1
$\begingroup$

I just want to confirm my understanding of energy levels of a hydrogen atom, since many times they introduce it without stating exactly what it is. It is given as $$E_{n} = -\bigg[\frac{m}{2 \hbar}\bigg(\frac{e^2}{4 \pi \epsilon_0}\bigg) \bigg]\frac{1}{n^2}~~~~~~\text{for }n=1,2,3...$$

Would I be right in stating that for each $n$, this is the total energy of the atom, which is basically the kinetic and potential energy of the orbiting electron? Secondly, would I be right in stating that in quantum mechanics since the idea of an electron orbiting does not really fit the idea of qm, a more accurate definition would be that this is the amount of energy you would need to remove the electron from the nucleus? Lastly, if my idea is correct that this is the total energy of the electron orbiting the nucleus then does include the rest mass energy of the nucleus and electron?

$\endgroup$
  • $\begingroup$ If there is an electron in the $n^{\rm th}$ energy level then $E_{\rm n}$ is the energy required to remove it completely from the influence of the nucleus. $\endgroup$ – Farcher Jan 24 '17 at 11:29
  • $\begingroup$ @Farcher So it does not make sense to think of it as the total kinetic and potential energy of the electron? $\endgroup$ – user100411 Jan 24 '17 at 11:30
  • $\begingroup$ I did not say that. Have a look at what HyperPhysics says about the Schrödinger equation. hyperphysics.phy-astr.gsu.edu/hbase/quantum/schr.html $\endgroup$ – Farcher Jan 24 '17 at 11:37
  • $\begingroup$ @Farcher Yeah I know you didn't say it, but I'm asking if you think that we can make that equivalence? I read the link. This is why I am asking the question. The link is also non-commital, it is stating that the Hamiltonian is analogous to the total energy (kinetic plus potential) of a classical system without stating that this is what it is? But then also in the case of the hydrogen atom it is given as the binding energy, which makes it more confusing... $\endgroup$ – user100411 Jan 24 '17 at 11:55
0
$\begingroup$

Would I be right in stating that for each $n$, this is the total energy of the atom, which is basically the kinetic and potential energy of the orbiting electron?

Yes and no. You should underline that the nucleus is supposed to be at rest. Then yes, $E_n$ is the total energy of the system, which corresponds to the sum of the kinetic energy of the electron and of the potential energy due to the electron/proton interaction.

Secondly, would I be right in stating that in quantum mechanics since the idea of an electron orbiting does not really fit the idea of qm, a more accurate definition would be that this is the amount of energy you would need to remove the electron from the nucleus?

Yes. More rigously: the energy you need to take the electron to an infinitely distant point from the nucleus.

Lastly, if my idea is correct that this is the total energy of the electron orbiting the nucleus then does include the rest mass energy of the nucleus and electron?

No, because the formula has been derived within the non-relativistic formalism, where the hamiltonian operator $H$ corresponds to: $$ E=\frac{p^2}{2m}+V $$ which is the classical mechanical energy.

$\endgroup$
  • $\begingroup$ Thanks for this nice answer. Generally though in physics is the binding energy and total energy equivalent in this way? From what I read on wiki about binding energy it seems that there is a lot more to take into account than just potential energy and kinetic energy? Is this just a good approximation to equate the two concepts for this particular case of the hydrogen atom? $\endgroup$ – user100411 Jan 24 '17 at 12:11
  • 1
    $\begingroup$ Yes, the binding energy indeed corresponds to the total energy in the hidrogen atom. Computing the ground energy level, E_1, lets you determine, e.g, the minimum energy a photon should have to ionize the atom. But be carefull. If you work in other fields, e.g. nuclear physics, you may be interested in determining how much energy you need to break a nucleus. Well, in this situation you need to use other models and other formulae. $\endgroup$ – AndreaPaco Jan 24 '17 at 13:44
  • 1
    $\begingroup$ @JohnDoe little nitpick: note the minus sign before the expression for $E_n$. It is normally assumed, that energy is zero for an electron at rest, infinitely far away from the proton. Thus $E_n$ tells us, how much energy an electron has, compared to what is necessary to get far away from the proton. The binding energy and/or the energy of a photon capable of ionizing the hydrogen atom is then $-E_n$. $\endgroup$ – LLlAMnYP Jan 24 '17 at 14:04
  • $\begingroup$ @LLlAMnYP Thanks I was wondering about that. So then would I be right in stating that as the total energy $E_{n}$ (kinetic and potential) decreases the binding energy $-E_{n}$ increases? $\endgroup$ – user100411 Jan 25 '17 at 12:25
  • 1
    $\begingroup$ @JohnDoe I suggest you "Quantum Physics of Matter", by Alan Durrant, pag 130. Here is the link: books.google.it/… $\endgroup$ – AndreaPaco Jan 25 '17 at 14:43

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.