Needing more than a spark test?

[RJSakowski]

The thorium decay is the rate limiting step in the decay chain. Any daughter products decay instantaneously as compared to the thorium. That meaning that the concentration of any of the daughter product, other than the stable lead isotope, is essentially zero.My CRC Handbook of Chemistry and Physics, 43rd Ed. has a complete tabulation of all isotopes (known at the time) including their decay modes. Here is an on line copy: https://edisciplinas.usp.br/pluginf...ook of Chemistry and Physics 95th Edition.pdf
You will find the table starting at section 11-2. I believe that radioactive decay follows the equation N = N0(1 - exp(-at). You can solve for a by setting t =1.0E10 years = 3.1E17 sec. and N = N0/2. Differentiating that equation will give you the decay rate in events/sec. It's late at night and my brain is somewhat foggy but I came up with a decay rate of arond 8,000 disintegrations/sec for your stated sample size. (The math needs checking.)_

 

[homebrewed]

I pretty much came to the same conclusion as RJ, given the _very_ long half life of the currently present Thorium isotope. At this point there isn't anything left except the one that has a billions-of-years half life. Presumably that means a relatively low decay rate

Since gammas can't reflect (because of tiny wavelength), and can't go around corners, the shield ring need only be as high as the top of the diode body, and the radioactives on the other side of the ring need only be such that no gammas can see a straight line to the diode. There needs to be shield underneath the diode, or underneath the whole circuit, to drop out most of the external background noise. There will remain some you cannot stop.

I'm not sure what geometry you're assuming here. I'm not sure we are on the same page because I don't understand why the shield ring only has to be as high as the diode body if the sources are canted toward the sample.

My scheme is based on using a highly modified PocketGeiger rather than a custom PCB with the detector on it, which may explain the differences between our two approaches. This approach was shown with my OpenSCAD model, although some of the specifics were not quite right. Key features are the lead aperture plate and tilted source disks, which will illuminate the detector with the primary gammas unless shielded by a shield ring on the source side of the plate.

Radiation from other sources, like cosmic rays, unstable isotopes in the construction materials etc. are something that all XRF measurement systems encounter. Some inputs can be blocked to some extent, but not all.

 

[graham-xrf]

I pretty much came to the same conclusion as RJ, given the _very_ long half life of the currently present Thorium isotope. At this point there isn't anything left except the one that has a billions-of-years half life. Presumably that means a relatively low decay rate

Yes. Using RJ's value 8kBq from Thorium in rod is less than a quarter of the 37kBq we get from just one Am241 smoke detector.
Including some Thorium may still be worthwhile, because what comes from it has high enough energy to be able to get responses from tungsten and the remaining interesting metals up to lead. This is why I was thinking of using it "as well as" rather than "instead of" Am241.

I do think that the daughter products have not somehow evaporated from the Thorium. They may hang around for a few years, or hours, or minutes, or milliseconds, but are being produced all the time. At any time, we have the backlog of most recently decayed by-products in there. Do I have this wrong?

Also, I thought to just use more rods. I wondered about getting the thorium out of (say) 20 rods, concentrated.

Lastly, I remembered the spectrum I was looking at was from Thorium gas mantle. The Thorium in welding rods is also beating on the Tungsten in them, delivering some 59.3keV, 67.2keV, 8.4keV, and 9.7keV, just not as often as what comes from Am241
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Also, big thanks to @RJSakowski for the book pdf link. It's the sort of reference that bring memories of one's educational era. The kind you end up having fallen asleep on with forehead amid the pages. I only have the hard copy 14th Edition 1974 "Tables of Physical and Chemical Constants" by Kaye & Laby, when it was withdrawn from the local library and sold off cheap.

I'm not sure what geometry you're assuming here. I'm not sure we are on the same page because I don't understand why the shield ring only has to be as high as the diode body if the sources are canted toward the sample.

If the sources are each within "their little container" of lead, they can still irradiate over 180° half a sphere. If the top of the container is slightly higher than the source, that becomes a smaller angle, say about 170°. If they are placed, canted towards the sample, the rays can maybe still see the diode, albeit through the sides of it's plastic case. Putting a ring of shielding around the diode ensures it does not get hit by source photons.

The diagonal of the 16.5 x 14.5mm diode package is 22mm, and the top of the diode is 0.9mm from the PCB.
A shield ring 22mm ID, and 27mm OD, 1.5mm thick, or more generously, 2mm thick, placed around the diode body, maybe held there with glue, keeps the diode shielded. If zero photons can get to the diode even without the ring shield, then you don't need it at all. I think we both do understand keeping the diode in the shadow of shielding pretty well. I was just trying to shield the diode from all except what comes back from the sample, and adding a piece of lead under the diode further quietens the background.

You are right about the physical mounting of the diode. In my scheme, the diode is at right angles to the PCB, stood on one end.

 

[graham-xrf]

In view of the fact I have 4.5 litres of 36% hydrogen peroxide kept in a cupboard in a cold and very dark place (cellar) ..
and then I found --> THIS --> How to make thorium dioxide
.. and allowing for the now fairly well known nature of me..
What do you suppose is going to happen next?

 

[WobblyHand]

I set that against being able to place grams of the stuff there, as against nanograms. On a YT video, I saw a pack of TIG rods placed on a Geiger counter. The background count went up from 80-100 at the start, to over 1000. I do agree that if it has been around so long, it cant be losing atoms often, but by my count, a gram or two has got way more than 14 billion decaying atoms for us to play with..

Let me try..
Avagadro's Number is 6.022140857×10^23 atoms per mole.
There is the problem that the atoms are big fat heavy things, so a mole is 232.038 grams
(Please somebody weigh a TIG welding rod for me)
Going on it being nearly all Tungsten (98%) at 19.3g/cm^3 and 2% thorium at 11.7g/cm^3
A welding rod is 6" x 2.4mm (howzat for mixing units)? Volume is 15.24 x 0.24 =3.6576 cm^3
Weight of a 6" rod = (0.98 x 3.6576 x 19.3) + (0.02 x 3.6576 x 11.7) = 70.036 grams
The welding rod should have (0.02 x 70.0357/232.038) x 6.022140857e+23 = 3.63530844446e+21 Thorium atoms in it

It takes 14.05 billion years to lose half of them.
So 3.63530844446e+21/(2 x 14.05E9) = 129370407276 atoms decay per year.
There are 31536000 seconds in that year.
Dividing 129370407276 / 31536000 yields 4102.30870359
Our TIG welding rod should be losing a paltry 4102 atoms every second.
We would be using about 5" of it, so make that 3418 per second.
Probably only about a quarter of them would be in the direction we find useful, so about 854 per second.

I do agree that that is not much! The photons that come out are very energetic. Just not very often.
Is it worth trying to use Thorium to get a glow out of stuff?
Should we be maxing out on Am241 little things instead?
How many can we fit around the diode?

-----
Once a Thorium has decayed, it becomes Radium, which has half-life 5.7 years.
Radium goes on to Actinium, wasting half of itself in 6.1 hours going back into a lighter version of Thorium.
Lighter Thorium then has another try at becoming Radium (lighter version), but half is lost in 3,6 days.
Becoming Radon, half lasts 55 seconds, and in 140 millisecond, it's radioactive lead.
Not many minutes later, visiting Bismuth212, Polonium212, and Thallium 212, it ends up as lead shielding.

Not trying to discourage anyone, but merely trying to keep us on a good course. Assuming your methodology is correct, there are some math errors above. Your calculated volume of the rod is incorrect. Maybe you mixed up units? I get 0.6894 cubic cm volume for your mixed unit rod of 6" and 2.4mm diameter. Volume is pi * r * r * h. h = 6 in * 2.54 cm/in = 15.24 cm, r = 2.4 mm diameter / 2 * 0.1 cm/mm = 0.12 cm. Consequently the mass of the rod, and the number of decayed atoms is off.

For a 6" rod, I get a rod mass of 13.201 gm and 1546 decays/second. Using your approximation of 25% in the right direction gives me 386 decays/second. This is assuming decays/sec is calculated properly.

As a sanity check, I went outside to the garage to grab a package of 2.4mm 2% ceriated TIG electrodes. (It's -12C with a bit of a wind right now) Don't have any thorium rods. Measured an electrode to be about 14.5 cm (it was used/ground to a point) and the measured mass was 12.2 gm. My calculated mass for this rod, using Cerium characteristics was 12.49 gm. Considering that I wagged the length measurement due to the ground point, a 2.4% error is sufficient to mostly validate my calculations.

Haven't gotten through RJs calculations just yet in programmatic form. Attached is a python script with my calculations. No doubt I may have made some errors myself, but, it's an attempt. I've renamed it to thorium.txt. If you have a python installation (and numpy) rename it to thorium.py and run it.

 

[RJSakowski]

Interesting. What is the tungsten product from the reaction? Most of the tungsten salts are insoluble in water which would imply that the tungsten is tied up in the precipitate, either as the dioxide or trioxide since no other anions are present.

 

[WobblyHand]

Differentiating the radiation decay equation, as recommended by @RJSakowski and using the number of Thorium atoms in a 6" long 2.4mm diameter 2% thoriated rod shows a total activity of 1072 Bq. This is compared to 37000 Bq for a typical smoke detector.

In [60]: run thorium.py
Rod length [inches] 6
Rod diameter [mm] 2.4
Rod volume in cubic cm 0.6894
Rod mass in gm 13.201
No of Thorium atoms in rod 6.852397408070073e+20

Analytic # of Thorium decays per second 1071.98 [Bq]

Volume and mass calculation sanity check using 2% Ceriated TIG rod
------------------------------------------------------------------

Approx Ceriated rod length 14.5 [cm]
Ceriated TIG rod volume 0.656 [cm^3]
Calculated Ceriated rod mass 12.49 [gm]
Measured 2% Ceriated TIG rod mass 12.2 [gm]
Approx mass error 2.4 %

Latest version of thorium.py (renamed to thorium.txt)

 

[graham-xrf]

Not trying to discourage anyone, but merely trying to keep us on a good course. Assuming your methodology is correct, there are some math errors above. Your calculated volume of the rod is incorrect. Maybe you mixed up units? I get 0.6894 cubic cm volume for your mixed unit rod of 6" and 2.4mm diameter. Volume is pi * r * r * h. h = 6 in * 2.54 cm/in = 15.24 cm, r = 2.4 mm diameter / 2 * 0.1 cm/mm = 0.12 cm. Consequently the mass of the rod, and the number of decayed atoms is off.

For a 6" rod, I get a rod mass of 13.201 gm and 1546 decays/second. Using your approximation of 25% in the right direction gives me 386 decays/second. This is assuming decays/sec is calculated properly.

As a sanity check, I went outside to the garage to grab a package of 2.4mm 2% ceriated TIG electrodes. (It's -12C with a bit of a wind right now) Don't have any thorium rods. Measured an electrode to be about 14.5 cm (it was used/ground to a point) and the measured mass was 12.2 gm. My calculated mass for this rod, using Cerium characteristics was 12.49 gm. Considering that I wagged the length measurement due to the ground point, a 2.4% error is sufficient to mostly validate my calculations.

Haven't gotten through RJs calculations just yet in programmatic form. Attached is a python script with my calculations. No doubt I may have made some errors myself, but, it's an attempt. I've renamed it to thorium.txt. If you have a python installation (and numpy) rename it to thorium.py and run it.

We are not discouraged, it's going to make the thread better!
It was late, and I was tired, and we now both get 0.68944 cubic cm.
I am going to edit the posting with some fun strike-throughs and corrections

One thing you might suspect is that if I actually mention the units, then you know I was aware of, and taking care of them.
What actually happened was messing up (π*D^2)/4, or π*r^2 if you like.
I also know that a new TIG rod is 7", but I went with 6".

Of that volume 2% is Thorium, and 98% is Tungsten
We know the densities are 19.3g/cm^3 for Tungsten, and 11.7g/cm^3 for Thorium.
so (0.98 x 0.68944 x 19.3) + (0.02 x 0.68944 x 11.7)
It has 13.0400 grams of Tungsten, and 0.161329 grams of Thorium => 13.201 total (which is what you said).

Our paltry 0.161 grams of Thorium has (0.161329 ÷ 232.038) x 6.022140857e+23 => 4.1870123097e+20 atoms
Half of them (4.1870123097e+20 ÷ 2 ) = 2.09350615485e+20 got lost in 14.05 billion years.
(2.09350615485E+20 ÷ 14.05 E9) = 14900399679 are lost in 1 year (31536000 seconds)
We end up with a paltry 472 per second
Likely about a quarter to a third of them are heading in the right direction.
So, about 118 per second.

This is great! We are all doing what engineers do! Each coming up with a different answer when doing the same calculation.
We will get it right, I am sure!

 

[homebrewed]

When I was doing failure analysis on IC's we often had to remove tungsten as part of the deprocessing step -- it's used as a conductive protection layer between silicon and aluminum metallization. We used 30% hydrogen peroxide with a small amount (~10%) of ammonium hydroxide. The H2O2 oxidized the top layer of tungsten to tungsten oxide, and the NH4OH reacted to form ammonium tungstate, which is water soluble. As a result the reaction proceeded fairly quickly. It was a "nice" etch because the only thing it attacked was the tungsten barrier metal. Selectivity is a Good Thing when you're deprocessing IC's.

Adding some ammonium hydroxide (or some other base like NaOH or KOH) to the peroxide being used to dissolve thoriated welding rods will speed things up, but only for a little while. This is because the low pH also destabilizes the peroxide so it decomposes into water and oxygen. If we mixed up too-large a batch of our tungsten etch the stuff would slowly decompose and also release heat. The heat would cause it to decompose faster, then faster.....until it boiled. Since we were doing this under a fume hood with PPE's that wasn't a disaster, just messy. But I wouldn't want it to happen in my kitchen!

But there's a better way to dissolve tungsten. I'm not recommending it as something anyone would really want to do, just as a thought experiment. I don't want anyone inhaling thorium dioxide dust. The peroxide is there as a source of monatomic oxygen to oxidize the tungsten. An electrolytic cell with a NaOH solution as its electrolyte can do the same thing. Make the tungsten rod the anode and use something like stainless steel or a carbon rod for the cathode and pass current through the cell. Oxygen is (momentarily) liberated at the anode, oxidizes the tungsten, and then the tungsten oxide is dissolved by the NaOH. This exposes fresh tungsten so the process can continue.

We used a similar process to sharpen our own probe tips for probing the nodes of IC's. I designed and built a circuit to automate the process, based on a scheme used to sharpen STM tips. When adjusted correctly, the circuit produced probe tips that still looked sharp at 100,000X magnification on our SEM (albeit not the most modern one out there). Our SEM operator was skeptical when I told her that I doubted she would be able to image the end of the probe in the SEM but she had to eat her words

To produce the best probe tips I found it necessary to use potassium hydroxide at a concentration of about 8M so it was a strong and hazardous solution. For the purposes of just dissolving tungsten it might be possible to use something a little less nasty like sodium carbonate A.K.A. washing soda.

The electrolytic cell also will release hydrogen at the cathode so the process shouldn't be done in an enclosed space.

According to Wikipedia, thorium dioxide is not soluble in alkali solutions so there's a chance that it will drop out of the tungsten as the rod is dissolved. When googling thorium + salts all I found were references to molten-salt nuclear reactors, not quite the same thing. But I wouldn't bet the farm on this -- there are a lot of unknowns (or perhaps known but classified) when it comes to these kinds of Strategic materials.

But, even if thorium were active enough in terms of decays/second, examination of its decay chain didn't reveal any fissions that result in gamma rays, just alpha and beta particles. But I admit to a very cursory examination of the Wikipedia article on the subject.

 

[WobblyHand]

We are not discouraged, it's going to make the thread better!
It was late, and I was tired, and we now both get 0.68944 cubic cm.
I am going to edit the posting with some fun strike-throughs and corrections

One thing you might suspect is that if I actually mention the units, then you know I was aware of, and taking care of them.
What actually happened was messing up (π*D^2)/4, or π*r^2 if you like.
I also know that a new TIG rod is 7", but I went with 6".

Of that volume 2% is Thorium, and 98% is Tungsten
We know the densities are 19.3g/cm^3 for Tungsten, and 11.7g/cm^3 for Thorium.
so (0.98 x 0.68944 x 19.3) + (0.02 x 0.68944 x 11.7)
It has 13.0400 grams of Tungsten, and 0.161329 grams of Thorium => 13.201 total (which is what you said).

Our paltry 0.161 grams of Thorium has (0.161329 ÷ 232.038) x 6.022140857e+23 => 4.1870123097e+20 atoms
Half of them (4.1870123097e+20 ÷ 2 ) = 2.09350615485e+20 got lost in 14.05 billion years.
(2.09350615485E+20 ÷ 14.05 E9) = 14900399679 are lost in 1 year (31536000 seconds)
We end up with a paltry 472 per second
Likely about a quarter to a third of them are heading in the right direction.
So, about 118 per second.

This is great! We are all doing what engineers do! Each coming up with a different answer when doing the same calculation.
We will get it right, I am sure!

Good, found an error in my calculations. Agree with the number of atoms now. N = 4.187e20. But my decay rate is 655 Bq. Before the 25%.

I am using the decay rate found from solving what "a" is from the decay equation. N = N0( 1 - exp(-at) ). I come up with
a = ln(2)/(halflife[yrs] * num_sec_per_year [sec/yr]) = 1.5643809900134361e-18 [events per atom/sec] ? guessing on the units

Therefore the activity rate A is: A = a * N [Bq], I now get 655 [Bq]. At 25% --> 163 [Bq]
Reference: radioactive decay look at the section called rates.