Yes!

No, because the proton is lighter than the neutron.

Stam Nicolis Is there no neutrino scattering similar to Compton Scattering, where the photon looses energy and the target gains impulse? Likewise with ibd, the proton would become a neutron, emits a positron and gains some impulse itself while the neutrino looses some or all of its energy and changes direction?

Actually non, The proton is the lightest baryon. Which means that it is impossible for it to decay. Even if it absorbs a photon (remember the charge conservation). However, it can be transformed to a neutron via the electro-weak interaction. If an electron with a high enough energy colides with the proton, they may exchange a W(minus) boson and the proton will become neutron and electron becomes electronic neutrino.

To conclue, don't confuse between this interaction and the beta decay. Inside the nucleous, we do not have free particles.

You can see the nuckeous as a sea of interaction and quark.

Once more: A proton can't transform into a neutron, because the proton is lighter than the neutron. It requires energy to compensate for this. An example is provided here: Article Development of a proton-to-neutron converter for radioisotop...

So the answer to the question, still, is No: Hydrogen gas is stable to inverse beta decay.

Within the nucleus protons and neutrons are bound and, if the binding is strong enough, inverse beta decay doesn't occur, because it's a weak interaction process, whereas the binding is due to the strong interactions.

If the binding isn't strong enough, so that the proton can be distinguished from the neutron, this means that the relevant property is that it's lighter, therefore it won't spontaneously transform into a neutron. That's why hydrogen is stable.

Stam Nicolis Thanks for answering. But still it is not completely clear tom me.

You want to say that within the nucleus of an atom from Deuterium on, (De,He,Li, etc) protons and neutrons are bound through the strong nuclear force. Therefore inverse beta decay will not occur, because it interacts with the weak force. If we look at the globe with Hydrogen we see H2 molecules. Analyzing what you say you continue to say that the relevant property is that P is lighter than N. Therefore P can't "spontaneously' change to N. Hence it needs energy to do that. Getting the energy from the neutrino is not possible?

Considering the homestake experiment. They choose the stable isotope of chlorine (37Cl) to interact with neutrinos. This chlorine captures a neutrino and is transformed into 37Ar ejecting a positron. As the theory goes. Then what I don;t understand yet is the difference between the proton converting to neutron in chlorine and the proton not converting in hydrogen H2? This because also in Chlorine P is lighter than N? And you say that is the relevant property. Therefore I get stuck.

If the nucleus is hit on by neutrinos, then the process has a finite probability of occurring; it just can't occur spontaneously. So

p + νe->n+ e+

can occur, where the neutrino from outside a nucleus interacts with a proton inside a nucleus.

Bu this doesn't mean that

p->n+e++νe

can occur (which would imply that hydrogen is unstable).

The first event is a scattering event, which takes place for energies of the incoming particles (here the neutrino) above a certain threshold; the second event is a decay event, that can't occur, since it's forbidden by energy conservation.

So if hydrogen is put in a neutrino beam, there is a finite probability of the first reaction occurring. But it's impossible for the second reaction to occur, spontaneously with hydrogen gas since the proton's rest energy is less than that of the neutron's (and the proton's rest energy in hydrogen is even less than that of the free proton, since the proton in hydrogen is bound-but the difference, 13.6 eV, isn't significant for this discussion; what is significant is that the rest mass of the proton is less than that of the neutron, so the second reaction can't take place in hydrogen spontaneously).

Stam Nicolis Thanks so you say:

If hydrogen is put in a neutrino beam, there is a finite probability of the first reaction occurring. That seems to be the answer to my question in reality.

p + νe->n+ e+ does occur and.

does not depend on the proton being part of clorine or some other molecule.

in H2 you would get

2p + νe->p + n+ e+

e+ + e- > 2 gamma rays

and what happens with the neutron afterwards we have two possibilities

if the neutron is captured by the other proton

p + n > Deuterium (radical)

meaning this is a natural source of deuterium in the water.

or if the neutron is launched away by some rest energy from the neutrino it will decay to a hydrogen radical proton with an unpaired electron after 880 seconds.

n > p + e-

n -> p + e-

doesn't occur; there's an anti-neutrino (electron) produced, also, from energy, momentum, angular momentum (hat includes spin) and lepton number conservation; there's a difference of living in 1922 when all this wasn't that well understood, and living in 2022, when these aspects are well understood.

So a neutron decays as

n -> p + e- + antineutrinoe

Recombination of a proton and a neutron to form deuterium doesn't occur spontaneously-it requires energy, to provide the binding energy.

Deuterum doesn't occur in water due to proton and neutron recombination, but because deuterium (and tritium) are hydrogen isotopes anyway.