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Since I have already gone to the trouble of computing this value to great detail, I might as well present it.

Matt Hanson cited a relevant Wikipedia article, but it links to another that is even more relevant, and it even states that the following only happens about 4 times in a million, i.e., extremely rarely (as should be expected).

I did not expect this case to be possible either, since there is actually a lot of energy available to ionise any such hydrogen atom (and as mentioned, it usually does so, i.e., proton, electron and antineutrino flying apart from each other), so it is worthwhile checking that this decay mode is actually possible, rare that it may be. The SR 4-momentum computation relevant to this decay is, with $c=1$ $$ \begin{align} \begin{pmatrix}m_H+\Delta\\0 \end {pmatrix}&= \begin{pmatrix}\sqrt{m_H^2+p^2}\\-p \end {pmatrix}+ \begin{pmatrix}p\\p \end {pmatrix} \end {align} $$ where the mass of the ground state H atom is $m_H=$ 938.7830735 MeV, and this is the mass of proton plus mass of electron minus the binding energy 13.6 eV between them. $m_H+\Delta$ is the mass of the neutron, which thus defines that $\Delta=$ 0.7823470157 MeV.

These will thus immediately fix the momentum = energy = $p$ of the anti-neutrino (I assumed it is massless, which is by far a reasonable approximation here because of how inconsequential it will be), and this gives us $p=$ 0.7820212976 MeV. This is the energy that the anti-neutrino carries away, leaving just 325.7180673 eV for the ground state H atom to recoil away with. The ground state H atom is thus going to recoil away nonrelativistically.

Very little will change in this computation if you assumed that the H atom is excited rather than ground state. It will just be slight differences in the $m_H$ value being used, thereby inducing small changes in all the other values, since the mass of the neutron is fixed.

Anyway, it is surprising that this decay mode is even possible, so thanks for the interesting question.

Since I have already gone to the trouble of computing this value to great detail, I might as well present it.

Matt Hanson cited a relevant Wikipedia article, but it links to another that is even more relevant, and it even states that the following only happens about 4 times in a million, i.e. extremely rarely (as should be expected).

I did not expect this case to be possible either, since there is actually a lot of energy available to ionise any such Hydrogen atom (and as mentioned, it usually does so, i.e. proton, electron and anti-neutrino flying apart from each other), so it is worthwhile checking that this decay mode is actually possible, rare that it may be. The SR 4-momentum computation relevant to this decay is, with $c=1$ $$ \begin{align} \begin{pmatrix}m_H+\Delta\\0 \end {pmatrix}&= \begin{pmatrix}\sqrt{m_H^2+p^2}\\-p \end {pmatrix}+ \begin{pmatrix}p\\p \end {pmatrix} \end {align} $$ where the mass of the ground state H atom is $m_H=$ 938.7830735MeV, and this is the mass of proton plus mass of electron minus the binding energy 13.6eV between them. $m_H+\Delta$ is the mass of the neutron, which thus defines that $\Delta=$ 0.7823470157MeV.

These will thus immediately fix the momentum = energy = $p$ of the anti-neutrino (I assumed it is massless, which is by far a reasonable approximation here because of how inconsequential it will be), and this gives us $p=$ 0.7820212976MeV. This is the energy that the anti-neutrino carries away, leaving just 325.7180673eV for the ground state H atom to recoil away with. The ground state H atom is thus going to recoil away non-relativistically.

Very little will change in this computation if you assumed that the H atom is excited rather than ground state. It will just be slight differences in the $m_H$ value being used, thereby inducing small changes in all the other values, since the mass of the neutron is fixed.

Anyway, it is surprising that this decay mode is even possible, so thanks for the interesting question.

Since I have already gone to the trouble of computing this value to great detail, I might as well present it.

Matt Hanson cited a relevant Wikipedia article, but it links to another that is even more relevant, and it even states that the following only happens about 4 times in a million, i.e. extremely rarely (as should be expected).

I did not expect this case to be possible either, since there is actually a lot of energy available to ionise any such Hydrogen atom (and as mentioned, it usually does so, i.e. proton, electron and anti-neutrino flying apart from each other), so it is worthwhile checking that this decay mode is actually possible, rare that it may be. The SR 4-momentum computation relevant to this decay is, with $c=1$ $$ \begin{align} \begin{pmatrix}m_H+\Delta\\0 \end {pmatrix}&= \begin{pmatrix}\sqrt{m_H^2+p^2}\\-p \end {pmatrix}+ \begin{pmatrix}p\\p \end {pmatrix} \end {align} $$ where the mass of the ground state H atom is $m_H=$ 938.7830735MeV, and this is the mass of proton plus mass of electron minus the binding energy 13.6eV between them. $m_H+\Delta$ is the mass of the neutron, which thus defines that $\Delta=$ 0.7823470157MeV.

These will thus immediately fix the momentum = energy = $p$ of the anti-neutrino (I assumed it is massless, which is by far a reasonable approximation here because of how inconsequential it will be), and this gives us $p=$ 0.7820212976MeV. This is the energy that the anti-neutrino carries away, leaving just 325.7180673eV for the ground state H atom to recoil away with. The ground state H atom is thus going to recoil away non-relativistically.

Very little will change in this computation if you assumed that the H atom is excited rather than ground state. It will just be slight differences in the $m_H$ value being used, thereby inducing small changes in all the other values, since the mass of the neutron is fixed.

Anyway, it is surprising that this decay mode is even possible, so thanks for the interesting question.

Since I have already gone to the trouble of computing this value to great detail, I might as well present it.

Matt Hanson cited a relevant Wikipedia article, but it links to another that is even more relevant, and it even states that the following only happens about 4 times in a million, i.e., extremely rarely (as should be expected).

I did not expect this case to be possible either, since there is actually a lot of energy available to ionise any such hydrogen atom (and as mentioned, it usually does so, i.e., proton, electron and antineutrino flying apart from each other), so it is worthwhile checking that this decay mode is actually possible, rare that it may be. The SR 4-momentum computation relevant to this decay is, with $c=1$ $$ \begin{align} \begin{pmatrix}m_H+\Delta\\0 \end {pmatrix}&= \begin{pmatrix}\sqrt{m_H^2+p^2}\\-p \end {pmatrix}+ \begin{pmatrix}p\\p \end {pmatrix} \end {align} $$ where the mass of the ground state H atom is $m_H=$ 938.7830735 MeV, and this is the mass of proton plus mass of electron minus the binding energy 13.6 eV between them. $m_H+\Delta$ is the mass of the neutron, which thus defines that $\Delta=$ 0.7823470157 MeV.

These will thus immediately fix the momentum = energy = $p$ of the anti-neutrino (I assumed it is massless, which is by far a reasonable approximation here because of how inconsequential it will be), and this gives us $p=$ 0.7820212976 MeV. This is the energy that the anti-neutrino carries away, leaving just 325.7180673 eV for the ground state H atom to recoil away with. The ground state H atom is thus going to recoil away nonrelativistically.

Very little will change in this computation if you assumed that the H atom is excited rather than ground state. It will just be slight differences in the $m_H$ value being used, thereby inducing small changes in all the other values, since the mass of the neutron is fixed.

Anyway, it is surprising that this decay mode is even possible, so thanks for the interesting question.

Since I have already gone to the trouble of computing this value to great detail, I might as well present it.

Matt Hanson cited a relevant Wikipedia article, but it links to another that is even more relevant, and it even states that the following only happens about 4 times in a million, i.e. extremely rarely (as should be expected).

I did not expect this case to be possible either, since there is actually a lot of energy available to destroy any such Hydrogen atom (and as mentioned, it usually does so), so it is worthwhile checking that this decay mode is actually possible, rare that it may be. The SR 4-momentum computation relevant to this decay is, with $c=1$ $$ \begin{align} \begin{pmatrix}m_H+\Delta\\0 \end {pmatrix}&= \begin{pmatrix}\sqrt{m_H^2+p^2}\\-p \end {pmatrix}+ \begin{pmatrix}p\\p \end {pmatrix} \end {align} $$ where the mass of the ground state H atom is $m_H=$ 938.7830735MeV, and this is the mass of proton plus mass of electron minus the binding energy 13.6eV between them. $m_H+\Delta$ is the mass of the neutron, which thus defines that $\Delta=$ 0.7823470157MeV.

These will thus immediately fix the momentum = energy = $p$ of the anti-neutrino (I assumed it is massless, which is by far a reasonable approximation here because of how inconsequential it will be), and this gives us $p=$ 0.7820212976MeV. This is the energy that the anti-neutrino carries away, leaving just 325.7180673eV for the ground state H atom to recoil away with. The ground state H atom is thus going to recoil away non-relativistically.

Very little will change in this computation if you assumed that the H atom is excited rather than ground state. It will just be slight differences in the $m_H$ value being used, thereby inducing small changes in all the other values, since the mass of the neutron is fixed.

Anyway, it is surprising that this decay mode is even possible, so thanks for the interesting question.

Since I have already gone to the trouble of computing this value to great detail, I might as well present it.

Matt Hanson cited a relevant Wikipedia article, but it links to another that is even more relevant, and it even states that the following only happens about 4 times in a million, i.e. extremely rarely (as should be expected).

I did not expect this case to be possible either, since there is actually a lot of energy available to ionise any such Hydrogen atom (and as mentioned, it usually does so, i.e. proton, electron and anti-neutrino flying apart from each other), so it is worthwhile checking that this decay mode is actually possible, rare that it may be. The SR 4-momentum computation relevant to this decay is, with $c=1$ $$ \begin{align} \begin{pmatrix}m_H+\Delta\\0 \end {pmatrix}&= \begin{pmatrix}\sqrt{m_H^2+p^2}\\-p \end {pmatrix}+ \begin{pmatrix}p\\p \end {pmatrix} \end {align} $$ where the mass of the ground state H atom is $m_H=$ 938.7830735MeV, and this is the mass of proton plus mass of electron minus the binding energy 13.6eV between them. $m_H+\Delta$ is the mass of the neutron, which thus defines that $\Delta=$ 0.7823470157MeV.

These will thus immediately fix the momentum = energy = $p$ of the anti-neutrino (I assumed it is massless, which is by far a reasonable approximation here because of how inconsequential it will be), and this gives us $p=$ 0.7820212976MeV. This is the energy that the anti-neutrino carries away, leaving just 325.7180673eV for the ground state H atom to recoil away with. The ground state H atom is thus going to recoil away non-relativistically.

Very little will change in this computation if you assumed that the H atom is excited rather than ground state. It will just be slight differences in the $m_H$ value being used, thereby inducing small changes in all the other values, since the mass of the neutron is fixed.

Anyway, it is surprising that this decay mode is even possible, so thanks for the interesting question.