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)

2

Answer 2

Mass number of helium $(A_{He})=4$ and mass number of sulphur $(A_s)=32$
Radius of nucleus, r = r$_0(A)^{1/3} ∝ (A)^{1/3}$ Therefore
$\frac{r_{s}}{r_{He}}=\left(\frac{A_{s}}{A_{He}}\right)^{1/ 3}=\left(\frac{32}{4}\right)^{1/3}=\left(8\right)^{1/3}=2$