why do we see characteristic colors of alkali metals in oxidizing flame?

Question

I am a high school student and I am a little confused in a topic, My confusion is that:

We know alkali metals and their salts imparts characteristic color in oxidizing flame, for example: Lithium gives crimson red color, sodium gives yellow , potassium gives violet and Caesium gives blue, My school textbook says that this is because the heat from the flame excites the outermost electron to a higher energy level and when it returns back it will radiate in visible range as I mentioned above, but if this is the correct reason then we know the outermost electron of lithium is the most attracted among all alkali metals .So, it means that it would take high energy to excite its outer electron so a short wavelength of light is needed i.e probably in violet region, so when it returns back it should emit the same wavelength. Then why do we see these weird pattern of colors? I think this is not the complete reason for why do we see those colors? please explain this in brief only so that I can easily understand.

Answer 1

I don't think it is a good idea to connect ionization energy with the color of flame emission, especially in a Bunsen burner.

Ionization energy means that you have separated the electron out of the nucleus attractive field. However, electronic transitions, corresponding to visible wavelengths, take place while the electron is still bound to the nucleus. Thus the flame color of alkali metals might correlated with the IP but not the causation.

Let us look at the resonance wavelengths of emission of Li, Na, K, Rb, and Cs. Resonance wavelengths means the transition from the $\mathrm{1s}$ to $\mathrm{2p}$ orbital or $n\mathrm s$ to $(n+1)\mathrm p$ orbital in the case of alkali metals.

$$ \begin{array}{r|ccccc} \hline \text{Element} & \ce{Li} & \ce{Na} & \ce{K} & \ce{Rb} & \ce{Cs} \\ \hline \lambda/\pu{nm} & 670 & 589 & 769 & 794 & 894 \\ \hline \end{array} $$

You cannot even see these wavelengths with your eyes after 700 nm which are in the deep deep red. Some people can see the red lines of K 769 nm. However, you can see the energy gap is indeed decreasing as we go down the group. The ionization energy also goes down the group. However, you can see, the colors which you see in the Bunsen burner have nothing to do with ionization energy.

The wavelengths have been extracted from a old book titled Graphische Darstellung der Spektren von Atomen und Ionen mit ein, zwei und drei Valenzelektronen by Dr. W. Grotrian, 1928. I think NIST atomic spectra database will be better today but I am not sure if they list resonance lines clearly. This book has graphically labelled transitions for almost all elements known in 1928.

Answer 2

The relationship between ionisation energy and emission spectra is complex

You make the assumption (implicitly) that the colour you see in emission spectra is from the single complete ionisation of an electron from the highest energy electron orbital in lithium. But the colour you see is not that simple for two reasons.

One is that you only see visible light (roughly the octave from about 400 nm to about 750 nm of wavelength). Many emission lines occur outside that range in the infra red or ultra violet, so the colour you see will rarely be a product of the complete range of possible emission lines.

Another is that there are far more transitions that the one that is related to complete ionisation of the outermost electron. All atoms have many possible electron orbitals higher than the lowest occupied one. Transitions between most pairs of energy levels are possible (there are complicated but irrelevant reasons why some can't happen related to quantum mechanics). And flames often have enough energy to put some electrons in many of those orbitals leading to a large number of possible emission lines when electrons can fall from any energy level to any other energy level. (see this introduction to the concept that explains the complexity for the simplest atom, hydrogen, which shows the calculated orbital energies and the possible transitions that result).

So colours result from filtering of the possible emissions to visible light only and the complexity of the transitions that are possible. So there is no meaningful relationship between colour and the ionisation energy.