Q. Radius of $\ _{2}{He}^{4}$ nucleus is 3fermi. The radius of ${\ _{82}Pb}^{206}$ nucleus will be
A) 5 fermi
B) 6 fermi
C) 11.16 fermi
D) 8 fermi
Explanation
(c) We have, $r\ \propto \ A^{1/3}\ $
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{r_{2}}{r_{1}} = \left\lbrack \frac{A_{2}}{A_{1}} \right\rbrack^{1/3} = \left\lbrack \frac{206}{4} \right\rbrack^{1/3}$
$\therefore$ $r_{2}\ = 3\left\lbrack \frac{206}{4} \right\rbrack^{1/3}$= 11.16 fermi
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.