Radius of ${ }_2^4 He$ nucleus is 3 Fermi. The radius of ${ }_{82}^{206} Pb$ nucleus will be
Answer & explanation
Correct answer: option 3
We have $r \propto A^{1 / 3} \Rightarrow \frac{r_2}{r_1}=\left(\frac{A_2}{A_1}\right)^{1 / 3}=\left(\frac{206}{4}\right)^{1 / 3}$
∴ $r_2=3\left(\frac{206}{4}\right)^{1 / 3}$ = 11.6 Fermi