Please excuse me if this question has been asked and/or even partially answered here before... (I did search about, and did not see it specifically asked)

In particular, and being purposefully 'general' in my question, I was wondering how the 'energy accounting' 'works out' in the formation of Hydrogen from a single proton and single electron that are initially separated at infinite distance.

Allow me to explain somewhat with some math...

If we take an electron and a proton at infinite distance, the potential energy between them is:

$$ U=\frac{k_e q^2}{r}=\frac{k_e q^2}{\infty}=0 \:\mathrm{J} $$

If we take an electron and a proton at at the Bohr radius (mind you, I am not suggesting the electron 'orbits' the proton), the potential energy between them is:

$$ U=\frac{k_e q^2}{r}=\frac{k_e q^2}{a_0}=\frac{k_e q^2}{5.29\times 10^{-11}}=-4.36\times 10^{-18} \:\mathrm{J} =-27.2\:\mathrm{eV} $$

Now, considering the Virial theorem and that a photon of 13.6eV is emitted during the formation of the Hydrogen atom, and the necessity of the absorption of another photon of 13.6eV into this very same Hydrogen atom in order to separate the proton and electron that constitute it and bring them back (hypothetically) to infinity...

How is the 'rather' obvious fact that both the proton and electron are being accelerated toward each other, and hence there should be radiation emitted by both particles in the process of the proton and electron coming together not 'detract' from the initial potential energy between them?

I will say it another way...

If the initial potential energy is radiated away 'during formation' of the Hydrogen atom as a photon of 13.6eV, which actually makes sense per the Virial theorem, leaving the proton and electron with a combined kinetic energy of 13.6eV (somehow, again, not saying 'orbiting'), how then is this rather 'perfect' energy accounting happening when there should be some radiation from both the proton and electron occurring during the time in which they are approaching each other, and this radiation should 'detract' from the total initial potential energy?

Or, is the radiated photon of 13.6eV 'during formation' of the Hydrogen atom some sort of distributed sum of the increasing radiation power from both the proton and electron as they appoach each other?

Any insights into this would be most appreciated, and I thank you for your time and patience...


1 Answer 1


In short, the classical understanding that 'accelerated charges radiate' does not necessarily hold in quantum mechanics. In the classical limit you can get it back, but the fundamental reality as far as quantum mechanics is concerned is that you have an electromagnetic field with hamiltonian (energy) $$ H_\mathrm{field}=\frac12 \int (\mathbf E^2+\mathbf B^2)\mathrm dV $$ which couples via its vector and scalar potentials to the hamiltonian (energy) of the particles, $$ H_\mathrm{particles}=\sum_\alpha \left[ \frac1{2m_\alpha} (\mathbf p_\alpha -e\mathbf A(\mathbf r_\alpha))^2 + V_\mathrm{mech}(\mathbf r_\alpha) +\phi(\mathbf r_\alpha) \right]. $$ Everything else is extra.

In particular, in your case, you're mixing concepts from two levels: the 'accelerated charges radiate' intuition from classical electrodynamics, and the radiative transitions that come from a full quantum treatment. If you want to keep things hand-wavy, then you can simply attribute the emitted photon to the acceleration of the electron and proton as they change their state (but that's pretty much it; that's essentially all the detail that that picture will allow you). If you want to do things correctly, forget about accelerating charges and do the whole show quantum mechanically.

  • $\begingroup$ Thanks a lot for your answer, it certainly clears things up for me. I 'denoted' later that my question really was crossing 'concepts', if you will. Thanks, not only for editing my LaTeX, but, for the great answer... $\endgroup$ Jan 11, 2017 at 9:13

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