The ratio of the surface area of the nuclei ${ }_{52} \mathrm{Te}^{125}$ to that of ${ }_{13} \mathr

Question

The ratio of the surface area of the nuclei ${ }_{52} \mathrm{Te}^{125}$ to that of ${ }_{13} \mathrm{Al}^{27}$ is\

Professor PC Thomas Classes · Answer

Nucleus radius, $\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}$ \ Where $R_{0}$ is a constant and $A$ is the mass number\ $	herefore \frac{\mathrm{R}_{\mathrm{Te}}}{\mathrm{R}_{\mathrm{Al}}}=\left(\frac{\mathrm{A}_{\mathrm{Te}}}{\mathrm{A}_{\mathrm{Al}}}ight)^{1 / 3}=\left(\frac{125}{27}ight)^{1 / 3}=\frac{5}{3}$\ Assuming nucleus to be spherical\ $	herefore$ Surface area, $S=4 \pi R^{2}$\ $	herefore \frac{\mathrm{S}_{\mathrm{Te}}}{\mathrm{S}_{\mathrm{Al}}}=\left(\frac{\mathrm{R}_{\mathrm{Te}}}{\mathrm{R}_{\mathrm{Al}}}ight)^{2}=\left(\frac{5}{3}ight)^{2}=\frac{25}{9}$\