The ratio of the radii of the nuclei $$_{13}A{l^{27}}$$  and $$_{52}T{e^{125}}$$  is approximately

Solution :
Experimental measurements show that volume of a nucleus is proportional to its mass number $$A.$$ If $$R$$ is the radius of the nucleus assumed to be spherical, then its volume and mass no. relation is given by
$$\eqalign{ & {\text{Volume}}\left( V \right) \propto {\text{mass}}\,{\text{no}}{\text{.}}\left( A \right) \cr & {\text{or}}\,\,\left( {\frac{4}{3}\pi {R^3}} \right) \propto A \cr & {\text{or}}\,\,R \propto {A^{\frac{1}{3}}} \cr & \therefore \frac{{{R_{Al}}}}{{{R_{Te}}}} = \frac{{{{\left( {27} \right)}^{\frac{1}{3}}}}}{{{{\left( {125} \right)}^{\frac{1}{3}}}}} = \frac{3}{5} = \frac{6}{{10}} \cr} $$