From the relation R = R_0 A^{\frac{1}{3}} , where R_0 is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of A).

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Why can an electron not exist inside the nucleus? - Quora
Electrons are orbiting around the nucleus i.e. out side the nucleus, it is an experimental fact. ... Why does the nucleus of an atom stay together and the electron stay orbiting around it ... One of the applications is to prove that electron can not exist inside the nucleus. ... [math]P = {2 \over 3} {q^2 a^2 \over 6 \pi \ epsilon_0[/math].
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Atomic nucleus - Wikipedia
The atomic nucleus is the small, dense region consisting of protons and neutrons at the center ... The diameter of the nucleus is in the range of 1.7566 fm (1.7566× 10−15 m) for ... In principle, the physics within a nucleus can be derived entirely from ... power is required to accurately compute the properties of nuclei ab initio.
For more information, see Atomic nucleus - Wikipedia
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Semi-empirical mass formula - Wikipedia
In nuclear physics, the semi-empirical mass formula (SEMF) is used to approximate the mass and various other properties of an atomic nucleus from its number ...
For more information, see Semi-empirical mass formula - Wikipedia
Density \rho = \frac{\text{ mass }}{\text{ volume }} = \frac{m}{V}
Given that
Radius of nucleus R = R_0 A^{\frac{1}{3}} where, R_0 - constant and A - mass number of nucleus
Step 1: demonstrating that the density of nuclear substances is constant
Volume of the substance(sphere) V = \frac{4}{3} \pi r^3 where, r - radius of nucleus
V = \frac{4}{3} \pi (R_0 A^{\frac{1}{3}})^3
V = \frac{4}{3} \pi R_0^3 A
Density of nuclear substance \rho=\frac{m}{V}=\frac{A}{\frac{4}{3}\pi R_0^3A}
\rho = \frac{3 A}{4 \pi R_0^3 A}
\rho = \frac{3}{4 \pi R_0^3}
\frac{3}{4 \pi R_0^3} is constant
Hence, the density of nuclear matter is independent of A. It is nearly constant.