The mass density of the nucleus is constant. But what is the electric charge density of the nucleus and the atom (any order of magnitude)?


So the naïve answer is that the charge density of nucleon is roughly $1\,e/\rm fm^3$, while an atom has (neglecting the spike at the nucleus) charge density roughly $-1\,e/\rm Å^3$. Each of these is pretty huge in SI units! The conversion factors are \begin{align} 1\,e &= 1.60\times10^{-19}\,\rm C \\ 1\,\rm fm &= 10^{-15}\,\rm m \\ 1\,Å &= 10^{-10}\,\rm m \end{align} so the nucleon charge density is something like $10^{26}\,\rm C/m^3$.

The charge density of a nucleus, of course, will be the superposition of the charge densities of all of its nucleons: positive, and more or less uniform.

If you look closer, though, the picture gets more complicated. For instance, then neutron has zero net charge and so you might expect it to have zero charge density. However, you can draw a Feynman diagram where the neutron spends part of its time as a proton-$\pi^-$ loop, which suggests the neutron should have a positively-charged "core" and a negatively-charged "skin." On the other hand, the quark model of the neutron has it made of quarks. The negatively-charged $d$-type quarks are slightly more massive than the positively-charged $u$-type quarks, so you might expect the neutron's negative charge to be concentrated at its center.

Apparently there is some truth to both these pictures: the neutron has negative charge density at its core, where the heavy $d$ quarks "live," and also negative charge density more than about 1 fm from the core, where the "pion cloud" is more important. These two regions are balanced by a shell of positive charge density (whose shape is almost perfectly spherical, since the neutron has miniscule electric dipole moment).

N.B. The paper I link above uses units of $\rm fm^{-2}$ for charge density; I haven't checked, but that's probably using $\hbar=c=1$ and CGS units for charge.


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