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

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(B) The radius of a nucleus 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 $^{27}_{13}Al$ nucl

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$\frac{5}{3} R_{Al}$

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(B) The radius of a nucleus 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 $^{27}_{13}Al$ nucleus,the mass number $A_{Al} = 27$. Thus,$R_{Al} = R_0 (27)^{1/3} = 3 R_0$.For the $^{125}_{53}Te$ nucleus,the mass number $A_{Te} = 125$. Thus,$R_{Te} = R_0 (125)^{1/3} = 5 R_0$.Taking the ratio of the two radii:$\frac{R_{Te}}{R_{Al}} = \frac{5 R_0}{3 R_0} = \frac{5}{3}$.Therefore,$R_{Te} = \frac{5}{3} R_{Al}$.

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

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