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 $\left(A_{H e}\right)=4$
and mass number of sulphur $\left(A_{s}\right)=32$
Radius of nucleus, $r=r_{0}(A)^{1 / 3} \propto(A)^{1 / 3}$.
Therefore $\frac{r_{s}}{r_{H e}}=\left(\frac{A_{s}}{A_{H e}}\right)^{1 / 3}=\left(\frac{32}{4}\right)^{1 / 3}=(8)^{1 / 3}=2$