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 1

Answer: (A)

Key Idea: The radius of nucleus is proportional to cube root of atomic mass. The nuclear radius is $R\propto A^{1/3}$ ?...(i) or $A\propto R^3$ or $\frac{A_1}{A_2}={\left(\frac{R_1}{R_2}\right)}^3$ Here, $R_1=R,R_2=\frac{R}{2},A=56$ $\therefore$ $\frac{56}{A^2}={\left(\frac{R}{R/2}\right)}^3=2^3=8$ or $A_2=\frac{56}{8}=7$ Thus, stable nucleus will be $L{i}^7$ .