If the nucleus {}_{13}^{27}Al has a nuclear r | vedclass

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(D) The nuclear radius $R$ is given by the formula $R = R_0 A^{1/3}$,where $A$ is the mass number and $R_0$ is a constant.For the given nuclei:For ${}

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$6$

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(D) The nuclear radius $R$ is given by the formula $R = R_0 A^{1/3}$,where $A$ is the mass number and $R_0$ is a constant.For the given nuclei:For ${}_{13}^{27}Al$,$A_1 = 27$ and $R_1 = 3.6 \, fm$.For ${}_{52}^{125}Te$,$A_2 = 125$ and we need to find $R_2$.Taking the ratio:$\frac{R_2}{R_1} = \left( \frac{A_2}{A_1} ight)^{1/3}$Substituting the values:$\frac{R_2}{3.6} = \left( \frac{125}{27} ight)^{1/3}$$\frac{R_2}{3.6} = \frac{5}{3}$$R_2 = \frac{5}{3} 	imes 3.6 = 5 	imes 1.2 = 6 \, fm$.Thus,the radius of ${}_{52}^{125}Te$ is $6 \, fm$.

If the nucleus ${}_{13}^{27}Al$ has a nuclear radius of about $3.6 \, fm$,then ${}_{52}^{125}Te$ would have its radius approximately as .......$fm$.

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