After spinning around the atomic mass equation for calculating neutron separation energies, I have run into somewhat of a conundrum. It appears that the mass of the simplest of the examples, hydrogen is not correctly produced, yet I cannot put my finger on what is the reason for this. Let me elaborate: the atomic mass of some nucleus $^A _Z X_N$ is defined as

$$m(A,Z) = Z\cdot (m_p + m_e)+ (A-Z)\cdot m_n -B/c^2,$$

where subscripts $m_p$, e and n mean proton, electron and the neutron masses, respectively. And where B is the net binding energy of a nucleus. If I take the hydrogen atom, and try to predict its mass this way, I find, unsurprisingly

$$m(^1 H) = m_p + m_e - B/c^2 $$

However, as there is only one nucleon in hydrogen, the contributions to B do not feature a any term from nuclear binding, only the binding energy of the single atomic state in hydrogen, i.e. the ionization energy of hydrogen is featured. The calculation does not add up, however.

Proton mass is 938.272 MeV/$c^2$, electron mass is 511 keV/$c^2$ and the ionization energy of hydrogen is 14 eV. Using these values gives an atomic mass of 938.8 MeV/$c^2$ for hydrogen, but the measured atomic mass of hydrogen is 939.0 MeV$/c^2$. Is there an explanation for the surprising 200 keV/$c^2$ discrepancy?

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    $\begingroup$ What is your source for the atomic mass being 939.0 MeV/$c^2$? As far as I know, the accepted value is 1.00784 amu = 938.8 MeV/$c^2$. $\endgroup$
    – Thorondor
    Commented Sep 30, 2020 at 13:19
  • $\begingroup$ The place to find the latest news in atomic mass is the Atomic Mass Evaluation 2012 papers, easily found via google or through the original in Chinese Physics C 36(12) 1287-1602 (part I) and 1603-2014 (part II with the mass tables). $\endgroup$
    – Jon Custer
    Commented Sep 30, 2020 at 13:43

1 Answer 1


Without knowing the source of your numbers, I suspect that you have the atomic mass of elemental hydrogen, which is a mixture of naturally-occurring isotopes of hydrogen, primarily $^1H$ and $^2H$, with a trace of $^3H$.

You need to find the mass of the isotope $^1H$. That will fix the discrepancy.

  • $\begingroup$ To confirm this: the natural abundance of deuterium is about $10^{-4}$. Including a component with mass 2-ish GeV in the average, with this weighting, shifts the average mass upward by 200 keV, as observed in the question. $\endgroup$
    – rob
    Commented Nov 1, 2022 at 2:35

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