## Chemistry and Chemical Reactivity (9th Edition)

$1.2\times10^9$ atoms
Since the radius is 145 pm, the diameter of the gold atom is $D= 145\ pm\times 2=290\ pm\times \dfrac{1\times10^{-12}\ m}{1\ pm}\times\dfrac{1\times10^2\ cm}{1\ m}=290\times10^{-10}\ cm$ For a distance of $L=36\ cm$, we would need $n=L/D$ atoms which would be $1.2\times10^9$ atoms.