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.