If the nucleus of ${ }_{13}^{27} \mathrm{Al}$ has a nuclear radius of about 3.6 fm , then ${ }_{52}^

Question

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

Professor PC Thomas Classes · Answer

(d): Here, $A_1=27, A_2=125, R_1=3.6 \mathrm{fm}$ \ $ \begin{array}{ll} 	ext { As, } & \frac{R_2}{R_1}=\left(\frac{A_2}{A_1}ight)^{1 / 3}=\left(\frac{125}{27}ight)^{1 / 3}=\frac{5}{3} \ 	herefore & R_2=\frac{5}{3} R_1=\frac{5}{3} 	imes 3.6=6 \mathrm{fm} \end{array} $