You are thinking in terms of atoms and molecules and you are mainly talking of solid state matter .
Solid state is another quantum mechanical phase, it has lattice structure with much smaller energies than atomic and molecular transition structures. Lattices have vibrational levels which are mainly responsible for the black body radiation solids emit, infrared is also photons.
A rule of thumb with radiation impinging on solids is that if the wavelength is smaller than the lattice dimensions the photons can penetrate easily the lattice, interacting only with direct scatters hence the higher penetration of X rays and gamma rays. Here is an article that discusses the penetration of radiation, X rays and higher.
For glass and optical frequencies there is a good answer here in this site., essentially the structure of the transparent materials is such that the photons pass through without loosing energy in the visible.
For infrared where the wavelengths are large in comparison with lattices or distances between molecules in liquids, the photon can give up its energy in collective excitations at the surface gradually heating up the material.
For ultraviolet, glass, depending on the type, has some absorptive bands, the photon energy transferred at the surface to collective modes or breaking molecular bonds and transformed to heat ( infrared) further in.
So your
Once you reach a critical frequency, however, the photons will begin to be absorbed because they have enough energy to excite the electrons (which is why glass is opaque in ultra-violet).
has small probability to happen until x-ray energies are reached which are the energies of bound electrons, and the link above gives the dependence in a simplified manner.
Small-value inductors and capacitors are possible, but you also need extremely small size else the lumped system approximation your equation relies on becomes invalid. Let's say the wavelength of the light to produce is 700 nm. A distance of half that (350 nm) will cause an inversion (180° phase change). Generally, you want to keep the maximum dimension to $\frac{1}{10}$ the wavelength to consider it a lumped system without having to worry about that further.
Your inductor and capacitor, therefore, have to all fit roughly within a 70 nm dimension. With everything that close to everything else, there will be a lot of parasitic capacitance. Getting a low enough capacitance and still have the deliberate capacitance across the ends of the inductor dominate over all the distributed stray capacitance won't be easy. Even if you could do that, you still have to drive it with a 430 THz signal somehow. Where are you going to get that from?
Best Answer
The radio frequency photons emitted by a Tesla coil are much too low energy to directly excite atoms to emit visible (~2 eV) or UV photons (~6 eV) . A 1 MHz radio photon ($h\nu=4\times 10^{-9}eV$) is also way too low in energy to ionize an atom which requires ~10 eV.
What actually happens is that the near electric field from the Tesla coil accelerates any free electrons in the tube's low pressure gas. These electrons pick up enough energy before they collide with another gas atom that they ionize that gas atom. The ionized electrons are then accelerated and ionize further atoms. When electrons fall back onto the ionized atoms, visible light (as in a neon filled tube) or UV light (from mercury vapor in a fluorescent tube) is emitted. The phosphor coating on the inside of the fluorescent tube is excited by the UV photons or by the accelerated electrons directly.
For some numbers, suppose the peak voltage between the ends of a 1 meter long tesla coil is $10^5$ volts. The near electric field is then $10^5 volts/meter$. The mean free path of an electron in the low pressure gas (pressure=100 um Hg) is ~1 mm. Thus the electron is accelerated to ~100 eV which is more than enough energy to ionize an atom.