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I'm a bit confused about this part since it doesn't specifically mention it in my book as far as I've read.

But let's say if I have to compare the ionization energy of the hydrogen atom to the ionization energy of Be$^{3+}$, I know that the ionization energy of the hydrogen atom is 13,6 eV and for Be$^{3+}$ it's about 218 eV, which in turn makes it about 16 times greater than the ionization energy for the hydrogen atom. Now here lies my confusion, what is meant by "The ionization energy is the energy difference between the $n→∞$ level energy and the $n =1$ level energy."

Where does "the n→∞ level energy" come from exactly?

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An electron that has been separated completely from 'its' nucleus is said to be at infinite distance from that nucleus, because at that distance the Coulombic potential $V(r)$ is zero (formula for a hydrogenic atom):

$$V(r)=-\frac{Ze^2}{4\pi \epsilon_0 r}\implies V(r)\to 0\text{ for }r\to +\infty$$

For hydrogen the energy spectrum is given by:

$$E(n)=\frac{E_1}{n^2}$$

Where $n$ is the principal quantum number ($=1,2,3,...$) and $E_1=-13.6\ \mathrm{eV}$.

Here too: $$n\to+\infty \implies E(+\infty)\to0$$

For an infinitely high quantum number the electron's energy is zero.

So the ionisation energy is for hydrogenic atoms (in the ground state i.e. $n=1$) is:

$$\Delta E=E(+\infty)-E_1=-E_1$$

For actual hydrogen that is $-(-13.6\ \mathrm{eV})=+13.6\ \mathrm{eV}$.

For $\mathrm{Be}^{3+}$ that ionisation energy is much higher due to $E_1$ being much more negative than for $H$.

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  • $\begingroup$ Ah ok so we set it to "zero" mostly for convenience sake essentially since well obviously taking the limit of when n goes to infinity then energy will "converge" to zero and therefore we get the difference in energy between the nth energy level and the ground state. $\endgroup$
    – Myzanthros
    Dec 5 '17 at 20:04
  • $\begingroup$ Does this also work between let's say n=4 to n=2? Or is it explicitly just for when it concerns the involvement of the ground state? $\endgroup$
    – Myzanthros
    Dec 5 '17 at 20:05
  • $\begingroup$ @Myzanthros: yes. $\Delta E_{4\to 2}=\frac{E_1}{2^2}-\frac{E_1}{4^2}$. Of course that wouldn't be an ionisation. It would be an excitation or relaxation. $\endgroup$
    – Gert
    Dec 5 '17 at 20:18
  • $\begingroup$ Thx, a bit clearer now :) $\endgroup$
    – Myzanthros
    Dec 5 '17 at 20:38
  • $\begingroup$ Made an edit in the first formula line. $\endgroup$
    – Gert
    Dec 5 '17 at 20:41

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