We now that from this equation: $$\hat {H} (r , \theta , \varphi ) \psi (r , \theta , \varphi ) = E \psi ( r , \theta , \varphi) $$ we can derive the energy of a system. Total electron energy for hydrogen atom is: $E_n = -\dfrac {m_e e^4}{8\epsilon_0^2 h^2 n^2}$ using central force potential.

But what is the kynetic energy of nucleus?

  • $\begingroup$ You mean the KE wrt. the center of mass frame because of reduced mass? $\endgroup$ – Qmechanic Sep 30 '17 at 7:09
  • $\begingroup$ Hi @Qmechanic, from the nucleus...; center of mass $\endgroup$ – santimirandarp Sep 30 '17 at 8:18
  • $\begingroup$ The kinetic energy doesn't commute with the Hamiltonian so the best you can do is get a distribution. $\endgroup$ – dmckee Sep 30 '17 at 18:54
  • $\begingroup$ @dmckee what do you mean? $\endgroup$ – santimirandarp Sep 29 '18 at 20:52

$E_n$ is the total energy of the electron-nucleus system because it contains a potential energy term which accounts for the electrostatic interaction between the electron and the nucleus.

The nucleus is much heavier than the electron so it's considered as staionary, hence it's kinetic energy is zero. But this is just an approximation (a good one though), both the electron and the nucleus rotate about their common center of mass, and one can easily attack this problem by using the center of mass coordinates and reduce this two body problem to a one body problem and introduce the reduced mass of the system.

The mathematical analysis is strightforward and you can find it in many references. you'll find a similar formula of the total energy: $$E_n = -\dfrac {\mu e^4}{8\epsilon_0^2 h^2 n^2}$$
where $\mu$ is the reduced mass which is given by $\mu={m_em_p \over m_e+m_p}$ with $m_e$ and $m_p$ are the mass of the electron and the proton (the nucleus). You can see clearly that $\mu \simeq\ m_e$ due to the fact that $m_p$ is much greater than $m_e$. Thus your formula is a good approximation.

One of the benefits of considering the motion of the nucleus is the theoretical prediction of the existing of neutrons! By analysing the spectra of the hydrogen gaz which may contain both hydrogen and deuterium, we observe a slight shift in the observed sepctral lines of the ordinary hydrogen due the presence of deuterium, the Ryidberg formula of these two atoms allow us to find the ratio: $${m_H \over m_D}\simeq\ 0.5$$

with $m_H$ is the mass of the hydrogen nucleus (the proton) and $m_D$ is the mass of the deuterium nucleus (proton+neutron).

In other words, the atomic nuclei don't contain only protons, but also a neutral particles with approximately the same mass as protons called neutrons which were discoverd experimentally many years later by Chadwick.

  • $\begingroup$ This is nice, but it falls short of answering the question. $\endgroup$ – garyp Sep 30 '17 at 12:57

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