A) \[{{\left[ \frac{13}{53} \right]}^{{}^{1}/{}_{3}}}{{R}_{Al}}\]
B) \[{{\left[ \frac{53}{15} \right]}^{{}^{1}/{}_{3}}}{{R}_{Al}}\]
C) \[\frac{5}{3}{{R}_{Al}}\]
D) \[\frac{3}{5}{{R}_{Al}}\]
Correct Answer: C
Solution :
\[\operatorname{Radius} of Nucleus \propto {{A}^{{}^{1}/{}_{3}}}\] \[\therefore A = mass number\] \[{{\operatorname{R}}_{Al}}=\left( 27 \right){{~}^{{}^{1}/{}_{3}}}\] \[{{\operatorname{R}}_{\operatorname{Te}}}=\left( 125 \right){{~}^{{}^{1}/{}_{3}}}\] \[\frac{{{\operatorname{R}}_{Al}}}{{{\operatorname{R}}_{\operatorname{Te}}}}=\frac{3}{5}\] \[{{\operatorname{R}}_{Te}}=\frac{5}{3}{{R}_{Al}}\]
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