How does H's ionization energy relate to its transition energy (Bohr's Model)?

Question

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?

Answer 1

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.

Answer 2

The visible transitions in the $\ce{H}$ atom were observed experimentally to follow the equation

$$\overline{\nu} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$

where $R_\ce{H}$ is the Rydberg constant and $n_1$ and $n_2$ are integer constants, $1, 2, 3, \ldots, \infty$. The Bohr model of the atom produces the same formulae and a way to calculate the Rydberg constant; in wavenumber units this is $109677\,\mathrm{cm^{-1}}$.

The ionisation energy of the $\ce{H}$ atom occurs when the electron is in its $n_1 = 1$ level and has just enough energy to be removed removed from the atom, this takes it to $n_2 = \infty$ and this energy is equal to the Rydberg. This is also $13.6$ electron volts which is often quoted. If the electron is in another energy level you can calculate how much energy is now needed to ionise it. Have a look in a phys chem text book (such as McQuarrie & Simon or Atkins) for a proper description of this.

Answer 3

How exactly an atom's ionization energy relate to its transition energy?

I think it is best to clear up what the terminology of the question is so that we can address the correct answer. The ionization energy is the direct measure of the electric interactions between the positively charged nucleus center of an atom, and the negatively charged electrons around it. Ionization energy is like that invisible force you can feel between two magnets close together. If you were to measure how much effort is required to split the magnet apart till they are effectively not interacting, that would be like measuring the ionization energy and an electron from an atom. One way to supply the energy to pull off (knock-out) an electron, can be by shining light (in the case of the first ionization energy for a hydrogen atom, $1312\ \mathrm{kJ\ mol^{-1}}$, where $\mathrm{kJ\ mol^{-1}}$ is amount of energy required = $15.60\ \mathrm{eV}$ (another unit of energy that express the amount of electric attraction the negative electron feels toward positive nucleus), and in terms of focused energy carried by light (electromagnetic waves, generated by moving charges or current) that is $= 91.18\ \mathrm{nm}$ electromagnetic waves, which we call X-ray light.)

What is a transition energy?

The intuitive way to think about a transition state is to try and recall a time when you were in danger or had to make snap decision that brief instant before hell breaks lose is like a transition state. Something you have to build up eneough energy to get over. In the Bohr model, the transition state is when an electron moves from one orbital to another. The HUGE thing about this model that made it work is that Bohr assumed the energy between transition states in atoms (i.e the differences between the energies measured for each electron in an atom and the next) is quantized, meaning that it take it can be certain values only. Like having a light switch with 4 positions. It can have different levels but switches can just hover in between, either one or off .The transition energy is the energy required to move electrons between transition state. It is like when water is being heated. You are delivering energy into it by heating and nothing seems to happen until it reaches $100\mathrm{ºC}$ and then it boils.

Bohr related the model for the hydrogen atom and thus approximation for inonization energies for electron in hydrogen energy levels is give by.

Bohr model is given by $[ \Delta E = \frac{m_ee^4}{8\varepsilon_0^2h^2} \lgroup\frac{1}{n_2} - \frac{1}{n_1} \rgroup$, where $n_i = 1,2,3,..]$

Where $n_1$ and $n_2$ are the quantum principle numbers between each quantized state. ie. $n_1 = 1$, means level 1, $n_2 = 2$, level 2 etc..

Even thought we know Bohrs model is incorrect, the quantization of energy as boundaries for electrons is the basis of all modern models. It is not atoms that absorb energy in the case of ionization energy,

Misconception is that atoms absorb energy when its actually the electrons of nucleus particles that do. In the case of ionization energy, it its the electrons themselves that are being energized to the point of boiling off. Thus the valence electrons, which the ones furthest from the nucleus with lowest ionization energies, have lower ionization energies than core electrons (those closer to nucleus)

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McQuarrie DA & Simon JD (2015) Physical Chemistry: A molecular approach . Viva Books