why can't a photon of appropriate energy enable P->N conversion?

Submitted by sosodank t3_yf0kz3 in askscience

i understand that the uud structure of the proton is the lowest allowed energy state of three quarks, and thus it can't decay (supersymmetric theories aside). the neutron, meanwhile, is udd and can (when by itself, or in the right kind of nucleus) undergo beta decay to a proton via spontaneous emission of a massive, charged W boson.

furthermore, i understand that for instance non-spontaneous fission occurs when a neutron of sufficient energy interacts with the nucleus, bringing it to an unstable energy level. u235 is close to this level, and thus fissions when interacting with a neutron of even thermal energy, whereas u238 needs a higher-energy neutron to fission. high-energy photons can furthermore knock nucleons out in photodisintegration.

2.78MeV differentiate the up and down quark. This energy ought be readily available; the excited Ni60 metastate decays with a 1.33MeV γ, about half this value. Why don't we see protons transmuting to neutrons when such gamma rays hit them, especially in e.g. a plasma where there are no electrons to get in the way (at nuclear distance scales)?

sorry, i am merely a dumb computer scientist.

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abeinszweidrei t1_iu1b6lo wrote

It would violate charge conservation. The proton (uud) has electric charge +1, while the neutron has electric charge 0. The photon also has charge zero. So if you imagine the process of

p + gamma -> n

and count the electric charges on both sides, they have to be equal. In this case they are not, thus this process is not allowed. Conservation of electric charge is one of the conservation laws that are never broken.

For all such kinds of processes, it's always a good start to check if the charges are consistent (and you might extend this to weak and color charges)

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ehj t1_iu1hwov wrote

Replying to sosodank (#267,080)

Momentum must be conserved, however this is not the explanation for the neutrino.

Neutrinos are part of the weak interaction. You correctly expect a positron on the right side of your reaction because of conservation of electric charge.. But why? Well conservation of electric charge is a consequence of the law of the force of electomagnetism. In radioactive decays the 'Weak' interaction comes into play and is in some ways similar to the electromagnetic force (and different in some ways and slightly more complicated). But it is similar in the way that there are also charges that must be conserved in the Weak interaction. In particular there is something called lepton number for each flavour of lepton which are the electron, muon and tauon. And this lepton number must be conserved.

But this comment on the neutrino and momentum conservation is indeed important, and it is an obvious question - why do we 'need' the neutrino. Historically we saw several types of radioactive decay and some of them (alpha and gamma) always had a fixed energy of the emitted particle e.g. gamma decay of an excited nucleus A*->A+gamma. Energy and momentum conservation of such a process where 1 particle turns into 2 predicts that the emitted particle will always have the same energy in every decay.

However for beta decays where an electron is emitted it was observed that the electron energy was a spectrum up to some upper energy - contradicting the 1 to 2 particle decay and therefore physicists postulated the existence of an additional very light particle being emitted in this decay such that the energy was split between the electron and the postulated 'neutrino' as this could explain the spectrum seen. This was how the neutrino was first discovered.

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sosodank OP t1_iu1i339 wrote

Replying to ehj (#267,884)

awesome expansion, and I finally appreciate the neutrino a bit after despising it all my life. thanks a lot for being the best part of reddit.

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sosodank OP t1_iu1ikra wrote

Replying to sosodank (#267,906)

let me add a question: why must the energy of the emitted electron have a non-trivial spectrum? when you emit an alpha, you're falling to some distinct energy level, hence the single value (or at least a discrete spectrum in the case of complex isotopes). what's different about the beta? is it due to the W boson intermediary?

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mfb- t1_iu2vln7 wrote

Replying to sosodank (#267,976)

You get a continuous spectrum for beta decays (or p + gamma -> n + e+ + nu) because you have three particles in the final state, so there are more options how to share the energy and momentum across the particles. An alpha decay only has two particles, which only leaves one option for the energy distribution because everything else would violate conservation of momentum.

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