If radius of the ²⁷₁₃Al nucleus is taken to be R_Al then… | Nuclear size

If radius of the ²⁷₁₃Al nucleus is taken to be R_Al then the radius of ¹²⁵₅₃Te nucleus is nearly:

1 2015 AIPMT-I Nuclei Nuclear size Easy

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

A $\left(\frac{53}{13}\right)^{1/3}R_{Al}$

B $\frac{5}{3}R_{Al}$

C $\frac{3}{5}R_{Al}$

D $\left(\frac{13}{53}\right)^{1/3}R_{Al}$

Detailed Solution

Nuclear radius $R \propto A^{1/3}$
$\frac{R_{Te}}{R_{Al}} = \left(\frac{125}{27}\right)^{1/3} = \frac{5}{3}$
$R_{Te} = \frac{5}{3}R_{Al}$

2 questions from this chapter have appeared across 2 exam years.