The density of copper is 8.95 g `cm^(-3)` .It has a face centred cubic structure .What is the radius of copper atom?
Atomic mass Cu =63.5 `gmol^(-1)N_(A)=6.02xx10^(23) mol^(-1)`

Explanation

Given, Atomic weight (M)= 63.55 g `mol^(-1)`
Density=8.93 g `cm^(-3)`
Edge length (a) = not given
Face centred cubic system has 4 atoms per unit cell (Z)
We know that -
Density, `d=((Z xx M))/(N_(A) xx a^(3))`
Substituting the values -
`8.93 "g cm"^(-3) =((63.55"g mol"^(-1) xx 4))/(6.023 xx10^(23) xx a^(3))`
Or, `a^(3) =((63.55"g mol"^(-1) xx 4))/(6.023 xx 10^(23) xx 8.93 "g cm"^(-3))`
`=4.72 xx 10^(-23) `
`=0.472 xx 10^(-24)`
`a=3.6 xx 10^(-8)cm`
Now, radius (r ) is fcc system is given by -
`r = ((sqrt(2)a))/(4)`
`=((1.414 xx 3.6 xx 10^(-8) cm))/(4)`
`=1.27 xx 10^(-8) cm`
= 0.127 nm