Physics - Modern Physics Question with Solution | TestHub

Nuclear radius, $r\propto A^{1/3}$ where $A$ is mass number $$\frac{r_{\mathrm{Cu}}}{r_{\mathrm{Al}}}={\left(\frac{A_{\mathrm{Cu}}}{A_{\mathrm{Al}}}\right)}^{1/3}={\left(\frac{64}{27}\right)}^{1/3}$$ $$\Rightarrow r_{\mathrm{Cu}}=3.6\left(\frac{4}{3}\right)=4.8\mathrm{Fermi}$$