Question

The mass number of $He$ is $4$ and that of sulphur is $32$. The radius of sulphur nucleus is larger than that of helium by the factor of

• A. 4
• B. 2
• C. 8
• D. $\sqrt{8}$

Answer 1

Answer: (B)

Mass number of helium $\left(A_{He}\right)=4$
and mass number of sulphur $\left(A_s\right)=32$
Radius of nucleus, $r=r_0(A{)}^{1/3}\propto (A{)}^{1/3}$.
Therefore $\frac{r_s}{r_{He}}={\left(\frac{A_s}{A_{He}}\right)}^{1/3}={\left(\frac{32}{4}\right)}^{1/3}=(8{)}^{1/3}=2$