Blog: What is the ionization energy required to raise a hydrogen atom from the n=2 state to the n = infinity state, using the Bohr model?

The ionization energy is simply the difference in energy between two states, in this case the $\displaystyle{n}={2}$ state and the $\displaystyle{n}=\in{f{\in}}{i}{t}{y}$ state.

In the Bohr model (which is a simple but very good approximation for the hydrogen atom), all electron states are modeled as having angular momentum that is some whole number $\displaystyle{n}$ times Planck's reduced constant $\displaystyle\frac{{h}}{{{2}\pi}}$ :

$\displaystyle{L}={n}\frac{{h}}{{{2}\pi}}$

This results in energy levels defined by n, such that the energy of each is inversely proportional to $\displaystyle{{n}}^{{2}}$ , with a constant derived from more fundamental constants (but we can just take it as given at -13.6 eV):

$E = - {13.6 eV}/{n^2}$

Then, the n = 2 state has this energy:

$E = - {13.6 ev}/{2^2} = - 3.4 eV$

And there is in fact an n = infinity state, the limit at which the electron's energy reaches zero:

$E = - {13.6 eV}/{infty^2} = 0$

The ionization energy is the difference between these two, which is 3.4 eV. But we are asked for the energy in kJ/mol, so we need to do a unit conversion. There are $1.602*10^{-22}$ kilojoules per electron-volt, and $\displaystyle{6.02}\cdot{{10}}^{{23}}$ electrons per mole.