Ionization Energy and Bohr

Question

I was having problems with ionization energy - but all it is is the energy required to remove first electron that pops off right? In the case of Bohr's atom, there's only one electron, so it's that electron in it's ground state. In the case of a multielectron atom, it's the energy required to remove the outer electron, in it's $n = 1$ state, right?

Answer 1

That's right except the $n = 1$ part. The outer (aka "valence") electron is always in a higher-$n$ state, except for hydrogen (electron configuration $1\mathrm{s}^1$) and helium ($1\mathrm{s}^2$). For example, lithium's three electrons are often denoted $1\mathrm{s}^22\mathrm{s}^1$. The two $1\mathrm{s}$ electrons are more tightly bound than the $2\mathrm{s}$ electron, and so the ionization energy refers to the liberation of the latter.

Where I think you might be confused is that we do require all the electrons -- in particular the valence electron -- to be in the ground state configuration for the system as a whole. Consider lithium again. You could have an excited state of lithium such as $1\mathrm{s}^23\mathrm{s}^1$ or $1\mathrm{s}^12\mathrm{s}^14\mathrm{p}^1$. Though still neutral lithium, these configurations are easier to ionize than ground-state lithium. Note, though, that we generally don't index these excited states with a single integer (e.g. $n$), even though we in theory could, because that glosses over the internal structure. Thus the ground state of the system being ionized is simply $1\mathrm{s}^22\mathrm{s}^1$, while the outer $n = 2$ electron is in the lowest shell it can occupy.