{2}\]
\[ = \dfrac{{\text{8}}}{{\text{8}}}{\text{ + }}\dfrac{{\text{6}}}{2}\]
$ = 1 + 3 = 4$
So, there are four atoms in the face centered cubic cell.
Atomic mass of copper is \[63.5\].
Here,
M is molar mass of copper is \[63.5\].
The Avogadro number is \[6.022 \times {10^{23}}\].
a is the edge length of its unit cell is \[2 \times \sqrt 2 \times 128 \times {10^{ - 10}}\]
n is the number of atoms in the unit cell is \[4\].
Density of the unit cell,
\[{\text{Density of the unit cell = }}\dfrac{{{\text{nM}}}}{{{{\text{a}}^{\text{3}}}{{\text{N}}_{\text{A}}}}}\]
Volume of unit cell is a3
\[{\text{Density of the unit cell = }}\dfrac{{63.5 \times 4}}{{{{{\text{(}}2 \times \sqrt 2 \times 128 \times {{10}^{ - 10}}{\text{)}}}^{\text{3}}} \times 6.022 \times {{10}^{23}}}}\]
On simplification we get,
\[ = 8.9gc{m^{ - 3}}\]
According to the above discussion and calculation the density of the copper crystal is \[8.9gc{m^{ - 3}}\].

So, the correct answer is “Option C”.

Note:
There are seven types of primitive crystal systems. There are cubic, tetragonal, orthorhombic, hexagonal, monoclinic, triclinic and rhombohedral. This classification of crystal is based on the crystallographic nature of angles, unit cell, primitive lattice and axes. These cubic crystals are further divided into three types. There are simple cubic crystals, face centered cubic crystals and body centered cubic crystals. This classification is based on the atoms arrangement in the cubic structure in the unit cell.