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2

The Universe is indeed electrically neutral at the cosmological length scales which means that the total charge of the positively charged particles is equal to (minus) the total charge of the negatively charged particles. However, one must be more careful what these particles are. Electrons and protons are two dominant charged particle species. However, ...


2

For some reason I get 28.296 MeV for binding energy of $^4He$. Coulomb potential energy (for a charged sphere) can be written as $E=\frac{3}{5}(\frac{1}{4\pi\epsilon_0})\frac{Q^2}{R}$. This term is actually plugged into a liquid drop model formula (the Coulomb term $-a_C\frac{Z(Z-1)}{A^{1/3}}$) and from both theory and experiment give you a similar value ...


3

I would like to answer your second and third questions first. Although there exist good approximations of the force due to strong nuclear force, a much easier option would be to compare the binding energy with the potential energy created due to electrostatic repulsion. To make an estimation of the potential energy, I assume the radius of Helium nucleus to ...


1

Yes this is how proton cosmic rays are produced... quasars, supernovae, gamma rays bursts... also anti proton cosmic rays originate from proton anti proton creation in proton cosmic ray collision with nuclei in interstellar medium (otherwise anti protons cannot leave the vicinity of a matter dominated (as opposed to anti-matter) source)


0

I will answer question 2. Let $f$ be a smooth test function, $$\int_0^1 \frac{4}{3}\frac{1+z^2}{(1-z)_+}f(z) + 2 \delta(1-z) f(z) dz = \int_0^1 \frac{4}{3}\frac{(1+z^2)f(z) - 2f(1)}{(1-z)} + 2 \delta(1-z)f(z) dz $$ using the definition of a $+$-distribution. Continuing after some simplification; one tries to force the two bits together by cleverly adding ...


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No, they don't. Only some of negative particles (electrons) on the top shell could do that. Atomic nuclei together with inner electronic shells do not move (almost). In metals they form kind of crystalline cubic grid.



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