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

If R is the radius of the nucleus, the corresponding volume $\frac{4}{3}{\mathrm{\pi R}}^3$ has been found to be proportional to A.

The relationship is expressed in inverse form as 

$\mathrm{R}={\mathrm{R}}_0{\mathrm{A}}^{1/3}$

The value of ${\mathrm{R}}_0 \mathrm{is} 1.2 \mathrm{x} {10}^{-15} \mathrm{m}, \mathrm{i}.\mathrm{e}., 1.2 \mathrm{fm}$

Therefore $\frac{{\mathrm{R}}_{\mathrm{AI}}}{{\mathrm{R}}_{\mathrm{Te}}}=\frac{{\mathrm{R}}_0{\left({\mathrm{A}}_{\mathrm{AI}}\right)}^{1/3}}{{\mathrm{R}}_0{\left({\mathrm{A}}_{\mathrm{Te}}\right)}^{1/3}}$

$\frac{{\mathrm{R}}_{\mathrm{AI}}}{{\mathrm{R}}_{\mathrm{Te}}}=\frac{{\left({\mathrm{A}}_{\mathrm{AI}}\right)}^{1/3}}{{\left({\mathrm{A}}_{\mathrm{Te}}\right)}^{1/3}}=\frac{{\left(27\right)}^{1/3}}{{\left(125\right)}^{1/3}}=\frac{3}{5}$

$\mathrm{or} {\mathrm{R}}_{\mathrm{Te}}=\frac{3}{5}{\mathrm{xR}}_{\mathrm{AI}}=\frac{3}{5}\mathrm{x}3.6=6 \mathrm{fm}.$