Needing more than a spark test?

Although I am in awe of you guys when it comes to physical chemistry and nuclear physics, I do attempt some calculation myself, if only to learn how. Now I am out of my depth. It is about deciding how long, or thick, the scintillator crystal needs to be.

Consider these --> LYSO crystals available from epic-scintillator.com.
For $22, we get something 7mm long, and at 1mm x 1mm, about enough to fit on the end of a SiPM photodiode. I can see why they are put into arrays. Before we even consider the probability of of a X-ray photon going right through without encountering an atom of Sodium, or Iodine, or Thallium (or hygroscopic water), there is the low probability of the photon arriving hitting the end of the tiny 1mm^2 crystal anyway. The rest, at slightly wrong angle, or just plain off to one side, and the remaining approximately 800mm^2 shower of incoming hits are wasted!
OK - we get it that you can use an array - or a bigger entry area crystal and a perspex light pipe down to the diode.

The $90 offering is 30 x 30 x 0.5mm. The area looks respectable, but is that even useful?

So do I have this right?
Among the atoms it is mostly very empty space. If the X-Ray enters a crystal made too thin, there is some cross-section probability of it not encountering the atoms anyway to make a scintillation before it exits the other side, said crystal being too thin. If the crystal is thick enough, it eventually stops all but the (unluckiest?) X-Ray photons. With thick, you always get a light flash - yes?

So now we consider all the other scintillator opportunities. CsI(Ti), NaI(Tl), whatever.
Even before one thinks about end area, how do we figure the crystal length needed?

Sources of scintillators
I looked at the low cost eBay source --> HERE
4 x 4 x 22mm, but apparently, the seller does not ship to the UK, though I could ask.

You can get an eBay array --> like THIS
At $395, I would not consider it anyway. My point is that it is 22mm thick. Enough to scintillate nearly everything?

Do the X-rays go through aluminum?
Looking at NaI(Tl) on the end of a PMT tube.
I get these from the very informative video by bionerd23
--> a close look INTO a scintillation crystal (radiation detector) / sodium iodide + thallium / NaI(Tl)

Particularly, about how is it sealed up from water ingress?
In one form, the crystal is "deeper down", or cylindrical, because she sticks her finger in there.

NaI(Tl)1.png

In another version, The NaI(Tl) crystal is in a aluminum can.
The window at the other end is glass, put up against the PMT

NaI(Tl)2.png

Turn the can over, and you see the glass end that butts up against the PMT tube

NaI(Tl)3.png

.. and compare to one which has had exposure to air water vapor by leak.

NaI(Tl)4.png

So I guess aluminum end is where X-rays just go right on in regardless, and scintillate away.

I like the PMT vs SiPM because of the area to fit up against all sorts of crystals, big or small.
I think you need a reasonable crystal area, say about 25mm diameter, or a square equivalent.

You don't really need a PMT if you can pipe the light onto a SiPM avalanche photo-diode, but the thing that needs to be large enough to justify setting up a whole ring of Am241 pellets is the scintillator. Also long enough, so as not to let them X-ray photons get away.

I think there are risks in acquiring the hygroscopic types, unless they are very low cost, and with a reasonable assurance they are either new, or "new old stock" not degraded. I would not ever consider "pre-owned" for NaI(Tl) or CsI(Tl). At least, if one is using scintillator crystals designed to fit on the end of a PMT, they are likely to already be a suitable length.

So is there even a rough approximation way of deciding the crystal length

Sorry if my take on this is a bit neophyte. This is all stuff I never got to grips with before.

[Edit: I just realized that in the video by bionerd23, whenever she switches on the X-ray beam, photons from somewhere manage to hit the CCD in the camera making the video]
 
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Graham, I didn't want to quote your entire post because it's pretty long so I'll just try to answer or at least point to some directions to explore for answers.

Regarding the length of scintillator crystal, most appear to be meant for much higher energy x-rays. Higher energy photons have longer penetration depths so that (probably) is the primary reason many are quite long relative to their other axes. Some indications regarding the length required for a particular photon energy can also be obtained by looking at the same NIST table of x-ray absorption edges for the major component(s) in the scintillator. Basically what you want is a crystal that is long enough to absorb 90%+ of the incident photons -- I'm assuming that if they're absorbed, they will contribute more signal to the PMT.

1.
Let's be aggressive and say we want to absorb 99% of the photons that enter the long axis of the crystal. Going to the NIST table for Lutetium, I see that the "alpha", or absorption factor per cm of depth, is 3,129 at 6Kev. If 99% of the photons are absorbed, we can use this formula: .01 = exp(-alpha*d), where alpha = 312.9 and d is the required depth. Taking the log of both sides we get: 4.605 = 3129*d (I removed the minus signs from both sides). Solving for d we get d = 1.47e-2 cm, or .147mm -- 147 microns!!! The actual penetration depth will be more than that because we're talking about a compound that contains other elements, but....it's hard to believe that a few mm's won't suffice, eh?

Sodium has an "alpha" around 70 at 6Kev so it would take .66mm to absorb 6Kev photons. Not too bad there, but...iodine, the other component, has an alpha of 617 @6Kev so it is the one that really defines the penetration depth. Of course, we're talking about a compound so the actual penetration depth will be somewhere between the two, but in any case it doesn't appear to be a limiting factor for us.

2.
I've stayed away from NaI just because many of the vendors on ebay caution about opening the package without taking precautions about moisture. However, the assemblies that are designed to directly interface with a PMT probably are OK w/regard to moisture, if they are new/unused. An aluminum front window would have to be thin enough to be transparent to ~6Kev photons. The NIST tables can tell you what that thickness is, and I just did that. Here's the deal: at 6Kev, aluminum's "alpha" is 115. So if you want 99% of the photons to go THROUGH an aluminum window you don't want it to be thicker than 8.7E-5mm .....that's 87 nanometers....ugh! Probably nowhere close to what them there aluminum windows are. Given the fact that we're looking at a relatively low gamma ray flux from the Americium, I don't want to lose any more photons than absolutely necessary.

Just to restate the basic equation I'm using: I = I0 * exp(-alpha*t) (that's "I nought"), where t is the thickness and alpha is the absorption factor obtained from the NIST tables. So to calculate thickness for topic #1: .01 = exp(-alpha*t). Taking the log of both sides we get: -4.61 = -alpha*t, or t = 4.61/alpha. If alpha is in 1/cm units, t will be in centimeters. Topic #2: Calculating the thickness of aluminum to only attenuate the x-rays by .01, we just calculate ln(.99) and proceed as before, using the alpha value for aluminum.

I've corrected myself several times to get orders of magnitude right but STILL may have messed up. Everyone is more than welcome to check my math!
 
Graham, I didn't want to quote your entire post because it's pretty long so I'll just try to answer or at least point to some directions to explore for answers.

Regarding the length of scintillator crystal, most appear to be meant for much higher energy x-rays. Higher energy photons have longer penetration depths so that (probably) is the primary reason many are quite long relative to their other axes. Some indications regarding the length required for a particular photon energy can also be obtained by looking at the same NIST table of x-ray absorption edges for the major component(s) in the scintillator. Basically what you want is a crystal that is long enough to absorb 90%+ of the incident photons -- I'm assuming that if they're absorbed, they will contribute more signal to the PMT.

1.
Let's be aggressive and say we want to absorb 99% of the photons that enter the long axis of the crystal. Going to the NIST table for Lutetium, I see that the "alpha", or absorption factor per cm of depth, is 3,129 at 6Kev. If 99% of the photons are absorbed, we can use this formula: .01 = exp(-alpha*d), where alpha = 312.9 and d is the required depth. Taking the log of both sides we get: 4.605 = 3129*d (I removed the minus signs from both sides). Solving for d we get d = 1.47e-2 cm, or .147mm -- 147 microns!!! The actual penetration depth will be more than that because we're talking about a compound that contains other elements, but....it's hard to believe that a few mm's won't suffice, eh?

Sodium has an "alpha" around 70 at 6Kev so it would take .66mm to absorb 6Kev photons. Not too bad there, but...iodine, the other component, has an alpha of 617 @6Kev so it is the one that really defines the penetration depth. Of course, we're talking about a compound so the actual penetration depth will be somewhere between the two, but in any case it doesn't appear to be a limiting factor for us.

2.
I've stayed away from NaI just because many of the vendors on ebay caution about opening the package without taking precautions about moisture. However, the assemblies that are designed to directly interface with a PMT probably are OK w/regard to moisture, if they are new/unused. An aluminum front window would have to be thin enough to be transparent to ~6Kev photons. The NIST tables can tell you what that thickness is, and I just did that. Here's the deal: at 6Kev, aluminum's "alpha" is 115. So if you want 99% of the photons to go THROUGH an aluminum window you don't want it to be thicker than 8.7E-5mm .....that's 87 nanometers....ugh! Probably nowhere close to what them there aluminum windows are. Given the fact that we're looking at a relatively low gamma ray flux from the Americium, I don't want to lose any more photons than absolutely necessary.

Just to restate the basic equation I'm using: I = I0 * exp(-alpha*t) (that's "I nought"), where t is the thickness and alpha is the absorption factor obtained from the NIST tables. So to calculate thickness for topic #1: .01 = exp(-alpha*t). Taking the log of both sides we get: -4.61 = -alpha*t, or t = 4.61/alpha. If alpha is in 1/cm units, t will be in centimeters. Topic #2: Calculating the thickness of aluminum to only attenuate the x-rays by .01, we just calculate ln(.99) and proceed as before, using the alpha value for aluminum.

I've corrected myself several times to get orders of magnitude right but STILL may have messed up. Everyone is more than welcome to check my math!
Oh yeah, I just DID find an error. Several times I misstated the absorptivity for Lutetium, it IS 312.9 not 3129. But I used 312.9 to calculate the thickness needed to absorb 99% of 6Kev photons. Sorry for the confusion.
 
I really think if you just embedded manganese dioxide in epoxy the resin will be so thick that you would get a nice even distribution.
Robert

Edit: I should have said "so viscous".
 
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I really think if you just embedded manganese dioxide in epoxy the resin will be so thick that you would get a nice even distribution.
Robert
Dunno, haven't tried anything like this (in so many ways). It should be interesting! But a vital way to evaluate energy filters is a good set of elemental references. Without that we really have no way to determine how our various gizzies are working.
 
Roll your own (scintillators)?
I was just reading Wikipedia on organic, and then plastic scintillators.

Napthalene:
Smelly stuff, anisotropic with problems of energy resolution - but hang in there for now. It gets used.

Polyethylene napthalate:
This is sailcloth and drinks bottles plastic. Scintillates by itself without adding other fluors.
"Is expected to replace existing plastic scintillators due to higher performance and lower price".

Base plastics + fluors
Polystyrene:
As a base. A good thing for which to melt down those non-degradeable transparent disposable spoons.

Polymethylmethacrylate (PMMA):
i.e. Acrylic has advantages. High ultraviolet and visible light transparency. Transparent to it's own radiation.
Add some napthalene with the solvent.

Fluors wavelength changers.
Used to catch UV and render it visible blue or green. Generally yuk polyphenyl hydrocarbons, but only needed in small amounts. I would not know how to come across oxazole or n-terphenyl (PPP).

I would not be past melting up up some acrylic or polystyrene, stirring in some re-purposed mothballs, making into a scintillator and a taper lightpipe all in one. Like all simple-sounding schemes conjured in ignorance, this notion may have some downsides, but is the idea completely mad?
 
I am surprised those non metallic substances have the density to absorb x-ray at the energies we are talking about. Are they useful in this range?

"But a vital way to evaluate energy filters is a good set of elemental references. " Are you suggesting this will be hard to come by? I was thinking that would be the easy part. The elements do have to be of super high purity since the signal from any minor contaminant would probably be below our noise level?

Am I understanding your notation correctly? I = I0 * exp(-alpha*t) . What does "I" represent represent? Attenuation as a fraction? Just off hand the absorption distances seem too small. You are certainly correct that the thicker crystals are intended for higher energies. 80- 140 KeV would be common in medical imaging. Still, I am guessing a crystal should be at least 1 cm thick for this.
As you probably know, aluminum is typically used to filter an x-ray tube. Typically 0.5mm thick. At 100 KeVP (peak KeV so the mean is probably near 50 KeV) this preferentially attenuates a large portion of the lower energy x-rays (to reduce patient dose.) An aluminum window is probably a bad idea here since these are the energies we want to see. Plastic would be a much better choice and just as easy to use.

The crystal in the video Graham shows is designed for higher energies so the aluminum is not a problem. Graham- I think that probe is used to measure gamma sources. Those energies would not be well attenuated by aluminum.

I hope this doesn't sound critical, I am really enjoying learning and thinking about this as we go.

Robert
 
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Re: Non-metallic substances.
They may be non-metallic, but I think the atoms on them play as hardball as any other.

Re: Calibration.
I must fully learn the detail for the plotting software, but I think the height of the signal peaks in the pulse depends only on the photon they release when provoked. The brightness of a scintillation signal peak is not related to the proportion of other cluttering substances.

The height of a software display peak is a different sort of peak, it being a number count of the received pulses, previously sorted by height. This one is very definitely is related to the substance proportions.

I would think that, so long as you know the calibration substance has enough of one element, say more than 30%, so as to show it's count well above the other atoms in there, it will serve OK for setting the wavelength/frequency scale on the X-Axis.

Other materials, even those atoms having scintillation pulses very much higher, will not necessarily show higher displays, because any one of their (possibly huge) scintillations still only delivers one count. It is the accumulated count of complete proper pulses, sorted by energy, so betraying belonging to only one element, that lets you scale the proportions. Divide the bucket count by the total of the whole lot in a measure interval to get the percentage.

There may be some mileage in using a 2-element compound that you can get quite pure. Just about any chloride or oxide. Salt, titanium dioxide (i.e. white paint), ferric chloride, zinc sulfate, whatever. Those old NiCad batteries that won't holds a charge anymore, leaving you with several electric drills that don't have a life is full of two proper transition metals. Who cares if the energy of the chlorine emission is off the scale, or happens at all.

Batteries seem to be a good source of useful elements. If you can learn the care you need to dismantle them without fire-related lithium runaway accidents, those old laptop (notebook?) computer batteries that you replaced from eBay, have got cobalt in them.

In some ways, using fully oxidized compounds can be good. It makes them safer, no longer toxic, and you can get the salts pure. When I get this going, I am bound to put all sorts of stuff in under the thing.

High Energy?
Not to be confused with "lots of".
For the range of materials we want to see, are some of the emitter X-Ray wavelengths "high energy" enough to choose one of those scintillators that fit the PMT tubes, or maybe a LYSO crystal that is more than 0.5mm thick?

I was dreaming up the homebrew acrylic plastics concoction around a "gun" type, or probe-type gadget. All much smaller. Based on a Si-PM photodiode, the trick bit of tapered plastic, and some Am241 "buttons". That may have to be Mk2. I am going to try photomultiplier tubes first.
 
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