Question

Calculate the density of a nucleus in which the mass of each nucleon is $1.67 \times {10^{ - 27}}kg$ and ${R_0} = 1.4 \times {10^{ - 15}}m$ .
(A) $2.995 \times {10^{17}}kg/{m^3}$
(B) $1.453 \times {10^{19}}kg/{m^3}$
(C) $1.453 \times {10^{16}}kg/{m^3}$
(D) $1.453 \times {10^{17}}kg/{m^3}$

Answer 1

Hint : The nucleus is spherical in shape so the volume of a nucleus will be calculated by the formula for the volume of a sphere. Density is given by the mass per unit volume. So on substituting the mass of all the nucleons and then dividing it by the volume, we will get the density of the nucleus.

Formula Used: The formulae used to solve this have been stated as,
$\Rightarrow {\rho _{nucleus}} = \dfrac{{{m_{nucleus}}}}{{{V_{nucleus}}}}$ where ${\rho _{nucleus}}$ is the density of the nucleus, ${m_{nucleus}}$ is the mass of the nucleus and ${V_{nucleus}}$ is the volume of the nucleus of an atom.
$\Rightarrow {V_{nucleus}} = \dfrac{4}{3}\pi {R^3}$ where ${R_{}}$ is the radius of the nucleus and $R = {R_0}{A^{\dfrac{1}{3}}}$ , $A$ is the mass number of the nucleus.

Complete step by step answer
The nucleus of an atom is the small, dense region in the atom that consists of protons and neutrons. The nucleus was discovered by Ernest Rutherford and observed that the mass of the atom is mainly concentrated in the nucleus, and has a net positive charge due to the presence of protons in the nucleus. The protons and neutrons are bound together by nuclear force and these particles are collectively known as nucleons.
We know that the nucleus is a tiny spherical region located at the centre of an atom. Given in the question that, the radius of the nucleus is ${R_0} = 1.4 \times {10^{ - 15}}m$ .
From our knowledge of mathematics, we know that the volume $V$ of a spherical body of radius $r$ is equal to-
$\Rightarrow V = \dfrac{4}{3}\pi {r^3}$
Therefore, the volume of the nucleus is thus given by the formula given below.
$\Rightarrow {V_{nucleus}} = \dfrac{4}{3}\pi {R^3}$ where $R$ is the radius of the nucleus but $R = {R_0}{A^{\dfrac{1}{3}}}$ , $A$ is the mass number of the nucleus.
$A$ is the total number of protons and neutrons in the nucleus.
$\therefore {V_{nucleus}} = \dfrac{4}{3}\pi {\left( {{R_0}} \right)^3}A$
Assigning the value of ${R_0} = 1.4 \times {10^{ - 15}}m$ in the above equation, we get
$\Rightarrow {V_{nucleus}} = \dfrac{4}{3}\pi \left[ {{{(1.4 \times {{10}^{ - 15}})}^3} \times A} \right]{m^3}$
$\Rightarrow {V_{nucleus}} = 11.4940A \times {10^{ - 45}}{m^3}$ .
It is known to us that, density is mass per unit volume, i.e. \rho = {\raise0.7ex\hbox{ m } \\!\mathord{\left/ {\vphantom {m V}}\right.} \\!\lower0.7ex\hbox{ V }} where $\rho$ is the density, $m$ is the mass and $V$ is the volume.
The density of the nucleus is thus given by the formula.
$\Rightarrow {\rho _{nucleus}} = \dfrac{{{m_{nucleus}}}}{{{V_{nucleus}}}}$ where ${\rho _{nucleus}}$ is the density of the nucleus, ${m_{nucleus}}$ is the mass of the nucleus and ${V_{nucleus}}$ is the volume of the nucleus of an atom.
It has been calculated, $\Rightarrow {V_{nucleus}} = 11.4940 \times {10^{ - 45}}{m^3}$
Given in the question, ${\text{mass of each nucleon}} = 1.67 \times {10^{ - 27}}kg$ .
The mass number of an atom is equal to the total number of protons and neutrons in the nucleus. The mass of the nucleus is the mass of all the nucleons.
Thus, ${m_{nucleus}} = A \times 1.67 \times {10^{ - 27}}kg$ where $A$ is the mass number.
$\therefore {\rho _{nucleus}} = \dfrac{{1.67 \times {{10}^{ - 27}}}}{{11.4940 \times {{10}^{ - 45}}}}{{kg} \mathord{\left/ {\vphantom {{kg} {{m^3}}}} \right. } {{m^3}}}$
$\Rightarrow {\rho _{nucleus}} = 0.14529 \times {10^{18}}{{kg} \mathord{\left/ {\vphantom {{kg} {{m^3}}}} \right. } {{m^3}}}$
Therefore, the density of the nucleus,
$\Rightarrow {\rho _{nucleus}} = 1.453 \times {10^{17}}{{kg} \mathord{\left/ {\vphantom {{kg} {{m^3}}}} \right. } {{m^3}}}$
Hence the correct answer is option D.

Note
The radius of a nucleus is always given in terms of the number of nucleons. The concept of nuclear densities finds applications where high density situations are included, such as those of neutron stars occur.