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

  • A
    $(\frac{53}{13})^{1/3} R_{Al}$
  • B
    $\frac{5}{3} R_{Al}$
  • C
    $\frac{3}{5} R_{Al}$
  • D
    $(\frac{13}{53})^{1/3} R_{Al}$

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