# A question about Helium spectrum

I'm starting to read about two-electron atoms from physics of atoms and molecules by Bransden.

Analyzing Helium spectrum I noticed that the energy of $He^+(1s)+e^-$ is greater than the energy of He, where the energy of $He^+(1s)+e^-$ is calculated using the $E_n$'s formula of hydrogenic atom with Z=2 (-2.0 a.u.). Obviously in order to obtain $He^+(1s)+e^-$ I have to excite He, so it is natural that the energy of $He^+(1s)+e^-$ is bigger that energy of He. But how is it possible that an atom with Z=2 when it has 1 electron this is less bound and when it has 2 electrons these are more bound? It appears to me that the one-electron should be more bound because the nuclear charge is Z=2 and there aren't repulsive effects.

Sorry for bad English and thanks you all.

• <I have to excite He, so it is natural that the energy of He+(1s)+e− is bigger that energy of He.> ..why you say that as for excitation you must have put in work/energy and it will get less bounded. Aug 20, 2018 at 16:22

The first ionisation energy is $24.59 \, eV$, experimentally. This makes it go from $He$ to $He^+ + e^-$.
The second ionsation energy is $54.42 \, eV$, which can be calculated theoretically because the Helium ion is a hydrogenic atom. This makes it go from $He^+$ to $He^{++} + 2e^-$.
So you see that it is easier to remove the first electron, when you start from a 2 electron situation. This is because the second electron is more loosely bound to the nucleus, since the nuclear charge $Z$ is shielded by the first electron and becomes an effective $Z_{eff} < 2$.
You start with the energy of the second electron to be $-54.42\, eV$ (from the hydrogenic atom formula) and you add the postive repulsive energy that it gets from $e-e$ interaction. This gives you an energy of $-24.39 \, eV$, which means you have to provide $24.39 \, eV$ in order to free (ionise) the first electron.