Question

The radius of the nucleus is proportional to, (if A is the atomic mass number)
A. $A$
B. ${A^3}$
C. ${A^{\dfrac{1}{3}}}$
D. ${A^{\dfrac{2}{3}}}$

Answer 1

For this question, we need to find the relation between the atomic mass number and the radius of the nucleus for an atom. Use result from Rutherford’s experiment stating relation between volumes of nucleus and the atomic mass number. Substitute the value for the volume of the nucleus and find the relation between the radius of nucleus and atomic mass number.

Complete step by step answer:
Using Rutherford's Experiment, we have that the volume of nucleus is proportional to atomic mass number.
$\therefore \,V\,\alpha \,{\rm A}$
Where $V$ is the volume and ${\rm A}$ is the atomic mass number;

But, the volume of the nucleus is $\dfrac{4}{3}\pi {R^3}$ where $R$ is the radius of the atom.
$\Rightarrow \dfrac{4}{3}\pi {R^3}\,\alpha \,{\rm A}$
$\Rightarrow {R^3}\,\alpha \,{\rm A}$
$\Rightarrow R\,\alpha \,{{\rm A}^{\dfrac{1}{3}}}$
$\therefore R\, = {R_0}\,{{\rm A}^{\dfrac{1}{3}}}$
Where, ${R_0}$ is a constant having value ${R_0} = 1.2 \times {10^{ - 15}}m$
The value of ${R_0}$ is the same for every nucleus.
From the above equation, we can conclude that $R\,\alpha \,{{\rm A}^{\dfrac{1}{3}}}$ .

Therefore, option C is the correct option.

Additional details:
The above relation gives the average radius of the nucleus. It was Ernest Rutherford who discovered that all the positive charge of an atom was located in a tiny dense centre which was later called as nucleus. All the nuclei have nearly the same density. Nucleons combine to form a nucleus. The nuclear forces treat protons and neutrons in the nucleus equally, it does not differentiate between a proton and a neutron.

Note: The relation between the atomic mass number and the radius of nucleus can be used to find the radius of some unknown nucleus, if the atomic mass number is known. Similarly, we can find the atomic number, when the radius is given. Most nuclei are approximately spherical.