If radius of the \({^{27}_{13}\text{Al}}\) nucleus is taken to be RAl, then the radius of \({^{125}_{53}\text{Te}}\) nucleus is nearly:
1
\(\left(\dfrac{13}{53}\right)^{1/3} \text{R}_{\text{Al}}\)
2
\(\left(\dfrac{53}{13}\right)^{1/3} \text{R}_{\text{Al}}\)
3
\(\dfrac{5}{3} \text{R}_{\text{Al}}\)
4
\(\dfrac{3}{5} \text{R}_{\text{Al}}\)
5
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