Q. 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
A) $4\ Fermi$
B) $5\ Fermi$
C) $6\ Fermi$
D) $8\ Fermi$
Explanation
(c) $r \propto A^{1/3} \Rightarrow \frac{r_{1}}{r_{2}} = \left( \frac{A_{1}}{A_{2}} \right)^{1/3}$
$\Rightarrow \frac{3.6}{r_{2}} = \left( \frac{27}{125} \right)^{1/3} = \frac{3}{5} \Rightarrow r_{2} = 6\ fermi$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.