# Where is this extra energy in the Hydrogen Atom coming from?

The mass of hydrogen is 938.783073804(56) MeV/c^2
The mass of a proton is 938.272088163(35) MeV/c^2
The mass of an electron is .510998949(97) MeV/c^2
The mass defect is thus   -.000013308(84) MeV/c^2
The ionization energy is   .000013598(43) MeV      (with relativistic and nuclear corrections)


Data from NIST and CODATA. Hydrogen should be equal in mass to a proton plus an electron minus the ionization energy, but acccording to the above, a hydrogen atom is 0.290 electronvolts heavier/more energetic than that would suggest, and this difference is statistically significant.

The ionization energy takes into account the kinetic energy of the electron, and I would assume the kinetic energy of the nucleus in the center of mass frame is zero, so where is this extra energy coming from? For reference, this difference is about 3 times larger than the maximum predicted mass of the electron neutrino, so not an insignificant clerical error!

• "I would assume the kinetic energy of the nucleus in the center of mass frame is zero" How is that consistent with conservation of momentum? Oct 16 at 1:25
• Because the electron in the ground state is distributed spherically symmetrically about the nucleus. Any kinetic energy in the nucleus would just be the kinetic energy of the atom as a whole, no? Oct 16 at 2:09
• You mean you are considering the nucleus to be stationary because of its mass compared to the electron, and so it has zero kinetic energy. Oct 16 at 2:31
• Yes, but even we did not neglect the nucleus's KE, it would be on the order of the ratio of electron to proton mass times the electron's KE (13.6 eV), which would be only about 0.007 eV, clearly not the source of the 0.3 eV that's "missing". Oct 16 at 3:02
• Please link to source page where these numbers are given if possible, there may be some relevant information on how these were determined. Oct 16 at 4:41

It is difficult to measure mass of neutral hydrogen atom to similarly high accuracy as the mass of the proton, because the atom is electrically neutral and quite unstable (forms $$H_2$$ molecules or ions). I do not think digits in 938.783073804 MeV/c2 after 6th decimal are certain to be correct.