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I am a beginner in this topic but I am aware that when flame test (for alkali and alkaline earth metals) is done there is a s-s transition of electrons from the ground to their excited state.

But I don't understand why the electron needs to transit back to their ground state if I consider that the temperature of the oxidising or non luminous flame (that is the energy supplied to the electron) remains constant.

So my question is

Why does the electron release the absorbed energy and transit back to the ground state even though a constant external energy is supplied to it?

In other words, what is the need for the electron to absorb the energy , then reach the excited state and release the energy to transit back to the ground state. Instead why does it not remain in the excited state after absorbing the required amount.

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    $\begingroup$ I'm voting to close this question as off-topic because it belongs on Physics.SE $\endgroup$
    – A.K.
    Oct 22, 2018 at 2:53
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    $\begingroup$ @A.K. I was not sure like I had said I just began reading this topic. But if it fits better in Physics.SE...can the question be migrated? $\endgroup$
    – user 33690
    Oct 22, 2018 at 14:26
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    $\begingroup$ Questions can indeed be migrated, but I disagree that this question belongs on Physics.SE. $\endgroup$
    – a-cyclohexane-molecule
    Oct 22, 2018 at 18:32
  • $\begingroup$ Related but probably not a dupe: chemistry.stackexchange.com/q/31476 $\endgroup$
    – Jan
    Dec 21, 2018 at 5:22

2 Answers 2

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The electron releases its energy because it can. It happens because it has a non-zero probability to do so. If it would never happen it would have zero probability.

Whether the electron just got excited prior to emitting some light does not matter. Particles do not have a memory. Instead such processes simply happen based on whether the requirements are met (e.q. absorbing energy is only possible if energy is available) and on probabilistic rules (quantum mechanics).

Consider some particles in thermal equilibrium: Even the though the temperature does not change on a macroscopic level, the particles still constantly exchange heat with each other. This is called dynamic equilibrium.

The difference to the flame test is, that the particles exchange energy not only by heat, but by light as well.

Exchange of heat is basically a collision of two atoms, so this is a 2-particle process: its rate dependents on the likelihood of those 2 particles "hitting" each other.

Emission of light is a 1-particle process: it can just happen spontaneously. However, absorbing light is a 2-particle process: a photon needs to intercept an atom in order for the atom to absorb its energy.

Heat is the source of energy, and it allows for redistribution between all particles. But for the light, the emission happens much more often than absorption, because 1-particle processes are much more likely to happen than 2-particle processes. Therefore we see the overall emission of light until the heat source is gone.

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    $\begingroup$ This may also go into the direction of Fermi's golden rule, where they realized that using Bohr's atomic model and simple physics could not predict why some emission lines tend to be brighter than others, or in other words why some transitions seem to happen more often than other's. This is also where the transition probability or transition rate is introduced. $\endgroup$
    – Justanotherchemist
    Oct 22, 2018 at 11:03
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Thermal energy is not constant on an atomic level. Heat energy is due to the motion of particles, and is transferred through collisions, or through radiation, which is also quantized. Each collision has a chance of raising electrons to an excited state, where it may linger for a while in a metastable state, or fall back to a lower state quickly. In either case, energy is released on falling back (such as the yellow sodium line).

This might be more easily understood in the case of a glow discharge. Electron collisions ionize atoms, which then revert to a lower state.

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