According to Bohr model of hydrogen atom, the electron orbits in circular orbits around the nucleus. when I'm asked to find the energy needed to ionize the hydrogen atom, I calculate the energy stored in the electron. this energy is consisted of kinetic energy and electric potential energy.

But, I was said that in order to get more precise results, I I have to calculate the "Reduced Mass" of the problem because the nucleus also orbits around the center of mass.

My question is why do we care about the velocity/movement of the nucleus in calculating the energy of the electron? solving the "reduced mass" problem gives my the energy of "the whole system" instead of the single electron. so why do we need to do that? is the energy of the electron changes somehow when the nucleus also orbits around the center of mass?

  • $\begingroup$ The hydrogen ground state has ${\cal l } =0$ so the Bohr model is not correct. Then why bother working with it? $\endgroup$
    – my2cts
    Dec 16, 2021 at 23:01

1 Answer 1


The hydrogen atom is a bound system of two particles. If one of these particles is removed (e.g. by ionization), the other doesn't continue to orbit the former center of mass: it's no longer bound to the removed particle.

So you can't remove only the electron, ionization applies to the whole atom. Treating it as if the whole energy is in the electron is an approximation, where proton mass is assumed to be much larger than electron mass: $m_{\text p}\to\infty$.

  • $\begingroup$ So according to what you say, if i want to ionize an atom i have to “give” him the energy of the whole system instead of just the energy of a single electron. $\endgroup$
    – mcr0yal
    Dec 17, 2021 at 7:52
  • $\begingroup$ @mcr0yal right. A single electron, when in an atom, doesn't have a definite energy, because the state is entangled with that of the proton. $\endgroup$
    – Ruslan
    Dec 17, 2021 at 8:13
  • $\begingroup$ thank you very much! $\endgroup$
    – mcr0yal
    Dec 17, 2021 at 9:47
  • $\begingroup$ @mcr0yal if the answer resolved your question, you can accept it by clicking the tick mark to the left of it. $\endgroup$
    – Ruslan
    Dec 17, 2021 at 10:02

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