Question

A nucleus with mass number $A = 27$ has a radius equal to $3.9fermi$. What will be the radius of a nucleus with mass number $A = 125$?
(A) $5.1fermi$
(B) $6.5fermi$
(C) $7.8fermi$
(D) $9.0fermi$

Answer 1

Hint: The mass number of an atom is proportional to the cube of its radius. The value of proportionality constant is given by the information provided in the question, i.e., the mass number and the radius of the atom.
Formula Used: The formulae used in the solution are given here.
$R = {R_0}{\left( A \right)^{1/3}}$ where ${R_0}$ is a constant and $A$ is the mass number.

Complete Step by Step Solution: The nucleus of an atom is the central region of an atom where the majority of the mass is concentrated. Through the scattering of alpha particles experiment by Rutherford, we learned that the nucleus of an atom contains a majority of the mass of the atom. Numerically speaking, the nucleus of an atom occupies almost ${10^{14}}$ times the volume of the atom but contains $99.99\% $ of the atomic mass. The nucleus of an atom is so small that if you expanded an atom to fill up a room, the nucleus of an atom would still be no larger than a pinhead!
An element’s mass number ($A$) is the sum of the number of protons and the number of neutrons.
The nuclear radius is the distance from the centre of the nucleus at which the density of nuclear material decreases to one-half of its maximum value at the centre. Nucleus has no definite boundary, but its radius $R$ is given by following relation:
$R = {R_0}{\left( A \right)^{1/3}}$, where ${R_0}$ is a constant and $A$ is the mass number.
It has been given that a nucleus with mass number $A = 27$ has a radius equal to $3.9fermi$.
Thus, when $R = 3.9fermi$ and $A = 27$,
$3.9 = {R_0}{\left( {27} \right)^{1/3}}$
$ \Rightarrow {R_0} = \dfrac{{3.9}}{3} = 1.3$
Now, it has been given that, $A = 125$, and we have calculated that, ${R_0} = 1.3$. We substitute these values in the equation to find the radius of a nucleus, $R = {R_0}{\left( A \right)^{1/3}}$.
Thus, we have,
$R = 1.3 \times {\left( {125} \right)^{1/3}}$
$ \Rightarrow R = 1.3 \times 5 = 6.5fermi$
Thus, the radius of the nucleus of the atom with mass number $A = 125$, is $6.5fermi$.

Hence, the correct answer is Option B.

Note: The nucleus of an atom consists of a tightly packed arrangement of protons and neutrons. These are the two heavy particles in an atom and hence $99.99\% $ of the mass is concentrated in the nucleus. Of the two, the protons possess a net positive charge and hence the nucleus of an atom is positively charged on the whole and the negatively charged electrons revolve around the central nucleus. Since the mass concentration at the nucleus of an atom is immense the nuclear forces holding the protons and the neutrons together are also large. The protons are in such close vicinity to each other inside the tiny nucleus and therefore the electrostatic forces of repulsion also act inside the nucleus. Nuclear energy relies on nothing but releasing the energy trapped in the nucleus of an atom. The total number of protons in a nucleus is equal to the number of electrons revolving around the nucleus and hence the atom, on the whole, is electrically neutral.