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 If the nuclear radius of ²⁷ Al is 3.6 Fermi, the approximate nuclear radius of ⁶⁴Cu  in Fermi is

Answer:

Explanation:

 Nuclear radius , $r\propto A^{1/3}$

 Where A is mass number

 $\frac{r_{Cu}}{r_{Al}}=\left(\frac{A_{Cu}}{A_{Al}}\right)^{1/3}=\left(\frac{64}{27}\right)^{1/3}$

$\Rightarrow r_{Cu}=3.6\left(\frac{4}{3}\right)=4.8 Fermi$