If the radius of the _{13}^{27}Al nucleus is | vedclass

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(C) The radius $r$ of a nucleus is related to its mass number $A$ by the formula $r = R_0 A^{1/3}$, where $R_0$ is a constant.Therefore, the ratio of

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$6$

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(C) The radius $r$ of a nucleus is related to its mass number $A$ by the formula $r = R_0 A^{1/3}$, where $R_0$ is a constant.Therefore, the ratio of the radii of two nuclei is given by $\frac{r_1}{r_2} = \left( \frac{A_1}{A_2} ight)^{1/3}$.Given for $_{13}^{27}Al$, $r_1 = 3.6 \, 	ext{Fermi}$ and $A_1 = 27$.For $_{52}^{125}Te$, $A_2 = 125$.Substituting the values: $\frac{3.6}{r_2} = \left( \frac{27}{125} ight)^{1/3}$.$\frac{3.6}{r_2} = \frac{3}{5}$.$r_2 = \frac{3.6 	imes 5}{3} = 1.2 	imes 5 = 6 \, 	ext{Fermi}$.

If the radius of the $_{13}^{27}Al$ nucleus is estimated to be $3.6 \, \text{Fermi}$, then the radius of the $_{52}^{125}Te$ nucleus will be nearly .......... $\text{Fermi}$.

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