Question

If the nuclear radius of Al7 is 3.6 fm, the approximate nuclear radius of Cu64
in fermi is

Answer 1

$r=r_0A^{\frac 13},$

$\frac{r_{Cu}}{r_{Al}}=(\frac{{A_{Cu}}}{A_{Al}} )^{\frac 13}=(\frac{64}{27})^{\frac 13}=\frac 43,$

$r_{Cu}=\frac 43\cdot 3.6=4.8~\text{fm}.$