Electronic Water Dissocation Via Radiolysis

[ScooterGuy]

Hi, all.

Ok, sort of a weird question, and I'm not even sure if I'm asking it correctly, but here goes:

First, some background:
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According to:
Hydrogen Bonding and Orbital Models
In ambient atmosphere the O—O in the water dimer is 2.985 angstrom (calculated by JMW); the short O—H bond is 0.948 angstrom and the long bond is 2.037 angstrom.

That's .2985 nm, .0948 nm and .2037 nm.

2.037 angstrom corresponds to a frequency of 1.4717e+18 Hz or 1.4717 ExaHertz. X-ray range.

2.985 angstrom corresponds to a frequency of 1.0043e+18 Hz or 1.0043 ExaHertz. X-ray range.

0.948 angstrom corresponds to a frequency of 3.1624e+18 Hz or 3.1624 ExaHertz. X-ray range.

Water Radiolysis - Dissociating Water with Radio Waves
"Guenther and Holzapfel irradiated water with X-rays in contact with a large free volume in a vacuum system and found large continuing yields of hydrogen gas."

According to:
https://en.wikipedia.org/wiki/Electr...ption_by_water
Water has its highest absorption at 65 to 70 nm, which is 4.2827e+15 to 4.4087e+15 Hz. That's ultraviolet range. But that'd just cause the water to heat up, I think.

So, if we hit the water with:
1.0043 ExaHertz
1.4717 ExaHertz
3.1624 ExaHertz
that should excite all the points of vibration in the molecules (and between the molecules) at their resonant frequencies, forcing the molecules to dissociate.
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Notwithstanding that it'd be difficult to electronically generate that high a frequency with a tank circuit, I think I've hit upon a solution... use a lithium niobate piezoelectric crystal to eject high-eV electrons.

If we eject a high-eV electron directed toward lead shielding, when that electron impinges upon that lead shielding, it'll give off a photon in the X-ray range, with the frequency of that X-ray photon dependent upon the kinetic energy of the originating electron. The kinetic energy of the originating electron is dependent upon the voltage and frequency used to emit it.

So... how do I calculate the frequency of that resultant X-ray radiation, given only the voltage and frequency used to emit the electron toward the lead shielding, such that I can adjust the input voltage and frequency to the lithium niobate PZ (to adjust the kinetic energy of the emitted electrons), and thus arrive at the three frequencies above?

 

[mfb]

2.037 angstrom corresponds to a frequency of 1.4717e+18 Hz

Then you have a wavelength corresponding to the atomic distance. So what? The energy of the photons at that frequency is significantly above the scale of the electron energies.

Those frequencies do not have a special meaning for water.