Needing more than a spark test?

[graham-xrf]

According to the Wikipedia entry for X-ray fluorescence, the mechanism is as follows: The incident photon excites an inner-shell electron and as a result it is ejected from the atom. An outer-shell electron drops down into the lower orbital and emits another x-ray due to the energy change. This secondary photon's energy is characteristic of the atom, and of which shell it came from.

If one electron is ejected from the atom, only one electron is needed to refill the vacancy and just one photon will be emitted as a result. So far I haven't found any discussion regarding the possibility of a gamma ray ejecting more than one electron from a single atom. That doesn't mean it can't happen -- unless there's some quantum-mechanical reason for it. I note that photons, just like electrons, have momentum (p = E/c): but can one photon "share" its momentum among multiple electrons? Momentum has to be conserved......and it's a vector.....and what about the possibility of multiple electron-hopping events as the electron "hole" moves from orbital to orbital? Now my head is starting to hurt

But in a practical sense I don't think it is an issue. There's no reason to believe that if multiple photons can come from a single atom that they all are emitted in the same direction, so "pileup" due to that is quite unlikely, if it's even possible.

Thanks much for that.
I find it interesting that an inner shell electron gets hit first. I thought the outer shell electrons took the least energy to shake loose.

Don't worry about head-hurting stuff. The apparently temporary existence of the internal bits with 2/3 charges, up, down, strange, etc will take it to migrane level. Charmed quarks they may be, but the Quantum Standard Model, fairy story though it may be, is 10 decimal places accurate.

I did not think all the energies would be emitted in the same direction. I did just wonder if an excitation with more than enough energy to account for them all, would provoke the whole lot to come out, maybe not together, but all within a reasonable interval.

Head-hurting stuff would include..
Bell's Inequality.
The Delayed Choice Quantum Eraser - Sabine Hossenfelder
Young's Double-Slit Experiment. - (Lots and Lots of stuff on that).

 

[homebrewed]

Pileup can occur simply if charge packets due to any cause are received within the bandwidth of the detector circuit. We don't need to know their source, or from which original x-ray it came from. If two or more charge packets are received too close in time, then a malformed pulse will result. If we can recognize it's malformed, it can be rejected, accepted or further processed. In the case of a near coincident big and little pulse, it may be hard to separate them, or even recognize there were two pulses. Just a real world condition that need to be handled in a graceful, predictable fashion.

One of the simplest discriminators is based on pulse width. Pulse overlap will most likely result in a wider-than-normal pulse. Too wide? Don't put it in an MCA bin.

 

[WobblyHand]

One of the simplest discriminators is based on pulse width. Pulse overlap will most likely result in a wider-than-normal pulse. Too wide? Don't put it in an MCA bin.

Yes, KISS is a good approach. One can go mad searching for optimal solutions when far simpler "sub-optimal" solution is more than good enough.

 

[graham-xrf]

Yes, KISS is a good approach. One can go mad searching for optimal solutions when far simpler "sub-optimal" solution is more than good enough.

That is also the one I was going for. Too high, or too wide, is grounds for rejection.
You can easily do it in software, but..

.. at least one of my circuit thoughts had a triggered one-shot, making a "validity" timeout, which was also a "newcomers" lockout. In there also was a high speed peak detector, with a hold-reset operated by the one-shot. Old-school analog!

Very little for the computer to do, until I realized letting the computer work the enable after it had safely saved the value was an obvious way to do it.

 

[whitmore]

Thanks much for that.
I find it interesting that an inner shell electron gets hit first. I thought the outer shell electrons took the least energy to shake loose.

The rate for gamma absorption is highest for just-above-the-fluorescence energies; a quick look at photoelectric
cross sections shows big jumps in absorption at each inner-electron threshold energy, like here, in ironIron absorption of Xrays
The cliff-like jump (circa 7 keV) is at the eject-an-inner-shell electron energy. There's a lot more outer-shells electrons, but
by a factor of nearly ten, all the outer shell contributions are less than the inmost-shell interactions

You get the fluorescence output as the inner-shell void gets refilled.

 

[graham-xrf]

The rate for gamma absorption is highest for just-above-the-fluorescence energies; a quick look at photoelectric
cross sections shows big jumps in absorption at each inner-electron threshold energy, like here, in ironIron absorption of Xrays
The cliff-like jump (circa 7 keV) is at the eject-an-inner-shell electron energy. There's a lot more outer-shells electrons, but
by a factor of nearly ten, all the outer shell contributions are less than the inmost-shell interactions

You get the fluorescence output as the inner-shell void gets refilled.

Thanks for the explanation. There is a little pause, but the energized shells don't stay that way very long.

At a more prosaic level, and indeed more practical, I have two candidate circuits, and I am trying to press it to the point I can catch up a bit with Mark. I was going after the current that would get injected into the transimpedance amplifier. Designing these things is tricky. Yesterday, I read most of a whole book about it. I needed some idea of what goes in, and what we want out.

Taking a few shortcuts, I try for turning the charges collected into currents. The pulse that comes from a photodiode, and the process that generates them in the depletion layer is very complicated, resulting in a summation of two current waveforms, one rather faster than the other. I just skipped the detail, and took a gross approach. I only need ball-part values to decide a gain.

Iron will deliver 4 energies, but only two are in a range the detector can do anything with. The last two are outside of the ability of the photodiode to respond to. We know that to create an electron-hole pair in silicon, it takes 3.66eV +/- 0.03eV. They will try for recombination, and some are lost this way, but very few. There are other efficiency losses, but for a ball-park estimate, we can consider the first two energies.

6.40384keV would make 1749 electrons, so 2.8022E-16 Coulombs
7.05798keV would make 1928 electrons, so 3.08899E-16 Coulombs

Now the quick 'n dirty estimate.
Of course, these are only accumulations of charge. In order to become currents, they have to move their energy, (though not themselves) in a certain time, and we already know the rise-time is about 2uS, and the whole thing is over in about 13uS. If we consider that the most of it was in about 8uS, and pretend it was a square wave, then we get 35.02pA and 38.6pA. Then estimating what the amplitude would have been if that energy had been in a roughly triangular waveform, we are talking of peaks of around 70pA, for roughly 7keV. This is something to go on.

We look back to magnesium(1.25keV) and aluminium 1.48keV. Now we are talking about 12pA. That's near the bottom of what we can deal with. In most TIA amplifiers, it can get buried in noise, unless one designs carefully, but it looks like we can get close. I know this is sloppy, inexact, estimation, but it's all we need to make circuit decisions.

If we can ever get something from tungsten, it has to deliver 59.31keV from a 59.48keV Am241 excitation. Maybe?
The 16207 electrons is as much as 324pA. That's still less than the approx 2.5nA dark current, but it is above the noise on that dark current.
We need a huge gain, but I am now more hopeful of this.

 

[rwm]

I suspect you will get Tungsten for the reasons Whitmore laid out.
R

 

[graham-xrf]

I suspect you will get Tungsten for the reasons Whitmore laid out.
R

Yes - I did see that. The thing that removes any confusion about responses from zinc would be the additional presence of any high energy 59keV response. Regardless that it's count will be low, if there is a smidgen of count there, then it's tungsten.

I was noting the closeness of 59.54keV from our smoke detectors, as compared to W(74) tungsten x-ray emission at 59.32keV.
That is just barely enough to, in theory, get the Kα charged up enough to deliver that one, but no others. I am thinking if it is "just above", then excitation is inevitable.

By the explanation in @whitmore 's link, our Am241 has not got enough kick to get any other tungsten L-shell energised. We get one peak only. If we see a high energy peak there, regardless the low count, it's either tungsten, or some unshielded smoke detector direct from Am241. Zinc's K-shell energies are coincidentally close to tungsten's L-shell energies, but if we see any around that region at all, then there really is some zinc in there.

Now mulling over "dead-bug" prototype construction vs proto-PCB. I think I will be going PCB from the start. I allow I might have to mangle (er.. "modify") it somewhat, later down the line, but I will try and make it "development friendly".

 

[homebrewed]

As I've mentioned previously, I have been planning on determining a de-noised peak value by doing a least squares fit on the data around the peak. A simple find-the-maximum function, whether it's done in S/W or H/W, is unavoidably affected by noise. Noise is our enemy because it will broaden the spectrum and make it harder to distinguish elements like iron and cobalt.

However, I'm now thinking there's another approach that can produce even better noise rejection with less computational overhead than a least squares fit. Just integrate the pulse, once we have decided it's a good one to use and stuff that value into an MCA bin. We don't really need to know the peak value, as long as we use the same measurement approach on known pure elemental standards to calibrate the system. If the noise is uncorrelated then our SNR should improve as sqrt(N), but without the need to calculate the polynomial coefficients and back-calculating the "actual" peak voltage. We also can use more data points because the 2nd-order polynomial fit really is only valid around the peak value.

The integration approach isn't trying to squeeze the data into any kind of "fit" box so all of the pulse can contribute to our output value.

The integration scheme also is much easier to understand than my tweaked least-squares fit code (even if I did try to comment the heck out of it). That's a big plus right there

An ADC and a circular buffer still would be a good front end for this approach because we can discriminate and then use the entire pulse for the integration.

 

[homebrewed]

Along with some semi-theoretical hand waving, I also have been working on making the focus and shield rings. It's gotten a little iffy because my lathe started acting up in a manner that suggests the high/low transmission gears are failing. So far so good, but I can tell they're going downhill pretty fast. Replacement parts are on order. Since I have to remove the spindle and at least one of the spindle bearings, I also will upgrade the lathe with angular contact bearings.

Anyway, the aluminum focus ring is done. The shield ring is an interesting challenge because it's made out of lead, notoriously difficult to machine -- so I have avoided that approach. Instead, I cut a ~.150" wide strip out of my 1/16" thick lead sheet using metal snips. The ring needs to be .140" tall so I had to remove about .01 inch from the strip. The sides of the strip weren't all that flat so that was an issue as well. To address that, I got a 1 inch thick piece of HDPE from my scrap bin, about 3 x 5 inches on the other dimensions. I faced one side with the biggest end mill I've got to get a nice smooth surface. Then I turned it vertical so the 3 inch dimension was on the mill's X axis and, using a 1/16" slitting saw I cut several slots in the HDPE. One was about .145" deep and the other was .140" deep. Once that was done, I clamped the HDPE piece in my woodworking vise and inserted a 2.5" length of my lead strip in the .145" slot. I used a file to flatten that side, then removed the strip, turned it over and inserted it in the .140" slot. Again, I used the file to remove lead down to the top surface of the HDPE. This approach gave me a smooth top and bottom for the shield ring. I used HDPE because it is very abrasion resistant so it wasn't affected much, if at all, by the filing work.

I filed a 45 degree bevel on one end of the lead strip, then wound it around a Delrin mandrel I made on my (ailing) lathe. The strip was longer than necessary so I used a single-edged razor blade to slice through the lead, doing my best to match the bevel angle on the other end.

The focus ring doesn't quite fit inside the focus ring so I need to refine it a little bit. I'll do that with some small jewler's files I have.

Photos will be posted soon. I want to use my DSLR, which can get a lot closer to subject and stay in focus -- but I need to charge the battery first.