If radius of the ${ }_{13}^{27} Al$ nucleus is estimated to be 3.6 Fermi then the radius of ${ }_{52}^{125} Te$ nucleus be nearly

6 Fermi

$r \propto A^{1 / 3} \Rightarrow \frac{r_1}{r_2}=\left(\frac{A_1}{A_2}\right)^{1 / 3}$

$\Rightarrow \frac{3.6}{r_2}=\left(\frac{27}{125}\right)^{1 / 3}=\frac{3}{5} \Rightarrow r_2$ = 6 Fermi