Determining ratio of competing fragmentation processes in mass spectrometry
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.