Copper crystallizes with face centered unit cells. If the radius of the copper atom is 127.8 pm. Calculate the density of the copper metal (atomic mass of Cu = 63.55 u).

Hint: FCC means face centered cubic packing. FCC has six faces and eight corners. There is a relationship between edge length of the unit cell and density of unit cell in FCC and it is as follows.
\[d=\dfrac{Z\times M}{{{a}^{3}}\times {{N}_{A}}}\]
Here d = density of the atom
Z = number of the atoms in the unit cell
M = Molecular weight of atomic mass of the atom
a = edge length of the unit cell
${{N}_{A}}$ = Avogadro number

Complete step by step answer:
> In the question it is given that Copper crystallizes with a face centered unit cell and given the radius of the copper atom and we have to calculate the density of the copper metal.
> First we have to calculate the edge length of the unit cell after that we have to calculate the density of the copper atom.
> The edge length of the unit cell can be calculated as follows for FCC.
\[a=\dfrac{4r}{\sqrt{2}}\to (1)\]
Here, r = atomic radius = 127.8 pm = $1.278\times {{10}^{-8}}cm$
a = edge length of the unit cell.

> Substitute all the known values in the above formula (1) to calculate the edge length of the unit cell in FCC and it is as follows.
\[\begin{align}
 & a=\dfrac{4r}{\sqrt{2}} \\
 & a=2\sqrt{2}r \\
 & a=2\sqrt{2}\times (1.278\times {{10}^{-8}})cm \\
\end{align}\]

> Now substitute the edge length of the unit cell in the below formula to get the radius of the copper metal.
\[d=\dfrac{Z\times M}{{{a}^{3}}\times {{N}_{A}}}\to (2)\]
Here d = density of the atom
Z = number of the atoms in the unit cell = 4 atoms
M = Molecular weight of atomic mass of the atom = 63.55
a = edge length of the unit cell
${{N}_{A}}$ = Avogadro number = $6.023\times {{10}^{23}}$

> Substitute all the known values in the above formula (2) to get the density of the copper in FCC and it is as follows.
\[\begin{align}
& d =\dfrac{Z\times M}{{{a}^{3}}\times {{N}_{A}}} \\
& d =\dfrac{4\times 63.55}{{{[2\sqrt{2}\times (1.278\times {{10}^{-8}})]}^{3}}\times 6.023\times {{10}^{23}}} \\
 & d=\dfrac{254.2}{4.723\times {{10}^{-23}}\times 6.023\times {{10}^{23}}} \\
 & d = 8.94g{{m}^{-3}} \\
\end{align}\]