# Determining ratio of competing fragmentation processes in mass spectrometry

In a photoionization experiment, three photons of wavelength 266 nm were absorbed by a neutral molecule of 6-methyl-2-heptanone. If the kinetic energy of the electron removed during ionization is 1.4 eV, what is the ratio of the simple cleavage to the rearrangement produced fragment ions? The following plot describes the fragmentation of the molecular ion of 6-methyl-2-heptanone. The IE of this compound is 9.10 eV. My intuition would just be to use the value of 9.10 eV provided in order to look at the rates of cleavage to rearrangement in order to determine ratios. I'm not sure how the 1.4 eV and the 266 nm photons come into play.

## 1 Answer

Your three photons represent the total energy added to the system:

$$E_\mathrm{photon} = 3 \cdot \frac{hc}{\lambda} = \pu{13.98 eV}$$

That, less the sum of ionization energy and kinetic energy of the removed electron, leaves the change of the internal energy of the cation:

$$E_\mathrm{int} = \pu{13.98 eV} - \pu{9.10 eV} - \pu{1.4 eV} = \pu{3.48 eV}$$

Finding the value of $$\log k$$ for both curves at this energy will give the relative rates of simple cleavage vs rearrangement.