If you consider the classical states of hydrogen, one in which the electron is rotating at an orbital distance of $r$ and then take $\lim_{r\to\infty}$ one obtains that the $r_\infty$ state has more energy, hence more mass from the energy mass relationship $E=MC^2$. Then when is a proton lighter than standard hydrogen, where one could take the $r_\infty$ to be equivalent to the stand-alone proton state, and the other to be standard hydrogen.

EDIT: I asked about Deuterium previously, but wish not to discuss it further.

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    $\begingroup$ The answer here would obviously be Deuterium since Deuterium has an extra neutron which has a mass that would overshadow any kind of "binding energy" induced effects. Perhaps you would like to remove Deuterium from contention? $\endgroup$ – enumaris Dec 5 '18 at 23:02
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    $\begingroup$ The binding energy of an electron to the proton in H is 13.7eV. The mass of an electron is 511keV. $\endgroup$ – Jon Custer Dec 6 '18 at 0:08
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    $\begingroup$ A proton does not have more mass than a hdrogen atom, so either your calculation or your memory is wrong. $\endgroup$ – G. Smith Dec 6 '18 at 4:36

There's currently another answer which addresses the question that you asked, but not (I think) the question that you meant to ask:

Does a bound hydrogen atom have the same mass as a free election plus a free proton at rest, or does the binding energy $E=mc^2$ change this mass?

The answer here is that the masses are different: the bound atom is less massive than the sum of its constituents by the (mass equivalent of the) binding energy, which is $13.6\rm\,eV$.

That's a tiny mass-energy, only a few parts per billion of hydrogen atomic mass. I don't believe the difference in mass has been measured directly, and I don't expect that it will be anytime soon, because the methods for measuring the masses of charged and neutral particles are quite different.

In nuclear physics, where the energies involved are much larger, you can have mass differences which can be measured directly. For example, the binding energy of the deuterium nucleus is about 0.1% of its mass. You can imagine separately measuring the neutron mass, proton mass, deuteron mass, and the energy of the photon released in $\rm np\to d\gamma$ to check this directly. However, again because of electric charge, it turns out that there is (so far) no other comparably-precise measurement of the neutron mass, so the deuterium-formation experiment is currently effectively a neutron-mass measurement.

(Note that the binding energy of the deuterium atom is still about $13.6\rm\,eV$; it's different from the hydrogen binding energy starting in the fourth or fifth significant figure, because the nuclear mass makes a little bit of a difference.)


Mass of proton: 1.007276 u. Source: https://en.m.wikipedia.org/wiki/Proton

Mass of hydrogen: 1.007825 u. Source: https://en.m.wikipedia.org/wiki/Isotopes_of_hydrogen

Mass of deuterium: 2.014102 u. Source: https://en.m.wikipedia.org/wiki/Isotopes_of_hydrogen

As commenter @enumaris pointed out, deuterium is the heaviest because it has a neutron in addition to a proton and an electron.

Hydrogen is heavier than a proton because it also has an electron. The kinetic energies and electrostatic potential energies of the proton and electron in hydrogen are small compared with their mass-energies.

Gravitational energies are completely negligible in particle physics, nuclear physics, atomic physics, and molecular physics.


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