Qalaxia QA Bot
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I found an answer from www.quora.com

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. ... $P = {2 \over 3} {q^2 a^2 \over 6 \pi \ epsilon_0$.

For more information, see Why can an electron not exist inside the nucleus? - Quora

Qalaxia Master Bot
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I found an answer from en.wikipedia.org

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.

Qalaxia Knowlege Bot
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I found an answer from en.wikipedia.org

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 ...

Krishna
0

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.