Chemiluminescence is the reaction of two chemicals to produce a high energy intermediate that decomposes to release energy in the form of light. My question is why is the energy not released as heat? There is nothing that I have found online that says why it can't be released as heat or why it is specifically released as light. The reason I am asking is if there is a way to completely emit energy in a singular form and if we could control the proportions with which one reaction releases heat we could make certain reactions universally more efficient.
Chemiluminescence – Why Energy Is Not Released as Heat
physical-chemistrythermodynamicsvisible-light
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I believe your puzzlement comes from confusing two frameworks: the quantum mechanical (photons) and the classical mechanics one, waves.
When one is calculating in terms of classical electromagnetic waves there are classical considerations : refraction, absorption, reflection with their corresponding constants .
When one is zooming in the microcosm and talking of photons, a wave is composed of zillions of photons which go through, each at the velocity of light.
The bulk of the target material is in effect the electric and magnetic fields holding the atoms together to form it: the nuclei are tiny targets and the electrons are small zooming targets. The probability of a single photon to scatter on a nucleus or an electron is miniscule. It interacts/scatters out of its optical ray path with the electric/magnetic fields that are holding the glass or crystal together. The scattering angles are very small in transparent materials thus preserving the optical path, enormous in opaque ones . It is those fields that one has to worry about, not the individual atoms and their excitations.
The photons scatter mostly elastically with the fields holding the solids together with tiny or high cross sections depending on the frequency of light and spacing of the materials. In crystals and glasses the optical frequencies have small probability of interaction.
x rays find most materials transparent because the photons' energy is much larger than the energies available by the fields holding the crystals together, and the scattering angles with the fields are very small, except when they hit the atoms, and then we get x ray crystallography.
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Here is the sequence as I see it:
A classical electromagnetic wave is made up of photons in phase according to the wave description.
There is an enormous number of photons in the wave per second making it up. Here is a useful article which explains how a classical wave is built up from a quantum substrate.
Each photon does not change the atomic or crystalline energy levels going through a transparent material in the quantum mechanical way by emitting a softer photon. It scatters quantum mechanically elastically, through the medium, changing the direction infinitesimally so that it keeps the quantum mechanical phase with its companions and displays transparency. Since a medium has a composite collective electric and magnetic field it is not a simple "electron photon going to electron photon" QED diagram. In the case of a crystal one could have a model of "photon crystal photon crystal" scattering amplitude for example.
The higher the sequential probability of scattering going through a medium the larger the final deflection through it will be, and the higher the over all probability of losing the phase with its companions in the wave.( the thicker the glass the less transparency and image coherence).
The transparency of the medium depends on the ordering of the atoms and molecules composing it so that it allows to keep the coherence between individual photons of the beam. The lower the density the better chance to keep the transparency, viz water and air.
hope this helps conceptually.
"So, why can't we do the reverse?" Because of the Second Law of Thermodynamics! Very, very roughly, heat is 'thinly spread' energy and won't spontaneously organize itself into the 'concentrated' energy that we want (in the same way that a drop of ink released into a tank of water won't spontaneously gather itself up into a drop again). Advice: if you're really interested, read up about thermodynamics!
Best Answer
The mechanism of light emission in chemiluminescence is identical to fluorescence/phosphorescence. The high energy intermediates in chemiluminescence are in their electronically excited states and relax back to the ground state by emitting a photon. The excited state is just reached differently (via a chemical reaction and not an excitation by light). Thus, the question of why the energy is released in the form of light and not heat is similar to the question of why some materials fluoresce/phosphoresce.
We will now focus on molecules, which is the important part for chemiluminescence. Whether an excited molecule emits light or gives off heat has to do with its specific electronic structure. There are certain geometries in a molecule (if you deform the molecules just right) where the electronic ground state and excited state come very close to each other and the coupling between them is very strong. At these geometries, the molecules can relax to the ground state without emitting a photon. If the molecule can reach these points by wiggling around in the excited state (at a normal temperature) it will give off energy as heat and will not luminesce. This is called internal conversion/intersystem crossing and happens in most molecules. If a molecule can't reach these points it is trapped in the excited state and will give off the energy slowly as light. Only then can we see luminescence.