How does H's ionization energy relate to its transition energy (Bohr's Model)? I am currently reviewing some material about orbital energy levels. In my review book there is a short snipet that reads:


  The IE of $\ce{H}$ from its ground state ($n=1$) is $1312\ \mathrm{kJ/mol}$. Because of the squaring of the principle energy level, the ionization of an $\ce{e-}$ in $\ce{H}$ from the $n=2$ level is $1/4$ of that value ($328\ \mathrm{kJ/mol}$). The transition energy from the $n=1$ level to $n=2$ level is the difference between the two values, $984\ \mathrm{kJ/mol}$.


I am confused about how exactly an atom's ionization energy relate to its transition energy. If IE is the energy required to pluck off the first e- from an atom and TE is the energy required to change between different orbitals, does this mean that an atom with greater IE will need to absorb a greater amount of energy to jump orbitals? 

Lastly, is there a equation that directly ties in both ionization energy with transition energy?
Ionization energy of some arbitrary atom is not related whatsoever to its transition energy. Also, there are many different transitions (even in a hydrogen atom), each with its own specific energy. It is just that this particular transition of this particular atom happens to have the energy of 3/4 its ionization energy, one of the few things the Bohr model managed to get right. You can't generalize that to other atoms and other transitions.