The distance covered by each atom of copper will be twice the radius of each atom, which is its diameter
it will be like as shown in diagram....
Let the number of atoms required is "n"
[tex]distance to be covered=nX(2r)=nX(2X4.65)[/tex]
thus total atoms required will be
[tex]n=\frac{4.65mm}{2(128pm)}[/tex]
Let us convert pm to mm for unit cancellation.
[tex]10^{-12}m=1pm[/tex]
[tex]10^{-3}m=1mm[/tex]
Thus
[tex]1mm=10^{-9}pm[/tex]
hence
[tex]number of atoms=\frac{4.65mm}{2X128X10^{-9}}= 1.81X10^{7}[/tex]
