# Why is a hydrogen atom lighter than a electron and a proton

• Johnahh
In summary: It takes 27.2 eV to move an electron from the Bohr radius to infinity ! That's why the hydrogen atom is heavier than the electron and proton separately- the electron has to add that energy to escape.
Johnahh
What exactly is it that makes the energy/mass of a hydrogen atom be lower than that of the electron and proton separately?
I am aware that this process is what causes stars to emit light, during nuclear fusion deuterium has a smaller mass than a proton and a neutron so its extra mass/energy is emitted as light.

The binding energy of the electron in hydrogen is 13.6 eV. By Einsteins E=mc^2 this corresponds to a tiny mass difference.

The binding energy of the electron in hydrogen is 13.6 eV. By Einsteins E=mc^2 this corresponds to a tiny mass difference.

So some energy is taken from the electron and used as binding energy?

Johnahh said:
So some energy is taken from the electron and used as binding energy?

No, that's not it.

When you combine an electron and a proton to form a hydrogen atom, the reaction gives off energy. At least part of that energy comes from the electrostatic attraction between the two component pieces.

Energy is conserved.

That means that once this extra energy has radiated away (as heat, photons or whatever), the resulting hydrogen atom will have less total energy than the electron and the proton from which it was assembled.

By E=mc^2, this lost energy manifests as lost mass. The composite has less mass than the sum of the parts from which it was assembled.

The term "binding energy" refers to how much energy is released when a bond is formed or, equivalently, how much energy would need to be injected to break a bond. If were able to measure carefully enough you would find that binding energy is always accompanied by a "mass defect".

No, the energy is taken from the electric field surrounding both the electron and the proton.
The electromagnetic field is smaller in the atom than for the isolated electron and proton.

thank you both :).

41burhan said:
the mass of Hydrogen is less then that of electron and proton because of mass defect and binding energy.
And that "binding energy" is due to electric fields, in the case of an atom, as stated already. Inside a nucleus there are other forces, so the binding energy is due to them also.

Inside a nucleus there are other forces, so the binding energy is due to them also.

So gluons, W and Z particles are also the cause?

Johnahh said:
What exactly is it that makes the energy/mass of a hydrogen atom be lower than that of the electron and proton separately?
I am aware that this process is what causes stars to emit light, during nuclear fusion deuterium has a smaller mass than a proton and a neutron so its extra mass/energy is emitted as light.

The hydrogen atom is NOT LIGTER.

Starting with an electron and proton at near infinity, the coulomb potential force causes the electron to accelerate (gain energy) toward the proton and vis-versa and the electron/proton system gains 27.2 eV (e^2/Bohr_R). Only half that energy 13.6 eV, is emitted by the hydrogen stom. So the hydrogen atom is heavier by 13.6 eV. Remember any energy in an object, even heat energy makes the object heavier.

Unfortunately the measured hydrogen-1 mass (that I have seen) is not accurate enough by several orders of magnitude to measure this difference in mass.

enotstrebor said:
The hydrogen atom is NOT LIGTER.

Of course it is. I have to add energy to separate the electron from the proton.

enotstrebor said:
So the hydrogen atom is heavier by 13.6 eV.

Then why do I not see hydrogen atoms spontaneously emitting a 13.6 eV photon as they decay into an unbound proton and electron?

enotstrebor said:
The hydrogen atom is NOT LIGTER.

Starting with an electron and proton at near infinity, the coulomb potential force causes the electron to accelerate (gain energy) toward the proton and vis-versa and the electron/proton system gains 27.2 eV (e^2/Bohr_R). Only half that energy 13.6 eV, is emitted by the hydrogen stom. So the hydrogen atom is heavier by 13.6 eV. Remember any energy in an object, even heat energy makes the object heavier.

The rest mass of the proton and electron compared to the rest mass of the hydrogen atom is what we mean by mass here. In your example you could have the extra energy left over as kinetic energy, but when we slow it down and measure it, it's mass is indeed less.

Of course it is. I have to add energy to separate the electron from the proton.
Nugatory said:
Then why do I not see hydrogen atoms spontaneously emitting a 13.6 eV photon as they decay into an unbound proton and electron?

Do the calculation to move an electron from the Bohr radius to infinity, i.e e^2/bohr_R = 27.2

It takes 27.2 eV to move an electron from the Bohr radius to infinity !

The photon energy you are adding is only 13.6 eV. Where do you think the other 13.6 eV comes from?

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Drakkith said:
The rest mass of the proton and electron compared to the rest mass of the hydrogen atom is what we mean by mass here. In your example you could have the extra energy left over as kinetic energy, but when we slow it down and measure it, it's mass is indeed less.

The initial center of mass is not changed by the in-fall of the electron and proton as both accelerate toward each other. As the center of mass does not change, the hydrogen atom itself has no velocity nor kinetic energy.

Thus all of the extra 13.6 eV is internal to the hydrogen atom, i.e. 13.6 eV heavier

enotstrebor said:
Do the calculation to move an electron from the Bohr radius to infinity, i.e e^2/bohr_R = 27.2

It takes 27.2 eV to move an electron from the Bohr radius to infinity !

The photon energy you are adding is only 13.6 eV. Where do you think the other 13.6 eV comes from?
The electron bound in a hydrogen atom already has 13.6 eV of kinetic energy. So you only need to add 13.6 eV to remove the electron.

Redbelly98 said:
The electron bound in a hydrogen atom already has 13.6 eV of kinetic energy. So you only need to add 13.6 eV to remove the electron.

Exactly, what I have been saying.

Just as nuclear kinetic energy makes the nucleus heavier than the sum of the proton and neutron masses, likewise the bound electron kinetic energy makes the hydrogen atom heavier than the mass of the isolated electron and proton.

enotstrebor said:
Exactly, what I have been saying.

Just as nuclear kinetic energy makes the nucleus heavier than the sum of the proton and neutron masses, likewise the bound electron kinetic energy makes the hydrogen atom heavier than the mass of the isolated electron and proton.
Now you're forgetting about the hydrogen atom's -27.2 eV of potential energy. Add that to the 13.6 eV of KE, you get a total bound energy of -13.6 eV. So the hydrogen atom has lower energy than and is lighter than the isolated electron and proton by 13.6eV/c2

enotstrebor said:
Just as nuclear kinetic energy makes the nucleus heavier than the sum of the proton and neutron masses, likewise the bound electron kinetic energy makes the hydrogen atom heavier than the mass of the isolated electron and proton.
No, it is the other way round. Helium-4 is lighter than 2 protons plus 2 neutrons, for example.

Proton mass: 1.007 276 u
Neutron mass: 1.008 665 u
Helium-4 mass: 4.002 603 u

enotstrebor said:
Exactly, what I have been saying.

Just as nuclear kinetic energy makes the nucleus heavier than the sum of the proton and neutron masses, likewise the bound electron kinetic energy makes the hydrogen atom heavier than the mass of the isolated electron and proton.

My apologies about nuclear mass, which is not a negative coulomb attraction binding process (which happens to add more energy than it emits) and is a true energy emission and removal binding process which does make the nucleus lighter than its constituents.

enotstrebor said:
Exactly, what I have been saying.

Just as nuclear kinetic energy makes the nucleus heavier than the sum of the proton and neutron masses, likewise the bound electron kinetic energy makes the hydrogen atom heavier than the mass of the isolated electron and proton.

My apologies on nuclear binding energy. Unlike the negative coulomb "binding" situation of the electron and proton of the hydrogen atom, in which the attractive forces that bring the electron and proton together and adds more energy than it emits, the nuclear process of bringing both protons and neutrons together adds no energy and thus the energy loss on nuclear binding of protons and neutrons does reduce mass.

You may find somewhere the masses of He2+, He+ and e- with high precision (Ions are easier to weigh than atoms). He+ should be lighter than it's constituents.

@enotstrebor: It is the same situation for both atoms and nuclei - you have an attractive force, you bring particles together and release energy corresponding to the lower potential energy afterwards. As you already mentioned, the kinetic energy is just 1/2 of the potential energy for electrons - the other 50% are emitted (usually as light), reducing the total energy content of the system.

mfb said:
@enotstrebor: It is the same situation for both atoms and nuclei - you have an attractive force, you bring particles together and release energy corresponding to the lower potential energy afterwards. As you already mentioned, the kinetic energy is just 1/2 of the potential energy for electrons - the other 50% are emitted (usually as light), reducing the total energy content of the system.

You talk about the total energy. The issue is about the mass energy. Some forms of energy do not have the property termed mass. Potential energy does not have the property of mass. Kinetic energy does have the property of mass.

Thus although the starting e/p system's total energy (mass+potential) is less than total energy of the orbital e/p system, the mass of the orbiting e/p system is greated than in the initial state.

Redbelly98 said:
Now you're forgetting about the hydrogen atom's -27.2 eV of potential energy. Add that to the 13.6 eV of KE, you get a total bound energy of -13.6 eV. So the hydrogen atom has lower energy than and is lighter than the isolated electron and proton by 13.6eV/c2

Potential energy does not have the property of mass, so you are confusing total energy with mass energy. Kinetic energy does have the property of mass. A hot object ways more than a cool object.

Thus the electron's orbital kinetic energy adds mass while the potential energy does not.

See prior post.

Potential energy is energy in electromagnetic (or other) fields and does contribute to inertial and gravitational mass.

enotstrebor said:
See prior post.
It does not help to confirm your own wrong posts.

mfb said:
@enotstrebor: It is the same situation for both atoms and nuclei - you have an attractive force, you bring particles together and release energy corresponding to the lower potential energy afterwards. As you already mentioned, the kinetic energy is just 1/2 of the potential energy for electrons - the other 50% are emitted (usually as light), reducing the total energy content of the system.

The two situations are not at all equivalent as they involve two different forces.

What we know about coulomb potential energy is that it does not have the property of mass. In the nuclease there is not an attractive coulomb force situation but a tremendous repulsive potential energy.

Remember two things. First the nuclear force is known not to be like a coulomb force. Second although one can speculate that mass is related to the Higgs field and in theory can give mass, the coupling constant between the Higgs and the electron and between the Higgs and the proton quark structure are selected, and there is no theory about the nature of the coupling between a particle and the Higgs field

The nature and relationship of nuclear binding via interactions between the quarks in two different nuclear particles is (to my knowledge) not well understood, in addition to the unknown nature of coupling magnitude to the Higgs field.

If related to quarks, then potentially, for example, there is a weakening of the intra-particle quark binding releasing energy and reducing mass while increasing inter-particle binding sufficient to oppose the repulsive coulomb forces, i.e. a net reduction in mass.

So no the two situations are not at all alike.

What we do know is that kinetic energy adds mass as measured in the laboratory. As there is more kinetic energy in a hydrogen atom than in the non-velocity mass of the electron and proton, the hydrogen atom should be heavier.

enotstrebor said:
The two situations are not at all equivalent as they involve two different forces... So no the two situations are not at all alike.
Perhaps we should have said that the two situations are "analogous" instead of "equivalent"? The mathematical descriptions of the energy transfer as potential energy is released are the same; only the shapes of the curve of potential energy as a function of distance from the center are different.

What we do know is that kinetic energy adds mass as measured in the laboratory. As there is more kinetic energy in a hydrogen atom than in the non-velocity mass of the electron and proton, the hydrogen atom should be heavier.

(Quantum mechanically we're on shaky ground talking about the kinetic energy of a bound electron... But for present purposes the non-QM description is good enough so I'll stick with it).

There is mass associated with the Coulomb potential energy. As the electron falls towards the nucleus some of this mass turns into kinetic energy, and some of it is radiated away as electromagnetic energy. There's a net loss of mass, corresponding to the energy that is radiated away.

By definition, mass is the energy of a compound system in the system where the center of energy is at rest (up to proportionaliy 1/c^2). Clearly Coulombic potential contributes to mass.

DrDu said:
By definition, mass is the energy of a compound system in the system where the center of energy is at rest (up to proportionaliy 1/c^2). Clearly Coulombic potential contributes to mass.

An electron proton in-fall system gains 27.2 eV of kinetic energy and associated momentum on reaching the Bohr radius.

What are you saying the coulomb potential contribution to the system is at the Bohr radius and how it "contributes to mass" at this point?

enotstrebor said:
An electron proton in-fall system gains 27.2 eV of kinetic energy and associated momentum on reaching the Bohr radius.
The electron is not localized at a specific radius.

What are you saying the coulomb potential contribution to the system is at the Bohr radius and how it "contributes to mass" at this point?
The energy stored in the electric field is reduced, and energy in electric fields contributes to the mass of the system.

mfb said:
The energy stored in the electric field is reduced, and energy in electric fields contributes to the mass of the system.

I would agree that with this statement that the field energy decreases and becomes particle kinetic energy and momentum and higher effective mass.

As far as I know the field energy itself however does not weigh anything nor has the associated property of mass.

Correct?

enotstrebor said:
I would agree that with this statement that the field energy decreases and becomes particle kinetic energy and momentum and higher effective mass.

As far as I know the field energy itself however does not weigh anything nor has the associated property of mass.

Correct?

I don't think that's correct. If you measure the mass of the system with the proton and electron before capture and after capture the mass will be less once the electron has been captured by the proton. The missing mass will be radiated away as energy in the EM field, aka light. Light does contribute to gravitation and to the mass of the system.

However, if you measure the mass of the system with the electron nearly an infinite distance away from the proton, and then again right before the electron is captured by the proton, there will be no change in the mass. (Unless the acceleration of the two particles emits EM radiation)

Does that make sense?

enotstrebor said:
An electron proton in-fall system gains 27.2 eV of kinetic energy and associated momentum on reaching the Bohr radius.

What are you saying the coulomb potential contribution to the system is at the Bohr radius and how it "contributes to mass" at this point?

No, it's kinetic energy only rises by 13,6 eV. There is the virial theorem which states that 2T=- V

I haven't followed this thread in detail, but to make it very clear: The mass of a hydrogen atom in its ground state is
$$m_{\text{hydrogen}}=m_{\text{p}}+m_{\text{e}}-E_{\text{B}}/c^2,$$
where $E_{\text{B}} \simeq 13.6 \; \text{eV}$ is the binding energy. The mass is smaller than the sum of the masses of the proton and the neutron.

Much more pronounced is this socalled "mass defect" for atomic nuclei, i.e., their mass is considerably smaller than the sum of the masses of the protons and neutrons they consist of.

neilparker62
enotstrebor said:
Exactly, what I have been saying.

Just as nuclear kinetic energy makes the nucleus heavier than the sum of the proton and neutron masses, likewise the bound electron kinetic energy makes the hydrogen atom heavier than the mass of the isolated electron and proton.

As so many have said, this is wrong - isolated proton and electron weigh more than hydrogen atom.

I'll try one more way to justify it. Imagine a box with and isolated proton and electron far apart. As they get close, the electron has picked up lot's of KE, and forms an atom in an excited state. A photon is emitted as it falls to ground state leaving the (transparent) box. By your argument, the box + contents have gained net mass/energy as the excited atom forms, before the photon is emitted. I hope you can see that is absurd - trivial violation of conservation of energy.

Instead, the box + contents remain constant in mass/energy until the photon passes out. As the atom forms, PE is converted to KE. As a result, the excited atom weighs the same as the sum of isolated proton and electron. As it falls to ground state, emitting a photon, the atom then weighs less than isolated proton and electron by the mass equivalent of the photon emitted.

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