Question

The mass number of He is 4 and that of sulphur is 32. The radius of sulphur nuclei is larger than that of helium by
(A) $\sqrt{8}$
(B) 4
(C) 2
(D) 8

Answer 1

Hint: The relationship between the mass number and the radius of the nucleus is helpful in solving the given illustration.

Complete step by step solution:
Let us firstly study about the mass number of an atom and also the radius of the same to compare and relate them.
Mass number-
The total number of neutrons and protons in the atomic nucleus is the mass number of the atom. This is different for each isotope of an element.
Facts-
-Neutral atoms have equal number of electrons and protons.
-As, the number of neutrons is variable for the same atom it results in the formation of an isotope.
-Thus, isotopes of the same element will have the same atomic number but different mass numbers.
Radius of an atom-
It is the distance between the central nucleus of an atom to the outermost shell i.e. size of an atom is the atomic radius.
As the atomic number increases, the number of shells increases resulting in increase in atomic radius.
Interrelationship between atomic radius and mass number-
$\begin{align}
  & R\propto {{A}^{\dfrac{1}{3}}} \\
 & R={{R}_{o}}{{A}^{\dfrac{1}{3}}} \\
\end{align}$
where,
R = radius of an atom
A = mass number of an atom
${{R}_{o}}$ = constant having value of (1.2 to 1.7)$\times {{10}^{-15}}$

Thus, the radius of a nucleus varies as ${{A}^{\dfrac{1}{3}}}$.
Illustration-
Given that,
Mass number of helium is 4
Mass number of sulphur is 32.
The radius of sulphur nuclei is larger than that of helium by,
$\dfrac{{{R}_{s}}}{{{R}_{He}}}={{\left( \dfrac{{{A}_{s}}}{{{A}_{He}}} \right)}^{\dfrac{1}{3}}}={{\left( \dfrac{32}{4} \right)}^{\dfrac{1}{3}}}=2$

Thus, option (C) is correct.

Note: Do not get confused by seeing the option (A) as $\sqrt{8}$. Our answer can also be $\sqrt[3]{8}$ which is equal to 2.