How an Electron Snaps One Bond in a Molecule ✛

Original: "Strong Evidence for Formation of Hydroxyl Anion via 2-Particle-1-Hole Feshbach Resonances"

· Siddique Ali, Meenakshi Rana, Soumya Ghosh, Narayan Kundu, Aryya Ghosh, Dhananjay Nandi

arXiv:2609.07324v1 · 2026-09-07 · CC BY 4.0 · 1 min

Chemical Physics Atomic and Molecular Clusters Atomic Physics

Scientists have found that an electron with 8.6 eV energy breaks the C–OH bond in an alcohol molecule, acting like a saboteur who knows exactly where to strike.

An electron flies into an alcohol molecule like a saboteur who knows exactly where to strike. It doesn't destroy everything in its path; instead, it finds the weak spot, the bond between carbon and the hydroxyl group, and snaps it, leaving behind a negatively charged hydroxide ion.

The incoming electron carries 8.6 eV of energy, roughly the energy of a photon of ultraviolet light.

Previously, it was unclear how the electron chooses this bond. Researchers thought it simply stuck to the molecule temporarily, like a guest lingering at the doorstep. But calculations showed that the electron penetrates inside and rearranges the molecule's electron cloud. This creates special states that live unusually long. Six such states were found, each acting like a lever that directs energy into breaking precisely the C–OH bond.

The experiment confirmed the calculations: the yield of hydroxide ions shows a broad peak with a maximum at 8.6 eV. This peak looks like a single feature, but it is actually a superposition of dozens of states, of which only six survive. The rest quickly eject the electron back. Now it is clear that the electron saboteur operates through long-lived resonances, which helps assess the radiation resistance of molecules and interpret negative-ion mass spectra.

The resonance at 8.6 eV is so broad that it could be mistaken for a single peak, but it is actually a superposition of dozens of anionic states, of which only six survive.

P_s = \exp\left[-\int_{R_0}^{R_c} \frac{\Gamma(R)}{\hbar v(R)} dR\right]

Ps is the probability that the anion does not lose the electron before reaching the point Rc where autodetachment is closed

v(R) = \sqrt{\frac{2[E_{\text{total}} - V(R)]}{\mu}}

v(R) is the classical velocity of nuclear motion on the potential V(R) at total energy Etotal and reduced mass μ

\tau = \frac{\hbar}{\Gamma}

τ is the resonance lifetime, Γ is its width, and ħ is the reduced Planck constant

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How an Electron Snaps One Bond in a Molecule ✛