I'm not sure if this is more of a chemistry-type question, but my question focuses more on light spectroscopy than the chemical elements;

Why do some metals appear coloured in flame tests while they don't appear coloured at all at room temperature when in the presence of light? Is it just that flame provides far more energy than light could?

My understanding of colouring was that when a substance is exposed to wavelengths across the visible spectrum, it absorbs specific wavelengths for excitation and thus appears as the colours that are allowed to reflect/transmit, and the electrons de-excite via irradiative processes.

All the textbooks explain the flame tests as the emission of a specific wavelength during de-excitation, but as far as I was aware that results in a 'fluorescence' phenomena and not the colour that we observe.

Or is it just that some of the metal atoms vaporise and thus behave as gases, with emission spectrums being radiated in random directions and the absorption spectrum being negligible?

My thoughts right now;

  • the flame provides far more energy than light, which is why it doesn't appear coloured at room temperature
  • the specific energy absorbed doesn't change the spectrum incident in a non-negligible way, but the emission photons are much more intense relatively speaking, and thus the flame would appear as the emission colour

Any help would be appreciated, let me know if this is too chemistry-esque for this forum. Thanks!

  • $\begingroup$ I would emphasise that you're asking about the state of the metal as it interacts with the heated gas, otherwise you risk being closed as a duplicate of Possible colors of fire? $\endgroup$ Commented Jun 10, 2022 at 7:52
  • $\begingroup$ my question is more about why the emission is so much more significant than the absorption transmission, although you do provide a good answer there in general for flame colouring, but I do already know what you're saying here. I'm still going to save your answer to that question though because it is a good succinct response :) $\endgroup$
    – tgsweat
    Commented Jun 10, 2022 at 7:57
  • $\begingroup$ You have things muddled and need to edit so there is one clear question. With regard to absorbance, there is a world of difference between the shiny silvery surface of a piece of metallic sodium, for example, and excited sodium atoms in a flame. In my answer here, I show a simple alcohol lamp flame with excited sodium atoms making the flame yellow by emitting yellow light. The de-excited atoms can subsequently absorb that yellow light, from the hollow cathode lamp shown, and that can be used in atomic absorption spectroscopy (AAS). $\endgroup$
    – Ed V
    Commented Jun 10, 2022 at 11:48
  • $\begingroup$ Thanks for your comment; I agree its a bit muddled, my bad. Is it possible to close the question somehow? I thought it through myself and reached an acceptable answer for me, should I post that as a separate answer or just edit my post? $\endgroup$
    – tgsweat
    Commented Jun 10, 2022 at 12:26
  • $\begingroup$ Good! Posting answers to your own questions is fine and perfectly in line with stack exchange policies. Might as well give it a shot and see how people vote! $\endgroup$
    – Ed V
    Commented Jun 10, 2022 at 13:20

2 Answers 2


Or is it just that some of the metal atoms vaporise and thus behave as gases, with emission spectrums being radiated in random directions and the absorption spectrum being negligible?

Yes. In a chunk of metal the energy states of the electrons are very different from in individual metal atoms. In a metal the atoms are all close together and they interact to form energy bands. The electrons in the uppermost band (the conduction band) behave in a similar way to free electrons and as a result they reflect all wavelengths of light and don't have any particular colour.

In a flame there is enough energy to dislodge metal atoms, and those individual atoms then undergo electronic transitions that emit light. For more on this see Possible colors of fire?

  • $\begingroup$ I've also mentally answered it in this way which makes more accurate sense to me. To excite, the metal atoms collide with high-energy gas particles. At a high enough temperature, the gas particles (by Maxwell-Boltzmann dist.) will contain a number of particles with high enough kinetic energy to exchange the specific electron transition energy required via collision with the metal, i.e. the exchange of a 'virtual photon' which is then de-excited as a 'real photon' and hence the colour observed. I probably should've been clearer about this phenomena in my response. $\endgroup$
    – tgsweat
    Commented Jun 11, 2022 at 6:49

When a substance is put into a flame, the substance heats up. Some sources online say that 'electrons are excited by heat energy,' but this wasn't acceptable enough to me considering that by the quantum model of energy, electrons won't be excited gradually as excitation occurs as a one-quantum-one-electron interaction.

So, my conclusion (that satisfies me for the purpose of this question) is that in a heated environment (say ~$1000°$C for the outer regions of a bunsen flame) atoms/electrons will have a much larger average kinetic energy - of a vibrational/rotational degree - and thus the quantised energy to excite will be lower (as much more atoms have required 'activation energy' from collision theory) and can then excite by collisions with fast-moving gaseous particles (perhaps electron deflection in these gases?), which then de-excite via radiation of visible light photons (of course, depending on the electron transition difference).

This explanation would explain the immediate colouring shown as collision and then de-excitation would occur in a negligible time period at such a high kinetic energy. Also, the 'glow' that comes along with the colour would be the result of black-body radiation from the oscillation of vibrating electrons.

As to why it doesn't excite at room temperature and visible light, the circumstances required (high KE and particle collisions) aren't there, so it remains the shiny lattice as it would at room temp.

Anything I've gotten wrong here, please let me know!


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