Question
If radius of the $$_{13}^{27}Al$$ nucleus is taken to be $${R_{AI}},$$ then the radius of $$_{53}^{125}Te$$ nucleus is nearly
A.
$${\left( {\frac{{53}}{{13}}} \right)^{\frac{1}{3}}}{R_{AI}}$$
B.
$$\frac{5}{3}{R_{AI}}$$
C.
$$\frac{3}{5}{R_{AI}}$$
D.
$${\left( {\frac{{13}}{{53}}} \right)^{\frac{1}{3}}}{R_{AI}}$$
Answer :
$$\frac{5}{3}{R_{AI}}$$
Solution :
Radius of the nucleus is given by
$$\eqalign{
& R = {R_0}{A^{\frac{1}{3}}} \Rightarrow R \propto {A^{\frac{1}{3}}} \cr
& \frac{{{R_{Al}}}}{{{R_{Te}}}} = {\left( {\frac{{{A_{Al}}}}{{{A_{Te}}}}} \right)^{\frac{1}{3}}} = {\left( {\frac{{27}}{{125}}} \right)^{\frac{1}{3}}} = \frac{3}{5} \cr
& {R_{Te}} = \frac{5}{3}{R_{Al}} \cr} $$