Question

The density of copper metal is \[8.95gc{m^{ - 3}}\]. If the radius of the copper atom is \[127.8pm\], the copper unit cell is a simple cubic, a body centered cubic or face centered cubic structure? (Atomic mass of cu \[ = 63.54g/mol\] and \[{N_A} = 6.023 \times {10^{23}}mo{l^{ - 1}}\])

Answer 1

Hint: We have to know that copper is used as a good conductor of electricity and it is used as a stimulation of energy. The copper sphere is mainly used for decorative purposes and it is also used for cleaning rituals and healing. We must remember that copper is also known as lucky metal or healing metal. If two identical solid copper spheres are connected with each other, then the gravitational force is proportional to the fourth power of radius.

Complete answer:
According to the question, radius of copper atom\[ = 127.8pm\],
For FCC unit cell, \[a = 2\sqrt 2 r\]
Therefore, \[{a^3} = {\left( {2\sqrt 2 r} \right)^3} = 4.723 \times {10^{ - 23}}\]
The atomic mass of copper is equal to \[63.54g/mol\] and the density of copper metal is \[8.95gc{m^{ - 3}}\]
The formula of density of the unit cell is equal to, \[d = \dfrac{{Z \times M}}{{{N_A} \times {a^3}}}\]
Substitute the given values in the above equation we get,
\[8.95 = \dfrac{{Z \times 63.54}}{{6.023 \times {{10}^{23}} \times 4.723 \times {{10}^{ - 23}}}}\]
By rearranging the equation, will get the value of Z.
\[Z = \dfrac{{8.95 \times 6.023 \times 4.723}}{{63.54}}\]
On simplification we get,
\[Z = 4.00\]

Note:
We must have to know that the FCC and BCC are the different arrangements of crystalline structures. As we know, the face centered lattice which is represented as FCC and it has a tightly packed atomic arrangement having the atomic packing factor $0.74$. We need to remember that the face centered cubic lattice is generally defined as the cubic lattice. Here, the face position is completely equivalent to each of the eight corners. And the BCC indicates the body centered cubic lattice and the total number of spheres available in the BCC equal to $9$.