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**Hint**To prove that nuclear mass density is independent of the mass number $A$we will start with the understanding of what does nuclear mass density means. Nuclear mass density is the ratio of nuclear mass to nuclear volume and by using the formula of nuclear density we will compare it with mass number A.

**Formula used**

$ \Rightarrow R \propto {A^{\dfrac{1}{3}}}$

$ \Rightarrow \rho = \dfrac{A}{V}$

$ \Rightarrow Volume(V) = \dfrac{4}{3}\pi {R^3}$

**Complete step by step answer:**

We will start with the relation of the nuclear radius with the mass number which shows that the nuclear radius is directly proportional to the cube root of the mass number.

$ \Rightarrow R \propto {A^{\dfrac{1}{3}}}$

Now to remove the proportionality between the nuclear radius and mass number we will introduce a constant of proportionality.

$ \Rightarrow R = {R_0}{A^{\dfrac{1}{3}}}$ -------- Equation (1)

Where ${R_0}$is constant of proportionality, $R$is the radius of the nucleus and $A$is the mass number.

Now the nuclear mass density of the nucleus can be given by the ratio of nuclear mass and its nuclear volume. Given as

$ \Rightarrow Density(\rho ) = \dfrac{{mass}}{{volume}}$

$ \Rightarrow \rho = \dfrac{A}{V}$ --------- Equation (2)

For nucleus mass = $A$and Volume of the nucleus will be given by the formula of volume of a sphere, hence

$ \Rightarrow Volume(V) = \dfrac{4}{3}\pi {r^3}$ (For sphere)

$ \Rightarrow Volume(V) = \dfrac{4}{3}\pi {R^3}$ (For nucleus)

Now substituting the values of $mass(A)$ and $volume(V)$of the nucleus in equation (2), we get

$ \Rightarrow {\rho _{nucleus}} = \dfrac{A}{{\dfrac{4}{3}\pi {R^3}}}$ -------- Equation (3)

Now from Equation (1) and Equation (3), we get

$ \Rightarrow {\rho _{nucleus}} = \dfrac{A}{{\dfrac{4}{3}\pi {R_0}^3A}}$

$\therefore {\rho _{nucleus}} = \dfrac{3}{{4\pi {R_0}^3}}$

This shows that ${\rho _{nucleus}}$nuclear mass density is nearly constant so it is independent of mass number$A$.

**Note**We can further deduce the value of nuclear mass density as we know that ${R_0}$ is constant so by substituting the values of $\pi = 3.14$and the value ${R_0} = 1.25fm$ hence we can obtain the value of nuclear charge density as $2.3 \times {10^{17}}\dfrac{{kg}}{{{m^3}}}$, as an average.

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