Scintillation photon energy conversion?

[girts]

Hello,
I read that in a scintillator before the photomultiplier tube is a crystal or sometimes gas or even sometimes a plastic, in other words a material which exerts luminosity under ionizing radiation. Say in the doped crystal for example, a high energy photon of say several Mev hits the crystal and the crystal then responds by emitting photons but in the lower frequency visible spectrum, here's what I don't understand, clearly photons have certain energy levels per photon, then where does the extra energy go if the resultant photons from the crystal are of lower frequency hence lower energy per photon?
I take from the famous photoelectric effect that light was firstly understood as not just a wave but also a discrete particle hence the resultant emitted electron was of corresponding energy to the incoming photon, but here in the crystal the outcoming photon is of lower frequency/energy than the incoming one or the one that struck the crystal in the first place, so my question is where is the missing energy?
My own guess maybe the incoming single high energy photon results in the scintillation of outgoing multiple lower energy photons , in other words the crystal works similar to a transformer?

 

[mfb]

Photons don't directly produce photons. They produce electron/positron pairs (at high energy) or transfer their energy to electrons in the material (at lower energy). These electrons and positrons then transfer some of their energy to electrons in the material (a few eV per electron), and if they fall back they can emit some light.
A large fraction of the overall energy heats the crystal, but there is also some light emitted. A single MeV photon leads to many photons in the visible range.

 

[girts]

Oh right, I did not thought that one through. Surely photons themselves don't exist in the crystal as do protons, neutrons and electrons so the incoming ones first have to interact with the particles making up the crystal and then through those particles secondary photons of lower wavelength (visible spectrum) are produced right?
So given a situation where the crystal receives both high energy and medium/low energy photons the high energy ones produce electron positron pairs and the lower energy photons excite the electrons in the material directly to a higher energy state and while falling back a photon or a couple are emitted?
in the electron positron pair production in the crystal they must annihilate quite fast and also produce a bunch of photons correct?

So in terms of reactions how would this look like because I suppose a high energy photon creates the electron/positron pair which then turns into the outgoing photons but since their energy is much lower some of the energy must have gone somewhere else , where does it go?
You said some of the energy heats the crystal, by what means this happens , or should I say by what interaction, directly from the incoming photons or by some secondary means due to the incoming photons?Oh another question, would it be fair to say that the crystal somewhat (not fully) transforms a few high energy photons into more lower energy ones, for example if there are 3 high energy incoming ones one would get out say 9 or more visible spectrum ones?
By this question I wonder are there any physical means or process in which the opposite happens, you have multiple lower frequency photons entering and fewer high frequency/energy photons exiting? Essentially a sort of a transformer we could say?

 

[mfb]

Oh right, I did not thought that one through. Surely photons themselves don't exist in the crystal as do protons, neutrons and electrons so the incoming ones first have to interact with the particles making up the crystal and then through those particles secondary photons of lower wavelength (visible spectrum) are produced right?
So given a situation where the crystal receives both high energy and medium/low energy photons the high energy ones produce electron positron pairs and the lower energy photons excite the electrons in the material directly to a higher energy state and while falling back a photon or a couple are emitted?

Right.

in the electron positron pair production in the crystal they must annihilate quite fast and also produce a bunch of photons correct?

Both particles fly their own way and ionize the material while they go through the material. They can also produce bremsstrahlung, and if this bremsstrahlung has enough energy it can produce more electron/positron pairs and so on - if the original photon has sufficient energy you get an electromagnetic shower.
Eventually all positrons are stopped and annihilate, sure.

but since their energy is much lower some of the energy must have gone somewhere else , where does it go?

A single high-energetic original particle leads to many low-energetic particles in the scintillator.
Some energy goes to photons, some energy goes to heat. Where "heat" in crystals means lattice vibrations (phonons) and maybe a few excited electrons. Most of the time the intermediate step is an electron with just a few eV of energy.

for example if there are 3 high energy incoming ones one would get out say 9 or more visible spectrum ones?

More. Per MeV of energy deposition, typical scintillator materials produce 100 (PbWO₄) to 40,000 (NaI) photons.

By this question I wonder are there any physical means or process in which the opposite happens, you have multiple lower frequency photons entering and fewer high frequency/energy photons exiting? Essentially a sort of a transformer we could say?

With laser light and nonlinear materials, yes. Photon upconversion. But that is off-topic here.