Answer
3.4 eV
Work Step by Step
The third electron in lithium is a 2s electron. It is outside the completed 1s shell. Using the hint, we assume this electron “sees” a nucleus with a net charge of +e. Find the energy of the state using equation 27–15b.
$$E_2=-\frac{-13.6eV}{2^2}=-3.4eV$$
This crude model predicts that it would take 3.4 eV of energy to remove the 2s electron from lithium.
The measured ionization energy is larger, so the 1s electrons do not actually shield the nucleus completely. In fact, the third electron “sees” a nucleus with a net charge greater than +e, and is held more tightly than our crude model assumes.
