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It’s not that protons can’t be transferred. It’s just that if we lived in a place where proton transfer was common, we would have a very different perspective on chemistry. …

Free neutrons in flight are not deflected by electric fields. Objects which are not deflected by electric fields are electrically neutral. The energy of the strong proton-neutron interaction varies …

When the neutron star forms, most of the protons and electrons combine together to form neutrons This is mostly correct. … The neutrino is an uncharged, very low-mass electron-like particle, in the same way that neutrons and protons are different charge states of the same sort of particle. …

Anything that produces a fragment of a nucleus could be interpreted in either of two ways: a certain number of protons and neutrons broke off of the clump, or the single particle divided into separate … The popular science material always talks about the number of protons and neutrons in a nucleus, but I've always wondered if that's a real thing nuclear physicists believe or if it is just a convenient …

Protons are indistinguishable particles; they all have the same mass, exactly. …

Yes, in this case what's meant by "energy" is the kinetic energy $K = (\gamma-1)mc^2$ that the incident particle can transfer to the target system. As you point out, it wouldn't make sense to talk ab …

You haven't linked to any of the articles that you've read, but the value that you've quoted seems to come from this Nature paper that was published this week. Note that the news there is the precisio …

I asked a question very much like this several years ago (in person, not online, but someone else asked it here): "what's the baryon asymmetry of the proton?" Thinking, of course, about the three vale …

Most stable nuclei have a "neutron separation energy" of eight or ten MeV --- that's the minimum cost to produce a neutron, before you factor in any inefficiencies. You can't get ahead spending 10 MeV …

You do not need the volume. You need the number density, $$ n=\frac NV=\frac p{kT} $$ Then the product $n\ell$, with $\ell$ the known thickness of your target, is the number of scatterers per unit are …

The SI unit for mass is the kilogram. Most organizations which insist on SI units will make an allowance for the fact that the kilogram is inconveniently large for discussing masses in atomic and suba …

However, that only works because protons, neutrons, and electrons are subject to “number conservation laws.” … Protons and neutrons are “baryons,” and there is no known process which can change the number of baryons in a closed system. …

The answer by annav is correct that the neutron decay products will travel in straight paths in a vacuum chamber; the electron and proton will interact with the walls of the chamber, slow down, and be …

This means, among other things, that neutrinos emitted from neutron decay at rest will never have enough energy to cause positron emission on protons at rest elsewhere. … For instance, the mass difference between postassium-40 and argon-40 is about $1.50\,\mathrm{ MeV}/c^2$, with potassium (19 protons) heavier than argon (18 protons), so the decay $$ \rm ^{40}_{19}K \to …

It looks like a typo for $$ \rm \rho^0 + n \to K^+ + \Sigma^- $$ where $\rho^0$ is the uncharged member of the isospin triplet with mass 770 MeV. According to the particle data group, the quark conten …