The normal technique for flame tests is to dip a clean nichrome or platinum wire into a solution of the relevant salt, and observe the resulting flame colour when inserting the wire into a non-luminous Bunsen flame.

But what is the species undergoing the electron transition that emits the light of characteristic frequency?

One explanation suggests vaporisation and then atomisation of the sample, with the light emission occurring from the excited atom, rather than ion. Another suggests that it’s the ions undergoing the electronic transitions. The Wikipedia article is somewhat vague on the question.

Now, I was doing some flame tests for a student, and in the absence of an available nichrome or platinum wire I simply took a pinch of the solid salts ($\ce{NaCl}$ and $\ce{KCl}$ respectively), and threw it into a non-luminous Bunsen flame.

The characteristic yellow and lilac flame colours were instantly and repeatably observed.

My question is this:

Given that the outer regions of a Bunsen flame are at less than $1000\ \mathrm{^\circ C}$, and that the boiling points of $\ce{NaCl}$ and $\ce{KCl}$ are $1413\ \mathrm{^\circ C}$ and $1420\ \mathrm{^\circ C}$ respectively (according to Wikipedia), it seems unlikely that the instantly generated characteristic colours originate from individual ions or individual atoms.

So what is the species exhibiting the electron transitions? Is it the metal ions still within the crystal lattice?

Edit: I should emphasise that the flame colour was observed in the outer regions of the flame, not the super-hot tip of the inner cone. So $1000\ \mathrm{^\circ C}$, not $1500\ \mathrm{^\circ C}$ by any stretch of the imagination. And I don’t at all buy the notion that the colours were simply the glow of hot solid particles emitting black body radiation. The colours were the characteristic sodium yellow and potassium lilac, instantly recognisable by anyone that’s familiar with simple tests like these.

  • $\begingroup$ Just because something doesn't boil doesn't mean that many atoms or ions from it won't get into the flame. If the emission is intense (like the bright yellow lines in the spectrum of sodium vapour) then you don't need many ions to be in the vapour phase to see it. For example, Fluorescent tubes are driven by emissions from mercury vapour and start working well at room temperature, >300°C below the metal's boiling point. $\endgroup$
    – matt_black
    May 31, 2015 at 10:29

2 Answers 2


There is a very detailed discussion of what cause the emission lines in flames with $\ce{NaCl}$ in The Origin of the Spectra (ironically by P. Darwin Foote et al.) 1922, starting around page 165.

They are saying that for $\ce{NaCl}$ most of the transitions are from atomic $\ce{Na}$, the $\ce{NaCl}$ having vaporized, and Na+ then captured an electron, and a small fraction of the transitions are from the act itself of Na+ capturing the electron.

They calculate that a Na atom emits 2000 D-line quanta per second.

They also calculate the fraction of Na atoms with electrons in the $3p$ state is $\ce{10^{-11}}$ at 1000K, rising to $\ce{10^{-7}}$ at 1500K. (Note: the book says "$2p$", but $2p$ in Fowler notation is $3p$ in modern notation)

Also, they say the vapor pressure of $\ce{NaCl}$ in a Bunsen flame is on the order of $\ce{10^{-6} \ mmHg}$.

So the color is from vaporized $\ce{NaCl}$ (vapor phase $\ce{NaCl}$ being molecular), that has dissociated and formed atomic $\ce{Na}$.

Also in The Absorption of Light by Flames Containing Sodium Phys. Rev. 38, 699 (emphasis added):

The dissociation probably occurs in two or more steps. Possibly the molecules break up into sodium and chlorine ions which then lose their charges becoming uncharged atoms. There may also be one or more intermediate steps in the recombination. But certainly the sodium giving the light which we are studying consists of uncharged atoms, as is evident from its spectrum. Evidence that there are comparatively few charged sodium atoms in the flame is given by a study of the conductivity of flames. Wilson, (p. 85) for example, estimates that only 1.6 percent of the sodium in a fame is ionized even when the concentration is small.

It also seems reasonable to assume that at the edge of the flame there is more complete dissociation of the salt than in the center, and that this is the cause of the difference between the concentration ratios obtained with light from the center and from the edges. A fairly definite proof that there is more dissociation at the edges is given by the appearances of a fame into which copper chloride is being sprayed. Such a fame gives almost no green color except at the edges, where it is a bright green. This is the more noticeable if there is a small amount of sodium impurity in the copper chloride. The center of the flame is then yellow with a distinct border of green.

The greater dissociation at the surface is no doubt in some way due to the greater amount of air there, but whether it is caused by a higher temperature or by a difference in chemical action can not at present be stated.


Bunsen flame is up to 1500 C hot depending on the region.

it seems unlikely that the instantly generated characteristic colours originate from individual ions or individual atoms.

It is, however, more or less generally accepted. More precisely, in some flames the color is derived from emission of excited atoms, while in other it is molecules. alkali metal colors are believed to be derived from atomic emission, while copper bluer, strontium red and barium green are emitted by monochloride molecules (well, barium green and strontium red can be, less efficiently, produced by other molecules, but professional pyrotechnic compositions are usually manufactured with chlorine donors).

The mechanism of excitation, however, is a bit less clear, but it is sometimes believed to be chemical and not thermal in nature.

Solid particles in flame usually produce black body radiation, which may be, depending on temperature, red, orange, yellow and whitish (at extremely high temperatures, well above any that a flame can produce it becomes bluish)

  • $\begingroup$ Thanks for the reply. However, I'm troubled by "more or less generally accepted" and "believed to be". See my edit - I'm throwing a sprinkling of solid salt into the part of the flame that's well below the boiling point. Are free sodium and potassium atoms really credible as the source of the flame colours? $\endgroup$
    – ChrisA
    Jan 15, 2015 at 9:06
  • 2
    $\begingroup$ @ChrisA Yes. Emission lines discoverable through flame test are ridiculously intensive, and, given that there is no much energy released in visible light by non-sooting flames, it is quite easy to color the flame. Non-zero vapor pressure may exist even over solid, say ice slowly evaporates (sic!) in very dry air even set below freezing point. However, since emission lines to be active in flame test must be ridiculously intensive, there is only a few elements active in flame test. $\endgroup$
    – permeakra
    Jan 15, 2015 at 12:22
  • $\begingroup$ @ChrisA As for excitation mechanism, it is plainly too hard to give a definitive proof, since there is a lot of flames. However, visible light quants have energies 1.6+ eV, which corresponds to 15 000 + Kelvin. $\endgroup$
    – permeakra
    Jan 15, 2015 at 12:27

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