If the radii of {}_{13}^{27}Al and {}_{30}^{6 | 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 ${}_{13}^{27}Al$,th

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$\frac{3}{4}$

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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 ${}_{13}^{27}Al$,the mass number $A_{1} = 27$.For ${}_{30}^{64}Zn$,the mass number $A_{2} = 64$.Taking the ratio of the radii:$\frac{R_{1}}{R_{2}} = \frac{R_{0} A_{1}^{1/3}}{R_{0} A_{2}^{1/3}} = \left(\frac{A_{1}}{A_{2}}\right)^{1/3}$Substituting the values:$\frac{R_{1}}{R_{2}} = \left(\frac{27}{64}\right)^{1/3}$Since $27 = 3^{3}$ and $64 = 4^{3}$,we have:$\frac{R_{1}}{R_{2}} = \left(\frac{3^{3}}{4^{3}}\right)^{1/3} = \frac{3}{4}$Thus,the correct option is $B$.

If the radii of ${}_{13}^{27}Al$ and ${}_{30}^{64}Zn$ nuclei are $R_{1}$ and $R_{2}$ respectively,then $\frac{R_{1}}{R_{2}} = $

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