Question
The stable nucleus that has a radius half that of $$F{e^{56}}$$ is
A.
$$L{i^7}$$
B.
$$N{a^{21}}$$
C.
$${S^{^{16}}}$$
D.
$$C{a^{40}}$$
Answer :
$$L{i^7}$$
Solution :
The relation between nuclei radius $$\left( R \right)$$ and mass number $$\left( A \right)$$ is given by
$$\eqalign{
& R \propto {A^{\frac{1}{3}}}\,......\left( {\text{i}} \right) \cr
& {\text{or}}\,\,A \propto {R^3} \cr
& {\text{or}}\,\,\frac{{{A_1}}}{{{A_2}}} = {\left( {\frac{{{R_1}}}{{{R_2}}}} \right)^3} \cr
& {\text{Given,}}\,\,{R_1} = R,{R_2} = \frac{R}{2},A = 56 \cr
& \therefore \frac{{56}}{{{A_2}}} = {\left( {\frac{R}{{\frac{R}{2}}}} \right)^3} = {2^3} = 8 \cr
& {\text{or}}\,\,{A_2} = \frac{{56}}{8} = 7 \cr} $$
Thus, required stable nucleus will be $$L{i^7}.$$