The maximum wavelength $\lambda$ capable of breaking a chemical bond can be estimated as $\lambda = hc/E.$ UV-A $(\pu{380 nm})$ can break bonds with the energy up to $\pu{315 kJ mol^-1},$ UV-B $(\pu{320 nm})$ can break $\pu{374 kJ mol^-1}$ bonds, etc.

So, the sunlight (at UV) can break both the $\ce{C-C}$ and the $\ce{C-O}$ bonds. Sunlight has enough energy to break up vitamin C molecules in juice. Supposedly plants have some excision repair. But on the chemistry side, when UV breaks a bond, what's the probability the bond can just reform back?

I'm going to assume there are multiple factors. What are factors where 50% of the time, the bond can just reform, as well as factors where ~90% and ~10% of the time, the broken bond can just reform back?

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    $\begingroup$ You are asking about geminate recombination (cage effect) if in solution, in the species can recombine rapidly depending on solvent viscosity or much more slowly (as 1/sqrt(time)) at longer times, again depending on solvent properties, so there is no answer as such as time for $50$% it depends on species etc. In the gas phase an isolated molecule will have no geminate recombination if it dissociates. $\endgroup$
    – porphyrin
    Commented Aug 3, 2023 at 14:01
  • $\begingroup$ @porphyrin by species, can be water? I generally have 2 questions for this, the 1st can just be vitamin C in water. $\endgroup$ Commented Aug 3, 2023 at 15:53
  • $\begingroup$ If the bond was instantly re-generated, how would you know it was ever split? In other words, what are you trying to find out here? With a significant probability for that process, it would just look like the interaction cross section of that compound with light of a given frequency would be somewhat smaller. Smaller than a theoretical value that is very hard to predict. $\endgroup$
    – Karl
    Commented Aug 3, 2023 at 20:56
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    $\begingroup$ @Karl, we can see bonds breaking under certain circumstances by using femtosecond resolved spectroscopy (pump-probe method) . A bond cannot break in less than half a vibrational period and it is possible to time resolve this. There is a famous example in NaI molecules by A. Zewail where the bond breaks in a step-wise manner, see fig 15 of his Nobel Prize lecture. $\endgroup$
    – porphyrin
    Commented Aug 4, 2023 at 6:44
  • $\begingroup$ @porphyrin what kind of vibrational period? $\endgroup$ Commented Aug 4, 2023 at 12:03

1 Answer 1


UV light creates an excited electronic state, on this electronic state the nuclei are no longer in an equilibrium position and begin to move, i.e. start to move towards breaking the bond. This is the simple semi-classical picture where we view nuclei as classical point particles. More correctly we should speak about the dynamics of the nuclear wavefunction on the excited electronic state potential. But the idea is the same. The initial nuclear wavefunction is no eigenfunction of the excited electronic state and thus "moves".

It is very difficult to describe the dynamics of this process as it involves nuclei and electrons at the same time. A bond can be broken if the wavepacket moves along a vibrational coordinate, elongating the bond. Another aspect that is the exchange of energy between electrons and nuclei. At some point the electronis typically relax via internal conversion to the electronic ground state. Energy is conserved during this step and the decrease of potential energy of the electrons is transferred to the nuclei as kinetic energy.

This excess energy can lead to bond breaking in the groundstate, since the nuclei are now in an excited vibrational state that may lie across a potential barrier that kept the initial bond stable.

This answer is also useful if you want to get a better understanding how bonds are broken.

Lastly we also need to consider exchange of energy with the environment, which allows the molecule to cool down by exchanging vibrational energy with other molecules. This will make the whole process irreversible and stop the dynamics eventually.

The question for a simple probability to reform a bond is thus very difficult and has no simple universal answer. You need to look at all the steps that I have described, which dependent on the molecule in question and the conditions, to work out an answer. You may be able to measure these processes but then again it will depend on the molecule and the conditions and there is no universal answer.

  • $\begingroup$ You mentioned sunlight hitting the nucleus, but yet for covalent bonds like C-C and C-O, the UV must hit the electrons. Perhaps you can divide it into sunlight hitting the electrons vs. hitting the nucleus? Also, another set of category is when sunlight hits ionic bonds. $\endgroup$ Commented Aug 6, 2023 at 14:07
  • $\begingroup$ @NealConroy I don't think I mention "sunlight hitting the nucleus". The very first sentence "UV light creates an excited electronic state" means that electrons are excited. This excitation of the electrons can also cause the nuclei to move. Bond breaking usually means that the nuclei forming the bond also move apart, or at least an increase in bond length. I attempted to describe this process from a physical/theoretical chemists point of view. $\endgroup$
    – Hans Wurst
    Commented Aug 6, 2023 at 19:45
  • $\begingroup$ Okay, you think bond reformation also happens more likely in liquids than in gas, as what porphyrin was suggesting? $\endgroup$ Commented Aug 7, 2023 at 5:39

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