PeterDonis said:
No, it isn't. 13.6 electron volts just means the ionization energy is the same, numerically, as the energy it would take to push one unit of electric charge through a potential change of 13 volts. It does not mean that ionizing a hydrogen atom is the same process, physically, as pushing one unit of electric charge through a potential change of 13 volts.
Also, the energy change of 13.6 electron volts for ionizing a hydrogen atom is the change between the electron being bound in the atom and separated from the atom. It is not the change between the electron being bound in the atom in its normal bound state and the electron being next to the proton, which is what "voltage between the electron and proton" would imply.
Spent another full day on this...
After reading this response again I think my original post was misunderstood since I had a big error in the separation of charges, so it probably didn't make much sense... so I'll try again...
(and moving forward assuming a classic Bohr atom, or point charges, or whatever is needed to humor my pedestrian model...)
Based on
this paper and
this thread it appears that the inward electrostatic force between the proton and electron is balanced by the outward centrifugal force of the electron. Based on the
Coulomb calculation at 53 pm (Bohr radius for hydrogen) the voltage is 27.2 volts between the proton and electron in the ground state. (27.2/2 = 13.6 volts) Makes sense assuming 27.2 volts of tension holding the electron in orbit, minus the outward centrifugal force maintaining a stable orbit, or some geometric wavelength limitation) Either way, it seems good enough to account for the 13.6 eV ionization energy
Then I found that
Lyman limit for the hydrogen atom (91.18 nm photon) and found a
Rydberg calculator. Assuming I did the calculation right, the Lyman limit is at n-41 on hydrogen, which puts the electron 88.9 nm from the proton. (based on 53pm*n-41 squared) This is why I originally said there would be 13.6 volts at a "certain distance of separation", which produces the 91.18 nm photon. (13.6 eV)
Now, assuming were holding the electron and proton apart by 88.9 nm...
1) The
Coulomb calculation tells me that voltage is only 0.016 volts... but I'm expecting to see a voltage of at least 13.6 volts... does the Coulomb calculation not apply here since we have two moving point charges that will be approaching closer and closer?
2) Once the hydrogen is in the ground state, there is no more "potential" since the charges are as close together as they can get. So can we call this 0 volts?
Brain fried... help needed...
thanks,