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

[enotstrebor]

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?

 

[Drakkith]

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?

 

[DrDu]

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

 

[vanhees71]

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.

 

[PAllen]

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.

 

[enotstrebor]

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?

From any measurements I have seen, the measured mass of the hydrogen atom is not sufficiently accurate to tell if it is 13.6 eV lower in mass. Can you give me an experimental measurement reference?

If you calculate the energy gained going to the Bohr radius, i.e. e^2/R_bohr you get 27.2 eV. This also means you have to put in 27.2 to move it back out to infinity, yet a photon of 13.6 eV will do the trick, this to me says 13.6 eV must already be in the spread out over space "orbiting electron" to make up the difference and this is in keeping with QM.

Though the specifics are different than using the Schrödinger equation/QM, you can use the simplistic Bohr model as a quick check http://en.wikipedia.org/wiki/Bohr_model which uses E=KE+PE seven equations down and see that the Kinetic Energy = .5M_e v^2. Substituting the previous equation for v the kinetic energy of the electron = .5 e^2/R_bohr, i.e 13.6 eV of kinetic energy.

Is there something wrong with this?

So you see my dilemma, if in the hydrogen atom the electron has more energy and momentum, how can the hydrogen atom be lighter?

 

[Drakkith]

Is there something wrong with this?

So you see my dilemma, if in the hydrogen atom the electron has more energy and momentum, how can the hydrogen atom be lighter?

I'd guess that potential energy is the difference, as has been explained already.