I saw photo-ignition of hydrogen-chlorine mixture using UV light on YouTube.

Chlorine molecule bond dissociation energy is $\pu{2.51 eV}$ (bluish colour). Bromine molecule bond dissociation energy is $\pu{2 eV}$ (orange colour).

Can I use any photon which has energies greater than $\pu{2 eV}$ to split bromine molecule, like green or blue or UV light? Or should I use only photons which has energies around $\pu{2 eV}$?

My concern is about absorption spectrum of gases. And this question is not just for bromine gas, but for any diatomic molecule like hydrogen or Chlorine.


1 Answer 1


If the photon's $E=\hbar\omega=h\nu=\frac{hc}{\lambda} $ energy is greater than the energy of the bond which you want to break up it will break it up. It's always best to use higher energy photons but not too high because it may break up bonds which were unwanted to break up in your reaction . So the best for you is to use blue or higher energy UV light for your experiment.You need to deal with the rate of the reformation of bromine molecules, there will be an equlibrium between the production and decomposition of bromine molecules depending on the intensity, not the energy of the light. If the photons has higher energies than a given bond they will always break it up, this is why gamma radiation is so dangerous. The absorbtion spectrum doesn't have a lot connections to this phenomena. You can use the Maxwell-Boltzmann distribution $$f_{v}(v_{i})=\sqrt{\frac{m}{2\pi k_{B}T}}exp\left[-\frac{mv_{i}^{2}}{2k_{B}T}\right]$$ to determine the Doppler-effect's impact on the absorption spectrum depending on the temperature and to construct an absorbtion function explicitly. The doppler effect will affect the percieved frequency and energy of the light by gas particles so if you want to be as accurate as possible, you need to calculate it.The Maxwell-Boltzmann distribution is only suitable in low temperatures (below several thousand Kelvins).


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