Question

If radius of the ${}_{13}^{27}Al$ nucleus is taken to be $R_{Al}$, then the radius of ${}_{53}^{125}Te$ nucleus is nearly

• A. ${\frac{53}{13}}^{1/3}{R}_{Al}$
• B. $\frac{5}{3}{R}_{Al}$
• C. $\frac{3}{5}{R}_{Al}$
• D. ${\frac{13}{53}}^{1/3}{R}_{Al}$

Answer 1

The correct option is B $\frac{5}{3}{R}_{Al}$

We know that for a nuclei with mass number $A$ its radius $R\propto A^{1/3}$

So $\frac{R_{Al}}{R_{te}}=\sqrt[3]{3\frac{27}{125}}=\frac{3}{5}$

Or $R_{Te}-\frac{5}{3}{R}_{Al}$

Option B is correct.