# Where does 13.6 $\rm eV$ of energy go in the process of bringing an electron from infinity to the Bohr ground state?

In the answer to this SE question about the Relation of potential energy and total energy in Bohr Model of the hydrogen atom, it is clearly explained by using the Virial Theorem that:

"In the Bohr ground state the potential energy is -27.2 eV. Note that as described above this energy is negative. The kinetic energy is +13.6eV, so when we add the two together we get the total energy to be -13.6eV."

Initially, an electron is at rest, infinitely far away from a proton at rest, so the total energy is zero. Assume that the proton is fixed, and that the electric field accelerates the electron towards the proton. Eventually, the electron reaches the Bohr ground state in the 1s orbital where the total energy is -13.6eV, so the electron proton system has lost 13.6eV of total energy. Where would this energy go? Would it get converted into radiation due to the electron's accelerating towards the proton, go into the energy required to keep the proton fixed, or somewhere else?

• It would go into the electromagnetic field, just as a falling electron stores energy in the gravitational field. To what extent we should describe the former in terms of photons depends on what else you want out of your physical model.
– J.G.
Sep 7, 2022 at 16:26
• Sep 7, 2022 at 16:47

"go into the energy required to keep the proton fixed, or somewhere else?" This doesn't actually make sense. There's nothing holding the proton fixed. In fact, when you examine spectra from stars you see something called Doppler broadening, where the spectrum is spread out by the random motions of the protons and electrons as they move around due to heat. That said, some of the energy can go into changing the kinetic energy of the center of mass of the combined hydrogen. Not much, though, since the photons produced by the collision don't have much momentum so they can't produce much of a kick. A single Lyman limit photon has a momentum of 13.6 eV/c. The change in velocity that represents for a proton is 4.3 meters per second. Even at room temperature, the root-mean-square velocity of a proton is about 1000 m/s. That means that the change in energy for the proton from that kick would be about $$50\,\mu \mathrm{eV}$$.