Question
If the nucleus $$_{13}^{27}Al$$ has nuclear radius of about $$3.6\,fm,$$ then $$_{32}^{125}Te$$ would have its radius approximately as
A.
$$9.6\,fm$$
B.
$$12.0\,fm$$
C.
$$4.8\,fm$$
D.
$$6.0\,fm$$
Answer :
$$6.0\,fm$$
Solution :
It has been known that a nucleus of mass number $$A$$ has radius
$$R = {R_0}{A^{\frac{1}{3}}},$$
where $${R_0} = 1.2 \times {10^{ - 15}}m$$ and $$A=$$ mass number
In case of $$_{13}^{27}A\ell ,$$ let nuclear radius be $${R_1}$$ and for $$_{32}^{125}Te,$$ nuclear radius be $${R_2}$$
$$\eqalign{
& {\text{For}}\,\,_{13}^{27}Al,\,{R_1} = {R_0}{\left( {27} \right)^{\frac{1}{3}}} = 3{R_0} \cr
& {\text{For}}\,\,_{32}^{125}Te,\,{R_2} = {R_0}{\left( {125} \right)^{\frac{1}{3}}} = 5{R_0} \cr
& \frac{{{R_2}}}{{{R_1}}} = \frac{{5{R_0}}}{{3{R_0}}} = \frac{5}{3}{R_1} = \frac{5}{3} \times 3.6 = 6\,fm. \cr} $$