# Why isn't the ground state energy $-13.6 eV$ of electron in hydrogen atom also the rest energy $0.5 MeV$ of electron?

Why isn't the ground state energy $-13.6 eV$ of electron in hydrogen atom also the rest energy $0.5 MeV$ of electron?

That 13.6eV is the difference in the energy of a bound electron as compared to an electron at rest far from the proton. So it is an energy difference. We don't include the rest mass because that's a constant so it's the same for both the bound and free electron. When we take the difference in the energies the rest mass cancels out.

You may be interested to know the mass of a hydrogen atom is actually less than the mass of an isolated proton + the mass of an isolated electron. The difference in the mass is 13.6eV. That's because to make a hydrogen atom we have to start with an isolated proton and an isolated electron then remove 13.6eV to make them stick together.

• But I cant see where is that difference calculating (I mean the calculation using the uncertainty principle) : E=p^2/2m - αcp, then differentiate with respect to p and set equal to zero to get the minimum. E=-1/2.(α^2.mc^2)=-13,6 eV. – Zuzana Bardačová Sep 30 '17 at 16:11
• @ZuzanaBardačová I've never seen that calculation, can you give a reference to it please? – jim Sep 30 '17 at 19:04
• quantummechanics.ucsd.edu/ph130a/130_notes/node98.html – Zuzana Bardačová Sep 30 '17 at 22:29
• @ZuzanaBardačová: firstly that energy is also a difference because the potential energy $-e^2/r$ is the difference between the energy at infinity and the energy at a distance $r$. Secondly they are differentiating to find the minimum of $E(r)$ so adding a constant term $mc^2$ would disappear anyway when differentiated. – John Rennie Oct 1 '17 at 4:46

This is because the fine structure constant is equal to approximately $\frac{1}{137}$ instead of $2$. The reason why the fine structure constant takes this value is not known, but it has been argued that even small changes in the fine structure constant would make it more difficult for life to appear.

$E_{1}$ is the energy of the system in its ground state. This means that the system cannot be in a lower energy state. In the case of the hydrogen atom, $E_{1}=-13.6$ eV.

In order for a hydrogen atom to form, one electron must bound with a proton. When this bonding happens, there is a release of energy. Because a hydrogen atom is a bond system, breaking it in its constituent parts requires energy. The amount of energy needed to remove the electron from the proton is equal to the energy released when they formed a bound state.

For the hydrogen atom in the ground state this energy is $-13.6$ eV. It is also called ionization energy. The rest energy of the electron is around $511$ keV and the rest energy of the proton is around $938$ MeV.