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A.) $1:3$

B.) $3:1$

C.) ${{\left( 3 \right)}^{\dfrac{1}{3}}}:1$

D.) $1:1$

Answer
Verified

Hint: The radius of an atomic nucleus is dependent on the mass number (A) of the atom. The mass number gives the number of nucleons present in the nucleus of an atom. So the mass of the nucleus will also depend upon the mass number. The volume of the atomic nuclei is the ratio between the mass of the nucleus to the volume of the nucleus.

Complete step by step answer:

We know that the mass number (A) and the radius of the atomic nuclei are related by the relation,(assuming the nucleus to be a sphere of radius R)

$R={{R}_{0}}{{(A)}^{\dfrac{1}{3}}}$ …. Equation (1)

Where,

${{R}_{0}}$ is a constant whose value is $1.2\times {{10}^{-15}}m$ or $1.2\text{ fm}$

$R$ is the radius of the nucleus.

$A$ is the mass number.

So the volume of the spherical nucleus can be written as,

$V=\dfrac{4}{3}\pi {{R}^{3}}$

Substituting the values of R in the above equation, we get,

$V=\dfrac{4}{3}\pi {{\left( {{R}_{0}}{{A}^{\dfrac{1}{3}}} \right)}^{3}}$

$\therefore V=\dfrac{4}{3}\pi A{{R}_{0}}^{3}$ … equation (2)

The mass of the atomic nuclei is given by the product of the mass of a nucleon (proton or a neutron) and the mass number.

$M=mA$ …. Equation (3)

Where,

M=Mass of the atomic nuclei.

m is the mass of the nucleon.

A is the mass number.

So the density can be written as,

$Density\left( \rho \right)=\dfrac{M}{V}$

Substituting the values of M and V from equation (2) and (3) in the above equation we get,

$\rho =\dfrac{mA}{\dfrac{4}{3}\pi {{R}_{0}}^{3}A}=\dfrac{m}{\dfrac{4}{3}\pi {{R}_{0}}^{3}}$

So, the density of the atomic nuclei is independent of the mass number (A).

So, the ratio of densities between the atomic nuclei of various mass numbers will always be $1:1$.

So, the answer to the question is option(D).

Note: The nucleons are particles present in the atomic nucleus. The nucleus is composed of protons and neutrons. Both of their masses are similar, which is approximately $1.67\times {{10}^{-27}}kg$. While the proton is positively charged the neutron is neutral.

The magnitude of the charge of the proton is equal to the magnitude of the charge of the electron which is $1.6\times {{10}^{-19}}Coulombs$.

The nuclear mass densities is one of the highest densities found in nature, it is of the order of ${{10}^{17}}$.

Complete step by step answer:

We know that the mass number (A) and the radius of the atomic nuclei are related by the relation,(assuming the nucleus to be a sphere of radius R)

$R={{R}_{0}}{{(A)}^{\dfrac{1}{3}}}$ …. Equation (1)

Where,

${{R}_{0}}$ is a constant whose value is $1.2\times {{10}^{-15}}m$ or $1.2\text{ fm}$

$R$ is the radius of the nucleus.

$A$ is the mass number.

So the volume of the spherical nucleus can be written as,

$V=\dfrac{4}{3}\pi {{R}^{3}}$

Substituting the values of R in the above equation, we get,

$V=\dfrac{4}{3}\pi {{\left( {{R}_{0}}{{A}^{\dfrac{1}{3}}} \right)}^{3}}$

$\therefore V=\dfrac{4}{3}\pi A{{R}_{0}}^{3}$ … equation (2)

The mass of the atomic nuclei is given by the product of the mass of a nucleon (proton or a neutron) and the mass number.

$M=mA$ …. Equation (3)

Where,

M=Mass of the atomic nuclei.

m is the mass of the nucleon.

A is the mass number.

So the density can be written as,

$Density\left( \rho \right)=\dfrac{M}{V}$

Substituting the values of M and V from equation (2) and (3) in the above equation we get,

$\rho =\dfrac{mA}{\dfrac{4}{3}\pi {{R}_{0}}^{3}A}=\dfrac{m}{\dfrac{4}{3}\pi {{R}_{0}}^{3}}$

So, the density of the atomic nuclei is independent of the mass number (A).

So, the ratio of densities between the atomic nuclei of various mass numbers will always be $1:1$.

So, the answer to the question is option(D).

Note: The nucleons are particles present in the atomic nucleus. The nucleus is composed of protons and neutrons. Both of their masses are similar, which is approximately $1.67\times {{10}^{-27}}kg$. While the proton is positively charged the neutron is neutral.

The magnitude of the charge of the proton is equal to the magnitude of the charge of the electron which is $1.6\times {{10}^{-19}}Coulombs$.

The nuclear mass densities is one of the highest densities found in nature, it is of the order of ${{10}^{17}}$.

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