There isn't any neutrino field in this problem. However, since the electron and the proton of the hydrogen atom interact, not only, through electromagnetic inetractions, but through weak interactions, as well (the two particles interact, not only through the exchange of photons (for the electromagnetic interactions), but, also, through the exchange of the Z boson (for the weak interactions)), the exchange of the Z leads to parity violation effects and these have been studied, theoretically and experimentally.

Cf. http://www.adsabs.harvard.edu/abs/1997RPPh...60.1351B

for a review and http://arxiv.org/abs/physics/0503143 for more recent results.

Thank you very much Stam.

The inverse beta-decay of H-atom suggested by me in the question does involve the neutrino field. The interaction is through the gauge bosons but the final product is the single excitation of the neutrino field (which is the neutrino). In the H-atom spectra, we have H* ---> H + Photon(QED gauge boson)

Here, we'll have H(1s) ---> neutron + neutrino (perfectly allowed final state resulting from electroweak gauge boson exchanges). It will thus appear as if the atom has emitted a neutrino and made a transition to a state which is the neutron itself.

There will, of course, be suppression of the transition probability from the small finite range of weak interactions, the smallness of the weak coupling constant and the very high energy range.

That's not possible from energy conservation:

Rest energy of the neutron=939.565 MeV

Rest energy of the proton= 938.272 MeV

Rest energy of the electron= 0.511 MeV

Rest energy of neutrino is negligible in comparison--its kinetic energy makes the situation even more unfavorable.

So proton+electron cannot spontaneously become neutron+neutrino. You need energy from outside. The available energy has already gone into making the proton+electron bound state, i.e. the hydrogen atom. They have some kinetic energy, of course, but the final poducts won't be at rest either. So the hydrogen atom will not spontaneously decay into neutrons and neutrinos.

If you excite it, or ionize it, you're sending the proton and the electron even further apart, therefore rendering the reaction even less likely.

Dear Stam,

Your argument is correct as long as we don't take into account the weak interaction energies when the electron goes closer to the proton i.e. to within the weak interaction range. Uncertainty principle gives us \delta E~ hc/r ~an order of 10^6 increase in the kinetic energy (to a few MEVs) as the electron draws closer to the proton. The higher KE of the electron will be utilized to produce the neutrino. We must not forget that the electron even in the 1s state solution of Schrodinger equation has a finite probability of being found within the volume of the sphere with radius~ weak interaction range. Thus the possibility of K-capture (like in higher atomic no. atoms) cannot be ruled out.

No decay is truly spontaneous.The decay occurs as the result of interactions amongst the fields. And, they are all here to allow for the decay, though with less probability.

Have a look to this reference:

Parity violation in hydrogen revisited

R. W. Dunford, R. J. Holt

Journal of Physics G Nuclear and Particle Physics . 07/2007;

DOI:10.1088/0954-3899/34/10/001

Like Stam discribed is a decay into a neutron and a neutrino not possible.

But the electron binding energies of the S-states are a little bit modified compared

to QED only and the parity violation of electrons in the S-states should be possible.

Proton considered as a point source of electric field only on distances larger than 0.87*10^{-15} m, whereas the weak interaction has a very short range - around 10^{−17}–10^{−16} m.

Within radius 0.87*10^{-15} m proton (as it follows from the electron-proton scattering experiments) looks like as a smooth cloud without any singularities for small-energy electrons, but for electrons with large energies proton has three point targets (evidence of quarks existing?). This is big mistery why? Why protons which are combined from quarks dont have electrical dipole moments? Why for small-energy electrons proton look like as a smooth spherical cloud of charge?

Dear all,

The way the arguments are being given, it seems that electron capture by the proton is an impossibility !

Regards,

Rajat

Of course not. Nobody claimed that the proton cannot ``capture'' an electron-hydrogen is a bound state of an electron and a proton.

What *is* impossible is that the hydrogen atom is unstable, given what we know.

Dear Stam,

Than you very much.

The H-atom is stable because the free proton and the free electron are themselves stable and they are in the electromagnetic ground state. No position or energy is unattainable by the electron in the H-atom ground state. It has a small but non-zero probability of even going right through the proton also, although the most probable distance from the proton is the 1st Bohr radius. In that case, the capture should be possible. If it is not possible in H-atom, how is it possible in any other case?

Regards,

Rajat.

The statement ``No position or energy is unattainable by the electron in the H-atom ground state.'' is false. The energy for (in practice) the electron in the hydrogen atom is bounded from below, which expresses the stability of the ground state.

Furthermore, the energy levels of the electron are discrete-so not all, classically allowed, values are available either, as was explained by Niels Bohr one hundred years ago.

``Going through the proton'' can't be given any consistent meaning. Once more: the *definition* of a bound state implies that the particles do not escape to infinity, so are ``captive''.

If the bound state is stable, not, just, metastable, then that's it. The bound state of a proton and an electron as a hydrogen atom is stable, within the full Standard Model, not, just, within QED.

On the other hand, positronium, for example, is a bound state of a positron and an electron--but it's *metastable*: it decays into two, or three, photons, depending on the polarizations of the positron and electron.

dear Stam,

Was not Bohr's orbit model superceded by Schrodinger's orbital solutions. The wave function does not vanish at r = 0, i.e. at the position of the proton, rather it is a maximum there. This is why I say that even if it be for a very small time interval only, the electron certainly has a finite small probability of coming arbitrarily close to the proton to get captured.

Regards,

Rajat

Schrödinger's (and Heisenberg's and Dirac's etc) work *explained* Bohr's proposal, regarding the hydrogen atom, they didn't show it wrong.

The electron *is* captured by the proton-that's what a bound state means. What this *doesn't* mean is that the proton, interacting with the electron, can give a neutron and a neutrino. The reason is purely energetic.

Dear Stam,

In the first part, Bohr's model assigned one unit of angular momentum to the ground state and Schrodinger etc. proved him wrong by discovering that the ground state had ZERO angular momentum. They did prove Bohr wrong in the process of explaining him !!

In the second part is there no way of supplying the required energy for the proposed process to take place?

Thanks and regards,

rajat

No: the principal quantum number is 1 in the ground state-that's why the energy is finite--whereas, classically, it *would* be infinite, since the (classical) potential well is of infinite depth, for zero angular momentum.

Bohr had it right, of course. The orbital quantum number is zero-so the ground state is spherically symmetric.

Nobody's saying you can't have the reaction p+ e -> n + v, if you input energy. However, you can't have it happen in the hydrogen atom. The ionization energy of the hydrogen atom is 13.6 eV, (minus) the energy of the electron in its ground state. This energy is much less than the difference of rest energies I mentioned previously. Therefore you would ionize hydrogen if you gave the energy necessary for the reaction to happen. Therefore it can't happen within the bound state.

Dear Stam,

In the first part, the angular momentum in Bohr's theory is n h-bar, where n is principal quantum number. And the minimum value is n=1, for the ground state. It is true that orbital quantum no. l is 0, but that does not mean that Bohr took angular momentum to be 0. You are mixing up old and new quantum theory. Bohr's ground state had angular momentum integral multiple of h-bar. It is one of the basic postulates of Bohr. How are you saying otherwise?

I agree with you in the second part, but the question of the electron getting the energy from getting arbitrarily close to the proton is definitely a possibility, although with very low probability.

Thank you very much,

Regards,

Rajat

The relevant issue is the ground state energy, that is *finite*. History is fascinating, but a side issue here.

The energy will fluctuate-but you're interested in *on-shell* states, if you're trying to detect p + e -> n + v. The statement is that you can't make *on-shell* states this way.