Scintillator converts xray photons into visible photons

[IntuitioN]

I believe that a scintillator converts xray photons into visible photons through the process of exciting electrons, then allowing it to fall to ground state.

My q is: shouldn't the energy of the photon released be the same as the photon energy that is used to excite the electron?

Just on that matter, does a photon energy HAVE to correspond exactly to the different electron energy shells in an atom for the photon to be absorbed?(eg if n=0 is -26eV and n=1 is -14eV then what happens if a 13eV photon comes along. Would it be absorbed?)

 

[ZapperZ]

I believe that a scintillator converts xray photons into visible photons through the process of exciting electrons, then allowing it to fall to ground state.

No. A scintillator is a material that produces light when a charged particle passes through. Such material typically has a high index of refraction.

A device that detects light is a photomultiplier/photodetector.

Zz.

 

[jtbell]

We do use scintillators as photon detectors (X-rays and gamma-rays), at least in undergraduate labs, but they don't detect the photons directly. An incoming photon interacts with the crystal via the photoelectric effect, which produces an electron that receives almost all the photon energy; or it scatters off an electron in the crystal (Compton scattering), and the electron recoils, taking up part of the original photon energy; or (if it has enough energy) it produces an electron-positron pair via pair-production.

In each of these cases, the produced/recoiling electron then interacts with the scintillator material, giving up its kinetic energy and producing a bunch of visible photons. So what we're actually detecting is the electron's kinetic energy, not the photon energy. If the electron was produced via the photoelectric effect, its kinetic energy is practically equal to the incoming photon energy, because the work function (a few eV) is negligible compared to a few tens or hundreds of keV. If the electron was produced via Compton scattering, it has only a fraction of the incoming photon energy, which depends on the scattering angle.

 

[ZapperZ]

We do use scintillators as photon detectors (X-rays and gamma-rays), at least in undergraduate labs, but they don't detect the photons directly. An incoming photon interacts with the crystal via the photoelectric effect, which produces an electron that receives almost all the photon energy; or it scatters off an electron in the crystal (Compton scattering), and the electron recoils, taking up part of the original photon energy; or (if it has enough energy) it produces an electron-positron pair via pair-production.

In each of these cases, the produced/recoiling electron then interacts with the scintillator material, giving up its kinetic energy and producing a bunch of visible photons. So what we're actually detecting is the electron's kinetic energy, not the photon energy. If the electron was produced via the photoelectric effect, its kinetic energy is practically equal to the incoming photon energy, because the work function (a few eV) is negligible compared to a few tens or hundreds of keV. If the electron was produced via Compton scattering, it has only a fraction of the incoming photon energy, which depends on the scattering angle.

Gosh. This sounds like a roundabout way of detecting such a thing. Is the primary reason for doing it this way is because an undergraduate lab typically lacks an electron spectrum analyzer? Once the electron interacts with the scintillator material, you almost always lose a lot of info on the energy of the original photon. All you can do is say "Ah, there was a photon passing by".

But if that's all you want to do, why not directly use a photodetector?

<scratching head>

Zz.

 

[IntuitioN]

We do use scintillators as photon detectors (X-rays and gamma-rays), at least in undergraduate labs, but they don't detect the photons directly. An incoming photon interacts with the crystal via the photoelectric effect, which produces an electron that receives almost all the photon energy; or it scatters off an electron in the crystal (Compton scattering), and the electron recoils, taking up part of the original photon energy; or (if it has enough energy) it produces an electron-positron pair via pair-production.

In each of these cases, the produced/recoiling electron then interacts with the scintillator material, giving up its kinetic energy and producing a bunch of visible photons. So what we're actually detecting is the electron's kinetic energy, not the photon energy. If the electron was produced via the photoelectric effect, its kinetic energy is practically equal to the incoming photon energy, because the work function (a few eV) is negligible compared to a few tens or hundreds of keV. If the electron was produced via Compton scattering, it has only a fraction of the incoming photon energy, which depends on the scattering angle.

thank you so much. my lecturer didn't really explain how scintillators work (more on the lines of xray photon in, visible photon out in a nutshell) and this clarified that perfectly.

by any chance, do you know the answers to my 2nd question, the energy shells questions?

 

[jtbell]

Gosh. This sounds like a roundabout way of detecting such a thing. Is the primary reason for doing it this way is because an undergraduate lab typically lacks an electron spectrum analyzer?

This is for a couple of sophomore-level introductory modern physics labs in which photon spectroscopy as a practical tool isn't the point. They mainly give students an introduction to pulse-height analysis. First we use a single-channel analyzer, then a multi-channel analyzer. They also look at some characteristics of the Compton-scattered electrons and explain them using the Compton-scattering equation that they study in class about that time.

If we were using photon spectroscopy as a serious tool for studying something else, we'd definitely prefer to use something better than NaI(Tl) crystals!

 

[vanesch]

If we were using photon spectroscopy as a serious tool for studying something else, we'd definitely prefer to use something better than NaI(Tl) crystals!

Germanium xtals come to mind, no ?

cheers,
Patrick.

 

[vanesch]

Just on that matter, does a photon energy HAVE to correspond exactly to the different electron energy shells in an atom for the photon to be absorbed?(eg if n=0 is -26eV and n=1 is -14eV then what happens if a 13eV photon comes along. Would it be absorbed?)

In general, as already said before here, scintillators (and almost any particle detector) detect the kinetic energy of charged particles. These charged particles can be the incident radiation, but they can also be generated in a secondary way in the detector. This can be in the scintillator itself, or in another material.
For instance, we do some research on scintillating gas detectors for thermal neutrons. Here the "scintillator" is CF4 gas, while the "converter" is He-3 gas. Both are of course mixed together: upon (nuclear) interaction of the thermal neutron with He-3, it produces a triton (180 KeV) and a proton (570 KeV), which then ionize and excite the gas (essentially the CF4), because they slowly lose kinetic energy in all these small ionisation processes.
This excited gas then emits visible photons, which is the scintillation effect looked after.
The same happens in a solid scintillator and photon detection: there is a process which converts the incoming photons into charged particles with kinetic energy. The details of that process depend on the material and the incoming radiation. So at least the material at hand must absorb such radiation. If, as you describe, the incoming photon energy is below the lowest energy jump that is available, your detector is transparant to the incoming radiation and will not see it. But usually for such low energy photons you don't use a scintillator as a detector, but directly a photomultiplier or an avalanche photodiode.

cheers,
Patrick.